[inder k. rana] an introduction to measure and int(bookzz.org)

Any Introductionto Measure and'Integration -- - '

SEr:OND ED'IT'ION

1' nde r K. Ran a

Graduate Studieson Mathematics

Volume 49

An Introductionto Measure andIntegration

An Introductionto Measure andIntegrationSECOND EDITION

Inder K. Rana

Graduate Studiesin Mathematics

Volume 45

American Mathematical SocietyProvidence, Rhode Island

Editorial BoardSteven G. Krantz

David Saltman (Chair)David SattingerRonald Stern

2000 Mathematics Subject Classification. Primary 28-01;Secondary 28A05, 28A10, 28A12, 28A15, 28A20, 28A25,

28A33, 28A35, 26A30, 26A42, 26A45, 26A46.

ABSTRACT. This text presents a motivated introduction to the theory of measure and integration.Starting with an historical introduction to the notion of integral and a preview of the Riemannintegral, the reader is motivated for the need to study the Lebesgue measure and Lebesgue integral.The abstract integration theory is developed via measure. Other basic topics discussed in the textare Fubini's Theorem, Lp-spaces, Radon-Nikodym Theorem, change of variables formulas, signedand complex measures.

Library of Congress Cataloging-in-Publication DataRana, Inder K.

An introduction to measure and integration / Inder K. Rana.-2nd ed.p. cm. - (Graduate texts in mathematics, ISSN 1065-7339 ; v. 45)

Includes bibliographical references and index.ISBN 0-8218-2974-2 (alk. paper)1. Lebesgue integral. 2. Measure theory. I. Title. II. Graduate texts in mathematics ; 45.

QA312.R28 2002515'.42-dc2l 2002018244

Copying and reprinting. Individual readers of this publication, and nonprofit librariesacting for them, are permitted to make fair use of the material, such as to copy a chapter for usein teaching or research. Permission is granted to quote brief passages from this publication inreviews, provided the customary acknowledgment of the source is given.

Republication, systematic copying, or multiple reproduction of any material in this publica-tion is permitted only under license from Narosa Publishing House. Requests for such permis-sion should be addressed to Narosa Publishing House, 6 Community Centre, Panchscheel Park,New Delhi 110 017, India.

First Edition © 1997 by Narosa Publishing House.Second Edition © 2002 by Narosa Publishing House. All rights reserved.

Printed in the United States of America.

Q The paper used in this book is acid-free and falls within the guidelinesestablished to ensure permanence and durability.

Visit the AMS home page at http : //www. ams. org/

10987654321 070605040302

In memory of my fatherShri Omparkash Rana

(24th April, 1924-26th May, 2002)

Contents

Preface

Preface to the Second Edition

xi

xvii

Recipe for a one semester course and interdependence of the chapters xix

Notations used in the text xxi

Prologue. The length function 1

Chapter 1. Riemann integration 5

§1.1. The Riemann integral: A review 5

§1.2. Characterization of Riemann integrable functions 18

X1.3. Historical notes: The integral from antiquity to Riemann 30

X1.4. Drawbacks of the Riemann integral 36

Chapter 2. Recipes for extending the Riemann integral 45

§2.1. A function theoretic view of the Riemann integral 45

X2.2. Lebesgue's recipe 47

§2.3. Riesz-Daniel recipe 49

Chapter 3. General extension theory 51

§3.1. First extension 51

X3.2. Semi-algebra and algebra of sets 54

vii

viii Contents

§3.3. Extension from semi-algebra to the generated algebra 58

§3.4. Impossibility of extending the length function to all subsets ofthe real line 61

§3.5. Countably additive set functions on intervals 62

§3.6. Countably additive set functions on algebras 64

§3.7. The induced outer measure 70

§3.8. Choosing nice sets: Measurable sets 74

§3.9. The a-algebras and extension from the algebra to the generateda-algebra 80

§3.10. Un iqueness of the extension 84

§3.11. Co mpletion of a measure space 89

Chapter 4. The Lebesgue measure on R and its properties 95

§4.1. The Lebesgue measure 95

§4.2. Relation of Lebesgue measurable sets with topologically nicesubsets of R 99

§4.3. Properties of the Lebesgue measure with respect to the groupstructure on R 103

§4.4. Uniqueness of the Lebesgue measure 106

§4.5. *Cardinalities of the Q-algebras G and BR 110

§4.6. Nonmeasurable subsets of ][8 113

§4.7. The Lebesgue-Stieltjes measure 114

Chapter 5. Integration 117

§5.1. Integral of nonnegative simple measurable functions 118

§5.2. Integral of nonnegative measurable functions 122

§5.3. Intrinsic characterization of nonnegative measurable functions 130

§5.4. Integrable functions 143

§5.5. The Lebesgue integral and its relation with the Riemannintegral 153

§5.6. L1 [a, b] as completion of R[a, b] 158

§5.7. Another dense subspace of L1 [a, b] 163

§5.8. Improper Riemann integral and its relation with the Lebesgueintegral 168

Contents ix

§5.9. Calculation of some improper Riemann integrals 172

Chapter 6. Fundamental theorem of calculus for the Lebesgue integral 175

§6.1. Absolutely continuous functions 175

§6.2. Differentiability of monotone functions 179

§6.3. Fundamental theorem of calculus and its applications 191

Chapter 7. Measure and integration on product spaces 209

§7.1. Introduction 209

§7.2. Product of measure spaces 212

§7.3. Integration on product spaces: Fubini's theorems 221

§7.4. Lebesgue measure on ][82 and its properties 229

X7.5. Product of finitely many measure spaces 237

Chapter 8. Modes of convergence and LP spaces 243

§8.1. Integration of complex-valued functions 243

§8.2. Convergence: Pointwise, almost everywhere, uniform andalmost uniform 248

§8.3. Convergence in measure 255

§8.4. Lp spaces 261

§8.5. *Necessary and sufficient conditions for convergence in LP 270

§8.6. Dense subspaces of LP 279

X8.7. Convolution and regularization of functions 281

X8.8. Loo (X, S, µ): The space of essentially bounded functions 291

§8.9. L2(X, S, µ): The space of square integrable functions 296

§8.10. L2-convergence of Fourier series 306

Chapter 9. The Radon-Nikodym theorem and its applications 311

§9.1. Absolutely continuous measures and the Radon-Nikodymtheorem 311

§9.2. Computation of the Radon-Nikodym derivative 322

X9.3. Change of variable formulas 331

Chapter 10. Signed measures and complex measures 345

x Contents

§10.1. Signed measures 345

§10-2. Radon-Nikodym theorem for signed measures 353

§10-3. Complex measures 365

§10.4. Bounded linear functionals on L,(X, S, µ) 373

Appendix A. Extended real numbers 385

Appendix B. Axiom of choice 389

Appendix C. Continuum hypothesis 391

Appendix D. Urysohn's lemma 393

Appendix E. Singular value decomposition of a matrix 395

Appendix F. Functions of bounded variation 397

Appendix G. Differentiable transformations 401

References 409

Index 413

Index of notations 419

Preface

"Mathematics presented as a closed, linearly ordered, system of truthswithout reference to origin and purpose has its charm and satisfiesa philosophical need. But the attitude of introverted science is un-suitable for students who seek intellectual independence rather thanindoctrination; disregard for applications and intuition leads to isola-tion and atrophy of mathematics. It seems extremely important thatstudents and instructors should be protected from smug purism. "

Richard Courant and Fritz John(Introduction to Calculus and Analysis)

This text presents a motivated introduction to the subject which goes undervarious headings such as Real Analysis, Lebesgue Measure and Integration,Measure Theory, Modern Analysis, Advanced Analysis, and so on.

The subject originated with the doctoral dissertation of the Frenchmathematician Henri Lebesgue and was published in 1902 under the ti-tle Integrable, Longueur, Aire. The books of C. Caratheodory [8] and [9],S. Saks [35], I.P. Natanson [27] and P.R. Halmos [14] presented these ideasin a unified way to make them accessible to mathematicians. Because of itsfundamental importance and its applications in diverse branches of mathe-matics, the subject has become a part of the graduate level curriculum.

Historically, the theory of Lebesgue integration evolved in an effort toremove some of the drawbacks of the Riemann integral (see Chapter 1) .However, most of the time in a course on Lebesgue measure and integra-tion, the connection between the two notions of integrals comes up only

xi

xii Preface

after about half the course is over (assuming that the course is of one se-mester). In this text, after a review of the Riemann integral, the reader isacquainted with the need to extend it. Possible methods to carry out thisextension are sketched before the actual theory is presented. This approachhas given satisfying results to the author in teaching this subject over theyears and hence the urge to write this text. The nucleus for the text wasprovided by the lecture notes of the courses I taught at Kurukshetra Uni-versity (India), University of Khartoum (Sudan), South Gujarat University(India) and the Indian Institute of Technology Bombay (India). These noteswere slowly augmented with additional material so as to cover topics whichhave applications in other branches of mathematics. The end product is atext which includes many informal comments and is written in a lecture-notestyle. Any new concept is introduced only when it is needed in the logicaldevelopment of the subject and it is discussed informally before the exactdefinition appears. The subject matter is developed by motivating examplesand probing questions, as is normally done while teaching. I have tried toavoid slick proofs. Often a proof is either divided into steps or is presentedin such a way that the main ideas of the proof emerge before the detailsfollow.

Summary of the text

The text opens with a Prologue on the length function and its propertieswhich are basic for the development of the subject.

Chapter 1 begins with a detailed review of the Riemann integral and itsproperties. This includes Lebesgue's characterization of Riemann integrablefunctions. It is followed by a brief discussion on the historical development ofthe integral from antiquity (around 300 B.C.) to the times of Riemann (1850A.D.). For a detailed account the reader may refer to Bourbaki [6], Hawkins[16] and Kline [20]. The main aim of the historical notes is to make youngreaders aware of the fact that mathematical concepts arise out of physicalproblems and that it can take centuries for a concept to evolve. This sectionalso includes Riemann's example of an integrable function having an infinitenumber of discontinuities and a proof of the fundamental theorem of calculusdue to G. Darboux. The next section of Chapter 1 has a discussion aboutthe drawbacks of the Riemann integral, including the example due to VitoVolterra (1881) of a differentiable function f : [0, 1] --) R whose derivativefunction is bounded but is not Riemann integrable. These considerationsmade mathematicians look for an extension of the Riemann integral andeventually led to the construction of the Lebesgue integral.

Chapter 2 discusses two significantly different approaches for extendingthe notion of the Riemann integral. The one due to H. Lebesgue is sketched

Preface xiii

in this chapter and is discussed in detail in the rest of the book. The secondis due to P.J. Daniel [10] and F. Riesz [32], an outline of which is given.These discussions motivate the reader to consider an extension of the lengthfunction from the class of intervals to a larger class of subsets of R.

Chapters 3, 4 and 5 form the core of the subject: extension of measuresand the construction of the integral in the general setting with the Lebesguemeasure and the Lebesgue integral being the motivating example. The pro-cess of extension of additive set functions (known as the Caratheodory ex-tension theory) is discussed in the abstract setting in Chapter 3. Thischapter also includes a result due to S.M. Ulam [41] which rules out thepossibility of extending (in a meaningful way) the length function to allsubsets of R, under the assumption of the Continuum Hypothesis.

The outcomes of the general extension theory, as developed in Chapter 3,are harvested for the particular case of the real line and the length functionin Chapter 4. This gives the required extension of the length function,namely, the Lebesgue measure. Special properties of the Lebesgue measureand Lebesgue measurable sets (the collection of sets on which the Lebesguemeasure is defined by the extension theory) with respect to the topology andthe group structure on the real line are discussed in detail. The final sectionof the chapter includes a discussion about the impossibility of extending thelength function to all subsets of R under the assumption of the Axiom ofChoice.

In Chapter 5, the construction of the extended notion of integral isdiscussed. Once again, the motivation comes from the particular case offunctions on the real line. Lebesgue's recipe, as outlined in Chapter 2, iscarried out for the abstract setting. The particular case gives the requiredintegral, namely, the Lebesgue integral. The space L1 [a, b] of Lebesgue in-tegrable function on an interval [a, b] is shown to include 7Z [a, b], the spaceof Riemann integrable functions, the Lebesgue integral agreeing with theRiemann integral on R [a, b]. Also, it is shown that L1 [a, b] is the completionof R [a, b] under the L1-metric. The final section of the chapter discusses therelation between the Lebesgue integral and the improper Riemann integral.

Chapter 6 gives a complete proof of the fundamental theorem of cal-culus for the Lebesgue integral. (This theorem characterizes the pair offunctions F, f such that F is the indefinite integral of f. This removes oneof the main drawbacks of Riemann integration.) As applications of the fun-damental theorem of calculus, the chain-rule and integration by substitutionfor the Lebesgue integral are discussed.

The remaining chapters of the book include special topics. Chapter7 deals with the topic of measure and integration on product spaces, with

xiv Preface

Fubini's theorem occupying the central position. The particular case ofLebesgue measure on R2 and its properties are discussed in detail.

Chapter 8 starts with extending the concept of integral to complexvalued functions. The remaining sections discuss various methods of ana-lyzing the convergence of sequences of measurable functions. The Lp-spacesand discussion of some of their dense subspaces in the special case of theLebesgue measure space are also included in this chapter. The last sectionof the chapter includes a brief discussion on the application of Lebesgueintegration to Fourier series.

Chapter 9 includes a discussion of the Radon-Nikodym theorem. Asan application, the change of variable formulas for Lebesgue integration onJE are derived.

In Chapter 10, the additive set functions, which are not necessarilynonnegative or even real-valued, are discussed. The main aim is to provethe Hahn decomposition theorem and the Lebesgue decomposition theorem.As a consequence, an alternative proof of the Radon-Nikodym theorem isgiven. This chapter also includes a discussion of complex measures.

The text has three appendices. Appendix E gives a proof of the sin-gular value decomposition of matrices, needed in sections 7.4 and 9.3. InAppendix F, functions of bounded variation (needed in section 6.1) arediscussed. Appendix G includes a discussion of differentiable transforma-tion and a proof of the inverse function theorem, needed in section 9.3. (Inthe present edition four more appendices, A,B,C and D, have been added.)

The text is sprinkled with 200 exercises, most of which either include ahint or are broken into doable steps. Exercises marked with are needed inlater discussions. The sections and exercises marked with * can be omittedon first reading. Some of the results in the text are credited to the discoverer,but no effort is made to trace the origin of each result. In any case, nooriginality is claimed.

Prerequisites and course plans

The text assumes that the reader has undergone a first course in mathe-matical analysis (roughly equivalent to that of first five chapters of Apostol[2]). The text as such can be used for a one-year course. A recipe for aone-semester course (approximately 40 lecture hours and 10 problem discus-sion hours) on Lebesgue measure and integration is given after the preface.Since the text is in a lecture-note style, it is also suitable for an individ-ual self-study program. For such readers, the chart depicting the logicalinterdependence of the chapters will be useful.

Preface

Acknowledgments

xv

It is difficult to list all the individuals and authors who have influencedand helped me in preparing this text, directly or indirectly. First of allI would like to thank my teacher and doctoral thesis advisor, Prof. K.R.Parthasarathy (Indian Statistical Institute, Delhi), whose lectures at theUniversity of Bombay (Mumbai) and the Indian Statistical Institute (Delhi),clarified many concepts and kindled my interest in the subject. I learnedmuch from his style of teaching and mathematical exposition.

Some of the texts which have influenced me in one form or anotherare Halmos [14], Royden [34], Hewitt and Stromberg [18], Aliprantis andBurkinshaw [1], Friedman [13] and Parthasarathy [28].

I am indebted to the students to whom I have taught this subject overthe years for their reactions, remarks, comments and suggestions which havehelped in deciding on the style of presentation of the text.

It is a pleasure to acknowledge the support and encouragement I receivedfrom my friend Prof. S. Kumaresan (University of Bombay) at various stagesin the preparation of this text. He also went through the text, weeding outmisprints and mistakes. I am also thankful to my friend Dr. S. Purkayastha(Indian Institute of Technology Bombay) for going through the typeset man-uscript and suggesting many improvements. For any shortcoming still leftin the text, the author is solely responsible.

I thank C.L. Anthony for processing the entire manuscript in LA'IPX.The hard job of preparing the figures was done by P. Devaraj, I am thankfulfor his help. Thanks are also due to the Department of Mathematics, IITBombay for the use of Computer Lab and photocopying facilities. I wouldlike to thank the Curriculum Development Program of the Indian Instituteof Technology Bombay for the financial support to prepare the first versionof the manuscript. The technical advice received from the production de-partment of Narosa Publishers in preparing the camera ready copy of themanuscript is acknowledged with thanks.

Special thanks are due to my family: my wife Lalita for her help in moreways than one; and my parents for allowing me to choose my career and fortheir love and encouragement in pursuing the same. It is to them that thisbook is dedicated.

Finally, I would be grateful for critical comments and suggestions forlater improvements.

Mumbai, 1997

Preface to the SecondEdition

In revising the first edition, I have resisted the temptation of adding moretopics to the text. The main aim has been to rectify the defects of the firstedition:

Efforts have been made to remove the typos and correct the mis-matched cross references. I hope there are none now.

In view of the feedback received from students, at many places phraseslike `trivial to verify', `easy to see', etc have been expanded withexplanations.

Sequencing of topics in some of the chapters has been altered to makethe development of the subject matter more consistent.

Short notes have been added to give a glimpse of the link betweenmeasure theory and probability theory.

More exercises have been added.

Four new appendices have been added.

While preparing the first edition of this book, I was often questionedabout the `utility' of spending my valuable `research time' on writing abook. The response of the students to the first edition and the reviewers'comments have confirmed my confidence that writing a book is as valuableas doing research. I thank all the reviewers of the first edition for theirencouraging remarks. Their constructive criticism has helped me a lot inpreparing this edition.

xvii

xviii Preface to the Second Edition

I would like to thank Mr. N. K. Mehra, Narosa Publishers, for agreeingto copublish this edition with the AMS.

I take pleasure in offering thanks to Edward G. Dunne, AcquisitionsEditor, Book Program AMS, for the help and encouragement received fromhim.

I thank the editorial and the technical support staff of the AMS for theirhelp and cooperation in preparing this edition.

Once again, the help received from P. Devraj in revising the figures isgreatly appreciated. Thanks are also due to Mr. C.L. Anthony and ClarityReprographers & Traders for typesetting the manuscript in

I would greatly appreciate comments /suggestions from students andteachers about the present edition. I intend to post comments/correctionson the present edition on my homepage at

Mumbai, 2002 Inder K. Rana

Recipe for a onesemester course andinterdependence of thechapters

*Lebesgue measure and integration(40 lectures and 10 problem/discussion hours)

Prologue: Everything

Chapter 1: Sections 1.1 and 1.2 (depending upon the background of thestudents), 1.3 and 1.4 can be left for self study.

Chapter 2: Sections 2.1 and 2.2.

Chapter 3: Sections 3.1 to 3.3; 3.5 to 3.9; 3.10 and 3.11 (omitting theproofs).

Chapter 4: Sections 4.1 to 4.3; 4.4 and 4.5 (omitting the proofs); 4.6.

Chapter 5: Sections 5.1 to 5.6; Parts of 5.7 to 5.9 can be included de-pending upon the background of students and the emphasisof the course.

Chapter 6: Sections 6.1; 6.2 (omitting proofs); 6.3 (stating the theorem6.3.6 and giving applications: 6.3.8, 6.3.10 to 6.3.13, 6.3.16(omitting proof).

Chapter 7: Sections 7.1 to 7.4.

xix

xx Recipe for a one semester course and interdependence of the chapters

Interdependence of the chapters

PrologueThe length function

Chapter 1

Riemann integration

Chapter 2Recipes for

extending theRiemann integral

Chapter 3Chapter 3General extension 3.2,3.6and3.11

Chapter 4 theory only

The Lebesguemeasure on IR and Chapter 5

its properties Chapter 5 5.1 to 5.5 only

Integration

Chapter 6Fundamental theoremof calculus for theLebesgue integral

Chapter 10Signed measures andcomplex measures

Chapter 7Measure and Integration

on product spaces

Chapter 8Modes of

convergence andLp-spaces

Chapter 9Radon-Nikodymtheorem and its

applications Chapter 10

10.1 and 10.2 only

Notations used in thetext

The three digit system is used to number the definitions, theorems, propo-sitions, lemmas, exercises, notes and remarks. For example, Theorem 3.2.4is the 4th numbered statement in section 2 of chapter 3.

The symbol is used to indicate the end of a proof. The symbolA:= B or B =: A means that this equality is the definition of A by B. Thesymbol before an exercise means that this exercise will be needed in thelater discussions. Sections, theorems, propositions, etc., which are marked* can be omitted on first reading.

The phrase "the following are equivalent:" means each of the listed state-ment implies the other. For example in Theorem 1.1.4, it means that eachof the statements (i), (ii) and (iii) implies the other.

The notations and symbols used from logic and elementary analysis areas follows:

V

xEAx¢AAcBAcBP(X)

implies; givesdoes not implyimplies and is implied by; if and only ifthere existsfor all; for everyx belongs to Ax does not belong to AA is a proper subset of BA is a subset of Bset of all subsets of X

xxi

xxii Notations used in the text

0 empty setA \ B set of elements of A not in BA x B Cartesian product of A and Bn

Xi Cartesian product of sets X 1, ... , Xn.i=1inf infimumsup supremumU, U unionn,n intersectionA symmetric differenceAc complement of a set AE closure of a set EaE boundary points of a set Elim sup limit superior; upper limit

n--oolim inf limit inferior; lower limitn--oo

J X > Y f is a function from X into Y and f (x) = y.x ,--->x

ICI the set of natural numbersZ the set of integersQ the set of rational numbers][8 the set of real numbers][8* the set of extended real numbersC the set of complex numbersI[8n : n-dimensional Euclidean spacelxi absolute value of x

( a, b), (a, b], [a, b), [a, b], (-oo, oo), ;intervals in R.(-00, a), (-oo, a], (a, oo), [a, 00)

{an}n>1 : sequence with nth term an .(X, d) a metric space.

For the list of other symbols used in the text, see the symbol index givenat the end of the text.

Prologue

The length function

We denote the set of real numbers by R. Let ][8* denote the set of extendedreal numbers. (See appendix A for details.)

Let Z denote the collection of all intervals of R. If an interval I E Z hasend points a and b we write it as I(a, b). By convention, the open interval(a, a) = 0 V a E R. Let [0, +oo] := {x E ][8* Ix > Of = [O,+oo) U {+oo}.Define the function A : I ) [0, oo] by

A(I(a, b)) .- lb - al if a, b E 1[8,

+00 if either a = -oo or b = +oo or both.

The function A, as defined above, is called the length function and hasthe following properties:

Property (1): A(O) = 0.

Property (2): A (I) A (J) if I C J.

This is called the monotonicity property of A (or one says that A ismonotone) and is easy to verify.

Property (3): Let I E I be such that I = U=1 Ji, where Ji fl Jj = 0 fori = j. Then

A(I) A(Ji)i=1

This property of ), is called the finite additivity of )A, or one says that ),is finitely additive.

1

2 Prologue

To prove this, let I be an infinite interval. Then at least one of the Ji'sis an infinite interval and hence A(I) = +oo =

E'l\(J). Next, let I be a

finite interval with end points a and b, a < b. Then each Ji is a finite intervalwith end points, say, ai and bi where ai < bi. We may assume, without lossof generality, that a = al < b1 = a2 < b2 = = an <b n = b. Then

n n

AM = b - a = bn - al = E(bi - ai) = E A(Ji).i=1 i=1

Property (4): Let I E I be a finite interval such that I C U'i Ii, whereIi E I. Then

00

A (I) a(IZ).i=1

To prove this, we note that if Ii is an infinite interval for some i, then clearlyA(I) < +oo = >°°1 \(1I). So, assume that each Ii is a finite interval. Wefurther assume for the time being that I = [a, b] and that each Ii is anopen interval. Then by the Heine-Borel theorem, there exists some n suchthat I C U=1 I. Let Ii = (at, bi), i = 1, 2,... , n. Since a c I, there existssome i such that a c Ii. We rename this interval as I. If b E 11, thenA(I) < A(I1) < E001 A(Ii) and we are through. If not, then b1 2. Proceeding this way, we will have

al <a<b1 <b2 <...bm_1 <b<bm,for some m < n, and ai <b_1 for i > 2. Hence

A(I) = b-a< bm - a1

E(bi - bz-1) + bl - ali=2ME(bi - aZ) + (bi - al)

i=2

m 00

A (1j) < A (Ij).2=1 8=1

In the general case when I is an arbitrary finite interval, given any realnumber e > 1, we can find a closed interval J C I such that eA(J) = A(I)and an open interval Jn D In such that A(Jn) = EA(In) for every n > 1.Then J C U=1 Jn and by the earlier case, A(J) < J:I° Hence

00 00

A (I) A (J) < E E A (Jn) E2 EA (1n).n=1 n=1

The length function 3

Since this holds for every e > 1, letting e --> 1, we get00

A(I) < A(In).n=

Property (5): Let I E I be a finite interval such that I = U=1 In, whereIn E I and in fl L,,,, _ 0 for n : m. Then

00

A (I) = A (In).n=

To see this, we note that by property (4), A (I) < E001 A(In). To prove thereverse inequality, let I have left-endpoint a and right-endpoint b. Let Iihave left-endpoint ai and right-endpoint bi, i > 1. Since in n Im = 0 forn m, we can assume without loss of generality that d k > 1,

anal<b1<a2<...<bk<b.Then for every k > 1 we have

k

E A (In) _k

- an) < (b - a) < A (I) -n=1 n=

Hence E'1 A (In) <A (I).

Property (6): Let I E I be any interval. Then

00A(I) = E A(i n [n, n +n=-oo

If I is a finite interval, then I = U=k(I fl [n, n + 1)) for some integers kand 2, and by property (5) we have the required property. If I is an infiniteinterval, then A(I) = +oo and for an infinite number of n's we will haveI fl [n, n +1) = [n, n + 1). Thus

+00

E A(i n [n, n + 1)) = +oo = AM.n=-oo

Property (7): Let I E I be any interval such that I = U=1 In, In E I andin fl I 1 z = 0 for n : m. Then

00

A (I) = E A (In).n=i

This property of A is called the countable additivity of A, or one saysthat A is countably additive. If I is finite, we have already proved it in

4 Prologue

property (5). If I is infinite, we use properties (5) and (6) to obtain therequired equality.

Property (8): Let I E I and I C U=1 In, In E Z. Then

A (I) < E A (In).n=1

This property of A is called the countable subadditivity of A, or one saysthat A is countably subadditive. When I is a finite interval, we havealready proved it in property (4). The only case we have to consider is whenI is infinite and all the In's are finite. In this case, we write

+00

U (in [k, k + 1))

and note that for every k, I fl [k, k + 1) C U=1(Im fl [k, k + 1)). Usingproperties (5) and (6), we have

+00

E /\ (i n [k, k + 1))

< 1: 1: A (,Tn n [k, k + 1))k=-oo n=100 +oo

E E A(in n'[k, k + 1))n=1 k--oo00

E A (1n)n=1

The above properties show that A, the length function, is a nonnegative,monotone, countably additive and countably subadditive function. We re-mark that A({x}) = 0 for every x E ][8, for {x} = [x, x]. Finally, we state aproperty of A which depends upon the group structure on R.

Property (9): A (I) = A (I + x), for every I E I and x E R, where I + x{y+xly E I}.

This property of the length function is called translation invariance, orone says that A is translation invariant.

Chapter 1

Riemann integration

1.1. The Riemann integral: A review

The geometric problem that leads to the concept of Riemann integral is thefollowing: given a bounded function f : [a, b] -> II8, how to define the areaof the region bounded by the graph of the function and the lines x = a andx = b? For example if f (x) > 0 and a < x < b, one would like to find thearea of the region (Figure 1)

S(f):={(x,y)EII82ja<x<b, 0<y< f(x)}.

0 x=a

Figure 1: Area S(f)

Heuristically, this can be done by approximating the required area by theunion of rectangular areas from the inside and outside of S(f) (see Figures2 and 3). The required area is then captured between these approximatingareas. As the number of rectangles is `increased', one gets `better' approx-imations and one hopes to find the required area by a `limiting' process.These intuitive ideas can be made precise as follows.

5

1. Riemann integration

a=; x1.......... xi-1 x j ................. b = x x

Figure 2: Approximation from inside

a=xo xl.......... xt-1 x : ................. bx x

Figure 3: Approximation from outside

Throughout this section, f : [a, b] ---' JR will be a fixed bounded function.A finite set of points P = {x0, x1, ... , xn} is called a partition of [a, b] if

a=xo <xl <... <xn=b.

For a partition P = {xO, xl, ... , xn} of [a, b],

is called the norm of the partition P. Let P = {XO, xl, ... , xn} be apartition of [a, b]. Let

mi := inf If (x) I xi_1 < x < xi}

and

Mi := sup{ f (x) I xi_1 < x < xi}.

Definen

L(P, .f) mZ(xi - xZ-1)i=i

1.1 The Riemann integral: A review 7

and

U(P, f A (Xi - Xi-1) -

i=1Since f is assumed to be bounded, each m2 and M2,1 < i < n, exists andhence each of L(P, f) and U(P, f) is well-defined. The numbers U(P, f )and L(P, f) are called, respectively, the upper sum and the lower sumof f with respect to the partition P. Geometrically, L(P, f) approximatesthe required area from `inside' (see Figure 2), and U(P, f ) approximates therequired area from `outside' (see Figure 3).

If Pl and P2 are two partitions of [a, b] such that Pl C P2, i.e., everypoint in P1 is also a point in P2, then P2 is called a refinement of Pl.Given any two partitions Pl and P2, Pl U P2 is also a partition of [a, b]. Infact, P1 U P2 is a refinement of both Pl and P2 and is called the commonrefinement of Pl and P2. We first prove the intuitively obvious result: aswe refine a partition, the approximations U(P, f) and L(P, f) improve.

1.1.1. Proposition:

(i) For every partition P of [a, b],

m(b - a) < L(P, f) < U(P, f) < M(b - a),where m:= inf{ f (x) I a < x < b} and M:= sup{ f (x) I a < x < b}.

(ii) For any two partitions Pi and P2 of [a, b], if P2 is a refinement of P1,then

L(Pij) < L(P2J) < U(P2J) < U(Pij).(iii) For any two partitions Pl and P2 of [a, b],

L(Pj f) < U(P2 f).(iv) Let

and

a := sup{L(P, f) I P a partition of [a, b] }

,3 := inf{U(P, 1)1 P a partition of [a, b]}.

Then both a and ,3 exist, and a < ,3.

Proof: (i) Let P = {xO, xl, , xn} be any partition of [a, b]. If m2inf If (x) I xi_1 < x < xi} and Mz := sup {f(x) I xi_1 < x < x2}, thenm < m2 < Mi < M for 1 < i < n. Thus (see Figure 4)

m(x2 - xi-1) < mi(xi - xZ-1) < Mi(xi - xi-1) < M(xi - xi-1).

Hencen n

m(b - a) < Emi(xi - xi-1) < EMi(xz - xz-1) < M(b - a).i=1 i=1

1. Riemann integration

yl

Mi

mi

0 xo = a Xi_i xi

Figure 4

This proves (i).

(ii) Let Pl and P2 be any two partitions of [a, b] such that P2 is a refine-ment of Pl. First consider the case that the partition P2 has just one pointmore than Pl (see Figure 5).

A

0 xo = a xi -1 y xi xn= b x

Figure 5

Let P1 = ,xi-1,xi,... ,xn} and P2 = {xo,xi,' .,xi--1,y,xi, , xn}. It is easy to see that

suP{ f (x) I xi-1 < x < y} < sup{ f (X) I xi-1 < x < xi}

and

sup{f(x)Iy < x < xi} < sup{f(x)I xi_1 < x < x2}.

Hencen

U(Pi,f) - ESUPlf(x)I xj_1 G x < xjI(xj-xj-1)j=1

n

supff W I xj_1 < x < x.7I (x.7 - x.7-1)

j#i+ sup{ f (x) I xz-1 < x < xi}(xz - xi-1)

n

SUPlfW lxj_1 < x < xjI(xj - xj-1)

j=1j #i

+ sup{ f (x) I x2_1 < x < x2}(xi - y)

+ sup{ f (x) I x2_1 < x < x2}(y - xi-1)n

1: SUPlf(x) I xj_1 < x < xi I (Xj - x.7-1)j=1j#2

+ supff (X) I Xi ! X <- YYXi - Y)

+ supff (X) I Xi-1 -< X <- 0(y - Xi-1)U(P25 f)

In case P2 has k extra points, the above argument repeated k times willshow that U(Pl, f) > U(P2, f) whenever P2 is a refinement of Pl. Similararguments will prove that L(P2, f) > L (Pi, f ). This proves (ii).

(iii) Let P1 and P2 be any two partitions. Then by (ii),

L(Pi>f) < L(Pl UPa>f) < U(Pl UPa>f) < U(Pa1 f)

This proves (iii).

(iv) Let

L:= {L(P, f) I P a partition of [a, b] j

and

U:= {U(P, f) I P a partition of [a, b]}.

Since G and U are nonempty bounded subsets of ][8, a and 3 exist. Clearly,a <)3.

10 1. Riemann integration

1.1.2. Definition: Let f : [a, b] -> I[8 be a bounded function. The realnumbers

fand

b

f(x)dx :=sup{L(P, f) I P a partition of f}

fam

b

f(x)dx := inf {U(P, f) I P a partition of fj

are called the lower integral and the upper integral of f, respectively.The function f is said to be Riemann integrable on [a, b] if

fb

ff(x)dx = f(x)dx.

The common value in that case is called the Riemann integral of f over[a, b] and is denoted by

fbLb1 ff(x)dx.dx =

1.1 .3. Examples:

(i) Let f : [a, b] II8 be the constant function, f (x) - c. Then for everypartition P of [a, b], U(P, f) = L(P, f) = c(b - a). Hence f is Riemannintegrable and

Ia

b

f (x)dx = c(b - a).

x

Figure 6


(ii) Let f : [0, a] -> lI8 be defined by f (x) = x. Let us consider the partitionsPn :_ {0, a/n, 2a/n, , (n - 1)a/n, a}, n > 1. Clearly,

U(P .f)

Similarly,

y

2 nan2 Ej

(jn)n

j=1

In(n+i)n2 2 ,

2 nn

Figure 7

It is easy to see that {U(P, f)}n>1 is a decreasing sequence andJL(Pn, is an increasing sequence. Hence

sup{L(Pn, f) } =lima n - 1

n>1 n--+oo 2 ( n

and2ainf {U(P,f)} =

n moo 2 (n n 1) _

1

a2

2

a2

2

Thus

a2= sup{L(Pn,f)}

2 n>1

sup{L(P,f)} _P

< infjU(Pj)j

ia

f(x)dx

< infjU(Pnjf)j

a2

Hence f is Riemann integrable and

fnf (x)dx =

a2

(iii) Let f : [a, b] R be defined by

f(x):=< 0 if x is a rational, a < x < b,1 if x is an irrational, a < x < b.

It is easy to see that for any partition P of [a, b], U(P, f) = (b - a) andL(P, f) = 0. Hence

ff(x)dx6 = 0< (b - a) = ff(x)dx.6

Thus f is not Riemann integrable.

In general, it is not easy to compute the upper and the lower integrals ofa function and verify its integrability. The following theorem is very usefulfor such verifications.

1.1.4. Theorem: Let f : [a, b] ][8 be a bounded function. The followingstatements are equivalent:

(i) f is Riemann integrable.

(ii) For every e > 0 there exists a partition P of [a, b] such that

U(PI f) - L(PI f) < e.

(iii) There exists a unique real number a such that for every partition Pof [a, b]

L(P, f) < a < U(P, f).

Proof: We shall prove that (i) t (ii) and (iii) (i).

Suppose (i) holds and e > 0 is arbitrary. By definition, there existpartitions Pl and P2 of [a, b] such that

fb

U (P2i f) - /2 < f (x)dx < L(P, f) + e/2.

Let P Pl U P2. Then

U(P,f)-L(P>f) U(Pa7f)-L(Pi7f) < e.

This proves that (ii) holds. Hence (i) =:>. (ii).

The implication (ii) (i) is easy and is left as an exercise. Next, supposethat (i) holds. Let

f(x)dx:= a.L1dx = f b

Then a = fb f(x)dx and it satisfies the required property. To prove theuniqueness, suppose 3 ,Q E Il8 such that for every partition P of [a, b],L(P, f) < 0:5 U(P7 f). Then

f (x) dx = Ceff(x)dxa

Hence a = ,Q. This proves (iii). Finally suppose (iii) holds but f is notintegrable. Then for every partition P of [a, b],

b

L(P, f) < f(x)dx < Jb

f(x)dx < U(P, f )Jna. a

But then every number a between fa f(x)dx and fa f(x)dx has the propertythat L(P, f) < a < U(P7 f ). This contradicts (iii). Hence (iii) implies thatf is Riemann integrable, i.e., (iii) (i).

1.1.5. Theorem: Let f [a, b] I[8 be a bounded function. Then thefollowing statements are true:

(i) If there exists a sequence {P}>' of partitions of [a, b] such that

lim (U (Pn7 f) - L (Pn7 f)) = 07n--+oo

then f is Riemann integrable and

b

lim L(Pn, f) = f (x)dx = lim U(Pn, f).n--+oo a n--+oo

(ii) If f is Riemann integrable, then there exists a sequence {Pn}n>1 ofpartitions of [a, b] such that each Pn+1 is a refinement of Pn, I I 0as n --+ oo and

Jim (U(P,f)-L(P,f))= 0.n->oo

Proof: (i) Suppose {P}>1 is a sequence of partitions such that

lim (U(PnI f) - L(Pnj ,f)) = 0.

Let e > 0 be given. We can choose no such that d n > no

U(Pn f) - L(Pnj f) < 6.

Hence by theorem 1.1.4, f is Riemann integrable. Further, d n > no

U(P.J)-Hence

Similarly,

fb

f (x)dx f) < e.

f6urn U`Pn, f(x)dx.

a

lim L (Pn, f) _n->oo

6

f(x)dx.

(ii) Suppose f is Riemann integrable. Since

fb

f (x)dx = sup{L(P, f) I P a partition of [a, b]}

= inf{U(P, f) I P a partition of [a, b]},

we can choose sequences {P}>1 and f P, }n>1 of partitions of [a, b] suchthat

b

lim L(Pn, f) = f (x)dx = lim U(Pn, f ).n->oo a n->oo

Let Pn = {xO, x1, ... , xn}, where xo := a, Xk := xk_1 + (b - a)/n for 1 <k < n. (Pn* is called the regular partition of [a, b].) Let Po {a, b} andput

n

Pn : = Pn_ 1 U Pn U U (Pn U Pn) , V n > I.(k=1

Then j< I /n V n and each Pn+1 is a refinement of Pn. Further, since

U(Pn, f) - L(Pn, f) < U(Pn , f) - L(Pn, f),clearly

Jim (U(P,f)-L(P,f))=0.n-+oo

1.1.6. Examples:

(i) Let f : [a, b] ) Il8 be any monotonically increasing function, i.e., f (x) >f(y) if x > y. Let Pn = {xO, xl, ... , xn} be the regular partition of [a, b],i.e., xi - xi_1 = (b - a)/n, l < i < n. Then

U(P ,f) =

and

LAP .f)_n

n

f I a+ Z(bn a) / \b na/i=

Hencei=

U(Pnj f) - L(Pnj f)

a)1 (b_ai(a+_1_ n)n

b - a n - i(b - a)(a +

n f(b - a)

(f (b) f (a))n

a(i-1)(b-a)11

n lJ

It follows from theorem 1.1.5 that f is Riemann integrable. A similar ar-gument will prove that f is Riemann integrable when f is monotonicallydecreasing.

(ii) Let f : [a, b] ][8 be a continuous function. We show that f is Riemannintegrable. Since f is continuous, it is bounded and is uniformly continuous.Let e > 0 be arbitrary. We choose 8 > 0 such that

If (x) - f (y) I < e/(b - a) whenever Ix - yj < J.Let P be any partition of [a, b] such that IIPII < J. Let P = {xo, xl,... , xn}.Since f is continuous, it attains its maximum and minimum values Mi andmi, respectively, on the interval [x2_1, x2] at some points, say f (y2) = Miand f (zi) = mi, l < i < n. Clearly,

(Mi - mi) < e/(b - a) for every iand

U(P, .f)-L(Pj .f)n

i=1

E

b-

Thus f is integrable, by theorem 1.1.4.

Mi - rn) (xi - xi-1)

n

i=1- xi_ 1) = E.

1.1.7. Exercise: Analyze whether the following functions are Riemannintegrable or not:

(i) 1(x) ;- r 1 if 0 < x < 2, x 41,l 2 if x=1.

(ii) f : [a, b] -+ R, f has only a finite number of discontinuity points.

(iii) f : [0, 1] --) II8,

f (X) .=

1.1.8. Exercise:

0 if x is irrational, 0 < x < 1,x if x is rational, 0 < x < 1.

(i) Let f : [a, b] -- Il8 be a nonnegative continuous function and let f (c) > 0for some c c [a, b]. Show that fa f(x)dx> 0.

(ii) Let f [a, b] Il8 be continuous and f (x) > 0 V x c [a, b]. Iff 6 f(x)dx = 0, show that f (x) =- 0.

1.1.9. Exercise: Let f, g : [a, b] I[8 be bounded functions. Prove thefollowing assertions:

f(x)dx+f g(x)dx.b(1) L (f + 9)(x) ? f b b

a

b

(ii) J (f+g)(x)dx <no.

(iii) (a f) (x)dx =L

6

fbf(x)dxfg(x)dx.+

f(x)dx if a > 0,faa

6

f (x)dx if a < 0.

1.1.10. Exercise:x E [a, b],

and

Let f [a, b] ) I[8 be a bounded function. Define for

I (x)

{f(x) if f(x) > 0,

0 if f(x)<0

-f (X) if f (X) < 00 if

f fand the function f.

(i) Show that f = f + - f - and If I = f + + f(ii) Let f be Riemann integrable. Show that both f+ and f - are also

Riemann integrable.

(iii) From (i) deduce that If I is also Riemann integrable andfb

f(x)dx 5fb

l

1.1.11. Exercise: Construct a function f [a, b] I[8 such that If I isRiemann integrable but f is not Riemann integrable.

1.1.12. Exercise: Let f : [a, b] ) II8 be Riemann integrable. Show thatf is also Riemann integrable on [a, x] V x E [a, b] and the function

fXF(x) := x E [a, b],

is uniformly continuous.

1.1.13. Exercise: Let f : [a, b] II8 be a bounded function. Show that fis Riemann integrable if

for

b b

f(x)dx = fa (-f)(x)dx

fb b

af (x) dx fa (- f ) (x) dx.

We end this section with the definition of integrability as given by Bern-hard Riemann in 1854.

1.1.14. Definition: Let f [a, b] --> II8 be a bounded function and letP = {a = xO < xl < < xn = b} be any partition of [a, b]. Let ti E[x_ 1, xi],1 < i < n, be chosen arbitrarily and let

n

S(P) f f (ti) (Xi - Xi- 1)i=1

S(P, f) is called a Riemann sum of f with respect to the partition P.We say that the function f is R-integrable if there exists a real number Lhaving the property that for every e > 0 there exists some 6 > 0 such thatif P is any partition of [a, b] with IIPII < S and S(P, f) is any Riemann sumof f with respect to the partition P, then IS(P, f) - LI < e. We write this as

lim S(P, f) = L.IIPII--+o

1.1.15. Theorem (G. Darboux): Let f [a, b] ) II8 be a boundedfunction. Then f is R-integrable if f is Riemann integrable, and in that

caseb

f f(x)dx= II1 m S(P, f)Q

Proof: See Apostol [2].

1.2. Characterization of Riemann integrablefunctions

Let R[a, b] denote the set of all functions f : [a, b] ) I[8 which are Riemannintegrable. Our next theorem describes some properties of R[a, b] and themap f H fa f(x)dx for f E R[a, b].

1.2.1. Theorem: Let f, g : [a, b] ) ][8 be bounded Riemann integrablefunctions and a be any real number. Then:

(i) f + g is also Riemann integrable and

fJ 6(f+g)(x)dx=f 6 f(x)dx+fba(ii) a f is Riemann integrable and

fb 6

(af)(x)dx=af

(iii) If f (x) < g(x) V x E [a, b], then6 b

f f(x)dx < g(x)dx.a

Proof: (i) Since f, g are integrable, given e > 0 we can choose partitions Pland P2 of [a, b] such that

U(Pij f) - L(Pij f) < e/2

and

U(P2,g) - L(P2, g) < e/2.Let P = Pl U P2. Then clearly

U(Pj f) - L(Pj f) < e/2

andU(P, g) - L(P, g) < e/2.

Since for any set A C [a, b],

sup{ f (x) + g(x) I x E Al < sup{ f (x) I x E Al + sup{g(x) I x E Al

and

inf { f (x) + g(x) I x c Al > inf { f (x) I x E Al + inf {g(x) I x E A},

1.2 Characterization of Riemann integrable functions

we have

and

Hence

U(PI f + 9) < U(PI f) + U(PI 9)

L(PI f + g) > L(PI f) + L(PI g).

U(PI f + 9) < U(PI f) + U(PI 9)L(P, f) + L(P, 9) + EL(PI f + g) + c.

Thus f + g is Riemann integrable by theorem 1.1.4. Further,

f6

(f + g) (x) dx 0 is arbitrary,we have

. (1.1)J b(f+g)(x)dx < bf(x)dx+f bg(x)dxa a

Similarly,/6

J(f + g)(x)dx

aL(P, f + 9)

U(P1f)+U(P19)-6

g(x)dx - c.a

f(x)dx+f

Again, since f, g and f + g are integrable and e > 0 is arbitrary, we have

g(x)dx. (1.2)J 6(f+g)(x)dx > J b f(x)dx+f ba a

Equations (1.1) and (1.2) complete the proof of (i). The proofs of (ii) and(iii) are easy and are left as exercises.

The above theorem tells us that 7Z[a, b] is a vector space over R and themap f F fb f dx is a linear, order preserving map from R [a, b] to R. In viewaof examples 1.1.6, 7Z[a, b] includes the class of all monotone functions andthe class of all continuous functions. Thus the space 7Z[a, b] is quite large.In view of exercise 1.1.7(ii), it is natural to ask the following question:

Let f : [a, b] ) R be a bounded function which is discontinuous at aninfinite set of points in [a, b]. Is f Riemann integrable?

Example 1.1.3 (iii) shows that the set D of discontinuities of f cannotbe too large, e.g., D cannot be equal to [a, b]. What about if D is countablyinfinite? (See example 1.2.14.) To analyze integrability of f on an intervalwhich includes points of discontinuity of f, we should look at the pointsof discontinuity in a more quantitative way. To do that we introduce thefollowing concept.

1.2.2. Definition: Let f [a, b] ][8 be a bounded function and J asubinterval of [a, b]. Let

w(f, J) := sup{ f (x) : x c J} - inf{ f (x) : x E J}

and for x c [a, b], letw(f, X) lim w(f, J),

xEJwhere A(J) denotes the length of the interval J. The function cv(f,J) iscalled the oscillation of f in the interval J and w(f, x) is called theoscillation of f at x. Clearly w(f, x) > 0 for every x c [a, b].

1.2.3. Examples:

(i) Let f (x) := r x + 2 if x > 0,0 ifx<0.

Then w(f,x)=0 V x Oandw(f,0)=2.

(ii) Let f (x) :0 if 0 < x < 1,x if 1 < X < 2.

Then w(f,x)=0if0<x<1, w(f,l)=landw(f,x)=0if1<x < 2.

Let = J sin 1/x if x 0,(iii) f(x)

0 ifx=o.Then w (f , 0) = 2.

1.2.4. Exercise: Let f : [a, b] -) R.

(i) Show that f is continuous at c E [a, b] if w(f, c) = 0.(ii) Let w(f, x) < E for every x c [a, b]. Show there exists S > 0 such that

If (x) - f (y)l < e whenever Ix - yj < S.(Hint: Use the fact that [a, b] is compact.)

The main aim of introducing the concept of oscillation of a function ata point is that it helps us to analyze the set of discontinuities of f. In some

1.2 Characterization of Riemann integrable functions 21

sense, oscillation gives a quantitative measure of discontinuity. For the restof the chapter, we fix f : [a, b] ) Il8, a bounded function. Let

D :_ {x E [a, b] I f is not continuous at x}.

In view of exercise 1.2.4, we can write00

D x c [a, b] I w(f, x) > 01 = U D,

whereDn:= fX G [a,b] I w(f,x) > I/nj,

i.e., Dn is the set of those points x at which the oscillation of f exceeds 1/n.For any e > 0, let us analyze the set

D, := f x c [a, b] I w(f, x) > ej,

when f is Riemann integrable. Since f is Riemann integrable, given any,q > 0 we can choose a partition P = {a = xO < xl < < xn = b} of [a, b]such that

Let

U(P, f) - L(P, J) 1

Then

S:= {iI1i < n and (x_1,x)flDO}.

De c Uxi_i,xiliES

and for every i ¢ S, (x_, xi) n D, _ 0. Note that for i E S, 3 someci E [x_i,x]flD. Thus

(M-m) > w(,f, ci) > E,where

Mi suP{ f (x) I x E [x_1,x]} and mi := inf{ f (x) I x c [x_,xJ}.Hence

,q> U(P, ,f) - L(P, f) ? (Mj - mz)(xi - xi-i)iES

(x-x_).iES

Thus for any i > 0, we have intervals [x_, xi], i E S, such that

D e c U[xi_1, xi] and E(xj-xj-j)<,q1E-iES iES

In a sense, what we have shown is that if f is Riemann integrable, then DEcan be covered by closed intervals of total length as small as we want. Letus record this in the next proposition.

1.2.5. Proposition: Let f : [a, b] ) ][8 be a Riemann integrable functionand DE :_ {x E [a, b] I w(f, x) > E}. Then for every 77 > 0 there exist closedintervals Il , 1 2 , . , IkC [a, b] such that

k k

DE C Uli and EA(Ij) <,q.j=1 j=1

1.2.6. Exercise: Let f, DE be as in proposition 1.2.5. Show that for anygiven q > 0 there exist open intervals I1, ... , Ik such that DE C U=1 Ij andEk=1 A (Ij) < 77.

1.2.7. Corollary: Let f [a, b] ) ][8 be Riemann integrable. Then forevery q > 0 there exists a sequence of closed intervals Jl, J2, J3, ... in [a, b]such that

00 00

D := {x I f is not continuous at x} C U A and E A(JO < n.k=1

Proof: Note that D = lJ1 Dn,i where

Dn := {x E [a, b] I w(f, x) > 1/n}.

Now, by proposition 1.2.5 applied to Dn for every n, we can find closedintervals In , I2, ... jkn, such that

kn n

Dn C- U Ijn and E A (Ijn) < 2n.j=1 j=1

For n = 1, 2, ... and 1 < j < kn, let the intervals Ijn be renamed asJ1, J2, ... . Then

00 00 00 n

D = U1n C UJn and EA(Jk) < EEA(Ijn) < ,q.

n=1 n=1 k=1 n=1 j=

In view of the above corollary, we have the following definition.

1.2.8. Definition: A subset E C JR is said to be a null set if, givenE > 0, there exists a countable family of intervals {In In > 11 such that

:`0 < E.E C U=In and >1A(I)

1.2.9. Remark: It is not important in definition 1.2.8 as to what is thenature (i.e., open, closed, etc.) of the intervals. In fact, if E is a null set wecan choose intervals of the type we like to satisfy the required properties.

1.2.10. Examples:

(i) Clearly every singleton set {x}, x E R, is a null set, for {x} = [x, x] := Iand A (I) = 0. It is easy to check that every finite set is a null set.

(ii) Any countably infinite set S = {Xi, x2, x3i ... } is a null set. To see this,let c > 0 be given. Let Ii := (xj - e/2j+1, xj + E/2i+1), j = 1, 2, ....Then

00 00

S C U Ii and 1:/\(Ii) < E.i =1 i=

(iii) Q, the set of rational numbers, is a null subset of R.

1.2.11. Exercise:

(i) Show that every subset of a null set is also a null set.00

1 An is a null set.(ii) Let A1i A2, ... , An,*** be null sets. Show thatlJ,

(iii) Let E C [a, b] be any set which has only a finite number of limit points.Can E be uncountable? Can you say E is a null set?

(iv) Let E be a null subset of ][8 and x E R. What can you say about thesets E+x:={y+xI yEE}andxE:={xyI y E E}?

1.2.12. Exercise:

(i) Let I be an interval having at least two distinct points. Show that I isnot a null set.

(ii) If E contains an interval of positive length, show that it is not a null set.Is the converse true, i.e., if E C R is not a null set, then does E contain aninterval of positive length?

1.2.13. Example (Cantor sets): Let I := [0, 1] and let 0 < a < 1 befixed. We remove from the interval [0, 1] successively a collection of openintervals as follows: (i) Remove from [0, 1] the central open interval I' oflength a/3. Then

1 1 a 1 aIl' 2

2.312+2.3

This will give two disjoint closed intervals ill and J2 (Figure 8), each oflength (1 - a/3) /2.

1 _ a 1 a2 2.3 + 2.3

0 Ji Ii JZ 1

Figure 8: Step one

(ii) Remove the central open intervals Ii and 122 each of length a/9 fromthe intervals ill and J2 , respectively, to get four disjoint closed intervalsJi, J2, J3 and J4 (Figure 9) each of length (1 - a/3 - 2a/9)/4.

__Y 1A

0 J2 J21 2 J3 J4

Figure 9: Step two

(iii) Proceeding as above, suppose at the nth step we have obtained 2ndisjoint closed intervals J1', ... J21 such that each has length(1 - a/3 - - 2'2-la/3n)/2n. We remove central open intervals II+1,

1 < k < 2n, each of length a/3n+1 from each of the J,n to get 2n+1 disjointclosed intervals J, +1 1 < k < 2n+1. Let

2n-1

An := U In-j=1

2n 00 00

Bn : = U J and Ca : = fl Bn =[0,1]\ An .

k=1 n=1 (n=1

The set An consists of those points which are removed at the (n + 1)th stagefrom the leftover of [0, 1] after the nth stage, and Bn is the set of pointsleftover after the (n + 1)th stage removal has been completed. The set Ca isthe set of points left after all the intervals Iin,1 < j < 2n and n = 1, 2, ... ,have been removed from I. The sets Ca are called Cantor sets.

Note that Ca 0; for example, it includes the endpoints of all the inter-vals removed. Further, the sum of the lengths of all the intervals removedfrom I is

00

E a2n/3n+l = a.n=0

If 0 < a < 1, then Ca cannot be covered by a countable collection ofsubintervals of total length less than 1 - a (see property (8) of the lengthfunction A in the prologue). Hence Ca is not a null set for 0 < a < 1.

For a = 1, Ca is a null set. To see this, let c > 0 be arbitrary. We choosea positive integer n sufficiently large so that

00

E 2k-1/3k < e.k=n+1

Then the finite number of intervals JJ,1 < k < 2n, will cover C1. Thus fora = 1, Ca is a null set. The set C1 is more popularly known as Cantor'sternary set. The reason for this is that points of C1 are precisely thosereal numbers x in [0, 1] that admit ternary expansions 0 a1 a2 ... an--- , i.e.,

00

x = an/3n,n=1

where each an is either 0 or 2. This fact can be used to show that theset C1 is uncountable. For this, consider the map f : C1 -- ) [0, 1] definedas follows: for x c C1, if x = E 1 an/3n is its ternary expansion, thenf (x) := y c [0, 1] is the point having the binary expansion

00

Y = > bn/2n,

n=1

where bn = an/2 for every n. It is easy to check that f is a bijective map.Thus C1 is an uncountable, null subset of R.

The sets Ca, 0 < a < 1, are all very useful in constructing examples, aswe shall see later. This is mainly because of the following nice topologicalproperties that the sets Ca have.

First of all, each Ca is a closed subset of R, as it is intersection of theclosed sets Bn, n = 1, 2, .... We show next that Ca is nowhere dense in[0, 1]. To see this, note that Ca includes no nonempty open interval. For if(a, b) C Ca, then (a, b) C Bn V n. But then (a, b) C JJ for every n and forsome k, 1 < k < 2n. Thus V n,

(b-a) < A(Jj) < 1/2n

)

and hence (a, b) _ 0, a contradiction. That Ca is nowhere dense followsfrom the fact that if a closed set includes no nonempty interval, then it isnowhere dense.

Finally, we show that Ca is a perfect set. To prove this we shall showthat Ca is dense in itself (since Ca is closed). Let x E Ca be the left endpointof an interval II for some n > 1 and 1 < k < 2n-1. Then x E JJ V n, forsome 1 < k < 2n, and x will always be its right endpoint. If xn denotes

the left endpoint of this JL , then clearly x,z E C« and x,z converges to x. Asimilar argument applies if x E C« is the right endpoint of some II. Finally,suppose x c C« is such that x is not the endpoint of any II ; then x willbe an interior point of JJ for each n and some k. Clearly the sequence ofendpoints of these JJ will converge to x. Hence C« is dense in itself.

In the terminology of null sets, corollary 1.2.7 states that for a Riemannintegrable function f, the set of discontinuities of f is a null set. It isnatural to ask the question: Is the converse true, i.e., if f is a boundedfunction on [a, b] and the set of discontinuities of f is a null set, can we sayf is Riemann integrable? The answer to this question is in the affirmative.Before we prove this, let us look at an example of a function on [0, 1] whichhas an infinite number of discontinuities in every subinterval of [0, 1], yet isRiemann integrable.

1.2.14. Example (Popcorn function): Let f : [0, 1] [0, 1] be definedas follows:

1 if x = 0,f (x) := 1/q if x c (0, 1] is a rational x = p/q in lowest terms,

0 if x E [0, 1] is an irrational.

The function f is continuous at every irrational and discontinuous at everyrational point. Let us analyze f for integrability. First note that if I is anyopen interval that intersects [0, 1], then f (x) = 0 for some x E [0,1] n I.Thus L(P, f) = 0 for every partition P of [0, 1]. Let e > 0 be arbitrary. Let

A := f x c [01] 1 f (x) > 6/21.

We note that A is a finite set, since it consists of only those rationals x = p/qwhere 1/q > E/2 and 0 < p < q. Now we can cover these points of A by opensubintervals of [0, 1] such that the total length of these subintervals is lessthan E/2. Without loss of generality, we may assume that these subintervalsare disjoint and are [xk, Yk], 1 < k < n, where yk > xk, k = 1, 2,... , n - 1.Consider the partition P of [0, 1] given by

Let

and

P={02J0<xl <yl <...<xn<iJnCxn+1=1}.

Mk:=sup{f(x)I xk <x<yk}, forl<k<n

< n + 1.Mk :=sup{ f (x) Iyk_1xxk}, for 1<k

1.2 Characterization of Riemann integrable functions

Then for each k, Mk < 1 and Mk < e/2. Thus

U(PI f) - L(Pj f) = U(PI f)

27

n n+l

- 1: Mk(yk - xk) + 1: Mklxk - Yk-1)k=1 k=1

n n+l< 1: Mk(Yk - xk)+ El2 j:lxk - Yk-1)

k=1 k=1n

<k=1

< E/Z + E/2 = E.

Hence f is integrable on [0, 1] and

f f(x)dx = 0.

1

A closer look at the above example tells us that f becomes integrablemainly because the set {x E [0, 1] 1 w(f, x) > e/2} = {x E [0, 1] 1 f (x) > e/2}is a null set. Thus in order to analyze the integrability of f we shouldanalyze the sets DE :_ {x E [a, b] I w(f, x) > e} in detail and expect f to beintegrable if DE is a null set for every e > 0.

1.2.15. Proposition: Let f : [a, b] - ][8 be Riemann integrable. Then DEis a compact set for every e > 0.

Proof: Since [a, b] is compact and DEC [a, b], it is enough to show that DEis a closed subset of [a, b]. Let t E [a, b] be any accumulation point of D.Suppose t V DE, i.e., w(f, t) < e. Then there exists some St > 0 such that

w(f, I) < e, where I = (t-6t,t+6t)n [a, b].

But then for every interval J such that t E J C I, we have w(f, J) < e.Hence

w(f, y) < e for every y E I.

Thus, I n D, _ 0, a contradiction to the fact that t is an accumulation pointof DE . Hence t E DE and DE is a closed subset of [a, b].

1.2.16. Theorem: Let f [a, b] ) ][8 be a bounded function such thatD :_ {x E [a, b] I w(f, x) > Of is a null set. Then f is Riemann integrable.

Proof: Let e > 0 be given. In view of theorem 1.1.4, we have to constructa partition P of [a, b] such that U(P, f) - L(P, f) < e. For this we proceedas follows:

Step 1: Since D = {x E [a, b] I w(f,x) > Of is a null set, DE, := {x E[a, b] I w (f , x) > E' j is also a null set for every E/ > 0. Let E' be fixedarbitrarily. We choose open intervals I1, I2, ... , In, ... such that

00 00

DE, C UIn and EA(In)<E'.n=1 n=1

Since DE, is compact (proposition 1.2.15), we can find a positive integerm such that D C U'=1 In. Let us arrange the endpoints of the intervalsInn [a, b] , n = 1 , 2, ... , m, as a = x 1 < X2< < x,. = b. Then these pointsgive us a partition P of [a, b]. Further, note that if

S:={1 0 such that If (x) - f (y) l < ewhenever I x-yI < S. Without loss of generality, we assume that the partitionP has the property that 11PM < S, for if required we can consider a refinementof P. Thus for every i V S, if x, y E [x_i, xi], then If (x) - f()I < e'.

Step 3: Let f(x)I < M d x E [a, b]. Let

MZ := sup{ f (x) I xi_1 < x < xi} and m2 := inf{ f (x) I xz_1 < x < xi}.

Then using steps 1 and 2, we get

U(P, f)-L(P, f) _ 1:(Mi-Tni)(xa-xs-i)i=l

j:(Mi - Tni) (xi - xi- 1)iES

+ j:(MZ - mi)(xi - xZ-1)ivs

< 2Mi (xi - xi-1) + E j:(Xi - xi-1)iES i¢S

T

< E (xi - xi-1)i=1

(2M + b

Since E' is arbitrary, we could start with E/ such that (2M + b - a)E' _proving that f is Riemann integrable.

E)

1.2.17. Exercise:

(i) Let D C [a, b] be a null set. Show that [a, b] \ D is dense in [a, b].

(ii) Let D C R be a null set. Can we say R \ D is dense in R?

(iii) Let f E -R[a, b] be such that the set {x E [a, b] I f (x) # 0} is a null set.Use (i) to show that

f1.2.18. Exercise:

b

f (x)dx = 0.

(i) Let f and g be Riemann integrable on [a, b]. Show that f +g, f -g, fg anda f are all Riemann integrable for a E R. In particular, R[a, b] is a vectorspace over R. Further, the map

6if f(x) dx, f

is a linear map from R[a, b] to R.

(ii) Let f c R[a, b] be such that f > 0. Show that fQ f(x)dx > 0. Further,

fb

f (x)dx = 0 if {x c [a, b] I f (x) > 0} is a null set.

(Hint: Use exercise 1.2.17(iii) for the `if' implication and proceed as in theproof of proposition 1.2.5 for the `only if' implication.)

(iii) Let f, g be bounded functions on [a, b] such that f c 7Z [a, b] and {x E[a, b] I f (x) =A g(x)} is a null set. Is g Riemann integrable? If yes, is

fna.

1.2.19. Exercise:

b

f(x)dx = f b g(x)dx?a

(i) Let g E -R[a, b] and assume m < g(x) < M for all x E [a, b]. Let f[m, M] --+ R be a continuous function and h = fog. Show that h E 7Z [a, b]

(ii) Let f be as defined in example 1.2.14 and g(x) = 1 if 0 < x < 1,g(0) = 0. Show that h = 9o f is not Riemann integrable on [0, 1], althoughboth f E R[0,1] and g E R[0,1].

1.2.20. Exercise: Let f be any monotone function on [a, b]. Show thatf E IZ[a, b]. (Recall a monotone function has only countably many disconti-nuities. See also example 1.1.6 for a direct proof.)

1.2.21. Exercise: Let X be any nonempty set and A be any subset of X.The function XA : X --) {0, 1} defined by

XA1 if xEA,0 if x ¢ A,{

is called the characteristic function or the indicator function of theset A. Prove the following:

0) XAnB - X,gXB.

XAUB - XA + XB - XAnB111 XAAB - IXA - XBI'where

AAB = (A U B) \ (A n B).

1.2.22. Exercise: Let S C [a, b] be any nonempty set and let aS be theboundary of S. Show that XS E R[a, b] if aS is a null set.

1.2.23. Exercise: Let 0 < a < 1 and let f = Xca be the characteristicfunction of the Cantor set C« (see example 1.2.13). Show that f c R[0,1]ifF a = 1, and in that case fo f(x)dx = 0. Compute fo f(x)dx and fo" f(x)dxwhen f=Xc,O<a<l.

1.3. Historical notes: The integral fromantiquity to Riemann

The origin of the integral lies in the works of the Greek mathematicians.Euclid (300 B.C.) in his book `The Elements' and later Archemedes (287 -212 B.C.) in his treatise `The Method' gave techniques of computing areasand volumes which essentially amounted to computing definite integrals likefQ xdx and fa x2dx. However, the techniques were purely geometric and nogeneral method of computing the areas and the volumes was given.

Till the seventeenth century, concepts like area and volume were notdefined. They were only understood geometrically, and particular methodswere given to compute them. The discovery of calculus in the seventeenthcentury, which culminated in the work of Issac Newton, Gottfried WilhelmLeibniz and Leonhard Euler, defined the integral only as the inverse of thederivative, i.e., one wrote j¢ f (x)dx = F(b) - F(a), if one could find F suchthat F' = f. Again, this gave only a method of computing the area belowthe curve y = f (x), between x = a and x = b, provided one could find F(called an antiderivative of f) such that F' = f. The concept of whatshould be called the area below the curve was not defined. It should bementioned here that at that point of time even the concept of a function (as

1.3 Historical notes 31

we understand it today) was not clear. In fact, the need to define a functionmore precisely had a great hand in the evolution of the integral concept.

An introduction to the notion of function was made by Rene Descartes(1596-1650), who gave relations between variables in terms of algebraic equa-tions. The need to define a function precisely was felt more strongly duringthe second half of the eighteenth century, when the works of Jean d'Alembert(1747), Leonhard Euler (1748) and Daniel Bernoulli (1758) on the vibratingstring problem started a debate on the concept of a function. By the endof the eighteenth century, the concept of function as given by Euler in 1755became acceptable: "If some quantities depend upon other in such a way asto undergo variation when the latter are varied, then the former are calledfunctions of the latter." However, it was still believed that the graph of afunction could be traced with a "free motion of the hand", i.e., functionswere still believed to be piecewise smooth. In the year 1807, Joseph Fourierrekindled interest in the definition of function. He considered `arbitraryfunctions': a function which takes any values and which may or may not begoverned by a common law, i.e., not given by a single formula everywhere(but still believed to be differentiable except at a finite number of pointsin any finite interval). For such a function f defined on [-7r, 7], Fourierclaimed that one can have the following representation:

a00

f (x) = 2 +) , [ a, cos nx + bn sin nx ], (1.3)

where the constants ao, a1, ... , and b1, b2, ... are given by

C6n = 1 J f (x) cos nx dx and b,,,, = 1 Jir

f (x) sin mx dx, (1.4)7r 7r -

for every n = 0, 1, ... and for every m = 1, 2,... .

To prove his claim, Fourier implicitly assumed that the representation(1.3) is possible, and then to compute an and bn, he assumed term by termintegration of the series in (1.3) when multiplied with cos nx or sin nx. Hefurther assumed that the integrals in (1.4) made sense. In fact, integra-bility of f (x) cos nx and f (x) sin mx presented no difficulty, since for himf had at most a finite number of discontinuity points in [-7r, 7]. So hecould claim the existence of integrals by the eighteenth century viewpoint,i.e., area below the curve - the area concept still being undefined. ThoughFourier had broadened the concept of function to include piecewise con-tinuous function, it was only in 1823 that Augustein-Louis Cauchy gave aprecise definition of a continuous function and defined the integral of sucha function as limllpll,ooS(P, f) (see definition 1.1.14). Not only did Cauchygive the definition of the definite integral and show that for a continuous

function it exists, he also showed the existence of antiderivative for contin-uous functions. To be specific, for the first time a rigorous proof of whatcame to be known as the `fundamental theorem of calculus' was given byCauchy.

The series in (1.3) came to be known as the Fourier series and theconstants an, bn (given in (1.4)) the Fourier coefficients of the function f.The problem of finding conditions under which the Fourier series convergesto f (x) attracted many mathematicians, and it played an important role inthe development of the concept of integral. Even though Cauchy's notionof integral was able to give meaning to the Fourier coefficients an, bn, it wasstill believed that functions can have only a finite number of discontinuitiesin any finite interval, and hence Cauchy's notion of integral can be extendedto such functions also. However, Peter Gustav Lejeune-Dirichlet in 1829gave an example of a function (f (x) := 1 if x is rational and f (x) := 0 ifx is irrational, now known as the Dirichlet function) which had an infi-nite number of discontinuities in every finite interval. Here was an exampleof a function which was not defined by a formula, nor could its graph bedrawn. Later, in 1837, Dirichlet gave the definition of the function which isemployed most often now. The example given by Dirichlet not only broad-ened the concept of a function, but also made mathematician feel the needto treat continuous and discontinuous functions with equal vigor. Also, itmade mathematicians think about extending the notion of integral fromcontinuous functions to a more general class of functions in order to analyzethe problem of convergence of Fourier series. It was Bernhard Riemann (in1854) who extended the notion of integral to bounded (not necessarily con-tinuous) functions on intervals, as given in definition 1.1.14. To show thatthe class of functions for which integral fa f(x)dx existed was indeed biggerthan the class of continuous functions, he gave the following example (seealso example 1.2.14).

1.3.1. Example (Riemann): Let f : [-1/2,1/2] IlS be defined by

fW ._ r x if -1/2 < x < 1/2,0 if x = 1/2 or x = -1/2.

Let f be extended to Il8, such that

f (x) := f (x + 1), x c R.

The graph of f is given in Figure 10 below. Let the function R : 1[8 -> IlSbe defined by

00 f(nx)R(x -

n2x C R.

n=

1.3 Historical notes

Figure 10: Graph of f

Clearly, for every x,

33

00

JR(x)l <2E

n2< + oo,

n=ishowing that R is a bounded function. Also it is easy to see that R is con-tinuous at x if for every k, kx is not an odd multiple of 1/2, i.e, x = m/2k,m and 2k being relatively prime. Let x = m/2k, where m and 2k are rel-atively prime. Then nx = mn/2k will be an odd multiple of 1/2 if n is anodd multiple of k, i.e., n = k(2P+ 1), P = 1, 2, ....And at these points, i.e.,at k(V + 1)x, = 1, 2, ... ., thfunction f(mx)/m2 has a jump of magnitude1/2n2 from the right and 1/2n2 from the left. Hence for x = m/2k,

°O1

2

R(x+) = R(x) -2k2 (2l + l)2 = R(x) -

16k2e_1

and

)( +e=i

R(x-) = R(x) +2k2

L2

= R(x) +16k2'2l 1

°O 2

(Here we have used the result due to Euler : J:' 1 2t = g For a proofsee corollary 8.10.8.)

Hence R is discontinuous at x if x = m/2k, m and 2k being relativelyprime. We note that such points are dense in R. Finally, we show that R(x)is integrable. Let e > 0 be any real number. Then

D, x E [a, b] w(R, x) > EJ

_ {x E [a, b] x = m/2k, m and 2k relatively prime and 7r2/8k2 > e}.

Thus DE is at most a finite set, and hence is a null set. Now by theorem1.2.16, R is integrable.

Riemann's original definition of integrability of a function f on an in-terval [a, b] is as given in definition 1.1.14. The equivalent definition ofintegrability (i.e., definition 1.1.2) is due to Gaston Darboux (1875). Healso proved theorem 1.1.15 and the following theorem.

1.3.2. Theorem (Fundamental theorem of calculus for Riemannintegrable functions): Let f : [a, b] -- ][8 be a Riemann integrable func-tion. Let F(a) := 0 and F(x) := fa f (t)dt, x E [a, b]. Then the followinghold:

(i) F is uniformly continuous.

(ii) If f is continuous at x E [a, b], then F is differentiable at x andF1 (x) = f (x).

(iii) If G is any differentiable function on [a, b] such that G'(x) = f (x),for every x E (a, b), then d x, y E [a, b]

G(y) -G(am) =J

y f(t)dt.x

Proof: (i) Since f is bounded, let If(x)I < M for every x E [a, b]. Then fory > x, using exercise 1.1.10(iii), we have

I F(y) - F(x) I = J y f(t)dtx

y<

X

T If(t)Idt c May-xI.

From this the uniform continuity of F follows.(ii) Let f be continuous at x. Then for y 4 x,

f(x) -F(Y)-F(x) _ 1 -(Y - T)fW - f f(t)dt]Y - T 7/-X

1LJ

y f (x)dt -J

y f (t)dt(Y X

1 1

1Jy[f ( ')-f(t)]dtJ.y-xThus

f(x) - F(yy- x

F(x)<

Iy-xl fy I- f(t)I dt.Since f is continuous at x, given e > 0 we can choose S > 0 such that

11(x) - f (t) l < E whenever Ix- tj < S. Thus for y such that Ix - yj G S,

f WF(y) F(x)

y-xE

IY-XIdty = E.

1.3 Historical notes 35

Hence F (x) exists and

F/

x limF(y) - F(x)

= f W.Y--+X y - x

(iii) Since G'(x) = f (x) V x E [a, b], G'(x) E R[a, b]. Let a < x < y < bbe fixed and c > 0 be given. Choose a partition P = {x = xo < xl <

xn= y} of [x, y] such that

U(P, G') - L(P, G') < e. (1.5)

By the mean value theorem for G on each of the intervals [Xk_1, xk], we canchoose Ck E (xk_1,xk) such that

G(xk) - G(xk-1) = (xk - xk-1)G'(Ck)

Thusn

G(y) - G(x) _ [G(xk) - G(xk-i) ]

n

- E C-' (Clc)lxk - xk-1Jk=1

Since

we have

Also,

LAG > P) E G (Ck)(xk - xk-1) < U(GP))k=1

L(G', P) < G(y) - G(x) < U(G', P).

L(G/

, P) < f G (t)dt = J f (t)dt U(G', P).x

From (1.5), (1.6) and (1.7), we have

f (t)dt - [G(y) - G(x)Jf

Ix

yf (t)dt = G(y) - G(x). 0

In the exercises below, we give some of the well known consequences ofthe fundamental theorem of calculus.


1.3.3. Exercise:

(i) (Integration by parts): Let F, G be differentiable on [a, b] such thatF', G' E R[a, b]. Then

F(x)G'(x)dx = F(b)G(b) - F(a)G(a) - Jb

F'(x)g(x)dx.fJb

a a

(ii) (Leibniz Rule): Let f : [a, b] -- R be continuous and u, v : [c, d] -- [a, b]be differentiable. Then V a E [c, d],

d V(X)

dx u(x)f (t)dt

x=a= f (V (CO) V, (CO - f (U (CO) U, (CO -

1.3.4. Exercise:

(i) (Direct Substitution): Let g : [c, d] -- 1[8 be a differentiable functionwith g' E R[c, d]. Let f : g([c, d]) -- 1[8 be continuous. Then the followingintegrals exist and are equal.

fd g(d)

f (g(x))g'(x)dxfn

f (t)dt.

(ii) (Inverse Substitution): Let g [c, d] -- 1[8 be a continuously dif-ferentiable function with g'(x) 4 0 V x E [c, d]. Let g[c, d] = [a, b] andf : [a, b] -- 1[8 be continuous. Then

6 91

(b)Ina. f (t)dt = f f (9(x))9 (x)dxg

1(a)

(Hint: g'(x) # 0 V x E [c, d] implies that g is strictly monotone.)

1.4. Drawbacks of the Riemann integral

Using theorem 1.3.2, Hermann Hankel (1871) constructed the function

G(x) := R(t)dt,LX

where R(t) is the function considered by Riemann as in example 1.3.1. Thefunction G(x) provided an example of a function which is continuous every-where but is not differentiable at an infinite set of points, contrary to thepopular belief in the 18th century that a continuous function is differentiableeverywhere except at a finite number of points. It was Karl Weierstrass(1872) who finally dispelled all doubts by constructing a function on thereal line which is continuous everywhere but differentiable nowhere. Thesedevelopments raised the question about the existence of the derivative for

1.4 Drawbacks of the Riemann integral 37

non-continuous functions. The second fallout of theorem 1.3.2 was that onecould no longer claim that integration (in the sense of Riemann) is the re-verse of differentiation. One could no longer say that a pair of functions F, fon [a, b], f E R[a, b], are related by the equation

fy

F(y) - F(x) = f (t)dt, a < x < y < b,

iff F is differentiable and F'(x) = f (x). The beauty of the fundamentaltheorem of calculus, as exhibited by Cauchy, lost its charm. This motivatedmathematicians to look at differentiability properties of functions more crit-ically on one hand, and on the other to look for differentiable functionsf [a, b] ) R such that f ' is bounded but not integrable. Ulisse Dini(1878) introduced four different concepts of derivative of a function at apoint. Using these concepts, he showed that for a bounded integrable func-tion f : [a, b] ][8, the function F(x) := fa f(t)dt has all the four deriva-tives, they are all bounded, integrable and F(x) - F(a) = f(DF)(t)dt,where DF(t) is any one of those derivatives. He further pointed out thatthe condition f' be integrable in theorem 1.3.2(iii) is necessary. He alsoconjectured that it should be possible to construct functions f : [a, b] -) Il8which have bounded, non-integrable derivatives. The first example of sucha function was given by Vito Volterra (1881). We point out here that it iseasy to construct an example of a function f [a, b] -) R such that f isdifferentiable but f' is not bounded and hence not integrable, for examplef (x) = x2 sin(1/x2) if x and f(0) = 0 if x = O, x E [0, 1], is one suchfunction.

1.4.1. Example (Volterra): Step 1: Let a E R. Define fa: ][8 ) ][8 by

fa(X)(x - a)2 sin 1 if x -7 a,x-a

0 ifx=a.{The graph of the function fa is as given in Figure 11. It is easy to see that fais differentiable everywhere and fa vanishes at an infinite number of pointsnear x = a.

Step 2: Let a, b E R with a < b. Let

as := sup{x I a < x < (a + b)/2 and f(x) = Of.

Define fa,b : (a, b) ][8 as follows:

fa(x) if a < x < aa,fa,b(x) fa(°a) if aQ <x < b - (ca - a))

fa(b + a - x) if b - (aa - a) < x < b.

Figure 11: The function fQ

Note that fa,b(X) is defined as fa(X) in (a, aQJ, constant with values fa(aa)in [aa, (a + b)/2), and in ((a + b)/2, b) it is just the reflection of fa(x) aboutthe line (a + b)/2. Since fa,b(X) = - fb(x) for x E [b - (aa - a), b), we have

Ifa,b(X)I < max{(b - x)2, (a - x)2}, d x E (a, b).

It is easy to check that fa,b is differentiable everywhere and

fa,b(X) 2(b - a) +

Further,

fa,b(X) = 2(x - a) sin(1/(x - a)) - cos(1/x - a), for x near a.

Thus by choosing n sufficiently large, we can always select x close to asuch that 1/(x - a) = n7r and hence f,b(x) = fl. Similarly near b, fab(X)oscillates infinitely often between +1 and -1.

Step 3: Let Ca be the Cantor set, 0 < a < 1, as constructed in example1.2.13. Recall that

100 2^-1

U U 13nLn=1 j=1

where f o r each n = 1, 2, ... and 1 < j < 2n-1, I7-1 is one of the openintervals removed from [0, 11 at the nth stage. Define F : [0,11 -' R by

F(x) 0 if x E Ca,fa,b(X) if x E In-1 :_ (a, b) for some n and j, 1 < j < 2n-1.

If x ¢ Ca, then clearly, F is differentiable and F'(x) = fa,b(X) for some(a, b) C [0, 1]. Let x c Ca and let r > 0 be arbitrary. Let y E [0, 1] be suchthat Jy - xJ <,q. If y E Ca, then

F(y) - F(x)y-x

In case y ¢ Ca, let y c I -1 :_ (a, b) for some n, j with 1 < j < 2n-1

Without loss of generality, let I a - x I < I b - x 1. Then

F(y) F(x) F(y) (y - a)2 sin(1/(y - a))y _x < ly-x1 <

Thus F' (x) = 0 for x V Ca. Hence F is differentiable everywhere and

F, (x) I C IC 3, V x c [0,11.

Finally, F' is not continuous at each x E Ca . To see this, fix x c C. andlet b > 0 be arbitrary. Then we can choose y V C. such that I x - Y J < band F'(y) = f,b(Y) = +1; here we use the facts that C. is nowhere denseand that f,b(x) fluctuates infinitely often near a and b between +1. ThusF' is not continuous on C. and )(Ca) = a > 0. Hence F' is not Riemannintegrable, by corollary 1.2.7.

1.4.2. Remark: We recall that Fourier in his works had implicitly assumedthat a series of functions can be integrated term by term. Under whatconditions this can be justified? The equivalent questions is: Let {f}>ibe a sequence of Riemann integrable functions on an interval [a, b]. Letfn(x) -> f (x) V x c [a, b]. Can we say f is Riemann integrable on [a, b]and

b 6lim Ina, fn(x)dx J f (x)dx?n-'O° a

The following exercises show that the answer in general is in the negative.

1.4.3. Exercise:

(i) Let fn(x) = ne-", for x c [0, 1] and n = 1, 2, ....Show that each fn isRiemann integrable and lim fn(x) = f (x) = 0 V x E (0, 1], but

n-+oo

Ifol

1

JO

1

x)dx.fdx does not converge to f(

(ii) Let {rl, r2, ... } be an enumeration of the rationals in [0, 1]. Define d n =1,2,...,

1 if x E {rl,r2,... , rn},0 if x E [0,1] \ {r1, r2, ... , rn}.

Show that {f}>i is a convergent sequence of Riemann integrable func-tions and f (x) lim fn(x) is a bounded function which is not Riemannintegrable.

A sufficient condition that allows the interchange of the limit and inte-gration is given by the following theorem.

1.4.4. Theorem: Let {f}>i be a sequence of Riemann integrable func-tions on [a, b] and let {f}>i converge to a function f uniformly on [a, b].Then f is also Riemann integrable on [a, b] and

6/6

ff (x)dx = limna.J

Proof: We ask the reader to supply the proof as follows:

(i) Using the uniform convergence of {fn}n>i, deduce that f is boundedon [a,bJ.

(ii) Let En {x E [a, b] I fn is not continuous at x} and E U=1 EnShow that E is a null set and f is continuous on [a, b] \ E. Hencef E R[a, b]

(iii) Show that

limn--oo fb 6

f (x) dxf(x)dx_f = 0.

This will prove the theorem.

1.4.5. Exercise: Let fn(x) = xn, 0 < x < 1 , n = 1, 2, ... . Show that fnis Riemann integrable on [0, 1]. Show that {f(x)}>1 converges for everyx E [0, 1]. Is the limit function f Riemann integrable? Can you concludethat

1 1lim in fn(x)dx J f (x)dx?n-'O° in

Another sufficient condition for convergence of Riemann integrals isgiven by the following theorem.

1.4.6. Theorem (Arzela): Let {f}>i be a sequence of Riemann inte-grable functions on [a, b] such that for some M > 0, Ifn(x)l < M V x E [a, b]and b n = 1, 2, ... . Let fn(x) f (x) V x E [a, b] and let f be Riemannintegrable on [a, b]. Then

6lim Jna. fn(x)dx J b f (x)dx.n-'O° a

Proof: We refer to Luxemberg [24] for a proof. We shall deduce this the-orem as a consequence of a more general theorem: Lebesgue's dominatedconvergence theorem (see exercise 5.5.9).

As we see from theorems 1.4.4 and 1.4.6, one has to impose quite strongconditions to ensure the validity of interchanging of the limit and the inte-gral sign. For example, in theorem 1.4.6, one has to assume that the limitfunction is also Riemann integrable. That this condition cannot be droppedfollows from exercise 1.4.3(ii) and the next exercise.

1.4.7. Exercise: Let [0, 1] ) R be defined by

5 1 if x7 k/2n,fn W' 0 if x=k/2"forsome k,l<k<2n-l.

Note that for all n we have 0 < fn(x) < 1 V x E [0, 1]. Show that fn E R[a, b]V n = 1, 2, .... Let f : [0, 1] ][8 be defined by

f(x).-{1 if x 54 k/2n for any n,1 < k < 2n,0 if x = k/2" for some n,1 < k < 2n.

Show that fn(x) --+ f (x) V x E [0, 1], but f R[a, b].

We discuss next another drawback of Riemann integration.

Let C[a, b] denote the set of all continuous functions f : [a, b] R. Forf, g E C[a, b], define

b

d(f,g) := Jna. If (x) - g(x) I dx.

We leave it to the reader to check that d(f, g) is indeed a metric on C[a, b],i.e., it has the following properties: for f, g and h E C[a, b],

(i) d(f, g) > 0, and d(f, g) = 0 if f = g.(ii) d(.f,9) = d(9> .f)

(iii) d(f) h) < d(f) g) + d(g) h).

The metric d(f, g) is called the L1-metric on C[a, b]. One would like toknow: is C[a, b] complete under this metric? Let {f}>i be a Cauchysequence in C[a, b] in the L1-metric, i.e., such that d(fn, 0 asn, m ) oo. The question is, does there exist some function f E C[a, b]such that d(fn, f) ---) 0 as n -> oo? Let us consider the following par-ticular situation: let a = 0, b = 1. For every n > 1, consider the functionfn : [0, 1] -> R defined by

0 if 0<x<1/2,fn(x) n(x - 1/2) if 1/2 < x < 1/2 + 1/n,

1 if 1/2 + 1/n < x < 1.The graph of fn is given in Figure 12. Suppose that there exists a functionf [0, 1] - 1(8 which is continuous and d(fn, f) 0 as n`dn> 1)

J1/2 = 1/2I JIn

Thus

f1/2

f (X) = 0 V 0 < X < 1/2.

x

Figure 12 : The function fn

oo. Then

On the other hand, if 1 > x > 1/2 and the integer no is such that 1/2 + 1/no < x,then b n > no,

f I I - f (x) I dx1 pl

JxIdx

< fa

Hence

IL1 - f (x) I dx = 0 for every 1/2 < x < 1.

1

This implies, by exercise 1.1.8,

f(x)=1 for every 1/2<x<1.Thus f V C[0,1]. This shows that the space C[a, b] is not complete withrespect to the L1-metric.

1.4.8. Exercise (Incompleteness of R[a, b] in the L1-metric):

(i) Show that d(f, g) makes sense for f, g E R[a, b], but is not a metric onR[a, b]. (Hint: Properties (i), (ii) and (iii) hold, except that d(f, g) = 0need not imply that f (x) = g(x) V x E [a, b]. It is only what is calleda pseudo-metric.)

(ii) For f, g E R[a, b], we say f is equivalent to g and write f r--., g if{x E [a, b] If (x) =A g(x)} is a null set. Show that r,-,, is an equivalencerelation on R[a, b] and

af b f(x)dx=f 6 9(x)dx if

Let R[a, b] denote the set of all equivalence classes and let f denotethe equivalence class which contains f E R[a, b]. Let

b

d(f,j) := faa I f (x) - g(x) I dx,

for f , g E R[a, b]. Show that d is a well-defined metric on R[a, b],called the L1-metric.

(iii) Show that R[a, b] is not complete under the metric d as follows: Con-sider the Cantor set Ca, 0 < a < 1, as constructed in example 1.2.13.Recall that Ca = n°°_1 Bn = [0, 1] \ (U=° 1 An). Let fn = XBn for ev-ery n > 1.

(a) Show that fn E R[a, b] with

f .1 f(x)dx>1-a d n, and lim J 1 I

0

(b) Show that there does not exist any function f E R[0,1] such thati

lim I fi(x) - f (x) I dx = 0,c- '°° fn

as follows:Suppose there exists f E R[a, b] such that (1.8) holds. Then


f1 f1(bl) I f(x)dx= lim J f(x)dx > 1 -a.

0

(b2) There exists a function g E R[a, b] such that g(x) = 0, for everyx E UO_1 An and f g. Using this, deduce that

f 1 f(x)dx= I lg(x)dx = 0,0

which is in contradiction to (bl).

The above exercise shows that R[a, b] is not a complete metric spacewith respect to the metric d. Since every metric space can be completed,one wonders about the completion of the metric space 7Z[a, b]. A concreterealization of the completion of R[a, b] is one of the outcomes of Lebesgue'stheory of integration (see section 5.6).

Some other drawbacks of the Riemann integral are: it is defined onlyfor bounded functions; it is defined for functions on bounded intervals only;functions, even if they are defined on bounded intervals and have finite range(e.g., Dirichlet's function) are not necessarily integrable; and so on.

From the year 1854 onwards, when Riemann extended the definitionof integral due to Cauchy, many mathematicians contributed in the effortto extend the notion of integral in order to remove these drawbacks. Theefforts of these mathematicians: Camille Jordan, Emile Borel, Rene Baire,Johann Radon to name a few, culminated in the work of Henri Lebesgue,who in 1902 announced a generalization of the Riemann integral. In 1920,Friedrich Riesz gave an alternative method of constructing this extendedintegral. The approach of Lebesgue, which was later (1914) made abstractby Constantin Caratheodory, is more set theoretic, while the approach ofRiesz is purely function theoretic. We outline these approaches in the nextchapter.

Chapter 2

Recipes for extendingthe Riemann integral

2.1. A function theoretic view of the Riemannintegral

Let us recall that R[a, b], the set of Riemann integrable functions on theinterval [a, b], has the following properties:

(i) R[a, b] is a vector space over R.

(ii) The map f H fa f (x) dx, f E R[a, b], is a nonnegative linear mapfrom R[a, b] to R.

(iii) Whenever functions fn, g E R[a, b], n = 1, 2, ... , are such thatI fn(x)1 < 1 g(x)I and lim f(x) = g(x) V x E [a, b], then

blim J f(x)dx= J 6 g(x) dx.

(The property (iii) follows from theorem 1.4.6. Let us call it as dominatedconvergence property.)

In view of the discussion in section 1.4, our aim is to extend the notionof Riemann integral to a class of functions bigger than R[a, b], keeping theproperties (i), (ii), (iii) intact. We state it as the following:

Extension Problem: Construct a class £ of functions from lit to JR anddefine a map I : £ - ) lit with the following properties:

(i) G is a vector space over ][8 and R[a, b] is a subspace of G.

45

46 2. Recipes for extending the Riemann integral

(ii) I : G ]I8 is nonnegative and linear, i.e.,

I(f)>0 b'f EC with f >0and

I(af +Q9) = al(f) +,3I(9), d f, 9 E G; E R.

(iii) For f E R[a, b],b1(f)

= f f(x)dx.a

(iv) 1(f) is free from the defects of the Riemann integral.

(In (i) and (ii), even though f is defined only on [a, b], we can regard it asa function on I[8 by putting f (x) = 0 V x E 1[8 \ [a, b].)

In view of (iii) we can call 1(f) the extended integral of f .

To solve the extension problem, let us look at the Riemann integral froma different viewpoint. Let f [a, b] I[8 be a bounded function. Given apartition P = {a = xO < xl < - xn = b} of [a, b], for x c [a, b], let

n

4)p(x) := m1 x[Xo,Xl] (x) + E m2 X(xi_1, ](x)

i=2

and

Here

n

Tp(x) := M1XjX"'Xl] (x) + E Mi x(Xi-i)Xi] (x)'

i=2

ml := inf{ f (x) I xo < x < xl},

Ml sup{ f (x) I x0 < x < x1},

mi inf { f (x) I xi < x < xi+l },

Mi := sup{ f (x) I x2 < x < xi+l}, 1 < i < n.

Then for every x E [a, b], 4PP(x) < f (x) < TP(x) and 4PP, TP E 7Z [a, b].Further,

fbG

fP(x)dx L(P, f) and

where L(P, f) and U(P, f) are the lower and upper sums of f with respectto the partition P.

In case f is Riemann integrable, by theorem 1.1.4 there exists a sequenceof partitions such that each Pn+1 is a refinement of Pn,asn -*oo and

IIPnhI -- o

fbf(x)dx= lim L(Pn, f) = lim U(Pn, f ).

n->oo n->oo

2.2 Lebesgue's recipe 47

Note that d n > 1 and b xE [a, b],

"PPn W 41'Pnl1 W < f(x) < 'FPnl1 W <_ 'FPn WThus {J?pn }n>1 is an increasing sequence of functions bounded above by fand {'T!p is a decreasing sequence of functions bounded below by f atevery point in [a, b]. Further,

Ia

b b b

f (x)dx = limn-*oo J pn (x) dx = lim

a n--+ooJna,

T Pn (x) dx. (2.1)

The functions p and T p are functions of a special type:

2.1.1. Definition: A function (D : [a, b] -+ R is called a step function ifthere exists a partition P = Ja = xo < x1 < < xn = b} of [a, b] such that

is constant, say Ci, on every open interval (xi, xi+1 ), 0 < i < n - 1.

a = xo x1 . . . . xi-1 xi . xn-1

Figure 13 : Graph of a step function

b = xn

Thus pn and ' pn , as defined above, are examples of step functions,and in view of (2.1) we can treat the integrals of the step functions as thebuilding blocks for the Riemann integral of f. In order to enlarge the spaceR [a, b] and to extend the notion of integral, it is natural to extend the spaceof building blocks itself. There are two well-known methods for doing this.We give an outline of each method.

2.2. Lebesgue's recipe

Let us note that a step function takes only finitely many values, and theseare taken on disjoint open intervals. As far as the integral is concerned, wecan disregard the values of the step function at the partition points. In fact,at these points the step function can be given an arbitrary value, withoutchanging its Riemann integral. Thus if 1 : [a, b] -- * JR is a step functionsuch that (D(x) = Ck for x E (xklxk), where a = xO < x 1 < < xn = b,

then we can represent 1 as follows:n

(P (x) = E CkXI, (x), x E [a, b], (2.2)k=1

where Ik = [xk_1, xk) for 1 < k < n - 1 and In = [Xn_i, xn] . Note that[a, b] = U=i Ik, with Ii n Ij _ 0 for i j. In order to extend the classof building blocks for our new integral, we can consider functions of thetypes : _ 1 Ci XAi , where n is a positive integer and the Ai's are subsetsEi=of [a, b] such that Ai n Aj _ 0 for i j and U=1 Ai = [a, b]. Let us callsuch functions generalized step functions or simple functions. Theadvantage of such functions is that they take only finitely many values (likestep functions) on disjoint sets (which are not necessarily intervals, as thosefor the step functions are). Also, they make sense for any subset E of R.For example, if E = U=1 Ai, where Ai n Aj _ 0 for i j, then we can calls = Em 1 CixAi a simple function on E. Let us denote by Lo the collectionof simple functions on R. Note that a step function on [a, b] can be treatedas a simple function on R by defining it to be zero outside [a, b]. Thus Loincludes the class of step functions on R. Further, noting that for a stepfunction (P on [a, b] as given by (2.2) above, since

Ia

6 n(P (x) dx Ci A (1i) (2-3)

i=1

we can extend the notion of integral to simple functions as follows: fors E Lo with s = E 1 CiXAi , let the extended integral be

f n

sdA Ci A (Ai). (2.4)i-1

Of course in (2.4) the quantities A(Ai),1 < i < n, are undefined. Comparingequation (2.4) with equation (2.3) and keeping in mind that we have to havef (MA = fa (x)dx in case (1) is a step function, it is natural to expect A(Ai)to be the `length of the set Ai'. Thus the first step in our extension programshould be: try to extend the notion of length from intervals to a bigger classof subsets of R. This we will discuss in chapters 3 and 4. Supposing we areable to define the notion of length for a class A of subsets of Il8 such that Aincludes intervals, we can consider functions of the type >1'=i CiXA2 , witheach Ai E A. Such functions are called simple measurable functions.For such a function we define the new integral by (2.4). Since A(Ai) couldbe +oo, to make the sum in (2.4) meaningful, we assume that Ci > 0 V i.That is, we consider only nonnegative simple measurable functions. Since weexpect our integral to behave nicely under limiting operations, we considerfunctions f : ]I8 -> ]I8 for which one can construct sequences {Sn}n>i of non-negative simple functions such that 0 < sn < f and sn(x) ) f (x) V x E R.

2.3 Riesz-Daniel recipe 49

Note that this automatically forces f to be nonnegative. For such functionsf, we can define

fd\ := lim s,zdA.n- +oo

This raises many questions: Is f f dA well-defined, i.e., is it independentof the choice of the sequence {sn}n>1? For what nonnegative functionsf ][8 --) Il8, is construction of such sequences possible? A function forwhich construction of such a sequence is possible is called a measurablefunction and will be discussed in chapter 5. Note that the integral f f dAcould be +oo for f > 0 and f measurable. For a general function fR --) Il8, one considers the positive part f + and the negative part fdefined by f+(x) := f(x) if f(x) > 0 and f+(x) = O if 1(x) < 0. Letf(x) := f +(x) - f (x). Then f = f+ -f-. One notes that f + and f - areboth nonnegative functions. If both f+ and f - are measurable, we say f isa measurable function. For a measurable function f, the required linearityproperty of the integral demands that we should define

f fdA:= f f +dA - f f -dA.

A problem arises in the case f f +dA = f f -dA = +oo. To overcome this, onesays f isLebesgue integrable if f f +dA < +oo and f f -dA < +oo. Thenthe class of Lebesgue integrable function is the required class of functions forwhich f f dA is well-defined. The detailed construction of this integral andthe proof that this integral really works, i.e., indeed extends the Riemannintegral and removes its drawbacks, will be analyzed in chapters 5 and 6.

2.3. Riesz-Daniel recipe

There is another way of characterizing the Riemann integrability of func-tions. Since sup A = - inf (-A) for any subset A of ][8, f is Riemann inte-grable on [a, b] if and only if

fb

I(as the expression on the right side is the lower integral of 1). Thus amongall functions on [a, b] we consider the subset of those which satisfy equation(2.5). This subset turns out to be a vector space, and the integral is linearon it.

One way of extending the definition of Riemann integral would be not toinsist on being able to approximate from both ìnside' and òutside', but becontent with approximating the area either from ìnside' or from òutside'only.

Let us call a function (D : I[8 ) I[8 a step function on ][8 if 3 a, b E I[8with a < b such that (D (x) = 0 for x [a, b] and (D restricted to [a, b] is astep function as defined in 2.1.1. For a step function (D on R, let f (DdA bedefined as in equation (2.3). With step function on ][8 as building blocks, letus try to approximate the `area' from `inside' only. Thus we are tempted todefine (tentatively) the extended integral f f dA as follows:

f f dA := sup f (DdA)

where the supremum is taken over all step functions (D such that (D < f. Wemay then call f integrable if f f is finite. But this naive approach has aproblem: when f is integrable -f need not be integrable, and in particularf fd\ = - f (- f ) dA may not hold. To overcome this problem, we proceedas follows:

Let ,C+ denote the set of all those functions f for which there exists asequence {(%},>1 of step functions such that {m}m>1 increases to f andlimn--+o,o f (Dnda exists. Note that G+ includes the class of step functions.We then define f f dA for f E G+ by setting

Jfd:= lim

Jn-+oo

One then shows that this is well-defined. It is easy to see that G+ is closedunder addition and scalar multiplication by positive reals. To get a vectorspace out of this, we simply take G := G+ - L. That is, G consists offunctions h which can be written as h = f - g, where f, g E G+, and extendthe integral in an obvious way:

f h dA = J f dA - J g dA.

Of course, one has to show that this integral is well-defined.

Notice that if we take fn E G+ such that {f}>i increases to f withf fn bounded by some constant b n, then it is almost clear that f E G+and that we have r

ffd.J f d= limn-+oo

This is a special case of the so-called monotone convergence theorem. Witha little more effort, one can prove a similar result with no restrictions onfn except that fn E G. This shows that this method cannot enlarge G anyfurther.

We refer the reader for more details about this way of developing theLebesgue integral to Daniel [10], Riesz [31], [32], [33] and Stone [39].

Chapter 3

General extensiontheory

3.1. First extension

3.1.1. In chapter 2 we saw that the first step towards extending the notionof Riemann integral is to extend the length function A from intervals to alarger class of subsets of R. Suppose E C R is such that it is union of afinite number of pairwise disjoint intervals. Then it is natural to define thelength of E to be the sum of the lengths of these intervals. Mathematically,let E = U=1 Ik, where Ik E I and Ik f Ie _ 0 for 1 c k£ <n. Since theextended notion of length of E, denoted by A(E), is expected to be finitelyadditive, we will have

n n

A (E) (Ik) A(Ik)./c=1 k=1

Thus, if E = U=1 'k, where the Ik's are pairwise disjoint intervals, theabove forces us to define

A (E) A (Ik). (3.2)

Before we proceed further, we should check that as in equation (3.2) iswell-defined. Let E = U=1 Ik = Um 1

Je, where Ik, JQ E I with Ik1 0for 1 < k 1 = , 4 k2 < n and JQI n JQ2 = 0 for 1 < f1 4 P2 < m. ThenEn

k=1 A(Ik) = Em, A(Je) To check this, we note that Ik = Ui(Ik n if)for every k and JQ = U=1 (Ik fl if) for every 2, where the intervals Ikn if

51

52 3. General extension theory

are pairwise disjoint. Now using finite additivity of A, we have

n n M M

1: A(1k) - I: EA(ikn it) - E A(Jt).k=1 k=1 t=1 t=1

Note that here we have used the fact that for 'k, JJ E I, Ik n it c I. Thus,\, given by equation (3.2), is well-defined on the collection

n

E= U Ik,Ik EI,IkniI=Ofork f,nEN .

k=1

Clearly, I C F(l), and (E) _ A(E) if E E Z. Further, A :F(Z) [0, oo]

has properties similar to that of A, i.e., is monotone, countably additiveand countably subadditive. Some of the properties of the classes I and F(l)and the function are given in the following propositions.

3.1.2. Proposition: The class I has the following properties:

(i) 0,1[8 E Z.

(ii) If I, J E Z, then I fl J E Z.(iii) If I E Z, then I° = Jl U J2, where Jl, J2 E I and Jl fl 12 = 0.

Proof: Exercise.

3.1.3. Proposition: The class F(l) has the following properties:

(i) Z C F(Z).

(ii) 0 E F(l) and 1[8 E .F(l).

(iii) El, E2 E F(Z), then El n E2

(iv) If E E F(l), then E' E .F(l).(v) F(l) is the smallest class of subsets of I[8 such that (i), (ii), (iii) and

(iv) hold.

Proof: (i), (ii) and (iii) are straightforward. If E E F(Z) and E = Unk=1 Ik

with Ikn IP = Q1 for 1 < k , Q < n, then

n

Ec = flit.k=1

Also, V k,

Ikc = Jk U Jk ,

3.1 First extension 53

where J , J E Z and J n J _ 0. Thus

E°= fl(JuJ)k=1 i j=1 k,1=1

2 n

U U (4nJ))(

Thus Ec is a union of a finite number of pairwise disjoint intervals. Hence,Ec E T(Z). This proves (iv).

Finally, let C be any other collection of sets such that I C C and C hasproperties (ii), (iii) and (iv). If E1, E2 E C, then

El U E2 = (Ei n E2)c

and hence ElUE2 E C. Let E E T(Z) and E = U=1 Ikwith Ik E Z, IknIP =0 for k 4 2. Then Ik E C and hence E E C, showing that T(Z) C C. Thisproves (v).

3.1.4 Proposition: The function A :T(I) --+ R, as defined in equation(3.2), has the following properties:

(i) A(I) = a(I) V I E Z.

(ii) A is countably additive, i.e.,

00

A (E) (Ei)i=1

whenever EET(Z) is such that E = U°__1 EZ with each Ei E T(Z),EZ fl F, = 0 for i 4 j.

Proof: (i) is obvious. To prove (ii), let `d i, Ei = U=1 I2, where I1, ... , I.2

are pairwise disjoint intervals. Then JIj' 1 1 < i < n, 1 < j < ki } is acollection of pairwise disjoint intervals. Let

fE = U KT,

T=1

where K1,... , KK are pairwise disjoint intervals. Then for every r

00 ki

KT = U (KT n i; )i=1 j=1

and for every iki

Ei= U U(Kr1).

r=1 i=1

Thus, using the countable additivity of A and the definition of A, we have

(E) =e

00

A (Kr)r=1

oo ki

E E a(Kr (11 )r=1 i=1 j=1

00

ki f

E /\(K. n ii)j=1 r=1

A (Ei). 0i=1

3.1.5. Exercise: Show that A as in proposition 3.1.4 has the followingproperties:

(i) A is finitely additive, i.e.,

A (E) (Ei)i=1

whenever El, ... , En are pairwise disjoint sets in F(I) and E _U=1 Ez

(ii) A is monotone, i.e., if E, F E F(I) and E C F, then (E) < (F).(iii) A is countably subadditive, i.e.,

00

A (E) < (Ei)i=1

whenever El, E2 ... are pairwise disjoint sets in T(I), E E F(I) andECU°°lEi.

3.2. Semi-algebra and algebra of sets

Motivated by the properties of the collections I and T(I), we have thefollowing:

3.2.1. Definition: Let X be a nonempty set and let C be a collectionof subsets of X. We say C is a semi-algebra of subsets of X if it has thefollowing properties:

(i) O, X CC.

(ii) A n B C C for every A, B C C.

(iii) For every A C C there exist n E N and sets C1, C2,... , Cn C C suchthat CinCj =0 for i74 j and A'=U 1Ci.

3.2 Semi-algebra and algebra of sets 55

3.2.2. Definition: Let X be a nonempty set and j a collection of subsetsof X. The collection 2 is called an algebra of subsets of X if 2 has thefollowing properties:

(i) 0, X E F.(ii) A n B E T, whenever A, B E T.

(iii) A' E T, whenever A E T.

3.2.3. Examples:

(i) The collection I of all intervals forms a semi-algebra of subsets of R. Fora, b E R with a < b, consider the collection Z of all intervals of the form(a, b], (-oo, b], (a, oo), (-oc, boo). We call Z the collection of all left-open,right-closed intervals of R. It is easy to check that Z is also a semi-algebraof subsets of R.

(ii) The collection T(Z), as discussed in 3.1.1, is an algebra of subsets of R.So is the class {E C R I E = U=iIk with Ik E I and Ik n IQ = 0for 1 <

(iii) Let X be any nonempty set. The collections {0, X I and P (X){E I E C X I are trivial examples of algebras of subsets of X. The collection'P(X) is called the power set of X.

(iv) Let X be any nonempty set. Let

C:= V C X either E or EC is finite}.

Then C is an algebra of subsets of X. In case X is a finite set, then C = P (X) .

Suppose X is not finite. Clearly 0, X E C, and Ec E C if E E C. Finally,suppose El, E2 E C. If either E1 is finite or E2 is finite, then obviouslyEl n E2 E C. If both Ei and E2 are finite, then (E1 n E2)c = E1 U E2 isfinite and thus E1 n E2 E C. Hence C is an algebra.

(v) Let X and Y be two nonempty sets, and ,F and 9 semi-algebras ofsubsets of X and Y, respectively. Let

.FXc= {FxG I FET,GE g}.

We show that ,Fxg is a semi-algebra of subsets of XxY. Clearly XxY E,F x c . Next, let A, B E F x C . Let A = F1 x G i and B = F2 x G2 , whereF1, F2 E ,F and G1, G2 E 9. Then

AxB = (F1xC1) n (F2xC2) _ (F1 n F2)x(Gl n G2).

Since T, G are semi-algebras, Fl fl F2 E T and G1 fl G2 E G. Hence it followsthat A x B E .Fx G. Next, let A= F x G E Tx G, where F E T and G E G.


Since F, 9 are semi-algebras, we have pairwise disjoint sets F1, ... , F,' E Fand pairwise disjoint sets G1, ... , Gm E 9 such that

n m

Fc = Fi and Gc= UGi.

i=1 j=1

Thus

(FxG)c = (FCxY) U (FxGc)n

((uF)i xY U (Fx UGH

(UFxY) U (U(FxCi)).i=1 j=1

Clearly, for every i and j, Fi x Y E Fxc and FxC3 E .'xC. Further, allthese sets are pairwise disjoint. This proves that Fxg is a semi-algebra.

3.2.4. Exercise:

(a) Let F be any collection of subsets of a set X. Show that F is an algebraif the following hold:

(i) 01 X E F.

(ii) Ac E F whenever A E F.

(iii) A U B E F whenever A, B E F.

(b) Let ,F be an algebra of subsets of X. Show that

(i) If A, B E .F then ADB := (A\B)U (B\A) E .F.(ii) If E1, E2, ... , En E F then F1, F2,. . . , Fn E F such that Fi C Ei

for each i, Fi n F3=0 for i j and U=1 Ei = Uj 1 Fj .

The next exercise describes some methods of constructing algebras andsemi-algebras.

3.2.5. Exercise:

(i) Let X be a nonempty set. Let 0 : E C X and let C be a semi-algebra(algebra) of subsets of X. Let

Show that C fl E is a semi-algebra (algebra) of subsets of E. Note that C fl Eis the collection of those subsets of E which are elements of C.

3.2 Semi-algebra and algebra of sets 57

(ii) Let X, Y be two nonempty sets and f X Y be any map. ForE C Y, we write f -1(E) :_ {x E X I f (x) E E}. Let C be any semi-algebra(algebra) of subsets of Y. Show that

f -'(C) := f f -'(E) I E G Q

is a semi-algebra (algebra) of subsets of X.

(iii) Give examples of two nonempty sets X, Y and algebras T, G of subsetsof X and Y, respectively such that T x g := {A x B I A E T, B E 9} isnot an algebra. (It will of course be asemi-algebra, as shown in example3.2.3(v).)

(iv) Let {Fa}i be a family of algebras of subsets of a set X. Let Tn.c-.[.Fa. Show that T is also an algebra of subsets of X.

(v) Let {T,, }n>1 be a sequence of algebras of subsets of a set X. Under whatconditions on .fin can you conclude that T:= U=1 .fin is also an algebra?

3.2.6. Exercise: Let C be a semi-algebra of subsets of a set X. A setA C X is called a a--set if there exist sets CZ E C, i = 1, 2, ... , such thatCZ n C3= 0 for i j and U°° 1 CZ = A. Prove the following:

(i) For any finite number of sets C, C1, C21... , Cn in C, C \ (U=1 CZ) isa finite union of pairwise disjoint sets from C and hence is a o.-set.

(ii) For any sequence {C}>1 of sets in C, U=1 Cn is a a--set.(iii) A finite intersection and a countable union of a--sets is a a--set.

3.2.7. Proposition: Let C be any collection of subsets of a set X. Thenthere exists a unique algebra T of subsets of X such that C C T and if A isany other algebra such that C C A, then T C A.

Proof: Note that there exists at least one algebra of subsets of X whichincludes C, namely P(X). Consider A, the intersection of all those algebrasof subsets of X which include C. Then it follows from exercise 3.2.5(iv) thatA is the required algebra.

3.2.8. Definition: Let C be any collection of subsets of a set X. Then theunique algebra given by proposition 3.2.7 is called the algebra generatedby C and is denoted by T(C).

3.2.9. Example: The algebra generated by Z, the class of all intervals,is {ECRI E=U=iIk,Ik El, Iknit=0if i< k £<n}, as proved inproposition 3.1.3.

A similar result holds in general. See the next exercise.

3.2.10. Exercise:

(i) Let C be any semi-algebra of subsets of a set X. Show that F(C), thealgebra generated by C, is given by {E C X I E = U=1C, Ci E C andCiflC,=0 fori= j,nEN}.

(ii) Let X be any nonempty set and C = {{x} I x c X} U {0, X}. Is C a semi-algebra of subsets of X? What is the algebra generated by C? Does youranswer depend upon whether X is finite or not? (See example 3.2.3(iv).)

(iii) Let Y be any nonempty set and let X be the set of all sequences withelements from Y, i.e.,

X = {x = {xn}n>1 IxnEY,n=1,2,.}.For any positive integer k let A C yk, the k-fold Cartesian product of Ywith itself, and let it < i2 < < ikbe positive integers. Let

c(21, 22> ... , Z/c; A) :_ {x = (xn)n>1 E X I (x1,... , xiti) E Al.

We call C(il, i2i ... , ik; A) a k-dimensional cylinder set in X with baseA. Prove the following assertions:

(a) Every k-dimensional cylinder can be regarded as an n-dimensionalcylinder also for n > k.

(b) Let A = {E C X I E is an n-dimensional cylinder set for some n}.Then A U {0, X} is an algebra of subsets of X.

(iv) Let C be any collection of subsets of a set X and let E C X. LetC n E :_ {C n E I C E Cl. Then the following hold:

(a) C n E C ,F(C) n E fAnE I A E F(C)}. Deduce that F(C n E) C.F(C) n E.

(b) Let A = {A C X A n E E F(C n E)}. Then A is an algebra ofsubsets of X, C C A and A n E = T(C f1 E).

(c) Using (a) and (b), deduce that T(C) f1 E = T(C n E).

3.2.11. Remark: Exercise 3.2.10(i) gives a description of.F(C), the algebragenerated by a semi-algebra C. In general, no description is possible for F(C)when C is not a semi-algebra. See also theorem 4.5.1.

3.3. Extension from semi-algebra to thegenerated algebra

In section 3.1, we showed that the length function, which is initially definedon the semi-algebra Z of all intervals, can be extended to a set functionon ,F(Z) , the algebra generated. Further, the extended function has the

3.3 Extension from semi-algebra to the generated algebra 59

properties similar to that of the length function. We want to do the samefor functions defined on arbitrary semi-algebras. We make the followingdefinitions.

3.3.1. Definition: Let C be a class of subsets of a set X. A functionj L: C - [0, +oo] is called a set function. Further,

(i) µ is said to be monotone if µ(A) < µ(B) whenever A, B E C andA C B.

(ii) µ is said to be finitely additive ifn n

µ U Az = j:µ(Az)i=1 i=1

whenever Al, A2i ... , An E C are such that Ai fl Aj =0 for i 4j andU=1A i E C.

(iii) µ is said to be countably additive if

n=1 n=1

whenever A1, A2,--- in C with Ai r1 Aj _ 0 for i j and U=1 An CC-n=(iv) ,u is said to be countably subadditive if

00

(A) < 1: ,u(An)n=1

whenever A E C, A = U=1 An with An E C for every n.

(v) ,u is called a measure on C if 0 E C with ,u(O) = 0 and ,u is countablyadditive on C.

3.3.2. Exercise:

(i) Let C be a collection of subsets of a set X and µ C -* [0, oo] be aset function. If µ is a measure on C, show that µ is finitely additive. Ismonotone? Countably subadditive?

(ii) If C be a semi-algebra, then ,u is countably subadditive if d A E C withA C U°°1 Ai, Ai E C implies

00

(A) < (Ai).Z=1

Our next theorem is an abstract version of the extension of the lengthfunction given in proposition 3.1.4.

3.3.3. Theorem: Given a measure µ on a semi-algebra C, there exists aunique measure µ on T(C) such that µ(E) = µ(E) for every E E C.

The measure µ is called the extension of µ.

Proof: We note that every E E .F(C) can be represented as E = U=1 Eifor pairwise disjoint sets E1, ... , En E C C. Thus we can define

00

A (E) -. = µ(Ez)i=1

Now proceeding as in 3.1.1 and proposition 3.1.4 for A, we can show thatis well-defined and has the required properties.

3.3.4. Exercise: Let X = (0, 1] and let A = F(I fl (0, 1]) be the algebragenerated by all left-open, right-closed intervals in (0, 1]. For every x E (0, 1],let x = En=1 °xn/2n, xn E {0,1}, be its binary expansion with the convention that we always choose the expansion which does not terminate in zeros.For example, the number 1/2 has two expansions:

00

1/2 = xn/2n =

n=1

1

00

n=1

where x1 = 0, xn = 1 V n > 2 and y1 = 1, yn = 0 V n > 2. We choosethe first one. Then we can identify x with the sequence {xn}n>1. Let n > 1be fixed and let u1, U2,. .. , un E {0,1}. Let A = {x = {x}>1 E X I xi =ui,1 1 E X

nyn 2 I

n

Exi = k .

i=1

Show that B(n, k) E A for every n and k, k < n. Show that A(A) = 1/2nand (B(n,k)) = ()/2fl, where is the extension of the length function Aas given in proposition 3.1.4. Let

Show that

C = x = {xn}n>1 E X

00 00 CIO

n

lim - Exi = 1/2n-*oo n

C= n u n x= {xn}n>l E Xk=1 m=1 n=m

where

i=1

In E f(x) I < ki=1

f(x) +1 if xi = 1,-1 ifxi=0.

Note that fi(x) = 2xi - 1 if x = {xn}n>1. Does C E A?

3.4 Impossibility of extending the length function to all subsets 61

3.4. Impossibility of extending the lengthfunction to all subsets of the real line

In section 3.1 the notion of length was extended from the semi-algebra Iof intervals to the collection the algebra generated by Z. We pointout that the length function is countably additive and A({x}) = 0 for everyx E R. In this section we present a result due to S.M. Ulam (1930) [41]which, under the assumption of the "continuum hypothesis", implies that itis not possible to extend the length function to all subsets of R. Recall thatthe continuum hypothesis implies that it is possible to define a well-order <on ][8 such that d y E ]I8, the set {x E ][8 I x < y} is countable (see appendixc)

3.4.1. Theorem (Ulam): Let µ be a measure defined on all subsets of ][8such that µ((n, n + 1]) < oo d n E Z and µ({x}) = 0 for every x E R. Thenµ(E) = 0 for every E C R.

Proof: It is enough to show that µ((n, n + 1]) = 0 for every n E Z. Letno E Z be fixed and let X :_ (no, no + 1]. The main idea of the proof is toexpress X as a countable union of sets each having µ-measure zero.

Since X also has cardinality that of ][8, by definition, there exists a well-order < on X such that for every x E X, the set {y E X I y < x} is countable.Let o,, : jyEXjy < x} -) N be any one-one map. Then for x, y E X withy < x, ¢,, (y) is a natural number. Further, for x, y, z E X, if x G y < z then0z(y) :,-A 0z(x). For x E X and n E N, define

xFn :={yEX jy>xandoy(x)=n}.

Then for every fixed x E X,

X= FnU F: U {y X I x}.n=1

Since the set {y E X I y < x} is countable and µ({y}) = 0 for every y E Y,µ{y E X I y < x} = 0. To complete the proof we have only to show thatthere exists some x E X such that b n E N, 0. To show this, notethat for x :,-A y and n E N, if z E Fy fl F , then 0, (x) = 0, (y) = n. Sincex :,-A y, either x < y or y < x and hence either x < y < z or y < x < z.In either case, 0,(x) = 0,(y) = n will be contradicted. Thus for everyn E N, the family {F'}x is a pairwise disjoint family of subsets of X.Since µ(X) < +oo and X is uncountable, 0 is possible only for acountable number of x's. Thus {x E X I 0 for some n} is at mostcountable. Hence there exists x E X such that 0 for every n E N.Thus µ(X) = 0.

3.4.2. Remark: Ulam's theorem shows the impossibility of extending thelength function from intervals to all subsets of I[8, assuming the continuumhypothesis. We shall see later in section 4.6 that similar results can be provedif one assumes the `axiom of choice' (see appendix B). Ulam's result uses theproperty of A that A({x}) = 0 V x E 1I8, and the fact that A([n, n+ 1]) < +oofor every n E Z. In the later results we shall use the translation invarianceproperty of the length function A.

3.5. Countably additive set functions onintervals

We have shown in the prologue that the length function is a countablyadditive set function on the class of all intervals. One can ask the question:do there exist countably additive set functions on intervals, other than thelength function? The answer is given by the following:

3.5.1. Proposition: Let F : Ilg -> lt8 be a monotonically increasing func-tion. Let µF [0, oo] be defined by

AF(a,b(-oo b

F(b) - F (a),li [F(b) - F(- )],AF

(a o )

m x ,

li [F( ) - F( )], oAF

A F (- 00, 00)

m x a ,

lim [F(x) - F(-x)].

Then µF is a well-defined finitely additive set function on Z. Further, µF iscountably additive if F is right continuous.

One calls µF the set function induced by F.

Proof: Clearly, AF : Z [0, oo] is a well-defined function. The otherfacts about uF can be proved on the lines that the length function A hasthese properties, as proved in the prologue. (Note that the length functioncorresponds to the case F(x) = x, V x, the intervals involved are left-openand right-closed, and F is right continuous.) We leave the details as anexercise.

3.5.2. Exercise: Let F(x) _ [x], the integral part of x, x E R. Describethe set function µF.

The converse of proposition 3.5.1 is also true.

3.5.3. Proposition: Let [t: T [0, oo] be a finitely additive set functionsuch that µ(a, b] < +oo for every a, b E R. Then there exists a monotonically

3.5 Countably additive set functions on intervals 63

increasing function F : ][8 ) ][8 such that µ(a, b] = F(b) - F(a) V a, b E R.If µ is also countably additive, then F is right-continuous.

Proof: Since the function F : R ) ][8 that we are looking for has to havethe property that (a,b]=F(b)-F(a) b a, b E 1[8, we fix a E Il8, say a = 0,and let b E ][8 vary. This motivates the definition of a function F as follows:

µ(0, x] if x > 0,F (x) 0 if X = 01

-[L(X, 0] if X < 0.

It is easy to check that F is a monotonically increasing function and µ(a, b] _F(b) - F(a) for every a, b c R. We need only to check that F is rightcontinuous if µ is countably additive. For this let x E R and let {xn}n>1be a decreasing sequence in ][8 with x. Then {F(x)}>1 is adecreasing sequence and is bounded below by F(x). Thus lim F(xn) exists.To compute this limit we consider two cases.

If x > 0, then

Using the countable additivity property of µ, we have

F(xl) = [t (0, xj] = [t (0, x] + [t (x, xj]00

= F(x)+µ U (xn+n=1n=1

00

F(x) + E A (Xn+1, Xn]n=1

= F(x) + limk-+oo

A (xn+1) xn]

F(x) + lim [F(xn)

n=

- F''(xn+l)l

F(x) + F(xi) - lim F(Xk+l).

Hence

lim F(Xk) = F(x).

In case x < 0, since {x}> 1 decreases to x, we may assume without lossof generality that x1 < 0. Then

X < ... < xn+.1 < xn < ... < X1 < 0

and by the countable additivity of µ we have

-F(x) = p (x, 0] = p (x, xj] + p (xi, 0]

= µ U (xx_l]) - F(xl-rc=2-rc

00

n=2

n=2

r_lim [F(xi) - F(Xk)] - F(xi)tiyoo- lim F(xk).

-00

n=2

k

lim [F(x_1) - F(xn)] - F(xi)

1 F(xi)

(Xn, Xn-11 - F(xi)

k

This proves that F is right continuous.

3.5.4. Remarks:

(i) In case µ(][8) < +oo, a more canonical choice for the required function Fin proposition 3.5.3 is given by F(x) := µ(-oo, x], x E R.

(ii) Propositions 3.5.1 and 3.5.3 completely characterize the non-trivial count-ably additive set functions on intervals in terms of functions F : ][8 ) III

which are monotonically increasing and right continuous. Such functionsare called distribution functions on R. The set function µF induced bythe distribution function F is non-trivial in the sense that it assigns finitenon-zero values to bounded intervals.

3.5.5. Exercise: Let F be a distribution function and a c R. Show thatF1 := F + a is also a distribution function and AF = /LF1 . Is the conversetrue?

3.6. Countably additive set functions onalgebras

In this section we give some general properties of a set function ,a definedon an algebra A of subsets of an arbitrary set X. These properties do notdepend upon the particular choice of X, A or i,t.

3.6.1. Theorem: Let A be an algebra of subsets of a set X and let µA -> [0, oo] be a set function. Then the following hold:

3.6 Countably additive set functions on algebras 65

(i) If µ is finitely additive and µ(B) < +oo, then (B-A) = µ(B)-µ(A)for every A, B E A with A C B. In particular, µ(Q1) = 0 if µ is finitelyadditive and µ(B) < +oo for some B E A.

(ii) If µ is finitely additive, then µ is also monotone.(iii) Let µ(Q1) = 0. Then µ is countably additive iffy is both finitely additive

and countably subadditive.

Proof: Proofs of (i) and (ii) are straightforward and are left as exercises.To prove (iii), let jL be countably additive and let A = U=1 Ai, Ai E A,with Ai n Aj _ 0 for i j. Then A = U°° 1 Ai, with each Ai = 0 for i > n,and

00 n

p(A) = >jL(Ai) = LjL(Aj).i=1 i=1

To prove the countable subadditivity, let A E A be such that A = U°_1 Aiwith Ai E A, V i. We write

n-1

Bl := Al and Bn := An - U Ai n A for n > 2.2=1

Then Bn E A for every n and Bn n 0 for n m with00 00

U Bn = UAn .n=1 n=1

Now using (ii) and the countable additivity of jL,00 00 00

Bye = 1: (B) C E µ(An)jL (A) Un 1'n=1 1= n'n=1 1 n== n=

This proves that jL is countably subadditive.

Conversely, let jL be finitely additive and countably subadditive. LetA = U=1 An, where A, An c A for every n and An n Am = 0 for n m.By countable subadditivity,

00

µ(A) < E µ(An).

Ai C A for every n. Since U=1 Ai E A, by (ii) and finite addi-Also, U=1i= j=tivity, we have for every n,

n \ n

µ(A) ? µ U Az _i-1i=1 i=1

lL(Az).

Letting n --+ oo, we get µ(A) > E°_1 µ(A2). This proves that µ is countablyadditive.

3.6.2. Remark: The proof of theorem 3.6.1 includes a proof of the followingfact which is used in many other arguments: for any sequence {A}>1 ofelements of an algebra A, there exists a sequence {B}>1 of pairwise disjointsets in A such that Bn C An b n and U=1 An = U=1 Bn. (See exercise3.2.4.(b).)

In theorem 3.6.1 we saw a necessary and sufficient condition for a finitelyadditive set function defined on an algebra to be countably additive. An-other characterization of countable additivity of set functions defined onalgebras is given in the next theorem.

3.6.3. Theorem: Let A be an algebra of subsets of a set X and let µA -> [0, oo] be such that µ(0) = 0.

(a) If µ is countably additive then the following hold:(i) For any A E A, if A = U=1 Ate, where An E A and An

An+1 d n, then

µ(A) = lim (A).n- +oo

C

This is called the continuity from below of M at A.(ii) For any A E A, if A= nnoo=1 An, where An E A with An D

An+1 V n and p(An) < +oo for some n, then

lim p (An) = M(A).n---oo

This is called the continuity from above of M at A.

Conversely,

(b) If µ is finitely additive and (i) holds, then µ is countably additive.

(c) If µ(X) < +oo, µ is finitely additive and (ii) holds, then µ is countablyadditive.

Proof: (a) Suppose tL is countably additive and let A E A, A = U=1 An,where An E A with An C An+1 V n. Let B1 := Al and Bn := An- An-1for n > 2. Then {B}> 1 is a sequence of pairwise disjoint sets in A suchthat An = U=1 Bk b n and A = U=1 Bn B. Thus using countable and finiteadditivity of M, we have

00 n

µ(A) p(Bk) = lim E (Bk) = lim µn-,oo n-,oo

k=1 /c=1(UBk)=

k=1k=1

lim (A).n-4oo

This proves (i).

(ii) Let A= nnc)o 1 An, where An E A with An An+1 for every n,and let p Ano) < --boo. We write Bn := An0 - An for every n > no. Then

3.6 Countably additive set functions on algebras 67

Bn E A, Bn C Bn+i for every n > no, and U°° n0 Bn = Ano - A. Thus using(i) and theorem 3.6.1 (i), we have

µ(Ano) - µ(A) p(An0 -A)lim p(Bn)

n-, 00

lim p(An0 - An)n-, 00

limo [p(A0)- N'(An)]

p(An0) - lim p(An).n-,00

Hence ,u(A) = limn--+oo ,u(An) . This proves (a) completely.

(b) Let tt be finitely additive and let (i) hold. We have to show that ttis countably additive. Let A E A, A = U=1 An, where An E A for every nand An n A,n = 0 for n m. Then A = U=1(U=1 A 0, and by the givenhypothesis, since U=1 Ak is increasing, we have

,u (A) = lim ,un- ,00 U Ak lim

n->oo

This proves (b).

(c) Again, let A E A, A = U=1 An, where An E A for every n andA n n A, = 0 for n m. Put B n := A - (U=1 Ak). Then Bn E ABnBn+i V n and n00

1Bn = 0. Thus by the given hypothesis and theorem

3.6.1(i), we have

0 = µ(0) = lim (B)n-+oo

lim pn-30o

p(A) - limn-3 00

n 00

E /-t (Ak) Ik-

Ak

n

n

JAkk

n

k=1

limn-, 00

= tt (A) -00

k=1

tt(Ak)-

Hence µ(A) = E 1 This proves the theorem completely.

3.6.4. Exercise:

(i) In the proofs of part (ii) and part (c) of theorem 3.6.3, where do you thinkwe used the hypothesis that µ(X) < +oo? Do you think this condition isnecessary?

(ii) Let A be an algebra of subsets of a set X and µ :.A --> [0, oo] be afinitely additive set function such that µ(X) < +oo. Show that the followingstatements are equivalent:

(a) lim µ(Ak) = 0, whenever {Ak}k>1 is a sequence in A with Ak

Ak+1 V k, and nk,= i Ak = 0-

(b) µ is countably additive.

3.6.5. Exercise: Extend the claim of theorem 3.6.3 when A is only asemi-algebra of subsets of X.

(Hint: Use exercise 3.2.6 or theorems 3.3.3 and 3.6.3.)

3.6.6. Exercise: Let A be an algebra of subsets of a set X which isalso closed under countable unions (such an algebra is called a a-algebra,see definition 3.9.1), and , t : A --+ [0, oo] be a measure. For any sequence{E}>1 in A, show that

(i) , t (lim infn,,,o En) < lim infn,,,o (E).(ii) ,u (limsup,z,,,o En) > lim supn,,,o ,u (En) .

(Hint: For a sequence {E}>i of subsets of a set X,00 00 00 00

lim inf En U ('I Ek C lim sup En := Ek.n-,oo

n=1 k=n n--+oo n=1 k=n

Here are some more examples of finitely/countably additive set func-tions:

3.6.7. Example: Let X be any infinite set and let xn E X, n = 1, 2,... .

Let be a sequence of nonnegative real numbers. For any A C X,define

M (A) pi.{ijX2EA}

It is easy to show that µ is a countably additive set function on the algebraP(X). We say µ is a discrete measure with `mass' pi at xi. The measureµ is finite (i.e., µ(X) <--boo) if E', pi < +oo If E°° 1 pi = 1, the measureµ is called a discrete probability measure/distribution. Note thatµ({xi}) = pi b i and µ({x}) = 0 if x xi. So, one can regard µ as a set

3.6 countably additive set functions on algebras 69

function defined on the subsets of the set Y := {xn : n > 1}. Some of thespecial cases when X = {0, 1, 2.... }are:

(a) Binomial distribution: Y := {0, 1, 2,... , n} and, for 0 0.

(c) Uniform distribution: Y := {1, 2,... , n},

Pk := 1/k d k.

3.6.8. Exercise: Let X be any countably infinite set and let

C={{x}I xEX}.Show that the algebra generated by C is

.F(C) := {A C X I A or A' is finite}.

Let µ :T (C) -> [0, oo) be defined by

0 if A is finite,`4 1 if A°is finite.

Show that µ is finitely additive but not countably additive. If X is anuncountable set, show that µ is also countably additive.

3.6.9. Exercise: Let X =ICY, the set of natural numbers. For every finiteset A C X, let #A denote the number of elements in A. Define for A C X,

An (A)#fm: < m < n,m cz Al

nShow that An is countably additive for every n on P(X). In a sense, µn isthe proportion of integers between 1 to n which are in A. Let

C = {A C X I lim An (A) exists}.n-+oo

Show that C is closed under taking complements, finite disjoint unions andproper differences. Is it an algebra?

3.6.10. Exercise: Let µ : Z fl (0,1] [0, oo] be defined by

µ(a, b]b-a 0,0<a<b<1,

: +oo otherwise.

(Recall that Z fl (0, 1] is the class of all left-open right-closed intervals in(0, 1].) Show that µ is finitely additive. Is µ countably additive also?

3.6.11. Exercise: Let X be a nonempty set.

(a) Let µ : P (X) ) [0, oc) be a finitely additive set function such thatµ(A) = 0 or 1 for every A E P(X). Let

U = JA E 'P(X) I tt(A) = 11.

Show that U has the following properties:(i) 0 V U.(ii) If A E X and BD A, then B E U.

(iii) If A, B E U, then A f1 B E U.(iv) For every A E P (X), either A E U or A° E Lf.

(Any U C P(X) satisfying (i) to (iv) is called an ultrafilter in X.)(b) Let U be any ultrafilter in X. Define t t: 'P(X) [0, oc) by

1 ifAELf,0 if A VLf.

Show that µ is finitely additive.

3.6.12. Exercise: Let A be an algebra of subsets of a set X.

(i) Let µl,µ2 be measures on A, and let a and 3 be non-negative realnumbers. Show that aµ1 +Qµ2 is also a measure on A.

(ii) For any two measures µl,µ2 on A, we say µl < µ2 if µ1(E)µ2(E), d E E A. Let {ufl}fl>i be a sequence of measures on Asuch that µn < tcn+l> V n > 1. Define V E E A,

µ(E) := lim (E).n->oo

Show that µ is also a measure on A and V E E B,

µ(E) = sup {µn(E) I n > 1}.

*3.6.13. Exercise: Let X be a compact topological space and A be thecollection of all those subsets of X which are both open and closed. Showthat A is an algebra of subsets of X. Further, every finitely additive setfunction on A is also countably additive.

3.7. The induced outer measure

In section 3.1 we have seen how to extend the notion of length from theclass of intervals to the algebra generated by intervals. This idea was madeabstract in section 3.3, where we showed how to extend a countably additiveset function ,u from a semi-algebra C of subsets of a set X to .F(C), thealgebra generated by C. In order to extend the notion of length further,we first try to approximate the size of any arbitrary subset A of R usingsets whose size (i.e., the length) is already known, i.e., the intervals. To do

3.7 The induced outer measure 71

this, given a set E we cover it by intervals and calculate the total of thelengths of these covering intervals. This will give us an approximation ofthe size of E. We take the infimum of such approximate sizes and call itthe outer measure of E. Since dealing with the length function on Z or anarbitrary measure µ on an algebra A of subsets of a set X does not makeany difference in the process, for the rest of the section µ will be assumedto be a given measure on an algebra A of subsets of a set X. Our aim is totry to extend µ to a class of subsets of X which is larger than A. Intuitively,sets A in A are those whose size µ(A) can be measured accurately. Theapproximate size of any set E C X is given by the outer measure as definednext. Recall (see appendix A) that for any nonempty set A C [0, +oo],we write inf (A) inf A fl [0, +oo) if A fl [0, +oo) 4 0, and inf(A) +oootherwise.

3.7.1. Definition: Let A be an algebra of subsets of a set X and :.A -[0, oo] be a measure on A. For E C X, define

00 00

µ (E) := inf Y µ(A2) AZ CA, U AZ D E2=1i=1 2=1i=1

The set function µ* is called the outer measure induced by µ.

3.7.2. Remarks:

(i) Given any E C X, there exists at least one covering {A}>1 of E byelements of A, namely {X}. Thus µ* (E) is well-defined.

(ii) The set function µ*(E) can take the value +oo for some sets E.

3.7.3. Exercise: Show that00 00

µ* (E) = inf Eµ(AZ) Ai E A, AZ n Aj =0 for i 4j and UAj DEi=1 i=1

3.7.4. Proposition (Properties of outer measure): The set functionµ : P (X) [0, oo] has the following properties:

(i) µ*(QJ) = 0 and µ*(A) > 0 V A C X.(ii) µ* is monotone, i. e.,

µ*(A) < µ*(B) whenever A C B C X.

(iii) µ* is countably subadditive, i. e., _

00 00

µ*(A) < 1: (A) whenever A= U Ai:i=1 i=1

(iv) µ* is an extension of µ , i. e.,

M* (A)= tL(A) if A cz A

Proof: Properties (i) and (ii) are obvious. To prove (iii), let A = U001°A.

If µ*(Ai) = +oo for some i, then clearly

00

E M* (Ai) +oo > M* (A).i=1

So, suppose M* (Ai) < +oo for every i. Then given e > 0, we can find sets{A}>1 such that Ai C U'l A with each A e A and

M*(Ai) + e/2' > J:M(A').j=1

But, then A = Ui',Ai C U°° 1 U°° 1 Ai and

00 00 00 00

[L* (Ai) + e/2' > 1: E M(A') > p* (A),i=1 z=1 i=1 j=1

00

(A) < (Ai) +i=1

Since this holds for every e > 0, we get

00(A) C (Ai).i=1

This proves (iii).

To prove (iv), let A E A. Clearly µ*(A) < µ(A). To complete the proofwe show that µ*(A) > µ(A). Clearly, µ*(A) > µ(A) in case µ*(A) = +oo.If (A) < +oo and e > 0 is given, we can choose pairwise disjoint setsAn E A, n = 1, 2, ... , such that A C U=1 An and

00

M*(A)+c > I: M(An) (3-3)n=1

3.7 The induced outer measure 73

Since {u=1An n A) }k>1 increases to A, using theorem 3.6.3(i) we have

00 k

E /L (An)n=1

lim 1: /C(An)k-*o0

n=1k

lim IL U Ann-+oo

n=1

µ U (An f1 A)n=1n=1

00

µ U(AnnA)(n=1

µ(A)From (3.3) and (3.4) we have b e > 0,

p* (A) + c > p (A).

Since e > 0 is arbitrary, letting e - 0, we get

/-t* (A) > I-L (A). 0

3.7.5. Remarks:

(i) A set function v defined on all subsets of a set X is called an outermeasure if v has properties (i), (ii) and (iii) in proposition 3.7.4. The outermeasure µ* induced by µ is characterized by the property that if v is anyouter measure on X such that v(A) = µ(A) V A E A, then µ*(A) > v(A).In other words, µ* is the largest of all the outer measures which agree withµ on A.

(ii) In the definition of µ* (E) the infimum is taken over the all possiblecountable coverings of E. T o see that finite coverings will not suffice, con-sider E fl (0, 1), the set of all nationals in (0, 1), and let 1 1 ,1 2 , .. . jnbe any finite collection of open intervals such that E C U=1 I. Then it iseasy to see that E 1A(IZ) > 1. This will imply A*(E) > 1 if only finitecoverings are considered in the definition of A*, which contradicts the factthat A*(E) = 0, E being a countable set.

3.7.6. Example: Let

A:= {A C ][8 I Either A or A' is countable}.

It is easy to see that A is an algebra of subsets of R. (In fact A is whatis called a a-algebra, see example 3.9.2.) For A E A, let µ(A) = 0 if A is

countable and ,u(A) = 1 if A' is countable. We claim that ,u is a measure on,A. For that, let A E A be such that A = U°° 1 Ai, where each Ai E A andAi n A = 0 for i j. We want to show that ,u(A) ,u(Ai) . If each Aiis countable, then A is countable and hence

00

,u(A) = 0 = ,u(Ai).i=1

Next, suppose that Ai is countable for some io. Since Aio n Aj _ 0 for j 4 io,we have A C A o and hence ju(Aj) = 0 V j 4 io. Also A' C A A. Thus A'is countable and ,u(A) = 1. This proves that ,u is a measure on A. Let ,if bethe outer measure induced by ,u on P(R). It follows from proposition 3.7.4that ,u* is countably subadditive on P(TI ). Is ,if countably additive? To findan answer to this, let us first understand ,c* better in this case. If A C T iscountable, then clearly A E A, and hence ,u* (A) = ,u(A) = 0. If A C T is anuncountable set, then clearly

jL*(A) < M* (R) = 1.

Further, if Ai E A, i > 1, are such that A C U°° 1 Ai, then A being uncount-able implies that for some io, Aio is not countable. Since Aio E A, we havethat A o is countable. Thus

00

(A) ? µ(Aio) > 1.i=1

Hence µ* (A) > 1. Thus µ* (A) = 1 if A is uncountable. Now Il8 = (-oo, 0] U(0, oo) and

(R) = 1 < 2 = M(-oo, 01 + M(O, oo).

This shows that µ* need not be even finitely additive on all subsets.

3.7.7. Exercise: Let X be any nonempty set and let A be any algebra ofsubsets of X. Let xO E X be fixed. For A E A, define

µ(4)0 if xo ¢ A,1 if xo E A.

Show that µ is countably additive. Let µ* be the outer measure inducedby µ. Show that µ* (A) is either 0 or 1 for every A C X, and µ* (A) = 1 ifxo E A. Can you conclude that µ* (A) = 1 implies xo E A? Show that thisis possible if {xO} E A.

3.8. Choosing nice sets: Measurable sets

We come back to our construction: given a measure on an algebra A ofsubsets of a set X, try to define a measure µ on some class S of subsetsof X such that A C S and µ(A) = µ(A), d A E A. As a first step in this

3.8 Measurable sets 75

direction, we defined the notion of ,u* , the outer measure induced by µ onall subsets of X. We saw that µ* (A) = µ(A), V A E A, but in generalp* need not be even finitely additive on P (X) . So, as a consolation, let ustry to identify some subclass S of P (X) such that µ* restricted to S willbe countable additive. This is the class S which we call the class of `nice'subsets of X. But the problem is how to pick these `nice' sets?

Since µ* is already countably subadditive, to ensure its countable addi-tivity on a collection s, which can be assumed to be an algebra, it is enoughto ensure its finite additivity on it (see theorem 3.6.1). Let us suppose thatthere is a class S of subsets of X such that S D A and µ* is finitely additiveon S with µ* (A) = ,u(A), V A E A Then for any set Y C X and E E s,by the subadditive property of µ*,

µ* (Y) < tz* (Y n E) + tz* (Y n E'). (3.5)

Also if µ* (Y) = +oo, then

µ* (Y) = +oo > µ* (Y n E) + µ* (Y n Ec) . (3.6)

In case µ* (Y) < +oo, given c > 0, we can choose pairwise disjoint setsA1, A2,-.- in A such that Y C UOO,Ai °and

(Y) + > 1: tz (Ai).00

i=1

Now using the facts that µ*(A) = µ(A) V A E A and that µ* is finitelyadditive on S, we have b i

tz* (Ai) = p* (Ai n E) + tz* (Ai n E').

Thus, using (3.7) and (3.8) we have b Y C X

W (Y) + C >-

00

A*(Ai)

i=100

Ll.t*(Ai n E) + lz*(Ai n Ec)]i=100 00

tz*(Ai n E) + l-z*(Ai n Ec)i=1 i=1

00 00> U Ai n E + U Ai n Ec

> tz* (Y nE)+ j.L* (Y nEc) .

Hence, V Y C X and E E S, from (3.5), (3.6) and (3.9) we have

(3.9)

µ* (Y) = µ*(Y n E) + µ*(Y n E°). (3.10)

Thus the `nice' sets satisfy the above property. We can put this propertyas the definition of `nice sets'. Thus a set E C X is a `nice' set if we use itas a knife to cut any subset Y of X into two parts, Y n E and Y n Ec, sothat their sizes µ* (Y n E) and µ* (Y n E') add up to give the size µ* (Y) ofY. Thus a `nice' set is in a sense a `sharp' knife. This motivates our nextdefinition.

3.8.1. Definition: A subset E C X is said to be iu*-measurable if forevery YCX,

µ* (Y) = µ* (Y n E) + µ* (Y n Ec). (3.11)

We denote by s* the class of all µ*-measurable subsets of X. Note thatE E s* if Ec E s*, due to the symmetry in equation (3.11).

3.8.2. Theorem: Let E C X. Show that the following statements areequivalent:

(i) E E S*.

(ii) For every Y C X,

A* (Y) > A* (Y nE) + A*(Y nEc).

(iii) For every Y C X, with µ* (Y) < +oo,

A* (Y) > A* (Y nE) + A* (Y n Ec).

(iv) For every A E A,

A(A) > A* (A n E) + A* (A n Ec).

Proof: Clearly (i) is equivalent to (ii), since µ* is subadditive. The impli-cations (ii) implies (iii) implies (iv) are also obvious. To complete the proof,we prove (iv) (ii). Let Y C X. If µ*(Y) = +00, then clearly

µ*(Y) = 00 > A*(Y nE)+ it* (Y nEc) -

Next suppose µ* (Y) < +oo. Let c > 0. Choose sets Ai E A, i = 1, 2, ...such that Y C U00 1 Ai and

00

(Y) + (Ai).i=1

Since, by the given hypothesis, V i > 1

it (A) > µ* (Ai n E) + µ* (Ai n Ec),

we have

[L* (Y) + ci=100

> >,Ii*(Ai nE) + (Ai nE)i=1 i=1

> I,* (Y nE) + p*(Y nEc).

Since e > 0 is arbitrary, we have

te (Y) > te (Y nE) + [L*(y nEc).

Hence (ii) holds.

We finally check that S* is indeed the required collection of `nice' sets.

3.8.3. Proposition: The collection S* has the following properties:

(i) ,ACS*.(ii) S* is an algebra of subsets of X, and µ* restricted to S* is finitely

additive.

(iii) If An E S*, n = 1, 2, ... , then U?=1 An E S* and [c* restricted to S*is countably additive.

(iv) Let Al :_ {E C X I 1*(E) = 0}. Then Al C S.

00

00

Proof: (i) Let E C X be arbitrary and let A E A. Let Ai E A, i > 1, besuch that U'l Ai 3 E and Ai n Aj _ 0 for i j. Then, using additivity of[L, we have

00 00 00

>,[L(Ai) _ >,µ(Ai n A) + µ(Ai n Ac)i=1 i=1 i=1

> [L* (EnA) + [L*(EnAc).

Since this holds for any pairwise disjoint covering {A}>1 of E by elementsof A, we have

,a* (E) > ,a* (E n A) + ,a* (E n Ac) .

Also, since [c* is countably subadditive, for any A E A we have

,a* (E) = ,a* (E n A) + ,a* (E n Ac).

Hence A E S*. This proves (i).

(ii) Clearly 0 E S* , and if A E S* then Ac E S*. Finally, let A1, A2 E S.To show that Al U A2 E S*, we have to show that for every E C X withii*(E) < +oo,

µ*(E) = µ*(E n (A1 U A2)) + µ*(E n (A1 U A2)C ). (3.12)

Since Al E S*, we have, b E C X with µ(E) < +oo,

µ*(E) = µ*(E n Al) + µ*(E n Ai).

Changing E to E fl (A1 U A2), we get

,u*(E n (A1 U A2))

= µ*((E n (A1 U A2)) n Al) + µ*((E n (A1 U A2)) n Ai)

= µ*(E n Al) + µ*(E n A2 n Ai). (3.13)

Also, since A2 E S*, we have

µ*(E n Ai) = µ*(E n Ai n A2) + µ*(E n Ai n A2). (3.14)

From (3.13) and (3.14), we get

µ*(E n (A1 U A2)) + µ*(En A; n AZ) = µ*(EnA1) + µ*(E n A;)y* (E).

Thus (3.12) holds, i.e., Al U A2 E S*. In particular, if Ai, A2 E S* andAl n A2 = 0, then by equation (3.13) with E = Al U A2, we have

A*(Ai U A2) = y* (Al) + y* (A2)-

This proves (ii) completely.To prove (iii), let Ai E s* , i = 1, 2, ... , and let A := U°° AZ. We notei=1

that, since s* is an algebra, A can be expressed as a countable union ofpairwise disjoint elements of s* (remark 3.6.2). Thus to show that A E s*,

we may assume without loss of generality that Ai n Aj = 0 for i L j. Now,using the fact that each Ai E s* and Ai n Aj _ 0 for i L j, we have for everyE C X,

tL*(E) == I-t*(EnA,) + I-L*(EnAc)µ*(EnAI)+µ*(EnAC nA2)+µ*(EnAi nA2)µ*(E n A1) + µ*(E n A2) + µ*(E n Ai nAZ)

n

i=1n

i=

(E n Ai) + µ* E n n Aii=1

(UAi)c)*(EnA)+* Eni=1

n 00 Cn

> (EnAi) + /,t* En U Aii-i= i=1i=1

Since this holds V n, letting n -* co and noting that /-c* is countablysubadditive, we have V E C X5

00

*(E) > F*(EnA)+*i=

00

> En U Aii=1i=1

00 C

(En (1Aj)c)C

+ En (OA)ii=1

(3.15)

This proves (in view of theorem 3.8.2) that U'l Ai E S*. In particular, ifwe take E = U'l Ai in (3.15), then

00

U Ai > E /-t * (Ai).i=1 e=1i

00

Since /f (U1 Ai) < E', /f(A) holds because of the countable subad-ditivity property of /*, we get that /.c* is countably additive on S. Thisproves (iii).

The proof of (iv) is easy and left as an exercise.

We have proved in the above proposition that s* A and s* is notonly an algebra, it is also closed under countable unions. Algebras withthis additional property are called a-algebras, and are going to play animportant role in the rest of the discussion. We analyze them in detail inthe next section. We close this section by giving an equivalent definition ofmeasurable sets when (X) < +oo.

3.8.4. Theorem: Let µ(X) < -boo. Then E C X is /f-measurable if

µ(X) = µ(E) + (EC)

Proof: Suppose E is /-t* -measurable. Then V Y C X

/-,* (Y) = /-,* (Y nE) + /-t*(y nEc).

In particular, for Y = X, we have

[t* (X) = /-t* (E) + /-t* (EC).

Conversely, let E C X be such that above equation holds. Since every setA E A is measurable, we have

/-t* (E) = /-t* (E n A) + /-t* (E n Ac)

and

,-t* (EC) _ , L* (EC n A) + ,-t* (EC n AC) .

Adding these two equalities and using the given hypothesis, we have

(X) M* (E) + M* (E')

_ [*(E n A) + µ*(Ecn A)j + [µ*(E n Ac) + µ*(E° n Ac)J> M*(A)+M*(Ac) > M*(X).

The last inequality follows because µ* is subadditive. Hence

µ* (A) + µ* (Ac) = µ* (E n A) + µ* (E° n A) + µ* (E n A°) + µ* (E° n Ac).

But

Thus

,u* (Ac) < ,u* (E n Ac) + ,u* (Ec n Ac) .

[t* (A) > [t* (EnA) + M*(Ec nA).Now it follows from theorem 3.8.2 (iv) that E is measurable.

3.8.5. Exercise: Identify the collection of ,u*-measurable sets for ,u as inexample 3.7.6.

Theorems 3.7.4. (iv) and 3.8.3 give us a method of constructing an exten-sion of a measure ,u defined on an algebra A to a class s* D A. We analyzethe properties of the class s* in the next section.

3.9. The cr-algebras and extension from thealgebra to the generated cr-algebra

Properties of s* , the class of ,u*-measurable subsets (as discussed in theorem3.8.3, motivate the following:

3.9.1. Definition: Let X be any nonempty set and let s be a class ofsubsets of X with the following properties:

(i) 0 and X E s.(ii) Ac E s whenever A E S.

(iii) U"O1 Ai E s whenever Ai E s, i = 1,2 , ...

Such a class S is called a sigma algebra (written as a-algebra) of subsetsof X.

3.9.2. Examples:

(i) As proved in proposition 3.8.3, the class s* of all ,u*-measurable subsetsof X forms a a-algebra.

(ii) Let X be any set. Then {0, X j and P(X) are obvious examples ofa-algebras of subsets of X.

3.9 Extension from the algebra to the generated a-algebra 81

(iii) Let X be any uncountable set and let S = {A C X I A or A' iscountable}. Then S is a a-algebra of subsets of X. To see this we note thatclearly Q l , X ES, and A E S iff A° E S. Let An E S, n = 1, 2, .... If eachAn is countable, then A := U °_ An is countable and hence A E S. In casesome Ak is such that A is countable, then (U=1 An)° C Ak implies thatU=iAk E S.

(iv) Let X be any set and let C be any class of subsets of X. Let S(C) := fl 5,where the intersection is taken over all a-algebras S of subsets of X suchthat S D C (note that P(X) is one such a-algebra). It is easy to see thatS(C) is also a a-algebra of subsets of X and S(C) D C. In fact, if S is anya-algebra of subsets of X such that S D C, then clearly S D S(C). ThusS(C) is the smallest a-algebra of subsets of X containing C, and is calledthe a-algebra generated by C. In general it is not possible to representan element of S(C) explicitly in terms of elements of C. See also theorem4.5.2.

3.9.3. Exercise:

(i) Let S be a a-algebra of subsets of X and let Y C X. Show that S fl Y :_fEny I E E S} is a a-algebra of subsets of Y.

(ii) Let f : X --+ Y be a function and C a nonempty family of subsets of Y.Let f-1(C) := {f-1(C) I C E C}. Show that S(f-1(C)) = f'(S(C)).

(iii) Let X be an uncountable set and C = {{x} I x c X}. Identify thea-algebra generated by C.

3.9.4. Exercise: Let C be any class of subsets of a set X and let Y C X.Let A(C) be the algebra generated by C.

(i) Show that S(C) =S(A(C)).(ii) Let C n Y:= {E n Y I E E C}. Show that S(C n Y) c S(C) n Y.

(iii) LetS:= {EU(BflYC) S E c S(C n Y), B E C}.

Show that S is a a-algebra of subsets of X such that C C S andSnY=S(CnY).

(iv) Using (i), (ii) and (iii), conclude that S(C fl Y) = S(C) fl Y.

3.9.5. Proposition: Let C be a class of subsets of a set X such that 0 E C.Then E E S(C) iff 3 sets Cl, C2,... in C such that E E S({Cl, C2, ... }).

Proof: Let

,t3 :_ {E E S(C) I E E S({Cl, C2i ... }) for some Cl, C2,... E C}.


Clearly C C B. Also, if E E B and E E s ({ Cl , C2, ... }), then by definition,EC E s ({ Cl , C2 , ... }). Hence Ec E B. Finally, let Ej E B, j = 1, 21 .... IfEj E s({C1,C2,...}), CZ E C V i,j, then Ej E s{CZ I j = 1,2,... ,i =1, 2,... }. Hence

00

UEjE8{CZ Ij = 1,2.... ; i = 1, 2,..j=1

Thus U'l E3 E B. This proves that B is a Q-algebra. Since C C B, bydefinition we have S(C) C B. Hence S(C) = B.

3.9.6. Note: The technique used in the proof of proposition 3.9.5 is veryuseful and is often used to prove various properties of a--algebras under con-sideration. The sets satisfying the required property are collected together.One shows that this collection itself is a a--algebra and includes a subfamilyof the original a--algebra which generates it. The claim then follows by thedefinition of the generated a--algebra. We call this the a--algebra tech-nique. Another example of this technique is found in the next exercise.

3.9.7. Exercise: Let X be any topological space. Let U denote the classof all open subsets of X and C denote the class of the all closed subsets ofX.

(i) Show that

s(u) =s(c).This is called the Q-algebra of Borel subsets of X and is denotedby BX.

(ii) Let X = R. Let I be the class of all intervals and Z the class of all left-open right-closed intervals. Show that I C S(Lf), I C S(Z), Z C S(Z)and hence deduce that

s(z) = s(i) = BR .

3.9.8. Exercise:(i) Let Z, denote the class of all open intervals of ][8 with rational endpoints.Show that S(ZT) = BR.

(ii) Let Id denote the class of all subintervals of [0, 1] with dyadic endpoints(i.e., points of the form m/2' for some integers m and n). Show that S(Id) _aRn [0, 1].

Combining proposition 3.7.4(iv), proposition 3.8.3 and example 3.9.2(i),we have the following proposition.

3.9 Extension from the algebra to the generated a-algebra 83

3.9.9. Proposition: Let A be an algebra of subsets of a set X and p a mea-sure on A. Then there exist a a-algebra S* (the a-algebra of µ*-measurablesubsets of X) and a measure µ* on S* such that A C S* and µ*(A) =p(A)for every A E A.

The above proposition tells us that a* is a measure and is an extension ofthe measure p from an algebra A to a a-algebra s* . In particular s* D S (A),the a-algebra generated by A. At this stage, the following natural questionsarise:

Question 1: Is the extension of µ from A to S(A) given by proposition3.9.9 unique?

Question 2: We know that S(A)CS*. Is there some other relation betweenthe two a-algebras S(A) and S*? Is S* much larger than SS(A)?

In the remaining sections of this chapter we shall answer these questions.The next example shows that in general the answer to our first question isin the negative.

3.9.10. Example: Let A = A(Z), the algebra generated by left-open right-closed intervals in R. For A E A, let µ(A) = +oo if A : 0 and µ(0) = 0.Then µ is a measure on A. Let E ][8 be chosen arbitrarily and fixed. LetA denote the algebra of subsets of ][8 generated by A and Define forA E

(A)+oo

if either A = 0 or A =

It is easy to check that µ is a measure on A. If A # 0 and A E A(Z), thenclearly A\ 0 and hence µ(A) By proposition 3.9.7, µ(and hence µ) can be extended to a measure, which we denote by µ again,on S(.A). Note that b x E X, {x} E S(A) with {x}) = +oo if x #and 0. Thus for 6, 6 E ][8 if 1 < 6, then 0, whereas

({i}) 0. Hence for every ][8, µ is an extension of µ and µ,l :for i :

3.9.11. Exercise: Let p and A be as in example 3.9.10. Show that d A EBR there exists a measure µ,y which extends µ to BR and µA : µB forA # B.

3.9.12. Exercise: Let X denote the set of rationale in (0,1] and let A beas in example 3.9.10. Show that the a-algebra X fl S(A) = P(X) and thatevery nonempty set in the algebra X fl A has an infinite number of points.For any E E X fl S(.A) and c > 0, define µ, (E) = c times the number of

points in E. Show that it, is a measure on X n S(A) = S(X n A) and itextends the measure it on XnA given by µ(A) = +oo if A : 0 and µ(O) = 0.Further, µ,l : µ,2 if cl : c2.

Example 3.9.10, exercises 3.9.11 and 3.9.12 show that in general a mea-sure y on an algebra A can have more than one extension to 8(A), thea-algebra generated by A. In all these examples, the common feature is thatthe measure it assigns a large value, namely (boo), to nonempty sets. Soone asks the modified question: under what conditions on it is the extensionunique? We analyze this problem in the next section.

3.10. Uniqueness of the extension

To look for the condition on it which may ensure the uniqueness of theextension, we analyze for the particular case of it, namely the length functionA on A(Z), the algebra generated by all the intervals in R. The lengthfunction, though it assigns the value boo to unbounded intervals, has theproperty that we can write T = U;(n, n 1] and d n, A(n, n + 1] _1 < boo. In other words, the set III can be decomposed into a countableunion of pairwise disjoint sets each having finite length. This motivates thefollowing definition.

3.10.1. Definition: Let C be a collection of subsets of X and let y :C -) [0, oo] be a set function. We say it is totally finite (or just finite) if,a(A) < boo V A E C. The set function it is said to be sigma finite (writtenas a-finite) if there exist pairwise disjoint sets Xn E C, n = 1, 2,... , suchthat,a(Xn) <+oo for every nandX=U'1Xn.

3.10.2. Example: The length function A on the class of intervals is a-finite,whereas none of the set functions considered in example 3.9.10 or exercises3.9.11 or 3.9.12 are a-finite. In general, a measure it defined on an algebraA of subsets of a set X is finite if µ(X) < +oo.

The problem that we want to analyze is the following: Let it be a a-finite measure on an algebra A of subsets of X. Let ,a1 and A2 be twomeasures on 8(A), the a-algebra generated by the algebra A, such thatit1(A) = ,a2(A) V A E A. Is it1 = ,a2? Since it is a-finite, we can writeX = LJ1 Xi, where each Xi E A, Xi n Xj _ 0 and pi (X) < boo V i = 1, 2and j = 1, 2, .... Thus for E E S (A),

pi (E) = µi E fl Xj _ (EnX).(3c1)) j=1

3.10 Uniqueness of the extension 85

Thus µ1(E) = µ2 (E) will hold provided we can say that µ1(E fl Xj) =µ2(Ef1Xj) V j. Equivalently, to have µl = µ2 on S(A), we have only to showthat µl and µ2 are equal on each of the Q-algebras S(A) f1 Xj = S(Af1 Xj),given that µl and µ2 are equal on the algebra A fl Xj. We note that therestriction of µl and µ2 to each Xj is totally finite. Thus we will have apositive answer to our question provided we can give a positive answer tothe question under the additional condition that µl and µ2 are totally finite.In other words, we may assume without loss of generality that µl and µ2are totally finite.

So, let us analyze the situation: µl and µ2 are totally finite measureson S(A) such that µ1(A) =µ2(A) for every A E A. We want to claimthat µ1(E) =µ2(E) for every E E S(,A). Let us try to use the o--algebratechnique as mentioned in note 3.9.6. Let

,A4:= fE Cz S(A) I MI(E)= A2(E) 1.

We are given that A C M, and we have to show that S(A) C M. It isnatural to try to prove that M is a o--algebra, so that A C M would thenautomatically imply, by the definition of S(.A), that S(A) C M, and wewill be through. Let us analyze the properties of the class M. If E E .M,can we conclude that E° E M? Since A C M, we have X E M and thus1_11(X) =µ2(X). Also E E M implies µ1(E) =µ2(E). Since µ1,µ2 are finite,we have

A1(EC) = M 1(X) - M 1(E) = P2 (X) - A2 (E) = P2 (EC) .

Hence EC E M. Next, let El, E2 E M. Can we conclude that El U E2 EM? Apparently, we cannot conclude this in general (we ask the reader toconstruct examples), i.e., M need not be a a-algebra. Thus the a-algebratechnique does not work here. Let us look for some other properties of M.For example, let En E M, n > 1, be such that En C En+1 V n. Can wesay that U=1 En E J4? Well, since /21 and /22 are measures, using theorem3.6.3(a), we get

00

Al U En = lim [ii (En)n- +oo

n=1

lim /12 (En)n--+oo

/12

Thus U=1 En E M. Also, if En E M and En En+1 V n, then, again not-ing that Al, ,u2 are totally finite, by theorem 3.6.3(b) we have fl 1 En E M.Thus we have proved the following:


3.10.3. Proposition: Let µl and µ2 be measures on a a-algebra S. Thenthe class M = {E E S I µ1(E) =µ2(E)} has the following properties:

(i) 0 E M.(ii) If An E M, n = 1, 2, ... , and An C An+l V n, then U=1 An E M.(iii) Assume further that µl and µ2 are totally finite. Then for An G M

with An D An+l for n = 1, 2, ... , we have n°°_1 An E M.

This motivates the following definition.

3.10.4. Definition: Let X be a nonempty set and M be a class of subsetsof X. We say M is a monotone class if

(i) U=1 An E M, whenever An E M and An C An+1 for n = 1, 2, ... ,

(ii) nn,jAnEM,wheneverAnEM andAnDAn+1 for n=1,2,... .

3.10.5. Examples:

(i) If µl and µ2 are totally finite measures on a a-algebra S and M {E ES I µ1(E) =µ2(E)}, then proposition 3.10.3 says that JVl is a monotoneclass.

(ii) Clearly, every a-algebra is also a monotone class.

(iii) Let X be any uncountable set. Let M :_ {A C X I A is countable}.Then ,M is a monotone class but not a a-algebra.

(iv) Let X be any nonempty set and let C be any collection of subsets of X.Clearly P(X) is a monotone class of subsets of X such that C C P(X). Let.M (C) := n .M, where the intersection is over all those monotone classes Mof subsets of X such that C C M. Clearly, M (C) is itself a monotone class,and if ,M is any monotone class such that C C ,tit, then A4 (C) C M. Thus,M (C) is the smallest monotone class of subsets of X such that C C A4 (C).The class M (C) is called the monotone class generated by C.

3.10.6. Exercise: Let C be any class of subsets of X.

(i) If C is an algebra which is also a monotone class, show that C is a a-algebra.

(ii) C C M(C) C S(C).

(This exercise will be often used in the sequel.)

Let us go back to our original problem which motivated us to define theconcept of a monotone class. We had two totally finite measures µ1 and µ2

3.10 Uniqueness of the extension 87

on S(A), the a-algebra generated by an algebra A of subsets of a set X. Wewere given that µ1(A) = µ2 (A) for every A E A. To prove that µl = µ2,we had to show that S(A) C M :_ {E E S(A) I µ1(E) =µ2(E)}. We sawthat A C M and M is a monotone class. Thus M (,A) C M, where M (,,4)is the monotone class generated by A. This leads us to a natural question(again motivated by our requirement): is M (A) = S(A)? Since every a-algebra is a monotone class and A C S(,,4), clearly ,M (,A) C S(A). Thus thequestion is, is S(A) C ,M(A)? The next theorem answers this question inthe affirmative.

3.10.7. Theorem (a-algebra monotone class): Let A be an algebra ofsubsets of a set X. Then S(A) = M(A).

Proof: As stated above, .M(,.4) C S(,.4). We only have to show that S(.A) C.M (A). Since A C ,M(,,4), to show that S(A) C .M (A) it is enough to showthat .M (A) is a a-algebra. We know .M (A) is a monotone class, and henceto show it is a a-algebra, it is enough to show that M (A) is an algebra (seeexercise 3.10.6).We use the Q-algebra technique.

We first show that ,M (,A) is closed under complements. Let

L3:= fE C X I E' E _A4(A)j-

We have to show that ,M (,A) C B. We note that if E E A, then E° E A C,M (A). Hence A C B. Further, it is easy to check that ,t3 is a monotone class.Hence .M(,,4) C B. Next we check that .M(,,4) is closed under unions, i.e., ifE, F E .M (.A), then E U F E .M (.A). For F E M (A), let

L (F) : = f A C X I A U F E M (A) I -

Now, we have to show that G(F) D .M (,A) for every F E .M(A). Let us firstcheck it for F E A. In that case, for every A E A, the sets A U F, A fl F° EA C .M (A), proving that A C G(F) if F E A. It is easy to check that ,C (F)is a monotone class for every F C X. Thus for F E A we have A C L(F), amonotone class, and hence.M(,A) C G(F). Now we note that the class L(F)is `symmetric' in the sense that E E ,C(F) if F E ,C(E). Thus for everyF E A and E E.M(A), since E E .M(,.4) C G(F), we have F E ,C(E). Thus,,4 C G(E) for every E E .M (.A). Once again exploiting the fact that ,C (E) isa monotone class, we have .M (.A) C G(E) for every E E .M(A). This provesthat M is also closed under finite unions. This completes the proof.

As an application of this theorem (as pointed before the statement ofthe theorem) we have the following:

3.10.8. Theorem: Let µ be a measure on an algebra A of subsets of a setA. If µ is a-finite, then there exists a unique extension of µ to a measure µon S(A), the a-algebra generated by A.

Proof: Recall that starting with µ, we constructed µ* on P(X) and foundS*, the a-algebra of µ*-measurable sets such that <S* D A, with µ* beingcountably additive on S* and µ*(A) = µ(A) V A E A. Thus µ(E) :=µ*(E), E E S(A), gives a measure with the required properties. We onlyhave to show that this is the only one. Let µl and µ2 be two a-finiteextensions of µ from A to i.e., µ1(A) =µ2(A) V A E A. We haveto show that µ1(E) =µ2(E) V E E S(A). As noted in the discussion afterexample 3.10.2, we may assume that µl and µ2 are in fact totally finite. Let

.M :_ JE C X I µ1(E) =µ2(E)}.

Then A C M and M is a monotone class (proposition 3.10.3). ThusM C M. Now by theorem 3.10.7, S(A) = .M (A) C ,/V(, proving thetheorem.

3.10.9. Remark: A a-finite measure A on an algebra A has a uniqueextension to 5(A), as shown above. In fact, A has extension to S*, the a-algebra of A* measurable sets, and we know that s* D 8(A). The naturalquestion arises: what is the difference between the sets in 8(A) and s*? Wehad raised this question in the last section also (question 2 after proposition3.9.9). We shall examine this question in the next section.

3.10.10. Note: In the proof of theorem 3.10.8, though the a-algebra tech-nique didn't work, we used the monotone class theorem (3.10.7) to provethat µ1(E) =µ2(E) V E E whenever the sets in A has this property.This technique is also very useful, and we call it the a-algebra monotoneclass technique.

3.10.11. Exercise: Let X = [a, b] and let S be the a-algebra of subsetsof X generated by all subintervals of [a, b]. Let µ, v be finite measures on Ssuch that µ([a, c]) = v([a, c]), d c E [a, b]. Show that µ(E) = v(E) V E E S.

3.10.12. Exercise: Let µF be the measure on the algebra .A(Z) as givenin proposition 3.5.1. Let µF itself denote the unique extension of µF to GF,the a-algebra of µF-measurable sets, as given by theorem 3.10.8. Show that

(i) BR C GF.

(ii) µF({x}) = F(x) - F(y). Deduce that the function F is continuous

at x iff AF(jXj) = 0.

3.11 Completion of a measure space 89

(iii) Let F be differentiable with bounded derivative. If A C R is a nullset, then µF(A) = 0.

The measure µF is called the Lebesgue-Stieltjes measure induced by thedistribution function F.

3.11. Completion of a measure space

In theorem 3.10.8 we showed that, given a a-finite measure p on an algebraA of subsets of a set X, p can be extended to a unique measure jc* on thea-algebra $* of ,a*-measurable subsets of X, and $* D S (A) . In remark3.10.9 we raised the following question: what is the difference between thesets in $* and the sets in S (A) ? We answer this question in this section.Let us fix a a-finite measure p on an algebra A for the rest of the section.Let A, denote the collection of sets of the form U'l Ai, Ai E A. Our nextproposition gives equivalent ways of describing ,c* (E) for any set E C X, jc*being the outer measure induced by p.

3.11.1. Proposition: For every set E C X,

µ*(E) = inf {µ*(A) I A E Aa, E C Al= inf {µ*(A) I A E S(A), E C Al= inf {µ*(A) I A E S*, E C Al.

Proof: By the definition of µ*,00

jc* (E) = inf (Aj)i=1

Let Ai E A, i > 1, be such that

00

Ai E A, E C U Ai .

i=1

00

A:= U Ai D E.i=1

Using remark 3.6.2, we can write A = U'l Bi, where Bi E A, Bi C Ai `v' i

and Bi f1 Bj = 0 for i L j. Thus00 00 00

j:jL(Aj) >_ EjL(Bj) _ /.L* (Bi) > p*(A) .

i=1 i=1 i=1

Note that A E A. Thus

µ* (E) > inf {µ* (A) I A E Ao, E C A}

> inf {µ* (A) I A E S(A), E C Al

> inf {µ* (A) jA E S*, E C A}

> jL* (E). M

3.11.2. Proposition: For every E C X, there exists a set F E S(A) suchthat E C F, µ*(E) = µ*(F) and µ*(F \ E) = 0.

The set F is called a measurable cover of E.

Proof: Since M is a-finite, we can assume that E = U'l EZ, where EZ n Ej =0 for i j and a* (E,) < +oo for every j. For every fixed j , by proposition3.11.1, V E > 0 there exists a set FE E S (A) such that E C F, and,u* (Ej) +E > M* (FE) . In particular for every E = 1/n we can choose Fn E 8(A)such that

E j C Fn and ,* (Ej) + 1/n > ,u* (Fn), n = 1, 2, ... .

Let Fj : = n°° 1 Fn. Then Ej C Fj E S (A) and, by theorem 3.6.3,

*(Ej) > limsup,u*(Fn) _ ,u*(Fj) > ,u*(Ej).n-*oo

Hence µ*(Ej) = µ*(Fj). Further, let G E S(A) be such that G C Fj - Ej.Then Ej C Fj - G and we have

W (Ej) < p* (Fj - G) =: M* (Fj) - p* (G) < M* (Fj) (Ej).

Hence µ*(G) = 0 V G E S(,A) with G C Fj - Ej. Now it follows fromproposition 3.11.1 that µ* (Fj - Ej) = 0. We construct Fj for every j andput F:= U'l F. Then E C F E S(,A). Since

00 00 00

F-E = UFO-UE. C U(F3-E3),j=1 j=1 j=1

we have

µ* (F - E)00

Fj - Ej) = 0.

Finally, since

µ*(E) < (F) < µ*(E) +µ*(F - E) = µ*(E)it follows that µ* (E) = µ* (F).

3.11.3. Corollary: Let E C X. Then there exists a set K C E, K E S(A),such that µ*(A) = 0 for every set A C E \ K.

The set K is called a measurable kernel of E.

Proof: Let G be a measurable cover of E and N be a measurable cover ofG \ E. Let K:- G \ N. Then

K = (G \ N) C (G \ (G \ E)) E.

Further, if A C (E \ K), then

A C (E\(C\N)) = E n N C (N\(C\E)).


Since N is a measurable cover of G \ E, the above implies that µ* (A) = 0.Hence K is a measurable kernel of E.

As an application of proposition 3.11.2, we have the following extensionof theorem 3.6.3(i).

3.11.4. Proposition: Let El C EZ C E3 C ... be subsets of X. Then00

,u U En = lim ,u* (En)n-+oo

n=1

Proof: Let E = U=1 E. By proposition 3.11.2, we can choose sets A, An ES(A) such that E C A, En C An C A with p* (E) _ ,u* (A) and ,u* (An) =1f (En) V n > 1. In order to be able to exploit the fact that ,u* is countablyadditive on S(A), we put

Bn :=

Since

00

m=nAm, n > 1.

EnCEmCAm Vm>n,we have En C Bn C An. Thus

,cc* (Bn) = [u* (En) _ ,u* (An) V n > 1

and {B}>1 is an increasing sequence of sets in S (A) . Now by the countableadditivity of ,u* on S(A) and theorem 3.6.3, we have

lim ,u* (En) = lim ,u* (Bn) = ,u*n-+oo n- oo

Since00 00 00

E UEn C UBn C U An C An=1 n=1 n=1

and p* (E) = p* (A), we have00

A*(E) = A* U Bnn=1

The required claim follows from (3.16) and (3.17).

3.11.5. Exercise:

(3.16)

(i) Let E C X, and let Gl, GZ be two measurable covers of E. Show thattL*(G,AG2) = 0-

(ii) Let E C X, and let K1, K2 be two measurable kernels of E. Show thatp* (Ki AK2) = 0-

(iii) Let j V:= JE C X I µ*(E) = 0}. Show that N is closed under countableunions and

<S* = S(A) U N :_ {EuN I E E S(A), N EN},

where S* is the a-algebra of µ*-measurable sets. Further, V A E S*

µ*(A) = µ*(E), if A = E U N, with E E S(A) and N E Al.

3.11.6. Definition: Let X be a nonempty set, S a a-algebra of subsets ofX and µ a measure on S. The pair (X, S) is called a measurable space andthe triple (X, S, µ) is called a measure space. Elements of S are normallycalled measurable sets.

Till now what we have done is that, given a measure on an algebra A ofsubsets of a set X, we have constructed the measure spaces (X, S(A), µ*),(X) S*, µ*) and exhibited the relations between them. The measure space(X, S*, µ*) has the property that if E C X and µ*(E) = 0, then E E S*. Thisproperty is called the completeness of the measure space (X, S*, µ*).The measure space (X, S(A), µ*) need not be complete in general. However,S* is obtainable from S(A) and N := JE C X I µ*(E) = Of by

S* = U N :_ {EuNEe$(A), N E N}.One calls (X, S*, µ*) the completion of (X, S(A), µ). This construction canbe put in a general context as follows.

3.11.7. Definition: Let (X, <S, µ) be a measure space and let N :_ IE CX I E C N for some N E S with µ(N) = 0}. One says (X, S, µ) is completeif Al C S. Elements of N are called the µ-null subsets of X.

The abstraction of the relation between the measure spaces (X, 5(A), ,a*)and (X, s* , ,u*) is described in the next theorem.

3.11.8. Theorem: Let (X, S, µ) be a measure space and let N be the classof µ-null sets (as in definition 3.11.7). Let S D N :_ {E A N I E E S, N EN} and <S U N :_ {E U N I E E S, N E N}. Then S A N= S U N is aa-algebra of subsets of X. Let µ(E A N) = µ(E), V E E SIN E N. Thenµ is a measure on S U N and (X, S U N, )7) is a complete measure space,called the completion of the measure space (X, S, µ). (The measure space(X, S U N, µ) is also denoted by (X, S, µ).

Proof: If E E S and N C A E S with µ(A) = 0, then the identities

EuN=(E-A)A(Afl(EuN))

and

ESN=(E-A)u(An(EA N))show that S U N = S A N. Now, using this fact, it is easy to show thatSUN = S A N is a a-algebra of subsets of X.

Let µ(E A N) := µ(E) for every E E S, N E N. To prove that µ iswell-defined, let El A Nl = E2 D N2, where Ez E S and Ni c N, i = 1, 2.Then, using the fact that A is an associative operation, it follows that

0 = (E1AN1)A(E2AN2) = (E1AE2)A(N1AN2).Thus

El A E2 = Nl D N2 C Nl U N2.Clearly N1UN2 E N, and hence µ(E1A E2) = 0. Thus µ(E1) = µ(E1f1E2) _µ(E2). Equivalently, µ(E1 A Nl) = -(E2 A N2). Hence µ is well-defined.Further, suppose E E S U N. Let E = E U N, where E E S, N E N andN C A E S with µ(A) = 0. Then

EuN=(E-A)A (An(EUN)).Thus

-(E U N) = p(E - A) = p(E).Finally, let E U N = U1(Ei U Ni), where E, Ei E S and N, Ni E N forevery i, with (Ei U Ni) fl (Ej U Nj) = 0 for i 4 j. Then µ(E U N) = µ (E).Also,

E U N= (QE)i U U Nii=1 i=1

Since u:1 Ni E N, we have

µ(EUN)=µ UEj2-1

also. Hence

µ(E U N) _0000

p(Ei) = L - (Ei),i=1 i=1

proving that )i is a measure. That (X, S U N, )C) is a complete measurespace, is easy to check.

Finally we describe the relation between µ* on P(X) and µ on A.

3.11.9. Proposition: Let µ be a measure on an algebra A of subsets ofa set X and let µ* be the induced outer measure. Let E E S* be such thatµ*(E) < +oo and let e > 0 be arbitrary. Then there exists a set FE E A suchthat e.

Proof: Since ,u* (E) < +oo, by definition of ,a*, we can find pairwise disjointsets Fn eA,n=1,2,...,such that ECU°°1Fnand

00

(E) + > , p* (Fn) > p* (E)n=1

But then E00 1 p* (Fn) < +oo, and hence we can find some no such that00

E p* (Fn) <n=no +1

If we put FE =U=1

Fn, thenno

E\FE = E\ U Fn

Thus

Also,

Hence

00 no

C (uF)(uF)n= 1

00

(E - FE) < 1: p* (Fn) <n=no +1

FE\E C (OF)n -E.(n=1

c'Op*(Fe-E) < Y_"p*(Fn)-A*(E),n=1

since ,if is countably additive on S*. Thus

Fn F.

µ*(E A FE) = µ*(E \ FE) + µ*(FE \ E) < 2e.

(Though we wanted to find F such that µ*(E A FE) < e, we obtained onlyµ* (E A FE) < 2e. With appropriate corrections at suitable places we will getthe required inequality.)

3.11.10 Note: Whenever (X, S, µ) is a finite measure space with µ(X) = 1,it is called a probability space and the measure µ is called a probability.The reason for this terminology is that the triple (X, S, µ) plays a funda-mental role in the axiomatic theory of probability. It gives a mathematicalmodel for analyzing statistical experiments. The set X represents the setof all possible outcomes of the experiment, the a -algebra S represents thecollection of events of interest in that experiment, and for every E E S, thenonnegative number µ(E) is the probability that the event E occurs. Formore details see Kolmogorov [21] and Parthasarathy [28].

Chapter 4

The Lebesgue measureon R and its properties

4.1. The Lebesgue measure

We now apply the extension theory of measures, developed in chapter 3, tothe particular case when X = Itt, A = A(I), the algebra generated by allintervals, and p on A is the length function A as described in section 3.1.The outer measure A , induced by the length function A, on all subsets ofIt is called the Lebesgue outer measure and can be described as follows:forECR,

00 00

A (E) := inf EA(ij) Ii E Z d i, Ii n Ij= 0 for i j and E C Iii=1 i=1

The a-algebra of A -measurable sets, as obtained in section 3.8, is called thea-algebra of Lebesgue measurable sets and is denoted by L R, or simplyby L. The a-algebra S (I) = S (A) := BR, generated by all intervals, is calledthe a-algebra of Borel subsets of R. We denote the restriction of A to L or13R by A itself. The measure space (Itt, L, A) is called the Lebesgue measurespace and A is called the Lebesgue measure. We note that since A on Iis a-finite (e.g., R = U. (n, n + 1] ), the extension of A to BR is unique bytheorem 3.10.8. We recall that the a-algebra BR includes all topologically`nice' subsets of Itt, such as open sets, closed sets and compact sets. Also,for E E BR, if we transform E with respect to the group operation on Itt,e.g., for x E Itt, consider E + x := f y + x I y E E}, then E + x E z3Ilg . For this,note that the map y F-- * x + y is a homeomorphism of It onto Itt, and henceE + x E B for every open set E. We leave it for the reader to verify (using

95

96 4. The Lebesgue measure on JR and its properties

a--algebra techniques) that this is true for all sets E E B1 . The relation ofA on £ with A on topologically nice subsets of JR will be explored in section4.2, and the question as to whether E + x c £ for E E £, x c III, i.e., do thegroup operations on JR preserve the class of Lebesgue measurable sets, willbe analyzed in section 4.3. The relation between the cr-algebras £ and B1will be discussed in section 4.5. We give below some properties of A whichare also of interest.

4.1.1. Exercise:

(i) Let Zo denote the collection of all open intervals of R. For E C X,show that

00

A (E) = inf {(i)2=1

00

Ii c Zo `d i, for i j and E C U Ii .

i=1

(ii) Let E C Il8 and let c > 0 be arbitrary. Show that there exists anopen set UE D E such that S(UE) < A (E) + e. Can you also conclude thatA (U,6 \ E) < 0

(iii) For E C ][8, let

diameter(E) := sup{ Ix - yj I x, y c E}.

Show that A (E) < diameter(E).

(iv) Show that for E C IIB, A* (E) = 0 if E is a null set as in definition 1.2.8.

(v) Let E C [0,1] be such that A* ([0,1] \ E) = 0. Show that E is dense in[0,1]

(vi) Let E C ]l8 be such that A (E) = 0. Show that E has empty interior.

As a particular case of proposition 3.11.4, we have the following.

4.1.2. Proposition: Let {E}>1 be any increasing sequence of subsets(not necessarily measurable) of R. Then

00

A U En = lim A (En).n-4oo

n=1

We give below an application of the above proposition. For a functionf b] --> II8, we denote by f '(x) the derivative of f at x, whenever itexists.

4.1 The Lebesgue measure 97

*4.1.3. Proposition: Let f : [a, b] ) ][8 and for any real number a > 0,let E,,, {x E [a, b] I f '(x) exists and I f '(x) I < a}. Then

A (f (Ea)) aA (E.).

Proof: Let e > 0 be chosen arbitrarily and fixed. For every n = 1, 2, ...let En denote the set of points x E Ea such that b y E (a, b),

0 < Ix - yj < 1/n implies If (x) - f (y)l < (a + E)Ix - yl.

Clearly, En C En+l d n and U=1 En = E, Thus {f(E)}>1 is an in-creasing sequence and f (E,) = U=1 f (En). By proposition 4.1.2, we get

A (f (Eo,)) lim A (f (En)). (4.1)

To compute A (f (En)), let n > 1 be fixed. We choose a sequence {Ik}k>1 ofintervals such that for each k, A(1k) < 1/n with En C U001 Ik and

00

1: A(1k) < A* (En) +k=1

Then by the definition of En, b x, y E En f1 Ik,

If(x)-f(Y)1 < (a+E)Ix-YI-Thus

diameter (f(EflIk))n < (Ik)).Hence

Therefore,

A (f (En n ik)) (a + c) A (1k).(f(EnIk))

00

nA (/(En)) A*(f(EflIk))k=1

00(a +c) E A(1k)

k=1

(a + c) (A (En) + (4.2)

From (4.1) and (4.2), using proposition 4.1.2, we have

A (f (Ec,)) (a + c) (A (Ec,) +

Since e > 0 was chosen arbitrarily, we get A (f(Ea)) < aA (Eo,).

*4.1.4. Theorem (Saks): Let f : Il8 Ilk and let E = {x E 1[8 I f'(x)exists and f '(x) = 0}. Then A* (f (E)) = 0.

98 4. The Lebesgue measure on JR and its properties

Proof: It is enough to prove that A (f (E n [a, b])) = 0 for every a < b. Leta 1,

En [a, b] C En {x E {a,b]flEI If'(x)I < 1/n} .

Hence by proposition 4.1.3,

A (f(En [a, b])) A (f (En)) A (En)/n < (b-a)/n.

Since this holds b n, A (f (E fl [a, b]) = 0.

4.1.5. Note: Saks' theorem tells that the critical values (the values f (x),x c lI8 such that f'(x) = 0) of a function f on lI8 constitute a null set. Thishas a converse, which we prove next.

*4.1.6. Theorem: Let f ][8 --> ]I8 have a derivative on a set E andA (f (E)) = 0. Then

A (jx C: E I f (x) 7 01) = 0.

Proof: Let B :_ {x E E I If '(x) I > 0} and for n = 1, 2, ... ,letBn := {x E B I If (x) - f()l > Ix - yI/n d y with 0 < Ix - yj < 1/n}.

Then B = UM1 B. Thus to show that A (B) = 0, it is enough to show thatA (Bn) = 0 V n. We fix an integer n and note that to show A (Bn) = 0, itis enough to show that /\ (I fl Bn) = 0 for any interval I of length less than1/n. Let A = I fl Bn. We have to show that A(A) = 0. For this, first notethat f (A) C f (E) implies that A* (f (A)) = 0. Thus given e > 0, we can findintervals {Ik}k>1 such that A(1k) < 1/n d k with

"0

00

f (A) C U Ik and E A (Ik) G E.k=1 k=1

Let Ak := An f -1(Ik). Then A C U'l Ak and for x, y E Ak, since Ix - I <1/n, we have 11(x) - f()l > Ix - yl/n. Hence

sup{Ix - yIIx, y E Ak} < nlf (x) - f (y)l< n (sup{Is - tIIt, s E f (Ak)}) .

Thus

Finally,

A (Ak) < nA (f (Ak)) nA (1k).

00 00

A (A) < E A* (Ak) < E A* (1k) < nE.k=1 k=1

4.2 Lebesgue measurable sets and nice sets 99

Since e > 0 is arbitrary, A (A) = 0.

An immediate corollary of the above theorem is the following:

*4.1.7. Corollary: If f : ][8 -> ][8 has a derivative on a set E and f isconstant on any subset F of E, then

A (Jx EE F I f'(x) 4 01) = 0.

4.2. Relation of Lebesgue measurable sets withtopologically nice subsets of III

As a particular case of proposition 3.11.9, we have the following.

4.2.1. Theorem: Let E C III and ) (E) < +oo. Then, given c > 0, thereexists a set FE which is a finite disjoint union of open intervals and is suchthat

(EAF) <

Proof: By proposition 3.11.9, given c > 0 there exists a set F E .EL(I)such that A (E 0 F) < E. Let F :=

U=1

Ii, where Ii E I and I1, ... , In are2

disjoint. Let JZ denote the open interval with the same endpoints as Ii andlet FE : = U=1 JZ . Then

A (E A Fe) = A (E A F) < c. N

We give next some more characterizations of Lebesgue measurable sets.

4.2.2. Theorem: For any set E C ][8 the following statements are equiva-lent:

(i) EEG, i.e., E is Lebesgue measurable.(ii) For every e > 0, there exists an open set GE such that

E e GE and a (GE \ E) <c.

(iii) For every e > 0, there exists a closed set FE such that

FE C E and A(E \ FE) <c.

(iv) There exists a Gb-set G such that

E C G and a (G \ E) = 0.

(v) There exists an FQ-set F such that

FCE andA(E\F)=0.Proof: We shall prove the following implications:

100 4. The Lebesgue measure on R and its properties

and

(;+) (iv)

W M W

(1) (ii) Let E E L and let e > 0 be given. If A (E) = A (E) < +oo, thenby definition, we can find intervals Il, 12,... , In, ... such that

00 00

E C U In and A(E) + E/2 > E A(In).n=1 n=1

For every n, choose an open interval Jn D In such that

A(Jn) < c/2n+1 + A(In).

Let Q. °O Jn. Then GE is an open set with GE D E and GE < +oo.U=1 ( )Further,

A*(GE\E) = A(GE\E) A(GE) - A (E)00

E A (Jn) - A (E)

00 00

1: /2n+l + E A(In) - A(E)n=1 n=1

< 6/2 + c/2 = c.

In case A(E) = A (E) = +oo, we can write E = U=1 En, where 0

for n m and A (En) < +oo for every n. Choose, by the earlier case, openset Gn D En such that for every n = 1, 2, ...

(Gn \ En) < E/2n.

Put G. U=1 Gn. Then GE D E, GE is open and (GE\E) C U=1(G\E).

Thus00

A(G,6\E) < EA(Gn\En) <n=1

(i) (iii) Let E E L. Then EC E £ and, by (ii) above, there exists anopen set GE D EC such that A (GE \ Ec) < E. Put KE := GE . Then KE C E isa closed set, and since GE \ Ec = E \ K, we have A (E \ KE) < E. Hence (iii)holds.

(ii) ===> (iv) By (ii), for every n, we choose an open set Gn D E such thatA (Gn \ E) < 1/n. Put G:= nn, 1 Gn. Then G is a Ga-set, G D E, and forevery n,

A(G\E) A(Gn\E) < 1/n.

4.2 Lebesgue measurable sets and nice sets 101

Hence A (G \ E) = 0. Thus (iv) holds.

(v) By (iii), for every n choose a closed set Kn C E such thatA(E \ Kim) < 1/n. Put K = J 1 Kim. Then K is an F,-set with K C E..Also, for every n,

A* (E \ K) < A* (E \ Kim) <n1

.

Thus A* (E \ K) = 0. Hence (v) holds.

(iv) (i) By (iv), there exists a GS-set G D E such that A* (G \ E) = 0.Thus both G, G \ E E ,t3RC GR. Since E = G \ (G \ E), we have E E GR.

(v) (i) Let E C R be such that there exists an Fo-set F C E withA (E\F) = 0. Since E = FU (E\F) with F E BR C G and E \ F is a A-nullset, both E \ F, F E G, and hence E E G.

4.2.3. Exercise: Let E C R. Show that the following statements areequivalent:

(i) EEG.

(ii) A (I) _ A (E fl I) + A (Ec fl I) for every interval I.

(iii) E fl [n, n + 1) E G for every n E Z.

(iv) A*(E fl [n, n + 1)) + A(E° fl [n, n + 1)) = 1 for every n E Z.

4.2.4. Exercise: Let A E G and x E R. Using theorem 4.2.2, show that(i) A+x EL, where A+x := {y+xI(ii) -A E G, where -A:= {-y I y c Al.

4.2.5. Exercise: Let E E G with 0 < A(E) < oo, and let 0 < c < 1. Showthat there exists an open interval I such that A (E fl I) > cA(I).(Hint: Either assume that the claim is not true and get a contradiction. Oruse theorem 4.2.2(ii) to get an open set U D E such that A(E) > A(U) -(1 - c)A(U), and use the fact that U is a countable union of disjoint openintervals.)

4.2.6. Note: Theorem 4.2.2 tells us the relation between £, the class ofLebesgue measurable sets, and the topologically nice sets, e.g., open setsand closed sets. The property that for E E £ and c > 0, there exists anopen set G D E with A (G \ E) < E can be stated equivalently as:

A(E) = inf{A(U) I U open, U D E}.

102 4. The Lebesgue measure on Ilk and its properties

This is called the outer regularity of A. Other examples of outer regularmeasures on ][8 (in fact any metric space) are given in the following exercise:

4.2.7. Exercise: Let (X, d) be any metric space and let µ be a measureon ,t3X, the Q-algebra generated by open subsets of X, called the v-algebra ofBorel subsets of X. The measure µ is called outer regular if d E E XiX,

µ(E) = inf{µ(U) U open, U D E},= sup{µ(C) C closed, C C E}. (4.3)

(i) If µ(X) c +oo, show that µ is outer regular if for every E E Bx ande > 0 given, there exist an open set UE and a closed set CE such that

UE D E D CE and µ(UE - CE) < e.

(ii) For A C X, let

d(x, A) := inf {d(x, y) I y E Al.

Show that for every A C X, x --> d(x, A) is a uniformly continuousfunction.(Hint: I d(x, A) - d(y, A) I < d(x, y) b x, y.)

(iii) Let µ(X) < +oo and

S := {E E ,t3X 1 (4.3) holds for E}.

Show that S is a Q-algebra of subsets of X.(iv) Let C be any closed set in X. Show that C E S.

(Hint: C = fl1{x E X : d(x, C) < 1/n}.)(v) Show that µ is outer regular on ,t3X.

Another topologically nice class of subsets of III is that of compact subsetsof R. It is natural to ask the question: does there exist a relation betweenG and the class of compact subsets of ][8? Let K be any compact subset ofR. Since K is closed (and bounded), clearly K E BR C G and A(K) < +oo.It is natural to ask the question: can one obtain A(E) for a set E E BR, ifA(E) < +oo, from the knowledge of A(K), K compact in ][8? The answer isgiven by the next proposition.

4.2.8. Proposition: Let E E G with 0 < 1\(E) < +oo and let e > 0 begiven. Then there exists a compact set K C E such that A(E \ K) < E.

Proof: Let En := En [-n, n], n > 1. Then {En}n>1 is an increasing se-quence of sets such that U=1 En = E. Since (E) < +oo, we can choosen sufficiently large so that µ(E) - /\(En) < e/2. Now, using theorem 4.2.2,we can find a closed set K C En C [-n, n] such that A(En) - A(K) < e/2.Then K is compact, K C E and A(E) - A(K) < E.

4.3 Properties of the Lebesgue measure 103

4.2.9. Note: Let (X, d) and Z3X be as in exercise 4.2.7. A measure µ on 13Xis called inner regular if b E E X3X with 0 < µ(E) < oo, and b e > 0, Ela compact set K such that µ(E\K) < e. Exercise 4.2.7 and proposition 4.2.8tell us that the Lebesgue measure is both inner and outer regular. Exercise4.2.7 says that on a metric space (X, d) every finite measure is outer regular.If X is also complete and separable, then every finite measure is also innerregular (see lemma 9.2.5).

4.3. Properties of the Lebesgue measure withrespect to the group structure on R

On the set ][8, we have the group structure given by the binary operation ofthe addition of two real numbers. We analyze the behavior of A on G underthe map y l--> y + x, y E R and x ER fixed.

In exercise 4.2.4, we saw that A + x is a Lebesgue measurable set when-ever A is Lebesgue measurable and x E R. It is natural to ask the question:for A E G and x E R, is A(A + x) _ A(A)? The Lebesgue outer measureA obviously has the property that A (E + x) = A (E) for every E C R andx E R. Next, suppose E E G. Then for every A C X and x E ]Eg,

A(A) = A(A-x) = A((A-x)nE)+A((A-x)nEc)-Since

(A - x) n E = (An(E+x))-x and (A- x) n E° = (An (E + x)°) - x,

we haveA(A) = A(An(E + x)) + A (An(E + x) c).

Thus E + x E G and A (E + x) = A (E), i.e., A(E + x) = A(E). Hence wehave proved the following:

4.3.1. Theorem (Translation invariance property): Let E E G. ThenE + x c G for every x c R, and A(E + x) = A(E).

4.3.2. Exercise: Let E E BR. Show that E + x E Big for every x E R.

4.3.3. Exercise: Let E E G and x E R. Let

xE :_ {may I y E E} and - E:= {-x I x E E}.

Show that -E, xE E G for every x E E. Compute A(xE) and A(-E) interms of A(E).

Here is another property of the Lebesgue measure which utilizes its reg-ularity properties.

*4.3.4. Theorem (Steinhaus): Let E, F E G be such that A(E)-}-oo, A(F) < -I-oo. Then x H A(E fl (F + x)), x E ][8, is a continuousmap.

Proof: First suppose that E, F are intervals. Without loss of generality wemay assume that E = (a, b) and F = (c, d). Suppose a < c < d < b. Then

(Q if x<a - d,(a,d+x) if a-d<x<a-c,

(Efl(F+x))= (c+x,d+x) if a-c<x<b-d,(c + x, b) if b - d < x A(Efl (F+x)) is continuous in other cases also when E, F are intervalswith A(E) < +oo, A(F) < +oo.

Next, let E and F be open subsets of R with A(E) < +oo, a(F) < +oo.We can express E = U=1 In) F = U=1 Jm where Imo,, Jm are open intervalssuch that Ijn Ik= 0 and Jj fl Jk= 0 for j = k. Note that A(Lm) < +oo andA(Jn) < +oo d n, m. Thus

00 00 00 00

E n(F -I- x) = U InI I V (Jm + x) = U U (Ian (Jm + x)).

n=1 m=1 n=1 m=1

Hence

c)oA(En(F + x)) = 1: A (in n (im + x))n,m=1

In fact, the series on the right hand side of the above equality convergesuniformly by Weierstrass' M-test because for every n and m,

00

A (Infl(Jm+)) < A (In) and E A (In,) = A (E) < -}-oo.rt=1

Since each A(In fl (Jm + x)) is continuous, A(E fl (F + x)) is continuous asa function of x, for any open sets E, F with a(E) < +oo and A(F) < +oo.

Finally, let E, F be arbitrary sets in ,C with A(E) < +oo and A(F)+oo. Let c > 0 be arbitrary. We can choose (by theorem 4.2.2) open setsU,V such that EC VF C C U and

A(V - E) < c/2) A(U - F) < c/2.

Now d x c Il8,

Vn(x+U) c (x+(U-F))U(V-E)U((x+F)flE)and

(x+F)flE c ((x+U)nV)u(U-F)u(V-E).

4.3 Properties of the Lebesgue measure 105

Thus

I

Since A(V fl (x + U)) is continuous as a function of x, for x E ][8 fixed, wecan find a b > 0 such that

ix - yj < 6 implies A(V fl (x + U)) - A(V fl (y + U)) < e.

But then for y with Ix - yj < S, we have

I A(E n (x + F)) - A(E n (y + F)) I

JA(V n (x + U)) - A(E n (x + F)) I

+I a(vn(x +U))-a(vn(y+u))I+I A(V n (y+U)) - A(En (y+F)) I

3e.

This proves the continuity of the map x H A(E fl (F + x)) for E, F E Gwith A(E) < +oo, A(F) < +oo.

*4.3.5. Corollary: Let E E G be such that 0 < A(E) < oo. Then thereexists a > 0 such that

Pa, +a] c E - E := fx - y I x, y cz Ej.

Proof: Since 0 < A(E) < oo, the set E : 0, and hence 0 E E - E. SinceA(E fl (x + E)) > 0 for x = 0, by the continuity of the map

X 1 ) A(En(E + x)),

there exists a > 0 such that for every y c [-a, +a], A(E n (E + y)) > 0.But then E fl (E + y) 0 for every y c [-a, a]. Thus y c E - E for everyY G [-a, a]. M

*4.3.6. Exercise: Let E E G with A(E) < +oo. Show that the functionx H A(E D (E + x)) is continuous on R.

*4.3.7. Exercise: Prove corollary 4.3.5 as follows:

(i) If E is an open interval or includes an open interval, the claim isobvious.

(ii) In general, use exercise 4.2.5 to show that there exists a bounded openinterval I such that A(E fl I) > 3A(I)/4.

(iii) Let a = A(I)/2. Show that for every x c [-a, +a], I U (I + x) is aninterval with

A(I U (I + x)) < 3A(I)/2

and(EflI)U[(EflI)+x] c (IU(I+x)).

(iv) For x E [-a, +a], if (En I) fl ((E fl I) + x) = 0, show that (ii) abovewill be contradicted. Deduce that (E n I) n ((E fl I) + x) 4 0 forx E [-a, +a], and hence [-a, +a] C E - E.

*4.3.8. Exercise: Let f : ][8 ) R be a measurable map such that f (x +y) = f (x) + f (y) V x, y E R. Prove the following statements:

(i) f (rx) = r f (x), d x E ][8 and r E .(ii) For every e > 0, A (f (0, e)) > 0.

(Hint: Use (i).)

(iii) Show that f is continuous at 0 E R.(Hint: Use corollary 4.3.5.)

(iv) Show that f is continuous everywhere and deduce that f (x) = x f (1)V x E R.

4.4. Uniqueness of the Lebesgue measure

We saw in section 4.1 that Lebesgue measure is the unique extension of thelength function from the class I of intervals to BR, the a-algebra of Borelsubsets of R. This gave us a measure A on BR with the following properties:

(i) For every nonempty open set U, A(U) > 0.

(ii) For every compact set K, A(K) < +oo.(iii) For every E E BR,

A(E) = inf{A(U) I U open, U D E},= sup{A(C) I C C E, C closed}.

If A(E) < +oo, then we also have

A(E) =sup{A(K) I K C E, K compact}.

(iv) For every E E BR and x ER, E + x E X3R and A(E + x) _ A(E).

Thus the Lebesgue measure is a translation invariant a-finite regularmeasure on X3R. The question arises: are there other a-finite measureson BR with these properties? Obviously, if c > 0 then cA defined by(cA)(E) := cA(E), E E BR, is also a a-finite measure and is translationinvariant. Suppose µ is any a-finite measure on BR such that µ is alsotranslation invariant and 0 < c := µ(0,1J < +oo. Then by the translationinvariance and countable additivity of µ,

µ (m,n]=c(n-m) for every m, n c 7L, n > m.

4.4 Uniqueness of the Lebesgue measure 107

Thus b m, n E 7L with n > m,

M (m, n] = cA (m, n].

Hence µ(I) = cA(I) if I E Z has integral endpoints. Using again translationinvariance and countable additivity, we get

µ(0,1/n] = µ(0,1]/n = c/n = cA(0,1/n].

Using these two properties again, we get

µ(0, m/n] = cA (0, m/n] for every m > 1.

In fact, if I = (m/k, n/k], where m, n c Z and k c N, then

µ(I) = (0,(n-rn)/k] = cA(0, (n - m)/k] = cA(I).Thus p(I) = cA(I) for every I E I with rational endpoints. Finally, let Ibe any finite interval, I = (a, b]. Then we can choose sequences of rationalsIrnin>1 and {Sn}n>i such that increases to a and sn decreases tob with rn < sn V n. Noting that M (J) < +oo for every finite interval, bytheorem 3.6.3,

(a,b] = lim p (rn, sn] = c lim A(rn)rm] = CA(I).n-*oo n-*oo

Also, `d b E R we have

lim 1 - 1/n, 1 + 1/n

= lim cA(1 - 1/n, l + 1/n ] = 0.n-*oo

Hence µ and A are Q-finite measures with µ(I) = A(I) for every I E Z. Now,using theorems 3.3.3 and 3.10.8, µ(E) = cA(E) for every E E BR. We notethat the condition 0 < µ(0,1] < oo will automatically hold if µ(V) > 0 forevery nonempty open set U and µ(K) < +oo for every compact set K. Thuswe have proved the following:

4.4.1. Theorem: Let µ be a measure on ,t3R such that

(i) µ(U) > 0 for every nonempty open set U C R.(ii) µ(K) < -I-oo for every compact set K C R.(iii) µ(E + x) = µ(E), d E E BR and d x E R.

Then there exists a positive real number c such that µ(E) = cA(E)V E E BR.

4.4.2. Note: In fact the above theorem has a far-reaching generalizationto abstract `topological groups'. Let us recall that the set of real numbersIE is a group under the binary operation +, the addition of real numbers.

Also, there is a topology on R which `respects' the group structure, i.e.,the maps (t, s) 1 ) t + s and t --f -t from R x R ) R and R - * R,respectively, are continuous when R x R is given the product topology. Inan abstract setting, if G is a set with a binary operation and a topologyT such that (G, ) is a group and the maps G x G -* G, (g, h) 1 ) g . h andG -* Gig i ) g-1 are continuous with respect to the product topologyon G x G, one calls G a topological group. Given a topological group,let L3G denote the a-algebra generated by open subsets of G, called the a-algebra of Borel subsets of G. The question arises: does there exist a a-finitemeasure ,u on G such that it has the properties as given in theorem 4.4.1?A celebrated theorem due to A. Haar states that such a measure exists andis unique up to a multiplicative (positive) constant if G is locally-compact.Such a measure is called a (right) Haar measure on G. Theorem 4.4.1 thenstates that for the topological group R, the Lebesgue measure A is a Haarmeasure. Consider the group (R \ {0}, ), where R \ {0} = {t E R I t 0}and is the usual multiplication of real numbers. Let R \ {0} be given thesubspace topology from R. It is easy to show that R \ {0} is a topologicalgroup and, V E E 13R\{0}

[t (E) : -1

dA(x) (4.4)

is a Haar measure on R \ {0}. (For a meaning of the right hand side ofequation (4.4), see Chapter 5.)

We close this section by proving a result, theorem 4.4.5, which has manyimportant applications in measure theory. Recall that given a covering ofan interval [a, b] by open sets, we can choose a finite number of them, stillcovering [a, b]. This is the well-known Heine-Borel theorem. The theoremwe want to prove essentially says that given a set E C ][8 with A (E) < +ooand a cover of E by intervals of `arbitrary small Lebesgue measure', one canalmost cover E by a finite number of pairwise disjoint intervals from thegiven cover. This is made precise below.

4.4.3. Definition: Let E C ][8, and let C be a collection of intervals ofpositive length. We say G is a Vitali cover of E if for every e > 0 and anyx e E, there exists an interval I E G such that x E I and A(I) < E.

4.4.4. Example: Let {r}>i be an enumeration of the rationale in [a, b].Let C :_ {[rn - 1/k, rn + 1/k] I n = 1, 2, ... ; k = 1, 2, ... }. Then G is aVitali covering of [a, b].

4.4.5. , Theorem (Vitali's covering): Let E C R be such that A (E) <+00, and let 9 be a Vitali cover of E. Then, given e > 0, there exists a finite

4.4 Uniqueness of the Lebesgue measure 109

pairwise disjoint collection {Ii,... , IN} of intervals in g such that

N

A E\ U In(n=1

E.

Proof (Banach): Without loss of generality we may assume that all theintervals in g are closed. Since A (E) < +oo, using theorem 4.2.2 we canfind an open set U D E such that A(U) < +oo. Let go = {I E 9 11 C U}.Obviously, Co is also a Vitali cover of E. Let I1 E Co be arbitrary. IfE C 12, then we are through. If not, suppose I1, I2, ... J n E go have beenselected such that they are pairwise disjoint. If E C U=1 Ik, then again weare through. If not, 3 x E E \ (U=iik). Since U=1 Ik is a closed set, thedistance between x and U=1 Ik, say q, is positive. Thus we can chooseIn+1 Ego such that x E In+1 and A(In+1) < ii /2. Then In+1 fl Ik _ 0 fork = 1, 2, ... , n. In fact, we can select In+1 with the additional property thatA(In+1) > an/2, where

an := sup{A(I) I I E go, I fl Ik= 0 for k = 1, 2, ... , n}.

By induction, we have a sequence {In}n>i of pairwise disjoint intervals fromgo such that A(In+1) > an/2 for every n > 1. Since each In C U,

00

1: A (1n) < A (U) < + oo.n=1

00Thus A(In) -* 0 as n -> oo. We claim that A (E \ (U1

Ik )) = 0. For this,let x E E \ U=iIk. Then x E E\ U=iIk V n. Let N E N be fixed. Onceagain, since x U=iIk as shown above we can choose I E go such thatI f1 Ik _ 0 for k = 1, 2, ... , N and x E I. Since A(In) -* 0 as n - ooand an < 2A (In), there exists some integer n such that an < A (I). But thenI fl In 0. Let no be the smallest integer such that I f1 Ino 0. Clearlyno > N. Since I fl Ino 0 and x E I, the distance of x from the midpoint ofIno , say xno , is at most A (I) + A(I0)/2. Note that

A(I) + A(Ino)/2 < ano-1 + A(Ino)/2 < 5A(Ino)/2.

Let Jno denote the closed interval with midpoint xno and A (Jno) . = 5 A (Ino).

Then x E Jno. Thus V x E U=1 Ik, 3 n> N such that x E Jn with A (Jn) _5A(In). Hence

00 00

< A (Jn) < 5 A (In)n=1 n=N+1 n=N+1

Since this holds for every N,00

A E\UInn=1

lim A (E\0I)Inn-1 )

= 0.

4.4.6. Exercise: Let E C R, and let C be a Vitali cover of E. Show that,given E > 0, there exists a countable family {I}> 1 of pairwise disjoint setsfrom C such that (E\U1 In) <(Hint: A is a-finite.)

We give next an application of the Vitali covering theorem. More appli-cations are given in Chapter 6.

4.4.7. Proposition: Let {I« I a E J} be any arbitrary collection of inter-vals of positive length. Then E := U,,Ej I« is Lebesgue measurable.

Proof: Assume without loss of generality that A (E) < +oo. Let

9:_ {ICRIIis bounded and I C Ia for some a}.

Then G is a Vitali covering of E. By theorem 4.4.5, El a sequence {Ia}i>iof pairwise disjoint intervals such that A (E \ LJ?0-1 °Ice,) = 0. Thus

E=

and hence E E L.

00 00

(E\UIai) U U Iaii=1 i=1

4.5. *Cardinalities of the a-algebras ,C and BR

As a special case of corollary 3.11.3, we have L= ,t3RU N, where

N:={NC 1[B IN CE E X3R,A(E)=0}.

Thus,t3RC ,C C P(R). The question arises: are there sets in ,C which arenot inX3R, i.e., is BR a proper subclass of G? First of all, we note that aset A C R is a null set as per definition 1.2.8 if A E N. Thus the Cantorternary set C E N C G. If E C C, then A* (E) = 0 and hence E E G. In otherwords, P(C) C L. Thus the cardinality of G is at least 2` (here c denotes thecardinality of the real line, also called the cardinality of the continuum).Since G C P(R), we get the cardinality of ,C to be 2`. On the other hand, BRis the Q-algebra generated by all open intervals of R with rational endpoints(exercise 3.9.8). What is the cardinality of BR? To answer this, we givebelow a description of the algebra and the Q-algebra generated by a class ofsubsets of a set X.

4.5 Cardinali ties of the a-algebras C and BR 111

Let X be any nonempty set with at least two points and let C be anyclass of subsets of X such that X E C. Let

d:= JA C X I A or A' G Cj

and

C*:= E C Xn

for some n > 1, E = U Ai, where each Ai E Ci=1

Then we have the following:

4.5.1. Theorem: Let C be any class of subsets of a set X such that X E C.Let Cl := C* and for every n > 1, let Cn+l := C. Let .F U°_1 C. Then,F is an algebra of subsets of X containing C. In fact .F _ .F(C), the algebragenerated by C.

Proof: Clearly, Cn C Cn+1 for every n. Hence C C F = U=1C. Next,if E E Cn for some n, then clearly EC E Cn+ 1. Thus for E E F, EC E.F. Also, if E1, ... , E Cn for some n, then Um 1 Ei E Cn+ 1. Thus forE1, E2,- .. , E F, we can assume without loss of generality that for every1 < i < m, Ei E Cn for some n, and hence Um 1 Ei E Cn+1 C F. This provesthat 1 is an algebra. To show that F = F(C), let A be any algebra ofsubsets of X such that C C A. Then clearly Cn C A for every n, and hence.FcA.

The corresponding result for a-algebras is the following: For any familyC of subsets of a set X, let

00

C* := U Ei for every i, either Ei E C or EZ E C .

i=1

Let 1 denote the first uncountable ordinal number. Let Co := C U {X}, andlet a be any ordinal number, 0 < a < Q. We use transfinite induction todefine Ca as follows: Suppose for every,3 < a, C,Q has been defined. Define

Ca := U C,a .

(0<)3<a

Then we have the following:

4.5.2. Theorem: Let C be any family of subsets of a set X, and for everyordinal number 0 < a < S2, let Ca be defined as above. Then S := Uo<«<s-t C«is a a-algebra of subsets of X. In fact, S = S(C), the a-algebra generated byC.

Proof: Clearly, if a <,3 < S1 then Ca C Ca C CO and, for A E Ca, AC E Ca,and hence Ac E Ca for every,3 > a. Thus A E s would imply that Ac E S.Next, let Ai E Cai for i = 1, 2, .... Then

00

Ui=1

00

Ai E U Cai(i=1

C Co

for every /3 such that /3 > a,, for every n. Hence Ai E s, i = 1, 2, ... , wouldimply that, U2=1 Ai E S. Finally, 0 = X c E C1 C s and hence X E S. Thisproves that s is a o--algebra. Let B be any o--algebra of subsets of X suchthat B D C. Then clearly B D C . Further, if Ca C B, then Ca C B and henceC,Q C B for every ,(3 < Q. Thus S C B, showing that s = 8(C).

4.5.3. Corollary: The o--algebra BR of Borel subsets of R has cardinalityc, that of the continuum.

Proof: Since each singleton set {x} is a closed set, {x} E .C3R. Thus BR hascardinality at least c. On the other hand, let C denote the class of all openintervals in ][8 with rational endpoints. Then by exercise 3.9.8, S(C) = BR-By theorem 4.5.2,

s(c) U Co.o<0<0

Note that Co has countably infinite elements, i.e., Co has cardinality o (readas aleph nought) and to construct an element of Co we have to choose o

elements with each element having two choices, i.e., either it is in Co or itscomplement is in Co. Thus Co = C1 will have at most 2° < c elements. Nowassume that each Ca has cardinality at most c for a < /3. Then

Clearly, {a I a < Q} has cardinality at most t2o and hence U1<c,<aCc,has cardinality at most c.No = c. Thus Cp will have cardinality at mostc. By transfinite induction, each C,6,,3 < S2, has cardinality at most c.Hence S(C) = U1<Q<0 Ca will have cardinality at most c.c = c, proving that.C3R= S(C) has cardinality exactly c elements.

4.5.4. Note: As stated in the beginning of this section, the class £ of allLebesgue measurable sets has cardinality 2'. We showed just now that thecardinality of BR is c. Since 2` > c, there exist sets which are Lebesguemeasurable but are not Borel sets. The actual construction of such sets isnot easy. One such class of sets is called analytic sets. An analytic set isa set which can be represented as a continuous image of a Borel set. For adetailed discussion on analytic sets, see Srivastava [38], Parthasarathy [29].

4.6 Nonmeasurable subsets of Ilk 113

4.6. Nonmeasurable subsets of R

We showed in section 4.5 that £, the class of all Lebesgue measurable sub-sets of Ilk, has 2` elements, i.e., the same as the number of elements in P(IR).The natural question arises: is £ = P(R)? We saw in 3.4 that if we assumethe continuum hypothesis, it is not possible to define a countably additiveset function ,u on P (R) such that ,u ({ x }) = 0 V x E R. In particular, ifwe assume the continuum hypothesis, we cannot extend A to all subsets ofR. Hence £ $ P(R). What can be said if one does not assume the contin-uum hypothesis? To answer this question, one can either try to construct aset E C Ilk such that E 0 £, or, assuming that such a set exists, try to seewhether one can reach a contradiction. G. Vitali (1905), F. Bernstein (1908),H. Rademacher (1916) and others constructed such sets assuming the `ax-iom of choice' (see appendix B). The example of Vitali used the translationinvariance property of the Lebesgue measure, and that of Bernstein usedthe regularity properties of the Lebesgue measure. Rademacher proved thatevery set of positive outer Lebesgue measure includes a Lebesgue nonmea-surable set. Even today, more and more nonmeasurable sets with additionalproperties are being constructed. For example, one can construct nonmea-surable subsets A of TI such that A (A l I) _ A (I) for every interval I C R.Of course, all these constructions are under the assumption of the `axiomof choice'. Lebesgue himself did not accept such constructions. In 1970, R.Solovay [37] proved that if one includes the statement "all subsets of T areLebesgue measurable" as an axiom in set theory, then it is consistent withthe other axioms of set theory if the axiom of choice is not assumed. We givebelow the construction (due to Vitali) of a nonmeasurable set, assuming theaxiom of choice.

4.6.1. Example (Vitali): Define a relation on [0,1] as follows: for x, y E[0, 1], we say x is related to y, written as x N y, if x - y is a rational. Onechecks that N is an equivalence relation on [0, 1]. Let {E}EI denote theset of equivalence classes of elements of [0, 1]. Using the axiom of choice,we choose exactly one element xa E Ea for every a E I and construct theset E := {xc,a E I} . Let Ti, r2, ... , rn, ... denote an enumeration of therationals in [-1, 1]. Let

En :=rn+E, n = 1,2,... .

It is easy to check that En fl Em = 0 for n m and En c [--1, 2] for everyn. If x E [0, 1], then x E Ea for some a E I, and hence x N xa, xa E E. Butthen x - xa is a rational and -1 < x - xa < 1. Hence x E En for some n.Thus

00

[0, 1] C U En Cn=1

114 4. The Lebesgue measure on II and its properties

Now, suppose E E G. Then En E G for every n, and A(En) _ A(E). On onehand, A(E) > 0, in which case A(En) _ A(E) > 0 for every n. But then bythe countable additivity of A,

00 00

00 - E A (En) - A U En < 3n=1 (n=1

which is absurd. On the other hand, if A(E) = 0, then A(En) = 0 V n andhence

00

1 = A([Ol 1]) < E A(En) = 0)n=1

again not possible. Hence E is not Lebesgue measurable.

4.6.2. Exercise: Using the axiom of choice, choose one element from eachof the sets x + Q, where x E ][8 and Q is the set of rationals in ][8, andconstruct the set E. Show that E is not Lebesgue measurable using thefollowing steps:

(i) Assume that E is Lebesgue measurable. Show that for some r EQ5 A(E + r) > 0.

(ii) Use corollary 4.3.5 to show that (E - E) fl Q # 0, and hence thedefinition of E is contradicted.

4.6.3. Note: As shown in section 4.1, we can extend the notion of lengthfrom the class of intervals to a class L of Lebesgue measurable sets whichincludes all topologically nice sets and which is invariant under the groupoperations. We also saw that unless some extra hypothesis/ axiom is assumedin set theory, it is not possible to show the existence of nonmeasurable sets.So, the question arises: can one extend the Lebesgue measure beyond theclass of Lebesgue measurable sets, preserving all its properties? The answeris: yes, it is possible. The interested reader may refer to Kakutani andOxtoby [19], or Hewitt and Ross [17].

4.7. The Lebesgue-Stieltjes measure

In the previous sections we have seen how the general extension theory, asdeveloped in chapter 3, can be applied to the particular situation when thesemi-algebra is that of intervals and the set function is the length function.More generally, if we consider the semi-algebra Z of left-open right-closedintervals in II and consider F : II -* II as a monotonically increasingright continuous function, then we can construct a countably additive setfunction p on the semi-algebra Z, as in proposition 3.5.1. Using theorem3.3.3 and theorem 3.10.8, we can construct a complete measure ,uF on aa-algebra of subsets of II which includes BR. This measure /1F is called

4.7 Lebesgue-Stieltjes measure 115

the Lebesgue-Stieltjes measure induced by the function F. Note thatµF has the property that p F (a, b] < +oo d a, b E IR, a < b. Conversely,given a measure p on 8R such that ,u(a, b] < +oo d a, b c IR, a < b, we canrestrict it to Z and, using proposition 3.5.3, define a monotonically increasingright continuous function F : III -* III such that the unique Lebesgue-Stieltjes measure µF induced by F is nothing but p (by the uniquenessof the extension). Thus measures p on 8R which have the property thatµ(a, b] < +oo d a < b can be looked upon as a Lebesgue-Stieltjes measureµF for some F. We point out that it is possible to find different F1, F2 :

III -* III such that both are monotonically increasing and right continuousand µF1 = P F2 . If p is finite measure, i.e., (IR) < +oo, then it is easyto see that F(x) := (-oo, x], x IR, is a monotonically increasing rightcontinuous function such that µ = µF. This F is called the distributionfunction of A. When (IR) = 1, p is called a probability and its distributionfunction F, which is monotonically increasing and is right continuous withlim [F(x) - F(-x)] = AF(IR) = 1, is called a probability distribution

X--+00function.

Chapter 5

Integration

Let us recall that our aim is to define the notion of integral for a class offunctions on IR so as to extend the class of Riemann integrable functions.Of course, we expect the new notion of integral to be linear and behavereasonably well with respect to limiting operations. As we saw in section2.1, in order to do this, our idea is to enlarge the class of building blocks.For Riemann integrable functions, the building blocks were step functions,i.e., functions which are representable as E'i=, aixli (x), where the ai's arereal numbers and the intervals Il , I2, ... , I,z form a partition of [a, b], thedomain of the function. For our new integral, we consider functions of thetype s = En

1 aiXAi 'where n is a positive integer, the ai's are real numbersand the Ai's are pairwise disjoint sets such that U=1 Ai = R. Further, wedemand that Ai E £ for every i, i.e., the Ai's are sets for which the notion oflength has been defined. Since we expect our new integral to be linear andto be an extension of the Riemann integral, we should define the integralof s to be f sd,\ := En 1 aiA(Ai) . However, \ (Ai) could be +oo for somei. To avoid unpleasant situations (e.g., oo - oo), we start with functionss = En

1aiXAi , where ai > 0 for each i. In fact, we can allow ai = +oo for

every i. Since the definition and properties of the new integral are not goingto depend on the fact that the Ai's are subsets of llt, but only on the factthat they are sets for which the notion of `length' has been defined, we shallproceed with the process of integration when lit is replaced by any set X, £is replaced by a a-algebra S of subsets of X, and A by a a-finite measureu on S. In section 5.1 we shall define the integral for the class of functionsof the type s =

1aiXAi , which serve as building blocks for our integral.Ei=

Then, keeping in mind the limiting property that the integral should have,we will extend it to a larger class of functions in section 5.2. The class of

117

118 5. Integration

functions for which this integral can be defined will be discussed in sections5.3 and 5.4.

For the rest of the chapter, unless stated otherwise, we shall work on afixed a-finite measure space (X, S, A). We shall further assume that (X, S, A)is a complete measure space.

5.1. Integral of nonnegative simple measurablefunctions

5.1.1. Definition: Let s : X -) [0, oo] be defined by

n

aixAi (x), x E X,i=

where n is some positive integer; a1, a2, ... , an are nonnegative extended realnumbers; Ai E S for every i; A i f Aj _ 0 for i 54- j ; and U=1 Ai = X. Such afunction s is called a nonnegative simple measurable function on (X, S)and En aZxAZ (x) is called a representation of s. We say aixAi isi=1

Eni=1

the standard representation of s if a1, a2, ... , an are all distinct. Wedenote by Lo the class of all nonnegative simple measurable functions on(X,S).

Note that s E Lo if s takes only a finite number of distinct values, saya1, s2, ... > an) the value az i being taken on the set Az i E s> i = 1> 2> ... > n.

And in that case its standard representation is Ei=, aixAi . Also note thatthe class Lo depends only upon the set X and the a-algebra s; the measureA plays no part in the definition of functions in ILo .

5.1.2. Examples:

(i) Clearly, if s(x) c for some c E [0, +oo], then s E ILo .

(ii) For A C X, consider XA : X [0, +oo], the indicator function of theset A, i.e., XA(x) = 1 if x E A and XA(x) = 0 if x A. Then XA E ILp ifA ES, for XA = aXA + bXAc with a = 1 and b = 0.

(iii) Let A, B E S. Then s = XAXB E ILo , since s = XAnB

(iv) Let A, B E S. If A fl B = 0, then clearly XA + XB = XAUB E Lo

5.1.3. Exercise: Let A, B E S. Express the functions IXA - XB J and

XA + XB - XAna as indicator functions of sets in S and hence deduce thatthey belong to ILo .

5.1 Integral of nonnegative simple measurable functions 119

5.1.4. Definition: For s E ILo with a representation s = EnaixAi , we

define f s(x)dµ(x), the integral of s with respect to µ, by

Js(x)dµ(x)

n

aip (Ai).z=1

The integral f s(x)dµ(x) is also denoted by f sdµ.Before we proceed further, we should check that f s(x)dµ(x) is well-

defined, i.e., ifn m

s = LaixAi = LbjxB,,i=1 j=1

where {A,... , An j and {Bi,... , B,n } are partitions of X by elements ofs, then

na(A) =

i=1

For this, we note that we can writen m

m

p(Bj).

m n

s = )ai)XAnBj = )°i)XAnBi=1 j=1 j=1 i=1

Thus if Ai n Bj 0, then ai = bj. Hence, using finite additivity of µ,n n m m n m

a(A) = E az E µ(Ai n Bj) = E bj E µ(Ai n Bj) = E b.7I-z (Bj)>i=1 i=1 j=1 j=1 i=1 j=

Thus f s(x)dµ(x) is independent of the representation s(x) = E?i azXA2

The properties of f sdµ, for s E lLp ,are given by the next proposition.

5.1.5. Proposition: For s, 81, 82 E ILo and a E R with a > 0, the followinghold:

(i) 0 < f sdp < +oo.(ii) as E lLp and

J (as)dµ = a J sdµ.

(iii) sl + 82 E lLo and

f(si+s2)d µ = fsid+fs2d.(iv) For E E S we have sXE E ILo , and the set function

E H v(E) :=J sXEdµ

is a measure on S. Further, v(E) = 0 whenever µ(E) = 0, E E S.

120 5. Integration

Proof: Statements (i) and (ii) are obvious. For (iii), let

Si = E aiXAz and s2 = EbjXBj .i=1

Then we can writen m n m

81 = EE aiXAZnBj and 82 = E

j=1

i=1 j=1 i=1 i=1

Thusn m

j XAZ nBj

sl + 82 = L >,'(ai + bj)XAzns;i=1 j=1

Hence Si + 82 E ]Lo and, using these representations, clearly

f(si+s2)d µ = fSid+fS2d.

This proves (iii). To prove (iv), let s = X: 1 aiX,yz. Then for E E S,

n

5XE =L aixAinE + aXEc , where a = 0.i=1

Hence 5XE E L+ and

v(E) = f SXEdµ aZµ(E n Az).i=1

Using this, it is easy to check that v is a measure. Clearly µ(E) = 0 impliesthat v(E) = 0.

5.1.6. Exercise: Let Si, 82 E ILo .Prove the following:

(i) If Si > 82, then f sldµ > f s2dµ.(ii) Let b x E X,

(Si V 82) (X) := max{sl(x), 82(x)} and (Si n 82) (X) := min{sl(x), s2(x)}

Then sl n 82 and Si V S2 E ]Lo with

f(siAs2)d µ < Jsidµ < f(SivS2)d, i = 1, 2.

(iii) Express the functions XA A XB and XA V XB , for A, B E s, in termsof the functions XA and XB

n

5.1.7. Exercise: Let X = (0, 1], s = X3(0, l] , the a-algebra of Borel subsetsof (0, 1] and µ = A, the Lebesgue measure restricted to s. For x E (0,1 ]

1

5.1 Integral of nonnegative simple measurable functions 121

if x has non-terminating dyadic expansion x = I:n001 xn/2n (as in exercise

3.3.4), let

+1 if x= 1,f(x).-1 -1 if xi = 0, i = 1, 2, ... .

Show that for every i, there exists simple function si c Lo such that fi =si - 1. Compute f sidA.

5.1.8. Exercise:

(i) Lets : X Ilg* be any nonnegative function such that the range of sis a finite set. Show that s E ]Lo if s-1{t} E S for every t c 1[8*.

(ii) For S1, S2 E ]Lo show that {x I sl(x) > 82(x)} E S. Can you say that the_sets {x E X I sl(x) > 82(x)}, {x E X I sl(x) < s2(x)} and {

82 (X) j are also elements of S?

5.1.9. Exercise: Let sl, s2 E ]Lbe real valued and sl > s2. Let 0 =81-82-0

Show that E ]Lo .Can you say that

Odtt Sldµ - I 82dµ?.

5.1.10. Proposition: Let s E ]Lo . Then the following hold:

(i) If {s}>i is any increasing sequence in ]Lo such that limn, sn(x)= s(x), x E X, then

fsd= lim f5nd/t.

(ii) f sdµ = sup { fs/

dµ 10 < s < 818' E 1Lo }

Proof : (i) Since 0 < sue, < s, by exercise 5.1.6 for every n we have

f5ndµ < fsd.Hence

lim sup J sndµ < f sdµ.n-*oo

Let 0 < c < 1 be arbitrary and let

Bn :_ {x E X I Sn(x) 1 CS(x)l.

122 5. Integration

Then Bn E S (exercise 5.1.8) and Bn C Bn+1 V n with U=1 Bn = X. Thusby proposition 5.1.5 (iv) and theorem 3.6.3, we have

c f s(x)dµ(x) = limJ c s(x)dµ(x)

n--+oo Bn

< liminfJ sn(x)dµ(x)n--+oo B

< liminf fn->oo

Since this holds d c with 0 < c < 1, we have

fs(x)d(x) < 1 m f J sn(x)dµ(x). (5.2)

From (5.1) and (5.2) we get

fsd/t= lim fsd/.n--+oo

This proves (i). The proof of (ii) is obvious.

5.1.11. Exercise: Let {sn}n> 1 and {s}> 1 be sequences in Lo suchthat for each x E X, both {s(x)}> 1 and {s(x)}> 1 are increasing andlim sn(x) = lim s' x). Show that

n--+oo n--+oo n (

lim f sA,c = lim sndjL.n--+oo n--+oo

(Hint: Apply exercise 5.1.6 and proposition 5.1.10 to f sn A s;n In for all fixedm to deduce that f s;ndtC < f sA,c. )

5.1.12. Exercise: Show that in general Lo need not be closed under limit-ing operations. For example, consider the Lebesgue measure space (R, £, A)and construct a sequence {s}> 1 in Lo such that lim sn (x) = f(x) exists

n--+oo

but f ¢ 1Lo

5.2. Integral of nonnegative measurablefunctions

Having defined the integral for functions s E 1Lo , i.e., nonnegative simplemeasurable functions, we would like to extend it to a larger class. Let us re-call that, given a Riemann integrable function f : [a, b] --) R, in section 2.1we constructed a sequence {f}> 1 of step functions such that {fn(x)}ni

5.2 Integral of nonnegative measurable functions 123

increases to 1(x) b'x c [a, b] and fQ f(x)dx = limn---,,. fQ fn()dX. In the ex-tended integral, step functions are replaced by nonnegative simple measur-able functions, whose integral is already defined. Let f : X -> 1[8* be a func-tion such that there exists a sequence {sn}n>1 of nonnegative simple measur-able functions increasing to f . Then one can define f fdµ := limn,o. f sndµ.This motivates the following definition.

5.2.1. Definition:

(i) A nonnegative function f : X -> ][8* is said to be S-measurable ifthere exists an increasing sequence of functions {sn}n>1 in ILo such thatf(x)= lim sn(x) d x E X.

nyo0If the underlying Q-algebra is clear from the context, an S-measurable

function is also called measurable. We denote the set of all nonnegativemeasurable functions by L+.

(ii) For a function f c ]L+, we define the integral of f with respect to it by

f(x)dp(x):= lim 8n(x)dp(x).f n---+oofIt follows from exercise 5.1.11 that for f E lL+, f fdµ is well-defined.

Clearly, ]Lo C L+ and f sdµ for an element s c lLo is the same as f sdµ,for s as an element of L+. The next proposition gives a characterizationof functions in ]L+ and the integrals of its elements. Another (intrinsic)characterization of L+ will be given in the next section.

5.2.2. Proposition: Let f : X --> Il8* be a non-negative function. Thenthe following hold:

(i) f E ]L+ if there exist functions sue, E ILo , n > 1, such that 0 < sue, <f d n and f (x) = lim sn(x) d x E X.

(ii) If f E IL+ and s c lLo is such that 0 < s < f, then f sdµ < f fdµand

f fd= sup {fsd

Proof: (i) Let f E L. By definition, there exists a sequence {sue,}n>1 in]Lo such that {sn(x)}n>i increases to 1(x) V x E X. Hence the directimplication holds. Conversely, let there exist a sequence {sn}n>1 in ]Lo suchthat 0< sue, < f d n and f (x) = lim sn(x) V X G X. Put

n-+oo

snW := max{sl(x), , sn (x)}, X E X.

124 5. Integration

Then sn E LO +' 0 < sn < f and {s(x)}n>i increases to 1(x) V x E X. Thisproves (i).

(ii) Let f E ]L+, and let {sn}n>1 be a sequence in ILo such that {sn}n>1increases to f (x) V x and

fd= lim Sndµ.n-->oo

Let

,Q :=sup { J sdµl

O<S< fjsE]Lp

and let sELo be such that0<s< f. Let

Bn := J X E X I S (X) < Sn (x) }, n > 1.

Then Bn E S (by exercise 5.1.8) and V n, Bn C Bn+1 with U=1 Bn = X.Thus by proposition 5.1.5 (iv) and theorem 3.6.3, we have

fsdµ = lim sdµ < lim Sndµ = ffd.n-+oo fn n-+oo

Further, since this holds V s E Lo with 0 < s < f, we have

/ f dM.

Conversely, if f fdµ = +oo, then V N > 0, we can choose no such thatf sno dµ > N. Since 0 < Sno < f, we have 3 > N, V N. Thus

Q = +oo = f f dM.

In case f fdµ < +oo and e > 0 is given, we can choose no such that

J fd_fSnodµ < E'

i.e.,.e.,

J fdµ < J snodµ + e < ,3 +

Since this holds b e > 0, we have

f f dµ < ,3.

5.2.3. Exercise: Let f E IL+ and let {sue}n>1 be in 1Lo and such thatsn (x) = f(x). Can you conclude{sn(x)}n>1 is decreasing and V X E X,

n-+oothat

fd= lim Sndµ?n-+oo

5.2.4. Definition: Let (X, S, µ) be a measure space and Y E S. We say aproperty P holds almost everywhere on Y with respect to the measureµ if the set E = {x E Y I P does not hold at x} E S and µ(E) = 0. Wewrite this as P for a.e. x(µ) or P for a.e. (µ)x E Y. If the set Y andµ are clear from the context, we shall simply write P a.e. For exampleif f X ) ][8 is a function, then f (x) = 0 for a.e. x(µ) means thatE=Ix EXI f(x)#0}ESand µ(E)=0.

5.2.5. Exercise: Let f E L. Show that

J fdµ = sup { J sdµl0 < s(x) < 1(x) for a.e. x(µ), s c ILp 5l

We describe next the properties of f f dµ, for f E L.

5.2.6. Proposition: Let f , fl, f2 E L. Then the following hold:

(i) f fd,u > 0 and for fi > f2

f fidµ ? ff2d.µ

(ii) For 0 we have (afl +,3f2) G L+ and

f(afi+/3f2)dµ = a f fidµ +,3f f2dµ

(iii) For every E E S we have XE f E L. If

v(E) := fXEfdµ> E E S,

then v is a measure on S and v(E) = 0 whenever µ(E) = 0.The integral f fxEdµ is also denoted by fE fdµ and is called theintegral of f over E.

(iv) If fl(x) = f2(x) for a. e. x(µ), then

f fd = fProof: (i) Let f c L. Clearly by the definition of the integral, f fd,u > 0.For fl, f2 E 1L+, if fl < f2, then using proposition 5.2.2(ii) and exercise5.1.6.(i) it follows that f fid,u < f f2d.

(ii) Let Ii, f2 E 1L+, and let a,,3 E ]I8 be nonnegative. Let {s}>i and182 }n>1 be increasing sequences in ILo such that d x E X,

lim s'(x) =f(x),i = 1, 2.ny

126 5. Integration

Then I sue + s2n}n>1 is an increasing sequence in ILo and

lim (csn + Sn) (x) = afl(x) +,6f2V x E X.n - >oo

Thus by exercise 5.1.11 and proposition 5.1.5,

J (af, +,6f2) dµ = limn->oo CYSn +,38n dla lim snd1.c +,3 lim fsd/i

n--- >oo n-400

= a ffid+ff2d.(iii) Let f E IL+ and E E S. Let {sn}n>1 be an increasing sequence in

ILo such that 1(x) = limn, sn(x) V x E X. Then clearly, {SnxE}n>1 is anincreasing sequence in ILo and

Y XE) (x)- b x E X.

Hence f E L+ and, by exercise 5.1.11,

fXd/ = nmoosnXE d p.

Further, if p (E) = 0, then it follows from the above equality and proposition5.1.5 that

v(E) :- Effd= ffxEd=0.

To prove that v is countably additive, we first show that v is countably sub-additive. Let {En}n>1 be a sequence in S and E U=1 En. Let {sn}n>1be an increasing sequence in ILo such that f (x) = limn,oo sn(x) V x E X.Then, using proposition 5.1.5 (iv) and proposition 5.2.2 (ii), we have

Efd[c = lim sndp

n--- >oo E

limn--- >oo

< limn-400

00

i=1

00

SndµfETi

This proves that v is countably subadditive. To prove that v is a measure,we only have to show that it is finitely additive (see theorem 3.6.1). Let

El, E2, ... , Em be pairwise disjoint sets in S and E = UT1 Ei. Then, againusing proposition 5.1.5 (iv),

fdic = lim sndi.cE n--+oo (1E

M

lim snditn--+o° Eii=1M

(limf Atn--*oo

ii= 1

fEj

This proves (iii). The proof of (iv) is easy and is left as an exercise.

Since the class ILo is not closed under limiting operations (exercise5.1.12), we defined the class IL+ by taking limits of sequences in ILo . Natu-rally, we expect L+ to be closed under limits. The next theorem discussesthis and the behavior of f fdµ under increasing limits, extending proposition5.1.10 to functions in L.

5.2.7. Theorem (Monotone convergence): Let {f}>i be an increas-ing sequence of functions in L+, and f (x) := lim f(x), x E X. Then

n--+oof E IL+ and

fd=lt n--+ooffd.µ

Proof: Since fn E L+, there exists a sequence {s}>i of functions in ILosuch that {s(x)}>1 increases to fn(x) V x. We represent it as an array(see Figure 14). In each row functions are increasing from left to right, andin the last column the functions are increasing from bottom to top. Forn = 1, 2, ... ,define

9n(x) := maxisn(x), Sn(x), ... , Sn(x)I.

By exercise 5.1.6(ii), gn E ILo , and for every x E X, {gn(x)}n>i is anincreasing sequence in R*. Let

9W := lim ynW.n--+oo

Then by definition g E L. Since {f}>i is increasing and

sn < h < f for all n and for all j,

128 5. Integration

s1 ns2 Si Sn

f

r

fn

Si S2 S Sn fi

S1

S2 S S2f2

1sl s

12 S

3

1Sn f,

91 g2 9i gn 9

Figure 14: Definition of gn

we have

Hence < f Also

gn = max ISO < A < f1<j<n

g . ,

sn < gn < g whenever j < n and n > 1.Thus, letting n -* oo, we have fj < g, V j > 1. Hence f < g, proving thatf = g E L. Also, by definition

gdiL = lim fgnd.n--+oo

Since for every ngn < fn < f,

we get

Hence

lim fgndjc < lim fndA < f dec.n--+oo n--+oo

fdjc = f gdjc = lim f gndjc < lim fndjc < f djc,n--+oo n--+oo

proving the theorem.

5.2.8. Remark: If is a sequence in L+ decreasing to a function f EIL+, then the equality f fdµ = limn--->oo f fdµ need not hold. For example,let X = ][8, S =,C and µ = A, the Lebesgue measure. Let fn = Then

fn E lLo C ]L+, and {f}>i decreases to f -= 0. Clearly, f fd\ = +oo forevery n and f f dA = 0. In fact, at this stage it is not clear whether f E lL+whenever {f}>i decreases to f, with each fn E L+. That this is true willbe shown as a consequence of the characterization of L+ proved in the nextsection (see corollary 5.3.15).

5.2.9. Exercise: Let be an increasing sequence of functions in L+such that f (x) := 71i fn(x) exists for a.e. x(µ). Show that f E L+ and

f dp = lim ffnd[L,n->oo

where f (x) is defined as an arbitrary constant for all those x for whichlim fn(x) does not converge.

5.2.10. Exercise: Let µ(X) < oo, and let f E lL+ be a bounded function.Let P:= {El, E21... , En} be such that U=1 EZ = x, E,, fl E, = 0 for i jand EZ E S V i. Such a P is called a measurable partition of X. Given ameasurable partition P = {E1,... , En}, define

MZ := sup{ f (x) I x E Ei} and mi := inf If (x) I x E EZ}.

Letn n

4Pp := E mixEz and 'p := MixE. .

i=1 i=1

Prove the following:

(i) For every partition P, show that DP, GYP E ]Lo and p < f < gyp.

(ii) f f dµ = sup {fpdµ I P is a measurable partition of X j ,

= inf {f'Tipd,u I P is a measurable partition of X } .(This gives an equivalent way of defining f f dµ, in a way similar tothat for the Riemann integral.)

(iii) Let

a=sup( fnlsdµ I s E ]Lo , s < f and (3 = inf { J sdµ I s E ]Lo , fs}.

l

a = sup { J 4DPdµ I P is a measurable partition of Xl

130 5. Integration

and

,6 = inf { J XFPdµ I P is a measurable partition of X } .

l 1

(iv) Deduce that f E 1L+ implies a = f fdµ = 3.

5.2.11. Note: Exercise 5.2.10 tells us that for f E lL+, in defining f fdµ itis enough to consider approximations of f from below, as the approximationsfrom above will also give the same value for f f dµ. This is because f E L+,i.e., f is nonnegative measurable. The converse is also true, i.e., if a = ,6,for f : X -- [0, oo], then f E L. To see this, first note that

a = ,Q < M(µ(X)) < ooI

where M is such that If (x) I < M V x E X X. Thus for every n we can choosefunctions On, On E ILK such that

On C f <_ On and f (On -On) dµ < 1n

Let 0 sup On and 0 := inf On. Then ¢, 0 E ]L+ (see corollary 5.3.13) andV n,

Thus

(0 - O)dp < (On -On) dp <1

f(_)dµ = 0

and by proposition 5.3.3, = ¢ = f a.e. it. Hence f E L.

5.3. Intrinsic characterization of nonnegativemeasurable functions

Recall, f E L+ means f X ) lI8* is a nonnegative function with theproperty that there exists a sequence {s}>1 of functions in 1Lo such that{sn(x)}n>1 increases to f (x) for every x E X. Note that the measure itplays no part in the definition of the functions in L. It is only to definethe integral of functions f E L+ that we need the measure it. We want tocharacterize the functions in L+ intrinsically. Consider the particular casewhen f = XA for A C X. Then A = {x E X I XA(x) = 1} = f-1{1}, A° _f'{O} and f'{t} = 0 d t E lI8*, t 0, 1. Thus

f = XA E II-o if xAl {t} E S Vt E Il8*.

Since any function s : X - ][8 which takes only finitely many values canbe written as En 1 aiXAi, where {Al, ... , is a partition of X, we havethe following proposition:

5.3 Intrinsic characterization of nonnegative measurable functions 131

5.3.1. Proposition: Let s : X ) Il8* be such that s takes only finitelymany distinct nonnegative values. Then the following are equivalent:

(i) s E IL+ (and hence s (E ILo ).

(ii) s-1{t} E S b t E II8*.

(iii) s-1 [t, oo] E S V t E I[8*.

(iv) s-1(I) E S for every interval I in ][8*.

Proof: Let the distinct values of s be a1, a2, ... , an and Ai : = s-1 { ai } .for i j.Then s = 1i=1aZXA, where U=1 Ai = X and Ai rl Aj = 0

(ii): Since s E lL+, let {sn}n>1 be a sequence in 1Lo such that{sn(x)}n>1 increases to s(x) for every x. Then for ai E ][8,

00 00 00AZ := s-1{ai} -nun {x I sn(x) > a2 - 1/m}m=1 r=1 n=r

and hence Ai E S. Thus s-1(t) E S if t = ai. If t E ]I8 and t # ai for any i,then clearly s-1(t) = 0 E S. Also, s-1{-o0} = 0 E S. Finally, if t = +oo,then

00 00

8 ii+ooi = n n (sm [n, +c)o])n=1 m=n

Thus s-1{+o0} E S. Hence (ii) is proved. Conversely, if (ii) holds thenclearly s E Lo C ,+

(iii): For t E R, s [t, oo] = U Ai, where the union is over onlythose i's such that ai E [t, oo]. Hence s [t, oo] E S for every t E R*. Thisproves that (ii) (iii).

The implication (iii) (iv) is an application of the `a-algebra tech-nique', and the implication (iv) (i) is easy to verify.

In view of the above proposition it is natural to ask the following ques-tion: does proposition 5.3.1 remain true if s E 1Lo is replaced by any f E L+?.The answer is yes, as proved in the next proposition.

5.3.2. Proposition: Let f : X R* be a nonnegative function. Thenthe following are equivalent:

(i) fELt.(ii) f (c, +oo] E S for every c E R.

(iii) f [c, +oo] E S for every c E R.

(iv) f [-oo, c) E S for every c E R.

(v) f -1 [-oo, c] E S for every c E R.

132 5. Integration

(vi) f-1{+00}, f{-oo} and f -1(E) E S for every E E B.

Proof: We shall prove the following implications:

(i) (ii) #==> (iii) -< (iv) -< (v) (vi)

(i) (ii): Since f c L+, there exists a sequence {sn}n>1 of functions in1Lo such that b x E X, increases to 1(x). Now, for c c ]E8

(C' +00] jX : f(X) > C100

U IX: Sn(X) > C1

00

n=1 n=1

C) 00]) .

By proposition 5.3.1,

Sn. l CC, oo] = (s'[c,oo]\s'{c}) E S.Thus f'(c,+oo] E S.

(ii) b (iii): Note that b c E ]I8,00 00

[c, +ooJ = n (c - 1/n, +oo J and (c, +oo ] = U [c + 1/n, oo] .

n=1 n=1

Thus

and

00f'[c,+oo] = n f -1 (c - 1/n, +oo1

00

f -1 (c, +oo ] = U f -1 [c + 1/n, +oo].n=1

From these identities, the implications (ii) 4===> (iii) follow.

(iii) b (iv): Since(c,+oo ] =1[8* \ [-oo, c] and [-oo,c ) = R* \ [c, +oo],

we have

f'(c,+oo] = X \f'[-oo,c] and f'[-oo,c) = X \ f -1[c, +oo].From this the required implications follow.

(iv) b (v): This is similar to the proofs of the implications (ii) -< (iii)and is left as an exercise.

(v) (vi): Note that

1+001 =00 00

fl(n,+oo]n=1

and { -001 n I - 00, -n).

Hence00 00

f-1({+00}) = n f -1(n, +oo] E S and f'{-} = n f -i [-oo, -n] E S

Let

n=1 n=1

L3:= JE E BR If -'(E) E Sj.

By (v) (and hence using (ii) and (iii)), it is easy to show that I E ,Ci wheneverI C R is an interval and B is a v-algebra of subsets of R. Hence B = BR,proving (vi).

(vi) (i): Let f : X Il8* be a nonnegative function such that (vi)holds. Since f is nonnegative, the range of f is a subset of [0, +00]. Forevery n, consider the partition of [0, +oo] given by

7t2n

A +001 = U [(k - 1)/2", k/2"`) U [n, +oo].(k=1

Then we get a partition of X, the domain of f, given by,22n

X = f -1 [0, +00] = U f -1 [(k - 1)/2n , k/2n) U f -1 [n, +oo].(k=1

For 1 < k < n2n, let

Xk f -1 [(k - 1)/2n, k/2n) and X := f'[n,+00].

For every n = 1, 2.... and 1 < k < n2n, by (vi) we have that Xn E S andX n E S. For every n > 1, we define function sn on X by

,22n k-1sn 2XXn, + nXX .

Clearly, Sn E ILK ,and it is easy to check that for every n,

Sn (X) _< Sic+l(x), b x E X.

For any x E X, if f (x) = +oo, then x E X for every n, and hencesn(x) = n. Thus lim sn(x) = f (x). Also, if f (x) < +oo, then for every n,x E Xk for some k, 1k271. Thus

nSn(x) = (k - 1)/2

Since f (x) E [(k - 1)/2", k/2n), we have sn (x) < f (x) and

f(x) - Sn(x) < 1/2n.

In other words, lim sn(x) = f (x). This proves that f E lL+.

134 5. Integration

The statements (ii) to (vi) of proposition 5.3.2 describe the elements ofL+ intrinsically. This proposition is used very often to check the measura-bility of nonnegative functions. Once again we emphasize the fact that fora nonnegative function f : X ) ]I8* to be measurable, i.e., for f to be inL+, the measure µ plays no part. As is clear from the above proposition, itis only the a-algebra S of subsets of X that is important (see also exercise5.3.27). The notion of measurability is similar to the concept of continuityfor topological spaces. As an immediate application of proposition 5.3.2, wehave the following:

5.3.3. Proposition: Let f E lL+ and E E S be such that fE fd,u = 0. Thenf (x) = 0 for a. e. (µ)x E E.

Proof: Let Xo := fxGX I f(x) > 0}, and d n # 1, let Xn :_ fxGX I f (x) > 1/n}. Then by proposition 5.3.2, Xn E S and hence En :=E fl Xn E S for n = 0, 1, 2, ....Since Eo = U=1 En, if p(Eo) > 0 thenµ(En) > 0 for some no, and we will have

Ef (x)dp(x) > f (x) dp (x) > p (Eno) /no > 0fi0

which is not true. Hence (E0) = 0, i.e., f (x) = 0 a.e. (µ)x c E.

5.3.4. Exercise:

(i) Let f E ]L+ and E E S be such that f (x) > 0 for every x E E andµ(E) > 0. Show that fEfd,u > 0.

(ii) Let f, g c lL+ be such that

J fd= J gdµ < +oo and JE fd= Efgd,VEES.

Show that f (x) = g(x) a.e. (x)µ.

(iii) Let f , g be nonnegative measurable functions on (][8, ,C) such that

Ia

show that then

b bfd=fgd< +oo for every a < b

Efd= Jfgd, d E E ,C,

E

and deduce that f (x) = g(x) a.e. (x)A.

Next, we extend the concept of measurability of nonnegative functionsto functions f : X -* R* which are not necessarily nonnegative. For this

we consider f+ and f-, the positive and negative part of the function frespectively, defined as follows:

f +(x)

and

f (x) if f (x) > 00 if f (x) < 0,

0 if f (x) > 0,_f (x) if f (x) < 0.f - (X) . =

Clearly, f+ and f - are nonnegative functions on X with

fff- and fI=f+f.Since we would like functions to have the property that their sums anddifferences are also measurable, we are motivated to make the followingdefinition:

5.3.5. Definition: Let (X, S) be a measurable space.

(i) A function f : X R* is said to be S-measurable if both f + andf - are S-measurable.

We denote by L the class of all S-measurable functions on X. If the un-derlying a-algebra is clear from the context, we call a S-measurable functionto be measurable.

(ii) If f E L is such that both f+, f L0+ , we call f a simple measurablefunction.

We denote the class of all simple measurable functions by ILo. Note thats E Lo iff both s+, s- E ILo .

5.3.6. Proposition: Let (X, S) be a measurable space and f : X --+ R* beany function. Then following statements are equivalent:

(i) f is S-measurable.(ii) There exists a sequence {s}>i of real valued functions on X such

that b n, s, and sn are both nonnegative simple measurable functionson (X, S) and lim 1(x) V x E X.

n-->oo

(iii) f satisfies any one (and hence all) of the statements (ii) to (vi) ofproposition 5.3.2.

Proof: We ask the reader to extrapolate the proof of proposition 5.3.2 toprove the required claim.

5.3.7. Exercise: Let f : X ---> R* be a bounded measurable function.Then there exists a sequence {sue,}n>1 of simple measurable functions suchthat {s}>i converges uniformly to f.

136 5. Integration

(Hint: If 0 < f(x) < M V x, then I sn (x) - f(x)I < 1/2n d n > no, whereno > M and the sn's are as in proposition 5.3.2.)

5.3.8. Exercise: Let f : X -) ][8* be a nonnegative measurable function.Show that there exist sequences of nonnegative simple functions {s}>iand {}>i such that

0 < ... < gn(n) C Sn+l (x) < ... < f W C ... < Sn-f-1 (x) _< Sn(x) ...

and lim sn(x) = f (x) = lim sn(x) V x E X.n-+oo n->oo

5.3.9. Examples:

(i) Let f : X - R* be a constant function, f (x) = a Vx E X. Then f ismeasurable Vc E R, since

fXEX I f(X) >Cj= 0 ifc<a,X if c > a.

(ii) Let f : X -4 R* be measurable, and a E R. Then a f is also measurable,since

fx c X I a(x) > cl

fX C X I f(X) < C/aj if a > 0,X ifa=0,c<0,0 if a = 05C > 0,

fX C X I f(X) < C/aj if a < 0.

(iii) Let X be a topological space and S be the Q-algebra of Borel subsetsof X, i.e., the a-algebra generated by the open sets. Let f : X -) ][8 beany continuous function. Then f is S-measurable. To see this, we firstnote that by continuity of f , f (U) is open whenever U C ][8 is open, andf (U) E y,t3X. Thus, if

S:= fE C R I f-'(E) C BXI)

then every open subset of ][8 belongs to S. Further, it is easy to checkthat S is a v-algebra of subsets of X. Hence by definition, ,C3X C S, i.e.,f -1(E) E ,t3X V E E BR, proving that f is measurable with respect to 13X.

We next describe some more properties of measurable functions.

5.3.10. Proposition: Let f,g be measurable functions. Then each of thesets {x E X lf (x) > g(x)}, {x E X1 { f (x) < g(x)}, {x E X lf (x) < g(x)j,{> g(x)} and {xEXf(x)=g(x)}eS.Proof: Since f, g are measurable, b c E ][8, the sets {x E X I f (x) > c},{x c X I g(x) > c} E S. Note that for x c X, f (x) > g(x) iff there exists a

rational r such that f (x) > r > g(x). Hence

jXEXjf(x)>g(x)j = U{x E X jf (x) > r> g(x)}rEQ

= U({xExlf(x)>r}n{xExIrEQ

r > g(XM-

Since f,g are measurable, {x E X I f (x) > r}, {x E X I r > g(x)} E S forevery r. Thus the right side in the above equality, being a countable unionof sets from S, implies that {x E X I f (x) > g(x)} E S. Interchanging fand g, we get {x E X I f (x) < g(x)} E S. Taking complements, we have{x E X I f (x) > g(x)j, {x E X I f (x) < g(x)} E S. Finally,

{xEXf(x)=g(x)}={xEXIf(x) > y(x)} n {x E x lf(x) < y(x)}

implies that {x E X I f (x) = g(x)} E S.

5.3.11. Proposition: Let f, g : X --+ R* be measurable functions and let/3 E R* be arbitrary. Let

A:= {x E X I f (x) = +oo, g(X) _ -oo}U{x E X jf (x) = -oo, g(x) = +oo}.

Define d x E X

f (x) + g(x) if x A,,(3 ifxEA.

Then f + g : X -4 R* is a well-defined measurable function.

Proof: Clearly f + g is well-defined and A E S. For any c E Il8,{if ,6>c,0 if ,(3 < c.

Thus {x E X I (f + g) (x) > c} n A E S. Also by examples 5.3.9 (i), (ii),

fx E X I (f+g)(x)>c}nAc _ fx E X I f (x) > c- 9 (x)j n Ac E S.

Hence fxE X I(f + g)(x) > c} E S.

-->5.3.12. Proposition: Let f : X ) I[8* be measurable and let (D : R*R* be such that ][8 fl {x E Il8* 1> a} E 13R, d a E ]I8. Then 4Do f is alsomeasurable.

Proof: Let a E R. Let

A+ := {x E R* 1> a}fl{+oo} and A_ :_ {x E R* (x) > a}f1{-oo}.

138 5. Integration

Since f is measurable, it is easy to check that both f (A+) and f (A) E S.Further

(4p 0 Plce' +C)O1= f -1 (4p-i [Ce5 +001)

= f -1 [ ({x c Il8* I (x) > a} f1 I[8) U A+ U A_]

= f'( {x E R* I (x) > a} nR) U f-1 (A+) Uf-1(A_).

Now it follows from the given hypothesis and the above observations that(4P o f)-1 [a, +oo] E S. Hence by proposition 5.3.6, (4P o f) is measurable.

5.3.13. Exercise: Let f and g : X ) R* be measurable functions, pand a E I[8 with p > 1, and let m be any positive integer. Use proposition5.3.12 to prove the following:

(i) f + a is a measurable function.

(ii) Let Q and y c I[8* be arbitrary. Define for x c ][8,

(f(x))m if f (x) E llg5f '(X) if f (X) +005

if f (X) -00.

Then f ' is a measurable function.

(iii) Let If IP be defined similarly to f', where p is a nonnegative realnumber. Then If IP is a measurable function.

(iv) Let 0,,y, S E ][8* be arbitrary. Define for x c Il8,

(11f)(X)

11f (x) if f (x) V {0, +oo, -oo},,6 if f(x)=O,

if f (X) = -005

if f (X) = +00.

Then 11f is a measurable function.

(v) Let 0 E ][8* be arbitrary and A be as in proposition 5.3.11. Define forxER,

(fg)(x) - r f (x)g(x) if x A,0 if x E A.

Then f g is a measurable function.

5.3.14. Proposition: Let f n : X -p Ilk*, n = 1, 2, ... , be measurable func-tions. Then each of the functions sup fn, inf fn, lim sup fn and lim inf fn

n n n-*oo n-, o0is a measurable function. In particular, if {f}>i converges to f, then f isa measurable function.

Proof: Let a E R. Then

(sup f)'[-oo,a} _ {x E X I sup fn(x) < a}

00

nn=1

and

(inff)'[a,+oo] _ {xEXI inf fn(x) > a}00

nix E x I fn(x) > al.n.=1

Since each fn is measurable, by proposition 5.3.6 {x E X f(x) < ae}, {x EX fn(x) > al E S. Thus the above equalities show that inf fn and sup fn

n nare measurable. Using this, clearly lim sup fn, lim inf fn are measurable.

n--+oo n-400

5.3.15. Corollary: Let {f}>l be a sequence in L. Then each of thefunctions sup fn, inf fn, lim sup fn and lim inf fn is in L. In particular, if

n n n-400 n-*ooJimfn =: f exists, then f E L.

n-400

Proof: Since fn E L+ C L, by the above proposition each of sup fn, inf fn,n n

lim sup fn, lim inf fn and f is in L. Further, since all of them are nonnegative,n-400 n-*oo

it follows that they are in L+.

5.3.16. Exercise:

(i) Let f : X --> R* be S-measurable. Show that If I is also S-measurable.Give an example to show that the converse need not be true.

(ii) Let (X, S) be a measurable space such that for every function f : XR, f is S-measurable if If I is S-measurable. Show that S = P(X).

5.3.17. Exercise: Let fn E L, n = 1, 2, ....Show that the sets {x EX I {f(x)}n is convergent} and {x E X I {f(x)}>1is Cauchy} belong toS.

Recall that in theorem 5.2.7 we analyzed the limit of f fdµ for an in-creasing sequence of nonnegative measurable functions. We analyze next thebehavior of f fdµ when is not necessarily an increasing sequenceof nonnegative measurable functions.

140 5. Integration

5.3.18. Theorem (Fatou's lemma): Let be a sequence of non-negative measurable functions. Then

fProof: By definition,

(urn inf fn dM < lim inf fndp.n--+oo n--+o0

lim inf fn := sup{ inf fk}.

n-*oo n>1 k>n

We have already shown in corollary 5.3.15 that lim inf fn E L+. Further,n->oo

{ inf is an increasing sequence in L+, converging to lim inf fn. Thusk>n n--+oo

by the monotone convergence theorem (5.2.7), we have

J (liminffn) dµ = lim J (inf fk I d. (5.3)n-oo noo k>n

Also, (inffk) < fn b n. Using proposition 5.2.6, we have

f(yk)dµ <

Hence

lim I I inf fk I < lim inf / fd,a. (5.4)n->oo k>n / - n-+oo

The required inequality follows from (5.3) and (5.4).

5.3.19. Exercise:

(i) Give an example to show that strict inequality can occur in Fatou'slemma.

(ii) Let {fTh}>i be a sequence of functions in lL+ and let E', fn(x)f (x), x E X. Show that f E ]L+ and

00f f dµ = E ffnd.µ

5.3.20. Exercise: Show that each of the functions f : ][8 -> ][8 definedbelow is G-measurable, and compute f f dA

W f W0 if X < 0

)1/x if x > 0.

(ii) f (x) := xQ(x), the indicator function of Q, the set of rationals.

0 if x > 1 or x < 0 or x E [0, 1] and x is rational,n if x is an irrational, 0 < x < 1 and in the decimal(iii) f(x)

expansion of x, the first nonzero entry is at the(n + 1)th place.

5.3.21. Exercise: Let f, g E lL+ with f > g. Show that (f - g) E IL+ andf fdµ > f At. Can you conclude that

f(f_g)dµ = ffd_ fgd?µ

5.3.22. Exercise: Let f , In E IL+, n = 1, 2, ... , be such that 0 < fn < f .

If lim fn(x) = f (x), can you deduce that

f dM = lim ffnd?n---oo

5.3.23. Exercise: Let f E L. For x E X and n > 1, define

1 1(x) if lf(x)l<n,fn(x) n if f (x) > n,

-n if f (x) < -n.


(i) fd n and d x E X.(ii) lim fn(x)f(x) b x E X.

n--+oo

(iii) I fn(x)I min{I fn(x)1,n} (I f I A n) (x) is an element of IL+ and

lim j fnId/t = f in-+oo

5.3.24. Note: For f E L, the sequence as defined in exercise5.3.23 is called the truncation sequence of f. The truncation sequenceis useful in proving results about functions in the class L. See for example,proposition 5.4.6.

5.3.25. Exercise: Let f E L and

v(E) := µ{x E X I f (x) E E}, E E BR.

Show that v is a measure on (I[8, ,t3R). Further, if g : I[8 II8 is any nonneg-ative ,CiR-measurable function, i.e., g-1(A) E BR V A E BR, then g o f E Land

Jg dv =

J(g o f) dµ.

142 5. Integration

The measure v is usually denoted by µf -land is called the distribution ofthe measurable function f.

5.3.26. Exercise: Let f c IL+ be a bounded function, say f (x) < N V x EX and for some N E N. Show that

N2n

J fd µ = yli 1: k2n 1 µIx

I

k2n1 < 1(x) < 21Y } . (5.5)

k=1

m

a b

Figure 15: Area below the curve y = f (x) from inside

Note that here V n E N we are taking the partition {O, 1/2n, 2/2n, ... , N}of the range of f and using it to induce the measurable partition Pn of thedomain I X of f, given by

{XIi <k < N2n}, where X := f -1 [(k - 1)/2n, k/2n).

Using this partition Pn, we form the sums on the right hand side of (5.5),approximating the area below the curve y = f (x) from `inside' (Figure 15).Thus in our integral we use partitions of the range, whereas in Riemannintegration we use partitions of the domain. In the words of H. Lebesgue,the situation is similar to the problem of finding the total amount in a cashbox containing say n coins, the ith coin being of denomination a2,1 < i < n.One method of finding the total amount in the box is to add the value ofeach coin, i.e., the sum En 1 a2. The other method is to separate out coinsof similar denominations. If there are kj coins of denominations aj, thenthe total amount is E, k3a3 .

5.3.27. Exercise: Let (X, S, µ) be a measure space and let (X, S, µ) beits completion (see theorem 3.11.8). Let f : X --> II8 be an S-measurable

5.4 Integrable functions 143

function. Show that there exists an S-measurable function f Xsuch that f (x) = f (x) for a. e. x(µ).(Hint: Use theorem 3.11.8 and proposition 5.3.6.)

5.3.28. Proposition: Let (X, S) be a measurable space and let f : X ->II8* be S-measurable. Let µ be a measure on (X, S). Let g : X ]Eg* be

such that {x c X I f (x) = g(x)} is a µ-null set. If (X, S, µ) is a completemeasure space, then g is also S-measurable.

Proof: To prove that g is S-measurable we have to show that d t c R,g-1[t,oo] E S (see proposition 5.3.6). Let

N:= fx c X I f (x) =/- g(x)j.

By the given hypothesis, N is a µ-null set, i.e., N C A E S with µ(A) = 0.But then µ* (N) = 0, and since (X, S, µ) is complete, N E S. Further, bythe same reasoning, if E C N, then E E S. Now for t c R fixed,

g-i[t,oo] = (g-1 [t, oo] nN) U (g -1 [t, ool nNc).

Since g-1 [t, oo] n N C N, we have g [t, oo] n N E S. Also, since f is alsoS-measurable,

g-1 [t, oo] n N° =f -1 [t, oo] n Nc c S.

Hence g-1 [t, oo] C -S.

5.3.29. Exercise: Let (X, S, µ) be a complete measure space and {ffl}fl 1

be a sequence of S-measurable functions on X. Let f be a function on Xsuch that f (x) = lim fn(x) for a.e. x(µ). Show that f is S-measurable.

5.4. Integrable functions

Given a measure space (X, S, µ) in section 5.3, we defined f f dµ, the integralfor functions f c L+, i.e., f X -> ][8*, f nonnegative measurable. If fis measurable, but not necessarily nonnegative, we can write f = f + - f -.Since both f + and f - are nonnegative measurable, f f +dµ and f f -dµ aredefined. Thus it is reasonable (as the integral is expected to be linear) todefine the integral of f to be f fd,u := f f +dµ - f f -dµ. The problem canarise in the case f f -dµ = f f +dµ = +oo. To overcome this difficulty, weintroduce the following definition:

5.4.1. Definition: A measurable function f : X ) ][8* is said to be µ-integrable if both f fd,u and f f -dµ are finite, and in that case we define

144 5. Integration

the integral of f to be

f fdµ . -J

f +dµ - f f dµ.

We denote by Li (X, S, µ) (or simply by L1(X) or L1(µ)) the space of allµ-integrable functions on X.

5.4.2. Exercise: For f E L, prove the following:

(i) f E L1(µ) if If I E L1(µ). Further, in either case

Jf dp :5 fit. dµ

(ii) If f E L1(µ), then If(x)I < -I-oo for a.e. x(µ).

The properties of the space Ll (X, S, µ) and of the map f --- f f dµ,f E Ll (X, S, µ), are given in the next proposition.

5.4.3. Proposition: For f, g E L and a, b c- R, the following hold:

(i) If I f (x) 1 < g(x) for a. e. x(µ) and g E L1(µ), then f E L1(µ)

(ii) If f (x) = g(x) for a. e. x(µ) and f E L1(µ), then g E L1(µ) and

f fdµ = fd/2.

(iii) If f E L1(µ), then a f E L1(µ) and

f (af)dµ = a J f dµ.

(iv) If f and g E L1(µ), then f + g E L1(µ) and

f(f+g)d/2=ffd/2+fgd/2.

(v) If f , g E L1(µ), then (af + bg) E L1(µ) and

f(af + bg)dµ = a f fd/2+ b f gdµ.

Proof: (i) Since If (x)1 < g(x) for a.e. x(µ), it is easy to check that

f+(x) g(x)I for a. e. x(µ) and f(x) g(x)I for a. e. x(µ).

Hence

f f +dµ < ftId/2<+00 and ff-d/2 < fIId/2 < +oo.

Thus f E L1(µ).

(ii) Since f (x) =g(am) for a.e. x(µ), it is easy to see that

f+(x) = g+(x) for a.e. x(µ) and f(x) = g-(x) a.e. x(µ)

Using proposition 5.2.6,

J g+dµ = ff+d µ < +oo and fgdµ = ff-d µ < +oo.

Thus g E L1(µ) and

f 9dµ -ffd.

(iii) is easy and is left as an exercise. To prove (iv), let fl, f2 E L1(µ).We note that Ifl + f21 fil + If2j, and hence, by proposition 5.2.6(i),

f I + f2 I dp <_ fIfiId+fIf2ld/-t < +oo.

Thus (fl + f2) E L1(µ) by exercise 5.4.2. Also

(f1+f2)-(f1+f2Y = f1+f2=f+f-f-f.a4.2, all the functions involved are finite a.e. (µ). ThusBy exercise 5.

(f1+f2)+f+fa = f+f+(f1+f2)- a. e. (µ)

S ince both sides of the above equality represent nonnegative measurablefunction, using proposition 5.2.6 we have

f (fl + f2)+dl-t+f .fi dµ+ f fz-dµ = ff+d+ff+d+fui+f2-d.Since all the terms in the above equality are finite, we get

f (f, + f2) dl-t = f (fi + f2)+d/.t - f (f, + f2) - dl-t

= f fdµ - f fi dlt + f fdµ - f fa dµ

= ffid+ff2d.This proves (iv). The proof of (v) follows from (iii) and (iv).

5.4.4. Exercise:

(i) Let µ(X) < +oo and let f E ]L be such that If (x) I < M for a.e. x(µ) andfor some M. Show that f E L1(µ).

(ii) Let f E L1(µ) and E E S. Show that XE f E L1(µ), where

:=J

xE f dµ.Lfd/2

146 5. Integration

Further, if E, F E S are disjoint sets, show that

IEuF f Cl/A, = fEfF f d/A,.

(iii) Let f c L1 (/c) and EZ E s, i > 1, be such that EZnEj = 0 for i j. Showthat the series E°° 1 fE, f dit is absolutely convergent, and if E U°° 1 EZ,then

00

E2 f dµ = fE f dµ.2-1

5.4.5. Exercise:

(i) For every e > 0 and f c L1(µ), show that

A ix < fiiidµ < oo.

This is called Chebyshev's inequality.

(ii) Let f c L1(µ), and let there exist M > 0 such that1

µ(E) ffdMfor every E E S with 0 < µ(E) < oo. Show that If (x) I < M for a.e. x(µ).

5.4.6. Proposition: Let f E L1(µ). For every E E S, let

dµ.v(E) := fxEIfIdµ and v(E) := fxEf

Then the following hold:

(i) If µ(E) = 0, then v(E) = 0.(ii) If µ(E) = 0, then v(E) = 0.

(iii) limµ(E)--+a v(E) = 0, i.e., given any e > 0, there exists 8 > 0 such thatv(E) < e whenever, for E E S, µ(E) < J.

(iv) If v(E) = 0 V E, then f (x) = 0 for a. e. x(µ) on E.

Proof: (i) and (ii) are easy, and are left as exercises.To prove (iii), let us first assume that f E L1(µ) is also bounded, say

If (x) I < M V x c X. Then for a given e > 0, any 0 < 8 < e/M will satisfythe required claim. For the general case, we consider fn :_ If I An (as definedin exercise 5.3.23). Then it follows from the monotone convergence theoremthat

f Ifli = lim ffd.

Thus, given e > 0, we can choose no such that

j (If I - fno)dp < c/2.

Thus `d E E s,

v(E) f XE I.f I dµ

f XE(I.f I - fno) dM + XEfno dtz

f (fI-fn0)d+no(E)< e/2 + no µ(E)

If we choose b such that nob < e/2, then for µ(E) < S we will have v(E) < C.This proves (iii).

To prove (iv), let X+ = {x E X I f (x) > 0}. Then, by the given hypoth-esis, d E E S,

f+=fE nx+andhence by proposition 5.3.3, f +(x) = 0 for a. e. x(µ). Similarly, f (x) = 0

for a.e. x(µ). Thus for A :_ {x : f+(x) = 0} and B = {x : f-(x) 0}, wehave µ(A) = µ(B) = 0. Since {xf(x) = 0} C A U B, we have µ({x1f (x)0}) = 0, i.e., 1(x) = 0 for a.e. x(µ).

5.4.7. Remark: It is easy to see that (iii) in the above proposition implies(ii). In fact (ii) also implies (iii). To see this, suppose (iii) does not hold.Then there exist an e > 0 and sets En E S, n > 1, such that µ(En) < 2-nbut v(En) > E. Let An = lJ'n Ek. Then {An}n>1 is a decreasing sequencein S and µ(An) < µ(En) < 2-"`. Thus by theorem 3.6.3,

00

µ nAn = lim µ(An) = 0.n-+oo

n=1

On the other hand, µ(An) > µ(En) > e, V e, contradicting (ii).

5.4.8. Exercise: Let A be the Lebesgue measure on Ilk, and let f EL1(R, L, A) be such that

f (t)dA(t) = 01 V x c R.

Show that f (x) = 0 for a.e. (A)x E R.

We prove next the most frequently used theorem which allows us tointerchange the operations of integration and limits.

148 5. Integration

5.4.9. Theorem (Lebesgue's dominated convergence theorem): Let1fn1n>1 be a sequence of measurable functions and let g E L1(µ) be suchthat d n, Ifn(x)l < g(x) for a. e. x(µ). Let {f(x)}>i converge to f (x) fora. e. x(µ). Then the following hold:

(i) fEL1().(11) f f dµ = limn---+oo f fn dµ.

(iii) limn--+oo f I fn - f I dµ = O.

Proof: Let us first suppose that Ifn(x)l g(x) and f (x) = lim fn(x)for every x E X. Clearly, since Ifn(x)l < g(x) V x, fn E L1(µ) for everyn. Also If (x) I < g(x) V x E X and hence f E L1(µ). We only have toshow that f fd,a = lim f fd,a. (Since the only tools available to us arery

the monotone convergence theorem - which obviously we cannot use here- and Fatou's lemma (theorem 5.3.18), we should try to get a sequence ofnonnegative functions). For this, we consider the sequence {g - f}>i.Clearly, g - fn c IL+ for every n, and by Fatou's lemma,

lam f J[(g - f)dµ > [(lim

oo( 9 - f))dµ = [(g - f)d.

Thus

lim sup f fndµ < ffd. (5.6)n-+oo

Similarly, {g + f}>i is a sequence in IL+, and by Fatou's lemma again,

lim inf f (g + f)dµ f (g + f)dtt.

Thus

lim inf j fdµ > JIf dµ. (5.7)n-+oo

The inequalities (5.6) and (5.7) together imply that

f fdtt = n mooffd.

To prove the theorem in the general case, we put

N:= {x E X I Ifn(x)l > g(x) for some n} U {x E X I f (x) # Eli

Then N E S and µ(N) = 0. Since fn E IL and lim fn(x) = f (x) for a.e. x(µ),n-,oo

by exercise 5.3.29, f E L. Further, by our earlier case applied to the sequence{xNcfn}n>1, we have

f fdc = lim fndtt.Nc n-'oo Nc

Also

INfd1t =

INfd=O, V n.

Thus by exercise 5.4.4, we have

fdtc = lim ffnd.n-+oo

This proves (ii).

Finally, (iii) follows from (ii) and the fact that b n, Ifn - f I < 2g.

We state another version of this theorem, which is applicable to seriesof functions.

5.4.10. Corollary: Let {f?}>1 be a sequence of functions in L1(µ)such that En=1 f I< +oo. Then f (x) O° fi(x) exists for a. e.x(µ), f c L1(µ) and

f fdlz = E ffnd.l,n=1

Proof: Let (x) := E°°_1 I fn(x)1, x E X. Then by the monotone conver-gence theorem (5.2.7), 4P E 1L+ and

fM1

k

J (x)dµ = I-lim Eyoo

7L=1

00

E (ffd) < +co.

Hence 4P E L1(µ) and, by exercise 5.4.2, 14D(x)l < +oo for a.e. x(µ). ThusE, 1 f?(x)l < --oo for a.e. x(µ). Let f(x) :_ E, 1 f?(x) if the right handside is finite, and f (x) := 0 otherwise. Clearly, f E L and I < 4P (x) fora.e. x(µ). Thus f c L1(µ). Since

E fk(x) fi(x)k=1

for every n and for a.e. x(µ), by the dominated convergence theorem (5.4.9),00

E (f fad) nlimoo (f dµ = f f dµ.

An extension of the Lebesgue dominated convergence theorem is givenin the next exercise.

5.4.11. Exercise: Let I C R be an interval and d t E I, let ft c L. Letg c L1(µ) be such that d t, II < g(x) for a. e. x(µ). Let to E ][8* be any

150 5. Integration

accumulation point of I and let f (x) lim ft(x) exist for a. e. x(µ). Then

f E L1(µ) and

f dp lim ft (x) dp (x).t--+to f

Further, if d x the function t H ft(x) is continuous, then so is the functionh (t) : = f ft (x) dp, t E 1.(Hint: Apply theorem 5.4.9 to every sequence to --+ to.)

As an application of the dominated convergence theorem, we exhibit thepossibility of interchanging the order of integration and differentiation inthe next theorem.

5.4.12. Theorem: Let ft E L1(µ) for every t E (a, b) C R. Let to E (a, b)be such that for a. e. x(µ),t H ft(x) is differentiable in a neighborhood U

of to and there exists a function g E L1(µ) such that t (x) < g(x) for

a. e. x(µ) and for every t E U. Then fi(t) f ft(x)dµ(x) is differentiable atto and

(to)dft

W dp(x).ito)

Proof: To show that 0 is differentiable, let us consider, for t to,

m(t) - d(to) =Jt - to

C ft(x) - ft0(x)\t_to )d/J(x).

Since0 (ft(x)-ft0(xY\

t -t-

[ dt WJ

for a. e.to l t-to

to be able to apply the dominated convergence theorem and deduce therequired claim, we only have to show that for all t in a neighborhood ofto, t =A to, the inequalities

ft(x) - ftoW < 9(x) for a. e. x(µ) (5.8)t-toholds for some E L1(µ). By the given hypothesis, since ft (x) is differen-tiable in a neighborhood U of to, by the mean value theorem for derivatives,we have for all fixed t E U,

ft W - fto (x) = dit (x)t - to dt ] t=C

for some real number c between t and to. Hence (5.8) holds, for all t E Uif we take g, the function given by the hypothesis. This completes theproof.

We shall see some more applications of the dominated convergence the-orem in the remaining sections. In the next section we look at some specialproperties of L1(X, s, /_c) in the particular case when X = III, S = £, thea-algebra of Lebesgue measurable sets, and = A, the Lebesgue measure.

The next exercise gives yet another variation of the dominated conver-gence theorem.

5.4.13. Exercise: Let {f}>i and be sequences of measurablefunctions such that I fn I < gn d n. Let f and g be measurable functions suchthat lim = f(x) for a.e. x(µ) and Eli gn(x) = g(x) for a.e. x(µ). If

lim J gn dµ = J g dµ < +oon-+oo

show that

lim J fn dµ = J f dµ.

(Hint: Apply Fatou's lemma to On - fn) and On + fn).)

5.4.14. Exercise: Let {f}>o be a sequence in L1(X, S, µ). Show thatIf fd}>l converges to f foIdµ if If Ifn - foIdi}>1 converges to zero.

(Hint: Use exercise 5.4.13.)

The following is a variation of the dominated convergence theorem forfinite measure spaces.

5.4.15. Theorem (Bounded convergence): Let (X, S, µ) be a finitemeasure space and f, fl, f2,... be measurable functions. Suppose there existsM > 0 such that I f(x) I < M a.e. x(µ) and fn(x) -* f (x) a. e. x(µ). Thenf, fn E L1(X, S, µ) and

f dl-i = lim fnd/-t.n--+oo j

Proof: Let g(x) = M V x E X. Then g E L1(X, S, µ), and the claimfollows from theorem 5.4.9.

5.4.16. Exercise: Let (X, S, µ) be a finite measure space and {f}>i bea sequence in L1(µ) such that fn --+ f uniformly. Show that f E L1(µ) and

lim j/fn-fId=0.n-+oo

Can the condition of µ(X) < +oo be dropped?

152 5. Integration

5.4.17. Notes:

(1) The monotone convergence theorem and the dominated convergence the-orem (along with its variations and versions) are the most important theo-rems used for the interchange of integrals and limits.

(2) Simple function technique: This is an important technique (similarto the Q-algebra technique) used very often to prove results about integrableand nonnegative measurable functions. Suppose we want to show that acertain claim (*) holds for all integrable functions. Then technique is thefollowing:

(i) Show that (*) holds for nonnegative simple measurable functions.

(ii) Show that (*) holds for nonnegative measurable / integrable functionsby approximating them by nonnegative simple measurable functionsand using (i).

(iii) Show that (*) holds for integrable functions f, by using (ii) and thefact that for f E Ll, f = f + - f - and both f +, f - E Ll.

We give below an illustration of this technique (see also theorem 5.6.2).

5.4.18. Proposition: Let (X, S, µ) be a a-finite measure space and f EL1(X, S, µ) be nonnegative. For every E E S, let

dµ.v(E) .=JEF,

fThen v is a finite measure on S. Further, fg E L1(X, S, µ) for every g ELl (X, S, v), and

ffdv = ffgd.

Proof: Since f is nonnegative and f fdµ < +oo, it follows from proposition5.2.6(iii) that v is a finite measure on S. Let g E Ll(X, S, v). If g = xE, forE E S, then v(E) < +oo and

f XEdv = v(E) _E

fdµ = fXEfd.

Hence XE f is µ-integrable, and the required equality holds. By linearity ofthe integrals, it follows that the required claim holds for nonnegative simplemeasurable functions. If g E L1 (X, s, v) is nonnegative, by the definition ofmeasurability, there exists a sequence of nonnegative simple func-tions increasing to f . Since gn < f `d n, we have gn E L1 (v) by proposition5.4.3(i), and hence

f 9n dv = fgnfd.

5.5 Relation between the Lebesgue and Riemann integrals 153

Since {gf}>i is a sequence of nonnegative measurable functions increasingto g f, by the monotone convergence theorem,

I gdv = lim fgdvn-4oo

lim fgfdlc = fgfd.n-4oo

Thus the required claim holds for nonnegative functions g E L1 (X, s, v).For general g E L1(X, s, v), g = g+ - g- implies g+, g- E L1(X, s, v). Thusthe required claim holds for g+, g- and hence for g.

5.4.19. Exercise: Let (X, S) be a measurable space and f : X ][8 be

S-measurable. Prove the following:

(i) So {f'(E) I E E 13R } is the a- algebra of subsets of X, and So C S.

(ii) If ][8 ---> R is Borel measurable, i.e., 0-1(E) E BR V E E BR, then0 o f is an So-measurable function on X.

(iii) If 0 : X --> R is any So-measurable function, then there exists aBorel measurable function ¢ : IlS ) R such that 0 = ¢ o f .(Hint: Use the simple function technique and note that if 0 is a simpleSo-measurable function, then En

1 for some positiven

integer n, ai E R for each i, and Ei E BR. Thus b _ (>1 ai XEz) o f . )i=1

5.4.20. Exercise: Let be a decreasing sequence of nonnegativefunctions in L1(µ) such that fn (x) --+ f(x). Show that

lim fffd=O if f (x) = 0 a. e. x(µ).

5.4.21. Exercise: Let µ, v be as in proposition 5.4.18. Let Sv denote theQ-algebra of all v*-measurable subsets of X. Prove the following:

(i) S C Sam,(Hint: Use the theorem 3.8.4.)

(ii) There exist examples such that S is a proper subclass of S. Showthat µ*{xEXIf(x)=0}=0.

5.5. The Lebesgue integral and its relation withthe Riemann integral

In this section we analyze the integral, as constructed in the previous section,for the particular situation when X = R, S = L (the a-algebra of Lebesguemeasurable sets) and µ = A, the Lebesgue measure. The space L1(JR, L, A),

154 5. Integration

also denoted by L1(][8) or Ll (A), is called the space of Lebesgue integrablefunctions on R, and f f dA is called the Lebesgue integral of f. For anyset EEG, we write Ll (E) for the space of integrable functions on themeasure space (E, G fl E, A), where A is restricted to c fl E. In the specialcase when E = [a, b], we would like to show that the new notion of integralfor f E Ll [a, b] indeed extends the notion of Riemann integral. To be precise,we have the following theorem:

5.5.1. Theorem: Let f [a, b] -> ][8 be a Riemann integrable function.Then fELi[a,b] and

f fb

J f dA =

Proof: Since f is bounded and A([a, b]) < +oo, by exercise 5.4.4 f E L1 [a, b]if f is measurable. To show that f is measurable and f f dA = fb f(x)dx,we go back to the motivation for the Lebesgue integral (as in section 2.1).Since f is Riemann integrable, using theorem 1.1.5, we can choose a se-quence {P}>1 of partitions of [a, b] such that b n, Pn+1 is a refinement ofPn, JjPnjj ) 0 as n > oo and

fb

f(x)dx = lim L(Pn, f) = lim U(Pn, f ).n-,oo n-,oo

Let P n = fa=xo < xl < < xkn =b}. and for every 1 1, and x c [a, b], letn

XFnW - mZ X[Xz-11 Xi) Wi=1

and

Then

n

4bn (x) : = E Mi x[Xi_ 1 )Xi) (X)

V n we have XYn, Pn c Ll ([a, b]) with

f `I'nda = L(Pn> f) and f PndA = U(Pn, f )

Further,

T n (X) _< f W _< 'D n W

for x c (a, b). Since b n, Pn+l is a refinement of Pte, it is easy to see thatf XFnln>l is an increasing sequence and {n}n>i is a decreasing sequence.

Thus the sequence { 'n }n> 1 is a sequence of nonnegative simple mea-surable functions, and, by Fatou's lemma (5.3.18),

I lim inf (4n - TO dAn--oo

lim inf (fin - 'Fn) dAn--oo

lim inf 4ndA - lim sup OndAn-- oo n-- o0

lim inf U(Pn, f) - lim sup L(Pn, f )n--oo

0.

n--oo

Thus by proposition 5.3.3

lim inf 41X,(x) = limsup Tn(x) for a.e. x(A).

Since Tn(x) < 1(x) < 4N(x) for x c (a, b) and both {(x)}> 1 andf Tn (X) In> 1 , being bounded monotone sequences, are convergent, we have

f (x) = lim 4N(x) = lim Tn(x) for a.e. x(A).

Hence f is measurable by exercise 5.3.29. Further, since f i, we get

lim f WndA = f f dA.n--oo

Hence

fb

f(x)dx = lim L(Pn, f) = lim l/J ndA = JIf dA.

This proves the theorem.

5.5.2. Remark: In fact, the proof of the above theorem includes a proof ofthe following: If f E R[a, b], then f is continuous a.e. x(A). This is becausef (x) = lim Tn(x) = lim 4N (x) a.e. x(A). Thus if we put

E {x c [a, b] I 1(x) - nmoo (x) - 1 0 Tn(x) }>

then E is a Lebesgue measurable set and A([a, b] \ E) = 0. For x E E, givenan arbitrary e > 0, we can choose no such that

ono (x) - ono (x) < c. (5.9)

Further, if x is not a point in any partition Pn, then we can choose S >0 such that whenever y E [a, b] and Ix - yj < b, then y belongs to thesame subinterval of the partition Pno to which x belongs. Thus by (5.9),If (x) - f (y) I < E, showing that x is a point of continuity of f . Thus the setof discontinuity points of f forms a subset of ([a, b]) \ E) U P, where P isthe set of partition points of Pn, n = 1, 2 , .... Hence f is continuous almost

156 5. Integration

everywhere. We had proved this and its converse in chapter 1. Anotherproof of the converse is given in the next exercise.

5.5.3. Exercise: Let f [a, b] R be bounded and continuous fora. e. x(A).

(i) Let {P71}71>1 be any sequence of partitions of [a, b] such that each Pn+lis a refinement of Pn and I I Pn I I -) 0 as n -> oo. Let fin, Tn be asconstructed in theorem 5.5.1. Let x (E (a, b) be a point of continuityof f . Show that

lien 4)n (x) = f (x) = lim Tn (x).7t-,oo

(ii) Using (i) and the dominated convergence theorem, deduce that f EL 1([a, b]) and

ffd= lim 4)ndtt =n->oo

lim [Wnd.n->oo

Show that f E R[a, b] andbf

ff(x)dx.J f d=

5.5.4. Exercise: Let f : [0, 1] ---> [0, oo) be Riemann integrable on [e, 1]for all E > U. Show that f E L1 [0, 1] if limEyo fEl f (x)dx exists, and in thatcase

i[f(x)d(x) = E o f f(x)dx.

5.5.5. Exercise: Let f (x) = 1/xP if 0 < x < 1, and f (0) = 0. Findnecessary and sufficient condition on p such that f E L1[0,1] . Computef 16 f (x)dA(x) in that case.(Hint: Use exercise 5.5.4.)

5.5.6. Exercise (Mean value property): Let f [a, b] ) ][8 be acontinuous function and let E C [a, b], E E G, be such that A(E) > 0. Showthat there exists a real number a such that

EJ(x)a(x) = cea(E).

5.5.7. Exercise: Let f E L1(I[8), and let g : ][8 -) ]I8 be a measurablefunction such that a < g(x) < (j for a.e. x(A). Show that fg E L1(][8) andthere exists ry E [a, Q] such that

f fgdT = 7f .fIdA.


5.5.8. Exercise: Let f E L1(Il8, G, A) and let a E R be fixed. Define

F(x) :=ija,xI f (t) do(t) for x > a,

1 fx,ai f(t)dA(t) for x > a.

Show that F is continuous.(Hint: Without loss of generality take f > 0 and show that F is continuousfrom the left and right. In fact, using proposition 5.4.3 and proposition5.4.6, you can deduce that F is actually uniformly continuous.)

5.5.9. Exercise: Let f E L1(][8, G, A) and let c E R be a point of continuityof f . Show that

lien n f (x) dA (x) = f (c).n---'O° Cl C + l/n]

5.5.10. Note: We showed in theorem 5.5.1 that for a Riemann integrablefunction f : [a, b] -k R, f is also Lebesgue integrable and the two integralsare equal. A similar result can be proved for functions f : [a, b] --* R whichare Riemann-Stieltjes integrable with respect to F : [a, b] - ) R, where F ismonotonically increasing and right continuous. One can show that such afunction f is £F n [a, b], measurable and the Riemann-Stieltjes integral off isthe same as the integral f fd,ip, where £F is the a-algebra of MF -measurablesets.

5.5.11. Exercise (Arzela's theorem) : Prove theorem 1.4.6 using theo-rem 5.4.9 and theorem 5.5.1.

5.5.12. Exercise:

(i) Let f : [a, b] - R be any constant function. Show that f E L1 [a, b].

(ii) Let f [a, b] - ][8 be any bounded measurable function. Show thatf E L1[a,b].

(iii) Let f : [a, b] -j R be any continuous function. Show that f E L1 [a, b].

5.5.13. Exercise: Let f c L1(I[8) be such that fIt f dA = 0 for every com-pact set K C R. Show that f (x) = 0 for a. e. x(A).

5.5.14. Exercise: Let {f}>i be a decreasing sequence of nonnegativefunctions in C(a, b) and let fi E L1 (a, b) . If E---Jo 1(-1)1-1 --If,, e C (a, b), show

158 5. Integration

that0-0

f b ((_i)n-1f(X))n=1

dx =

00

n=1

(Hint: For every n, Ek=1 fk < f E L1 (a, b)).

fn (x)dx) ) .

5.5.15. Exercise: Give examples to show that analogues of the monotoneconvergence theorem and the dominated convergence theorem do not holdfor the Riemann integral.

5.5.16. Exercise: Let f E L1(I[8) and, d t E [0, oo),

g(t) = sup {J I+ y) - f(x)I dµ(x) - t < y < t } .

l J

Show that g is continuous at t = 0.(Hint: Use the simple function technique).

5.6. L1 [a, b] as completion of R[a, b]

Recall that, in section 1.4, we saw that R[a, b] is not a complete metricspace under the d(f, g) := fa f(x) - j(x)ldx, for f, g E R[a, b].Since every metric space has a completion, the question arises: what is thecompletion of R[a, b] under this metric? The reader might have seen theabstract construction of the completion of a metric space. We shall showthat for R[a, b] this completion is nothing but L1 [a, b]. Thus Ll [a, b] is theconcrete realization of the completion of R[a, b]. We shall first show thatLl [a, b] is a complete metric space, and then show that R[a, b] is dense inLi [a, b].

For f, g E Ll [a, b] we say f is equivalent to g, and write f N g, ifthe set {x E [a, b] I f(x) :A g(x)} has Lebesgue measure zero. It is easy tosee that the relation N is an equivalence relation on Ll[a,b]. We denote theset of equivalence classes again by Ll [a, b]. In other words, we identify twofunctions f, g with each other if f N g. Thus for g, f E Ll [a, b], f = g iff(x) = g(x) for a.e. x(A), as functions. For f E L1[a, b], we define

Ilf 11, := f I f (x) I dA (x).

Clearly Ill Iii is well-defined, and it is easy to check that the function fIIfIIi i f E L1 [a, b], has the following properties:

(i) 1 > 0 d f E L1 [a, b].(ii) l.

(iii) III1 = JaI If IIi b a E ]I8 and f E L1 [a, b].

5.6 Ll [a, b] as completion of R[a, b] 159

The function 11 ' Ill is called a norm on L1 [a, b]. (Geometrically it is thedistance of the `vector' f E Ll [a, b] from the `vector' 0 E L1 [a, b].) Forf, g E Ll [a, b], if we define

d(f,g) := Ilf -glll,then d is a metric on L1 [a, b]. This was the reason that it was called theL1-metric in section 1.4. The most important property of this metric isgiven by the next theorem.

5.6.1. Theorem (Riesz-Fischer): Ll [a, b] is a complete metric space inthe Ll -metric.

Proof: Let be a Cauchy sequence in L1 [a, b]. We have to showthat there exists some f E L1 [a, b] such that 11 fn - f 11 1 --+ 0 as n -> oo.Since is Cauchy, it is enough to find a subsequence {fflk }n>1 andf E L1 [a, b] such that llfn, - f 11 1 --+ 0 as k --+ oo. Let us first try to finda subsequence {fflk }k>1 and f E Ll [a, b] such that fn,. (x) --+ f (x) for a.e.x(A), as k --+ oo. Since for every nk,

k-1

fni (X) + : [fnj+l W - fnj (X)lj=1

we should find {fnk such that00

fn 1 W + L (fnj+ 1 (X) - fnj (X)) < + 00j=j

For this we note that, by corollary 5.4.10, the function00

f fix) f(x) + : Ynj + 1 Wj=1

will be defined for a.e. x(A) with the property that f E Ll [a, b] and

I f (x) dA (x) = f I1(x)dA(x) + [fnj+ 1 (x) - fnj (x) ] dA (x),

provided00

- .fns < +oo. (5.10)j=1

Thus to prove the required claim, we want to construct a subsequenceof the sequence {I}>i such that (5.10) holds. This can be

160 5. Integration

achieved using the fact that {f}>i is Cauchy as follows: given e = 1/2,we choose nl such that

Mfm-f1Ik < 1/2, d n > nl.Suppose nl < n2< ... < nj_1 have been chosen. We choose nj > nj_1 suchthat

Then by induction we have the sequence {fflk}k>i, and00 00

1

2j< +00.fni + f?li+l ?Ij fni +

j=lHence by corollary 5.4.10, we will have a subsequenceand f c L 1 [a, b] such that

00

(x) f, t l (x) + E (f+1 (x) - f?-,; (x)) for a. e. x(A)j=1

exists. Further, f E L1 [a, b] and

f (x)dl\(x) fn, (x)dA(x) + Ynj+ 1 (x) -- fnj (x)) dA (x).

But then, for a.e. x(A),00

f (X) - f?lj (X) = E Unk + 1 (X) - fnk (X))

j=1

Of {fn}n>i

Thus for every j > 1,DO DO

1/2k = 1/2j.If - f??M1 < IIf+i - fnk Mi=,j +1

Hence f - f,,, I 1, 0 as j )oo.

=j+1

We pi-o-\, -e next that C [a, b], the space of continuous function on [a, b] ,

(and hence 7Z[a, b]) is dense in L1 [a, b] .

5.6.2. Theorem: The space C[a, b] is a dense subset of L1 [a, b].

Proof: First note that if g E C [a, b], then g is bounded and by exercise 5.5.12(iii). it is a measurable function. Hence g E L1 [a, b] by exercise 5.5.12. ThusC[a, b] C L1 [a, b]. To show that C[a, b] is dense in L1 [a, b] we use the `simplefunction technique' as outlined in note 5.4.17(ii).

Let E L 1 [a, b]. Given E > 0, we have to find a function g c C [a, b] suchthat 11 f - g 11 1 < F. Since for f E L1 [a, b] we have f = f + - f - and bothf -i- , f -- E L 1 [a, b], it is enough to prove the theorem for f c L 1 [a, b] with

5.6 L1 [a, b] as completion of R[a, b] 161

f > 0. So we may assume f is nonnegative measurable and f f dA < boo.By definition, we can choose a sequence {Sn}n>i of nonnegative simple mea-surable functions such that {Sn}n>i increases to f and

urn sndA = f f dA.n->oo

Note that each sn E L1 [a, b], and we can choose no such that IIs,,,, - fMi <(This shows that simple measurable functions in L1 [a, b] are dense in it.)Thus it is enough to prove the theorem for f E L 1 [a, b], a nonnegative simplemeasurable function. Since any such function is of the form E"I aiXA, ,

where ai c R, Ai E £ with U=1 Ai = [a, b] and Ai n Aj _ 0 for i j, it isenough to prove the theorem for f = XA for A E £ with A C [a, b]. Now bytheorem 4.2.1, we can choose a set F C [a, b] such that F is a finite disjointunion of intervals and ,\(A A F) < E. Thus

IIXA XFIII = A(A A F) < E.

Let F = U=1 Ii, where the Ii are intervals with Ii c [a, b] and II n I,for i L j. Then

m

XF = LXi2i=1

To complete the proof, we only have to show that given I C [a, b], thereexists a continuous function g on [a, b] such that II XI - g I I 1 < E. This is easy.For example, if I = (c, d), a < c < d < b, consider g having the graph asgiven in Figure 16 below:

Figure 16: The function g

Then I I XI - g I I 1= 1 /n. Thus for n sufficiently large, I I X g I I 1<

Since C[a, b] C R[a, b], theorems 5.6.1 and 5.6.2 together prove the fol-lowing theorem:

162 5. Integration

5.6.3. Theorem: Ll [a, b] is the completion of R[a, b].

5.6.4. Notes:

(i) In the proof of theorem 5.6.1, we didn't use anywhere the fact that thefunctions are defined on an interval. One can define 11 ' Ii for functionsdefined on (E, ,C fl E, A), where E E ,C is arbitrary, and the proof of theorem5.6.1 will show that Ll (E) := Ll (E, G n E, A) is also a complete metricspace under the metric

I:=E

I f (x) - g (x) I dA (x).

A closer look will show that the metric d makes sense on L1(X, S, µ), where(X, S, µ) is any complete measure space and we identify functions whichagree for a. e. (µ). Further, theorem 5.6.1 also remains true for Ll (X, S, µ).

(ii) Parts of the proof of theorem 5.6.2 also exhibit the following facts whichare of independent interest. Let f E Ll [a, b] and e > 0 be given. Then thereexists a simple function 4D such that I I f - 4D I 1 1 < e and a step function h suchthat I If - hII l < E. Thus simple functions in L1 [a, b] are dense in it.

5.6.5. Exercise: Let f c Ll (][8) and let e > 0 be given. Prove thefollowing:

(i) There exists a positive integer n such that

Ilf - X[-n,n] f Ill < 6-

(ii) There exists a continuous function g on Il8 such that g is zero outsidesome finite interval and Ifx[_,1 - 911i < E.

(iii) For f : II8 -> Il8, let

supp (f) := closure fxGR I f(x) # 0}.

The set supp(f) is called the support of f and is the smallest closedsubset of I[8 outside which f vanishes. Let the space of continuousfunctions on ][8 be denoted by C(R) and let

C,(Il8) := ff : Il8 ) ][8 I f E C(][8) and supp(f) is compact}.

Then C, (II8) is a dense subset of L1 (I[8).

5.6.6. Exercise: Let f E L1(I[8) and f(x) := f (-x) V x E R. Prove thefollowing:

(i) ff(x)dA(x) = ff(x)dA(x).(ii) fg E L1(Il8) for any bounded measurable function g : ][8 -- R.

5.7 Another dense subspace of L1 [a, b] 163

Dense subspaces of L1(][8) are very useful in proving results about L1-functions. As an application of exercise 5.6.5 we have the following:

5.6.7. Exercise: Let f E L1(][8) and for every h, k E 1i8, let

fh(x) := f (x + h) and O(x) := f (kx + h), x E R.


(i) fh, 0 E L1(II8) with

f O(x)dA(x) = Jkl f f (x)dA(x) and J f11(x)dA(x) = ff(x)dA(x).

lim(ii) For every h E 1[8, fh E Ll (][8) and fh - f = 0(Hint : Use exercise 5.6.5 to get a function g E C,(II8) with II9 - f <e, and note that llh -9111 < 2(b-a+ 1) when supp(g) C [a, b] andJhJ is sufficiently small.)

(iii) The function h H llfhMl is continuous.

5.6.8. Exercise (Riemann-Lebesgue lemma): Let f E L1(II8), and letg : ][8 ) lI8 be any bounded measurable function such that g(x+p) -g(x) _0 for every x E ][8 and p E 1[8 fixed. Show that d t : 0,

ff(x)g(tx)dx < M JJfp/t - f 111)

where M = sup lg(x)1/2. Hence deduce (using exercise 5.6.7) that

lim Jf (x)g(tx)dx = 0.

ltl-goo

(In the special case when f E Ll [0, 27r] and g(x) = cos x or sin x, we have

Eli

27r

f(x) cos nxdx = 0 = lim J 2 f(x) sin nxdx.20 0

This finds applications in the theory of Fourier series.)

5.7. Another dense subspace of Ll [a, b]

While proving theorem 5.6.2 we also showed that, given f E Ll [a, b], f canbe approximated by a function XF, where F is a finite disjoint union ofintervals in [a, b]. We then approximated, for every interval I C [a, b], thefunction XI, by a polygonal continuous function to get an approximationof f by continuous functions in the Ll-metric. One can ask the question:If I C [a, b] is an interval and e > 0 is given, can we find a function 4D on[a, b] which is infinitely differentiable and is such that 11x1 - 4D ll < e? Ifthis is possible, then we can construct an approximation of f E Ll [a, b] by

164 5. Integration

infinitely differentiable functions. To do this we need functions (D with thefollowing properties: given c, d E JR and c < d,

(i)

(x) < 1 and (D E C' (R).

Here, C°° (III) denotes the set of infinitely differentiable functions on R.In other words, we want functions which interpolate between the value

0 at c and the value 1 at d in a smooth (infinitely differentiable) way. Wegive below the construction of such functions.

5.7.1. Lemma: For any real number a > 0, let

(X)exp

'Xif X > 01

0 ifx<0,

and

exp if X < 0)(X)

0 ifx>0.

Then the functions (D, T E C' (I[8).

Proof: We first note that (D is a continuous function and it is enough toprove the lemma for a = 1. For x > 0, (D'(x) = exp (- ] / x,) /x2 . In fact.do >0andx>0,

(D (n) =Pte, (1/x) exp (-1/x),

where P, is a polynomial of degree 2n. This can easily be proved by induc-tion. We next show that (D(n)(0) = 0 dn. Clearly, if n = 0, it is true.

y

------------------------------

x

Figure 17 : The function (x)

5.7 Another dense subspace of L1 [a, b] 165

Y

1

0 x

Figure 18 : The function T (x)

Assume (D (0) = 0. Then for x > 0,

I /xlimx-,0

x=

tl [tP_i(t)exp(-t)J = 0.

The last equality holds because for t > 0 and t large, exp(t) dominates t'for any n > 0. This proves that (D E C°°(][8). Similarly, T E C°°(Il8).

5.7.2. Corollary: Let c and d be real numbers such that c < d. Let

exp(-1/(x - c)(d - x)) if c < x < d,fd(x)0 otherwise.

Then fc,d E COO (R).

Figure 19 : The function f c,d

f(x) c))if x <

,

fax) :_ x))if x > d

166 5. Integration

By lemma 5.7.1, both f, and Id are in C°°(][8), and f,,d(x) = fc(x)fd(x) forevery x E R. Thus f,,d E C°°(Il8).

5.7.3. Corollary: Let c < d and fc,d be as in corollary 5.7.2. Let

fx

77c,d(x) = f,d(t)dt, x E R.

Then 77c,d E C°°(R), gc,d(X) = 0 for x < c and 'qc,d(x) > 0 V x > c. Let6c,d = llc,d/M, where M := f d

c fc,d(t)dt. Then 6c,d E C°°(R), 0 < 6c,d < 1,with 6c,d(x) = 0 for x < c and bc,d(x) = 1 for x > d.

c d x

Figure 20 : The function 6c,d

Proof: Clearly, 77c,d E C°° (III) and for x < c, 'qcd(x) = 0. For x > c, sincef,d(t) > 0 for every t,

JXTlc,d(x) := f,d(t)dt > 0.

Let M := fcd f,d(t)dt and 6c,d ?7c,dlM. Then the function 6c,d clearly has-the required properties.

5.7.4. Corollary: For real numbers c and d with c < d there exists afunction 7rc,d E Coo (R) such that 0 < 7rc,d < 1, 7rc,d(x) = 0 for x > d, and7rc,d(x) = 1 for x < c.

c a x

Figure 21 : The function 7rc,d

Proof: Let 7c,d := 1 - 6c,d. Then 7rc,d has the required properties.

5.7 Another dense subspace of Ll [a, b] 167

5.7.5. Corollary: Let a 0 be such that a + e < b - E. Thenthere exists a function 4PE E C°°(R) with the following properties:

(i) 0 < 4DE(x) < 1 V x E R.

(ii) 4PE(x) = 1 V x E (a + e, b - e)

(iii) 4PE(x) = 0 V x ¢ (a, b).

Proof: Let c : = a + E and d : = b - E. Define 4PE : Ilk ----> R by

0 if x V (a, 6),

(X)I if x E (C) d)

kc if x E (a, c)7rd,b if x E (d) b),

a a+E b-E

Figure 22 : The function 4PE

where Sa,c and 7rb,d are as given by corollaries 5.7.3 and 5.7.4. Then 4PE EC°° (Ilk) and has the required properties.

5.7.6. Theorem: The space C°°[a, b] of infinitely differentiable functionson [a, b] is a dense subset of Ll [a, 6].

Proof: As remarked in the beginning of the section, we only have to provethat V E > 0 and (c, d) C (a, b), 3 E C°° [a, b] such that

II xcd -0$ Iii <E.

Choose an integer n > 1 such that

c + 2/n < d - 2/n and 4/n< E.

Now consider 4D E C°O(Il8), as given by corollary 5.7.5 with the proper-ties 0 < fi(x) < 1, (x) = 1 for x E (c + 2/n, d - 2/n) and (x) = 0for x ¢ (c + 1/n) d - 1/n) . Consider 4D , the restriction of 4D to [a, 6]. Then0$ E C°° [a, b] and

c+ d-1X c d - (i 1 = f

+oP (x) dA(x) + oD(x) dA(x) < 4/n < E.

d- ?)) 11

168 5. Integration

5.7.7. Exercise: Let C°° (Il8) :_ f f E C°° (IlS) I supp(f) is compact}. Showthat C,00(R) is dense in L1(IlS).

(Hint: lim 11 f -f Ill = 0.)n-+oo

5.8. Improper Riemann integral and its relationwith the Lebesgue integral

Let us recall that the Riemann integral is defined only for bounded functionsover bounded intervals. The notion of Riemann integral for functions whichare either not bounded or are defined over unbounded intervals can be de-veloped as follows. Let I = [a, oo) be an infinite interval and let f : I --> Il8be a bounded function. Let {b}>i be any sequence of real numbers suchthat bn+l > bn > a for every n and limn-*oo bn = oo. Let In :_ [a, bn]. Fur-ther, suppose that f E R[In] V n and that the sequence {fa " f(x)dx}>iis convergent, say to a. If a is independent of the sequence {bn}n>1, thenwe can define the integral of f over I to be a, denoted by fl f (x)dx (alsodenoted by f °° f (x)dx), and call it the improper Riemann integral of fover the interval I. Let us look at some examples.

5.8.1. Example: Consider the function f : [0, oc) - R defined by

f (x) := (-1)Th/ri if n - 1 < x < n, n = 1, 2,... .

Clearly, f is bounded and is Riemann integrable on every closed boundedsubinterval of [0, oc). Let us take I , = [0, m], m = 1, 2, ....Then

ff,

and the limit

f (x)dx = E (_1)n/n,n=i

00

lim fIn

(-1)"/n = - In 2n In n=1

exists. In fact, it is easy to see that

Jim J f(x)dx=-Jn2.p

Thus the improper Riemann integral fo f (x)dx exists. However, it is easyto see that f ¢ Ll [0, oc). For example, if it were so, then by the dominatedconvergence theorem, since X[o nj f E L1 [0, oc) V n and IXto,n] f I < If 1, we

will have00

: 1/n = +oo.f I= lim f IX[p,nlf IdA(x) = 1n=1

5.8 Improper Riemann integral 169

The second situation can arise when we have a function f [a, b] -* ]I8which is not bounded, but f is bounded and is Riemann integrable on [c, b]for every c > a and lim,ya f,' f(x)dx exists. In that case, limcya f b f (x)dxis called the improper Riemann integral of f over [a, b] and is denotedby Ja f(x)dx.

5.8.2. Example: Let f : (0, 1] --> ][8 be defined by f (x) = 1/vrx-. Then,for every c > 0, f c 1Z[c, 1] and

dx = 2 (1 _ C1/2).

ThusC vX

lim 1 dx = 2.JHence the improper integral of f on [0, 1] exists and

J1

dx = 2.Y

The improper Riemann integrals in other situations, for example whenf is defined on say (-oo, b] or when f is defined on [a, b) and is not boundedbut is Riemann integrable on [a, c] for every c < b, can be defined similarly.We leave the following simple properties of improper Riemann integrals forthe reader to verify.

5.8.3. Exercise:

(i) Let j'° fidx exist for i = 1, 2. Show that d E R, j'°(af, f ,3f2)dxexists and is equal to a J'a° f,dx f,6 fa f2dx.

(ii) Let f : [a, oo) R be such that f c R[c, d] V a < c < d < oo. Showthat the improper Riemann integral fa'00 f (x)dx exists if d e > 0, 3 N suchthat for every c, d > N, f' d f(x)dx < e.

(iii) Using (ii), show that for a function f : [a, oo) R, if f c R[a, b] V b >a and if fa" If (x)ldx < +oo (in which case one says that the improperintegral exists absolutely), then fa" f (x)dx also exists.

(iv) Using (ii) above, show that the improper Riemann integral f00 sin x/xdxexists (see also theorem 5.9.1), but fO(OO Isinx/xldx = +oo, i.e., the improperintegral of sin x/x does not exist absolutely.

As we saw in example 5.8.2 and exercise 5.8.3(iv), for a function f its im-proper Riemann integral can exist, even though it is not Lebesgue integrable.

170 5. Integration

Conversely, even if a function is Lebesgue integrable on bounded intervals,its improper Riemann integrable need not exist. The exact relation is givenby the following theorem:

5.8.4. Theorem: Let f [a, oo) ---> I[8 be such that f c R[I] for everyclosed bounded interval I C [a, oo). Then f is Lebesgue integrable iff theimproper Riemann integral f °O If (x) I dx exists.

Proof: Let f be Lebesgue integrable. To show that the improper integralfacc I exists, let be any sequence of real numbers such thatnl o an = 00. Let f, = XjaQnj f . Then each fn E R[a,an] with I fThI C I f I

and fn(x) -> f (x) for every x c [a, oo). Thus by the dominated convergencetheorem

lim Jn--+oo

On the other hand, by theorem 5.5.1,

f I

itIf IdA(x).

fan

no.

If (x) Idx.

Thus/'an f

lim Ifn(x)ldx I f (x) I dA (x).

Hence fa" f(x)ldx exists. Conversely, suppose limb, fa Iexists.Let {an}n>1 be any increasing sequence such that lim an = oo. Let

n--+oo

gn := X[a,an]Ifl,n= 1,2,... .

Then is an increasing sequence of nonnegative measurable functionsand, by the monotone convergence theorem,

fd(x)= lfa,oo) fa,an]

As gn c R [a, an], again by theorem 5.5.1,an

I = Ifldx.[a, an] in,

Since f a" Ifldx < +oo, we have

If I dA(x) lim If IdA(x)n-+oo fr,f a a,,,

an 00

lim Ifldx= IIfldx<+oo.

n--+oo a a

5.8 Improper Riemann integral 171

5.8.5. Exercise:

(i) Let f (t) = 1/1 + t2, t > 0. Show that the improper Riemann integralf °O f (t)dt exists. Compute its value also. Does f E L1(0, oo)?

(ii) Let1 + (I + t)e-I

> 0.t l+t t

Show that g c L1 [0, oo).

(Hint: (1+t)e_t -- 0 as t -- oo).

5.8.6. Remark: As we have seen in theorem 5.5.1, the theory of Lebesgueintegration extends the theory of Riemann integration and does not includethe theory of improper Riemann integration, as shown in example 5.8.1. Thereason is that Lebesgue integration is absolute integration, i.e., f is Lebesgueintegrable if If I is integrable. On the other hand, as seen in exercise 5.8.5,for f (x) = sin x/x, 0 < x < oo, the improper integral fo' f(x)dx exists, butthe improper integral fo If does not exist. Another way of observingthis difference is that for a Lebesgue integrable function f,

E

00

fdµ = E fE fd, (5.11)

whenever E = U=1 En, where E and the En's are measurable sets withEn n Em = 0, for n m. However, this property is not true for improperRiemann integration, even when E and all the En's are intervals. To seethis, recall that the series E

1(-1)Th/n is convergent but is not absolutely

convergent. Thus, given any a E III, a - In 2, we can have a rearrangementof El 1 (-1)/n such that the rearranged series converges to a. Let thecorresponding rearrangement of the intervals {[n - 1, n) }n>, be denoted by{E}>1. Let

f (x) :_ (-1)n /n for x E [n-1,n),n= 1, 2, ... .

We saw in example 5.8.1 that f °O f (x)dx exists. However,00

-Qn2 = J[o, f(x)dx = E fE f(x)dx = a,n=1

even though [0, oo) = U=1 En, where each En is an interval and 0

for n = m. This example shows that any theory of integration in which ev-ery improper Riemann integrable function is also integrable, cannot have theproperty fE f = E'l fEn f, whenever E = U=1 En where En's are pair-wise disjoint measurable sets. Thus one cannot hope to have a theory ofintegration which extends both the theory of improper Riemann integrationand Lebesgue integration. Suppose we drop the demand of the property

172 5. Integration

in (5.11) and ask the question: can one develop a theory of integration inwhich every bounded function on finite intervals is integrable and every im-proper Riemann integrable function is also integrable? Such theories arepossible: Perron integral, Denjoy integral, Kurzweil-Henstock in-tegral, gauge integral, generalized Riemann integral. All of theseintegrals are equivalent. For more details see Bartle [3], DePree and Swartz[11], McLeod [25] and Saks [35].

5.9. Calculation of some improper Riemannintegrals

As applications of the Lebesgue dominated convergence theorem and theo-rem 5.5.1 we calculate the improper integral fI Six x dx, which exists (exer-cise 5.8.3). Normally, the calculation of this integral is done as an applicationof the residue theorem in complex analysis.

1000

sin x5.9.1. Theorem: dx = 7r/2.x

Proof: Since the required integral exists, by the dominated convergencetheorem we have

fo

Let

Then

f°O sin x dx = lim f(O).x n-4oo

To compute f(O), we note that d t > 0 and every x > 0, t (e_xt sin x

exists andd (_tsinxdt x

= le-xtsinxl < e-xt

Since e-xt E L1 [0, n] for every t > 0, f(t) is differentiable V t > 0, bytheorem 5.4.12, and

rnf(t) J e-xt sin xdx.

0

It is easy to see that fn(t) E R[0, n] and hence, by the fundamental theoremof calculus,

°O sin x n sin xdx = lim dx.x n-*oo o x

nxt sin x

fn(t) := -e dx `d t > O and n = 1, 2, .. .lox

fn (t) dt = fn (n) - fn (0).0

5.9 Calculation of some improper Riemann integrals 173

Also for every nn

I C n fThus, lim fn(n) = 0, and we have

e-nacdx < 1

nV n.

1nJimf(0) =

Jim- [f(t)dtl

n---+oo n---o0 0

Integrating twice by parts, we get

fn (t) = f n e-xt sin xdx - 1 - e-nt (t sin n + cos n)

J0 1 + t2

Thus nj (t) fn(t) converges to 1/(1 + t2) as n -* oo, and

X[p,n] (t)f(t) I C 1 + 1

e-t+(t + t) 9(t)

By exercise 5.8.5, g E Li [0, oo). Thus, using the dominated convergencetheorem, we have

Jim n 001

n->oo- fn(t)dt =

2dt = -/2.

0 1 + tHence

(°°sinx ndx =

Jimfn(0) =

Jimfn(t)dt = 7r/2.

0 x n->oo n---+ oo 0

5.9.2. Exercise: Let f (t, x) = xt-le-x, for every t > 0 and x > 0. Provethe following:

(i) Show that for all fixed t as a function of x, f E L1 [1, oo) and hencethe improper Riemann integral

00r(t) := ff(t,x)dxexits. The function r(t) is called the Gamma function.

(ii) Show that r(l) = 1 and I'(t + 1) = tI'(t) b t > 0.(iii) The function r(t) has derivatives of all orders, and

(n) (t) = fxt_1e_x(lnx)ndx.I'

Chapter 6

Fundamental theoremof calculus for theLebesgue integral

We recall that one of the aims of extending the notion of Riemann integralwas to make the fundamental theorem of the integral calculus valid, i.e.,to find the relation between the pair of functions f and F defined on aninterval [a, b] such that the relation

tF(t) ) ) = is f(x)dA(x). (6.1)s

holds for every a< s < t < b. To analyze the above relation, we fix f EL1 [a, b] and consider the function

fF(x) := f (t)dA (t)x E [a, b].

The function F is called the indefinite integral of f.Note that F and f satisfy the relation (6.1). We analyze the properties

of F in the next section.

6.1. Absolutely continuous functions

We had already shown in exercise 5.5.8 that F, the indefinite integral off E L1 [a, b], is a uniformly continuous function. In fact, it has the followingproperty which is stronger than uniform continuity:

175

176 6. Fundamental theorem of calculus

6.1.1. Theorem: For every e > 0, 3 S > 0 such that for mutually disjointsubintervals (ak, bk), k = 1, 2, ... n, of [a, b]

n n

1:(bk- ak) < 6 implies 1: I- F(ak) I < e.k=1 k=1

Proof: This follows from proposition 5.4.6(iii) with µ = A.

In view of the above theorem, we introduce the following definition:

6.1.2. Definition: Let g : [a, b] -) R. We say g is absolutely contin-uous if V E > 0, 3 6 > 0 such that for any finite collection of mutuallydisjoint subintervals (ai, bi),1 < i < n, of (a, b)

n n

(bi - ai) < 6 implies I g(bi) - g(ai) I < E.i=1 i=1

6.1.3. Example: In view of theorem 6.1.1, for f E L1 [a, b], the functionF(x) fa f (t)dt, x E [a, b], is absolutely continuous. The fact that essen-tially these are the only examples of absolutely continuous functions is theclaim of the fundamental theorem of calculus (theorem 6.3.6).

Let us analyze absolutely continuous functions. We keep in mind that inorder to prove the fundamental theorem of calculus, we have to analyze thedifferentiability of F(x). Before proceeding further, the reader is advised torecall functions of bounded variation (see appendix F).

6.1.4. Theorem: Let g : [a, b] --> ][8 be an absolutely continuous function.Then g has bounded variation.

Proof: By the absolute continuity of g, given E = 1, we can find a 6 > 0such that

n

Igbj_gajn <i=1

whenever (a2, bZ), i = 1, 2,... n, are pairwise disjoint subintervals of [a, b] with2 (b2 - ai) < S. But then for any subinterval [c, d] of [a, b] with (d-c) < b,

clearly Vd(g) < 1, where Vd(g) denotes the variation of g on the interval[c, d]. Let P = {a = xO < xl < ... < xk = b} be any partition of [a, b] suchthat JJPJJ < 6. Then

k

V b(g) = VVii 1(g) < k,2=1

i.e., the function g has bounded variation.

6.1. Absolutely continuous functions 177

6.1.5. Corollary: For f c Ll [a, b], let

F(x) :=J

f(t)dA(t), x E [a, b].a

Then F is a difference of two monotonically increasing functions on [a, b].

Proof: Since F is absolutely continuous, it is of bounded variation, andevery function of bounded variation is a difference of two monotonicallyincreasing functions (see Appendix F.2).

In view of the above corollary, one asks the question: if g : [a, b] -) lI8 isa monotonically increasing function, what kind of differentiability propertiescan g have? Recall that a monotone function can have only a countablenumber of points of discontinuity. Thus a monotone function is continuousa.e. (A). One wonders: Is a monotone function also differentiable a.e. (A)?The answer is yes, and is the content of Lebesgue's theorem (6.2.1), provedin the next section.

6.1.6. Exercise: Let f c Ll [a, b], and let F be the infinite integral of f .

Show thatb

a6(F) = f If

(i) Let g [a, b] - lI8 be a Lipschitz function, i.e., there exists someM > 0 such that

I9(x) - g(y) I < M Ix - yj d x, y c [a, b].

Show that g is absolutely continuous.

(ii) Let g : [a, b] ) R be differentiable with bounded derivative. Show thatg is Lipschitz and hence is an absolutely continuous function.

(iii) Let f, g : [a, b] ) I[8 be absolutely continuous functions. What can yousay about the functions f + g, f - g, a f , and f g, where a c I[8?

6.1.8. Exercise: Let f : [a, b] ) lI8 be an absolutely continuous function.For x c [a, b], and let h(x) := VQ (f ). Prove the following:

(i) The function h is absolutely continuous.

(ii) The functions h(x) and h(x) - f (x) are both (monotonically increas-ing) absolutely continuous. Hence f = h - (h - f) gives a repre-sentation of an absolutely continuous function as a difference of twomonotonically increasing absolutely continuous functions.

6.1.9. Proposition:Let g : [a, b] - ][8 be an absolutely continuous func-tion. Then the following hold:

(i) If N C [a, b] is a null set, then g(N) is also a null set.(ii) If E C [a, b] be Lebesgue measurable, then g(E) is also Lebesgue mea-

surable.

Proof: (i) Let E > 0 be given. Since g is absolutely continuous, we canchoose 6 > 0 such that whenever (ak, bk),1 < k < n, are disjoint subintervalsof [a, b], then

n n

1: (bk - ak) < S implies E 1- g(ak) < 6-k=1 k=1

Next, N being a null set, we can find intervals Ik_ [xk, yk], k > 1, such that00 00

N 9U yk] C [a, b] and >(Yk - Xk) < S.k=1 k=1

Since g is continuous, there exist points ak, bk E [xk, YkI such that g(ay) = Mkand g(bk) = Mk, where

mj _Then

Thus

Since V n,

we have

Hence

min{g(x) I x E [x, yd } and Mk := max{g(x) I x E [x, yd }.

9([xk,yk]) C [mj,Mj] = [g(ak),g(bk)].

00

00

g(N) C U 9[xk,ykI C [g(ak),g(bk)}.k=1 k=1

n n

1: Ibk- ak I 1: (Yk - Xk) < Sk=1 k=1

1: Jg(bk) - g(ak)l < 6-k=1

00 00

g(N) C U [g(ak),g(bk)] with E 1- 9(ak) < E)k=1 k=1

i.e., g(N) is a null set. This proves (i).To prove (ii), let E c [a, b] be Lebesgue measurable. Using theorem

4.2.2(v), we can find an FQ-set F and a null set N such that E = F U N.Then g(E) = g(F) U g(N). Since F C [a, b] is an FQ-set, it is a countableunion of closed subsets of [a, b], say Cn C. Then each Cn is a compact set

6.2 Differentiability of monotone functions 179

and, g being continuous, g(Cn) is compact. Hence g(F) = U=1 g(Cn) isLebesgue measurable. The set g(N) is Lebesgue measurable by (i). Thusg(E) = g(F) U g(N) is Lebesgue measurable.

6.1.10. Exercise: Let f (x) := x2 sin2(1/x) for x # 0, f (0) := 0, and letg(x) := VTx- for x > 0. Show that

(i) f is absolutely continuous.(Hint: Use exercise 6.1.7.)

(ii) g is absolutely continuous.

(Hint: g(x) = LX dt.)

(iii) g o f is not absolutely continuous.(Hint: Uo (g o f) = +oo.)

6.1.11. Exercise: Let g : [a, b] --> [c, d] be absolutely continuous andmonotone, and let f : [c, d] --> Il8 be absolutely continuous. Show that f o gis also absolutely continuous.

6.1.12. Exercise: Let g : [a, b] ) [c, d] be absolutely continuous. Letf c Ll [c, d] be bounded and F be the indefinite integral of f . Show thatF o g is also absolutely continuous.

6.1.13. Exercise: Let f (x) := xsin(1/x) for x E (0, 1], and f (0) := 0. Letg(x) := Vfx- for 0 < x < 1/2, and define g linearly on [1/2, 1]. Show thatboth f and g are not absolutely continuous.

6.1.14. Exercise: Show that a function f is absolutely continuous if bothf + and f - are absolutely continuous.

6.2. Differentiability of monotone functions

In this section all the a.e. statements are with respect to the Lebesguemeasure A on R. The aim of this section is to prove the following theorem:

6.2.1. Theorem (Lebesgue - Young): Let f : [a, b] --> Il8 be a monotonefunction. Then f is differentiable a. e.

The proof of the theorem is long, and we shall make preparations for it.Recall that, to analyze the continuity of a function at a point quantitatively,in chapter 1 we introduced the concept of oscillation of the function at apoint. Similarly, to analyze the differentiability of a function at a pointquantitatively, we consider what are called the Dini derivatives of f.

6.2.2. Definition: Let 4P : [a, b] -) R. Then for any c E [a, b), we define

lim nf (x) := sup { jc < h< c + S} I J> 0, c + 8 < b}

and

lim sup fi(x) := inf {sup{(h) I c < h < c + 8} 1b > 0, c + 8 < b} .

h1c

For any c c (a, b], we define

lim inf (x) := sup { inf I c - S < h < c} I (5> 0, a < c - b}

and

lim sup (x) := inf I supf(D(h) I c - S < h < c} 15 > 0, a < c - 51.hTc

The quantities lim inf (D (x), lim sup (D (x), lim inf (D (x) and lim sup (D (x) ex-hJ,c hJ,c hTc hTc

ist as extended real numbers and are called, respectively, the lower rightlimit, upper right limit, lower left limit and upper left limit of (Dat c.

6.2.3. Definition: Let f : [a, b] -) R. For any c E [a, b), we define

(D+f)(c) lim infIf(c + h) - f(c)]

(D+f)(c) := limi0upf (c + h) - f(c)1

hh

For c E (a, b], we define

(D_f)(c) limjo f[f(c+h) - f (c)1

J

and

(Df)(c) := limrOup[f(c + h - f

J

The extended real numbers (Df) (c), (Df) (c), (D_f) (c) and (Df) (c) arecalled, respectively, the lower right derivative, upper right derivative,lower left derivative and upper left derivative of f at c. (These arealso known as Dini derivatives of f at c.) If all of them are equal, we call itthe derivative of f at c and denote it by f(c). Note that in our definitionf(c) is an extended real number.

6.2.4. Examples:

(i) Let f (x) := x13, x E R. Then

D+.f(o) = D+f(o) = D-.f (o) = D f(o) = +oo.

(ii) Let f (x) := jxj, x E R. Then

(Df)(0) = (Df)(0) = 1 and (Df)(0) = (Df)(0) = -1.

(iii) Let

f (X)o if x E Q n [o, il,1 if

Here Q denotes the set of rationale. Clearly, for any x E [0, 1], if xthen

D+ f (x) = 0, D+ f (x) = +oo, D- f (x) = -oo and D- f (x) = 0.

If x E [0,11 but x ¢ Q, then

D+ f (x) = -oo, (Df)(x) = 0, (Df)(x) = 0 and (Df)(x) = oo.

(iv) The Lebesgue singular function: Let C denote the Cantor ternaryset as constructed in example 1.2.13. For x E C, let x = E° 1 an/3n denotethe ternary representation of x, i.e., the representation in base 3, where allthe ai's are either 0 or 2. Let

00 00

(D(x) := an/2n+1 whenever x E C and x = an/3n.n=1 n=1

Note that if y = EM 1 an /2n+1 then y E [0, 1] . Let Ik , 1 < k < 2n, denotethe open intervals removed at the nth stage in the construction of the Cantorset C. If II = (a, b), say, then

n-1 00 n-1a = >ak/3k + 2/3k and b = >ak/3k + 2/3n.

k=1 k=n+1 k=1

Thusn-1 00 n-1

ak/2k+i + 1/2 k ak/2k+i + 1/2n =fi(b).k=1 k=n+1 k=1

We define a function f : [0, 1] - [0, 1] as follows:

f(x) ._ r (x) if x E F,(a) if x E Ik :_ (a,b) for some n,1 < k < 2n-1

The function f is called the Lebesgue singular function.

Figure 23 : Part of the graph of f

(For a more geometric description of f, see exercise 6.2.5.) Note that fis constant on each interval Ik . If x, y E C with x < y and are not theend points of any Ik, then the ternary representations of x and y must beof the form x = E'l ai/3i and y = E', bi/3i, where for some n we haveai = bi, 1 < i < n, and 6n+1 = 2, an+l = 0. But then (x) < (D(y), andhence f (x) < f (y). Finally, if 0 < x < y < 1 with either x or y C, it iseasy to check that f (x) < f (y). Thus f is monotonically increasing on [0, 1].

Also f (C) = [0, 1]. Clearly, f is continuous on each Ik . To show that fis continuous everywhere, let x E C and let

Nn:= fy E [0, 111 IX - YJ < 1/3nj.

If y E C n Nn, then the first n terms in the ternary representation of x andy must be identical. But then f (x) and f (y) will have binary representationhaving identical first n terms, and hence If (x) - f (y) I < 1/2n. If y E Nom, \ C,then there exists z E C which is an end point of some 1k, and hence f (y) =f (z). Thus b y E Nn we have If (x) - f()I < 1/2n b n, proving thecontinuity of f everywhere.

Clearly b x E Ikn, f is differentiable at x and f '(x) = 0. Thus f '(x) _0 V x E (0, 1] \ C, i.e., f'(x) = 0 for a.e. x E [a, 6]. We show that f is notabsolutely continuous. For this, let us extend f to R by defining

f(x):=0 if x<0 and f(x):=1 if x>1.

Since C is a null set, given e > 0, we can find pairwise disjoint intervals{(ak,bk)}k>1 such that

00 00

C C U(ak, bk) and E(bk-ak) <6-k=1 k=1

Since 0, 1 E C, without loss of generality, let 0 E (al, bl) and 1 E (a2, b2)with al < 0 < bl < a3 < b3 < < a2 < 1 < b2. Since f is monotonicallyincreasing, for every n > 1 we have

n

1 = f(1)-f(0)=f(b2)-f(ai) ? [f(bk)-f(ak)].

Hence00

E [f (bk) - f (ak)] = J <- 1-

k=1

On the other hand, f(bk) - f(ak) 0 V k, for otherwiseThus

00

0 < E [f (ak)- f (bk)] < 1i

and we can find N such thatN

E [f (bk) - f (ak)] > J/2.k=1

f (x) = 0 V X.

This shows that f is not absolutely continuous on I[8, and hence not on [0, 1].

Finally, we compute (Df)(x), (D+f)(x), (Df)(x) and (D_f)(x) whenx E C. Let x E C be such that x is the left endpoint of one of the openintervals that had been removed from [0,1] to construct C. Then clearly(Df)(x) = (D+f)(x) = 0. Let us calculate (D_f)(x) at such a point.Since x is the left endpoint of some Ik , let

k+l

x aZ/3Z + 1/3ki=1

for some n,1 < k < 2n and ai E {0, 2} V i. Thus for all sufficiently small h,say,

1/3"''+1 < h < 1/3""', m > k, implies 1/2"'+1 < f (x) - f (x - h) < 1/2m.

Thus3m/2m+1 <

f(x)-f(x-h)< 3m+1/2m+2.

hHence (D_f)(x) = +oo = (D- f) (x). Similarly, one can check that if x isthe right endpoint of one of the open intervals removed, then

(Df)(x) = (Df)(x) = 0 and (Df)(x) = (D+f)(x) = +oo.

Finally, if x is not an endpoint of the open intervals which are removed,then x =1: 1 ak/3k, where ak E {0, 2} and ak 0 for infinitely manyk's. Let x,z :=E jk=1 ak /3c , n = 1, 2,.... Then { xn }n> 1 is a sequence of rightendpoints of open intervals removed, and {x}> 1 increases to x. Since

00

f (Xn) ak/2

we have

f (X) f (Xn)

00

ak/2k+1

k=n+100

E ak/3k

k=n+1

+2r1 1/3k

k Nk N3N-1/ 2N

where N is the smallest of all the integers k > n such that ak 0. SinceN-*oowhen n-*oo,we get

lim f W - f (Xn) = + 0().n->oo x - xn

Hence (Df)(x) = -boo. A similar argument shows that (Df)(x) = +oo.

6.2.5. Exercise: Let f be the Lebesgue singular function as defined inexample 6.2.4(iv). Show that

(i) f (x) _ (2k-1)/2Th if x E In) every n = 1, 2.... and 1 < k < 2n-1

(ii) Define fn : [0, 1] ) [0, 1] V n = 1, 2)... , continuously by

0 ifx=0,1 ifx=1,

(2k-1)/mifxeIj, 1<k<21-1,1<m<n,n 2'n-1

linearly on [0,11 \ U (U Ik)m=1 k=1

Show that {f}>1 converges uniformly to f.(iii) Using the fact that f (C.) = [0, 1] and proposition 6.1.9, deduce that

f is not absolutely continuous.

The next proposition follows easily from the definitions of the Diniderivatives.

6.2.6. Proposition: Let f : (a, b) ) R. Then

(i) DI f (x) > D+ f (x) and D- f (x) > D- f (x).

(ii) f is differentiable at x E (a, b) iff all the Dini derivatives exist, arefinite and are equal.

Proof: Exercise.

6.2.7. Exercise: Give an example of a function f (a, b) -- ][8 such thatf is discontinuous at xo E (a, b), but all the Dini derivatives exist and areequal.

6.2.8. Exercise: Let f : (a, b) -- R. Let S {x c (a, b) If is differentiableat x} and

A {x c (a, b) I D- f (x) < D+ f (x)},B {xE(a,b)D+f(x)<Df(x)},C fx c (a, b) I D+f (x) = +ool,D x EE (a, b) I D- f (x) = +ool.

Show that (a,b)\(AUBUCUD) C S.

We are now ready to prove theorem 6.2.1.

Proof of Theorem 6.2.1: Without loss of generality we may assume thatf is monotonically increasing, for otherwise we can consider -f. Let

E {x c (a, b) I D+f (x) > D- f (x)},F fx c (a, b) I D+f (x) < D-f (x)j,G fxC(a,b)I(f')(x)=+ooj.

To show that f is differentiable a.e., in view of proposition 6.2.6 it is enoughto show that the sets E, F and G are all null sets. We first show that Gis a null set. If x E G, then b',Q > 0, 3 arbitrarily small h > 0 such that[x, x + h] C (a, b) and

f (x + h) -f(x) > ate.Thus for ,Q > 0 fixed, the collection

{[x,x+hJxEC and f(x+h)-f(x)>3h,h>0}is a Vitali covering of G. By theorem 4.4.5, there exists a sequence {xn}n>1in (a, b) such that

Thus00

A (G) E A [xn> xn + Itn,]n=100

E hnn=1

f(x+h)-f(x)

<

Since this holds V,8 > 0, we have A(G) = 0.We prove next that 1\(E) = 0. First note that

E _ {x E (a, b) I D+ f (x) > D- f (x)}

= U {x E (a, b) I D+ f (x) > r > s > D- f (x)}.r,sEQ,r<s

Thus to show that E is a null set, it is enough to show that d r, s E withr < s we have A*(ET,S) = 0, where

Er,s :_ {x E (a,b)IDf(x) > r > s > D- f (x) J.

Since A*(Er,s) < (b - a) < +oo d r, s given e > 0 we can find an openset H D ET,S such that

A (H) < A (Er)s) + E.

Also for every x c Er,s, since D_ f (x) < s, we can choose an arbitrarilysmall real number h > 0 such that

[x-h,xJcH and f(x)-f(x-h)<sh.

Thus the collection

{[x - h, x] Ix E ET,S, h > 0 and 1(x) - f (x - h) < sh}

forms a Vitali covering of ET,S. By theorem 4.4.5, we can choose pairwisedisjoint intervals [Xi - hl, xl], , [xn - hn, xn] such that for every k

f(xk) - f(xk - hk) < shy,

and

Thus

n

A Er,s U([xk_hkxk])) < E.k=1

n

A (Eras fl U [xk - hk, xk] > A (Er,s) - E. (6.2)k=1

Further,

1: [f(xk)-f(xk-hk)J < S Ehkk=1 k=1

sA(H)

S (A* (Er,S) + (6-3)

Again, for each y E (U=i(xk - hk) xk)) n E,.,,,, since D+ f (y) > r, we canfind z > 0 arbitrarily small such that (y, y + z) C (xk - hk, xk) for some kand

f (y + z) - f (y) > rz.

Once again, the collection of intervals

n

(y, y + z) I Y C (U(xk - hk, xk)) n E,.,S, z > 0 and f(y + z) - f (y) > rzk-I

forms a Vitali covering of the set (U=1 (xk - hk, xk)) n Er,s . Thus by the-orem 4.4.5, we can find pairwise disjoint intervals (yi, yi + zi),1 rziand

n m

A U nEr>s U (yi, 2Ji + zi) < E.k=1 i=1

Thus

A Er,s n n U (yi, yi + zZ )Z=1 ).J

(Uxk - hk, xk) - e.k=1

(6.4)

We note that for every i, Ji :_ (y, yi + zi) C (xk - hk, xk) := Ik for somek. Thus for every fixed k, since the intervals (y, yi + zi),1 xk)

k=1

(f(y+z) - f(yi)) (f(xk) - f(xk - hk))Eis Js Clk

Therefore,n m

j:(f (Xk) - f(xk - hk)) >_ (f(yj + z2) - f (y2))

M

> r zi

i=1mm

rA (uYiYi+zi))z- 1i=11

> rA Er,sn (U (xk- h

n(i=1

Using (6.4), we get

1: (f (Xk) - f (Xk - hk))

n> A E,,, n U (Xk- hk, Xk) -6 (6-5)

L ( (k=1

From (6.2), (6.3) and (6.5) we haven

s(A (ET,S) + e) > r A (Ersfl U(xk-hk,xk))) - ek=1

> r A [(Er,s)_2E].

Since E > 0 is arbitrary, we have

s(A (Er,s)) > r(A (Er, S))

Because r > s, the above inequality implies that A (Er,s) = 0. This completesthe proof that E is a null set. The proof that F is a null set is similar. Thiscompletes the proof of the theorem.

We next analyze measurability and integrability of the derivative func-tion.

6.2.9. Theorem: Let f : [a, b] ][8 be a monotonically increasing func-tion. Then f E Ll [a, b] and

r6f 'dA < f (b) - f (a).

Q

Proof: First note that, being monotonically increasing, f is a measurablefunction. By theorem 6.2.1, f'(x) exists for a.e. x. Let

9n (x) :=[f(x + ( l/n)) - f(x)1 x [a,

1/n J

Here, f (x) - f (b) for x > b. Each gn is a measurable function, and gn(x)f'(x) for a.e. x E [a, b]. Thus f '(x), which is defined a.e., is a measurablefunction (with arbitrary values on the null set where it is not defined).Since f is monotonically increasing, is a sequence of nonnegativemeasurable functions. By Fatou's lemma, we have

Jb

na.

f/(x)dl\(x)

b

liminf g, (x)d.(x)n->oo ina.

lnm f Lnfb

f (x + nJ b fin.

b+lflim inf n f(x)dA(x) - n ina. 6

'-'O° a-Fnf6+1 /n

ina.

a-F 1/'n

lim inf n J f(x)dx-n f ('-'O° b

(a).f(b)_limsuP[n()] f (b) - f

a+l/nf(b)-limsup n f(x)dA(x)

'n-*oo ina.

6.2.10. Remark: The inequality fb

f '(x) dA (x) < f (b) - f (a) can be

strict. For example, take f to be the Lebesgue singular function, as inexample 6.2.4(iv).

In the next section we shall prove that the inequality in theorem 6.2.9becomes an equality if f is absolutely continuous (see theorems 6.3.1 and6.3.2). We conclude this section with an application of theorem 6.2.1 todifferentiation of a series of nondecreasing functions.

6.2.11. Theorem (Fubini): Let {f}>i be a sequence of real-valuednondecreasing functions on [a, b] such that 1=i fn(x) := 1(x) exists forevery x E [a, b]. Then f is differentiable for a. e. x E [a, b], and, for a. e.x E [a, b],

f , (X) = Y00

Proof: Without loss of generality, we may assume that fn(x) > 0 V n andV x. Since each fn is nondecreasing, it is differentiable for a.e. x, by theorem6.2.1, and f(x) > 0, whenever it exists. Let

00

En:_ {x E [a, b] I fn(x) does not exist} and E= U En.n=1

Then E is a null set and d x E [a, b] \ E, {>i_ 1 f(x)}nil is a nondecreas-ing sequence. Further,

fk(X+h)fk(X)L h lim1: fk(x + h) - fi(x)

n-+oo hk=1

f (x + h) - f (x)h

Since f also is a nondecreasing function, by theorem 6.2.1 again, f'(x) existsfor all x ¢ F c [a, b], where F is a null set. Thus d x E [a, b] \ (E U F), wehave

= lim Efk(X+h)fk(X)

h--+o hk1f (x + h) - f (x)

n--+o h

= fi(x).

Hence fEn=1 fi(x)} is a nondecreasing sequence and is bounded above

by fi(x), x c [a, b] \ (E U F). Thus k=1 fi(x) is convergent for a.e. x.To complete the proof we have to show that Ek=lfi(x) =fi(x) for a.e.x E [a, b]. For this, it is enough to show that there is a subsequence of

f Enk-1 f(x)}n>1 which converges to f'(x) for a.e. x. To prove this, we

shall show that there exist positive integers nl < n2 < n3 < < nk .. .such that EOO 1 [f' (x) - 1 f3 (x), is convergent for a.e. x E [a, b], forthen its general term will converge to zero. We note that for every x E [a, b]and every choice of positive integers nl < n2 < < nk < ... ,

f,(X)-E (X)= f(x)-f3j=1

andnk 00

kn

j=1

f(x)-Efj(x)= E f(x).

6.3 Fundamental theorem of calculus and its applications 191

sequence of nonde-Since fn(x) > 0 V n and d x, {>I=nk+1 fj(x)}k>l is a

creasing functions. Thus, whenever the series >Ii (Ejo_nk+l fi) convergespointwise, then, by above argument, the corresponding series of deriva-tives will converge a.e. x. Thus we only have to choose such thatd x E [a, b], E001 (flk+1 f(x)) < +oo. Since

00 00

E fj (x) :! E fj (b)j=nk-}-1 j=nk+1

we only have to choose such that E'k=1 [>iflk+1 fi(b)] < +oo.But that is possible, since f(b) _ >1i fj(b) and we can choosewith

00 nk

E f(b) = f(b)-f(b) < 2j=nk+1 j=1

Then

k=1 j=nk -I-1 k=1

This completes the proof.

6.2.12. Exercise: Let f : [0, 1] ) [0, 1] be the Lebesgue singular func-tion as constructed in example 6.2.4(iv). Extend f to R by f(x) = 0 if x < 0and f(x) = 1 if x > 1. Let {rn I n = 1, 2, ... } be any enumeration of therationals. Define

n=12n

Show that 4) is a well-defined, continuous, strictly increasing function on R.Further, 4)'(x) = 0 for a.e. x.(Hint: Use theorem 6.2.11.)

6.3. Fundamental theorem of calculus and itsapplications

As in section 6.2, in this section also all a.e. statements are with respect tothe Lebesgue measure A on R.

00 00 001E E f3 (b) :! E g < + oo.

6.3.1. Theorem (Fundamental theorem of calculus-I): Let f E L1 [a, b]and F(x) := fa f (t) do(t), x E [a, b]. Then F is absolutely continuous andis differentiable, with F'(x) = f (x) for a. e. x E [a, b].

Proof: That F is absolutely continuous has already been proved in theorem6.1.1. Since f =f+ -f- and both f+,f- E L1 [a, b], we have

f-(t)dA (t)x

F(x) = J f(t)dA(t) - faa

Fl(x) - F2 (X) -

Both F1 (x) and F2(x) are nonnegative and monotonically increasing (andalso absolutely continuous). Thus to show that F is differentiable for a.e.x and F'(x) = f (x), it is enough to show that each of F1 and F2 have thisproperty. For in that case we will have

F (x) =Fl(x) -Fa(x) = f+(x) - f (x) = f(x).

So, we assume, without loss of generality, that f > 0. Then F is monotoni-cally increasing, and by theorem 6.2.1, F is differentiable for a.e. x c [a, b].We only have to identify its derivative. First assume that f is bounded also,i.e., 0 < f (x) < M V x E [a, b] and for some M. Let

Fn(X) :=F(x + (1/n)) - F(x) a<x<b.

1/n

Then each Fn is a measurable function (in fact continuous) and Fn(x) con-verges to F'(x) for a.e. x c [a, b]. Further, d n,

Il =I n J f (t)dA(t) < M, x c [a, b].

Hence by Lebesgue's dominated convergence theorem, d c c (a, b),fC

F1 (t)dA(t) = limfC

= rli [faCn F(t + 1/n)dA(t) - n J F(t)dA(t)Ja

lim n Jc+l/n

F(t)dA(t) - n JC

F(t)dA(t)

fc-

c--1/n fa+l/nlimn F(t)dA (t) - J F(t)da(t) (6.6)

n-'O° a

Since F is a nonnegative monotonically increasing function, we have

F(c) < nf F (t) dA (t) :5 F (c + 1 /n)c+1/n

andfa+l/n

F(a) < nJ F(t)dA(t) < F(a + 1/n).a

Using these and the fact that F is continuous (in fact absolutely continuous),we get

lim nn-*oo f

a+l/nc+l/n -

fF(t)dA (t)

= fc

f (t)dA(t)- (s.7)

From (6.5) and (6.7), we get b c E (a, b),

fCF

(t) dA (t) =fC

f (t) dA (t)

Since this holds for every a < c < b, by exercise 5.4.8, we have

f'(t) = F(t) for a.e. t E [a, b].

This proves the theorem in the particular case when f is bounded.In the general case, we `truncate f' (see note 5.3.24) as follows: define

b'n=1,2,...fn W

f (x) if f (x) < n,n if f (X) > n.

Then is a sequence of bounded measurable functions such that{ f(x)}>1 increases to f (x) for every x c [a, b]. For every n > 1, let

Gn (x) :=JX

- f)(t)dA (t), x E [a, b].

Then each Gn is an absolutely continuous, monotonically increasing func-tion, and hence 0 for a.e. x E [a, b]. Further,

F(x) = Gn(x) + f(t)dA(t), x c [a, b].a

Thus F'(x) exists for a.e. x E [a, b] and, by the earlier case,

F '(x) = Gn(x) + fn(x) for a. e. x E [a, b] and d n.

Hence Fi(x) > f(x) for every n and for a.e. x E [a, b]. Thus

F '(x) > f (x) for a.e. x E [a, b].

On the other hand, by theorem 6.2.9,

IaThus

b

F (x)dA (x) < F(b) = Jb

f(x)dA(x).a

Iab [F T) -.f(x), dA(x) = 0.

Since F'(x) - f (x) > 0 for a.e. x c [a, b], this implies that F'(x) = f (x) fora.e. x E [a, b]. 0

As an application of theorem 6.3.1, we have the following theorem.

6.3.2. Theorem: Let f E Ll [a, b]. Then there exists a set E C [a, b] suchthat A([a, b] \ E) =0 and V x E E

fhlim 1 J If (x + t) - f(x)I do(t) = 0.

h o

(Elements of E are called the Lebesgue points of f in [a, b].)

Proof: For every E I[R fixed, consider

F(y)in

If (t) - j dA(t), y E [a, b].

By the fundamental theorem of calculus (6.3.1), F is differentiable for a.e.y E [a, b], and for such y,

If (y) - j = F'(y) =1

lim - If (t) - j dA(t)n.-+o h Iy+h

hlim 1 I f (y + t) - j do(t). (6.8)n.-+o h fn

Let E denote the set of points y E [a, b] such that (6.8) holds. ThenA([a, b] \

E0 < A([a, b] \ E) < 1: AQa, b] \ 0.

EQ

We shall show that d x E E, the required property holds. Fix any x E [a, b]and let e > 0 be arbitrary. Choose a rational such that If (x) - I < e. IfxEE, thenxEE and

if (X) - Ifh

lim 1 J Ih--+oh oLet us choose 6 > 0 such that d 0 < IhI < a,

1

h

Thenf

h

f(x + t) - floa(t) - 1(x) E/2.

h

f If (x + t) - j dA(t) < If (x) - j +,E/2 < c.

Finally,

1h

h f I hIf (x + t) dA(t)

jh(x) - dA(t)f+ h

E -f- I f (x) - S I < ZE. 0

6.3.3. Note: Let f E Ll [a, b] and let t E R. Then it follows from exercise5.5.9 that every continuity point of f is also a Lebesgue point of f.

Theorem 6.3.1 gives one part of the fundamental theorem of calculus: fora function f E Ll [a, b], its indefinite (Lebesgue) integral F(x) fa f(t)dA(t)is differentiable a.e. x E [a, b], and at these points F'(x) = f (x). Next welook at the converse question: when is a function F : [a, b] ) ][8 the in-definite integral of its derivative? If we recall the example of the Lebesguesingular function, it is not enough to assume that F is continuous, is differ-entiable a.e. and F E Ll [a, b]. Some additional property of F is needed toconclude that it is the indefinite integral of its derivative. The extra prop-erty should be absolute continuity, for if F is an indefinite integral, then itis going to be so. The precise statement is as follows:

6.3.4. Theorem (Fundamental theorem of calculus-II): Let F[a, b] ) Il8 be an absolutely continuous function. Then F'(x) exists fora. e. x E [a, b], with F' E L1 [a, b] and

fy

F(y) - F(x) = Ff(t)dA(t), d a < x < y < b.

For the proof we need the following lemma, which is another applicationof the Vitali covering theorem (4.4.5).

6.3.5. Lemma: Let 0 : [a, b] --> Il8 be an absolutely continuous functionsuch that 0'(x) = 0 for a. e. x E [a, b]. Then ¢ is a constant function.

Proof: We shall show that d c E (a, b], ¢(c) _ ¢(a). Let

E := f x E (a, c) I 0'(x) 01.

By the given hypothesis, A (E) = c - a. Let q > 0 be given. For every x E E,we can find an arbitrarily small h > 0 such that

[x,x+h] C (a, c) and 10(x + h) - O(x) I < rah.

Thus the collection of intervals

{[x,x+h] I x E (a, c), h > 0, 1 O(x + h) - O(x) I < qh}

forms a Vitali covering of E. By theorem 4.4.5, given any 8 > 0, there existpairwise disjoint intervals [x2, x2 + hi], i = 1, 2, ... , n, such that

U [Xi , xi + hi] C (a, c) with I O(xi + hi)i=1

for every 1 < i < n and

- O(xi)I < qhi

A E\U[xi,xi+hi] <J.i=1

Without loss of generality, let

xO+h0:=a<x1 <x1+h1 <X2 <X2+h2 <... <xn+hn <C:=xn+1.Then

n

1: [xk+1 - (xk + hk)l = c - 1: [xk + hk - xk] - ak=0 k=1

n

U xk + hk](1k

Also,

n

A E\U[Xk, Xk+hkl < J- (6-9)

I O(c) - O(a) I = I O(Xn+l) - O(xo + ho) I

[(x) - O(Xk + hk)]k=O

+ 1: [O(Xk + hk) - O(Xk)]

< E I Olxk+ll - O(xk + hk)k=0

n

+ 1: 1 O(Xk+hk) - O(Xk) . (6.10)

Let e > 0 be arbitrary. We choose and fix S > 0 given by the absolutecontinuity property of 0. Then for this 8 we have, by (6.9),

n

1: 1 O(Xk) - O(Xk+ hk) < 6-k=0

Thus from (6.10), we haven

O(c) - 0(a) I e + E phi < e + 77(b - a).

Since 77 > 0 and e > 0 are arbitrary, we have O(c) _ O(a).(Lemma 6.3.5 can also be deduced from proposition 4.1.3 and proposition

6.1.9 as follows: Let E = {x E (a, b) I f '(x) = 0}. By proposition 4.1.3 witha = 0, A(O(E)) = 0. Also, A* (O[a, b]) = 0, i.e., the range of 0 is a singleton.)

Proof of Theorem 6.3.4: Since F : [a, b] ) li8 is absolutely continuous,by theorem 6.1.4, F is a function of bounded variation and hence F(x) =Fl (x) - F2(x), where Fl and F2 are monotonically increasing functions on[a, b]. But then, by theorem 6.2.1, F'(x) exists with F'(x) = Fl '(x) - F2(x)for a.e. x. Further, by theorem 6.2.9, Fi, F2 E L1 [a, b] and

fXF1 (x) < Fl(t)dA (t), F2(x) < F2(t)dA

(t)inX

Thus F' E Ll[a,b]. Let

G(x) :=fX

F (t) dA (t), x E [a, b].

Then G(x) is absolutely continuous, is differentiable for a.e. x E [a, b], andG'(x) = F'(x) for a.e. x E [a, b] (by theorem 6.3.1). Now by lemma 6.3.5,G(x) F(x) V x. 0

Theorems 6.2.1 and 6.3.4 can be combined together as follows:

6.3.6. Theorem (Fundamental theorem of calculus): Let F : [a, b] -R. Then the following statements are equivalent:

(i) F is absolutely continuous.

(ii) F(x) = J f(t)dA(t) V x E [a, b] and for some f E Ll[a, b].a

(iii) F(x) is differentiable for a. e. x E [a, b], with F' E L1 [a, b] and

(t)dA (t) d x [a, b].F(x) = faX F

In fact, the above theorem can be extended to functions defined on I asfollows: Suppose F : IE -* I is such that

F(x) f (t)dA(t)

for some f E L1(IL8) and every x E R. Then clearly b a > 0,x

F(x) = f (t)dA(t) + F(-a), x E [-a, a] .f a

Thus F is absolutely continuous on [-a, a] V a > 0. Further, it is easy tocheck that lima_,,,,, Vas (F) is finite. One writes

V± (F) := lim Vaa(F),

called the variation of F on R. In fact,+00

(F) C f 1 f(x)IdA(x).00

Finally, an application of the dominated convergence theorem shows thatlim F(x) = 0. Conversely, suppose F : Il8 ][8 is such that F is ab-x>-oo

solutely continuous on [-a, a] V a > 0 with finite variation on IEB, andlien F(x) = 0. Then, clearly, Fi(x) exists for a.e. x E 1[8, and by thes-,-oo

monotone convergence theorem, we have

J+oo

IF (t)IdA(t) = lim J +n I F'(t) I dA (t).00 -n

Since by exercise 6.1.6,

f JI F'(t) I dA (t) = "n (F) < V±oo(F) b' n,

we have

1-00

+ooFtdAt

ThusF/

E Ll (][g), and by the Lebesgue's dominated convergence theorem,we have

x

lim _LaF'(t)t)Since Va>Oandx>-a,

x

- 00F1 (t)dA(t).

F(x) = F(-a) + fF'(t)dA(t),

we have

F(x) = lim F(-a) + lim f F'(t)dA(t)a

_ fF'(t)dA(t).

Thus we have proved the following theorem:


6.3.7. Theorem: Let F : I[8 R. Then the following statements areequivalent:

(i) F is differentiable for a. e. x E ][8 with F' E L1(I[8) and

F(x) = J-00 F1 (t) dA (t) x E R.

(ii) F is absolutely continuous on [-a, a] V a > 0 with lim F(-x) = 0,and F has finite variation on R.

As an application of the fundamental theorem of calculus we have thefollowing.

6.3.8. Corollary: Let f , g E L1 [a, b]. Let

X X

F(x)J

f(t)d(t) + a and G(t) :=J g(t)dA(t) +

a a

where E I[8 and x E [a,bJ. Then

f6(Gf)(t)dA(t) = [C(b)F(b) - C(a)F(a)]ina.

Proof: First note that since F and G are absolutely continuous, it fol-lows from exercise 6.1.7 that FG is absolutely continuous,differentiable for a.e. x. Further, by theorem 6.3.1,

(FG) (x) = (Fg) (x) + (f G) (x) for a. e.

Now by theorem 6.3.4,

fb

F(b)G(b) - F(a)G(a) = (FG)'(t)d(t)

and thus FG is

X.

= f 6(Fg)(t)d(t)+f b(!G)(t)dA (t).

6.3.9. Corollary (Integration by parts): Let F, Gabsolutely continuous functions. Then

[a, b] --+ li8 be

bF(b)G(b) - F(a)G(a) = J (F'C)(t)d(t) + J 6(FG)(t)dA (t).a a

Proof: Take f = F' and g = G' in corollary 6.3.8.

In view of theorem 6.3.7, and corollary 6.3.9, we have the following.


6.3.10. Corollary (Extended integration by parts): Let f, g : Il8

1[8 be such that both are absolutely continuous on [-a, a] for every a > 0,li f(-x) = 0 = g(-x), and both f, g have finite variation on R. Then

da(t)J

f(t)g'(t) d(t)+ f f'(t)g(t)

\uaf(t)d A (t))

( fIRR 9(t) d T(t))(urn

oof(x)) (urn oo9`x) .

As another application of the fundamental theorem of calculus, we lookat the extension of the integration by substitution formula from the Rie-mann integral to the Lebesgue integral. The integration by substitution forRiemann integration is as follows (see also exercise 6.3.20).

6.3.11. Theorem (Change of variable for Riemann integration):Let f : [c, d] ) Il8 and 0: [a, b] [c, d] be such that both have continuousderivatives. Then

fProof: Let

b 0(b)f(y)dy.

f(a)

f ( t)dt, a x b.xF(x) := faX

Then (F o 0) is differentiable and, by the chain rule,

(F o 0)'(x) = (F' o 0)(x)0 (x) = (f o 0)(x)0'(x).

Thus by the fundamental theorem of calculus (6.3.6),

af b(fo)(x)'(x)dx = J 6(Fo)'(x)dx

a

(F o 0) (b) - (F o 0) (a)

Z91M

F'(y)dy

f(y)dy.¢(a)

An immediate extension of theorem 6.3.11 is the following.

6.3.12. Theorem: Let ¢ : [a, b] ) [c, d] be absolutely continuous. Letf : [c, d] ) I[8 be a continuous function and F its indefinite integral. Then


(i) (F o 0) is differentiable a. e. on [a, b], and

(F o 0)'(x) = (f o 0) (x) 0'(x)

whenever x E [a, b] is such that 0'(x) exists.O(6) 6

(ii) f (Q) f(y)dy = f f (O(x))0 (x)dx.

Proof: Since F o 0 is absolutely continuous, it is differentiable a.e. (bytheorem 6.3.6 and exercise 6.1.12). In fact, F'(y) = f (y) V y E [c, d]. Thuswhenever 0'(x) exists and is finite, we have

(F o 0)'(x) = (F' o 0) (x) 0'(x) = (f o 0) (x) 0'(x).

Further, by theorem 6.3.6,

16(f o )(t)(t)dt = f 6 (F o

_ (F o 0)(6) - (F o 0)(a)F(O(b)) - F(O(a))

0(b)f(y)dy.

0(a)

6.3.13. Corollary: Let 0 : [a, b] -+ [c, d] be a monotonically increasing,absolutely continuous function and f [c, d] -+ ][8 a bounded measurablefunction. Then (f o 0)0' is integrable on [a, b], and

O(b) 6

ONc f(y)dy= f (foq)(x)q(x)dx. (6.11)

Proof: Since f is bounded measurable and 0 is absolutely continuous (thus0' E L1 [a, b]), it follows that (f o 0) 0' E L1 [a, b]. To show that (6.11) holdsfor every bounded measurable function, we proceed as follows: Firstly, notethat it is enough to prove that (6.11) holds when f is a nonnegative boundedmeasurable function.

Step 1: Suppose (6.11) holds for a sequence {f}>i of functions on [c, d]such that {f}>i is uniformly bounded and fn(x) --+ f (x), x E [c, d]. Then(6.11) holds for f .

To see this, let I fn (x) I < K V x E [c, d] and n = 1, 2, .... Then byLebesgue's dominated convergence theorem,

fO(b)

f(y)dy

O(b)

lim f(y)dyn->oo q(a)

fenlim

J(foq5)(x)q5'(x)dx

b

f (f ° )(x)'(x)dx.

Step 2: Let I be any subinterval of [c, d]. Then (6.11) holds for f = XI.For this, we show that XI is the pointwise limit of a uniformly bounded

sequence of continuous functions. The required claim will then follow fromtheorem 6.3.11 and step 1. For example, suppose I = (a, 3) C [c, d]. Con-sider the functions fn [c, d] ) Ilg as shown in Figure 24, given below.Then I f(x)I < 1 V x E [c, d] and d n. Further, fn(x) -- XI(x) V x.

1

c C± a

Figure 24 : The function fn

n d

Step 3: Using step 2, it is easy to check that (6.11) holds whenever f is astep function, i.e., az. where I2 i C d 1< i< n arep > f = En

i=1 XI2 [c, ] pairwisedisjoint open intervals with U=1 Ii = [c, d] and a2 E III V i.

Step 4: Equation (6.11) holds for f = XU and xF, where U C [c, d] is anopen set and F C [c, d] is any closed set.

To see this we note that U, being an open set in ][8, is a union of countablymany pairwise disjoint open intervals, say Then by step 3, (6.11)holds for each fn := En 1 XIz . Note that, for every n > 1, If(x)IV x E [c, d] and fn(x) -- XU (x) as n -- oo. Thus by step 2, (6.11) holds for

f = XU also. Since {1 - f,, },,>1 converges to XU,, (6.11) holds for f = XFalso, where F C [c, d] is any closed set.

Step 5: Let E C [c, d] be any Lebesgue measurable set. Then (6.11) holdsfor f = XE

For this, we note that since E is measurable, by theorem 4.2.2, we canchoose a closed set F,z and an open set G,z for every n > 1 such that

F,z C E C G,z and A (G,, \ F,z) < I /n.

Let00 00

F:= U F,, and G:= n Gn.n=1 n=1

Then A(G \ F) -0 and F C E C G. Thus

XF < XE < Xc and XF (x) = XE (x) = Xc (x) for a.e. x E [c, d].

HenceO(6) ¢(6) O(b)f XF(y) dy = f XE(y) dy = f Xc(y)dy

Since XF and XG are pointwise limits of {XU-1Fk

}n>1 and {XUnGkand those sequences are uniformly bounded, by steps 1 and 4 we have

¢(6) 6

(XF o 0)(x)0'(x)dxf XF(y)dy = fna.(a)

(6) 6a Xc(y)dy = f (Xc0c5)(x)cb(x)dx.

The function 0 is monotonically increasing, (x) > 0 whenever it exists.Thus the above equalities give us

(XF ° )(x)(x) = (XG ° 0)(x)0 (x) for a.e. x E [a, b].Since

(XF0)(x)(x) < (XE0)(x)(x) C (XGo)(x)(x) forthese functions are all equal for a.e. x E [a, b]. Hence

IO(b)

XE(y)dy = 1 XF(y)dyO(a)

fnae(XF ° (X) (x)dx

f b

(XE° )(x)(x)dx.

a.e.x E [a, b],

Thus (6.11) holds for f = XE


Step 6: The equation (6.11) holds for any bounded measurable function f.This follows from step 5, using step 1 and the `simple function technique'.

We leave the details as an exercise.

We note that in the proof of the theorem 6.3.11, there are two key steps:

(i) Chain rule: If F and 0 are differentiable at x, then F o 0 is differen-tiable at x and (F o 0)'(x) = (F' o 0)(x)0'(x).

(ii) Fundamental theorem of calculus, which gives

fb

J f(y)dy.o (a)

In order to extend this theorem to functions f E L1 [a, b], we have to firstanswer the following question: let 0 : [a, b] -) [c, d] and let F : [c, d] R.When can we say (F o 0) is differentiable and

(F o 0)'(x) = (F o 0)'(x)O'(x) for a.e. x E [a, b]?

And whenever equation in (i) holds, we should try to apply the fundamentaltheorem of calculus (theorem 6.3.6) to deduce the corresponding change ofvariable formula for f E Li [a, b]. Even if both 0 [a, b] ) [c, d] andF : [c, d] -) ][8 are differentiable for a.e. x E [a, b] and further (F o 0) (x) isalso differentiable for a.e. x E [a, b], the relation (F o 0)'(x) = F(O(x))O'(x)need not hold for a.e. x E [a, b], as the next example shows.

6.3.14. Example: Let 0 : [0, 1] ) R be the strictly increasing, continuousfunction with 0'(x) = 0 for a.e. x E [0, 1], as constructed in exercise 6.2.12,and let F = 0-1. Then F, being monotone, is differentiable a.e. (by theorem6.2.1) and (F o 0) (x) = x d x E [a, b]. Thus (F o is differentiableeverywhere and (F o 0)'(x) = 1 V x E [a, b], but (F' o (x) 0'(x) = 0 fora.e. x E [O, 1].

A sufficient condition for the chain rule formula to hold is given by thefollowing theorem.

6.3.15. Theorem (Chain rule): Let 0 : [a, b] - ) [c, d] and F : [c, d]

R be such that 0, F and F o 0 are differentiable a. e. Let F be such that itmaps null subsets of [c, d] to null subsets of R. Then

(Fo)'(x) = (f ° 0)(x)0 (x) for a. e. x E [a> b],

where f [c, d] R is any function such that Fi(x) = f (x) for a. e.x E [c, d].


Proof: We first note that by the chain rule for differentiation, if F is dif-ferentiable at O(x) for some x c [a, b] and 0 is differentiable at x, then(F o 0)'(x) = F'(O(x))O'(x). We fix any function f [c, d] ][8 such thatF'(x) = f (x) for a.e. x E [c, d], and define

A :_ {x E [c, d] I F '(x) exists and f (x) = F (x)}, B :_ -1(A).

Then for a.e. x E B, 0 is differentiable at x (because 0 is differentiablea.e. on [a, b]), and since O(x) E A, the function F is differentiable at O(x)with F'(O(x)) =f(i(x)). Thus the required claim holds for a.e. x E B. LetC := [a, b] \ B. We shall show that the required claim also holds for a.e.xEC.Let

D :_ {x E C I (F o 0) and are both differentiable at x}.

Since x E C \ D means either F o is not differentiable at x or 0 is notdifferentiable at x, we have, by the given hypothesis, A (C \ D) = 0. Thus tocomplete the proof we only have to show that for a.e. x E D, (F o 0)'(x) _(f o 0) (x) 0'(x). We note that

O(D) C O(C) = [c, d] \ O(B) = [c, d] A

and A ([c, d] \ A) = 0, by the given hypothesis. Thus A (O(D)) = 0, and bytheorem 4.1.6 we have ¢'(x) = 0 for a.e. x E D. Also, by the given propertyof F, A (F(q(D)) = 0, and once again by theorem 4.1.6,

(F o 0)'(x) = 0 for a. e. x E D.

Hence

(F o 0)'(x) = 0 = (f o 0)(x)O(x) for a.e. x E D.


Next we analyze the validity of theorem 6.3.11 for the Lebesgue integral,i.e., given 0 : [a, b] - [c, d] and f [c, d] ][8, when can we say thatb a, ,(3 E [a, b]

inf (a) f(x)dA(x)?

Clearly, 0'(x) should exist for a.e. x c [a, b], f should be integrable and(f o 0) (x) 0'(x) should be integrable on [a, b] for both sides of the aboveformula to make sense. Are these conditions also sufficient? Let us suppose(f o 0) (x) ¢'(x) E L1 [a, b] and the above formula holds. Let

fe

aF(x):= f (t) dA (t), x E [c, d].


Then by the fundamental theorem of calculus, F is absolutely continuous,F'(x) = f(x) for a.e. x E [c, d], and for a.e. y c [a, b],

F'(y)) - F(0 (a)) = f f (x)dA(x) = fna

Thus (F o 0) is absolutely continuous. Conversely, let us suppose 0' (x)exists for a.e. x, f is integrable and F o 0 is absolutely continuous. Then bytheorem 6.3.15,

(F o 0)'(x) = (fo)(x)'(x) for a. e. x c [a, b].

Also by the fundamental theorem of calculus, we have (f o 0) 0' E Ll [a, b]and d a, Q E [a, b]

fd\ = F(0(13)) - F(O(ce))

Ip(Fo)/

(x)dA(x)J

P

(f o 0) (x) 0/(x)dA(x).

Thus we have proved the following theorem.

6.3.16. Theorem (Integration by substitution): Let 0 : [a, b] ) [c, d]

be differentiable a. e. and F : [c, d] ][8 be absolutely continuous. Then thefollowing are equivalent:

(i) F o 0 is absolutely continuous.

(ii) Let f : [c, d] ) ][8 be such that f (x) = F'(x) for a. e. x E [c, d] . Then(f o 0)0' E Ll [a, b] and d E [a, b]

1(OW) Q

f (x) dxin

(f o 0) (x) 0/(x) dA (x)

0(a) Ce

6.3.17. Corollary: Let 0 : [a, b] [c, d] be absolutely continuous andf E L1 [c, d]. If 0 is monotone, then d a,,3 E [a, b]

(Q) Qf (x)dA(x) (f o 0) (x) 0

/(x)dA(x).

Proof: LetProof: LetF(x) := fr f (y) dA (y)) X E Ic I d].

Then F is absolutely continuous and, by exercise 6.1.11, F o 0 is also ab-solutely continuous. Now, the conclusion follows from theorem 6.2.1 andtheorem 6.3.16.


6.3.18. Corollary: Let 0 : [a, b] ) [c, d] be absolutely continuous andf : [c, d] - ][8 be bounded and measurable. Then d E [a, b]

f (x)dA(x) in (f o 0)(x)o/(x)dx.

Proof: Let

F (x) = fr f (y) dA (y)) X E [c) d].

Since f is bounded, F is well-defined, absolutely continuous, and F'(x) _f (x) a.e. x c [c, d]. By exercise 6.1.12, F o 0 is absolutely continuous. Theconclusion now follows from theorem 6.3.15.

6.3.19. Exercise: Let ¢ : [a, b] -> [c, d] be absolutely continuous andf E Ll [c, d] be such that (f o 0) 0' E L1 [a, b]. Show that d a, ,Q E [a, b]

Q mca>

f (foq)(x)(x)dA(x) = f (ce) f (x)dA(x).

(Hint: Approximate f by bounded measurable functions on [c, d] and usecorollary 6.3.18, along with Lebesgue's dominated convergence theorem.)

6.3.20. Exercise: Prove exercise 6.3.19 via the following steps:

(i) Show that the conclusion holds when f is the characteristic functionof an interval.

(ii) Approximate f E Ll [c, d] by step functions (see note 5.6.4 (ii)) anduse (i) together with Lebesgue's dominated convergence theorem toprove the required conclusion.

6.3.21. Exercise (Change of variable formula for Riemann inte-gration): Let h : [c, d] I[8 be Riemann integrable and g(x) := jQ h(t)dt,x E [c, d]. Let f : g[c, d] ][8 be Riemann integrable. Then, using exercise6.3.19, deduce that

f 9(d) f(x)dx = f d f[g(t)]h(t)dt.

(Compare this with theorem 6.3.11.)

6.3.22 Note: The proofs of the chain rule and integration by substitutionas given above are based on Serrin and Varberg [36].

Chapter 7

Measure andintegration on productspaces

7.1. Introduction

In chapter 4, we saw that the intuitive notion of length, originally defined forintervals in ][8, was extended to the class of Lebesgue measurable sets whichincluded not only the intervals and all the topologically nice subsets of R,but also ,t3R-the o--algebra of Borel subsets of R. In a similar manner, onewould like to extend the notion of area in R2 (volume in Il83, and so on) toa larger class of subsets which includes ,t3R2 (13R3) - the o--algebra generatedby open subsets of R2 (R3). In the abstract setting, given measure spaces(X, A, µ) and (Y, ,t3, v), one would like to define a measure 77 on the o--algebragenerated by sets of the form {A x B I A E A, B E X3} in the `natural' way:77(A x B) = µ(A)v(B). We call this natural for, when X = Y = R, A =X3 = X3R and µ = v = A, the Lebesgue measure, then for intervals I andJ, r7(I x J) = A(I)A(J) is the area of the rectangle with sides the intervalsI and J. Thus 77 will automatically be an extension of the notion of areain R2. We note that the collection R {A x B I A E A, B E .t3} is onlya semi-algebra of subsets of A x B in general. (See example 3.2.3(v) andexercise 3.2.5(iii).) Let A 0 B denote the o--algebra of subsets of X x Ygenerated by R.

7.1.1. Definition: Let (X, A) and (Y, X3) be measurable spaces. A subsetE C X x Y is called a measurable rectangle if E = A x B for some A E A

209

210 7. Measure and integration on product spaces

and B E B. We denote by 7z the class of all measurable rectangles. Theo--algebra of subsets of X x Y generated by the semi-algebra 7Z is called theproduct o--algebra and is denoted by A®B.

7.1.2. Proposition: Let pX : X x Y -> X and py : X x Y -> Y bedefined by

pX (x> y) = x and pY (x, y) = y,d x E X, y E Y. Then the following hold:

(i) The maps pX and pY are measurable, i. e., b A E A, B E B we havePP A) E ,A 0 C3 and pYl (B) E .A 0 B.

(ii) The o--algebra A 0 B is the smallest o--algebra of subsets of X x Ysuch that (i) holds.

Proof: Let A E A and B E B. Then

pX (A) =AxYER and pYl(B)=XxBER.Hence (i) holds. To prove (ii), let S be any o--algebra of subsets of X xYsuch that pX and pY are both S-measurable. We show that S C ,AO.t3. LetAEAandBE.t3. Then AxY=pX1(A)ESandXxB=pY1(B)ES.Since

AxB = (AxY) n (XxB),it follows that R C S. Hence A 0 B = S(R) C S, proving (ii).

7.1.3. Exercise: Let (X, A) be a measurable space. Let a,)3 E R andE E A 0 BR. Show that {(x,t) E X x IlS I (x, at +)3) E E} E A o BR.(Hint: Use the o--algebra technique.)

7.1.4. Exercise: Let E E X3R. Show that {(x, y) E 1182 1x + y E E} and{(x, y) E 1[82 I x - y E E} are elements of BR 0 C3R.

7.1.5. Proposition: Let X and Y be nonempty sets and let C, D be familiesof subsets of X and Y, respectively. Let C x D {C x D I C E C, D E D}.Then the following hold:

(i) S(C x D) C S(C) 0 S(D).(ii) Let C and D have the property that there exist increasing sequences

{C}>1 and {D}>1 in C and D respectively such that00 00

Ui= X and UDj=Y.i=1 i=1

Then

S(C x D) = S(C) 0 S(D).

7.1 Introduction 211

Proof: Let C E C and D E D. Since C C S(C) and D C S(D), we have

CxD E S(C)xS(D) C S(C)OS(D).

This implies that CxD C S(C)OS(D). Hence S(CxD) C S(C)OS(D), prov-ing (i). To prove (ii), we only have to show that S(C)OS(D) C S(CxD). Byproposition 7.1.2, S(C)OS(D) is the smallest a-algebra of subsets of XxYwith respect to which the projection maps pX and py are measurable. Soto complete the proof, we show that pX, py are both <S(CxD)-measurable.Let C E C. Then

pXl(C) = CxY

Cx

00

= UCxDj).j=1

Since each CxDj E CxD C S(CxD), we have

pXl (C) E S(C x D) V C E C.

Let

Lf :_ {E E S(C) I pXl(C) E S(C x D)}.

Then by the above arguments

c c u c S(C).

It is easy to check that U is a a-algebra of subsets of X. Hence S(C) = Lf.Thus pXl(E) E S(CxD) for every E E S(C), i.e., pX is S(C x D)-measurable.Similarly, py is S(C x D)-measurable. This completes the proof.

7.1.6. Exercise:

(i) Let X and Y be nonempty sets and C, D be nonempty families of subsetsof X and Y, respectively, as in proposition 7.1.5. Is it true that S(CxD) =S(C)OS(D) in general? Check in the case when C = {Q1} and D is a a-algebraof subsets of Y containing at least four elements.

(ii) Let ,t3Rz denote the a-algebra of Borel subsets of Il82, i.e., the a-algebragenerated by the open subsets of ][82. Show that

BR2 = BR ®BR .

(Hint: Use proposition 7.1.5.)


7.2. Product of measure spaces

For the rest of the chapter, let (X, A, ,u) and (Y, B, v) be fixed measurespaces. Subsets of X x Y of the form A x B, A E A, B E B, are calledmeasurable rectangles. As before, let R denote the class of all measurablerectangles. The a-algebra of subsets of X x Y generated by 7?, denoted byA ®8, is called the product a-algebra.

The problem we want to analyze in this section is the following: howcan we construct a measure 77 : A ® B -> [0, --boo] such that rj(A x B) _u(A)v(B) for every A E .A, B E B?

In fact, the property required of 77 for the elements of 7Z suggests itsdefinition on R. Further, if we can show that 77 is countably additive onR, then, observing that 7 . is a semi-algebra, we can extend 77, via the outermeasure r7* (see theorem 3.10.8), to A (9 B, the a-algebra generated by R.Further, this extension will be unique provided 77 is a-finite on R. These arethe contents of our next theorem.

7.2.1. Theorem: Let 77: R -) [0, oo] be defined by

77(A x B) := µ(A)v(B), A E A, B E B.

Then 77 is a well-defined measure on R. Further, if µ, v are a-finite, thenthere exists a unique measure q : A 0 B -+ [0, +oo] such that

(A x B) = 77(A x B) for every A x B E R.

Proof: Obviously, 7](O) = 0. To show that q is countably additive, letA E .A, B E B be such that

00

AxB= U(AnxBn),n=1

where each An E A and each Bn E B, and (An X Bn) n (A, X 0 forn m. We have to show that

00

7](A x B) = E 7](An x Bn).n=1

For this, let x E A be fixed. Then for any y c B, (x) y) E U=1(An X Bn),and hence y E Bn provided x E An. Thus d x c A,

B= UBn, where S:={nEi`jxEA,}.nES

Further, for m, n E S we have Bn n 0, for otherwise we would have(AnXBn)fl(AmXBm)74O. Thus

7.2 Product of measure spaces 213

v(B) = EnC:S v(Bn).Equivalently, for x E A,

00

xA(x)v(B) = )'v(B) - 'XA, (x) 1/ (Bn,). (7.1)nES n=1

If x 0 A, then XA(x)v(B) = 0. Also for x 0 A, we have x 0 An for every n,and thus xAn (x)v(Bn) = 0 V n. Thus equation (7.1) holds for x 0 A also.Hence d x E X,

00

XA(x) v(B) - > XAn (x)v(Bn)n=1

An application of the monotone convergence theorem gives us

,q (A x B) = µ(A)v(B) = (z)u(Bn) I dµ(T)

00

E J X,Q,n \x/v\L7t/dN'\x/

00

n=1

00

/-t(An)v(Bn) = Eq(An x Bn)-n=1

This proves that rq is a measure on R. By theorem 3.10.8, rq can be extendeduniquely to a measure q on the a--algebra generated by R. provided it is a-finite. To complete the proof we show rq is a--finite when ,u, v are a--finite.For this, let

00 00

X = Xi, Y = Yji=1 j=1

be such that Xi E A, Yj E 13, where the Xi's are pairwise disjoint and theYj's are pairwise disjoint, with 1u(Xi) < boo and v(Yj) < boo d i, j. Then

00 00

XxY= UU(xix)i=1 j=1

is a partition of X x Y by elements of R such that

,q (Xi X Yj) = /-t (Xi) V (Xj) < + 00.

Hence q is Q-finite.

7.2.2. Definition: The measure ? on A ® 13 given by theorem 7.2.1 iscalled the product of the measures 1u and v and is denoted by 1u x v. Themeasure space (X x Y, A ® 13, 1u x v) is called the product of the measurespace (X, A, p) and (Y, 13, v), or just the product measure space.

We note that µ x v is uniquely defined on A x 8 when µ, v are Q-finite.So from now on we shall assume that µ and v are Q-finite. As is clear, µ x v isdefined on A0I3 via the extension theorem 3.10.8, i.e., via the outer measureand so on. So, the natural question arises: how to compute (,u x v)(E) fora general element E E A 0 8? Let us recall the computation (or rather thedefinition) of areas of planar regions that we met in the calculus: let E C 1182be a bounded region, say

E:= {(x, y) E ][82 1 a < x < b, f (x) < y < g(x)j,

where f (x) and g(x) are continuous functions.

x= a x

Figure 25

Then the area of E (see Figure 25) is defined by

f6Area(E) := J (1(x) - g(x))dx.

a

If we write h(x) := 1(x) -g(x), then h(x) is nothing but the length of the setE, :_ {y E 1[8 I (x, y) E E}, and an elementary result from multiple integralstells us that

bfArea(E) := A(E,,)dx.

a

The question arises: can we carry over this idea to find (,u x v) (E) forE E A 0 8? Obviously, the following questions will have to be answeredfirst:

(i) Given E E A0 8 and x E X, is E,, E 13 (so that v(E,,) makes sense)?

(ii) To integrate v(E,,) with respect to µ, one has to answer the question:is the map x 1 v(E,,) measurable?

(iii) If (i) and (ii) make sense, can we compute -q(E) by

n(E) = v(Ex)dµ(x)?fX3

(iv) Can one interchange the roles of µ and v in the above steps, i.e., is

-q (E) _ µ(Ey)dv(y) also?Y

We shall answer these questions one by one.

7.2.3. Definition: Let E C X x Y, x E X and y c: Y. Let

Ex :={yEYI (x,y)EE} and Ey:=Ix EXI (x,y)EE}.

The set Ex is called the section of E at x (or x-section of E) and theset EY is called the section of E at y (or y-section of E).

7.2.4. Examples:

(i)LetE=AxB,whereAEAandBEB. Clearly, Ex= B if x E A andEx=OifxVA. Similarly, Ey=AifyEBandEy=QJifyVB.

(ii) Let (X, A) be a measurable space and let A E A. Let

E={(x,t)EX xl[810<t<XA(x)}.

It is easy to see that

E = (A x [0, 1)) U (Ac- x f 01).

Thus

Similarly,

f [0,1) if x E A,E l {0} if x V A.

X ify=0,Ey = A if y E (0, 1),

Ql if y V [0, 1).

7.2.5. Proposition: For E, F, Ei E A 0 t3 and i E I, any indexing set, thefollowing hold d x E X, y E Y

(1) (UEI Ei)x = UiEI(Ei)x and (UEI Ei)y = UiEI(Ei).

(ii) (fljEI Ei)x = fljEJ(Ej)x and (niEl Eir = fliEI(Ei).(iii) (E \ F)X = Ex \ Fx and (E \ F)y = Ey \ Fy.(iv) If E C F, then Ex C Fx and Ey C Fy.


Proof: Exercise.

Now we can answer the questions asked above.

7.2.6. Theorem: Let E E A 0 B. Then the following hold:

(i) Ex E B and Ell E A for every x E X, y E Y.(ii) The functions x i) v(E,,) and y H µ(Ey) are measurable functions

on X and Y, respectively.

(Ey)dv(y).IX v(Ex)dµ(x) == (µ x v) (E) = ly µ

Proof: The proof is a typical application of the `Q-algebra-monotone classtechnique'. Let

S:={EEAo BI Ex EX3and EyA b'xEX,yEY}.It is easy to see that S is a a-algebra and A x B E S whenever A E A, B E B.Thus .Ax,t3 C S and hence S = A 0 B. This proves (i).

To prove (ii) and (iii), let

P :_ JE E A 0 B I (ii) and (iii) hold}.

To prove the required statements, we have to show that P = A 0 B. Forthat we shall show that P is a monotone class, is closed under finite disjointunions and includes R= {A x B I A E A, B E X3}. Since R is a semi-algebra,using exercise 3.2.10(i), we will have F(R) C P and hence Nl(,'F'(R)) C P,where F(R) is the algebra generated by R and Nl (.F(R)) is the monotoneclass generated by F(R). By theorem 3.10.7, Nl(.F(R)) is the same as thea-algebra generated by F(R), i.e., A 0 B. Hence we will have A 0 B = P.We prove these claims one by one.

Let E = A x B E R. Then by example 7.2.4 (i), we have

v(Ex) = v(B)XA(x) b'x E X,µ(Ey) = µ(A) XB (y) d y c Y.

Clearly, x H v(Ex) and y H p(EY) are measurable functions, and

Ixv(Ex)dµ(x) = µ(A)v(B) = J (E)dv(y).

Y

Thus R C P.Next, let El, E2 E P be such that El fl E2 = 0. Then bx E X,

(Ei)m n(E2)x= 0 and (El U E2)x = (El)m U (E2)x.

Hence

v((Ei U Ea)x) = v((Ei)x) + v((Ea)x)


Thus by proposition 5.2.6(ii), x -- v((El U E2),,) is measurable and

13Xv((Ei U

x [v((Ei)x) + v((Ea)x)] dµ(x)

_ (µ x v) (El) + (µ x v) (E2)_ (ixv)(EiUE2).

Similarly, y -- µ((E1 U E2)y) is measurable and

fy µ((E1 U E2)y)dv(y) = (µ x v)(El U E2).

Hence P is closed under finite disjoint unions.

Finally, to show that P is a monotone class, let En E P, n > 1, be suchthat En C En+l V n and E = U=1 E . Then for every n > 1

(En)y C (En+1)x and (En)y C (En+1)y Vx E X and y E Y.

Thus {v((E7))}>1 and {µ((En)y)}n>1 are increasing sequences of nonneg-ative measurable functions. Using theorem 3.6.3(a)(i),

lim v((En)x) = v(Ex) and lim µ((En)y) = µ(Ey).n->oo n-+oo

Thus by the monotone convergence theorem (5.2.7), x -- v(Ex) and y --µ(Ey) are nonnegative measurable functions with

J v(Ex)dµ(x) = lim J v((En)x)dµ(x), 1x n-'°° x

f µ(Ey)dv(y)Since En E P d n > 1, we have

limn---boo ' lz((En)y)dv(y)

Xv((En) x) dp (x) _ (p x v) (En) -

Yµ((En)y)dv(y)

Thus using (7.2), we get

.Xv(Ex) dµ(x) = lim (µ x v) (En,) _ µ(Ey)dv(y).n-'°° Y

Since (µ x v) is countably additive, by theorem 3.6.3(a)(i),0o

lien (p x v) (En) _ (µ x v) U Enn-b oo

n=1

Hence for n > 1, if En E P and En C En+1, then U°° 1 En E P. Similarly,if En E P and En D En+1 Vn > 1, to conclude that nn, 1 En E P we cantry to argue as above. To deduce the relations

lim v((En)x) = v(Ex), lim p((En)') = p(EY)n-+oo n-+oo

and

(µ x v) (E,,) = (µ x v) (E),

we need the extra condition that µ, v are finite measures, for in that caseµ x v will be finite and we can apply theorem 3.6.3(a) (ii). Then the equalitiessimilar to (7.2) will hold by an application of the dominated convergencetheorem. Thus in the particular case when µ, v are finite measures, theabove arguments will prove that P is a monotone class and the conclusionof the theorem holds.

To prove the theorem in the general case, i.e., when µ and v are Q-finite, let X = UOO 1 AZ and Y = U=1 Bj , where b i, j, AZ E A, B3 E B withµ(A2) < +oo and v(Bj) < +oo. Then (µ x v)(AZ x Bj) < +oo d i,j. ForE E A (& ,t3, by the earlier discussion, we have d i, j

fv((En(Aj x Bj)),, ) dµ (x) _ (µ x v)(E n (Ai x B; ))

= JY[((Efl(Ai x Bj))y) dv(y).

Thus00 00

(xv)(E) _ EE(pxv)((AjxBj)nE)i=1 j=100

E E00 f

J v ((E n (Ai x Bj)) x ) dµ (x) (7.3)i=1 j=1 x00 00

E E f l ( (En(Ai x Bj))y) dv(y). (7.4)i=1 j=1 Y

Also,(En(Ai x Bj))x = Ex n Bj if x c Ai, and 0 otherwise.

Thus we have by the monotone convergence theorem,00

i=

00

v ((E n (Ai x Bj))x ) dµ(x)EIXj=1

j=1

.X

Ix

00

f v (Ex n Bj)dµ(x)i=1 Z

Ixv(Ex fl Bj)dµ(x)

00

E v((EnB)) dµ (x)j=1

v(Ex n 1B;)) dµ (x)

=J

v(Ex)dµ(x). (7.5)x

Similarly,00 00

E E f ((E n (Ax B)))dv(y) - lY (E)dv(y). (7.6)i=1 j-1 J

Equations (7.3), (7.4), (7.5) and (7.6) complete the proof.

7.2.7. Exercise: Let (X, A, µ) be a v-finite measure space. For any non-negative function f : X ) Il8, let

E*(f):={(x,t)EXxIl810<t< f(x)}and

E*(f):={(x,t)EX x][80<t< f(x)}.Then the following hold:

(i) If is an increasing sequence of nonnegative functions on Xincreasing to 1(x), show that {E(f)}>1 is an increasing sequenceof sets in AOX3Rwith

00

UE*(fm) = E*(fn.=1

(ii) Using (i) and the `simple function technique', show that E* (f) EA0 BR, whenever f : X ) I[8 is a nonnegative measurable function.Deduce that E* (f) E A 0 X3R.(Hint: E*(f) = n°°_1 E* (f + 1/n).)

(iii) Let f : X ) I[8 be a nonnegative function such that E*(f) E A(9 13R.Using exercise 7.1.2 and the following equality

A {(x,t)EXxI[8I f(x)>c,t>0}

= U{(x,t) E X x Il8 I (x, t/n + c) E E* (f ), t > 0},n=1

show that A E A (9 X3R, and deduce that f is measurable.

(iv) Let f X ) Il8 be any measurable function. Show that G(f) EA &13R, where

G(f) (x, t) E X x Il8 I f (x) = t}.

The set G(f) is called the graph of the function.(v) Let f c Ll (X, A, µ), where µ(X) < +oo. Show that

(µ x v)(E*If1) = f fldµ = (µ x v)(E*If1). (7.7)

(Hint: First prove this for the case when f is bounded. For the generalcase, consider fn = I f I A n and note that increases toE* If 1.)

(vi) Extend (iv) to the case when µ is a-finite.

The identity (7.7) shows that for nonnegative functions f fdµ representsthe area below the curve y = f (x).

* 7.2.8. Exercise: Let X be any nonempty set and P(X) be its power set.Let A := 'P(X) O'P(X) and let D := {(x, y) E X x X I x = y}. SupposeD E A. Prove the following statements:

(i) There exist sets Ai, Bi E 'P(X) such that D belongs to the a-algebragenerated by (Ai x Bi), i = 1, 2, ... .

(Hint: See proposition 3.9.5.)

(ii) Let B = S({Ai I i = 1, 2,... }). Then card (,t3) < c, by theorem 4.5.2.For every x, y E X, Dx : Dy if x : y and Dx E 13 V x E X. Deducethat card (X) < c.

(iii) If card (X) > c, then D V P (X) 0 P (X), even though Dx E P (X)and Dy E P (X), d x, y E X. Hence in general,

'P(x) 0 x).

7.2.9. Remark: Given a-finite measure spaces (X, A, µ) and (Y, Ci, v), weshowed in theorem 7.2.1 that µ x v is the unique measure on A0 13 such that(µ x v) (Ax B) = µ(A)v(B). Note that the measure space (X x Y, AOCi, µ x v)need not be a complete measure space even if the measure spaces (X, A, µ)and (Y, 13, v) are complete. For example, if A C X, A V A and 0 : B E ,t3with p(B) = 0, then (µ x v)*(A x B) = 0, but A x B V A 0 B. In fact,theorem 7.2.1 itself gives a complete measure space (X x Y, A (D Ci, µ x v),where A (D 13 is the a-algebra of q*-measurable subsets of X x Y, q being asin theorem 7.2.1, and µ x v is the restriction of q* to A 0 B. The measurespace (X x Y, A 0 ,t3, µ x v) is nothing but the completion of the measurespace (X x Y, A 0 13) µ x v). It is easy to see that theorem 7.2.6 holds forE E A 0 ,Ci also, as claimed in the next exercise.

7.2.10. Exercise: Let E E A o B. Then the functions x ---> v(Ex) andy ---> p(EY) are measurable and

Ixv(Ex)dµ(x) = (-ILX v)(E) = J (E)dv(y).

Y

(Hint: E = F U N, F E A (D ,t3 and (µ x v)*(N) = 0 by theorem 3.11.8.)

7.2.11. Exercise: Let E E AO13 be such that p(EY) = 0 for a.e. (v)y E Y.Show that µ(Ex) = 0 for a.e. (µ)x E X. What can you say about (xv)(E)?

7.3 Integration on product spaces: F u bini's theorems 221

7.3. Integration on product spaces: Fubini'stheorems

Let (X, A, µ) and (Y, B, v) be Q-finite measure spaces and (X x Y, ,AOXi, µ x v)the product measure space. Theorem 7.2.6 can be interpreted as follows: forevery E E .A O.Ci,

X v)(x,y) - \ xE(xy)dv(y)) d(x)fxXE() X Y

xE (x, y)dµ(x)) dv(y).

This allows us to compute the integral of the function XE (x, y) by integratingone variable at a time. So, the natural question arises: does the above holdwhen XE is replaced by a nonnegative measurable function on X x Y? Theanswer is yes, and is made precise in the next theorem.

7.3.1. Theorem (Fubini): Let f : X x Y -3 ][8 be a nonnegative A (9 ,t3-measurable function. Then the following statements hold:

(i) For xo E X and yo E Y fixed, the functions x H f (x, yo) andy H f(xo, y) are measurable on X and Y, respectively.

(ii) The functions y F--) fX f (x, y)dµ(x) and x H fy f (x, y)dv(y) arewell-defined nonnegative measurable functions on Y and X, respec-tively.

f (XI Y)dv(y)) dµ (x) = L y)d(x))fX f(x,

= f(x,y)d( µ x v) (x, y)f xY

Proof: The proof is yet another application of the `simple function tech-nique'.

When f = XE and E E A ® B, the required claim is just theorem7.2.6, as mentioned above. It is easy to see from this that the requiredclaim holds when f is a nonnegative simple measurable function. In thegeneral case, by theorem 5.3.2, let {s}>i be a sequence of nonnegativesimple measurable functions on X x Y such that {s(x, y) }n>1 increases tof (x, y), V (x, y) E X x Y. Then for x E X fixed, {s(x, .) }n> 1 is a sequenceof nonnegative simple measurable functions on Y such that {s(x, y) }n> 1increases to f (x, y) for every y E Y. Thus for x E X fixed, y +) f (x, y) isa nonnegative measurable function on Y, and by the monotone convergencetheorem,

Yf(x,y)dv(y) =

n moo fIrSn(x,2J)dv(ZJ). (7.$)

Thus by the earlier case and corollary 5.3.15, x 1--) fy f (x, y)dv(y) is anonnegative measurable function on X. Since {f, sn(., y)dv(y)}nil is anincreasing sequence of nonnegative measurable functions on X, by the mono-tone convergence theorem and (7.8),

1(1 f(xY)dv(Y))d(x) = lim f(f sn(x1y)dv(y) dµ(x).

By the earlier case,

X ly fXxY\ Sd(x) = S(x,y)d(x

By the monotone convergence theorem again, we have

I f(xY)dv(Y)) d(x) = lim f s(xl y)d(xX \ Y n-iO° X

f(x,y)d(,a x v).

Similarly,

Y fX)

f(x, Y)d(x))fxxY

f(x,y)d(,a x v).


7.3.2. Exercise: Let f : X x Y ) ][8 be A (9 ,C3-measurable. Show thatthe following statements are equivalent:

(i) f E L1(µ x v) := L1(X x Y, A 0 B, a x v).

(11) f I Ix If(x,y)I dµ(x)) dv(y) < +oo.

(iii) f I ly I dp (x) < +oo.

In view of theorem 7.3.1, it is natural to expect a similar result forf E L1(µ x v), µ and v being o--finite measures. This is given by thefollowing theorem.

7.3.3. Theorem (Fubini): Let f E L1(µ x v). Then the following state-ments are true:

(i) The functions x 1---) f (x, y) and y H f (x, y) are integrable for a. e.y(v) and for a. e. x(µ), respectively.

(ii) The functions

y !---) I f(X,y)dp(x) and x l-->JY

f (x, y)dv(y)

7.3 Integration on product spaces: Fubini's theorems 223

are defined for a. e. y(v) and a. e. x(µ) and are v,µ-integrable,respectively.

(iii)Y \x f (x, y)dµ(x)1 dv(y) -

fX3 X Yf(x, Y) d(µ x v)

f (X) y)d1j(y) ) dp(x)fX3 ly

Proof: Let f + and f -denote the positive and negative parts of the functionf. Since f+ is nonnegative and integrable on X x Y, by theorem 7.3.1,

\f+(xy)dv(y)) dµ(x) - (ff+(xY)d(x))

X Y Y

f + (x, y)d(p x v) (7-9)f xY

< f xY f(x,y)d(xv) < +oo. (7.10)

Thus it follows from exercise 5.4.2(ii) that

fyand

f(x,y)dv(y) exists for a. e. x(µ)

+(x, y)dµ(x) exists for a.e. y(v).fX3 f

By a similar argument, we will have

J (J r(x)dv()) du (x) = J (JI-(x,y)du(x)) dv(y) (7.1i)

X Y Y X

f(x,y)d(µ x v) < +oo. (7.12)f xYY

Thus once again it follows from exercise 5.4.2(ii) that

JYf(x,y)dv(y) exists for a.e. x(µ)

and

fr(x)(x) exists for a.e. y(v).

Hence, for a.e. x the functions y &j I+(x, y), y H f -(x, y) are v-integrable and for a.e. y the functions x H f +(x, y), x &--> f - (x, y)are µ-integrable. Hence y F--) f (x, y) is v-integrable for a.e. x(µ) andy H f (x, y) is µ-integrable for a.e. y(v). Further, since f E L1(µ x v), bydefinition

f f(x,y)d(xv) = f f(x,y)d(xv) - f - (X) y) d(µ x v).X Y X Y f X Y

This, together with (7.9), (7.10), (7.11) and (7.12) gives

f(x,y)d(p x v) f(XI Y)dv(Y)Jf-J lyx

(ff,Yd)Y)..Y

Theorem 7.3.1, exercise 7.3.2 and theorem 7.3.3 together give us thefollowing theorem, which enables us to check the integrability of a functionof two variables and compute its integral.

7.3.4. Theorem: Let (X, A, µ) and (X, B, v) be Q-finite measure spaces.Let f : X x Y ) R be an A (2) Z3-measurable function such that f satisfiesany one of the following statements:

(i) f is nonnegative.(ii) f E L1(µ x v).

(iii)

(ice)

Then

if (XI y)ldv(y) d[z (x) < + oo.

If (XI y) Idp(x) dv(y) < +oo.

f (XI Y)d(y x v) f (XI y)dv(y)) dp(x)IX X Y X f'YY

- f \Jxf(xY)d(x)) dv(y),

in the sense that all the integrals exist and are equal.

7.3.5. Exercise: Let (X, A, µ) and (Y, B, v) be complete Q-finite measurespaces and let (X x Y, A 0 B, µ x v) be the completion of (X x Y, AO.CC3, µ x v).Let f X x Y -> R be any nonnegative extended real valued A (2)X3-measurable function. Show that:

(i) The function x H f (x, y) is A-measurable for a.e. y(v) and thefunction y H f (x, y) is ,t3-measurable for a.e. x(p).

(ii) The function y &--) fX f (x, y)dtC(x) is ,Ci-measurable and the functionx H fy f (X, y)dv(y) is A-measurable.

(iii)fX

f(xy)d(l x vJ ( f'YY f (X, y)dv(y)J dµ(x)xY X

=J (J f(x,y)d(x)dv(y).Y X


We give next some examples which illustrate the necessity of the condi-tions on µ, v and f for the conclusions of theorem 7.3.4 to hold.

7.3.6. Example: Let X = Y = [0, 1] and A = B = Blo,j], the o--algebra ofBorel subsets of [0, 1]. Let µ be the Lebesgue measure on A and v be thecounting measure on B, i.e., v(E) := number of elements in E if E isfinite and v(E) := +oo otherwise. Let D :_ {(x) y) I x = y}. Let, for n > 1,

Dn:=U ( [(j - ')/n, j/n] x [(j - 1)/n, j/n]

n=1 Dn. Thus D E A (9 B. In fact, D is a closed subset ofThen D= n,[0,1] x [0, 1]. Further,

fxxD(x,Y)(x)=0 d y E Y and J X(x, y)dv(y) = 1 b x E X.Y

Hence

D (x, y)dµ(x) I dv(y) = 0 andf I 13X X 13X ( ly XD (Xly)dv(y) ) dtt(x) = 1.

This does not contradict theorem 7.3.1, since v is not a-finite.

7.3.7. Example: Let X = Y = [0, 1], A = B = X3[0,1], and let p = v be theLebesgue measure on [0,1]. Let

in

f (X, Y) := (X2 + Y2)2 if (x,y) # (0, 0),

0 if x = y otherwise.

Noting that for fixed x, f (x, y) is a Riemann integrable function on [0, 1] and

19 Y -= f (XI Y),y 2 y2

we get

Hence

and

x2_y2

1

f(x,y)dv(y) = 1/(1 + x2).

(

1

1f(x) y)dv(y)) dµ(x) = 7r/4

(x) y)dv(y)) d(x) _0 / o01 1 f(x, y)dµ(x)1 dv(y) f 1 (f1\

This does not contradict theorem 7.3.3, because f ¢ Ll (X x Y). To seethis, note that

[0,1 ] X [0,1 ]

If (X) y) I dv(y) ) dp(x)o JO

1

> (fxIf(x)Idv)d()fi 1 7r/4

Jcos29d9 dµ(x)

o x

1

2xdµ(x) = + oo.

7.3.8. Exercise: Let X = Y = [-1,1], A = B and let µ = v bethe Lebesgue measure on [-1, 1]. Let

(x2+y2)2

Show that

if (X1 Y) Id(p x v)

0 otherwise.

J-1 - f 1(f1f(x)dv())d(x) 0 (f1f(xY)d(x))dv(Y).

Can you conclude that

i (f'1

f(xY)dv(Y)) dµ (x) =IX X Y

f()d( µ x v) (x, y)?

7.3.9. Exercise: Let f E Ll (X, A, µ) and g E Ll (Y, B, v). Let

O(x, y) := f (x)g(y), x E X and y E Y.

Show that E L1(X x Y, A 0 B, p x v) and

O(x, y)d(p x v)IX X Y

(ffd)

7.3.10. Exercise: Let f E Ll (0, a) and let

dv)fy g

fg(x) := (f(t)/t)dA(t)0 < x < a.

QShow that g E Li (0 a), and compute f g(x)dA(x).0

7.3.11. Exercise: Let (X, A, M), and (X, B) v) be as in example 7.3.7.Define, for x, y E [0,1],

1 1 if x is rational,f(x,' 2y if y is irrational.

Compute

f(x,y)dv(y)) dµ (x) andJ (L'Is f in L1(µ x v)?

f1 1

\Jof(x)d(x)) dv(y)

7.3.12. Exercise: Let (X, A, µ) be as in example 7.3.7. Let Y = [l, oo), i3 =AR fl [l, oo), and let v be the Lebesgue measure restricted to [1, oo). Define,for (x,y) E X x Y,

f(x,y) := e-xy - 2e-2xy.

Show that f ¢ L1(µ x v).

7.3.13. Exercise: Let X be a topological space and let i3X be the Q-algebra of Borel subsets of X. A function f : X -> ]I8 is said to be Borelmeasurable if f -1(E) E Bx V E E B. Prove the following:

(i) f is Borel measurable if f -1(U) E ,t3X for every open set U C R.(Hint: Use the `a-algebra technique'.)

(ii) Let f : X -> ][8 be continuous. Show that f is Borel measurable.(iii) Let be a sequence of Borel measurable functions on X such

that f (x) := lim fn(s) exists b x c X. Show that f also is Borelmeasurable.

(iv) Consider ][82 with the product topology and let f , g be Borel measur-able functions on R. Show that the function 0 on ][82 defined by

O(X, y) := f(x)g(y), x E X, Y E Y,

is Borel measurable.

7.3.14. Exercise: Let f : 1R2 -> III be Borel measurable. Show thatfor x c X fixed, y f (x, y) is a Borel measurable function on R. Is thefunction x f (x, y), for y c Y fixed, also Borel measurable?

*7.3.15. Example: Assume the continuum hypothesis and let < be thewell-order on [0, 1] such that b'x E [0, 1], {y E [0,1] 1 y < x} is at most acountable set (see appendix C). Let

E := {(x, y) E [0, 1] x [0,1]Ix < y}.

Then V y fixed and V x E [0, 1], XE (x, y) = 1 if x < y and 0 otherwise.Hence XE (x, y) is the indicator function of a countable set for y E [0, 1]fixed. Thus it is a Borel measurable function. Similarly, for x E [0, 1] fixed,y i-- * XE (x, y) is Borel measurable, for it is the indicator function of [0, 1] \C,a countable set. However, XE is not Borel measurable on [0, 1] x [0, 1]. To seethis, note that B[o,l] x [0,1] = B[o,1] ® B[0,1] Thus if XE were Borel measurable,then by theorem 7.3.1 we should have

xE x, y d(A x A) (x, y)01x01 ( )

- J[o,i J[o,iXE(x, y) dA (x) da(y)

XE(xI y) dA (y) dA (x).[x,11 [x,11

(7.13)

However, it is easy to see that for x, y E [0, 1],

J[o,i]XE (x, y) dA (y) = 1 and XE (x, y)dA(x) = 0.fo i1

This contradicts (7.13). Hence XE is not Borel measurable.

7.3.16. Exercise: Let f : R2 Ilg be such that for x E X fixed, y -f (x) y) is Borel measurable and for y E Y fixed, x H f (x, y) is continuous.

(i) For every n > 1 and x, y c ][8, define

fn (x, y) := (i - nx) f ((i - 1)/n, y) + (nx - i + 1) f (i/n, y),

whenever x E [(i - 1)/n, i/n), i c Z. Show that each fn : ][82 ][8 is

continuous and hence is Borel measurable.

(ii) Show that f(x,y) --+ f (x, y) as n -> oo for every (x, y) E ][82, andhence f is Borel measurable.

7.3.17. Exercise: Let (X, A) and (Y, X3) be measurable spaces and letf : X x Y -) R be a nonnegative A (2) ,t3-measurable function. Let µ be aa-finite measure on (Y, B). For any E E B and x c X, let

77(x, E) := f (x, y) dp (y).fE

Show that 77(x, E) has the following properties:

(i) For every fixed E E B, i---> 77(x, E) is an A-measurable function.

(ii) For every fixed x E X, E 1 --> 77(x, E) is a measure on (Y, B).

7.4 Lebesgue measure on R2 and its properties 229

A function 7: X x ,t3 -> [0, oo) having properties (i) and (ii) above is calleda transition measure. If 7 is a transition measure with 7(x, Y) = 1 V x EX, then 7 is called a transition probability. Transition probabilities playan important role in the theory of probability. Fubini's theorems can beextended to transition measures. For details see Parthasarathy [28].

7.4. Lebesgue measure on R2 and its properties

We now specialize the construction of (X x Y, A® B, µ x v) to the particularcase when X = Y = R, A = 13 = L ij and µ = v = A, the Lebesgue measure.We have already seen that BR®BR = 8R2 and that (R2, L ®L , A X A) is notcomplete. The completion of this measure space, denoted by (R2, LR2 , AR2),

is called the Lebesgue measure space. Elements of L R2 are called theLebesgue measurable subsets of R2, and AR2 is called the Lebesguemeasure on R2. The following proposition ensures that AR2 is the uniqueextension of the natural concept of area in R2.

7.4.1. Proposition: Let I denote the collection of left-open, right-closedintervals in R, and let Z2 {I x J 1I, J E Z}. Then the following hold:

(i) Z2 is a semi-algebra of subsets of I[82, and S(ZZ) = ,CiRz

(ii) AR2 (I x J) = A(I)A(J), V I, J E Z.(iii) The measure space (I[82, GRa , AR2) is the completion of the measure

spaces (R2, GROGR, A x A) and (][82, 8CiRa , ARa ).

Proof: Statement (ii) is obvious. To prove (i), we note that ZZ = Z x Z and,by exercise 7.1.4 (ii), BR 0 BR = S(Z) 0 S(Z) = S(Z X Z) = S(Z2). Since,t3RO,CiR = BR2 , this completes the proof. To prove (iii), we note that GRa isthe class of AR2-measurable subsets of I[82, where AR2 = A x A is the measureon the semi-algebra ZZ given by AR2 (I x J) = A(I)A (J). Thus (I[82, GRa , ARa )is the completion of the measure space (I[82, ,t3Rz , AR2). Also AR2 = A x Aon the Q-algebra G® 0 L. Hence (I[82, ,CRz , AR2) is also the completion of(R 2 , LR 0 LR,AXA). 0

7.4.2. Exercise (Regularity of AR2): Prove the following:

(i) AR2 (U) > 0 for every nonempty open subset U of ][82.

(ii) A set E E GRz if d e > 0, there exists an open set U such thatECUandA(U\E)<e.(Hint: Proceed as in theorem 4.2.2.)

(iii) AR2 (K) < +oo for every compact subset K of R.

(iv) AR2 (U) =sup{ARz (K) I K compact , K C U}.

(Hint: Given U open, there exists a sequence of compact sets Kn suchthat Kn T U.)

(v) If E E BR2 is such that AR2 (E) < +oo, then

AR2 (E) =sup{ARz (k) I K C E, K compact}.

We describe next some properties of the Lebesgue measure ARa . ForECI[82andx E]Eg2,letE+x:={y+x I yEE}.

7.4.3. Theorem: The Lebesgue measure AR2 has the following properties:

(i) Let E E BR2 and x E II82. Then E + x E ,t3Rz and

AR2 (E) = AR2 (E + x).

(This property of AR2 is called translation invariance.)(ii) For every nonnegative Borel measurable function f on ][82 and y E ]Eg2

J f (x + y) daR2 (x) = J f (x) dAR2 (x) = f f (-x) dAR2 (x) .

(iii) Let µ be any a-finite measure on ,CiRz such that µ(E + x) = µ(E)V E E X3R2, XE II82. Suppose

0 0. Then

p (E) = CAR2 (E)5 V E C BR-

Proof: (i). The proof is once again an application of the `Q-algebra mono-tone class technique'. So, we only sketch the proof. Let

One shows that A is a a-algebra of subsets of II82 and A includes all opensets, proving A = ,CiRa . Next, let

.A4 := fE C: BR2 I A(E + x) = A(E)J.

One shows that M is a monotone class including R = {(A x B) I A, B E ,t3R}and M is closed under finite disjoint unions. Thus M includes Y(R), thealgebra generated by R, and hence includes the monotone class generatedby Y(R), i.e., the a-algebra generated by Y(R) (see theorem 3.10.7). Since,M(Y(R)) = BR2 , the proof of (i) is complete.

(ii) The proof is an application of the `simple function technique' and isleft as an exercise.

(iii) We first note that (i) and (ii) above hold when AR2 is replaced byany translation invariant measure µ on ,t3Ra . One can give a proof on linesas in the case of the Lebesgue measure on I[8, see section 4.4. We give

7.4 Lebesgue measure on R 2 and its properties 231

another proof which is an application of Fubini's theorem. Showing thatp(E) = CAR2 (E) b E E ,t3R2 is equivalent to proving that

aR2 (Eo)µ(E) = (Eo)AR2(E), V E E ,t3R2.

Since AR2 is translation invariant, using (ii) we have d E E BR2

AR2 (Eo)µ(E) = AR2 (EO) f XE (y)dµ(y)

= fAR2(Eo - y)XE(y)dµ(y)

(fxEOx+Y)dAR2x) XE (y)dµ(y)

Using Fubini's theorem (7.3.1),

Aua2 (Eo)µ(E) - f (fXE(Y)XEO(x+Y)d(Y)) dAR2 (x).

Now using translation invariance of µ and theorem 7.3.1 again, we have

AR2 (Eo)µ(E) - (fx(Y - x)XEo (y)dµ(y)) daR2 (x)

xE(y - X)dAR2 (x)) XEo (y)d(y)

p (Eo) AR2 (E). 0

7.4.4. Exercise: Show that for f E L1(II82, ,CRa , AR2 ), x E II82, the functiony --> f (x + y) is integrable and

f f (x + y)dAR2 (X) f (x)dAR2 (x).

(Hint: Use exercise 5.3.27 and theorem 7.4.3.)

7.4.5. Exercise: Let E E GR2 and x = (x, y) E Il82. Let

xE := {(xt,yr) I (t,r) E E}.


(i) xE E GR2 for every x c R, E E GRa, and AR2 (xE) = IxyJAR2 (E).

(ii) For every nonnegative Borel measurable function f : R2 ) II8,

f f (xt)dAR2 (t) = I XY I f f (t)dAR2 (t),

where for x = (x, y) and t = (s,r),xt (xs,yr).(iii) Let AR2 {x E ]I82 1 IxI < 11=:7r. Then

ARa {x E II821 IxI < 1 } _ 7r and ARa {XE II8211 CI < r} _ 7rr2.

(iv) Let E be a vector subspace of ][82. Then ARa (E) = 0 if E has dimensionless than 2.

The claim of exercise 7.4.5(i) can be reinterpreted as follows. Let T

][82 ][82 be the linear transformation whose matrix is given by-

0J .

Then for x _ (x, y) we have xE = T(E) and

ARz (T (E)) = I

In case x = 0 or y = 0, then T is a singular linear transformation andwe have ARz(T(E)) = 0 b'E E ,CiRz. If neither x = 0, nor y = 0, i.e.,IxyI = I det(T) 0, i.e., T is a nonsingular linear transformation, then

AR2 (T(E)) = IdetTI\I2(E),

where det T denote the determinant of the matrix of T with respect to thestandard basis of Il82. The question arises: given a linear transformationT : R2 ) Il82, can we say that AR2 (T(E)) = I det Tj ARz (E)? The answer isgiven by the following theorem.

7.4.6. Theorem: Let T : R2 R2 be a linear transformation and E EGRz . Then T(E) E GRz and

AR2 (T(E))

Proof: If T is singular, then T(][82) is a subspace of R2 and has dimen-sion less than 2. Thus using exercise 7.4.5(iv), ARz (T(I2)) = 0 and henceAR2 (T (E)) = 0 V E E GRa . Hence T (E) E GR2 and

ARz (T (E)) = 0 = I

Now suppose that T is nonsingular. Then T is bijective and both T andT-1 are continuous. Suppose E E ,t3R2 . Then, clearly, T(E)E ,t3Ra . Define

µT(E) ._ AR2 (T(E)), E E ,CiRz.

It is easy to check that µT is a a-finite measure on ,t3R2 and

µT(E + x) = µT(E) V x E ][82, E E ,t3R2 .

Further, if we take U :_ (0, 1) x (0, 1), then U is a nonempty bounded openset, and so is T(U). Thus 0 < AT (U) < +oo, and by theorem 7.4.3, thereexists a constant C(T) > 0 such that

µT(E) = C(T) AR2 (E) V E E ,t3R2.

Hence V E E B 2,

ARz (T (E)) = C(T) ARz (E).

7.4 Lebesgue measure on IR2 and its properties 233

Note that this holds for every nonsingular linear transformation T, and wehave to show that C(T) = I detT1. The map T C(T), T nonsingular,has the following properties:

(i) C(D) = I det DI for every diagonal transformation D.We have already seen this in exercise 7.4.5.

(ii) C(O) = 1, if 0 is any orthogonal transformation.This is because an orthogonal transformation in R2 leaves the setE:= {x E ][821 jxj < 1} invariant.

(iii) C(T1T2) = C(Tl)C(T2) V Tl5 T2 nonsingular.This is easy to verify using the definition of C(T).

Now, let T be any nonsingular linear transformation. By the singularvalue decomposition (see appendix E) for linear transformations, there existorthogonal transformations P and Q such that T = PDQ, where D is somediagonal transformation. Thus

C(T) = C(P)C(D)C(Q) = C(D) = I det D1.

But,Idet(T)l _ Idet(PDQ)I = Idet DI>

since P and Q are orthogonal and thus I det PI _ I det QI = 1. Hence

ARz (T (E)) = C(T)AR2 (E) _ I

This proves the theorem for sets E E 13R2 . It is easy to extend it to setsE E GRa (see the next exercise).

7.4.7. Exercise: Let T : ][82 --> ][8 be a linear map.

(i) If N C ][82 is such that (N) = 0, show that (T(N)) = 0.(ii) Use (i) above and proposition 7.4.1(ii) to complete the proof of the-

orem 7.4.6 for sets E E GRa .

7.4.8. Exercise: Consider the vectors (al, bl), (a2, b2) E 1[82 and let

P:= {(alal + a2a2, albs + a2b2) E ][82 I al> a2 E 1[8, 0 < ai < 1},

called the parallelogram determined by these vectors. Show that

AR2 (P) = jajb2 - a2bi 1.

7.4.9. Exercise: Let E E GR2 with 0 < A(E) < oo. Show that there existsa unique point (Cl, c2) E 1[82 such that

i XXE (X + C1, Y + C2)dAR2 (x,y) = 0

and

In fact,

and

(C = (C1, C2)

f YXE (X + C1, Y + C2) dAR2(x, y) = 0.

1Cl =

A(E) fxxE(xY)dAR2(xY)

C2 =1

A(E) J yXE (x, y)dAR2 (x, y)

is called the centroid of E.)7.4.10. Theorem (Integration of 'radial' functions): Let f : [0, oo)(0, oo) be a nonnegative measurable function. Then

c'O f (r)rdA(r).f2 f (Ixl)dAR2 (x) = 27r J0

Proof: The proof is once again an application of the `simple function tech-nique'. We give the proof in steps

Step 1: The theorem holds for f = X(a,b),

0 < a < b < +oo.

First note that, by exercise 7.4.5, for f = x(a,b)

f 2 X(a,b) (JXJ) dAR2 (x) = AR2 {(x,y) E ][82 Ia2 < x2 + y2 < b2 }

,7r(b2 - a2).

Also, by the monotone convergence theorem and theorem 5.5.1,00 n

2,7r X(a,b) (x)xdA(x) = lim 27r X(a,b) (x)xdA(x)0 n-+ o0 0

b

2,7r xdxJa

= ir(b2 - a2).

Step 2: Let {E}>1 be a sequence of sets from [0, oo) n L such thateither the En's are pairwise disjoint, or the En's are increasing. If thetheorem holds for each XEn , then the theorem holds for XE also, whereE=Un=1E

n-

For this, let F n := U=1 Ek and On = XFn , n = 1, 2, ....Then {q5n}n>lis an increasing sequence of nonnegative measurable functions, and by themonotone convergence theorem we have

f2 XE(JxJ) dAIIB(X) = nmoo f dAIIgz (x). (7.14)

2

7.4 Lebesgue measure on 1R2 and its properties 235

In case the En's are pairwise disjoint, using the monotone convergence the-orem, the right hand side of (7.14) is given by

n

lim f On (I x1) = lim J XEk (I xl) (x)n--+oo 2 n--+oo 2

k=1k=1=1

lim 1: f XEk(Ixl) dA2 (x)

k=1

27r limn-*oo

27r limn--+oo

k=

27-

Hence

fz xe(IxU

27r limn-*oo

n f

J XEk (r) r dA (r)

n

E XEk (r) r dA (r)k=1

00

On(r)r dA(r)

00

J XE(r)rdA (r).

= 27r iXE(r) rdA (r).00

When {E}>i is an increasing sequence, each Fn = En, and once again,using the monotone convergence theorem, the right hand side of (7.14) isgiven by

lim f 2 On(Ixl) dAR2 (x) = n fXE(lx)d2(x)f00

27r limJ

)(En (r)r dA(r)n--+oo 0

(r)r dA (r)in

27r XE

Thus the required claim holds for f = XE

Step 3: The theorem holds for f = XU, U being any open subset of [0,00).To see this we note that U is a countable union of pairwise disjoint open

intervals. The required claim now follows from step 1 and step 2.

Step 4: The theorem holds for f = XN, where N C [0, oo) and A(N) = 0.To prove this, first note that N = U=1(N fl [0, n)), and in view of step

2, it is enough to prove the claim for each XNn[o,n) Since A(N fl [0, n)) = 0,using theorem 4.2.2 we can choose a decreasing sequence of opensets such that N fl (0, n) C nk 1 Uk and lim A(Uk) = 0. Without loss of1

generality, we may also assume that Uk C (0, n) V k. Then V k, by step3, we have

f XNn(O,n) (jxj) dAR2 (x) < f XUk (jxj) dAR2 (x)

f00

27J XUk

(r) r dA (r)0

27rnA(Uk).

Since this holds b k, letting k --+ oo, we have

f XNI(O,n) (I x I) dAR2 (x) = 0.

Clearly,00

XNn(O,n) (r)r dA(r) = 0.21r in

Hence

fn

00 f00

Xnrn(o,n) (xI) dAR2 (x) = 27r J XNn(o,n) (r)r dA(r) = 0.

Since A (N n [0, n)) = A (N n (0, n)), the required claim follows.

Step 5: The theorem holds when f = XE, E E GRand E C [0, oo).For this, note that we can write E = U=1 E n [0, n). In view of step 2,

it is enough to prove the claim for En E n [0, n) V n. Fix n and choosea decreasing sequence of open sets of [0, oo), using theorem 4.2.2,such that

UkDEn b k with A(Uk) < +oo and lim A(Uk) = A(E).

Put F:= n,k=l Uk and N := F \ En. Then A(N) = 0 and F = En U N. Bythe dominated convergence theorem and step 3, we have

fXFIxIdAR2X ) =

271

fn

00

XF(x) x dA (x).

Since En n N = 0 and the theorem holds for XN by step 4, it follows thatthe theorem holds for every XEn and hence for XE

Step 6: The theorem holds for any nonnegative measurable function.This is an application of the `simple function technique'. First we note

that the required claim is linear in f, i.e., if it holds for functions fl, f2 anda,,3 are reals, then it also holds for a f 1 + 3f2. This together with step 5

lim 27r f XU (x)(x)x dA (x)

7.5 Product of finitely many measure spaces 237

tells us that the theorem holds for nonnegative simple measurable functions.Since every nonnegative measurable function is a limit of an increasing se-quence of nonnegative simple measurable functions, an application of themonotone convergence theorem will give us the required claim.

7.4.11. Exercise: Let f : [0, oo) - ) II be a measurable function. Showthat, if either of the two integrals in theorem 7.4.10 exists, then so does theother and the equality holds.

7.4.12. Exercise: Let f : (a, b) x (c, d) - I[g be such that

(i) f is continuous;

(ii)of

exists and is continuous;

(iii) for some t E (a, b), dy f (t, y) exists d y;

(iv) y (L) exists and is continuous.

Show that y and09 09f

exist. Moreoverax y ,

09

a (ay) ay (ax

(Hint: Use Fubini's theorem and the fundamental theorem of calculus toshow that d E (a, b), 77 E (c, d) and s < y

f 77) - f (t, 77) - f 8) + f (t, 8) = dx dy,ay

C

(at\x

and note that the inner integral on the right is a continuous function of .)

7.5. Product of finitely many measure spaces

The ideas developed in sections 7.2, 7.3 and 7.4 can be easily extended todefine the product of a finite number of measure spaces.

For n > 2, let (Xi, A ,,i), i = 1, 2)... , n, be a-finite measure spaces.Suppose we have defined the product measure space (Xi x . x Xn-1, A1®

®An-1 i Al x X An-1)) which we write as

n-1 n-1 n-1

flXj,(gAj,jjpi .

i=1 i=1 i=1

Using the ideas of section 7.2, we can define the product measure spacen-1 n-1 n-1

Xi x Xn, A (9 An) Ai X Ani=1 i=1 i=1

The sets (1_II1 Xi x Xn and f1

Xi : = X 1 x x Xn-1 X Xn can beidentified via the bijection

((Xi,... , xn-1), xn) F --* (x1,. . .xn), xi E Xi, 1 < i < n.

Let 7 n : = {A1 x A2 X x An Ai E ,Ai } . It is easy to verify that Rn isa semi-algebra of subsets of fln 1 Xi. Let ®i 1.Ai denote the a--algebra ofsubsets of fln 1 Xi generated by R. It is easy to check that

n-(:A) n

®A=i=1

kee in in mind the identification n-1 X, X X Xp g rii= l Z n- flni=1 Z.

Next, it is easy to see that the measure (1[i1µi x An on ®i A hasthe property that d Ai E .Ai, i = 1, 2, ... , n,

n-1 n

µi x An (A1 x ... x An) = µi(Ai)i=1 i=1

Moreover, that this is the only measure with this property follows from theuniqueness theorem (3.10.8). We denote the measure ([ihii µi x An byfln

i=1 lei and call it the product measure. The measure spacen n n

Xi A Iiii=1 i=1 i=1

is called the product of the measure spaces (Xi, Ai, pi) , i = 1, 2, ... n,and is usually denoted by f Ai, µi)

7.5.1. Exercise: Let (Xi, ,Ai, pi), i = 1, 2, ... n, be a--finite measure spacesand let 1 < m < n. Show that the product measure spaces

m n m n m n

HXi X Xi 0 ,Ai X ,Ai 11 pi x µii=1 i=m+1 i=1 i=m+1 i=1 i=m+1

and (fln 1 Xi, ®i 1 Ai, fl2 1 pi) are same if the underlying sets are identi-fied in the natural way.

In the special case when each Xi = R, Ai = C and pi = A, the Lebesguemeasure, the completion of the product measure space as obtained above


is denoted by (Rn, L n , ARn). The a-algebra LRn is called the a-algebraof Lebesgue measurable sets of Rn, and ARn is called the Lebesguemeasure on Rn. It is not difficult to show that L n includes BRn, the a-algebra of Borel subsets of R .

7.5.2. Exercise: Let E C Rn and let p E N. Show that ARn (E) = 0 ifV E > 0 there exists a sequence of open balls in Rn such that eachhas radius less than p, and

00 00

k=1 k=E C U Bk and

ERn (Bk) c e.

(Hint: Cover Rn by countably many disjoint `cubes' of side sufficientlysmall.)

7.5.3. Exercise: Extend exercise 7.4.2, theorem 7.4.3, exercise 7.4.4 andtheorems 7.4.6, 7.4.10 to (Rn) L n , aqn). (See also exercise 9.3.17.)

We can also extend Fubini's theorems (7.3.1 and 7.3.3) to finite productspaces. We merely state these theorems, and ask the reader to supply theirproofs.

7.5.4. Theorem (Fubini): Let f : fln1Xi -* R be a nonnegative func-

tion. If f is ®i 1 Ai-measurable and i1, i2.... in is any permutation of112) ... , n, then

(... (f (f f(xi,.. . dVia(x,x)J ... I dVe.,(Tan)

is well-defined (i. e., all the integrands are measurable with respect to theproducts of relevant Q-algebras) and is equal to

Iv

where X =n=1

Xi.

-

n

f(Xl,--- xn)d Ai (X1, ... xn),i 1

7.5.5. Theorem (Fubini): Let f E L1 (Ai, pi)). Then for everypermutation i1, ... , in of 1, 2, ... , n, the integral

. . f(x1,... , x2n )dp21 (x21) d,u22 (x22) ... d,uin (Xn)f27L fX322 fx3i1

exists (i. e., the integrands are integrable functions on relevant product spaces)and is equal to

pi (x1,... , xn).

7.5.6. Exercise: Let f E L1(Il8n+m, Gin+m , ARn+m ). Prove the following:

(i) For a.e. 7n) a E Rn, the function y --> f (a, y) is defined for a. e.7tRm )y E lIg' and is integrable.

(ii) For a.e. (,\Rm )b E 1[8"`, the function x --> f (x, b) is defined for a.e.('\Rn )x E I[8n and is integrable.

(iii) The functions

Y!JRRm f(x,y)dRn(x) and x --> JRMn f(x,y)dm(y)

are defined a.e. x (ARn) and a.e. y (,\Rm), respectively, and areintegrable.

(ice)

fn J m f(x,y)dRrn(y)dARn(x)Afn+m

f * y)d(Al[8n x Alf$m ) (x> y)

fRmfRnf > y)dARn (X)dARrn (y)

7.5.7. Exercise: For x E 1[8n and r > 0, let

B(x, r) := {yEIIThHx-yI<r}.Using theorem 7.4.6 (for general n), show that

ARn (B (x, r)) = rn ARn (B (x, 1)).

7.5.8. Product spaces and probability theory:

We remarked in note 3.1.10 that a probability space (X, s, p) gives a math-ematical model for the analysis of a statistical experiment. Consider twostatistical experiments (X, A, u) and (Y, B, v) such that the outcome of onehas no effect on the outcome of the other. Then a model that representsthem jointly is given by the produced spaces (X x Y, A®B, p x v). In general,given a finite number of statistical experiments (Xi, Si) pi), i = 1, 2, ... , n,a model that gives a joint representation of these n experiments is the prod-uct probability space, (Fln Xi, (gi

1S,' f 1 t,Gi). In many practical sit-

uations one is interested in analyzing a sequence of statistical experimentsf (Xi, Si, t,Gi)}i>1. For example, consider the experiment of tossing a coin (themost basic statistical experiment), which has two possible outcomes: heads


and tails denoted by 0 and I. Suppose the same coin is tossed n times.Thus each outcome is an n-tuple having 0 and 1 as entries, and there are2n such n-tuples. Thus the sample space of this experiment (or a joint rep-resentation of n experiments, each being tossing a coin) is X = flZ X2,XZ 0, 1} V i. The a- algebra S is P (X), all subsets of X, and V x E X,p({x}) = 1/2n, assuming that the coin is `unbiased', i.e., there is equalprobability of a head or a tail appearing in each toss. Now suppose one isinterested in analyzing the following question: what happens to the numberof heads appearing in n tosses as n becomes larger and larger? Of course,the intuitive answer to this question is: it is probable that the number ofheads equals the number of tails. To analyze this question mathematically,we should let n -+ oo. Thus the sample space for the joint representationof this experiment will be X, the set of all sequences {xn}n>1 where eachXn = 0 or 1. We should construct a suitable a-algebra S of subsets of Xand define a suitable probability measure p on S and try to analyze.

1 ntL x = {x}>1 E X lim - E Xkn-*oo n 2

For a detailed discussion of the product of an infinite number of proba-bility spaces and their application to probability theory and statistics, seeBillingsley [5] and Parthasarathy [28].

Chapter 8

Modes of convergenceand L-spaces

Throughout this chapter we shall work with a fixed a-finite measure space(X, S, M), which is assumed to be complete. As explained in chapter 5, thereason to assume that (X, S, µ) is complete is to have the property thatif f and g are functions on X such that f is S-measurable and µ* {x EX I f (x) g(x)} = 0, then g is also S-measurable (see exercise 5.3.27 andproposition 5.3.28). In situations when (X, S, µ) is not complete, one canalways consider its completion and work with it.

The main aim of this chapter is to analyze the convergence of sequencesof measurable functions. Given a sequence {f}> 1 of measurable functionson (X, S, µ), we have already come across concepts like pointwise conver-gence of to a measurable function f, i.e., when {f(x)} convergesto f (x) V x E X. When a sequence does not converge pointwise, one wouldlike to find other methods of analyzing the behavior of the sequence {f}>1

for large n. Analysis of these methods, and the relations between them, isthe main aim of this chapter. But before we do that, we extend the notionof measurability and integration to complex-valued functions.

8.1. Integration of complex-valued functions

Let (X, S, µ) be a measure space and let C denote the field of complexnumbers. For a function f : X C, consider the functions Re (f) andIm (f) defined by: V x E X,

Re (f) (x) := Real part of f (x)

243

244 8. Modes of convergence and Lp-spaces

and

Im (f) (x) := Imaginary part of f (x).

The functions Re (f) and Im (f) are called, respectively, the real part andthe imaginary part of the function f . Note that Re (f) and Im (f) arereal-valued functions on X.

8.1.1. Definition: A complex-valued function f : X ) C is said to bemeasurable if both Re (f) and Im (f) are measurable functions. We say fis µ-integrable if both Re (f) and Im (f) are integrable. In that case wedefine the integral of f, denoted by f f dµ, to be

f fdµ := fRe(f)d+ifIm(f)d.

We denote the set of all complex-valued µ-integrable functions on Xby Ll (X, S, µ) itself. Whenever we restrict ourselves to only real-valuedµ-integrable functions on X, we shall specify it by L' (X, S, µ). Our nexttheorem tells us that L1(X, S, µ) is as nice a space as L' (X, S, µ) is.

8.1.2. Theorem:

(i) Let f , g E Ll (X, S, µ) and a, ,Q E C. Then a f +,ag E L1(X, S, µ) and

f(af+g)d µ = aJ fdµ+,3 J gdµ.

(ii) Let f : X -) C be a measurable function. Then f E Ll (X, S, µ) ifIf I E L' (X, S, µ). Further, in either case,

ffdµl < f If Idµ

(iii) Let f E Ll (X, S, µ) and E E S. Then xE f E L1(X, S, µ). We write

Efdµ := fXEfd.

IfE1,E2 Es and ElnE2=0, then

fdL=f fd+fElUE2 l Z

(i v) Let f c Ll (X, S, µ), and let {E}>1 be a sequence of pairwise dis-joint sets from S. Then the series(fE

nfd,a) is absolutely con-

vergent. Also, (XEf) E L1 (X, S, µ), where E U=1 En, and00

fE

8.1 Integration of complex-valued functions 245

Proof: (i) is straightforward and is left as an exercise. To prove (ii), letf E L1(X, S, µ). Then Re (f ), Im (f) E Li (X, S, µ). Since

Ill < v/_2_CIRe(f)I+IIm(f)I)>

clearly, If I E Li (X, S, µ). Conversely, let f be a complex-valued measur-able function. Then Re (f) and Im (f) are measurable functions. If If I E

Li (X, S, µ), then

IRe(f)I < IfI and IIm(f)l < fl.Thus by proposition 5.4.3, we have Re (f) E L i (X, S, µ) and Im (f) ELi (X, S, µ). Finally, to prove that

If µ>I ffdu J <_ flfdlet a:= If fdp. Then O <a< oo, and there is a O < B < 27r such that

a = exp(-i8) (f fdµ)

Let exp(-i8) f := fl + if2, where fl, f2 are real-valued. Clearly, fl, f2 ELi (X, S, µ) and

a= J exp(-i8) f dµ = ffid+iff2d.

Since a is real, we have f f2dµ = 0 and a = f f1d. Thus

a = ffidµ = I ffidµ < f Ifildp < f If Idp.

To prove (iii), we note that for every E E S, I XE f I i be a sequence ofpairwise disjoint sets in S. Consider the series(fE

nfd). Then b m,

if A,,,, := U=1 En and E:= U=1 En, thenm m

E In=1

fdµJ < E J If Idp = JAm JEn=1 EThus f E, 1

1

fEn fdi}m>i is a monotonically increasing sequence and, be-ing bounded above, is convergent. In fact d m,

pµI

E fn fd_ffd[` fA,,,,fd_ffdn_1

246 8. Modes of convergence and LP-spaces

By the dominated convergence theorem, the right hand side of the lastinequality converges to zero as n --* oo. Hence (iii) holds.

8.1.3. Exercise: Let f E L1 (X, S, µ) be such that

ffdµWhat can you conclude about f?

f If dµ

That the integral of complex-valued functions behaves as nicely withrespect to limiting operations as the integral of real-valued functions is thecontent of the next theorem.

8.1.4. Theorem (Complex form of Lebesgue's dominated conver-gence): Let {f}>i be a sequence in L1(X, S, µ). Let f (x) Eli fn(x)

exist for a. e. x E X and let there exist a function g E Li (X) S, µ) such thatd n, f(x)I < g(x) for a. e. x(µ). Then f E L1(X,S, µ) and

lim J fndIL=f f dµ.

Proof: Let f (x) lim fn (x), whenever it exists. Since fn E L1(X, B, µ),V n, Re (fn) and Im (fn) are measurable functions. Further,

(Ref) (x) = lim Re (fn (x)) for a. e. x(µ)

and(Trnf)(x) _ n1' Im (f(x)) for a.e. x(µ)

Thus Re (f) and Im (f) are both measurable by proposition 5.3.14. Notethat f I is a real-valued measurable function and I f (x) I < g(x) for a. e. x.Thus fl E Li (X, S, µ), and hence f E Ll (X, S, µ). Furthermore,for all n, 1(x) - fn(x)l is a real-valued measurable function withlim f (x) - fn(x)l = 0 for a.e. x and

f (x) - fn (x) I < 2g(x) for a. e. x(µ).

Hence by Lebesgue's dominated convergence theorem (5.4.9), we have

n moo If fd_f fd l.l' < limo f If-fdµ = 0.

8.1.5. Exercise:

(i) Let f : X -* C. Show that f is measurable if f -1(E) E S for everyBorel set E C X.

8.1 Integration of complex-valued functions 247

(ii) Extend the claim of exercise 5.3.27 to complex-valued measurable func-tions.

8.1.6. Exercise: Let {f}>i be a sequence of S-measurable complex-valued functions on (X, S, µ). Show that the following statements are equiv-alent:

(1) (>I= 1 1 fE L1 (X, S,I-(ii) E001 f Ifnldl-L < +oo.

Further, if either of the above is true, then (E001 fn) E Li (X) S, µ) and

f00 00

fn )dµ = n=1EJ fn dl-L.

(En= 1

8.1.7. Exercise: State and prove the Riesz-Fischer theorem (theorem5.6.1) for complex-valued integrable functions.

8.1.8. Exercise: Let f E Ll (X, S, µ) and

Efd/2=O for every E E S.

Show that f (x) = 0 for a.e. x(µ) (see proposition 5.4.6).

8.1.9. Theorem: Let f E Ll (X, S, µ), where µ(X) < oo. Let S be a closedsubset of C such that b E E S with µ(E) > 0,

(_1[fd/2ES.

µ(E) E

Then f (x) E S for a. e. x E X.

Proof: We have to show that µ{x E X I f (x) V S} = 0. Since S' :_ C \ S isan open set, there exist countably many open balls {B}>i in C such thatSC = Un=1 °Bn. Thus{xEX 00

f(x)E UB}nn-,1=

00

U f'(B).n-,

Since f'(B) E S V n, see exercise 8.1.5, to prove the required claim itis enough to show that µ(f -1(Bn)) = 0 V n. Fix an integer n and let Bnhave center zo and radius ro. Let En := f'(B) and suppose, if possible,

0. Then

1 1fd - f )d( )A(En)E,, i zo

<

(x)-zo xA(En) In (

)- Jd )1 f f(x zo (xµ (En) En

< ro ,

i.e.,1

(En) fEnwhich contradicts the hypothesis. Hence µ(En) = 0.

8.2. Convergence: Pointwise, almosteverywhere, uniform and almost uniform

All the functions considered in this section are defined on a fixed a-finite,complete measure space (X, s, ,i) and are complex-valued, unless statedotherwise.

8.2.1. Definition: Let {f}>i be a sequence of measurable functions andlet f be a measurable function on X.

We say {f}>i converges pointwise to f if {f(x)}>i convergesto f (x) for every x E X, i.e., given x E X and e > 0, El no := no (X) e)such that

Ifn(x) - f (x) I < ej V n > no.(ii) We say {f}>1 is convergent to f almost everywhere if

N := {x E X I {f(x)}>i does not converge to f (x)} E S

and µ(N) = 0.(iii) We say {fn}n>i converges uniformly to f if given e > 0, El no

no(e) such that

fWewrite these as fn P f, fn f (or fn f a.e.) and fn u) f,

respectively.

8.2.2. Exercise: Show that fn -f f = fn -f f = fn of f. Constructexamples to show that the reverse implications need not be true.

We shall show that even though uniform convergence is not implied bypointwise convergence, the situation is not that bad.

8.2 Convergence: Pointwise, a.e., uniform and a. u. 249

8.2.3. Proposition: Let be a sequence of measurable functionssuch that fn f, f a measurable function. Let E E S be such thatµ(E) < +oo and let e, 6 > 0 be arbitrary. Then there exist a set EE E S anda positive integer no, depending upon e and 6, such that the following hold:

(i) EE C E and (E\E) < e.(ii) f(x) - f (x)l <6V n > no and d x G E.

Proof: Let N' {x E X I lim fn(x) = f (x)}. Then by the given hypoth-n-*oo

esis, µ(N) = 0. Further, for all fixed b > 0,00

N° C U {m=1

LetEm(b) := {x E X I Ifn(x) - f (x)l < 6 V n > ml.

Then {E,,,(6)},,,,>1 is an increasing sequence of sets in S and00

N° C U Em (6).m=1

Thus {E fl E;,,(6)},,,,>1 is a decreasing sequence of sets in S and00

n(EnEmc(6)) C E n N.

Since µ(E) < +oo, we have lim µ(E fl E;,(b)) = 0. Thus given e > 0, 3 non->oo

such that µ(E n Eno (b)) < e. Put E. := Eno (6) fl E. Then

EE C E with

Further, b x c E, by definition,

Ifn(x) - f (x) I < 6, V n > no. 0

8.2.4. Theorem (Egoroff): Let fn, n > 1, and f be measurable functionssuch that fn f. Let E E S with µ(E) < +oo. Then, given e > 0, there isa set EE E S such that

(i) EE C E and (EflE) < e.(11) .fn ) f on E.

Proof: Let e > 0 be fixed. Using proposition 8.2.3 for every 6 = 1/m, wechoose Em E S and a positive integer nm such that

Em C E with (EnE)</2m

andf< 1/m d x E En,, and b n > n,,,,.

Let EE := n,°°_1 Em. Then EE C E and

co

00

µ(E n E°) < E µ(E fl E,°t) < E /2"'' = e.M=1 M=1

Also V m,

f(x)-f(x)I < l/m d x E EE and d n > n,,,,.

Hence fn u) f on E.

8.2.5. Definition: Let f, fn, n > 1, be measurable functions and E E S.We say {f}>i converges almost uniformly to f on E if V E > 0, El EE ES such that

(i) EE C E with µ(E n EE) < e.(ii) fn converges uniformly to f on E fl EE.

We write this as fn a.u. f or fn f a. u.

In view of the above definition, Egoroff's theorem can be restated as: iffn ---> f a.e. on a set E and µ(E) < +oo, then fn ) f almost uniformly onE. As a particular case, if µ(X) < +oo, we have the following proposition.

8.2.6. Proposition: If µ(X) < +oo and fn - f a. e. on X, then fn -- falmost uniformly on X.

The converse of this proposition is also true.

8.2.7. Proposition: If µ(X) < +oo and fn -- f almost uniformly on X,then fn -* f a. e.

Proof: By the given hypothesis, for every integer n > 1 we can choose aset Fn E S such that

µ(Fn) < 1/n and {fTh}Th>i converges uniformly to f on F.

Let F:= n°° 1 Fn. Then

A(F) < A(Fn) < I/n, V n.

Hence µ(F) = 0 and, for x E F°, fn(x) ---> f (x) as n -- oo. Hence fna. e.

f

8.2.8. Exercise: Let fn = x(n,.), n = 1, 2, .... Then {f}>i is a sequenceof Lebesgue measurable functions on R. Show that {f}>i is convergent

8.2 Convergence: Pointwise, a. e., uniform and a. u. 251

pointwise but not almost uniformly with respect to the Lebesgue measure Aon IE .

8.2.9. Exercise: Let fn, n > 1, and f be measurable functions. Fore > 0and m = 1, 2, ... ,let

A':= fx c X1 If,(x) - f (x)l > El.Prove the following:

(1) fa.u. f if

00

for every c > 0, lim U A;,L = 0.n--oo

m=ri

(ii) A P ) f if

for every E > 0, A decreases to 0.n>1

a.e.(iii) f f if

for every E > 0, ,cc

00 00

flUn=1

(iv) If µ(X) < -f-oo, then fn a.e.r f if

for every e > 0, lim µ i x E X I sup 1f,,, (x) - f (x) 1 > e = 0.nm>nAs an application of Egoroff's theorem, we show that on the Lebesgue

measure space (Ill, .C, A) , every measurable function is almost continuous.We need the following class of special simple functions, as considered indefinition 2.1.1.

8.2.10. Definition: Let I C III be an interval and let 4D : I --R bedefined by

n

4D (x) aiXI,(x)i=1

where {I1, I2, ... , In} are pairwise disjoint bounded intervals with I = U=1 Iiand al.... , an are real numbers. Such a function 1 is called a step func-tion on I.

8.2.11. Exercise: Let So denote the class of step functions on an interval1. Let 4b, IF G So and let a e R. Show that 4D + IF, a4b, max{ 4b, if } e So. Dothe functions 14D 1, I+, I-, min{ 4b, if } belong to So?

8.2.12. Exercise: Let 4b be a step function on an interval I, and let e > 0be given. Show that there exists a continuous function g on I such that

A({x E I[8 I b(x) 7 g(x)J) < E.

8.2.13. Proposition: Let f : [a, b] ---> ][8* be any measurable function suchthat f (x) is finite for a. e. x(A), and let e > 0 be arbitrary. Then there exista step function (D and a continuous function g on [a, b] such that

A(f x E [a, b] I If (x) - (D(x) I > EJ) < E

andA(f x E [a, b] I If (x) - g(x) I > EJ) < E.

Proof: We may assume without loss of generality that f is real-valued.

Step 1: Let f = aXA , where a E TIg, A E L and A C [a, b]. Then by theorem4.2.1, we can choose pairwise disjoint open intervals 11, 12,. .. , In such that

A UIZ OA < e.

i=1

Let

i=n+ 1 i=1

where In+l,... , are disjoint intervals. Putn m

a E Xi2 + E aixi. ,

i=1 i=n+1

where each a2 = 0 for i = n + 1, ... , m. Consider the restriction of (D to[a, b]. Then 4b is a step function on [a, b]. Farther,

{x E [a, b] I If (x) - fi(x) I > e} _ 0, if e > 1,

and for0<E<1,

{x E [a,bJI If (x) - fi(x) I ? e} C [() AAJ

.

Hence

A(f x (E [a, b] I If (x) - (D(x) I > EJ) < E.

Using exercise 8.2.12 and (D, we can also find a continuous function g onI = [a, b] such that

mm Inn.

U I, := R U Iz

A(f x EE [a, b] I If (x) - g(x) I > EJ) < E.

8.2 Convergence: Pointwise, a. e., uniform and a. u. 253

Step 2: Let f be any simple measurable function on [a, b]. Let f have therepresentation,

nf : = ai XA2 ,

i=1

where a1, a2.... , an are real numbers and A1, A2, ... , An are pairwise dis-joint sets in £ with U=1 Ai = [a, b]. For each i = 1, 2,... , n, let (Di, gi be thestep function and the continuous function, respectively, on [a, b] such that

A({x E [a, b] I I - (x)I ? E/n}) < E/n

and

PutA({x E [a, b] I IaiX,gi (x) - 9(x)I ? e/n}) < e/n.

(D:=E(Di and g:=Egi.i=1 i=1

It is easy to show that (D and g have the required properties.

Step 3: Let f be a bounded measurable function. Then by exercise 5.3.7,there exists a simple function s on [a, b] such that

If (x) - s(x) I < e/2 V x E [a, b].

Let (D be the step function and g be the continuous function, as given bystep 2, such that

A(Jx c [a, b] I I s(x) - (D(x) I ! E/21) < E/2

and

Then

A({x E [a, b] I 1s(x) - g(x) I > e/2}) < e/2.

{x E [a, 6] I If (x) - (x)I ? c} c Jx E [a, b] Is(x) - (x) I ? E/21

and

{x E [a, b] I I f (x) - g(x) I > e} C Jx E [a, b] I Is(x) - g(x) I > e/2}.

Hence, (D and g have the required properties.

Step 4: Let f be an arbitrary measurable function. For n = 1, 2,... , let

En := Jx E [a, b] I If (x) I > nJ,

Then En is a decreasing sequence of measurable sets and n° 1 En = 0. Thuslim A(En) = 0, and we can choose no such that A(Eno) < e/2. Let

n-+oo

f (x) if x Eno,no ifxEEno.

By step 3 applied to f , we get a step function 4P and a continuous functiong on [a, b] such that

A (f x [a, b] (x) - 4P (x) I > E/21) < E/2

and

A ({x E [a, b] I If (x) - 9(x)I ? E/21) < e/2.Let

A:= fx [a, b] I If (x) - 'P(x)I >- E/21

andA x [a, b] I If (x) - D(x) I > E/21.

Then A(A) < e/2. Since

x G [a, b] I If (x) - 4)(x) I > Ej C Ac Eno u (Enco n A) = Eno u (Enco n A)>

we have

A (f x E [a, b] I I f (X) - 4P (X) I ! EJ) < A (Eno) + A (A) < E.

Similarly, it can be shown that A ({x E [a, b] I If (x) - g(x)I > e}) < E. Thiscompletes the proof.

8.2.14. Theorem (Luzin): Let f : I[8 ]I8 be a measurable function.Then, given e > 0, there exists a continuous function g : I[8 -3 ]I8 such thatA(jx G R I f g(x)j) < E.

Proof: Let us first consider the case when f [a, b] -3 R. Let e > 0 begiven. By proposition 8.2.13, we can find a continuous function gn on [a, b]such that

A ({x E [a, b] I If (x) - 9n WI ? e/3.2"}) < e/3.2".

Let

En := {x E [a, b] I If (x) - gn(x) I > e/3.2n} and E := U E.00

n=1

Then A(E) < e/3, and for x V E,

If W - gn(X) I < /3.2n V n.

Thus gn(x) -> f (x), b x E E, fl [a, b]. Now by Egoroff's theorem (8.2.4), wecan choose F C E° f1 [a, b] such that

A(F) < e/3, and gn(x) -> f (x) uniformly on Ec fl F` fl [a, b].

Using theorem 4.2.2, we choose a closed set C C Ec fl Fc fl [a, b] such that

A((Ec nFc n [a, b] \ C)) < E/3-

8.3 Convergence in measure 255

Then gn ) f uniformly on the closed set C, and

A ([a, b] \ C) < A (E) + A (F) + A ((Ec n F° n [a, b] \ C) < e.

Thus f is continuous on the closed set C. Since [a, b] \ C is open, it is adisjoint union of countably many intervals. We define g(x) := f (x) if x E Cand linearly on the intervals in [a, b] \ C so that g is continuous on [a, b].Clearly,

A{x E [a, b]I I

: 9(x)} < A([a, b] \ C) < E.This proves the theorem in the special case.

When f : ][8 ) ][8, we write ][8 = UOO In, n + 1) and choose, for everyn, a closed set Cn C (n, n + 1) such that f is continuous on Cn and, by theearlier case,

/\(In, n + 1] \ Cn) < /(2n + 1).

Put C := U=_ C. Then it is easy to see that C is a closed subset of][8 with A(][8 \ C) < e, and f is continuous on C. Once again we defineg : ][8 ) ][8 such that g(x) := f (x) V x c C and g(x) is defined linearly on][8 \ C. Then g satisfies the required properties.

8.3. Convergence in measure

Quite often, one comes across sequences of measurable functions on a mea-sure space (X, S, p) which do not converge pointwise or a.e. One would liketo know: do they converge in some other way? Here is one way of analyzingsuch sequences which is very useful in the theory of probability (see note8.3.15).

8.3.1. Definition: Let f, fn, n > 1, be measurable functions. We say thesequence {f}> 1 converges in measure to f if d e > 0,

lim µ({x E X I I A (x) - f(x)I > e}) = 0.

We write this as fn m) f .

8.3.2. Examples:

(i) Let fn,n > 1, be measurable functions on the Lebesgue measure space(][8, G, A), defined by fn := xJn,n+l] Then limn,oo A(x) = f (x) = 0 V x E R.But

A({x c ][8 I Ifn(x)l ? 1}) = /\([n, n + 1]) = 1 V n,and hence {f}>i does not converge to f in measure. Thus pointwiseconvergence or convergence a.e. does not imply convergence in measure.

(ii) Consider the measure space ([0, 1], ,C[0,1] , A), where C[o, l] := ,C l [0, 1] andA is the Lebesgue measure restricted to C[0,1j.

For every n E N, choose the

unique integer m E N such that 21 < n < 2m+ 1, and let 0 < k < 2"2 be suchthat n = 2' + k. The correspondence n (m, k) is one-one, and m --* 00asn -00. For n> 1, let

f n := x1k whenever n = k + 2"2, where Im [k/2m, (k + 1) /2m].m

Then {f}>i is a sequence of measurable functions on ([0,1], £[0,1] , A). Fur-ther, `d x E [0, 1] and given any n c N, let x E Im, for some m and k suchthat n = 2"2 + k and 0 < k < 2"2. Then x E [fo/2m+1, (fo + 1)/2m+1] forsome 2k < fo < 2m+1 - 1. Let n' := Lo + 2m+1. Then

n' =fo+2"2+1 > 2m+1 = 2m + 2m > k + 2m =n

and f / (x) = 1. Thus V x c [0, 1] and V n > 1, 3 n' > n such thatfns (x) = 1, i.e., {ffl}>1 does not converge pointwise to f - 0. On the otherhand, given any E > 0,

Ix E [01 1] 1 Ifn(x)l ! Ej 0 ifE>1,Im if E < l and n = 2"2 + k.{

Thus

A({x c [01 1] I IE}) < 1/2'n, if 2'n < 7L < 2'n+lHence

lim A({x E [0,1] 1 I> e}) = 0,i.e., {fn}n>i converges to f in measure.

The above examples show that convergence in measure neither impliesnor is implied by convergence pointwise (or a.e.). However, when µ(X) <+oo, convergence a.e. implies convergence in measure, as shown in the nextproposition.

8.3.3. Proposition: Let µ(X) < +oo, and let {fTh}Th>1 converge a. e. toThen {f}>i converges in measure to f.

Proof: Recall that {f}>1 converges in measure to f if d e > 0,

n ooµ(lx E XI I- f(x)j > E}) _ 0.

Let

AnW := IX C XI JAW - f (X)l >_ Ej.

Then

AnW C U Am(E)m>n.

J

Thus, limn._,oo p (An (E)) = 0 will be true if limn.._oo i (u> , Am (E) = 0.But, {U>n Am(E) }n>1 is a decreasing sequence of sets in S and it decreasesto n°° 1 U°°_n A,(6). Since p (X) < +00, we have

00

lim tL U Am(E) _m=r

Since fn --) f a.e. and

µ(n1 m-n

00 00

n U A,,,,(e) C {x E X I fn(X) does not converge to f(x)},n=1 m=n

we have

(n u Am E> o.

n=1 m=n

Hence Tli 0, i.e., converges to f in measure.

8.3.4. Exercise: Let and be sequences of measurablefunctions on (X, S, µ) such that converges to fo and con-verges to go in measure. Prove the following statements:

M converges to Ifol in measure.

(ii) {fn f gam,}n>1 converges to fo ± go in measure.

(iii) If fn(x) = gn(x) for a.e. x, then fo(x) = go(x) for a.e. x, i.e., forconvergence in measure, limit is unique a.e.

8.3.5. Exercise: Let (X, S, µ) be a finite measure space. Let {fn}n> o bea sequence of measurable functions such that d e > 0

00

E µ({xEXI I?E})G+oo.

=t

Show that {fn}n>1 converges to fo a.e.(Hint: the given condition implies that converges to fo in measure.)

In example 8.3.2 we showed that convergence in measure need not implyconvergence a.e. However, the following `partial' implication holds.

8.3.6. Theorem (Riesz): Let be a sequence of measurable func-tions converging in measure to a measurable function f. Then there exists asubsequence {fnk}k>1 of such that {fnk }k>1 converges to f almosteverywhere.

Proof: To construct the subsequence which converges a.e. to f,in view of exercise 8.2.9, we should construct such that

nUfX E X I If-k(X) - f W1>- 11kj 0. (8-1)j_1 k-)

For, if00 00

xV A := n U {x E X I Ifnk(x) - f(x)I > 11k},j=1 k=j

then x E fl0{x E X I I- f (x) I < 1/k} for some jo. Thus for everyk > jo, I fnk(x) -f (x) I < 1/k, i.e., fnk (x) -) f (x). So, to complete theproof we should construct a subsequence such that (8.1) holds.Let

00

Aj:= UfX E X I lfnk(X) - f W1 > 11kj- (8.2)k=j

Since A C Aj, we have µ(A) < (A). If we could construct suchthat (A) -) 0 as j --) oo, we will be through. For example, if we couldchoose the subsequence {fflk }n>1 such that

µ({x E X I I fnk (x) - f(x)I 1/k}) < 1/2k+1,

then

00 00

1/2j

and we will be through. So we have only to choose such that(8.2) holds. For this we proceed as follows: since {ffl}fl 1 converges to f inmeasure, given e = 1, we can choose nl such that

µ({x E X I f(x) - f(x)I > 1}) < 1/2.

Suppose nl < n2 < ... < nk_1 have been selected. Choose nk> nk_1 suchthat

µ({x E X I Ifnk (x) - f(x) I> 11k}) < 1/2k+1

using the fact that the sequence {ffl}fl 1 converges to f in measure. Therequired subsequence exists by induction, and this completes theproof.

8.3.7. Note: As an application of the above theorem, we show that thecondition "fn ) f a.e." in Fatou's lemma, the monotone convergencetheorem and Lebesgue's dominated convergence theorem can be replaced

by fn ) f in measure. For example, suppose f, f in measure and3 g E L1(µ) such that I fn I < g V n. Then

fdµ ---> j f dA.

To see this, let us choose any subsequence {fflk}k>1 of {fn}n>1. Then fnkf in measure. By theorem 8.3.6, we can choose further a subsequence{fnk3j >1 of {fnk }k>1 such that fki -) f a.e. But then by Lebesgue'sdominated convergence theorem (5.4.9), f fnkj dµ -> f f dµ. Thus any sub-sequence {f Ik d}k>1 of {f fd}>i has a subsequence converging tof f dµ. Hence

f fd= rl o jffd.8.3.8. Exercise: Let {fTh}Th>i converge in measure to f. Let Ifnj < g b nand for some g E L1(µ). Show that

f Ifn - .fJ0.(Hint: Use exercise 8.3.4 and the note above.)

8.3.9. Exercise: Let {fn}n>1 be a sequence of measurable functions on(X, S, µ). Show that {fn}n>1 converges in measure to a measurable functionf if every subsequence of {f}>i has a further subsequence which convergesa.e. to f.

8.3.10. Exercise: Let {f}>i be a sequence of measurable functions.We say {f}>i is Cauchy in measure if for every e > 0 there exists apositive integer no such that d m, n > no,

A(fx E X1 Ifn(x) - fm(x)l >_ EI) < E-


(a) If {f}>1 is convergent in measure, then {f}>i is Cauchy in mea-sure.

(b) (i) If {f}>i is Cauchy in measure, then there exist positive inte-gers nl < n2 < ... such that b k,

A({x E X1 IfThk(x) - fns+iW I > 1/2k}) < 1/2k.

(ii) The subsequence {fflk is Cauchy for a.e. x, and hencethere exists a measurable function f such that fk (x) _* 1(x)for a.e. x(µ).

(iii) The subsequence {fflk}k>1, and hence the sequence {f}>iitself, converges in measure to f.

8.3.11. Exercise: Let be a sequence of measurable functions suchthat {fn}n>1 converges a.u. to a measurable function f. Show thatconverges to f in measure.

8.3.12. Exercise: Let be a sequence of measurable functions suchthat {fTh}Th>i converges a.e. to a function f. Let 3 g E L1(µ) such thatIfnI < g b n. Show that converges to f a.u. and hence in measure.(Hint: {x E X I fn (x) - f(x)I > e} C {x : g(x) > e/2}, and the latter is aset of finite measure.)

* 8.3.13. Exercise: Let (X, S, µ) be a finite measure space and let A4 (X)denote the set of all measurable functions on X. Identify functions in A4(X )if they agree a.e. (µ). For f , g E JVl (X), define

d(f,g) If W - g(X)I dtz(x).1 -I- If W - 9WI

Show that d is a metric on A4 (X). Further, for a sequence {fTh}Th>' in A4 (X)and f E Nl (X ), d(fn, f) -> 0 as n -> oo if {f}>1 converges to f inmeasure. In view of exercise 8.3.10, (M(X), d) is a complete pseudo-metricspace.

J

(Hint:1 +aIa + bI

<1 +ajal + 1

.}blJbI' and for 0 < e < 1) if jal > e then

IaI

-} jal>

2

*8.3.14. Exercise: Let (M(X), d) be as in exercise 8.3.13. For everyn = 1, 2, ... ,define

Vn :_ f f c A4(X) I µ{x c X I If (x)l > 1/n} < 1/n}.


(a) The family {Vn}n>1 has the following properties:(i) If f E Vn and I AI < 1, then of E Un.(ii) For every n, m there exists k such that UkC Vn n v,,.

(iii) If µ({x E X I f (x) 54 0}) > 0, then 3 n such that f V.(iv) For every f E Nl (X) and b n, I A E R such that A f E V.(v) For every n, 3 m such that Vim, + Vm, C Vn, where

Vn + Vm.: _ {f +9 If, 9 E Vim, } .

(b) For every f c JVl (X), let

Vf := ff + Vn I n = 112, ... I

Then {Vf I f c JVl (X) } defines a local neighborhood base for aHausdorff topology on ,/11(X).

8.4 LP -spaces 261

(c) A sequence { fn}n>1 converge to f in this topology if fn --> f inmeasure.

(d) Clearly Nl (X) is a linear space over ][8 with addition and scalar mul-tiplication defined by

M (X) x M (X) M (X))(f,g) f + 9,

R x A4 (X) M (X))(A) f) Af.

Let Nl (X) x J14 (X) be given the product topology. Show that themaps defined above are continuous. One says that M (X) is a `topo-logical vector space' under the addition and scalar multiplicationdefined above.

8.3.15. Note: When (X, S, µ) is a probability space, convergence in mea-sure is called convergence in probability. Recall (see note 3.11.10) that aprobability space (X, S, µ) gives a mathematical model for `statistical exper-iments'. An observation about the outcomes of the experiment is a functionf on the sample space X. To be able to make probability statements aboutf , it has to be a measurable function. Measurable functions on the sam-ple space X are called random variables in probability theory. Thus asequence of observations in an experiment may not converge toan observation f at every possible outcome x E X. One would like toknow the probability that the observation fn is away from f by an errore, i.e., analyze µ{x I Ifn(x) - f (x) l > e}. So it is natural to ask: doesµ{x I Ifn(x) - f(x)I > e} --+ 0 as n --+ oo, i.e., does fn --+ f in probability?

8.4. LP spaces

In section 5.6, we analyzed Ll [a, b], the space of Lebesgue integrable func-tions on [a, b]. We saw that L1 [a, b] is a vector space over ][8, and that forf E L1 [a, b], we can define the notion of absolute value of f, i.e., I Fur-ther, this notion of absolute value can be used to define the L1-metric onL1 [a, b] so that L1 [a, b] becomes a complete metric space. In this sectionwe look at examples of a family of spaces of this type, called LP spaces.Throughout this section, we fix a measure space (X, S, µ) which is a-finiteand complete.

8.4.1. Definition: Let 0 < p < oo. Let LP(µ) := LP(X, S, µ) denote thespace of all complex-valued S-measurable functions on X such that

i IfPdµ < +oo.

The space Lp(µ) is called the space of pth-power integrable functions.

8.4.2. Proposition: The space Lp(X, S, µ) is a vector space over C.

Proof: Let a E C and f E LP(X, S, µ). Then clearly a f c LP(X, S, µ), asa f is S-measurable, and

J lafIdµ = lf Ifidµ < O+oo.

Also for f, g c LP(X, Xi, µ), it is easy to see that f + g is S-measurable.Further, since

If + gIP < (f+gI)p < 2Pmaxllflp, I9lp} 2p(IfIp+I9IP),

we have

f if + 9I pdµ < 2P (I If id µ + f I< +oo.Hence f + g c Lp(X, S, µ).

8.4.3. Definition: Let f c LP(X, r3,µ). Define Ilf liP, called the p th-normof f, as follows:

1/p

IfI(fufiPd)Since If l= 11g11p if f (x) = g(x) for a.e. x(µ), we treat such f and g as

the same element of Lp(X, 13, . To show that if lip has the properties of ametric (as was the case for p = 1 in section 5.6), we need some inequalities.

8.4.4. Lemma: For nonnegative real numbers a, b and 0 < t < 1, thefollowing inequalities hold:

(i) atbl-t < to + (1 - t)b.

a 2 bWtC

2 (al/t + bllt)

Proof: Let 0 < t < 1 be fixed and consider the function

f (X) := (I - t) + tX - Xt Ix > 0.

To prove (i) we have to show that f (a/b) > 0 = f (1). It is easy to see thatf (x) has a minimum at x = 1 and hence d x > 0,

0 = f (1) < f (x) = (1-t)+tx-xt.In particular, if x = a/b, we have

(I - t) + t(alb) - (a/b)t > 0)

8.4 LP-spaces 263

(a/b)' < t(a/b) + (I - t))

atb'-t < at + (I - t)b.

This proves (i). To prove (ii), we consider the function g(x) = xllt, x > 0.It is easy to show that g is a convex function. Thus b a, b > 0,

((a + b)/2)1/t < 1/2(al/t + blot).

8.4.5. Exercise: Let 0 < t < oo and 0 I + ti/P.

8.4.6. Theorem (Holder's inequality): Let p > 1 and q > 1 be suchthat 1/p+ 1/q = 1. Let f c LP(µ) and g c Lq(µ). Then fg E L1(µ) and

1/P 1/vf fgdlL C If lPdl.L/ (fgId)

Proof: LetlIP

A:= (ffPd)

f6_ rj9X)I,s

B

and B:=

If A = 0, then clearly f (x) = 0 for a.e. x(µ), and hence the required claimholds trivially (with the equality sign). Similarly, if B = 0, the requiredclaim holds. So, suppose A 4 0 and B 4 0. Then by lemma 8.4.4, for allfixed x with

a=

we have

1/q

CI9IgdlL)

and t = 1/p,

If (X)g(x) IIf (X)Ip lg(X)lq

AB - pAP qBq

Thus f g E L1(µ) and

Jf(x)g(x)dµ < (i/p+i/q)AB = AB.

8.4.7. Corollary: Let p, q > 1 be real numbers with 1/p + 1/q = 1. Letf an In> 1 and {bn}n>u be sequences of complex numbers such that

00 00

E IanIP < oo and J:Ibn I q < oo.n=1 n=

Then E', IabnI < +oo and

(

00

YJ I anbn

1/P 00 1/9

IP (>IbnI qn=1

In particular, if an = bn = 0 V n > k, then

k

E lanbnl

11P k \1/q

n (9E Ibnn=1n=1

p

Proof: Consider the special case of (X, S, µ) with X = N, S =1 and apply theorem 8.4.6.

8.4.8. Note: In the special case when p = q = 2, Holder's inequality isknown as the Cauchy-Schwarz inequality.

8.4.9. Theorem (Minkowski's inequality): Let 1 < p < oo and f, g ELP(µ). Then f+gEL() and

lif + glip <_ lif lip + 11911P.

Proof: When p = 1, the inequality is obvious. So, suppose 1 < p < oo. Wehave already seen in proposition 8.4.2 that f +g c LP(µ). Since p = (p- 1)q,it follows that If +gIP-1 E Ly(µ). By Holder's inequality, both I ff + g I P-1and I9I If + 9I P-1 belong to L1(µ) with

1/q

P-1dµ MI IIP ([If + JI9(P-1))= MfMMf + 9II

P19

and

f I9IIf +9IP-1dµ

Thus

1/q

MMP (f If +9I9(P-1)1 = IIMMf +9IIPl9

< (If I + Igl)(If + gIP-')dp

< If Ilf +9p idµ+ f IgIlf +gIP 1dµ

,< (HfMP+ i9liP)i.f +91p/a

f If +9lpdµ

f if +9lf+9P 'dµ

8.4 Lp-spaces 265

HenceIlf + gllp-p,q < Ilf lip + llgllp

lif + glip :! lif lip + 11911P. M

8.4.10. Theorem (Riesz-Fischer): The space LP(X, S, µ), for 1 < p <oo, is a complete metric space with the metric defined by

dP(f, 9) := lIf-glI p b f,g E LP(X, S, µ)

Of course, here we identify f, g E LP(X, S, µ) if f (x) = g(x) for a.e.x(µ)

Proof: That dP(f, g) is a metric on LP(X, S, µ) follows from our identifi-cation and the Minkowski's inequality. The proof of the completeness ofLP(X, S, µ) can be based on the ideas of the proof of theorem 5.6.1. Weoutline the steps below and ask the reader to verify them. Let {f}>i bea Cauchy sequence in LP(X, S, µ).

(i) Choose nl < n2 < < nk < such that b k,

k

(11) Let 9k := I fni I+ lfn+i -fnsj=1

IIIskIPll1

k > 1. Then

00 P

(iiiiiP + E iifflk+1 - f< +0()lk=1

and hence 9k E LP(X, S, µ) V k > 1.

(iii) Let g(x) :_ E X. Using monotone convergence theorem

and (i), deduce that g E LP(X, S) µ), i.e.,

fP

00

Ifni I + E Ifnj+l - fnj I dp < oo.=j=1

(iv) From (iii), using exercise 5.4.2(ii), deduce that00

1(x) fn(x) - (fn+(x) - f(x))j=1

exists for a.e. x(µ), i.e., fflk(x) -> f(x) for a.e. x(µ).

(v) Deduce that f E LP(X, S, µ) by showing that I f,,+, IP < (g) V k,applying Lebesgue's dominated convergence theorem and using (ii)above. Finally, using the inequality lfki - f IP < gP, deduce thatllfflk-flIP--Oask---o0.

(For an alternative proof, see also exercise 8.4.15.)

8.4.11. Note: In theorem 8.4.10, we showed that L(), 1 < p < oo, is acomplete metric space under the metric

dn(.f, 9) := Ill -9In

It is natural to ask the same question for 0 (I + ty V t > 0)

it is easy to see that dP is a metric on LP(µ). Also, proceeding as in theorem8.4.10, we can show that LP(µ) is a complete metric space for 0 0, and Ill IIP = 0 if f = 0.

(ii) Ilaf 11p = IaIIIf 11P

(iii) I I f + iI P C If lip + iigll, called the triangle inequality.

Such a function can be defined on any vector space, and is called a norm.A vector space with a norm is called a normed linear space. Thus for1 < p < oo, the LP-spaces are examples of normed linear spaces whichare also complete under the metric Ilf - glIP, the metric induced by thenorm. Such normed linear spaces are called Banach spaces. However,when 0 < p < 1, f H If IIP is no longer a norm, as it fails to satisfy thetriangle inequality. To see this, consider a measure space (X, S, µ) such that3 A5 B E S with A fl B = 0 and 0 < µ(A) < oo, 0 < µ(B) < oo. Let a,,3 bepositive real numbers. Then

lIPIIaXAIIP -

I

= a(p(A)) 1/P.

8.4 LP -spaces 267

Similarly, II/3XBIIp = Q(1_t(B))l1p. Further,

1/p

[fxA+xB)Pd]IX,q + Xg III

[f(PxA11/p

= +,3pXB) dµJ

(app(A) +,3P1_t(B))11P.

Now, using exercise 8.4.5(ii) with t = (,31ce)P(p(B)1p(A)), we have

+0 P p(B) 1/P > 1+ 0 ( p (B) 1 IP

(a ) lp (A) ) ( cae ) pt -(A) )

(al/pµ(A) +)311p1_t(B))11P > a(1_t(A))11p+Q(µ(B))11p5

McXA+i3XBllP > llXAIP+IXBMPThis is the reason that LP -spaces for 0 < p < 1 are not very interesting tostudy.

8.4.12. Exercise:

(i) Let 0 G p G 1 and -oo <q < 0 be such that 1/p -}- 1/q = 1. Let f,g bepositive functions such that f E LP(µ) and g E L9(µ). Show that

If lill9 < ffgd.(Hint: Apply Holder's inequality to (fg)P E L11P and g-P E L_Pl9.)

(ii) If µ(X) < oo and 1 G p G q < oo, show that LQ(µ) C Lp(µ).(Hint: Apply Holder's inequality to If if E LQ1p(µ) and 1 E L(y_p)/q(µ).)

8.4.13. Definition: Let {f}>i be a sequence of functions in LP(-),1 i converges in LP (or inthe pth mean) to f if IIfn - f Ilp --+ 0 as n -+ oo.

The next theorem describes the relation between convergence in the pthmean and other modes of convergence.

8.4.14. Theorem:

(i) If {f}>i is a sequence in LP and it converges in LP to f E LP, then

IIfnIIP If M.

(ii) In general, convergence in the pth mean does not imply any one ofuniform convergence, or almost uniform convergence, or convergencea. e.

(iii) Convergence in the pth mean always implies convergence in measure.

(iv) In general, none of uniform convergence, or almost uniform conver-gence, or convergence in measure, or convergence a. e. imply conver-gence in the pth mean.

(v) If the underlying measure space is finite, then uniform convergenceimplies convergence in the pth mean, but none of almost uniform con-vergence or convergence a. e. need imply convergence in the pth mean.Also, convergence in the pth mean need not imply almost uniformconvergence or convergence a. e.

Proof: The proof of (i) follows from the inequality

iIfn li - IIfIII <_ lifn - flIP

(ii) Consider the measure space ([0,1], Goo 1j, A) and the sequence fas in example 8.3.2 (ii), i.e.,

fn X k , Im := [k/2m, (k + 1)/2m], with n = k + 2'n.Irn

Since each fn is the indicator function of a subinterval of [0, 1], fn E Lp(A).It is easy to see that for 1 i converges to f - 0 in the pth mean. However, we havealready seen that {fn}n>i does not converge to f at any point of [0,1] .

Clearly, convergence in the pth mean cannot imply uniform or almost uniformconvergence, because on finite measure spaces both imply convergence a.e.Also, convergence in the pth mean does not imply convergence a.e., as shownabove.

(iii) Let {fn}n>i and f E LP(X, S, µ), and let converge to f inLP. Let e > 0 be arbitrary and

E:= {x c- X I I? E}.

8.4 Lp-spaces

Then

fn (x) - f (x) lPdp (x) I fn(x) - f (x)lPdp(x)fE

Ifn(x)-f(x)lPdp(x)E

Epp(E) + I fn (x) - f (x) lPdpfEc

> Epp (E).

Thus µ(E) < IIfn - f II/c. This implies that {f}>imeasure.

269

converges to f in

(iv) Consider the Lebesgue measure space (][8, G, A) and let

fn(x) := n-1/PXtpnj (x), x E R.

Then { fn}n>1 converges to f - 0 uniformly. However,

f IfnlpdA = 1 V n.

Hence {fn}n>1 does not converge to f in the pth mean.

Next consider the measure space ([0, 1], Gto,l], A) and let

gn(x) -.= nl/p XtO,l/nl (X) I X C 101 11 -

Then it is easy to see that {gn}n>i converges almost uniformly, in measureand pointwise to f - 0. However,

l9nlPd = 1 V n,fo,l]

and hence {gn}n>i does not converge to f in the pth mean.(v) Let (X, S, µ) be a measure space such that µ(X) < oo, and let

Ifnln>l be a sequence of functions in Lp(X, S, µ) converging uniformly toa measurable function f on X. Then given e > 0, we can find no such that

I

IfI P = (Ifn - f I + Jfnj)p< (2maxjjfn - f 1, Jfnjj)P

< 2p(Ifn - f Ip + Ifnlp))

we have

f If jPdy < 2p J Ifno - .f lPdµ + 2P J Ifno I pop

< 2pEpµ(X) + 2pII fnll < +00

Hence f c Lp (X, S, µ) and, for n > no,1/p

IIfn - f II P - \ Ifn - f I < E(A(X))1/P.

Thus {f}>i converges to f in the pth mean. That none of almost uni-form convergence or convergence in measure or convergence a.e. need implyconvergence in the pth mean follows from the second example in (iv) above.Also, even if µ(X) < +oo, convergence in the pth mean does not implyconvergence a.e. (as given by the example in (ii)). Thus convergence inthe pth mean cannot imply either uniform convergence or almost uniformconvergence.

8.4.15. Exercise: Prove that LP(X, S, µ) is complete under the metric dpas follows: Let {f}>i be a sequence in LP(X, S, µ) which is Cauchy withrespect to the metric dp. Show that

(i) {f}>i is also Cauchy in measure.

(ii) sup f IfnlPdµ < +oon

(iii) Using exercise 8.3.10 (b), find a subsequence {fflk}k>1 of {f}>iwhich is convergent a.e. to, say, f. Use (ii) to deduce that f ELp(X, S, µ). Show that {fflk}fl>1, and hence {f}>i itself, convergesto f in LP (X, S, µ).

8.4.16. Exercise: Let f E Lp(X) s, µ), where 1 0,

µ({x E XI I ? 0}) < If IIp/'.This is called Chebyshev's inequality for LP functions (see also exercise5.4.5.).

8.5. *Necessary and sufficient conditions forconvergence in LP

As we saw above, in general convergence in measure or convergence a.e. neednot imply convergence in the pth mean. One would like to find sufficientconditions for convergence in the pth mean. Let f, fn c Lp (X) ,t3, µ), n > 1.Let {f}>i converge to f in Lp(X, S, µ). Since

IIfItP - Ilf IIPI <_ Ilfn - f IIP>

8.5 Necessary and sufficient conditions for convergence in LP 271

it follows that {IfII}>i converges to Ilf 11P Conversely, suppose f, fn ELp(X, S, µ), n > 1, are such that fn --> f a.e. and I--+ Ilf 11p Can weclaim that {f}> 1 converges to f in LP? The answer is in the affirmativeand is given by our next theorem.

8.5.1. Theorem: Let 1 1. Letfn --> f a.e. and IIfIjn --> IThen IIfn - f II p --> 0 as n -> oo.

Proof: We first note that by lemma 8.4.4 (ii),

Ifn - f IP < 2P-'(IfnIP + If IP).

Since Ifn - f IP -> 0 a.e. and gn 2p-1(I fnI p + If I 2pIf IP withf If IPdi < +oo, it follows from exercise 5.4.13 that

lim If, - f IPd1_L = 0. 0n--+oo

In order to state some more necessary and sufficient conditions, we in-troduce the following definition.

8.5.2. Definition: Let C be a collection of integrable functions. We saythat 9 is equicontinuous at 0 if for any E > 0 and for any decreasingsequence {En}n>1 of measurable sets with n°° 1 En = 0, 3 no such that

lEnj9Idµ<e d gE9 and b n>no.

8.5.3. Examples:

(i) Let = {g1, 92, ... , g I be any finite collection of integrable functions.Then V j, 1 < j < k,

lim IgIdii=0n-9oolEn

whenever {E}>1 is a decreasing sequence of measurable sets such thatnn,=, En = 0 (see proposition 5.2.6 and theorem 3.6.3). Thus V j, 3 njsuch that

n > nj.nfEi

If no := max{nl, ... , nk}, then

lEnI9jIdµ<e b n>no and 1<j<k.

Thus 9 is equicontinuous.

(ii) Consider the Lebesgue measure space (R, L, A) and let

gn := X[n,n+1] , n = 1, 2, ... .

Consider the sequence where Ek = [k, oo) . Then is adecreasing sequence and n, 1 Ej = 0. But

I&IgIdA = 1 for n > k.

Thus G = {g, 92, ... } is not equicontinuous.

8.5.4. Definition: Let g be a collection of integrable functions. We say 9is uniformly absolutely continuous if, given any E > 0, El b > 0 suchthat for E E S

µ(E) < S implies J IgIdp < e V g E .E

8.5.5. Example: The collections 9 of examples 8.5.3(i) and (ii) both areuniformly absolutely continuous. Example 8.5.3(ii) shows that every collec-tion Gwhich is uniformly absolutely continuous need not be equicontinuous.However, the converse is always true, as shown in the next proposition.

8.5.6. Proposition: If a collection 9 of integrable functions is equicontin-uous, then G is also uniformly absolutely continuous.

Proof: Suppose is not uniformly absolutely continuous. Then d n,El e > 0, a set En E S and a function fn E 9 such that

µ(En) < 1/2n but fE I fId> e.n

Let Fn := U`0 n Ek, n = 1, 2, ....Then {F}>i is a decreasing sequenceof measurable sets. Let F n°°_1 Fn. Then

00 00

p(F) p(Fn) < >(Ek) < E 1/2n = 1/2n'-1

k=n k=rr,

Thus µ(F) = 0 and

9d= f 9d>_ f 9d>f \F n n

But, {Fn \ F}n>1 is a decreasing sequence of sets and nn,=, (Fn \ F) = 0.This contradicts the equicontinuity of G.

8.5.7. Exercise: Let (X, S, µ) be a measure space and µ(X) < oo. Let Gbe a collection of measurable functions. Show that 9 is uniformly absolutelycontinuous if g is equicontinuous.

8.5.8. Theorem: Let {f}>i be a sequence of functions in LP(X, S) µ),1 i converges to f c LP if {f}>i converges to fin measure and C :_ {IfIP I n = 1, 2.... } is equicontinuous at 0.

Proof: Suppose that {f}>i converges to f in L. We know, by theorem8.4.14, that {fTh}Th>i converges in measure to f. To check the equicontinuityof C :_ {IfIP I n = 1, 2,... }, let e > 0 be given. Since {fTh}Th>i is convergentin LP, it is also Cauchy in LP, and thus El no such that

f d n and m > no.

Let {Ek}k>1 be any decreasing sequence in S with na1 Ek = 0. Usingproposition 5.2.6 and theorem 3.6.3, d n we can choose a positive inte-ger kn such that

LkIfIdi < E/2P+1 V k 1 kn.

If ko : = max{11, k2, , kno }, then V n < no and k > ko, we have

LkAlso, since

IfnlPdµ < e/2P+i

IfnIP G 2Pllfnp1P + Ifnp - fnlPl

we have, for n > no and k > ko,

f- fjd)IfIdµ < 2P( f If0Id /L + fEk I.Ek Ek

< ZP (E/2P+1 + E/2p-f-1) = E.

Hence {IfIP I n = 1, 2.... } is equicontinuous.Conversely, let {fn}n>1 converge in measure to a measurable function f ,

and let C :_ {IfI° I n = 1, 2, ... } be equicontinuous. We have to show thatjfnjn>1 converges in LP. In view of theorem 8.4.10, it is enough to showthat is Cauchy in L. Let c > 0 be given. We consider two cases.

Case (i): Let µ(X) < +oo. Since C is also uniformly absolutely continuous(proposition 8.5.6), we can find S> 0 such that d o> 1 and V E E Swith µ(E) < 6,

EIfnlPd[L < c/2P .

Let

En,k x E XI I-.fk(x)I >_2

274

Then

8. Modes of convergence and Lp-spaces

I Ifn-fkIdµ < 2 µ(X)< E/2 .

n,k

Also, since {fn}n>1 is convergent in measure, it is Cauchy in measure andhence we can choose a positive integer no such that V n and k > no,

p(En,k) < 6.

Then it follows from (8.3) that V n > 1 and k > no,

f/2p+2.

fn,kHence V n>1 and k > no,

Lfl,kfrifkVdP <

c Jn,k ckTi

< 2p (E/2p+2 + e/2P+2) = e/2. (8.5)

From (8.4) and (8.5) it follows that {f}>i is Cauchy in L. This completesthe proof for case (i).

Case (ii): Let µ(X) = +oo. Since µ is a-finite, we can write X = U°O_1such that b n > 1, (X) < --oo. Let

Ej

00

U Xn,j =1, 2, ... .

n=j

Then {E3}1 is a decreasing sequence of sets with n _ 1 Ej = 0. From theequicontinuity of C at 0, we have an integer jo such that V n,

f IgnIPdp < e/2p+2go

Thus

Since

by case (i),

I 9- 9Pd2p f + 2p f 9pd< . (8.6)go go go

IEjcO

1 .fn lPdpf IfId µ+-2p f

j0-1

E A(X3) < -boo,

Ign - g k I Pdµ < e for all sufficiently large n, k.

Hence, combining (8.6) and (8.7), we see that {g7}7>i is Cauchy in LP andhence is convergent.

8.5.9. Corollary (An extension of the dominated convergence the-orem): Let be a sequence of measurable functions on (X, S, µ)such that gn -* g in measure, where g also is a measurable function. Leth E Lp(µ) be such that IgnI < h a.e. (µ) for all n. Then convergesto g in LP.

Proof: In view of theorem 8.5.8, we only have to show that 9 = { 1 gn I P I n =1, 2, ... } is equicontinuous at 0. Let {En}n>1 be any decreasing sequence ofmeasurable sets with nn, 1 En = 0. Then by theorem 3.6.3 and proposition5.2.6,

limIn-oo lEn

Thus, given e > 0, 1 no such that

lEn

Hence V n > no and V m,

IhIdµ<e, b n>no.

I dµ < IhId µ < e,fE fEn

proving equicontinuity of {gP I n = 1, 2.... } at 0.

In the case y(X) < +oo, another useful criterion for convergence in LPis described in terms of `uniform integrability' of functions, which we definenext.

8.5.10. Definition: A collection C of measurable functions on (X, S, µ) issaid to be uniformly integrable if

lim sup J I= 0.t +00 (fE9

8.5.11. Examples:

(i) Let {f}>i be any sequence of measurable functions such that b nIfnI < g a.e., g an integrable function. Then {f}>i is uniformly integrable.To see this, first note that since g is integrable,

limJ

g(x)dµ(x) = 0.t'°° {X I g(X) > t}

Thus, given e > 0, we can choose t := t(E) such that

g(x)dµ(x) < e.Ix I s(x) > 0

Since by the given hypothesis

{x I I>_ t} C {x 19(x) >_

I>_ t} U N,

where µ(N) = 0, we have

f I f(x)I d(x) f 9(x)d(x)X I I fn(X)I >_t} x I s(x)>t}

(ii) Let be any sequence of integrable functions such that {fk Ik > 1}is a finite set, say {f, 12,... , fn I. Then is uniformly integrable,since

IfkI s9lfjldk.j=1

(iii) If {f}>1 and are uniformly integrable, then { fn + gn}n>1 isalso uniformly integrable. For this, let hn := max{lfnl, 19n1}. Then we haveI fn + gn I < 2hn, and hence

1XI I + 9n(X)1 > 2t} C {x I ttn(x) > t}.

Thus

if I fn (X) + gn (X) I dtz (x)

< 2J

hn(x)dµ(x){X I hn(x)>t}

< 2 IfIx hn (x)d(x)

+2 Jf{x hn(x)dµ(x)

2I 1 fn(x) I dl-z (x)

+2J 1 9n W I dµ(x)

From the above inequality it follows that Un + gn}n>1 is uniformly inte-grable.

8.5.12. Exercise:

(i) Let µ(X) < +oo and let g be uniformly integrable. Show that 9 C L1(µ).

(ii) Consider the measure space ([0,1] ,,C[o,1], A) and let fn := (n/log n)X[Ol/]

n = 1, 2, ....Show that {f}>1 is uniformly integrable and f fndA -> 0,although {f}Th>1 is not dominated by any integrable function.

When µ(X) < oo, an equivalent description of uniform integrability isgiven by the following theorem.

8.5.13. Theorem: Let !9 be a family of measurable functions on (X, S, µ)where µ(X) < oo. Then the following are equivalent:

(i) !9 is uniformly integrable.

(ii) supyEg (f I9I dµ) < +oo, and !9 is uniformly absolutely continuous.

Proof: Suppose (i) holds. Then, given e = 1, we can choose t large enoughso that

supgE!9

Thus d g E !9,

f I9(x)I dµ (x)

Ig(x)dp(x) < 1.

I lg(x)l dlL(x) + f I 9(x)I dlL (x)

 0 be given. Since g is uniformly integrable, we can choose tsufficiently large so that b g E !9,

Then `d E e s,

Elg(x)l d[L =

lg(x)l dµ(x) < e/2.

{xI Is(x)I?t}nElg(x)l dµ(x) + f

xl 19(x)I<t}nE

c/2 + t1i(E).

Hence if we choose 8(e) = E/2t, then d E E S

µ(E) < b(e) implies J I9Idµ < e b g E .E

This proves that (i) ==>. (ii).

Conversely, suppose (ii) holds. Let

K := sup J Il dµgE9

Then d t> O and g E G,

9(x)I dµ (x)f I9I dµ = f l

9(x) I dµ (x)fx I I9(x)I fit}

> tA(fx I 1g(X) I > t1)

Thus

I9I dµ < K/t.µ({x I 19(x)I >_ t}) C tit

Let e > 0 be given. If we choose b > 0 as given by the uniform absolutecontinuity of C, then for t > K/8 we will have

A(fxl lg(x)l > tj) < Klt < J.

Thus for all t > K/S,

tXI I9(x)I>t}

This proves that (ii) =: (i).

I 9(x)I dl-L (x) < E.

8.5.14. Theorem: Let {g,,},,>1 be a sequence of measurable ,functions on(X, S, µ) with µ(X) < -boo. Let g be a measurable function on (X, S, µ).Then the following statements are equivalent.

(i) {gP}>1 is uniformly integrable and gn --> g in measure.

(11) {gPn = 1, 2, ... }(iii) {gPn = 1, 2, ... }

measure.

(iv) gn -49 in LP.

is equicontinuous and gn -f g in measure.

is uniformly absolutely continuous and gn -f g in

Proof: In view of exercise 8.5.7 and theorem 8.5.8, we have (ii) 4=: (iii) 4=:(iv). To complete the proof we show that (i) 4=: (iv).

Suppose is uniformly integrable. Then it follows from theorem8.5.13 that is uniformly absolutely continuous and hence equicon-tinuous at 0, by exercise 8.5.7. Thus (i) =: (ii) q (iv). Conversely, let (iv)hold. Then (iii) holds. Also, being Cauchy in LP, given e > 0 wecan choose no such that

19no-gnIPdµ<c, b' n>no.

8.6 Dense subspaces of LP 279

Thus b n > no,

f J9nj1op < 2P finn - gno Ildµ + 2' J Igno I pop

ZpE + 2p J Ig0 I dµ

Hence

SAP I I9nIpol < max S 2PE + 2p I I 9no I f j9i Ipdµ, .. .l

< +oo.

I9no -1 I pdp

It follows from theorem 8.5.13 that {gP I n = 1, 2.... } is uniformlyintegrable.

8.6. Dense subspaces of LP

In section 5.7 we showed that L1(R) has nice dense subspaces. In this sec-tion, we first construct dense subspaces of Lp(X, s, p) when (X, s, p) is anabstract measure space, and later specialize to (Rn)£i n , An), the Lebesguemeasure space, An denoting the Lebesgue measure on Rn. As before, let(X, s, p), a a-finite, complete measure space, be fixed. We shall call a func-tion s E Lp(X, s, p) simple if both Re (s) and Im (s) are simple functions.

8.6.1. Theorem: Let f E LP(X) S, µ),1 0. Then thereexists a simple function s E LP(X, S, µ) such that II f-sIlp < e and Isl < If I

Proof: First suppose that f > 0 and f E L. Then by proposition 5.3.2there is an increasing sequence {s}> 1 of nonnegative simple functions suchthat {sn(x)}n>1 is increasing and sn(x) -> f (x) for a.e. x(µ). Thus

0 < If - SnIP < fP a.e. (p).

Hence sn E Lp(µ) V n and, by Lebesgue's dominated convergence theorem,

lim Jif - snIpdµ = 0.

Thus, given e > 0, we can choose no such that

I SnoI C If I and Ilsno - f11p < E.

For f c LP (µ) arbitrary, we can write

f = (Ref )+ - (Ref )- + i((Im f)+ - (Imf)),where each of (Ref)+, (Ref)-, (Im f )+ and (Im f ) - is a nonnegative func-tion in Lp(µ). Thus, given c > 0, we can choose simple functions si, 1 < i < 4such that

I< (Ref), IS21 < IIs31 5 IIS41 5 I

withIlsi - Ref)+IIp < E/4, 1152 - Ref) III < E/4>

and

Put11S3 - (Imf)IIp C E/4, 11S4 - (Imf)11p < e/4.

S := Si - 82 + 253 - s4).

Then s is a simple function and s E LP(µ). Further, Ilf - sIIP < e Also,4

IS 12 = (S1 - 522 + (s3 - 542 < Sa C If 12,

i=1

as (Ref)(Ref)- = 0 and (Imf)(Imf)- = 0.

8.6.2. Theorem: Let 1 < p < oo, and let Cc(R') denote the space ofcontinuous (complex-valued) functions on Tn with compact support. ThenCc(Rn) is dense in Lp(W , I. Rn , An).

Proof: Let g E Cc (][8n) and let K := supp(g), the closure of the set f X EI[8n I g(x) 4 0}. Then K is compact and

IMP < (supI

Since An(K) < +oo, it follows that IgIp is integrable, and hence g E LP := LP(Wi, Gin, an). Thus Cc (I[8n) C L. To show that Cc(I[8n) is dense, let E > 0be arbitrary and let f E Lp. We have to find a function g E Cc(1[8n) suchthat If - hIp < E. For this, we may assume without loss of generality thatf is real-valued and f > 0. Further, by theorem 8.6.1, we can find a simplenonnegative function s such that I- slIP < E. Since s is a finite linearcombination of indicator functions of sets with finite Lebesgue measure (ass E LP), we have only to prove the theorem in the case when f = XA, A EGin and An(A) < +oo. Since An is regular, by exercise 7.5.3, we can find acompact set K and an open set U in I[8" such that

K C A C U and An (U \ K) <

Next, using Urysohn's lemma (see appendix D), we can find a continuousfunction g : I[8n -; [0, 1] with supp(g) C U, such that g(x) = 1 V x E Kand g(x) = 0 V x U. Clearly, g E LP and

Ig-xAI <XU - XK.

Hence

g - XAI PdAn < fixU - XK IpdAn < 2pAn (U \ K) < EP.

Thus IIg - XA Ilp < E. This completes the proof.

8.7 Convolution and regularization of functions 281

8.6.3. Corollary: Let 1 0 be given, and let f E Lp(I[8n). By theorem 8.6.2, 3 g EC,(][8n) such that

11f - gllp < c/3- (8-8)

Since g is continuous and has compact support, say K, g is uniformly con-tinuous. Thus 3 S > 0 such that

I9(x) - 9(y)I < e/3(2An(K))1/P, whenever Ix - yj < S.

Thus b h E I[8n with Jhj <J, we have1/p

Mg - 9h.IIp - C f I9(x) - 9(h + x) lPdAn(x)\

I

1/p

l9(x) - 9(x +

h C. I[8n with I hI < S, we have

If - fh.llp C 11f - 911p + Mh - 9IIp + IIh - fhMp

Since 119h - fhMp = I- f11p, from (8.8), (8.9) and (8.10) we have

11f - fhllp < e whenever 0 < IhI < J.

Hence, limh-+oI - f 11 = 0 0

8.7. Convolution and regularization of functions

(8.s)

(8-10)

In this section all a.e. statements are with respect to the Lebesgue measureon Ian . Also, we denote L1(TR', LRn , An) by L1(Rn), n > 1.

8.7.1. Example: Let f c Ll (IlS) and let a be a positive real number. Define

fxa+a

(Z'af)(x)1

f (t)dA(t), x E .

The function Taf is called the symmetric moving average (or smooth-ing) of f. It is easy to see that (Ta f) (x) is well-defined. Let

F(x) -.= f--00

f (t)dA(t) x c R.

Then

(Taf)(X) = F(x + a) - F(x - a)x E I[8.

2aSince F is absolutely continuous (see theorem 6.1.1), it is easy to check thatTaf is also absolutely continuous. Also,

1 rx+a

f(t)dA(t)2a x-a

1

Xfx-a,x+a] (t)f (t)dA(t)2 aj1RR

11

2a f X[-a,+a] (t - x)f (t)dA(t)

1

2a f x[-a,+a] (t)f(t + x)dA(t).

The last equality holds because of exercise 5.6.6. Thus

(Taf)(X) - 2a I f (t + x)X[-a,+a] (t)dA(t).

Hence

fIRI< JIRR l= 111111.

Thus Taf E L1(I[8) and

u =1

2a I

f f(t+x)x[_a,+a j (t)dA(t)

1

2a JIRR

If(t + a) - f(X)IX[_a,+aI(t)dA(t)

Hence

liTaf - fill < 2a fR (fmR If (t + x) - f(X)1X1+1 (t) dA (t) I dA (x).

Using Fubini's theorem, we have

lITaf fill

where

<2a Jg X[-a,+aj

(t) I fm l+ x)- f dA(t)

2a

if(t) h(t) dA(t),

h(t) :- JMR If (t + x) - f (x)IdA(x).

By corollary 8.6.3, h(t) --+ 0 as t --+ 0. Thus, given e > 0, we can choosea > 0 sufficiently small so that

h(t) < E5 V t E [-a, +a].

Hence for this choice of a, we have

IITaff Iii < 2a fRR (t) h(t) do(t) < E.

Thus, given e > 0, we have found an a > 0 such that Taf is continuous (infact absolutely continuous) and is close to f in the L1-metric. Note that(Ta f) (x) is nothing but the average of f over an interval centered at x oflength 2a. Even though f may not be continuous, its averaging is continuous.This example is a particular case of a general procedure, which we describenext.

8.7.2. Definition: Let f , g : Ian -* C be measurable functions. When-ever, for x E R',

f if (x - y) I I< -i-oo>denewe define

(f * g) (x) f (x - y)g(y)dAn(Y)-we

The function f * g is called the convolution of f with g.

8.7.3. Example: Consider Taf, for f E L1(][8) and a > 0, as defined inexample 8.7.1. Then for x E R,

X[x-a,x+aJ (t) 1(t) dA(t)(Taf)(X) - 2a f2a f X[-a,+aJ (t - x) f (t) dA(t)

2a f X[-a,+al lx - t) 1(t) dA(t)

-X[-a,+aj f W._(2a

8.7.4. Theorem: Let f, g be measurable functions on R. Then the follow-ing hold:

(i) If (f * g) (x) exists, then so does (g * f)(x), and

Y * OW = (g MX).

(ii) If f E L1(I[8n) and g is bounded a. e., then (f * g) (x) exists b x E I[8nand is a bounded, uniformly continuous function.

(iii) If f, g are in C, (I[8n), then so is (f * g).

(iv) Let p, q be real numbers such that 1 < p, q < oo and 1/q + 1/p = 1.If f E Lp(][8") and g E Ly(][8n), then (f * g)(x) exists d x E lI8n.Moreover, f * g is a continuous function and it vanishes at infinity,i. e.,

lim (f*g)(x)I=O.lxl->oo

(v) If f, g E L1(]I8n), then (f * g) (x) exists for a. e. x, and f * g E L1(IlSn)with

Proof: (i) Let (f * g) (x) exist, i.e.,

f if (x - y)I I< +00-Since

(f * g) (x) f (x - y)g(y)dAn(Y),

making a change of the variables, y to y + x followed by y to -y, and usingexercise 7.5.3, we have

(f*g)(x) = f f(X)g(Xy)dAn(y) = (g*f)(x).

(ii) Since g is bounded a.e., there exist a set N and a real number M > 0such that µ(N) = 0 and Ig(x) I <11 V x E N°. Then V x E IIBn,

f if I I=N

fif(x-y)I I<

M J If (x - 2J)IdAn(2J)

= Nlll.f II1 < +00.Hence (f * g) (x) exists and I(f * 9) (x) l < NI IIf II1i V x E 1[8". Thus it is abounded function. For x, z E ][8,

I- (,f * 9)(z)I f' i- .f (z - y)I I<

MJ

If (x - y) - f(z - y)I dAn(y)

= NlII.fX-z - fIl1.

Further, given E > 0, if we choose S > 0 by corollary 8.6.3 such that

Ix - zI < S implies 11fx-z - f 11 < e/Nl,

then

I (f * 9)(x) - (f * 9)(z) < E whenever lx - z < E.This proves (ii).

(iii) Let f, g E C, (R'). Let

Kl := supp(f) and K2 := supp(g).

Then f (x-y) = 0 V (x-y) V Kl and g(y) = 0 V y V K2. Thus f (x-y)g(y) _0 whenever either y ¢ K2 or x VK1 + y. Hence

(f * g)(x) = 0 if x V (K1 + K2), i.e., supp(f + g) c (K1 + K2).

That f * g is continuous follows from (ii), since f, g are bounded and f, g EL1(IIBn).

(iv) Let 1 < p, q < oo with 1/p + 1/q = 1. Let f E LP(I[8n) and g ELQ(II8n). Using Holder's inequality, we have for every x,

f if (x - y)I I9(y)I dAn(y) -< JjfyjjPIIgIIq = If IIPII.III9 < +00-

Hence (f * g) (x) exists for every x E 1[8'x, and

I (f * 9) (X) I -< 11 f Ilp 119 Ilq-

Thus f * g is a bounded function. To show that f * g is continuous andvanishes at oo, we first show that f * g is a uniform limit of a sequence fromCc(][8't). For this, since f c LP(Il8n) and g c Ly(II8n), by theorem 8.6.2 wecan choose sequences and in Cc (I[8n) such that

11fk - f 1P ) 0 and llk - llq ) 0

By (iii), A * A E Cc(][8n) V k, and by Holder's inequality we have d x E 1[8'x,

I (fk*gk)(x) - (f * 9) (x)

C I (fk * 9k) (x) - (fk * 9)(x)I + I Yk * 9)(x) - (f * 9)(x)I

< f l x - y)I I (9k- 9)(y)I dAn (J)

+ f I (fk- f)(x - y)l l9(y)IdA,I(y)

-< llfllllgk - 9IIq + 11A - fllpllglIqFrom this it follows that {(fk * gk)}n>1 converges uniformly to f * g. Thusf *g is continuous. Further, let e > 0 be given. Choose k such that d x E 1[8nsup Il < e.

x(z-dBn

Let M > 0 be such that (fk * gk)(x) = 0 V x c Il8n with lxi > M. Then forsuch x,

I (f * 9) (X) I I(A * 9k)(X) - (f * 9)(X)I + I(fk * 9)(X)II (fk * 9k) (X) - (f * 9) (X) I

< SUP I(A *gk)(X) - (f *9)(X)l E-xEOe^

Hence lim I (f * g) (x) I = 0, i.e., f * g vanishes at oo.IXlyOO

(v) Let f, g E Ll (][8n). Consider the function

(f , g) (x, y) : = f (x) g (y), (x, y) EE (Rn X Rn).

Clearly (f, g) E L2 (][82'x). Thus using Fubini's theorem we get

IIfII1IIgII1 I f (x) I dAn (X) ) (f lg(y)IdAn(Y)

1(1 If (x - Y)I I(Y)Idn(x)) dAn(y)

= fflfx-Y1 Ig(y)Id2(x,y)00.

Hence it follows from exercise 5.4.2 (ii) that

f if (x - y)I I9(y)I dAn(y) < +oo for a.e. x.

Thus (f * g) (x) exists for a.e. x, and

f I(f * 9)(x)IdAn(x) -< If If (x - y)I I9(y)I dAn(x)dAn(Y) = IIi lThus

f * g e L1(I[8n) with Ilf * 91l, < Ilf 1

8.7.5. Exercise: Let f, g, h E L1(][8n) . Show that

(i) f*(g+h)=f*g+f*h.(ii) f*(g*h)=(f*g)*h.

8.7.6. Definition: Let f be a measurable function on Rn. We say f islocally integrable if f is integrable on every compact subset of Rn.

Let Li C (Rn) denote the space of all locally integrable functions on IR .

8.7.7. Exercise: Prove the following:

(i) f E Li c(I[8n) iff b x E I[8n 3 a neighborhood V,, of x such that f isintegrable on V.

(ii) If f E Lp (][8n), then f vanishes at oo.

(iii) If f is a bounded measurable function on ][8n then f E Li°c(][8n)

8.7.8. Definition: Let U be an open subset of Rn and let f : U -) R. Wesay that f is a C°°-function on U if f has continuous partial derivativesof all orders at every point of U. We denote the set of all such functions by

CO°(U). Let C°°(U) denote the set of those functions from C°°(U) whichhave compact support.

8.7.9. Example: We have already seen in corollary 5.7.2 that f : III -* III,defined by

f(x) :- exP(1/(x2 - 1)) if lxi < 1,0 if 1xI > 1,

is a C'-function on R. In fact, f E C°°(][8) with supp(f) = [-1, +1]. Con-sider the function II8n ) R defined by

O(x) := f(lxl), x E ][8",

where Ix12 = (x+ +xn) for x = (xi,... , xn) E 1[8n. Clearly, 0 < O(x) < 1and supp(O) = {x E 1Ig'tj jxj < 1} := B(0,1). It is easy to check that 0 is aC°°-function on W.

8.7.10. Theorem: Let 0 E be a nonnegative function such that

(x) = 0 d x c Il8n with jxj > 1, and J ¢(x)dAn(x) = 1.

Let e > 0 be arbitrary andOe(x) :=

E-no(x/E)> x E Ilgn.

Then the following are true:

(i) ¢E is a nonnegative C°°-function on ][8n with

supp(oE) C B(O,e) {x E Ilgnj 1xI < e} and 110E111 = I.

(ii) Let f c Li °(1[8n) and fE 0E* f. Then fE C°° (1[8n). If f is uniformlycontinuous, then fE f uniformly as e 0.

(iii) Let f E Lp(Il8n) and fE :_ ¢E * f. Then fE E Lp(][gn) fl CO°(l[8n) andfe > f in L.

The functions fE are called regularizations of f.

Proof: (i) Since 0 is nonnegative and 0 E C°O(1[8) with supp(O) C B(0,1),clearly 0E is also a nonnegative CO°-function on 1[8n with supp(¢E) C B(0, e).That lc5EII = 1 follows from the fact that I10 111 = 1 and exercise 7.5.3.

(ii) Let f c Li°°(][8n) and fE :_ ¢E * f. Clearly, fE is well-defined and

f (x) f'E (Y) (O'E * RX) - (O'E * My)

= fc(x - z)f(z)dAn(Z) - fc(y - z).((z)dAn(z)

(0,,(x -i)-OjY-z))f(z)dan(z). (8.11)

Since 0E is uniformly continuous with

0E(x - z) - OE(y - z) = 0 if z 0 B(x, e) U B(y, c)

and f is integrable on compact sets, it follows from (8.11) that fE is uniformlycontinuous. Further, using (8.11) and the dominated convergence theorem,it is easy to show that fE is a C°°-function on ][8". In fact, d k

(9 k fe (9 koe

ax axe* f.

Finally, to see that fE f uniformly whenever f is uniformly continuous,we note that

h (X) J mE(=-y)f(y)dan(y)

O(X Y)f (y) dAn (Y)

,n j (f 0 (z)f(x -EZ)dAn(z)

Thus

I fE(x) - f (x) f (x - cz) 0 (z) dA,, (z) f f (x) 0 (z) dA,, (z)

j if (x - cz) - f (x) I O(z) dAn(Z)-

Since f is uniformly continuous, given q > 0 we can choose co such that'd E < eo, jzj< 1 aid d x E ][8n ,i< rThus V x,

Jf6(x)-f(x)j < O(z)dAn(Z)

Hence fE --+ f uniformly.

(iii) Let f E Lp(I[8n). Then, by exercise 8.7.7 and (ii), fE :_ 0E * f is welldefined, and fE E C°°(Rn). To show that fE E LP(I[8"), we note that

fE (x) = j f (x - cz)O(z)dAn (Z)

whereJ

f (x - Ez)dµ(z),

O(z)dAn (z), V E E Gin.

It follows from proposition 5.2.6 that µ is a measure on .CRn and µ(I[8') = 1.Now, using Holder's inequality, we have

f(x) < \i If (x - ez)IPdµ(z))1/P

(f1dµ)\ 1/v

where q > 1 is such that 1/p + 1/q = I. Since µ(Il8n) = 1, we have

f IfE(x)lPd/\n(x) < J I J f(x - ez)IPdµ(z)lI d/\n(x)

= J J If (x - ez)lPd/\n (x) ) dµ(z)

= flflld(z)lif 11P (8.12)

Hence fE E LP(I[8n). To show that fE --+ f in LP, let 77 > 0 be given. Usingtheorem 8.6.2, choose g c Cc(Il8n) such that if - 9IIp < 77/3. Then

llf - f lip < lf - 9E11P + ig - 9IIp + i- f lip. (8.13)

Also, using (8.12), we have

IIg'E-.fellP- Iilg -.f)eIIP < Mg - fM.

Further, since g is uniformly continuous, (i) implies that gE -- g uniformly.Since supp(gE) C supp(g) U B(0, e), it follows that gE -* g in Lp also. Hence

Ilg - 9IIp < 7/3 for all sufficiently small E.

From this and (8.13), it follows that for all sufficiently small e > 0 we haveIIfE - f lip < 7, i.e., fn -- f in Lp.

8.7.11. Corollary: The space C°(Il8n) is dense in LP(I[8n).

Proof: Let f E L(W) and let E > 0 be given. Then using Lebesgue'sdominated convergence theorem, we can choose a positive integer k suchthat

11 f - XB(o,k) f llP < e/2. (8.14)

Since XB(O,k) f c LP(I[8n) and has compact support, we can choose e > 0such that (xB(O,k)f)c E C°°(I[8n) and

iKxB(o,k)f)c - XB(p,k)f lip < e/2. (8.15)

Hence for g := (XB(Ok) f) E CcO°(I[8n), it follows from (8.14) and (8.'5) that

lif - glip < E.

290 8. Modes of convergence and Lu-spaces

In view of theorem 8.7.10, it is natural to ask the question: For x E ][8"and f a locally integrable function, when does f (x)?Since f 0E(y)dAn(y) = 1, we have

I fE(X) - f W I (O'E * f)(X) - f W I

0"(X - Y)f (y)dAn(Y) - f W

= I J 0" (x - y) (f (y) - f (x)) dAn (Y)

C (1/En) f 0 (x EY

) I - 1(x) I dAn(Y)

C (1/E") sup 10(y)j f f(y) - f(x)d(y)yEB(x,1) >E)

In case n = 1, the above inequality is

I fe W - f(x)I 1(x). Such points are the Lebesgue points of f. Since almostevery point of R is a Lebesgue point for f c L1(][8) (see theorem 6.3.2), wehave the following:

8.7.12. Corollary: If f c L1(]I8), then

f(x) := (0E * f)(x) ' f(x) for a. e. x(A).

8.7.13. Note: Corollary 8.7.12 is also true for f c L1(]I8n). See note 9.2.12.

8.7.14. Note: Let X be a Banach space over R (or C) under the norm 11 ' 11.

One calls X a Banach algebra if there exists a binary operation, referredto as multiplication, from X x X to X, denoted by (X) y) H may, such thatfor x, y, z E X and a E R (or C), the following hold:

(i) (Yz) = (xy)z.(ii) x(y + z) = xy + Xz; (y + z)X = yX + zx.

(iii) a(xy) _ (ax)y =m(ay).

A Banach algebra X is said to be commutative if the multiplication onX is commutative, i.e., xy = yx b x, y E X. We say X is a Banachalgebra with identity if it is a Banach algebra and there exists someelement e c X such that ex = xe = x, d x E X. If we consider the Banach

8.8 Lc, (X, V , ,u) 291

space L, (RI) over C with convolution of functions as multiplication, then itfollows from theorem 8.7.4 and exercise 8.7.5 that L1(][8') is a commutativeBanach algebra. It is easy to show that L1 (][8't) is a Banach algebra withoutidentity, i.e., there does not exist any g E L1(I[8n) such that g * f = f forevery f E L1(I[8n). For example if n = 1, suppose that for some g E L1(R)we have g * f = f, b f c L1(I[8). Choose any real number b > 0 such that

+2d

- 25l9(x)IdA(x) < 1,

which is possible by proposition 5.4.6(iii). Let f = Then f E Li(R)and b x E l[8,

x+b

f fix) = (g * f) W = f_d

9(t)dA(t).

In particular, for x E [-b, +S], 1 = f (x) = (g * f)(x) . ThusX+S

fXT.

x+d 25

1 I I -d g(t)dA

(t) ` -s19(t)I do(t) < f-25 I9(t)I do(t) < 1,

which is a contradiction. Hence Ll (l[8) is a commutative Banach algebrawithout identity. A family of functions in L1(I[8") is called an ap-proximate identity if it has the following properties:

(i) Each fE is nonnegative.

(ii) For each e > 0, 1.

(iii) For every neighborhood V of 0 E I[8n, lim fV ff(x)d)fl(x) = 0.

An obvious example of an approximate identity for L1(Rn) is given by thefollowing. Let

1

fe A (B (0, XB(O,,E)) > 0'

Example 8.7.9 and the functions 0E, e > 0, as defined in theorem 8.7.10 tellus that Ll(I[8n) has an approximate identity from C,(I[8n). Approximateidentities are useful because of their properties as given in theorem 8.7.10and corollary 8.7.12.

8.8. LOO (X, S, M) : The space of essentiallybounded functions

8.8.1. Definition: A measurable function f X -) ][8* or C is said tobe essentially bounded if there exists some real number M such thatm(fx Cz X I If W1 > MI) = 0.

We denote by L,,,. (X, S, µ) the set of all essentially bounded functions.For f E L,,,.(X, S, µ) define

If 1100 := inf {M I µ{x E X I I I > M} = 0}.

We call 1111100 the essential supremum of f.

8.8.2. Theorem: Let f, g E L,,. (X, S, µ) and let a,,3 be scalars. Then thefollowing hold:

(i) a f c L,,. (X, S, µ) and

(ii) (f + g) E Lc)o (X) B, µ) and

(iii) lif Iloo = 0 iff f (x) = 0 a.e. x(µ).(iv) There exists a set E E S such that µ(E) = 0 and

Ill ll= SUP If(x)IxEX\E

Thus µ({x c XI If (x)I > I= 0.(v) If 1100 =sup{N I µ({x E XI If (x)I > N}) > 0}.

Proof: (i) Since V M E R and a a scalar,

{x E X I if(x)1 > M} = {x E X I I(af)(x)I > M1al},

it is easy to see that (a f) E (X, B, µ) for every f c Lco (X, B, µ) and

Haf Iloo = Ial Ilf II.-

(ii) Let ,f, g E (X, ,t3,µ) and let e > 0 be arbitrary. We can chooseMl, M2 E lE8 such that

µ({xEXI I>Ml})=0 with l E/2

and

µ({x E X I I I > M2}) = 0 with 11g1100 > M2 - e/2.

Let

E1:={xEXI If(x)l >Ml} and E2:={xEXI Ig(x)I >M2}.Then µ(E1 U E2) = 0 and, for x ¢ El U E2i

I<- f(x)+g(x) I <- M1 +M2.

Thus{x E X I If (x) + g(x) l > (Ml + M2) } C E l U E2i

and henceµ({x E X I If (x) + g(x)l > (Ml +.A/12) }) = 0.

Further,

If + Ml + Nl2 < IIfII00+IIgII00+.

8.8 L00 (X, S, A) 293

Since e > 0 is arbitrary, we have

Ilf + g1loo < Ilf Iloo + Ilglloo-

This proves (ii). The proof of (iii) is easy. To prove (iv), let

E := Ix C- X I If (x)I > Ilf Iloof.

Then00

e= U {x E xl I(MI M00 +i/n)}.n=1

By definition of If Iloo, for every n > 1 we can find an Mn > 0 such that

Mill00 < Mn < Ilf Iloo + 1/n and µ({x c X I If (x) I > Mn}) = 0.

Thus for every n > 1,

tt(f x G X I If (x) I > Ilf Iloo + 1/nJ) = 0

and hence µ(F) = 0. Define

f * Wf(x) if x ¢ E,

0 ifxEE.Since f *(x) = f (x) for a. e. x(µ), we have

IIM00 = Mf*Moo =sup I I = sup f(x).xEX xEX\E

This proves (iv).

To prove (v), note that if N is such that µ({x E X I lf(x)I > N}) > 0,then N < If M00. For if not, then we will have by (iv), for a.e. x,

I I < If M00 < N,which is not true. Hence

a:= sup{N I µ{x E X I If (x) I > N} > Of < Ilf M00

On the other hand, clearly

tt(f x C- x I If W1 > Ilf Iloof) = tt(x) > 0.Thus a > I III also. This proves (v).

8.8.3. Corollary: For f, g c Loo (X, S, µ), let

doo(f, g) := Ilf - g1j.-

Then for f, g, h c LcO(X, S, µ), the following hold:

(i) dco(f, g) > 0, and = 0 if 1(x) = g(x) for a. e. x(µ).(ii) (f, g) < (f, h) + (h, g).

(iii) If {fn}n>i is a sequence in Loo (X, S, µ), then doo (fn, f) -- 0 as n0o if fn f uniformly a. e.

-3

(iv) If {fTh}>i is a Cauchy sequence in LOO (X, S, IL), i. e., d00 (fn, fm) --> 0as n, m --> 00, then 3 h E LQO (X, S, I,c) such that doo (fn, h) --> 0 asn --> oo.

Proof: (i) and (ii) are easy. To prove (iii), let doo (fn, f) --> 0 as n --> oo.Then by theorem 8.8.2(iv), there exists En E S such that µ(En) = 0 and

IIf-fIIoo= sup f(x)-f(x).IxEX\En

Let E = U=1 E. Then µ(E) = 0 and b x ¢ E,

f(x) -f (x)IC Sup J.f (x)j- jj.fn- .f jjoo.xEX\En

Hence fn --> f uniformly on X \ E.Conversely, if fn ---+ f uniformly a.e., let A E S be the set such that

µ(A) = 0 and fn -+ f uniformly on Ac. Then, given e > 0, 3 no such thatd x ¢ A and n > no, Ifn(x) - f (x)l <E. Thus d n > no,

fxEiXI lf(x)l >e}CX\A.Hence µ({x E X I Ifn(y) f (x) I > })=0. Thus jjfn - f < e V n > no,i.e., dc, (.fn) f) --> 0.

(iv) Let dc)o (fn) f,,,,) -- 0 as n, m oo. By theorem 8.8.2(iv), 3 En,En,m E S such that µ(En) = 0 and µ(En,m) = 0 with

IIfnfmIIoo = supIxEX\En,,,,

and

IfII= sup f(x)I.xEX\E.,

Put00 00

E:= U En U U Ek,m .

n=1 k,m=1

Then µ(E) = 0, and {f}>1 is uniformly Cauchy on X \ E. Thus there is ameasurable function h such that fn --> h uniformly on X \ E. We first showthat h is bounded on X \ E. Note that d x E X \ E,

l5 f

sup lh(x)l < sup I- h(x)l + Jj.fnjjoo.X\E X\E

8.8 Loo (X, s, µ) 295

Since f, --+ h uniformly on E and doo (fn, f,,,,) --+ 0, it is easy to see thatboth terms on the right hand side of the above equality are bounded for alln sufficiently large. Finally, if we define h(x) = 0 on E, then h E L. Since

IIfnhIIoo < sup Ifn(x) - h(x)l,xEX\E

it follows that doo (fn, h) -* 0.

8.8.4. Remarks:

(i) For f, g E Loo (X, S) µ), let us write

f ^' g if f (x) = g(x) for a. e. x(lt)

Then is an equivalence relation. We denote the set of equivalence classesby L,,(X, ,t3,µ) itself. For any element f c LOO(X, S, µ), which is in factan equivalence class, we can define IIfIIoo by choosing any element in theequivalence class f. Then If 1k0 is well-defined, and LOO(X, S, µ) is a Banachspace under this norm.

(ii) We showed in theorem 8.5.1 that for f, f E LP(X, S, µ), where 1p < oo and n = 1, 2, ... , then dP(fn, f) -* 0 whenever fn --+ f a.e.and IIfnjjP --+ If IIP The corresponding result is false in LOO(X, S, [t). Forexample, consider Loo([0,1],,C[o,l], A), fn := X(1/n,1), n = 1, 2, .... Thenfn --+ f - 1 and IIfIk, = 1 = IIfII. However, {f}>i cannot converge to

f,forIIf-fII=1 V n.

(iii) In theorem 8.6.2, we showed that C,(Ilgn) is dense in LP(Il8n) for 1 < p <oo. This is not true for Loo(II8n). The reason for this is simple. Note that forf E Cc(IIBn), llfII00 = supxERn I f (x)1. Thus if C, (Ilgn) were dense in LOO (I[8n),then every function in Loo(I[8n) would become continuous, which is obviouslynot true. The closure of C,(][8n) in the LOO(IIBn) topology is Co(][8n), the spaceof continuous functions vanishing at infinity. (See theorem 8.7.4(iv) for thedefinition of a function vanishing at infinity.)

8.8.5. Exercise: Let f E Lc,,,, (X, S, µ), where (X, S, µ) is a finite measurespace. Prove the following statements:

(i) For 1 0 is such that µ({x E X II

> N}) > 0, show that

N(p(X))'IP < 11f 11p.

(iii) Let {pn}n>i be any monotonically increasing sequence of real numberssuch that pn > 1 V n and is not bounded above. Using (i),(ii) above and theorem 8.8.2(v), show that

IIfIk0 IIfII.

8.9. L2(X, S, µ): The space of square integrablefunctions

The reason for analyzing the space L2(X, S, µ) in detail is the following. Wehave already seen in sections 8.4 and 8.8 that for 1 < p < oo, the spacesLp (X, S, µ) are linear spaces and have a notion of distance (norm) underwhich they are complete. Let us recall, for f c L2(X, Si µ), that

1/2If (X) 12 dp(x)Ilf 112 ` (I

If we treat f c L2(X, S, p) as a `vector' with uncountable components, f (x)being the xth component, x E X, then If 112 can be viewed as a generalizationof the Euclidean magnitude of a vector in R'z with summation being replacedby integration. We recall that on R'z, we have the notion of dot product ofvectors, which is related to the magnitude of vectors and helps us to definethe notion of angles on '. That such a dot product can also be defined onL2(X, s, µ) and enables one to do geometry on L2(X, S, µ), is the reason wediscuss this space in detail.

8.9.1. Definition: For f, g E L2(X, S, µ), we define

(f, g) = f f (x) 9(x) dµ(x)>

whenever it exists, where g denote the complex conjugate function: g(x)g(x) b x EX.

8.9.2. Proposition (Cauchy-Schwarz inequality): For every f, g EL2 (X, S, µ), (f, g) is a well-defined scalar and

The scalar (f, g) is called the inner product of f and g.

Proof: It follows from Holder's inequality (8.4.6) with p = q = 2.

8.9.3. Proposition: For f, g, h E L2(X, S, µ) and E C, the followinghold:

(i) (f, f) > 0, and equality holds iff f = 0.

11) Yl 9) = (9, f).(iii) (af +,(3g, h) = a(f, h) +,(3(g, h).(iv) (f, ceg + Oh) = d(f, g) + 0(f, h).

(V) Ilf I12 = (fi f)1/2.

8.9 L2(X> S, µ) 297

Proof: All the statements are easy to check and are left as exercises.

8.9.4. Note: In the Cauchy-Schwarz inequality, the equality holds if f =(f, g)g, i.e., f and g are linearly dependent.

8.9.5. Exercise: Let be sequences in L2 (X, S, A) suchthat 1 m MM fn - f 112 = 0, l 11gn 9112 = 0 for f , g E L2 (X, S, M). Show

0o nthat

(fn) h) -* (f, h), (h, gn) (h, g), (fn, gn) ' (f) g)for every h E L2 (X , s, M). In other words, the inner product map ( , )L2 x L2 -f C is continuous in each variable and jointly.

8.9.6. Note: Given an arbitrary vector space H over the field R (or C), ifthere is a map (., ) : H x H ) R (or C) having the properties (i) to (iv) asgiven in proposition 8.9.3, then it is called an inner product space. Onevery inner product space H, it is easy to show that 11U11 := (u, U)112, U E H,is indeed a norm on H, called the norm induced by the inner product.Further, the Cauchy-Schwarz inequality holds:

(u, v)i < IIuIi !vM, Vu,v E H.

This can be proved as follows: For u E H, let I= ((U> u))1/2. If eitherU = 0 or v = 0, then clearly I(U, v)1 = 0 = 1jull lvi!. So, let u,v E H besuch that u 4 0 and v 4 0. Then !ul > 0 and !!vM > 0. Let u = u/Ijull and

v-1 = v/jjvII Then (u , u) = 1 = (v', v') and

0 < (U' - (U'j V')V'l U, - (U V )v

(u', 71)+I(u,v')I2(v',v') -2I(u,v')I21 _ I (U'l V') 12.

Hence

(u> v)I < 1uMMvM.

If H is also complete under the norm induced by the inner product, thenH is called a Hilbert space. Thus L2(X, S, µ) is an example of a Hilbertspace. We shall not go into the general theory of Hilbert spaces. We discusssome results for L2 (X, S, µ) which are in fact true, without any change inthe arguments, for general Hilbert spaces also.

8.9.7. Exercise (Parallelogram identity): Let f,g E L2(X, S, µ).Then

If + g + MI - gI2 2l2 ° zlltll?+2119112

298 8. Modes of convergence and LP spaces

8.9.8. Definition: Let f, g E L2(X, S, µ). We say that f and g are or-thogonal, and write f 1 g, if (f, g) = 0. For a subset S of L2(X, X3, µ), wewrite f 1Sif(f,h)=0 bhES.

8.9.9. Exercise:

(i) Let f E L2(X, S, µ) be such that f 1 g b g E LZ(X, S, µ). What canyou conclude about f?

(ii) Pythagoras identity: Let f, g E L2(X, S, µ) and f 1 g. Show that

I 2 = IIfII + IIgII.

8.9.10. Exercise (Bessel's inequality): Let fl, f2, ... , fn be elementsof L2(X, S, µ) such that IIfiII2 =1 V i and fi 1 fj for i j. Show that

jyj fj)12 < 11f 112

j=1

(Hint: Consider (z, z), where z = f _Enj 1(f, fi)fi.)

8.9.11. Definition: Let S be a nonempty subset of L2(X, S, µ).

(i) We say S is a subspace of L2 (X, S, µ) if d a, ,8 E C and f, g EL2 (X, ,t3, µ), we have a f + /3g E S.

(ii) We say S is a closed subspace if it is closed under the II ' II2 metric,i.e., for every sequence {f}> 1 in S with lim IIfn - f 112 = 0 for some

f E L2 (X, S, µ), we have f E S.

8.9.12. Example: Let (X, S, µ) be a (complete) measure space. Let Sobe a a-algebra of subsets of X such that So C S and such that (X, So, µ) isalso complete. Let

M :_ f f E L2(X, S, µ) I f is So-measurable}.

Clearly M is a subspace of L2(X, S, µ). In fact, M is a closed subspace ofL2(X, S, µ). To see this, let {f}>i be a sequence in M such that fn --> fin L2(X, S, µ). Then it follows from theorems 8.3.6 and 8.4.14(ii) that thereexists a subsequence of {f}>i such that fnk --> f a.e. (,a).Since each fnk is So-measurable, it follows that f is also So-measurable, i.e.,f E M. Hence M is a closed subspace.

8.9.13. Proposition: Let S be any nonempty subset of L2 (X, S, µ) and let

S1 := {g E L2(X, S> µ) I (.f, 9) = 0 d f E S}.

8.9 L2 (X, S, µ)

Then S1 is a closed subspace of L2 (X, S, µ).

The set S1 is called the orthogonal complement of S.

299

Proof: It is easy to see that S1 is a subspace of L2 (X, S, p) . That S1 isalso closed, follows from exercise 8.9.5.

We next prove a result which seems geometrically obvious.

8.9.14. Theorem: Let f E L2 (X, S, µ), and let S be a closed subspace ofL2 (X, S, µ) . Let

a := inf {11 f- g11 2 19 E S}.

Then there exists a unique function fo c S such that a = I I f - foII2. Further,i f f S then 0 : (f - fp) 1 S.

Proof: We first prove the uniqueness of fo. Suppose fo and fi E S are suchthat

11f - f0112 = I- h112 = inf {IIf - 9112 19 E S} := a.Since fo, fi E S and S is a subspace, we have (fo + fi)/2 E S and thus

a <I

= I.

Also, by the parallelogram identity (exercise 8.9.7),11f, _ f0112 = 11(f, f) _ (fo _ f)112

2 2

= 211fi - f 112

2 2+ 2IIf - foII2 - I2

Hence 1- fo11z = 0, i.e., fl = fo a.e. This proves the uniqueness. To provethe existence, by the definition of a, 3 gn c S such that

lim hI9n - 1112 = a.

Since (9n + gm)/2 E S V n, m, we have llgn + g,112/2 > a, and the parallel-ogram identity gives us

II 9n - 9m II2 = 2II9n - .f II2 + 2II9m - .f II2 - II9n + 9m - 2f II22 2 2

< 2II9n - ,f II2 + 2II9m - .f II2 - 4cx2.2 2

Hence lim 119n - 9mII2 = 0. This shows that {g}>i is a Cauchy sequencen,M-,oo

in L2(µ). By the completeness of L2(µ), there exists fo c L2(µ) such that

lim 1ign-foll = 0n-+oo

Since S is closed, clearly fo c S. Further,

IIfo-f 1I2 - Ig-f 111 C I

0

Figure 26

Thus

I- 1112 = lim IL9n - f,, = a.Next, suppose f S. Then h := fo - f # 0. For every fixed g E S, considerthe function

11h + OgI121 (C.

Clearly, fo +Og E S, and hence

0(0) = 11h + OgI12 = 11(fo + 0g) _ f 112 > a2)2 2

i.e., a2 > 0. Since a2 = 11h 112, we have

0 (h, g) + 0 (g, h) + 10 12 (g, g).

In particular, if we choose 0 = t(h, g), t E ][8, then

0 < 2tj (h, 9)12 + t2l (h, g)I2II9II2dt E R.

The right hand side is a quadratic in t and it has a minimum at t = 0. Hencethe coefficient of the linear term must be zero, i.e., (h, g) = 0. Thus h I S.This proves the theorem completely.

8.9.15. Corollary: Let S be a proper closed subspace of L2(X,,Ci, µ). ThenS-L 71- fol.

8.9 L2 (X, Si µ) 301

Next we give two applications of theorem 8.9.14. Our first applicationsays that if S is any closed subspace of L2(X, S, µ), then L2(X, S, µ) _S + S1 :_ If + g1f c S, g c S1} with s f1 S1 = {0}.

8.9.16. Theorem: Let Sl, S2 be subsets of L2(X, S, µ). Then the followinghold:

(i) Si is a closed subspace of L2(X, S, µ) and Sl fl Si c {0}. If Si isalso a subspace, then Sl fl Si = {0}.

(ii) Si C S2 if S2 C Sl.(iii) Sl C (St)', with equality if Si is a closed subspace of L2(X, S, µ).

(iv) If Sl and S2 are closed subspaces and f I g d f c Si and d g c S2,then Si + S2 := If + g I f E Sl, g E S2} is also a closed subspace.

(v) If Si is a closed subspace, then Sl fl Si = {0} and L2 (X, S, µ) _Si + Si . Thus every f c L2 (X, S, µ) can be uniquely expressed asf = g + h, where g E Sl and h c Si .(This is also expressed as L2(X, S, µ) = Si ED Si ,and is called theprojection theorem.)

Proof: (i) and (ii) are easy to check.To see (iii), clearly Si C (8j-)'. Suppose Sl is a closed subspace and

f c (8)'. By theorem 8.9.14, 3 fo c Si such that (f - fo) E S.Also, fo c Sl C (8j )1 and f c (sL)I imply (f - fo) E (8L)I. Thus(f - fo) E Si f1 (8L)I = {0}. Hence f = fo E Si. This proves (iii).

µ).To prove (iv), it is easy to see that Si + S2 is a subspace of L2 (X, B,To show that it is also closed, let {f}>1 be a sequence in Sl -I-- S2 such thatIIfn - f 11 2 --+ 0 as n --+ oo for some f c L2(X, S, µ). Let fn =gam, + ham,, wheregn c Si and hn E S2 V n. Since (ham,, gn) = 0 V n, by Pythagoras' identity(8.9.9(ii)) we have

119n - 9m1122 +" 11hn - hmII2 - II (9n + ttn) - (9m -I- ttm) 112

Hence both and {h}>i are Cauchy sequences in L2 (X, S, ,cc) . Letg, h c L2 (X, S,1 t) be such that

lim Mg-gM2 =n-4oo n- +oo0 = lim 11hn- h112

But then g c Si and h c S2, as both Si and S2 are closed. Also,

11fn - (g+h)2 C llg-gll2+ 11hn - h112.

Hence lim I I fn - (g + h) 11 2 = 0. Thus f = g + h. This proves that Si + S2n-->oo

is a closed subspace of L2(X, ,t3, µ).

Finally, to prove (v), let Si be a closed subspace. That Si fl Si = {0}follows from (i). Also, Si + Si is a closed subspace of L2(X, ,t3, µ). Since

Sl C (Si + Si) and Si C (Si + Si ),we have

(Si + Si )1 c Si and (Si + Si )1 C (Si )1 = Sl.

Thusf ol c (s, + s-L)-L c si n s-L c f ol.

Hence (Si + Si )-L = {0}, and by corollary 8.9.15,

S, @ SiL = L2 (Xi S5

As our second application of theorem 8.9.14, we characterize all boundedlinear functions on L2, as defined next.

8.9.17. Definition: A map T : L2(X, S, µ) --> C is called a boundedlinear functional if it has the following properties:

(i) For every f,g E L2 (X, S, µ) and E C,

T(af +Qg) = aT(f) +,6T(g).

(ii) There exists a real number M such that

I7'(.f)I <- MIIfII21 b f c La(X,S) µ).

8.9.18. Proposition: Let T L2(X, S, µ) --> C be such that b f, g EL2 (X, S, µ) and a, Q E C

T(af +,8g) = aT(f ) + OT(g).

Then the following are equivalent:

(i) T is bounded, i.e., 3 M E R such that

IT(f)I <_ Mill 112, d,f c Lz(x>s,µ)

(ii) T is continuous.(iii) T is continuous at 0 E LZ(X, S, µ).

Proof: Suppose (i) holds and f E L2(X, S, µ). Let {f}>i be any sequencein L2(X, S, µ) such that Jjfn - f 112 --+ 0 as n --+ oo. Then

I7'(fn) -7'(f)I = I7'(fn - f)l C MIJfn - f 112.

Thus {T(f)}>i converges to T(f), showing that T is continuous (in factuniformly continuous), i.e., (ii) holds.

8.9 L2(X, S, µ) 303

Clearly, (ii) (iii). Finally, let (iii) hold. Then given e > 0, 3 b > 0such that JT(f) j < e b f E L2 (X) S, µ) with 111112 ! J. If g E L2 (X, B, µ) isarbitrary and IIM2 7 0, then

Thus

II2 = aII119112

T(IIsIIZ<E,1e,IT(s)I<Ellall2

Hence (i) holds with M = e/b.

8.9.19. Example: Let g E L2 (X, s, p) be fixed. Define the map

T9 : L2(X,B, t) --*C

as follows:

7's(f) = (f, 9) = ffdvfeL2(xs,).It is easy to see that T9 is linear, i.e.,

Tg (cef 1 + )3h) = CeTg (f 1) + OTg (h)

V E C and 11,12 E L2(X, S, µ). Also, by the Cauchy-Schwarz inequal-ity,

ITg(f)l = 1(f, 9)1 _< 11911211f 112-

Hence Tg is a bounded linear functional on L2 (X, S, µ). We note that g E(Ker(T9))', where

Ker(T9) := f f E L2(X, B, µ) I Ty(f) = 0}.

Ker(Ty) is called the kernel of Ty. Our next theorem tell us that everybounded linear functional on L2(X, S, µ) is of this type.

8.9.20. Theorem (Riesz representation): Let T : L2(X, B, µ) --) C bea bounded linear functional. Then there is a unique go E L2(X, S, µ) suchthat

T(f) = (f, go) V f EE L2 (X, S, A) -

Proof: We note that in case there does exist a go E L2(X, S, µ) such thatT(f) _ (f, go) V f, then clearly go E (Ker(T))', where

Ker(T) := f f c L2(X, B, µ) I T(f) = 0}.

So to find go, first note that Ker(T) is a subspace of L2(X, S, µ), because T islinear. Further, since T is continuous, Ker(T) is in fact a closed subspace ofL2 (X, ,t3,µ). Now, there are two possibilities. First, Ker(T) = L2 (X, S, µ),i.e., T(f) = 0 V f E L2(X, S, µ), in which case T(f) = (f, 0), b f E


L2 (X, S, µ). On the other hand, if T (f) # 0 for some f c- L2 (X, S, µ), thenKer(T) is a proper closed subspace of L2 (X, S, µ). Hence (Ker(T))' =A {0}by corollary 8.9.15. Let 0 # g E (Ker(T))'. This g may not be the requiredgo, for the required go has to have the property that

7'(90) = (90, go).

Let a := T(g). Then a # 0, for otherwise we will have g E Ker(T) fl(Ker(T))J- = f0j. i.e., g = 0.

(g, gl(9, 9)

Let go :_ (a/(g, g))g. Then d,Q E C,

T (,3g) = 3T (g) = a,3.

Also,

Hence

Wg' go) = 0 (g, go) = 0 (g,a

-

-9) = a,3.(g) g)

T(,8g) = (,3g, go), V,3 E C.

Also,

(8.16)

T(f) = 0 = (f, go), if f E Ker(T). (8.17)

For a general f E LZ (X, S, µ) with f Ker(T), note that (f -/3g) E Ker(T)if

0 = T(r - 09) = Tcn - OT (g))

T(g)T(f)'

Thus if f Ker(T), then

C f - T f g I E Ker(T)

and hence, by (8.16) and (8.17),

= T T(g) g +T ( T(g)T(f) (f - T(f) 7'(f)

((g))T(g) / T(g)f - T(f ) g' go ) + \ T(f ) 9)90

U, 90) -

The uniqueness of go follows from the fact that if (f, go) = (f, gi) V f EL2 (X, S, µ), then (f, go - gl) = 0 d f E L2 (X, S, µ). In particular, if wetake f = go - gl, then IIo _g,112

z=0 and hence go = 9i

8.9 LZ (X, S, µ) 305

8.9.21. Note: In theorem 8.9.16, we showed that, given a closed subspaceM of L2 (X, S, µ), every element f E L2(X, S, µ) can be written uniquely asf = g + h, where g c M and h E M1. Let us denote g by PM(f), calledthe projection of f onto M. Geometrically, PM(f) is the unique bestapproximator of f in M. The properties of the map PM L2(X, S, µ) ->L2(X, S, µ) are given in the next theorem.

8.9.22. Theorem: PM L2 (X, S, µ) --+ L2(X, S, µ), as defined in note8.9.21 above, has the following properties: V f, g E L2(X, S,

(i) P,yl is linear and PM(f) = f whenever f E M.

(ii) f - PA11(f) M' -

(iii) (PM(f), g) _ (f> PM(g)) = (PM(f), PM(g)).Ov) PM (PM Y )) = f-

(V) IIfII2 = I + 11f - PM(.f)II2(vi) PM is continuous.

Proof: The statements (i) and (ii) follow from the definition of PM. Forf, g E L2 (X, S, µ), using (ii) we have

(I'M (f), 9) = (PM(f), PM (9)) + (PM (f), g - PM (g))

(PM(f), PM (9)) -

Similarly,

(PM (f), 1'1v1(9)) = (f, PM (9)) = (PM(f), PM (g)) -

Hence (iii) holds. Statement (iv) follows from (i), as PM (f) E M. Finally,d f E L2 (X, S, µ), since P,yl (f) 1 (f - P,yl (f )), it follows from exercise 8.9.9that

11f II22 = IIPM(.f)IIZ +11f - PM(.f)II2

This proves (v). Also it follows from (v) that for f, g c L2 (X, S, µ),

I- PM(9)II2 If - 9112)

which implies that PM is continuous (in fact uniformly continuous).

We close this section with an application of theorem 8.9.22. This findsapplications in probability theory and statistics.

8.9.23. Theorem (Existence of conditional expectation): Let(X, S, µ) be a probability space and let So be a a-algebra of subsets of Xwith So C S. Then for every f c L2 (X, S, µ) there exists a function f ELZ (X, So, µ) such that b E E So,

f dp = f dp.E IE


The function f is called the conditional expectation of g given the sub-a-algebra So.

Proof: Let

MO := f f E L2(X) Si µ) If is So-measurable}.

It follows from example 8.9.12 that MO is a closed subspace of L2(X, S, µ).Let E E So. Since µ(X) = 1, XE E L2(X, So, µ). Thus using theorem 8.9.22,we have

f fdµ = (f. xE)

_ (f, PM(XE))_ (PMf, XE)

IE Pm (f)dlt

Thus f := PM(f) is the required function.

8.9.24. Note: In probability theory and statistics, the above theorem (andits generalization as proved in theorem 9.1.18) has great significance. Asmentioned in note 3.11.10, a probability space (X, S, µ) gives a mathemati-cal model for analyzing a statistical experiment. An observation on X, thespace of all `outcomes' in the experiment, is called a random variable inprobability theory. Mathematically, it is a measurable function on the mea-surable space (X, S). Let us consider the problem of `predicting' a randomvariable f given the knowledge of another random variable, say ¢. We notethat all the information about the random variable 0 is contained in So, thecollection of events defined by So :_ {q1(E)E E BR}. Clearly So C S andis a v-algebra of subsets of X. Thus the problem is to predict f given theQ-algebra So. In case we assume f is square integrable, i.e., f E L2(X, S, µ),we want to find f E L2 (X, S, µ) such that

lif - f 11 C Ill-gil V 9 E L2(X, So, µ).

The function f is called a conditional expectation of f given So, andtheorem 8.9.23 tells us that such a function exists. A generalization of thisis proved in theorem 9.1.18.

8.10. L2-convergence of Fourier series

As we pointed out in the historical notes (section 1.3), the problem of in-vestigating the convergence of Fourier series motivated mathematicians to

8.10 L2-convergence of Fourier series 307

look for an extension of the integral concept. It is natural to ask the ques-tion: did the extended integral, i.e., the Lebesgue integral, achieve successin this direction? One can say without any doubt that one of the pioneeringapplications of the Lebesgue integral lies in the study of Fourier series. Weshall not go into the general theory of Fourier series, but only prove an im-portant result (the Riesz-Fischer theorem) which has applications in manybranches of mathematics, physics and electrical engineering. For the generaltheory of Fourier series, one can consult Bhatia [4], Carslaw [7], Korner [22],Titchmarch [40] and Zygmund [42].

8.10.1. Definition: For f c L1 [-7r, 7r], let1

7r

an :_ - f (x) cos nx dA(x), n = 0, 1, 2, . .

7r -irand

17r

bn :_ - f(x)sinnxdA(x), n = 1, 2, .. .7r -ir

The scalars an, bn are called the Fourier coefficients of f, and the series

ao

+2

00

n=1(an cos nx + bn sin nx)

is called the Fourier series of f. Let

SnW._ ao2

n

(ak cos kx + bn sin kx), x c [-,7r, ,7r].k=1

The function sn(x) is called the nth_partial sum of the Fourier series of f atx. The main problem analyzed in the theory of Fourier series is the following:when does {s(x)}>i converge to f (x), x c [-,7r, ,7r]? This is known as thepointwise convergence problem of Fourier series. For f E L1 [-,7r, ,7r] andx E [-7r, 7r], if lim si(x) = f (x), we say that f has pointwise representation

n-->ooby its Fourier series at x. The answer to the pointwise convergence problemis not easy, and, given the nature and scope of this text, we shall not go intoit. For details one can consult any one of the texts listed above. We statebelow an important relation between functions and their Fourier coefficients,for a proof of which the reader can check any one of the texts cited aboveor Hewitt and Stromberg [18].

8.10.2. Theorem (Uniqueness of the Fourier series): If f E Li [-,7r, ,7r]is such that all its Fourier coefficients are zero, then f (x) = 0 for a. e. x,i. e., a function is uniquely determined by its Fourier coefficients.

Though the pointwise representation of a function f c L1 [-7r, 7r] by itsFourier series is undeniably of great intrinsic interest, it has its limitations.

First of all, not every function f c L 1 [-7r, 7r] can have a pointwise represen-tation one has to put extra conditions on f to get a pointwise representation.Secondly, f E L'[-,7r, 7r] need be defined only a.e. Thus pointwise represen-tation makes sense only a.e. Finally, from the point of view of many applica-tions, pointwise representations have very little utility. Such considerationmotivated mathematicians to look for some other methods of analyzing theconvergence problem. We consider below one such method, namely the con-vergence of Fourier series in the L2-metric. For other methods we refer tothe texts cited earlier.

Let f E L2[-7r, 7r]. Using the Cauchy-Schwarz inequality, it follows thatf E L'[--F, 7r] also. Let a, b,z be its Fourier coefficients and let sn (x) be thenth-partial sum of the Fourier-series of f. In the terminology of section 8.9,if we write

Ok(x) := cos kx and 'Ok(x) := sin kx,

then

ak = Y, Ok)/7r and bk = (f, 'Ok)/7r.

8.10.3. Lemma: For nonnegative integers n and m, the following relationshold:

= 0;

= 0 ifn#m and =,7r ifn=m;= 0 ifn#m and =,7r ifn=m.

Proof: Exercise.

8.10.4. Theorem (Bessel's inequality): For f c L2[-7r,7r],ao I2

2 7r

Proof: Using lemma 8.10.3, it is easy to check that V k,

00

(IanI2 +Ibn12) HIfIIn=1

Hence

1 1- 121

0 <7r 2

k

n=(I

111f112.laoI2+ (IanI2 + I C

ITn=1

0

The above theorem has a converse, as given in the next theorem.

8.10 L2-convergence of Fourier series 309

8.10.5. Theorem (Riesz-Fischer): Let {ate,}n>p and {bn}n>1quences of real numbers such that

lao 2

2

00

n=1

(I 2 + Ibn12) < -f-oo.

be se-

Then there exists a unique function f E L2 [-,7r, ,7r] such that an, bn are itsFourier coefficients.

Proof: Let us define for x E [-7r, 7r]

gn(u) := 2 + E (akcoskx+bksinkx).

Clearly, each gn E L2 [-7r, 7r], and using lemma 8.10.3 it is easy to verify thatfor n > m,

1

7r

119n_gm112n

E (Ianl2 + Ibz12).

k=m+1

Hence is a Cauchy sequence in L2[-7r, 7r], and, by the completenessof L2 [-7r, 7r], I f E L2 [-71,71] such that II9n -f 112 -k 0. To complete theproof, we show that f has an, bn as Fourier coefficients. First note that forn > k,

(9n, Ok) Pr = ak -

Also, using exercise 8.9.5, we have

liM (9n, Ok) Pr = (f , Ok) /7n--+oo

Hence (f, Ok)/7r = ak. Similarly, (f, V)k)/7r = bk. This proves that f hasFourier coefficients an, bn. Since f E L2 [-7r, 7r] implies that f E Li [-7r, 7r](see exercise 8.4.12(ii)), the uniqueness of f follows from theorem 8.10.2.

8.10.6. Corollary: Let f E L2[-,7r, ,7r]. Then the Fourier series of f con-verges to f in the L2-norm, i. e., 11sn - f 112 - 0 as n oo. Further,

00

n=

This is called Parseval's identity.

(Ianl2 + b

Proof: First note that by Bessel's inequality (8.10.4), we have

laol

2

2 00

n=1

Thus it follows from theorem 8.10.5 that there is a unique function g EL2 [-7r, 7r] such that g has Fourier coefficients a, b, and ll gn - g 112 -- f 0 asn -f oo, where

n

gn(x) := ao/2 + E(akcos kx + bksin kx).k=1

By the uniqueness theorem (8.10.2), it follows that f = g and 9n = sn, thenth partial sum of the Fourier series of f. Thus 11sn - f 112 -- 0 as n )oo.Since

I Ilsnllz - If 1121 <- IIsn - lllywe get 1I8n1I2 -> IIfII 2 as n oo. Since

2 n

HIsmM22 = Ia2l + (IakI2+IbkI2),

it follows that

2

00

E(Jak I2 + Ibk I2) _k=1

8.10.7 Note: A careful observation of the above arguments will tell thereader that Fourier coefficients can be defined for Riemann integrable func-tions also (as was done historically first), and Bessel's inequality remainsvalid. However, Riesz-Fischer critically uses the fact that L2 [-7r, 7r] is com-plete under the L2-metric, which is not true for Riemann integrable func-tions.

8.10.8. Corollary (Euler's identity): E'1 2e+17 = X2/8.

Proof: Consider the function

f (X)0 if _7r < x < 0)1 if 0 < x < 7r,

and apply Parseval's identity (8.10.6).

Chapter 9

The Radon-Nikodymtheorem and itsapplications

9.1. Absolutely continuous measures and theRadon-Nikodym theorem

One of the methods of constructing a new measure from a given measureis the following. Let (X, S, µ) be a given measure space and let f be anonnegative real-valued measurable function on X. For E E S, define

d.v(E) .= fE f

It was shown in proposition 5.2.6 that v is a measure on (X, S). In fact, themeasure v is related to µ by the property that v(E) = 0 whenever µ(E) = 0for E E S. Thus, if v is obtained from µ via integration, then every µ-nullset is also a v-null set. This motivates the next definition.

9.1.1. Definition: Let µ and v be two measures on (X, S). We say v isabsolutely continuous with respect to µ if v(E) = 0 whenever µ(E) _0, E E S. We write this as v <C µ.

9.1.2. Examples:

(i) Let (X, S, µ) be a measure space and f be a nonnegative measurablefunction on (X, S). As indicated above, v <C it, where

v(E) := fd, E E S.

311

312 9. The Radon-Nikodym theory and its applications

(ii) Let µ denote the counting measure on the Lebesgue measurable space(][8, ,CR), i.e., for E E GR, µ(E) := number of elements in E, if E is a finiteset, and µ(E) := +oo otherwise. Then A <C µ, where A is the Lebesguemeasure on (][8, GR).

(iii) Let X =ICY and S = P(N). Define µ(O) = v(O) = 0 and d E E S,E =A 0, let

IL(E) 2" and v(E) 1/2n.nEE nEE

It is easy to verify that µ(E) = 0 if v(E) = 0. Hence µ <C v and v <C µ.

An equivalent way of describing absolute continuity is given in the nexttheorem.

9.1.3. Theorem: Let µ, v be measures on (X, S). Then the following hold:

(i) If v is finite and v <C µ, then for every e > 0, 3 S > 0 such thatv(E) < e whenever p(E) < S, E E S.

(ii) If for every e > 0, 3 S > 0 such that v(E) < e whenever µ(E) <J, E E S, then v <C µ.

Proof: (i) Suppose the claim is false. Then -1 e > 0 and sets En E S suchthat

/L(En) < 2-n but v(En) > E, n = 1, 2, .. .

Let

Then V n,

00 00

An := U Ek and A := An,k=n n=1

IL (A) i is a decreasing sequence and v is a finitemeasure, by theorem 3.6.3

v(A) = lim v(An) > E.n-> o0

This is a contradiction, as v <C µ. This proves (i).

(ii) Let µ(E) = 0. Then µ(E) < S V 8 > 0. Thus by the hypothesis itfollows that v(E) < E, V E > 0. Hence v(E) = 0, i.e., v <C µ.

9.1.4. Exercise: Show that the conclusion of theorem 9.1.3(1) fails if v isnot finite.

(Hint: Consider the measures µ and v as constructed in example 9.1.2(iii).)

9.1 Absolutely continuous measures 313

Recall that in section 3.5 we showed that every measure µ on (]E8, 13R),which is finite on finite intervals is given by a monotonically increasing rightcontinuous function F, and conversely every such F gives rise to a measureµF on (R, BR). It is natural to ask the question: when is µF <G A (here Aas usual is the Lebesgue measure on R)? The answer is given in the nexttheorem.

9.1.5. Theorem: Let F : ][8 ) R be a monotonically increasing rightcontinuous function and µF be the measure induced by F on (][8, BR) . ThenAF < A if and only if F is absolutely continuous on every bounded interval.

Proof: Let µF be absolutely continuous and let [a, b] be any bounded inter-val. We want to show that F is absolutely continuous on [a, b]. We considerthe restriction of µF on [a, b] and let e > 0 be given. Since µF is finite on[a, b], by theorem 9.1.3 (i) we can choose a J > 0 such that

µF (A) < e whenever A (A) < S, A C [a, b].

In particular, if A is a union of a finite number of pairwise disjoint intervals[ai, bi], i = 1, 2, ... n, then

n n

i - az) = A(A) < S implies µF(A) = J .(F'(bi) - F(ai)) < E>

showing that F is absolutely continuous.Conversely, suppose F is absolutely continuous on every bounded inter-

val. Let E E Z3Rbe such that A(E) = 0. We have to show that µF(E) = 0. Itis enough to show that µF(E fl [a, b]) = 0 V a, b c R. Fix a, b c R with a < b.Since F is absolutely continuous on [a, b], given e > 0, we can choose b > 0such that whenever the [ai, bi],1 < i < n, are pairwise disjoint subintervalsof [a, b], then

n

i=1

n

- aZ) < S implies ) JF(bi) -F(ai)l < 6-(bz

i= 1

By theorem 4.2.2, we can find an open set U such that E C U and A(U) <J/2. Since U is an open set, it is union of a sequence of pairwise disjointintervals. Without loss of generality we can assume that we have a sequence{(a, b] }n>1 of left-open, right-closed intervals such that

00

En [a, b] C U(an,bn] C [a,b]flUn=1

and00

1: A(an, bn] < A(U) < 5/2.n=1

k

j:,\(anj bn) < J/2 < 5)n=1

k

1: [F (bn)- F(an) < E, V k.n=l

00

1:[F(bn) - F(an)] < E.n=1

Now00

µF(E n [a, b]) <E µF (anj bn I= 1: [F(b) - F(an)] < E-n=1 ra=1

Since this holds b E > 0, we have µF(E f1 [a, 6]) = 0. Hence µF <G .

9.1.6. Remark: Let µF be the Lebesgue-Stieltjes measure induced by anabsolutely continuous distribution function F. Let (][8, ,CiF, µF) denote thecompletion of the measure space (][8, 13R, µF). Clearly, 13F D Cit. Let EC GRand /\ (E) = 0. Then E C A for some AC CiRwith /\(A) = 0. But thenµF(A) = 0, as µF <G X Hence µF(A) = 0, i.e., A c BF. Now it follows fromtheorem 4.2.2 and theorem 3.11.8 that GRC 13F-

9.1.7. Exercise: Let µl,µ2, µ3 be measures on (X, S). Prove the following:

(i) Al < G (Al -I- µ2)

(ii) If Al <G µ2 and µ2 <G µ3> then µl <C µ3.

(iii) If Al <C µ3 and µ2 <C µ3, then (µl + µ2) <G µ3.

9.1.8. Exercise: For E E £R, the class of Lebesgue measurable subsets ofR, let

001 1\(E n n])A (E) : = E -

n=l 2n ( 2n

Show that p is a probability measure on (R, £R), i.e., it is a measure andp(III) = 1. Let /2 (E) := p(E -}- x), E E £R, X E R. Show that

px « p < i.Gy, V x, y c R.

As pointed in the beginning of the section, given a measure µ and anonnegative measurable function f, the measure v(E) := fE f dµ, E E S, isabsolutely continuous with respect to it. The questions arise: can one de-scribe all the measures µ which are absolutely continuous with respect to µ?Do absolutely continuous measures always arise only via integration? The

answers to these questions are given by the Radon-Nikodym theorem. Theproof we give below of this theorem uses the results of section 8.9. An alter-native proof, which requires the concept of signed measures and the Hahndecomposition (theorem 10.1.8), will be described in theorem 10.2.2. Wefirst prove a theorem, due to John von Neumann, from which the Radon-Nikodym theorem can be deduced.

9.1.9. Theorem (von Neumann): Let v and it be two a-finite measureson a measurable space (X, S). Then there exist mutually disjoint sets Xi ES,1 0 d x E X2, and B E E S with E C XZ we have

v(E) = fgd.

Proof: Case (i): Let it, v both be finite. Then it + v is a finite measureand

L2µ + v) C Li (µ + v) C L1(v).To see this, note that the constant function h - 1 belongs to L2 (µ + v) andby the Cauchy-Schwarz inequality, for f E L2(µ + v), we have

i/a

J I v) C ((µ + v) (X ))1/2 \J Ifµ + v) /Thus f E L1(µ + v). For f E L1(µ + v), it is obvious that

ffldvi.e., f E Ll(v). Let T :L2(µ + v) ) Il8 be defined by

T(f) = J fdv, f E L2(µ+v).

Then it follows from the above discussion that T is a well-defined map.Clearly T is linear and

I< (v(X))"2fffjf2, f E Lz(µ+v).

Thus T is a bounded linear functional on L2(µ + v). By the Riesz repre-sentation theorem (8.9.20), there exists a function fo E L2(µ + v) such thatb feL2(+v),

T(f) = i f fo d(tL + v),

f fdv = ffd(+v).Putting f = XE, E E S, we have

v(E) = IE A d(µ + v).

Since this holds b E E S and fo E L1(µ + v), it follows from theorem 8.1.9that fo(x) is real-valued for a.e. x and fo(x) > 0 for a.e. (,u + v), x c X.Further, V E E S,

(1i+v)(E) > v(E)

E

fo d(µ + v),fE

(I - fo) d(p + v) > 0.

Once again this implies, by theorem 8.1.9, that 0 < fo(x) < 1 for a.e.(+v)x c X. Let

N JxGXJfo(x)>1JUJxGXJfo(x)<0J,Xi {xEXIfo(x)=1},Xi:=XUN,X2 {xEXI0<fo(x)<1},X3 IX C X I fO(X) = 01-

Note that (µ + v) (N) = 0. Further, the Xi E S,1 < i < 3, and are pairwisedisjoint with U X 2 = X. Also,

v(X3) = fod(p + v) = 0X3

and

(Xi) + V) (Xi) (X') fo)d(p + v) = 0.f

Since µ(N) = 0, we get µ(X1) = 0. Finally, V E E S, E C X2,

U(E) = Lb0d+

v(E) - Lfodv = fE

(I - fo)dv =fE

ffodii.Since this holds b E C X2, it follows that

s(I - fo)dv = sfodp,IX2 IX2

for every nonnegative simple measurable function s. An application of themonotone convergence theorem gives

fX2fod,1(1 - fo)dv = fX2 .f

for every nonnegative measurable function f. In particular, for E C X2, E ES, let

if x ¢ Xl,

00 otherwise.

Then VEESwithECX2,

Define

v(E) =fo(x)

E 1 - fo(x)dfo W

i - fox

0 otherwise.

Then g is a nonnegative measurable function and, V E E S with E C X2,

v(E) = JE gdµ.

Case (ii): We now consider the case when ,c.L, v are both a-finite. We canwrite X = U'l Y, where the sets Y E S are pairwise disjoint, v(Y) < ooand with (Y) < oo V i > 1. By case (i), for every i > 1, we can findmutually disjoint sets Xi , X 2, X3 E S such that Yi = XZ U X i2 U X3 with,i(XZ) = v(X3) = 0. Further, for every i > 1 there exists a measurablefunction gi such that gi (x) > 0 for a.e. ()x c X 2, gi (x) = 0 for a.e.

()xEX'UX,and V E E s,

v(E n X 2) = gidA.fEnX 2

We put

and define

X1 :_00 00

forxEX2,

00

U Xi , X2 := U X2, X3:= U X3i=1 i=1 i=1

gi (x) if x c X? for some i,1 0 otherwise.

Then X1, X2, X3 and g satisfy the required properties.


9.1.10. Theorem (Lebesgue decomposition): Let µ, v be two a-finitemeasures on a measurable space (X, S). Then there exist a-finite measuresva and vs with the following properties:

(i) V = V" + V'.

(ii) There exists a nonnegative measurable function f such that

va(E) = J fd µ for every E E S.E

(iii) There exists a set A E S such that µ(A°) = vs(A) = 0.

Furthermore, such a decomposition is unique.

Proof: By theorem 9.1.9, we have disjoint sets Xl, X2 and X3 in S suchthat X = X1 U X2 U X3, and v(X3) =µ(X1) = 0. Further,

v(E f1 X2) = f gdµ, d E E S,nx2

for some nonnegative measurable function g on X with the properties g(x) >0 on X2 and g(x) = 0 on X2. Let A:= X2 U X3. Define V E E S,

vQ(E) := v(A fl E) and vs(E) := v(E fl Xl).

Then v = va + vs and vs(A) = µ(A°) = 0. Finally, V E E S,

va(E) = v(E n (X3 U X2)) = v(E n X2) = f gdµ.nX2

Define

f(x) g(x) if x E X2,0 ifxEX1UX3.

Then f is a nonnegative measurable function on X and

va (E) = f dµ V E E s.E

This proves the existence part of the theorem. To prove the uniqueness,suppose there also exist measures va and vs, a set A' E s and a nonnegativemeasurable function f such that

v = Va+vs,IL ((A') c) = vs(A) = 0,

vQ(E) = fEf'db E E S.Then

Further,((A' n A)°) = 0 = vs(A' n A) = vs(A' n A).

vQ((A/ n A)°) = 0 = vQ((A/

n A)°). (9.1)

SincevQ(E) + vs(E) = va(E) +

VS/

(E),we have, b E E S with v(E) < +oo,

va(E) - vQ(E) = vs(E) - vs(E).

Sinceva(E n (A/ n A)°) - va(En( A' n A)c) = 0,

the above implies that

vQ(En(Al nA))-va(En(A/ nA))=vs(En(A/ nA))-vs(En(Al nA)) = 0.

Thus, V E E S with v(E) < +oo,

va(E n (Al n A)) = va(E n(A/

n A)). (9.2)

From (9.1) and (9.2) we have va(E) = va(E), d E E S with v(E) < +oo.Now, using the v-finiteness of v, it is easy to show that va = va. Similarly,vs=vs.

9.1.11. Definition: Let µ, v be measures on (X,S). We say µ is singularwith respect to v if for some E E S, µ(E) = v(Ec) = 0. In that case wewrite µ 1 v.

9.1.12. Remark: The measure vs of theorem 9.1.10 is such that vs L vaand v, 1t i.

9.1.13. Exercise: Let µl,µ2 and v be measures on (X, S) and let a, 0 benonnegative real numbers. Prove the following:

(1) µi 1 µ2 iff µ2 1µl(ii) (aµ1 +,Qµ2) 1 v if µi 1 v, i = 1, 2.(iii) v = 0 if v <C µl and v L µi.

9.1.14. Exercise: Let µ be the counting measure on GR (see example9.1.2(ii)). Show that µ does not have a Lebesgue decomposition with respectto A, the Lebesgue measure. Why does this not contradict the claim oftheorem 9.1.10?

9.1.15. Theorem (Radon-Nikodym): Let µ, v be v-finite measures ona measurable space (X, S) such that v « µ. Then there exists a nonnegativemeasurable function f such that

v(E) = J f dµ, b E E S.E

Further, if g is any other measurable function such that the above holds,then f(x) =g(x) for a. e. x(µ).

Proof: Since v <C µ, in the Lebesgue decomposition theorem va = v andvs = 0. Further, there is a nonnegative measurable function f such that

f dpj E c S.v(E) =E

To prove the uniqueness of f, let there exist another nonnegative measurablefunction g such that

v(E) = JI gldµ, d E E S.E

Suppose there exists a set E E S such that µ(E) > 0 and f (x) > g(x) Vx E E. Since µ, v are v-finite, we can choose A E S such that µ(A) <oo, v(A) < +oo and (EflA) > 0. Then

0 < fnE(f

(x) - g(x)) dµ(x) = v(A n E) - v(A n E) = 0,

a contradiction. Thus f (x) < g(x) for a.e. x(µ). Similarly, f (x) > g(x) fora. e. x(µ).

9.1.16. Definition: Let µ, v be Q-finite measures on (X, S) such thatv « M. The unique measurable function f (as given by theorem 9.1.15) suchthat, d E E S,

v(E) = JE fdµiscalled the Radon-Nikodym derivative of v with respect to µ and is

denoted by dµ (x).

9.1.17. Example: Let F : R ) R be a monotonically increasing abso-lutely continuous function, and µF the Lebesgue-Stieltjes measure inducedby F on (][8, ,CiR). Then µF « A(by theorem 9.1.5) and

(x) = F (x), for a.e. x(A).

To see this, we note that by the fundamental theorem of calculus (6.3.6),for a < b,

Jb

F1 (x) dA (x) = F(b) - F(a)d

µF(a, b]

=Ia

6 d (x)dA(x).

From this, using exercise 5.3.4 (iii), it follows that

EF (x)dA (x) = E (x)dA(x), d E E L.

Hence, using exercise 5.3.4 (ii) and the uniqueness of the Radon-Nikodymderivative, we have

F '(x) = (x) for a.e. x(A).

As an application of the Radon-Nikodym theorem, we have the following:

9.1.18 Theorem (Existence of conditional expectation): Let (X, S, µ)be a probability space and f E Ll (X, S, µ). Let So be a a-algebra of subsetsof X such that So C S. Then there exists a function f : X -> ][8 such that

(i) f is Sa-measurable and µ-integrable.

(ii) For every E E So,

Efdµ = J fd.

E

Proof: Consider f +, the positive part of f . Then f + E L1(X, S, µ). Let,`dEESo,

v(E) :- fE f dµ.Then v is a finite measure on (X, So). Clearly v is absolutely continuouswith respect to µ. Thus, by the Radon-Nikodym theorem, there exists anonnegative So-measurable and µ-integrable function gl : X -> I[8 such thatb'EESo,

Ef+dv(E)fid (9.3)

Similar considerations applied to f -, the negative part of f , give a non-negative So-measurable and µ-integrable function 92 : X -* Iii such thatb'EESo,

Ef -dp = 92dp. (9.4)f2d.

9.1.19. Exercise: Let µ, v be a-finite measures on (X, S) such that v <GShow that for every nonnegative measurable function f on X,

Jf dv

= f f µdµ

Further, if f c L, (v), then f v c L1(µ), and the above equality holds.µ

(Hint: Use the simple function technique.)

9.1.20. Exercise: Let µ be the counting measure on the Lebesgue mea-surable space (I[8, GR) (see example 9.1.2(ii)). Show that A K µ but thereexists no measurable function f such that b E E GR,

A(E) = fgd.

This shows that the condition of µ being o--finite is necessary for the Radon-Nikodym theorem to hold.

9.1.21. Exercise: Let µl,µ2 and v be o--finite measures on (X, S). Provethe following:

(i) If µi <C v, i = 1, 2, then (Al + µ2) K v and

d(µi + µ2)W = dµi W +dµ2 fix) for a. e. x(v).

dv dv dv(ii) If Al « µ2 and µ2 <C Al, then

dtil ((x)) = 1 for a.e. x(µ1) and a.e. x(µ2).((x))

9.1.22. Exercise: Let F be the Lebesgue singular function (see example6.2.4(iv)). Show that µF 1 A.

9.1.23. Note: In 9.1.15 we derived the Radon-Nikodym theorem fromtheorem 9.1.9, which used the structure of linear functionals on L2-spaces. Amore direct proof is given in theorem 10.2.2, where we prove it also for `signedmeasures'. The Radon-Nikodym theorem can be proved for more generalmeasure spaces, called decomposable measure spaces, see for example Hewittand Stromberg [18].

9.2. Computation of the Radon-Nikodymderivative

The Radon-Nikodym theorem is an existence theorem. In general, it is

not possible to find the Radon-Nikodym derivativedtc when « v. In thisdv

section, we look at the problem of identifying the Radon-Nikodym derivativedfor measures on (W', BRn) such that tz << A, the Lebesgue measure

dAon IISn. The motivation for such a computation comes from the followingproposition. (In this section all the a.e. statements are with respect to theLebesgue measure A on W'.)

9.2 Computation of the Radon-Nikodym derivative 323

9.2.1. Proposition: Let µ be a finite measure on (IR, B1) and F(x)µ(-oo, x], x c R. Then the following are true:

(i) F(x) is differentiable for a. e. x, and at these points

F1 (x) = lim/-Z(X - r, X + r)

A(x - r, x + r)

(ii) If µ « A, then for a. e. x(A),

d1i(x) = F'(x) = lim /-Z(X - r, X + r)

dA A(x - r, x + r)

Proof: Clearly, F is monotonically increasing and hence F'(x) exists for a.e.x, by theorem 6.2.1. Next, observe that F is a right continuous function andµ = µF, the Lebesgue-Stieltjes measure induced by F. Now, it will follow

from theorem 9.1.5 and example 9.1.17 that (x) = F (x) for a.e. x ifµ <G A. Thus to complete the proof of both (i) and (ii), we only have toshow that for a.e. x,

F (x) = limI-L(x - r, x + r)A(x - r, x + r)

Let xo E ][8 be fixed such that F'(xo) exists and let e > 0 be given. ChooseS > 0 such that

F(x) - F(xo) _ F'(xo) < e whenever 0 0 be such that 2r < J. ThenF(xo + r) - F(xo - r) - F'(xo)

2r

F(xo + r) - F(xo) - rF'(xo)2r

1

2

F(xo) - F(xo - r) - rF'(xo)2r

F(xo + r) F(xo) _ F'(xo)r

-) - F(x r)1F (xo)

2 r< 6/2 + e/2 = e.

Hence b 0 < r < S/2,,4(XO - r, X0 + r) - F'(xo)A(xo - r, xo -I- r)

E.

324 9. The Radon-Nikodym theorem and its applications

This proves the required claim.

The above theorem motivates the following definition. For x E Rn andr > 0, let B(x, r) :_ {y E IIBnl Ix - Y1 < rl.

9.2.2. Definition: Let µ be a finite measure on (RTh, l3Rn). We say µ isdifferentiable at x c ][8n if

(Dp)(x) := limp (B (x, r))

r---+o A (B(x, r))

exists, and in that case (Dp)(x) is called the derivative of µ at x.

9.2.3. Examples:

(i) Let µF be any Lebesgue-Stieltjes measure on ][8 such that (µF)(][8) < +oo.Then, as shown in proposition 9.2.1, (DuF)(X) exists, if F'(x) exists, andthe two are equal.

(ii) Let µ be defined on 13R by

p(E) := A(En [-1, +1]), EC BR-

It is easy to see that µ is differentiable at every x c R. Further, (D)(x) = 1if Jxj < 1, and (D)(x) = 0 if Jxj > 1.

The aim of this section is to prove the following theorem.

9.2.4. Theorem: Let µ be a finite measure on (][8n, ,t3Rn ). Then the follow-ing are true:

(i) If A E 13Rn and µ(A) = 0, then (Du)(x) = 0 for a. e. (\)x E A.

(ii) If µ «A, then (D)(x) = (x) for a. e. x(A).

(iii) (Du)(x) exists for a. e. x(A).

The proof of the theorem needs several elementary results, which weprove first. Recall that ) is a regular measure on (Rn, BRn). The next lemmaimplies that every finite measure on (1R, BRn) is regular.

9.2.5. Lemma: Let (X, d) be a complete separable metric space and let µbe a finite measure on X3X, the a-algebra of Borel subsets of X. Then µ isregular, i.e., b E E X3X,

µ(E) = inf{µ(U) I U open, U D E}= sup{µ(C) I C closed, C C E}= sup{µ(K) I K compact, K C E}.

(In other words, every measure is regular on a complete separable metricspace. See also note 4.2.9.)

Proof: We first show that b E E 13X,

µ(E) = inf{µ(U) I U open, U D E}= sup{µ(C) I C closed, C C E}. (9.5)

The proof is an application of the `Q-algebra technique'. First note thatproving (9.5) is equivalent to showing that for any e > 0, there are an openset UE and a closed set CE such that

CE C E C UE and (Uf\Cf) < E.

Let S be the collection of all sets E E 13x for which this holds. We shall showthat S is a o--algebra and S includes all closed sets, proving that S = Z3X .

Clearly, E E S implies EC E S. Let {E}n>1 be a sequence of sets from Sand let e > 0 be given. Then V n, we can choose an open set Un and aclosed set Cn such that

Cn C En C Un and p(Un \ Cn) < /2n+l .

Let00 n

U:= U Un and Fn := U Ck,n > 1.n=1 k=1

Then U is open, Fn is closed for all n > 1, and f U \ Fn }n> 1 decreases toU \ U°_ 1 Cn. Since ,a is finite, we have

lim (U\F) _n-+oo

00

U\ U cam.

= (OU\OC )n=1 n=1

00

<_ U (Un \ Cam,)(n=1

00

<- 1: A (Un \ Cn) <n=1

Hence I n such that pc(U \ Fn) < E. Since Fn C U'l En C U, this provesthat U=1 En E S. Next, let C be any closed subset of X. Let

Un:={xEX I d(x,y) < 1/n V yEC}.

Note that it is an open set for every n, and c = nn,=1 Un, Since ,a is fi-nite, AP = M(Un). Hence, given e > 0, we can choose n such thatM(Un) - M(C) < E. Clearly, C C Un and p (Un - C) < E, i.e., C E S. In

particular, 0 E S. This proves that S is a a-algebra and includes all closedsets. Thus S = 8x . This proves (9.5). Note that till now we have not usedthe fact that X is a complete separable metric space.

Finally, we show that V E E Bx,

µ(E) =sup{µ(K) I K compact, K C E}.

Let e > 0 be given. Choose a closed set C C E such that µ(E \ C) < e. LetD be any countable dense subset of X. Then

00

X= U U B(x, l/n), where B(x, l/n) := {y c X I d(x, y) < 1/n}.xED n=1

Let

B(x,1/n) := {y E X I d(x, y) < 1/n}.

Then for every n, we can choose xn,- E D, j = 1, 2, ... such that00

C = U (B(Xnj, I /n) n q.j=1

Since µ is finite, for every n > 1 we can choose kn such that

kn

/_1 C\UB(Xnj, I/n) < e/2j=1

Let

K :=n U (B(Xnj i I/n) n c)n=1 j=1

Then K is a closed and totally bounded set. Since X is complete, K is acompact set and

µ(E \ K) < µ(E \ C) + µ(C \ K) < e/2 + e/2 = e.


An obvious corollary of the above lemma is the following.

9.2.6. Corollary: Let p be any finite measure on (Wi, BRn). Then V E EBRn ,

p(E) = sup{µ(U) I U open, U D E}= inf {µ(C) I C closed, C C E}= sup{µ(K) I K compact, K C E},

i. e., every finite measure on (Rn, ,13Rrz) is regular.

9.2.7. Lemma: Given open balls B1, B2, ... , Bk in W1, there exist amongthem pairwise disjoint balls Bn1, ... , Bnm such that

k re

U Bi < 3 ni-1 j=1

Proof: Let Bj have radius rj,1 < j < k. We may assume that r1 > r2 >> rk. Choose n1 = 1. Choose n2 to be the smallest integer, if there

is any, such that Bn2 is disjoint from Bn1. Let n3 be the smallest integer,if there is any, such that Bn3 is disjoint from Bn1 and Bn2. We continuethis process till the finite set {B1,... , Bk} is exhausted. This will giveus pairwise disjoint balls Bn1, ... , Bnm , if any, from the given balls. Notethat each Bi,1 i,1 < j < m.Also note that if any two open balls B (x, r) and B (x', r') intersect andz c B (x, r) n B (x , r) with r' < r, then V z' E B (x , r) ,

d (x, z') < d (x, z) + d (z, x') + d (x', z') < r +r/

+r/

< 3r.

Thus B (x', r') C B (x, 3r). Next, let z E Bi for some i, and let n j be suchthat Bi n Bnj 0. Then ri < rnj , and we have Bi C An j , where Anj is anopen ball with the same center as Bnj but with radius three times that ofBnj . Thus

k in

U Bi C U An j 5

i=1 j=1

and using exercise 7.5.7, we have

k finA U B2 <A UAnj

i=1 j=1

M:< A (Anj)

j=1A(Bnj)- M

j=13n

9.2.8. Definition: Let ,4 be a finite measure on ,t3Rn. For x E ][8, let

()(x) :=1i r)) Io<r<6l

and

li f {f tz (B (x, r))(B(x, r)) I 0 < r < S } .

l J

We call (D)(x) and (Du)(x) the upper and lower derivative of µ at x,respectively.

9.2.9. Lemma: The functions Dµ and Dµ are Borel measurable functionson ][8, and µ is differentiable at x iff (ii)(x) = (D)(x) < -boo.

Proof: That it is differentiable at x iff (Dµ) (x) _ (Dµ) (x) < +oo is easyto check. To show the Borel measurability of (D)(x), it is enough to showthat for any fixed S, the function

fa(x) := I0<,r < J x E ][8n

is measurable. Let a E R and let E :_ {x E IE8" 1 fa(x) > a}. We shallshow that E is in fact an open set. For this, let x E E. Then there exists0 < r < S such that

(B(x,r))A (B (x, r))

Let 77 > 0 be arbitrary and let I x - I < 77. Then B(x, r) C B(y, r+77). Thus

(B(y,r+'q))p (B (y, r + 77)) > p (B (x, r))µ(B(x, r))A (B (x, r))

A (B (x, r))

(B(x,r))A (B (x, r))

Hence for every y c B(x, r) with I x - I < 77, using exercise 7.5.7, we have

µ(B(y, r + n)) > (ii(B(x,r))'\ (_A(B(y,r))

A (B (y, r + 77)) A (B (x, r)) A(B(y, r + n))r "

(r+)(pB(x,r))A(B(x,r)Since

d 77 > 0, (r/(r + ,))" < 1 and it converges to 1 as 77 -* 0, we canchoose 77o such that

(r n (p(B(x, r))

> a.r+770) A(B(x,r)))

For this choice of 770 we have B(x, 770) C E. Hence E is an open set. Similararguments will prove that ()(x) is also Borel measurable.

9.2.10. Lemma: Let a E Il8, a > 0 and A E BRn. Then

(i) ({x E A I (D)(x) > a}) < 3"µ(A)/a.(ii) A({x E A I ()(x) > 0}) = 0 whenever µ(A) = 0.

Proof: Let E:= {x E AI(Dµ) (x) > a}. By lemma 9.2.9, E E ,CiRn . Let Kbe any compact set and V be any open set such that K C E C V. We shallshow that

p(K) < 3nN-(V)/a.

Let K and V as above be fixed. For x c K, since x E E,that B(x, rx) C V and

3 rx > 0 such

p (B(x,r))A(B(x,rx))

> a. (9.6)

The family {B(x, rx)}xEK forms an open cover of K, and hence there existpoints xl, ... , E K such that

m

K C U B (xi, rxi) . (9.7)i=1

By lemma 9.2.7, we get a pairwise disjoint subcollection, say B1, B2, ... , Bk,of {B(x, rxi) 11 < i < m}, such that

m k

UB(xirxi)) < 3n A (Bj). (9.8)i=1 i=1

From (9.6), (9.7) and (9.8) we havek 3n n

j=1 j=

3n1,(V)

a

Now it follows from the regularity of a (exercise 7.5.3) and the regularity ofµ (corollary 9.2.6) that

A(E) < 3n/.l(A)/cr.This proves (i).

To prove (ii), note that00

{x c A R ()(x) > 0} = U{x E A R ()(x) > 1/k},

and by (i), A({x E A I (D/2)(x) > 1/k}) = 0 V k if p(A) = 0.

9.2.11. Lemma: Let f E Ll (][8n) with f (x) > 0. Define

µ(E) := f f(x)d(x), E E ,t3Rn.E

Then µ is a finite measure on BRn, and (Dµ) (x) exists for a. e. x(A), and(D/2)(x) = f (x) for a. e. x(A).

Proof: Clearly, µ is a finite measure on ,t3Rn, and d x E I[8n,

0 < (D/2)(x) < (Dp)(x)

Thus to prove the required claim, it is enough to show that

A({x E ll8n I (D/2(x) < f (x) < ()/2(x)}) = 0.

We first show that A({x E ][8n I (D)(x) > f (x)}) = 0. Since

{x E ll8n I ()(x) > f(x)} = U {x E Ilg" I ()(x) > r > f(x)},rEQ

to prove the required claim it is enough to show that d r

A(jx E Rn I (Dp) (x) > r > f (x) 1) = 0. (9.9)

For r E Q fixed, let

E T :_ {x E Ilgn I (Dµ)(x) > r > f (x)} and AT :_ {X E Ilgn I f (x) > r}.

Then Ar is a Lebesgue measurable set with S(AT) < +oo and Er C A. Toprove (9.9), it is enough to show that (D)(x) < r for a.e. (A)x E A. Letxo E ][8n be fixed, and B any open ball with center xO. Then

(B) r A (B) + (p (B) - r A (B))

rA(B) + fB(f(x)

< rA(B) + (f (x) - ?-) dA(x).f nA,.

Let

v(E) := f (f(x) - r)dA(x), E E BRn. (9.10)rnE

Since A(Ar) < If 111/r < +oo, v is a finite measure on CiRn and

ft (B) < r A (B) + v (B).

Thusµ(B) < + v(B)A(B) - A (B)

Since this holds for every open ball with center xo, we have

(a)(xo) < r+(Dv)(moo)Thus (D)(x) < r, whenever (Dv)(x) = 0. In view of lemma 9.2.10, wewould be through if we could say v(Arc) = 0. If A, were a Borel set, thiswould follow from (9.6). Since Ar need not be a Borel set (even though Ar isLebesgue measurable), we cannot deduce that. However, we can find a Borelset F with Ar C F and A(F) _ A(Ar) (see exercise 7.5.3). Then v(Fc) = 0,and hence we have (Dv)(x) = 0 for a.e. x E F°. Since A(F \ AE) = 0, wehave (Dv)(x) = 0 for a.e. x E A. This proves that (D)(x) < r for a.e.x E A. Thus (9.9) holds, and we get

A(fx E Rn I Dp(x) > f (x)j) = 0.

A similar argument will prove that

A(fx E Rn 112A(X) < f (X) 01). 0

9.3 Change of variable formulas 331

9.2.12. Note: Let f E Ll(][8n). Applying lemma 9.2.11 to f+ and f-, weget, for a. e. x E R',

f (y)dA(y) (9.11)f (x) =TwoA(B(x> r)) B(x,T)

A particular case of this (for n = 1) is theorem 6.3.2. The points where(9.11) holds are called the Lebesgue points of f.

Proof of theorem 9.2.4: (i) Let A E Z3Rn be such that µ(A) = 0. Since0 < (Dµ) (x) < (Dµ) (x) V x E ][8n, to prove the required claim it is enoughto show that A({x E A I (Du)(x) > 0}) = 0, and this follows by lemma9.2.10.

(ii) Let f . Since it is finite, f E L1(µ) and

it (E) = Lf(x)(x),

Thus, by lemma 9.2.11, (Du)(x) exists for a.e. x E ]I8n and is equal to f (x).

(iii) That (D)(x) exists for a.e. x follows from the Lebesgue decompo-sition of it (theorem 9.1.10) along with (i) and (ii).

9.2.13. Exercise: Let F [a, b] ) ][8 be a monotonically increasingfunction. Show that F is differentiable a.e. (theorem 6.2.1), as follows:

(i) LetF(x+) if x E [a, b),F(x)F(b) if x = b.

Show that F is monotonically increasing and is right continuous on[a, b]. Further, F'(x) exists whenever k(x) does.

(ii) Let µF be the Lebesgue-Steiltjes measure induced on [a, b] by F.Apply theorem 9.2.4 and proposition 9.2.1 to deduce that (Dfr)(X)exists for a.e. x, and Fi(x) = (Dfr)(X).

We shall see an application of theorem 9.2.4 in the next section.

9.3. Change of variable formulas

A change of variable formula is one which enables us to shift the integrationfrom one space to another (in a hope that it will be easier to calculate it onthat space). We have already seen examples of such formulas earlier, e.g.,theorem 6.3.12, corollary 6.3.13 and theorem 6.3.16. In the abstract setting,the situation can be described as follows: let (X, B) and (Y, S) be measurablespaces and let T : X ) Y be a transformation. Let f : Y --j ][8 (or C) be


S-measurable on Y. Consider the composite function (f o T) : X ) R (orC). If T has the property that T-1(E) E B V E E S (in that case T is saidto be measurable), then it is easy to check that f o T is ,t3-measurable onX. Next, suppose µ is a measure on (X, B). Define

v(E) := p(T-1(E)) V E c S.

Then v is a measure on (Y, S). We denote it by µT-1, and call it the mea-sure induced by T on (Y, S). The following can be called the abstractchange of variable formula.

9.3.1. Theorem: Let T : (X, B) -) (Y, S) be a measurable function andlet µ be a measure on (X, B). Then for any S-measurable map f on Y,

fyf(y)d(T')(y) = f(foT)(x)d(x),

in the sense that if the integral on either side exists, then so does the integralon the other side, and the two are equal.

Proof: The proof is an application of the `simple function technique' andis left as an exercise.

9.3.2. Corollary: Let (X, B), (Y, S) and T be as in theorem 9.3.1. Letµ be a a-finite measure on (X, B) and v be a a-finite measure on (Y, S)such that v(T(E)) = 0 whenever p(E) = 0. Then there exists a nonnegativemeasurable function 0 on (Y, S) such that d f c L1(Y, S, v), (f o T) ELl (X, ,t3,µ) and d E E S

E(f o T) (x) (0 o T) (x) dp (x) -f (y)dv(y) = f-1 (E)

Proof: Let E E B and (µT-1)(E) = 0. Then, by the given hypothesis,

v(E) = v(T(T-1(E))) = 0.

Hence v « µT-1. Since µ is a-finite, the measure µT-1 is also a-finite,and by the Radon-Nikodym theorem (9.1.15), there exists a nonnegativeS-measurable function ¢ such that b f E Ll (Y, S, v), f 0 E Ll (Y, S, µT-1)and

fyBy theorem 9.3.1,

f(y)(y)d(T')(y).f(y)dv(y)=fr (9.12)

f f(y)(y)d(T')(y) = f(f °T)(x)(o (9.13)Y


Combining (9.12) and (9.13) and replacing f by IXE for E E S, we get therequired equality.

9.3.3. Remark: Let us look at a particular case of the above corollary.Let T [a, b] --) [c, d] be monotonically increasing, onto and absolutelycontinuous. Then A(T(E)) = 0 whenever A(E) = 0 (see proposition 6.1.9).Thus by corollary 9.3.2 we have d f E Ll [c, d],

fwhere

f (y)dA(y) = ina. (f o T) (x) (0 o T) (x) dA (x)

fi(x) := 1(x) for a.e. x(A).

Let us compute fi(x). From (9.14), for f 1, we have

fb

(b) - T(a) = d - c = (T(y))dA(y).T

Also d x e [a, b], [a, x] C T-1(T([a, x])). Thus, taking

f = XT[a,x] = X[T(a),T(x)]

in (9.14), we have

f

d

X[T(a,x)] (y) dA (y)c

b

XT[a,x] (T(z))(0 o T) (z) dA (z)

> o T) (z) dA (z)

Similarly, V x E [a, b] ,

(9.14)

(9.15)

(9.16)

b

T(b) - T(x) > JoT)(z)dA (z). (9.17)

From (9.15), (9.16) and (9.17) we have

f6

T(b) -T(x) = (0 oT)(z)dA (z), d x E [a, b]

Now by the fundamental theorem of calculus,

T '(x) = (oT)(x) for a.e. x(A).

d

T(x) - T(a) Jr X[T(a),T(x)] (y) dA (Y)


Thus by corollary 9.3.2 we get the relation

fd1(Y)dA (Y) = f 6(1o 7')(x)7'(T)dA (x),

which was also proved in corollary 6.3.13.

To extend this result to R', we have to look for transformations TR72 -* R72 which will ensure that A(T(E)) = 0 whenever A(E) = 0, ) here

dAbeing the Lebesgue measure on R', and hope to identify dAT_ 1. One such

situation is described by the next theorem. For a linear transformation Ton R', let det T denote the determinant of [T], where [T] is the matrixrepresentation of T with respect to a basis of R1.

9.3.4. Theorem (Linear change of variable formula): Let T : I[8'Ilg' be a nonsingular linear transformation. Then d f E L1(I18n), (f o T) EL1(Ilgn) and

f f(y)dA(y) = det T f (fo T)(y)dA (y)

Proof: For f = XE, E E £Rn, the required equality follows from theorem7.4.6 and exercise 7.5.3. The proof for the general case is an application ofthe `simple function technique' and is left as an exercise.

9.3.5. Exercise:

(i) Let S be a vector subspace of I[8n such that S has dimension less than n.Show that S E ,t3Rn and A(S) = 0.

(ii) Show that the Lebesgue measure A on ][8n is invariant under orthogonallinear transformations.

It is natural to try to extend theorem 9.3.4 to cases when T is not nec-essarily linear. We shall show that a change of variable formula is possiblefor transformations which can be approximated locally by linear transfor-mations. Before proceeding further, we ask the reader to recall the contentsof appendix G.

9.3.6. Lemma: Let T : V ) W be a Cl -mapping, where V and W areopen subsets of Ilgn. Let E C V be such that A(E) = 0. Then T(E) E Ginand A(T(E)) = 0.

Proof: Let E C V be such that A(E) = 0. Since T is differentiable at everyy E V, given c > 0 we can choose S > 0 such that b x E V with I x - I < b,

T- T (x) - T'(x) (x - y) I < e.Ix - yi

Thus

LetT- T(x) < (IIT'(x)II + E)Ix - y l.

M(x, E) := JIT'(x) 11 + E.

Let m E N be such that m > M(x, e), and let p E N be such that 1/p < J.Then

IT(y) - T(x) I < mix - yl, b y E B(x, l1p)Thus we have E = Um,p E,,,,,p, where for each m, p E ICY,

E'm,p :_ {x E E II - T(x)l < mix - yI V y E B(x, lip)}.

Since A(E) = 0, clearly A(Em,p) = 0 V m and p. To prove the requiredclaim, it is enough to show that T(Em,p) E Gin and A(T(Em,p)) = 0 V mand p. Since A(Em,p) = 0, given e > 0 we can find (by exercise 7.5.2) openballs {B(x, such that 0 < ri < 1/p d i and

00 00

E.,,t,p C UB(xi,ri) with AB(xi, r2) < e.z=1 i=1

For x E E,,,,,p, if x E B(xi, ri), then I x - xiI < ri < 1/p and hence

I< mix - xil < mri.Thus T(x) E B(T(xi),mri). Hence

00

T(Em,p) C UB(T(xi),mrj).i=1

Now using exercise 7.5.7 along with the translation invariance of A, we have00

00

i=1nA(B(xi, ri)) =

Hence A (T(Em,p)) = 0. Thus T(Em,p) E Gin and A(T(Em,p)) = 0, provingthe claim.

00

A (B (T (xi), mri) mnA (B (T (xz), ri))

9.3.7. Corollary: Let T : V -* W be as in lemma 9.3.6. Further,assume that T is also a homeomorphism. Then T (E) E 8Rn if E E BRn ,and T (E) E £Rn if E E £Rn

Proof: LetS:= {A C W I T-1 (A) E BRn n V }.

Clearly, S is a a-algebra of subsets of W, and, T being continuous, S includesopen subsets of W. Hence S includes B n n W. Now suppose that for E C Vwe have T(E) E BRn nW. Then E = T-1(T(E)), and hence E E B n nV. Theabove arguments applied to T-1 in place of T will show that E E BRn n Vimplies T (E) E BRn n W. Next, let E E £Rn . Since A is a-finite, we can write

00

E = U Em, where each Em E £Rn and A (Em) < boo.m=1

To show that T (E) E Gin , it is enough to show that T (Em) E ,CRn for everym. Since A(Em) < +oo, using exercises 7.4.2 and 7.5.3 we can choose a Borelset B C V such that

E,,,, C B and A(B \ E,,,,) = 0.

Then A (T(B \ En)) = 0, by lemma 9.3.6, and hence T(B \ En) E Gin. Since

T (Em) = T(B) U T(B \ E.,,,,)

and T (B) E 13Rn , by the earlier part, we have T (Em) E Gin . The reverseimplication follows from the above arguments applied to T-1.

9.3.8. Theorem: Let V, W be open subsets of Rn with a(W) < +oo, andlet T : V ---) W be a bijective Cl-map. Let Ti,1 < i < n, be the coordinatemaps of T such that

aT.Z (x)1 4 0 d x E R'.JT(x) := det La

Let µ(E) :_ A(T(E)) V E E 13Rn. Then µ is a finite measure on 13Rn, and

(D)(x) = JT(x)I V x E V.

Proof: First note that µ is a finite measure, and hence (Dµ) (x) exists fora.e. x, by theorem 9.2.4. We shall show that (D)(x) in fact exists b x E V.Also, JT(x) 4 0 V x implies by the inverse function theorem (appendix G)that T is a homeomorphism. Let dT(x) be the differential of T at x.

Case (i): Assume that 0 E V,T(0) = 0 and (dT)(0) = Id, the identitytransformation. We shall show that (Dµ)(0) = 1 = JLet

1 > e > 0 be fixed. By the differentiability of T at 0, 1 b > 0 suchthat

IT(x) - xj < clxl V 0 < IxI < 6.

Thus for lxi < r,

IT(x)l < elxl + IxI < (1 + e)r, i.e., T(B(O, r)) C B(0, (1 + e)r).

We claim that Bl B(0, (1 - e)r) C T(B(0, r)). To see this, note that

Bl = (B1 \T(B(O,r))) U (B1 n T(B(0, r))).

Since 0 E Bl fl T(B(0, r)) and T(B(0, r)) is open, Bl fl T(B(0, r)) is anonempty open set. On the other hand, if x E V with IxI = r, then

r = IxI < Ix -T(x)I + IT(x)l < elxl + IT(x)l = re -{- IT((x)l

Thus IT(x)l > (1 - e)r, i.e., no boundary point of B(O,r) is mapped intoB1. Hence

Bl \T(B(0,r)) = Bl \T(B(0,r)).Once again, T being continuous, T(B(0, r)) is compact and hence BlT(B(O,r)) is also open. Thus Bl is a disjoint union of two open sets. SinceB1 is a connected set, one of them should be empty. As Bl f1T(B(0, r)) 0,we have Bl = Bl fl T(B(0, r)) and hence Bl C T(B(0, r)). Thus d r < S,

B(0, (1 - E)r) C T(B(0, r)) C B(0, (1 + e)r).

HenceA(B(0, (1 - e)r))

<A (T(B(0, r)))

<A(B(0, (1 + e)r))

A (B(0, r)) A (B (0, A (B (0, r))

(1_E)12 < A lT lB(0, < (i+.A (B(0, r))

V r<SandO<e<l,wehave

limA (7'(B(0, r))) = 1

r-+0 A (8(0, r))

(D1)(0) = 1 = I

Case (ii): Assume that 0 E V, T(0) = 0 and A:= (dT)(0) is a nonsingulartransformation. We shall show that (Dµ) (0) = IConsider

S(x) := A-1(T (x)), x E V. Then S is also a Cl-homeomor-phism and (dS)(0) = Id. Thus, by case (i),

(Dv)(0) = 1, where v(E) :_ A(S(E)), E E ,t3Rn .

Since v(E) = I we have

(Dµ) (O) = Idet AI(Dv) (0) = I detAI = I

Case (iii): Assume that 0 E V, T(0) = 0 and A:= (dT)(0) is singular. Weshall show that (Dµ) (0) = 0 = I

Note that A(Il8') is a vector subspace of I[8n of dimension less than n andhence A (A(I[8n)) = 0, by exercise 9.3.5(1). Let q > 0 be given. Choose 6 > 0such that

IT(x) - A(x)l < qlxl, V 0 < IxI < 6.Thus for r < S, if IxI < r then IT(x) - A(x)l < qr. Hence

T(B(0, r)) C {y E ][8"I ly - A(x) I < qr d IxI < r}_ {77 V IxI < 1}.

LetEn :_ {z E Il8n I Iz - A(x)l <,q d IxI < 1}.

Then A(T(B(0, r))) < rnA(E.). Since the sets E71 decrease to A(B(0,1)) as77 -- 0, given e > 0 we can choose q > 0 such that A(E..) < e (recall thatA (A(Il8n)) = 0). For this q and b 0 < r < S, we have

A(T(B(O, r))) < re.

ThusA (7'(8(0, r))) rnE = CEO

a(B(0, r))<

A(B(0, r))

where C := r is a constant independent of r. ThusA (B(0, r))

(DA) (0) = limM (B (0, r))

0 = IJT(0)1.r--+O A (B (0 r))

Case (iv): Let x E V be fixed. We shall show that (D)(x) = IDefine, for y E V:= (V - x),

S(y) := T(y + x) - T(x).

Then S is also a Cl-homeomorphism with 0 E V and S(0) = 0. Since

S(B(O,r)) = {T(y + x) -T(x)ly E B(O,r)} = {T(z) -T(x)iz E B(x,r)},

we have

A (S (B (0 r))) = A (T (B (x , 1) + T (x))) = AT ((B (x, r))).

Hence by cases (i) and (iii),

(D)(x) = jJs(O)j = J

9.3.9. Theorem (Nonlinear change of variable formula): Let TV ) W be a bijective Cl -mapping from an open subset V of I[8" onto anopen subset W of I[8n such that JT(x) # 0 V x E V. Then the followinghold:

9.3 Change of variable formulas

(i) If E C V is such that E E Gin, then T(E) E GRn and

A (T (E)) = IE I JT (x) I dA (x).

(ii) If f E Ll(W), then (f oT)I .ITI E L1(V) and

JVWf (x)dA(x) = (f o T) (x) I JT(x) I dA (x).

IV

339

Proof: Since JT (x) 0 `d x E V, by the inverse function theorem, T is ahomeomorphism, and by corollary 9.3.7, T (E) E £Rn V E E 4n. Let

W,n=WnIx ERnI IXI <m},m> 1.

Then each Wm is an open set with A(Wm) < +oo. For each m > 1, define

,U,n(E) := A(T(E) n Wm), E E 4n.

Then Am is a well-defined finite measure. By lemma 9.3.6 it follows thatAm << A. Thus by the Radon-Nikodym theorem (9.1.15) and theorem 9.2.4,we have V EErRn,

µ.,,,,(E) = L(Dm)(X) dA (x). (9.18)

Note that µm (E) = 0 if E fl T-1(Wm) = 0. Also, for x E V such thatT(x) E Wm, we have T(B(x, r)) C W,,,, for all sufficiently small r > 0 (sinceT is continuous). Thus, for all sufficiently small r,

Am(7'(B(x> r))) = A (7'(B(x> r))). (9.19)

If we consider the restricted map T : T-1(Wm) Wm and apply theorem9.3.8, we have, for every x E T-1(Wm),

A (7'(B(x>r))) = IJT(x)I. (9.20)limr--*o A (B(x, r))

From (9.18), (9.19) and (9.20) it follows that

A(T(E) fl Wm,) = I JT(x) I dA (x).fNow letting m -> oo and using theorem 3.6.3 and the monotone convergencetheorem, we have d E E Gin,

A(T(E) n w) = IV XEW I JT(x) I dA (x).

In particular, if E C V and E E LRn, then

I JT(x) I dA (x).A (T (E)) =fE

This proves (i).

For (ii), we prove that the required claim holds for f = XE, EE ,GRn nW;the rest, being an application of the `simple function technique', is left as anexercise. So, let E E £jn n W. Then E = A U N, where A C W is a Borelset and )(N) = 0. Since T is continuous, T-' (A) E BRn C Cjn , and hence,by part (1),

A(A) A(T(T-'(A)))

I

I-1(A)- v(9.21)Since A(N) = 0, we can choose a Borel set B such that N C B and A(B) = 0.Then

(XBoT)(x) I JT(x) I dA (x).v

The integrand being a nonnegative function, we get (XB oT)(x) IJT(x)l = 0for a.e. x(A). Hence (XNoT)(x) I= 0 for a. e. x(A). Thus

Iv (XN o T) (x) I dA (x) = 0 = A (N \ A). (9.22)

From (9.21) and (9.22), we have

A(E) = A(A) + A(N \ A) = f(XE° T)(x) I JT(x)I d(x).

9.3.10. Corollary. Let X, Y E 4CRn and let T : X - Y be a map suchthat the following hold:

(i) There exist open sets V C X and W C Y such that

A(X \ V) = A(Y \ W) = 0.

(ii) T : V W is a one-one, C1 -mapping such that

JT(x)#0 V xEV.Then d f c Ll (Y), the function (f o T) I JT(x) I E Ll (X), being definedarbitrarily on X \ V, and

fyf (y) dA (y) = (f o T)(x) IJT(x) I dA (x).

X

Proof: This follows from theorem 9.3.9.

9.3.11. Example (Polar coordinate transformation): Let

v:={(r,B) ER2jr>o,o<e<27r} and W:=][82\ {(x,o) I x>o}.

Clearly, V and W are open subsets of II82, and T : V W, defined by

T(r, B) :_ (rcos,rsin),is a one-to-one mapping. T is called the polar coordinate transformation.It is easy to check that T is a Cl-mapping and

IJT(x)l =r b NEV.

Let

Since

X={(r,6)EII821 r>0, 0<0 <27r}.

X\V={(r,0)1 r>0}U{(r,2,7r)I r>0}U{(O,B)10<e<2,7r},it is easy to check that T(X) = ]I82 and A(X \ V) = 0, A being the Lebesguemeasure on ][82. Thus for any f E Ll (X ), by corollary 9.3.10 and Fubini'stheorem,

I(v)da(v) =f =

f(foT)(x)IJT(x)ldA(x)00 27r

(f o T) (r, 8) r dA (r) dA (B).0 fn

9.3.12. Exercise: Let

Ca := {(x, y) E ]I82 I x > 0, y > 0 and X2 + y2 < a2}

andS, := J(x,y) cz R2 10 < X < a, 0 < Y < al.

Then the following hold:

(i) Using example 9.3.11 and Fubini's theorem, respectively, show that

e-(x2+y2)da(x,

y) =7

(1 - e-a2ca 4

and

e-(x y) =JSc

C/

(ii) Using the inequality

e-(x2+y2

) dA (x, y) <fcca fa e-(x

2+y2)dA(x,y) < e

Ca

where a = a, deduce that°° -X2

e dx =0 2

-`x2+y2)d A (x, y),

(For a more general form of this exercise, see exercise 9.3.15.)

a2e-X2

dx

9.3.13. Example: As another application of the change of variable formu-las, let us compute Vn(a), the volume (i.e., the Lebesgue measure) of theball

B(O, a) := Ix E R' I jxj < alfor a E ][8, a > 0 and n > 3. Let An denote the Lebesgue measure on][8n and A the Lebesgue measure on R. First note that, using the linearchange of variable formula (theorem 9.3.4) for the linear transformationT(x) = ax, x E ][81, we have

Vn(a) = I det TI Vn(l) = CtinVn(l).

Thus to compute Vn(a), it is enough to compute Vn(1). Let us denote ageneral point of I[8n-2 (n > 3) by w. Then using Fubini's theorem, we have

Vn(1) _ Xvn(l) (x) dAnfn

L pgf, _ 2xv,(i) u', x, y) di-a(w) dA(x) da(y)

_ L Ln_2 Xvre_2(1_(x2+y2)1/2) (w) din-1 (w) dA (x) dA (y)

= J 2 (V_(1 - (x2 + y2)1/2)) dA(x) dA(y)

(x2 + 2J2))(n-2)/2Vn-2(1) d.\(x) dA(y)

27r 1Un-2(1)

0 in271 Vn_2 (1 )/n.

(I - r2) (n-2)/2 r drdO

Thus if n is even, say n = 2/c, then(27r)n/2Vn(1)

=n(n -

If n is odd, say n = 2k - 1, then

?Ck

k! '

22kgk-1 kt

n(n - 2)...3.1 2k!

9.3.14. Exercise: Let a > 0 and

Sn(a) := x E Rnnn

jxj < ai=1

Show that

(i) An(Sn(a)) = anAn(Sn(1))-

(ii) An(snM) = 2/12An(sn-1M)i b n > 2.

(111) A('Sn(a)) = 2nan/n.

9.3.15. Exercise: Let A be any positive definite n x n matrix (i.e., A = TtTfor some nonsingular matrix), and let m E I[8n be fixed. Show that

f exp{-1/2[(x - m)tA-1(x - I det Aln

(Here each x c 1[8n is written as an n x 1 column vector, and t denotes thetranspose of a matrix.)

9.3.16. Exercise: Let vi Iv2, ... , vn be n-vectors in Rn. Let

P :=nn

o<tj <1,1<j<n

called the parallelogram determined by the vectors v 1, ... , vn . Show that

An(P) = IdetTL

where T is the matrix with vi as the j th column.

9.3.17. Exercise (Integration of radial functions): Let 0 [0, oo)--) R be any nonnegative real valued measurable function. Show thatd n>3,

00(IxI)dn(x) = nVn(l) f 0(r)rn-ldrf n

s follows:a

Let T ][8n ) [0, oo) be defined by T(x) := JShow that T ismeasurable. Use theorem 9.3.1 to write

00x ) d = f ()d(n7'l)()f(ii) Using Fubini's theorem and example 9.3.13, show that b a E (0, oo)

T1) [0, a) = nV(a) = nV(1) rldr.f(

Deduce from this that00 00

0(,r)d(AnT-1)(,r) = nVn(1) O(r)rn-ldr.0

(We proved this for n = 2 in theorem 7.4.10, using the simple functiontechnique. The same technique can be used to prove this exercisealso.)

9.3.18. Exercise: Compute JT(x), x E I[83, for the following transforma-tions:

(i) Let

V:= {(r,8,z) E ][83 1 r > 0, 0 < 6 < 27r, -oo < z < +oo}

andW:= R' (x, 0, 0) 1 x > of.

The transformation T : V ) W is defined by

T(r, 8, z) (rcos0,rsin8,z).T is known as the cylindrical coordinates transformation and(r) 0, z) are called the cylindrical coordinates of a point in V.

(ii) LetV:= f(p,0,0) I p > 010 < 0 < 27r,O < 0 < 7rf

and

W:= Il83 \ ({(x, 0, 0) 1 x > Of U {(0, 0, z) I z E Il8}).

The transformation T : V W is defined by

T (p, B, ¢) :_

T is called the spherical coordinates transformation and (p, 0, ¢)are called the spherical coordinates of a point in V.

Chapter 10

Signed measures andcomplex measures

The aim of this chapter is to discuss the properties of set functions which arecountably additive but are not necessarily nonnegative or even real-valued.Such set functions arise naturally. For example, if we consider a linearcombination of finite measures, it need not be a measure, i.e., it need not benonnegative (of course, it will be countably additive). Another way in whichsuch set functions can arise is when we integrate an integrable function: forf E Ll (X, s, u)

fdy, E C S.fE

We shall study such set functions in the next section.

10.1. Signed measures

10.1.1. Definition: Let (X, S) be a measurable space. A set functionP: S R* is called a signed measure if it has the following properties:

(i) P (0) = 0 -

(ii) µ takes at most one of the values +oo or -oo.(iii) Whenever {E}n>1 is a sequence of pairwise disjoint sets in S with

E:= U01 En, then00

I-L (E) = 1: M (En) in=1

where the equality holds in the sense that every rearrangement of the seriesJ:'I µ(En) converges to µ(E) if I< -oo, and diverges properly to

345

346 10. Signed measures and complex measures

µ(E) otherwise. Note that the series J:n=1 °µ(En) is absolutely convergentwhenever l< oo.

A signed measure µ on (X, S) is said to be finite if l< +oo, anda-finite if there exist sets A, E X, n = 1, 2, ... such that X = U=1 Anand I < +oo for every n.

10.1.2. Example: Let µl,µ2 be measures on (X, S) such that at least oneof them is finite. Define V E E S,

v(E) := µi(E) -µ2(E).

Then v is signed measure on S. Clearly v(O) = 0. Let E = U=1 En wherethe En's are pairwise disjoint elements of S. Then µi(E) = J 1 µ2(En)5i = 1, 2. Suppose µ1(A) < +oo b A E S. If µ2(E) < +oo also, then theseries J:' _1(µl(En) - A2 (En)) is absolutely convergent to µ1(E) -µ2(E).Hence

00

00

(E) = pi (E) - A2 (E) _ E p1(En) - EA2(En)n=1 n=100

E (Al (En) - A2(En))

00

_ E v(En).

In case µ2(E) = +oo or -oo, clearly the series O_1(µl(En) - A2 (En)) isdivergent to -µ2(E) =µ1(E) -µ2(E). Thus v is a signed measure. If bothµl and µ2 are finite measures, then

J11(X)J _< JA1(X)J + JA2(X)J < +00,

i.e., v is a finite signed measure. Similarly, if both µl and µ2 are a-finite,then v will also be a-finite.

We shall show that the method described in example 10.1.2 is the onlyway of constructing signed measures (see theorem 10.1.12). Before we dothat, we give some properties of signed measures which are similar to thoseof measures.

10.1.3. Proposition: Let µ be a signed measure on (X, S). Then thefollowing hold:

(i) If A, B E S and A n B = 0, then µ(A U B) = µ(A) + µ(B).(ii) If A E S with l< oo and B E S with B C A, then l< +oo

and µ(A \ B) = µ(A) - µ(B).(iii) µ is finite iff l< +oo b A E S.

10.1 Signed measures 347

Proof: The proof of (i) is obvious. To prove (ii), let A E S and JIf B E S and B CA, then A = (A \ B) U B, and we have

< oo.

jL (A) = jL (A \ B) + jL (B).

Since jµ(A)I < +oo and µ can take at most one of the values +oo or -oo,we get I µ(A \ B) I < +oo and II < oo. Further,

µ(A \ B) = µ(A) - µ(B)

(iii) Follows from (ii).

10.1.4. Proposition: Let µ be a signed measure on (X, S) and {En}n>1be a sequence in S. Then the following hold:

(i) If En C En+1 for every n > 1, and E:= U=1 En, then

IL(E) lim jL(En).

(ii) If E,i+1 C En for every n > 1 and Jµ(En)I < +oo for some n, thenfor E:= nn,=

1En)

p(E) lim jL(En).

Proof: Proceed as in theorem 3.6.3, keeping in mind proposition 10.1.3.We leave the details as an exercise.

10.1.5. Definition: Let µ be a signed measure on (X, S). A set A E S iscalled a positive set for µ if

lc(E) > 0, V E C A, E E S.

Similarly, a set A E S is called a negative set for ,a if

IL(E)<0,VEcA,EES.A set A E S which is both a positive and a negative set for ,a is called ait-null set.

In order to show that example 10.1.2 is the only method of constructingsigned measures, we first try to find the largest set A E S which is positivefor lc. Of course, 0 is both a positive and a negative set for any signedmeasure j,c. One can consider the union of all possible positive sets for ,ain order to get the largest positive set. Two problems arise. First of allwe cannot ensure this union will be in the a-algebra S. Next, an arbitraryunion of positive sets need not be a positive set. Also, it is natural to ask thequestion: do there exist sets of positive li-measure which are also positivesets for lc? The answers are provided by the next lemma.

10.1.6. Lemma: Let µ be a signed measure on S. Then the following hold:

(i) If A is a positive set for µ and B C A, B E S, then B is also a positiveset for µ.

(ii) If {An}n>1 is a sequence of positive sets for µ, then U=1 An is alsoa positive set for µ.

(iii) If E E S and 0 < µ(E) < +oo, then there exists a set A C E, A E S,such that A is a positive set for µ and p(A) > 0.

Proof: The proof of (i) is obvious.(ii) Let {An}n>1 be a sequence of sets which are positive for µ, and let

A:= U00 An= I n. Let

n-1

BI:=AI and B n = An \ U A, , for n > 2.1k=1

Then the sets Bn, n = 1, 21... , are pairwise disjoint and A = U=1 Bn. LetE E S with E C A. Then E = U=1(Bn n E) . Since B n n E C An and Anis a positive set for p, p(B n n E) > 0 for every n. Thus

00

A (E) = >"

p (BnnE) > 0.n=

Hence A is a positive set for µ.(iii) Let E E S and 0 < µ(E) < +oo. Either E itself is a positive set for

µ, or it contains sets of negative measure. In the earlier case we are through.In the later case, let q, be the smallest positive integer such that there existsa set El C E with El E S and µ(E1) < -1/rel. Note that µ(E \ El) < oo.Thus we can apply the above argument to E \ El. Proceeding inductively,either we will be through after some finite number of steps, or we will havea sequence {7k }k>1 of positive integers and sets Ek E S, k > 1, with theproperties that d k > 1,

Ek (::E)CE\

and 'qk is the smallest positive integer such that µ(Ek) < -1/'qk. Put00

A:=E\ UEk

Then E = A U (U1 Ek) and these sets are pairwise disjoint. Thus00

µ(E) = (A) + (Ek).k=

Since p(E) < +oo, the series on the right hand side of the above equality isabsolutely convergent. Hence Ii 1/nk is convergent, and we have k00 as k --> oo. Also, since (Ek) < 0 V k and µ(E) > 0, we have µ(A) > 0.To complete the proof, we show that A is a positive set. Let B C A withB E S, and let e > 0 be given. Choose 'qk such that 1/(qk - 1) < E. SinceV BCE \ (1Jk =1 Ej) with B E S we have µ(B) > -1/(% - 1) (by the

defining property of the 'qk's) In particular, d B C A C E \ (lJk _1 Ej)with B E S, we have

µ(B) > -110% - 1) > -e.Since e > 0 is arbitrary, µ(B) > 0.

10.1.7. Exercise: Let {An }n>1 be a sequence of negative (null) sets for µ.Show that U=1 An is also a negative (null) set for µ.

10.1.8. Theorem (Hahn decomposition): Let µ be a signed measureon (X, S). Then there exist sets A, B E S such that the following hold:

(i) X= A U B and A n B= D.(ii) A is a positive set for µ and B is a negative set for µ.

Proof: Since µ takes at most one of the values +oo or -oo, we may assumewithout loss of generality that -oo < µ(E) < +oo b E E S (for otherwisewe can consider -µ). Our idea of the proof is to construct a set A which isa positive set for µ, for which µ(A) is largest and is such that B := X \ A isa negative set for µ. Let

13 := sup{µ(E) I E is a positive set for µ}.

Clearly 13 > 0, since 0 is a positive set. Let {Ek}k>1 be a sequence of setswhich are positive for µ and such that

13 = lim p(Ek).

Let A:= U"k=1 Ek. By lemma 10.1.6, A is a positive set for µ and henceµ(A) < 13. Also, A \ EkC A implies that µ(A \ Ek) > 0 V k. Thus d k,

p (A) = p (Ek) + p(A \ Ek) > p (Ek).

This implies that µ(A) > lim µ(Ek) = ,6. Thus µ(A) = 16 and, by our as-sumption that the value +oo is not taken by µ, we have Q < +oo. LetB = X \ A. To show that B is a negative set for µ, let E C B and E E S.If µ(E) > 0 then 0 < µ(E) < ,Q < +oo and, by lemma 10.1.6, there existsa set F E S, F C E, such that µ(F) > 0 and F is a positive set for µ. Butthen F U A is also a positive set and

13 > µ(F U A) = µ(F) + µ(A) > µ(F) + Q.

This implies that p(F) = 0, a contradiction. Hence V E E S with E C Bwe have µ(E) < 0, i.e., B is a negative set for it.

10.1.9. Definition: Let it be a signed measure on (X, s) . A pair of setsA, B E S is called a Hahn decomposition of X with respect to it ifX = A U B with A n B= 0, where A is a positive set for it and B is anegative set for it.

Theorem 10.1.8 shows that a Hahn decomposition of X always exists.However, a Hahn decomposition need not be unique. This follows from thenext exercise.

.10.1.10. Exercise: Let it be a signed measure on (X, S) and let A, Bbe a Hahn decomposition of X with respect to it. Let N E S be such thatµ(N) = 0. Show that (A \ N), B U N is also a Hahn decomposition of X.Further, if Al, B1 and A2, B2 are two Hahn decompositions of X with respectto it, then µ(A1 D A2) = µ(B1 D B2) = 0, and d E E S, µ(E fl A1) _µ(E n A2), µ(E n Bl) = µ(E n B2).

10.1.11. Exercise: Let (X, S, µ) be a measure space and f E Li(X, S, µ).Let

v(E) := f dit, E C S.E

Find a Hahn decomposition of X with respect to v.

Finally, we prove that every signed measure is the difference of twomeasures.

10.1.12. Theorem (Jordan decomposition): Let it be a signed measureon (X, S). Then there exist measures µ+ and µ- on (X, S) with the followingproperties:

(i) µ+ - µ- and at least one of the measures µ+ and µ- is finite.

(ii) I µ-, i.e., there exist disjoint sets A, B E S such that µ+(B) _µ-(A) = 0 and A U B= X.

(iii) If it = v - 77, where v and 77 are measures with at least one of thembeing finite and v L 77, then µ+ = v or 77 and µ- equals the other.In other words, the decomposition of it as a difference of two singularmeasures is unique.

Proof: Let X = A U B be a Hahn decomposition of X with respect to it,where A is a positive set for it and B is a negative set for it. Define p+ and

on S as follows:

µ+(E) := µ(A f1 E) and µ- (E) := µ(B fl E), E E S.

Clearly, µ+ and µ- are measures on S with µ = µ+ - µ-, and at least one ofµ+ and µ- is finite, since µ takes at most one of the values +oo or -oo. Thatµ+ I µ- is obvious. Finally, let µ = v - 77, where v and q are measures withv 1 77, and say v is finite. Then µ never takes the value +oo and hence µ+is also finite. Let C, D E S be such that v(C) = 77(D) = 0 with C fl D = 0and C U D = X. Clearly C is a positive set for µ and D is a negative set forµ. Thus C, D is also a Hahn decomposition of X with respect to µ. Henceby exercise 10.1.10 we have b E E S,

µ+(E) = µ(E n A) = µ(E n c) = v(E)

andp-(E) = p(EnB) = p(EnD) = q (E). 0

10.1.13. Definition: Let µ be a signed measure on (X, S). The measuresµ+ and µ-, as given by theorem 10.1.12, are called the upper and lowervariations (or positive part and negative part) of µ, respectively. Themeasure IpI := µ+ + µ- is called the total variation of µ.

10.1.14. Exercise: Let (X, S, µ) be a measure space and f E L' (X, S, µ).Let

v(E) := f dp, E E S.E

Show that d E E S,

v+(E) = fEf +dpj v -(E)

Ef d µ and IvI(E)

10.1.15. Exercise: Let µ be a signed measure on (X, S). Show that thefollowing statements are equivalent:

(i) µ is finite (a-finite).(ii) µ+, µ- are both finite (Q-finite.

(iii) IpI is finite (a-finite).

10.1.16. Exercise: Let µ be a signed measure on (X, <S). Show that

II < V E E S

and IpI is the smallest measure on S with this property, i.e., if v is anyother measure on S such that Iµ(E)I < IvI(E) V E E S, then IµI(E)v(E) V E E S.

The next theorem gives an alternate description of the measures p+,

and I p1. For E E S, we say {E1, E2, ... , E} is a measurable partition ofE if E1, E2, ... , En are pairwise disjoint and E = U=1 Ei.

10.1.17. Theorem: Let µ be a signed measure on (X, S). Then V E E S,the following hold:

(i) µ+(E) =sup{µ(F) I F C E, F E S}.(ii) µ-(E) =sup{-µ(F) I F C E, F E S}.

(iii)

n

IpI (E) = sup II {E1,... , En} is a measurable partition of Ei=1i=1

Proof: (i) Let A, B be a Hahn decomposition of X with respect to µ. ForE E S, clearly

µ+(E) = µ(E n A) < sup{ µ(F) I F C E, F E S}. (10.1)

Since B is a negative set for µ, if F C E with F E S, then

µ(F) = (FnA)+(FnB) < µ(F f1 A) = µ+(F) < µ+(E).Hence

sup{ µ(F) A F C E, F E B} < µ+(E). (10.2)

Thus (i) follows from (10.1) and (10.2). Also, observing that µ- = (-µ)+,(ii) follows from (i).

Finally, to prove (iii), let E E S be fixed and let a denote the right sideof the equality in (iii). Then for any finite collection {F1,... , Fn} of setsfrom swith E=U 1Fi andFir1F3=0fori j, we have

n

i=

n

(Fi) (Fi) (Fi)2=

< Jp+ (Fi) I + Ip- (Fi) I

n

Thus a I + I I = µ+(E) + µ (E) = I

Hence a = IµI(E). This proves (iii).

10.2 Radon-Nikodym theorem for signed measures 353

10.1.18. Exercise: Let (R, GR, A) be Lebesgue measure space and let

v(E) := xe-x2dA(x), E E GR.E

Show that v is a signed measure with (-oo, 0) as a positive set and (0, oo)as a negative set for v. Find all positive, negative and null sets for v. Alsofind µ+, µ- and liI

10.1.19. Exercise: Let {an}n>1 be a sequence of real numbers. Define µon (N,P(N)) by

jL(E) an-

nEE

Show that µ is a signed measure if En an is absolutely convergent. Find aHahn decomposition of µ and show that it is unique if an 7 0 V n-

10.2. Radon-Nikodym theorem for signedmeasures

In section 9.1, we characterized the measures which arise via integration.For example, theorem 9.1.15 said that for a-finite measures ,a and v we havev « 1i if there exists a nonnegative measurable function f such that

v(E) := L fd,a `BEES.

The proof of this theorem as given in section 9.1 required some specialproperties of L2(X, S, µ). We give another proof here which is based on theHahn decomposition theorem. First, let us consider a finite measure µ on ameasurable space (X, S), and let f be any nonnegative measurable functionon (X, S). Let

v(E) _E

f dµ, E E S.

If v(E) # 0 for some E, then clearly El n c N such that

v{x E E 11/2n < f (x) < 1/2"-1} > 0.

LetEn := {x E E I 1/2"-1 G f (x) < 1/2n}.

Then, V A E S with A C E,

v(A) = fA f djL > jL (A) / 2

Thus, for e = 1/2n,

v(A) - Eµ(A) > 0, b A C En and v(En) > 0.

We show that the above conclusion holds in general also when v <G µ.

10.2.1. Lemma: Let µ, v be finite measures on (X, S) such that v 54- 0 andv « µ. Then there exist a set E E S with v(E) > 0 and an e > 0 such thatv(F) > eµ(F) V F C E, F E S.

Proof: For every integer n > 1, consider the signed measure (v - µ/n). Ouraim is to show that there exists some n such that (v - µ/n) has a positiveset of positive measure. For this, consider a Hahn decomposition An, Bn ofX with respect to v - µ/n. Then d n,

(v - µ/n)(Bn) < 0 and (v - /n)(A) > 0-00Let A:= U=1 An and B:= nn, 1 Bn. Then d n,

v(B) < v(Bn) < tt(BnVn < tt(X)/n-Hence v(B) = 0. Since X = AUB, we get v(A) = v(X) > 0. Thus v(Ano) >0 for some no. Also, Ano is a positive set for (v - µ/no). Choosing e = 1/nocompletes the proof.

Next we prove the Radon-Nikodym theorem for finite measures.

10.2.2. Theorem (Radon-Nikodym theorem for finite measures):Let µ, v be finite measures on (X, S) and v <C µ. Then there exists a non-negative f E L' (X, S, µ) such that b E E S,

v(E) = ffd.Further, if g is any other function such that

v(E) = J gdµ, b E E S,E

then f (x) = g(x) for a. e. x(µ).

Proof: The idea of the proof is as follows. Consider the set S of all non-negative measurable functions g on X such that

Egdµ < v(E) for all E E S.

We shall show that the `largest' function in S is the required f. For this,first we note that S 0. For example g = EXE E S, where c and E are asgiven by lemma 10.2.1. Let

a:= sup gdA I g E s.

Clearly 0 < a < oo. Let gn E S, n = 1, 2, ... , be such that

Ce = lim fgndtt.n-+oo

Putf (x) supgn(x), x E X.

n>1

We shall prove that f is the required function. Note that for g, h E S,

(g V h) (x) := max{g(x), h(x)}, x c X,

is also in S. To see this, let E E S be fixed and El = {x E E I g(x) > h(x)j.Then

f (9 V h)dp = fl (gvh)d+f\El (9 V h)dµ

< fgd/t+fJ hdµE 1 E\E1

v(Ei) + v(E \ El)= v(E).

ThusgVhES. Let

hn:=g1Vg2V ...Vgn E S V n.

Then {h}> 1 increases to f, and, by the monotone convergence theoremand (10.3), V E E S

E

Hence

Since gn < f V n,

fd= 71i J hn dµ < v(E). (10.4)E

fES and ffd a.

a = lim /gnd/ <n-4oo

Thus a = f fd. To show that equality holds in (10.4) for all E, consider

Efd, E E S.

Then i is a finite measure and q « M. If q - 0, we are through. If not, i.e.,,q 0- 0, 3 E>0andasetEo ES such that,u(Eo) >0andEo isapositiveset for 77 - q (by lemma 10.2.1). Thus V E C E0 and E E S,

v(E) - f dM - cp(E) > 0,E

i.e.,

(E) > JE (f + c)dp.


Thus V E E S,

v(E) = v(E n Eo) + v(E n Eo)

>_ J(f+c)d+f1dEnEo

nEo

EoE(f + EXEo )d+ fnE(f + EXEo )dI n

(f + EXEo )dµ

Hence (f + eXEo) E ES. On the other hand,

f (f + exEa )dµ = ffd µ + eµ(Eo) = a + eµ(Eo) > a,

which is a contradiction. Hence q - 0, i.e.,

v(E) = J f dµ for every E E S.E

The uniqueness of f follows from exercise 8.1.8.

10.2.3. Exercise: Extend theorem 10.2.2 when µ, v are a-finite measures.

Our next aim is to extend theorem 10.2.2 to signed measures. Recallthat for f E Li (X, S, µ), if we define

v(E) := fdp, EcS,E

then v is a finite signed measure on (X, S) and has the special propertythat v(E) = 0 whenever µ(E) = 0. In fact, if f is a real-valued measurablefunction on f and has the property that either f + E L' (X, S, µ) or fL' (X, S, µ), then

d, E E S,v(E) :_ fdµ - fE fE

is a well-defined signed measure on (X, S) having the property that v(E) = 0whenever µ(E) = 0. This motivates our next definition.

10.2.4. Definition: Let (X, S, µ) be a measure space. A signed measure von (X, S) is said to be absolutely continuous with respect to µ if v(E) = 0whenever µ(E) = 0. We write this as v « µ.

An equivalent definition of absolute continuity for signed measure interms of its variation measures is given by the following proposition.

10.2.5. Proposition: Let (X, S, µ) be a measure space and v be a signedmeasure on (X, S). Then the following statements are equivalent:

(i) v <G µ.

(ii) v+ <C and v- <G µ.

(iii) IvI «µ.

Proof: (i) (ii) Let A, B E S be a Hahn decomposition of X with respectto v. Let µ(E) = 0. Thus

(AnE)=(BnE) = 0,and, by the the given hypothesis, we have

v(EnA) = 0 = v(EnB) = 0.Thus

v+(E) = v(A fl E) = 0 and v-(E) _ -v(B n E) = 0.

Hence v+ <C µ and v- <G µ.

The implication (ii) (iii) is obvious, as IvI = v+ + v-.To prove (iii) (i), let µ(E) = 0. Then by the given hypothesis,

IpI(E) = v+(E) + v-(E) = 0

and hence v+(E) = 0 = v-(E) = 0, i.e., v(E) = 0.

Jordan's decomposition theorem together with proposition 10.2.5 givesus the following extension of the Radon-Nikodym theorem for signed mea-sures:

10.2.6. Theorem (Radon-Nikodym): Let (X, S, µ) be a Q-finite mea-sure space and let v be a Q-finite signed measure on (X, S) such that v <C µ.Then there exists a real-valued measurable function f such that either f + ELi (µ) or f Li (µ) and

v(E) _ f dµ := f +dµ - J f -dµ, d E E S.fE IE E

Further, the function f is unique in the sense that if there exists some othermeasurable function g such that

v(E) = J gdµ BEES,E

then f (x) = g(x) for a. e. (µ).

This unique f is denoted by v (x) and is called the Radon-Nikodymdp

derivative of v with respect to µ.

Proof: By exercise 10.1.15, both the measures v+ and v- are a-finite.Thus, by proposition 10.2.5, v+ << ,i and v- << n. Now an application ofthe Radon-Nikodym theorem for a-finite measures (exercise 10.2.3) appliedto v+ and v- completes the proof.

10.2.7. Exercise: Show that f, as given by theorem 10.2.2, will be inL1 (X, S, ,u) if v is a finite signed measure.

10.2.8. Exercise: Let µ be a Q-finite measure and v be a Q-finite signedmeasure on (X, S) such that v <C M. Show that

dµ (x) _ µ (x) - dµ (x) for a.e. x(µ).

Next we define the integrability of functions with respect to signed mea-sures and use it to extend theorem 10.2.6 to the case when both µ and v aresigned measure (see theorem 10.2.14).

10.2.9. Definition: Let (X, S) be a measurable space and let µ be a signedmeasure on (X, S).

(i) For a nonnegative real valued measurable function f on X we write

j f dp := j fdp+ - f f dp-,

whenever either f E Li (X, S, µ+) or f E Ll (X, S, µ-). In case fbelongs to both of them, we say f is integrable with respect to thesigned measure µ.

(ii) If f is a real-valued measurable function on f such that either f + isintegrable with respect to µ or f - is integrable with respect to µ, wewrite

Jfdµ .- f f +dµ - j f dµ.

We say f is integrable with respect to µ if both f + and f - areintegrable with respect to µ.

We denote by L' (X, S, µ) the space of real-valued µ-integrable functions onX which are integrable with respect to µ.

10.2.10. Proposition: Let µ be a signed measure on (X, S). Then thefollowing statements are equivalent:

(i) feL(X,S,).(ii) f E Li (X, S, µ+) n Li (X, S, µ-).(iii) f E Li(X,S, IiI).

Proof: The implications (i) (ii) are just a restatement of the definition.The implications (ii) (iii) follow from the fact that

f fdµ = ffd++ffd-for every measurable function f, whenever either side exists.

10.2.11. Exercise: Let f E L1(X, S, µ), µ a signed measure. Show that

f dµ f If dµ

The notion of absolute continuity of a signed measure with respect toanother signed measure can be defined as follows.

10.2.12. Definition: Let µ, v be signed measure on (X, S). We say v isabsolutely continuous with respect to µ if lµl(E) = 0 implies v(E) = 0.We write this as v « µ.

10.2.13. Remark: Note that for signed measured v and µ, v <G µ is notdefined by requiring v(E) = 0 whenever µ(E) = 0 for E E S. See exercises10.2.17 and 10.2.18.

10.2.14. Theorem: Let µ, v be a-finite signed measures on (X, S) suchthat v « µ. Then there is a real-valued measurable function f on X suchthat b E E S,

v(E) = fEr' f dµ.

Further, f is unique in the sense that if g is any other real measurablefunction on X with

v(E) = Jgdµ, d E E S,

Ethen f (x) =g(am) for a. e. x(lµl)

Proof: Let A, B be a Hahn decomposition of X with respect to µ. ThenV EES

µ+(E) = µ(A fl E) and µ-(E) = -µ(B fl E).For E C A, µ(E) = µ+(E)+µ-(E) _ IThus µ+(E) = 0 = IµI (E) = 0implies that v(E) = 0. Hence v <C on (A, S fl A). Since both v and µ+are a-finite, an application of theorem 10.2.6 gives a measurable function fAon A such that d E E A fl s,

v(E) = f fA dµ+.E

Similarly, we will have a measurable function lB on B such that V E EB nS,

v(E) = fEfBdDefineIA = 0 on B and lB = 0 on A. Then f := fA + fB is a measurable

function on X and, d E E S,

v(A) = v(E f1 A) + v(E n B)

A E.fa d+ fnBf n

Efdµ+

+ Efdµ = E f dµ.

The uniqueness of f follows from the fact that fA(x) is unique for a. e.x(µ+), and the fact that fB(x) is unique for a.e. x(µ-), and the fact thatµ+ <C IMI, µ « IµI, together with exercise 9.1.7.

10.2.15. Exercise: Let µ and v be o--finite measures on (X, S) such thatv is also a measure. If v <C µ - v, show that v <C µ and

X dv(X) 0.

l J/

10.2.16. Exercise: Let µ be a o--finite signed measure. Since µ+ <C µ andµ+ <C I µ1, we have measurable functions f and g such that

IE.fdM = f 9 d1pl, d E E S.

E

What is the relation between f and g?

10.2.17. Exercise: Let µ be a signed measure on (X, S) and f a real-valued measurable function on (X, S) such that b E E S, v(E) = fE fd,ais defined. Does µ(E) = 0 imply v(E) = 0? For example, consider theLebesgue measurable space (TI, G). Let

µ(E) := IEX d(x), EEG, and f (x) = e-x.

Can you conclude that v(E) = 0 if 1µI(E) = 0?

10.2.18. Exercise: Let µl and µ2 be defined on ([0, 1], G fl [0, 1]) by

µ1(E) := 2A(E rl [0,1/2]) - A(E) and µ2(E) := Lx d(x),

where A is the Lebesgue measure restricted to Gfl[0,1]. Show that A 2 <C Al,but µ1(E) = 0 need not imply µ2(E) = 0.

10.2.19. Proposition: Let vl, v2i µ be a-finite signed measures such thatvl + v2 is also a a-finite signed measure on (X, S).

(i) If vi <C µ and v2< µ, then (111 + v2) <C µ and for a. e. x(µ)

d(vi + v2) dvj(x) x + dV2

W

(ii) If vl <C µ, then

dµ (x) _ (x) for a. e.

Proof: (i) We first assume that p is a measure. Let v2 = v2 - vz , i = 1, 2,be the Jordan decomposition of v2. Since vi << ii and vi- << i.c, by exercise9.1.21,

d(viµ

v2+)(x) = µ (x)

and

dv++ 2 (X)

dfor a.e. x(µ),

d(vi µdv2) (x) = d (x) + d (x) for a.e. x(µ).

Since vl + v2 is a signed measure, using the above equalities and exercise10.2.8, we have, for E E S with I + v2) (E) I < +oo,

(V1 + V2)(E) (vj + v2+) (E) - (vi + v2 ) (E)d(vl + v2) d(vl + v2 )

fE dµ dµ - E dµ dµ

Idvl dvi 1 fET

dv2 dv2 1

JE dµ dµ dµ + JE dµ dµ) dµ

E (dvi dv2dµ dµ+ dµ.

Since vl and v2 are a-finite, this equality extends to every E E S. Now,from the uniqueness of the Radon-Nikodym derivative, we have

d (x) for a.e. x(µ).

In the general case when µ is a signed measure, let A, B E S be a Hahndecomposition of X with respect to µ. Consider the restrictions of vl, v2 andµ to A and B respectively and apply the earlier case to get, b E E S,

(ill + va)(E n A) =dvl

+dv2

(x)d,u+ (10.5)lEnA

dµ+µ dµ

and

P(111 + v2)(E n B)

dvl= JEnB +

dv2- (x)d. (10.6)(dp- µ

From the uniqueness of the Radon-Nikodym derivative, it follows that

µ+ (x) = µ (x) for a. e. (µ)x c A

and

2 (x) = dµ (x) for a.e. (µ)x c B.

In view of this, the required result follows from (10.5) and (10.6). Thisproves (i).

(ii) First consider µ to be a a-finite measure such that vi <C µ. ThenV E E S.

v(E) = lEdp(x)dµ(x) and - IE

Let A, B be a Hahn decomposition of X with respect to v, then

v+(E) = p(E) = I V E C A

and

Thus

and

Hence

v-(E) = _A(E) = lpl(E), V E C B.

dv (x) _ l µl (x) for a.e. (µ)x c A

d (x) = dl l (x) for a.e. (µ)x c B.

d (x) = dµl (x) for a. e. x(µ).

In the general case, we can consider a Hahn decomposition C, D of X withrespect to µ and apply the above case to the restrictions of v and µ to Cand D respectively.

10.2.20. Exercise: Let p, v be a-finite signed measures on (X, S) such

that v <G µ and µ <G v. Show that dv(x) 0 for a.e. x(µ) and

-1dv(x) v (x) for a.e. x(µ).

10.2.21. Theorem: Let Mb(X, S) denote the set of all finite signed mea-sures on (X, S). For µ, v E JVIb(X, S) and a E R, let d E E S

(µ + v) (E) := µ(E) + v(E) and (a)(E) := a(µ(E)).

Then .Mb(X, S) becomes a vector space over 1[8 under these operations. Fur-ther, for µ E Nlb(X, S), let

IWII:= W(X)I.

Then jis a norm on 14b(X, S), and the latter is a Banach space underthis norm.

Proof: That Mb(X, s) is a vector space under the above operations ofaddition and scalar multiplication is obvious. That II,uM is indeed a norm isalso easy to check. We show that Mb(X, s) is a Banach space under thisnorm. Let {n}n>1 be a Cauchy sequence in Mb(X, s), i.e.,

limI I An - Am I In--oo

Since V E E s,

0 as n,m - oo.

An (E)-µ,a(E) < <=IlAn - AmIl,

fAn(E)In>1 is a Cauchy sequence of real numbers. Let

/2(E) lim An (E), E E s.n--oo

W e can claim that c M b (X , s) and lim IlAn - A l l = 0. For this, clearlyn-->oo

/,c(0) = 0. Let E = U=1 En be a disjoint union of elements of s. We shallshow that /C(E) = E 1 A(En). Note that b k and b n,

(E) (Ej) µ(E)-µn(E)+µn(E)3=

An (Ej) -j=1 7=

Let E > 0 be given. We choose ko such that

JjAn - AmIl < 6, V n, m > ko.

Then for m, n > ko, using (10.7), we have

I An (E) - /-tm (E) I < E.

j=Ej)

µ(Ej) I. (10.8)

Also d m, n > ko,k k

[An (E3) - Pm(Ej)] I JAn(Ej) - Am (Ei) I

j=1 j=1

k

< 1: lAn - Aml(Ej)j=1

= I

< IlAn - Amll -< E-

Thus, letting m -+ oo, we get d n > kok

1: [An A) - A A) Ij=1

< E. (10.10)

Finally, since µn(E) = Ekj=1 µn(Ej) for all fixed n > ko, we can choose kl

such that b k > klk

An (E) - 1: tin (Ej) I C e. (10.11)j=1

Thus for k > kl, using (10.8), (10.9), (10.10) and (10.11), we havek

µ(E) - E µ(Ej) I < 3e.j=1

This proves that µ(E) µ(Ej), and hence µ is a signed measure. Since

lp(E)l liM lPn(X)l < +00,

µ E .M6(X, S). Finally we show that IlAn - All - 0 as n --- oo. Note thatfrom (10.9) we have, d n, m > ko and E E S,

I An (E) - Am (E) I <

Letting m --- oo, we get b n > k,

l- µ(E)l < e for every E E S.

Combining this with theorem 10.1.17, we get

(µn - µ)+(X) =sup{(µn - µ)(F) I E E S} <Similarly, (µn - )X) < E. Hence d n > k,

IIin. - II = I/2n - PI (X) = (l-in - P)+(X) + (A - P)+(X) < 2e.Thus,

lim IlAn - All = 0n-+oo

10.3 Complex measures 365

10.2.22. Exercise: For µ, v E Mb(X, S), we say v is smaller than µ, andwrite this as v < µ, if v(A) < µ(A) V A E S. Prove the following:

(i) Show that d µ, v E Mb(X, S), there exists 77 E Mb(X, S) such that77 > µ, q > v and r < 6 whenever 6 E .Mb(X, S) with 6 > µ and6 > v. In a sense 77 is the smallest of all the signed measures whichare bigger than both µ and v. This 77 is denoted by µ V v.(Hint: µ V v = (1-L + v - 1- vl)/2.)

(ii) Show that d µ, v E Mb (X, S) there exists a finite signed measure,denoted by µ A v, which is smaller than both µ and v and is largerthan any signed measure which is smaller than µ and v both.

10.2.23. Note: The order relation < defined in exercise 10.2.22 respectsthe algebraic operations on Mb(X, s), i.e., for v, p and 77 E Mb(X, s), ifv<I-c,then v+i <I-c+77 andav<ap V a>0,aES.Also for I-C,vE,u,sup{I,c, v} :_ I,c V v and inf {I,c, v} I-,c A v exist. Mb(X, s) is an example ofa Banach lattice.

10.3. Complex measures

Let (X, S) be a measurable space and µ, v be two finite signed measured on(X, S). Consider 77: S ) C defined by

,q (E) : = p (E) + i v (E), E E S.

Clearly q (0) = 0. Let {E,, },,>1 be a sequence of pairwise disjoint sets inS and let E := U=1 E. Then the series ((E) + is abso-lutely convergent, and it converges to µ(E) + iv(E). Hence E 1 isindependent of any rearrangement of the series, and we may write

00

77(E) _ E 77 (En).n=i

Thus 77 is a countably additive complex-valued set function on (X, S).

10.3.1. Definition: Let (X, S) be a measure space. A set functionv : S C is called a complex measure on (X, S) if the following aresatisfied:

(i) IL (0) = 0 -

(ii) µ is countably additive in the following sense: if E := U=1 En, wherethe En's are pairwise disjoint sets from S, then E'n=1 µ(En) is abso-lutely convergent and converges to µ(E). We write this as

00

µ(E)=) .µ(En).

n=

10.3.2. Example: Clearly, every finite signed measure can be treated as acomplex measure. The set function 11 = p + iv, where p, v are finite signedmeasures, is a complex measure. Another way of constructing complexmeasures is to consider a complex-valued integrable function f on (X, S, cc)and define

v(E) := f dp, E c S.fE'Then v is a complex measure. In fact, the equality

v(E) = fERe(f) dµ + i fE Im (f)dft, E E S,

shows that this example is the same as the earlier one. However, this complexmeasure v has the property that if µ(E) = 0 for some E E S, then clearlyv(E) = 0 as a complex number. This motivates the next definition.

10.3.3. Definition: Let (X, S, µ) be a measure space and v a complexmeasure on S. We say v is absolutely continuous with respect to µ ifv(E) = 0 for all E E S for which µ(E) = 0. We write this as v <C Et.

The question we would like to answer is the following: does the Radon-Nikodym theorem hold when v is a complex measure and v « p, p beinga a-finite measure? The answer is yes. Before we prove this, we make thefollowing observations.

10.3.4. Proposition: Let v be a complex measure on (X, S). Then thereexist finite measures v1, v2, v3, v4 on (X, S) such that v1 1112, v3 1 v4 andV EES,

v(E) = vl(E)- v2(E) + iv3(E) - iv4(E).

Proof: For every E E S, consider

(Re (v))(E) := Real part of v(E)

and

(Im (v))(E) := Imaginary part of v(E).

Then (Re (v)) and (Im (v)) are finite signed measures on (X, S). Put vi(Re (v))+, v2 :_ (Re (v))-, v3 := (Imv)+ and v4 :_ (Im (v))-.

10.3.5. Definition: Let v be a complex measure on (X, S). A complex-valued measurable function g on X is said to be v-integrable if g EL1 (X, S, v2) V 1 < i < 4, and in that case we write

f gdv := f gdvl - f gdv2 + i f gdv3 - i f gdv4.

10.3.6. Theorem (Radon-Nikodym theorem for complex mea-sures): Let (X, S, µ) be a Q-finite measure space and let v be a complexmeasure on S such that v <C µ. Then there exists a complex-valued functionf E L1(X, S, µ) such that

f gdv = f gd,u

for all g E n41 Ll (X, s, v2), where the v2, 1 < i < 4, are as given by propo-sition 10.3.4.

Proof: Consider v1, v2, v3, v4 as given by proposition 10.3.4. Then v2,u V i = 1, 2, 3, 4. Thus by the Radon-Nikodym theorem (exercise 10.2.3),3 f2 E L' (X, S, A) such that

vi (E) _ fdp, b E E S. (10.12)

Let f := f 1- f2 +i f3 - if4. Then f is a complex-valued measurable function,f c L1(X, S, µ) and

v(E)E

fd, e E E S.

Next, using (10.12) and the simple function technique, it is easy to showthat

J gdvZ = ffigd, (10.13)

whenever g E L', (X, S, v2). Let g c (12 1 Li (X, S, vi). Then (10.13) holds forsuch a g and each i = 1,2,3,4. We have

J gdv = J gdvl - J gdv2 + i j gdv3 - i j gdv4.

= fg(fi - f2 + if3 - Zf4)dµ

= fgfd.u

Recall that in case v is real-valued, i.e., when v is a finite signed measure,we showed that

Ll (X, S, v+) n Ll (X, S, v-) = Ll (X, S, I vI) .

To prove a similar result for complex measures, we take a hint from theorem10.1.17. and make the following definition.

10.3.7. Definition: Let v be a complex measure on (X, S). For E E S,define

jvj(E) = sup { v (Ei) I {Ei,... , En} is a measurable partition of E

The set function jvj is called the total variation of v.

10.3.8. Theorem: Let v be a complex measure on (X, S). Let vz,1 < i < 4,be the finite measures given by proposition 10.3.4. Then the following hold:

< E4(i) jvj(E) Vi(E)5 V E G S.

(ii) I is a finite measure on (X, S).

(iii) jv(E)l < IvI(X), V E E S.

(iv) vi(E) < IvI(E), V E EE S.4 L, (X5 S, vi).(v) L, (X5 S, Jvj) = ni=1

(vi) For any measure µ on (X, S), v <C [tiff vk <C µ iff I VI <C µ.

Proof: The proof of (i) is easy. To prove (ii), first note that, I= 0. Toprove the countable additivity of Jvj, let A = U'l A. where Aj E S V jand Ai fl Aj = 0 for i 4 j. Let a j_3 1 1 Then

n 00 0000 na < E 1: i= l< 1j=1 k=

Since cx < I vI (A) is arbitrary, we have

00

k=

jvj(A) < Ljvj(Ak) (10.14)k=1

To prove the reverse inequality, we may assume without loss of generalitythat jvj(A) < +oo. It is easy to see that jvj(E) < jvj(F) whenever E C F.Thus I vI (Aj) < +oo d j. Let e > 0 be given. Choose d j, sets E E S,1 <k< kj, such that E f1 EQ =0 for k P, Aj = U1E and

Jv(Aj) -Ell'kj

j=1

Then V m,m

j=1

I (Aj) <m

j=1

< E+

j=1 k=1

00

(En) I + Ivy U Aj)7=M+1

2i +

m kj

j=1 k=1

< 6 + IvI(A).

Since this holds b m and e > 0, we have00

I(Aj) < IvI(A). (10.15)j=1

Equations (10.14) and (10.15) prove that IvI is a measure. Since

IVI(X) < >'Vi(X) < +00i=1

(by (i)), IvI is a finite measure. This proves (ii).To prove (iii), let E E S be fixed. Then X = E U E° and hence

I< Iv(E)I+Iv(E°)I < IvI(X)To prove (iv), let Al, Bl be a Hahn decomposition of X with respect to

Re M. Then V E E S,

vi (E)

Similarly,

(Re (v))(E)(Re (v))(E n Al)

(Re (v))(E n Al)Iv(E n A1)I

v(E n Al)I + II

v I (E).

v2(E) (Re (v)) - (E)

_ (Re(v))(EflBi)lcvI (E).

That v3(E) and v4(E) are both less than IvI(E) can be proved similarly,using a Hahn decomposition of X.

To prove (v), let us first consider f, a nonnegative simple measurablefunction on (X, S). Let

m

j=

Then, using (i), we have

f fdvIm

j=1

m(Y

j=1

j XEj .

4

E Uk(Ej)k=1

Also, using (iv), V k = 1,2,3,4, we have

m

j=

i vk(Ej)

E J fdvkl (10.16)

Eajlvl(Ej) = ffdlvl. (10.17)

From (10.16) and (10.17), the claim of (v) follows when f is a nonnegativesimple function. Also note that a function f on X is defined a.e. (if f isdefined a. e. (vk), k = 1, 2, 3, 4. Let f be any nonnegative measurable functionand let {sn}n>1 be a sequence of nonnegative simple measurable functionson X increasing to if I a.e. IvI. Application of the monotone convergencetheorem to equations (10.16) and (10.17) above, with f replaced by sn,gives

n

f If IdIvI < E (f If ldvand f < ftii.4

fEL1(X,S,IVI) if fEflLl(X,S,vk).

This proves (v).

Finally, if it is a measure and it <G v, then µ(E) = 0 implies v(E) = 0.In particular, for E E S fixed with µ(E) = 0, we have µ(F) = 0 for everyset F C E, F C :S. Thus v(F) =0 V F C E, and it follows that IvI(E) = 0.Hence v <C it implies IvI <C it. Also it follows from (iv) that vk <G IvI V k =1, 2A4. 4. Thus vk <C it V k. Finally, if d k, vk « it, then, noting thatv = vi - v2 + 2v3 - 2U4, it follows that v <G it. This prove (vi).

10.3.9. Exercise: Consider .M (X, S), the set of all complex measures on(X, S). Define the operations of addition, scalar multiplication and norm on,M (X, S) as follows: for µ, v E Vl (X, S) and a E C,

(i) ([z + v) (E) := µ(E) + v(E), E E S.

(ii) (c)(E) := aµ(E), E E S.(iii)

11l,11 := IShow

that M (X, S) is a Banach space with these operations.

10.3.10. Remark: In view of theorem 10.3.8(v), theorem 10.3.6 can bestated as follows: Let µ be a a-finite measure and v a complex measure on(X, S) such that v <C µ. Then El f E L1(X, S, µ) such that

j gdv = J f9dµ, b 9 E Li(XI S, I

10.3.11. Proposition: Let (X, S, µ) be a a-finite measure space and letf E L1 (X, S, µ) be a complex-valued function. Let

v(E) :_ fdi, E E S.E

Then v is a complex measure and, d E E S,

IvI(E) =fEIfIdµ

In particular,

f If Id[z = Ilf Ill.

Proof: Let E E s be fixed. Then for any measurable partition {F1,... , Fn 1of E,

n n

v (Fj) I =j=

Hence

j=1 IFifdu I C fE If I

IvI(E) < J If Idµ. (10.18)E

To prove the reverse inequality, let {sn}n>1 be a sequence of nonnegativesimple functions such that I sn I < 1 V n and sn increases to XE. Define for

372

xEX,

9n W

0 if f (x) = 0,

where f (x) denotes the complex conjugate of f (x). Then {gnf}n>i convergesto XE 1f I and 1gn f I < fl E Ll (X, <S, µ). Thus by Lebesgue's dominatedconvergence theorem, we get

E

10. Signed measures and complex measures

If I - limo f fgnd. (10.19)

Also, sn being a simple function, let sn = >1i a X where {E1,... Ei=1 3

is a measurable partition of X. Since sn < XE , sn (x) = 0 for every x E X \ Eand0<aj<1 V j. Thus

f gn dpE

if f(x)0,

fgndp

iaif f dnEj

< fd,u

(EnE)lj=1

(10.20)

From (10.18), (10.19) and (10.20), we get

f IdµI = IE I

In particular, with E = X, we get II = I010.3.12. Corollary (Polar representation): Let µ be a complex measureon (X, S). Then there exists a measurable function f such that If (x) I = 1for all x c X, and, d E E S,

µ(E) = fE f d Ipl.

Proof: Clearly µ <C lpl. Let f be the function, as given by remark 10.3.10,such that d E E <S

it (E) = f d lpl.fE

10.4 Bounded linear functionals on Lp (X, s, µ) . 373

LetA:={xEXI If(x)l <1} and B:={xEXI If(x)l >1}.

Then by proposition 10.3.11, we get

IA(1 - If I)d Ipl = Itil (A) - If Id Itil = 0.

A

L If IdIpl - Ipl(B) = 0(If I - 1)dlpl =lB

implies that I = 0. Hence

If (x) I = 1 for a.e. ([I)xeX.We can redefine f to be such that If (x) I = 1 V x E X and still have,b EES,

µ(E) =E

fdI.U

10.3.13. Exercise (Hahn decomposition for finite signed mea-sures): Let µ be a finite signed measure on (X, S). Show that there existdisjoint sets A, B E S with the following properties:

(i) AUB=X.(ii) (E)>o V E E S, E C A.

(iii) (E)<0 V E E S, E C B.

(Hint: Use corollary 10.3.12 and A := {x E X I f (x) = 1}, B := {x EX I f W = -11.)

10.4. Bounded linear functionals on Lp(X, s, .c)

Let (X, s, µ) be a a-finite measure space. Recall that for 1 < p < 00,the space Lp(X, s, µ), as defined in section 8.4, is the space of all real-valued measurable functions on (X, s) such that f If V'di.t < +oo. The spaceLp(X, s, ,u) is a vector space over III, and if we identify functions which agreea.e. (µ), then, for f E Lp(X, s, µ),

1/p

lif lip If IPd[t)

defines a norm on Lp(X, S, µ) under which LP (X, S, µ) is complete. Let1 < p < oo. By Holder's inequality, for f E LP (X, S, µ) and g E LQ(X, S, µ),where 1/p + 1/q = 1, we have fg E Li (X, S, µ) and

fgdµ < If llpllgllq-

Let the map T : Lp (X, S, p) -f R be defined by

T(f) = f fgd.Then T is a well-defined linear map and

J7'(f ) I ll9llellf lip, d f c LP (X, S, µ).

In the case p = 1 and g E Lr00(X) S, µ), let

T(f) f gdp, f Cz Lr (X, S)J I

Once again, T is a linear map from L' (X, S, µ) into R, and

JC lIgIlllf111, V f E Li(X,SFor 1 < p < oo, let q be such that 1/p + 1/q = 1 when 1 < p < oo andq = oo when p = 1. We call q the conjugate of p. The conclusion of theabove discussion can be stated as follows:

10.4.1. Proposition: Let 1 < p < oo and g E L9 (X, S, µ), where q is theconjugate of p. Then T : LP (X, S, µ) -- R defined by

T(f) fgdµ, feL(X,S,),is a linear map and

IT(f)i iIgiiqIifiip, V f E LP (X, S) µ).

Further,sup J= IIgiiq

Il.fllP=1

Proof: Clearly, T is linear and

sup JT(f)l < iigiL. (10.21)IlfllP=1

We may assume, without loss of generality, that ll9llq : 0. If 1 < p < 00,put

folglq-lllgll

q /pThen

Thus

Further,

I A I Plglp,q-l)llgll

qq = lglqllgll

qq

(fIgid) llgllq-q= 1.

T(fa) - 9°dµ I (Il9llq 9IP) - iiIie( °)

10.4 Bounded linear functionals on LP '(X, s, P). 375

Thus

sup IT(f)l > 11911q- (10.22)Ilfllp=i

From (10.21) and (10.22), it follows that

sup IT(f)l = 11911a, if 1 jjgjjOO - cl) > 0.

LetE:= fx (E X I lg(x)l > 11g1loo - cl.

Since µ is a-finite, we can assume without loss of generality that 0 < µ(E)oo. Define d x E X,

fox)

where a(x) = 1 if g(x) >0 and a(x) = -1 if g(x) <0. Then foIi = 1 and

7'(fo) =µ(E)

fE l> IgMoo - e.Since this holds d e > 0, we have

7'(fo) > MI loo

From (10.21) and (10.23), we get

supI I = IgM M

IIJ'111=1

(10.23)

The aim of the rest of the section is to show that every linear mapT Lp(X, S, µ) I[8 which satisfies the condition IT(f) l < M 11f 11p for

some M > 0 and for every f E LP" (X, S, µ) is of the form T(f) = f fgdp forsome g E L9(X, S, µ). The proof is an application of the Radon-Nikodymtheorem. (Note that we have already proved this result independently forp = 2 in theorem 8.9.20 and used it in theorem 9.1.15 to deduce a proof ofthe Radon-Nikodym theorem. An independent proof of the Radon-Nikodymtheorem was also given in theorem 10.2.2 and exercise 10.2.3.) Before weprove this, we observe the following.

10.4.2. Proposition: Let T : Lp(X, S, µ) -> I[8, 1 < p < oo, be any linearmap. Then the following statements are equivalent (and if any one of them

µ)).is satisfied, we call T a bounded linear functional on LP(X, B,

(i) There exists a constant M > 0 such that

IT(f)l < M Ilf Ilp V f E Lr(XISIIt).

(ii) T is a continuous map.

If T satisfies either (i) or (ii), then

11ThI sup IIlfllp=i

exists and

Further,

I I = inf {M I IZ'(f )I < MII f IIP d f E LP(X, <S, µ)}.

The nonnegative number JIT11 is called the norm of the linear map T.

Proof: The implication (i) (ii) is obvious. Conversely, suppose T iscontinuous but (i) does not hold. Then b n, 3 fn E Lp(X, S, µ) such that

I> lSince

T is linear, T(f) = 0 if f (x) = 0 for a.e. x(µ). Thus j nand

Let

T fn(nllfnll) > 1.

Ingn l P

Then gn E LP(X,<S,µ) and ilgnhhp = 1/n -- 0 as n -- oo. But IT(gn)I >1 V n, which contradicts the continuity of T. Hence (ii) =* (i) also.

Next, let either (i) or (ii) hold, and let

K := inf { M I I< MII f IIP> d f E LP(X, S, µ) }.

Clearly, by the definition of K,

JIT(f)II < KlIf 11p) V f E Lr(X, B, /_Z).

Also, since (i) holds, 3 M such that

IT(f)l < M, V f E Lp(X,S,µ) with if liP = 1Hence

J 1 sup IIltllP=1

10.4 Bounded linear functionals on Lr (X, s, ,cc) . 377

exists. Further, V f E Li (X, S, µ) with I If I I P # 0, since tHfIIP

= 1,

T(II IIP)` MTII, i.e., IT(f)I < ITMIIfIIp

Hence 11ThI > K. On the other hand,

I< KII.f IIP, V ,f E LP (X, S, µ),

implies that I:5 K. Thus IThI = K. 0

p

10.4.3. Theorem (Riesz representation): Let (X, S, µ) be a a-finite,complete measure space and let 1 < p < oo. Let T : LP(X, S, µ) ][8 be acontinuous linear map. Then there exists a unique g E Lq(X) S) µ), where qis the conjugate of p, such that

T(f) = ffgd, b fEL(X,S,),

and JTIJ = l9la-

Proof: The idea of the proof is as follows. Suppose there exists someg c L9 (X, S, µ) with

T(f) fg dp, V f G Lr (X, S, tz).

If µ(X) < +oo, then we should have

v(E) := T(XE) = fgd, b E E S.

This suggests an application of the Radon-Nikodym theorem for v to locateg. Thus we shall prove the required claim when µ(X) < +oo by proving thefollowing steps:

Step 1. Considerv (E) : = T(XE) , EGS.

Then v is a well-defined finite signed measure, and v <C µ.

Step 2. The Radon-Nikodym derivative v (x) := g(x) E Lq(X, Si µ).

Step 3. The map T is given by

T(f) = /fgd, b fEL.

These steps will prove the claim in the case when µ(X) < +oo. Finally,we shall extend the claim when µ is Q-finite. We prove these steps one byone.

That v is well-defined, follows from the fact that for E E S we haveXE E LP (since µ(X) < +oo). Clearly

v(O) = 0 and Jv(X)I = IT(1)I < oo.

The finite additivity of v follows from the fact that T is linear. To prove thecountable additivity of µ, let E = U=1 En be a union of pairwise disjointsets from S. Let An := U=1 Ek. Then {XA, }n>1 increases to XE and, bythe dominated convergence theorem, we have

lim [IXAn - XE Jpdp = 0.n->00

Also,T(XE - X,qn J I C M II XE XAn IIP

Thus using (10.24) we have

limo I T(XE - Xqn I = 0.

(10.24)

Using the fact that T is linear, the above gives

v(E) = T(XE) = lim (10.25)

Since XAn = I:'k=l XEk> from (10.25) we have

v(E) = lienn->oo

n

T(XEk)

n

= lim L v(Ek)n-+oo

00

Hence v is a well-defined signed measure on (X, S). Suppose, µ(E) = 0.Then IIXE IJ7 = 0 and hence

I< MIIXEIIP = O.Thus I = 0. Hence v <C µ, proving step 1.

To prove step 2, let

g(x) := dµ (x) for a. e. x(µ).

Note that g is real-valued and

gdu, V EES.T(XE)E

10.4 Bounded linear functionals on Lp (X, s, it). 379

Thus, using the linearity of T and of the integral, we have

T(s) = fsgd, (10.26)

for every nonnegative simple measurable function s on (X, S). We show thatthis implies the required claim of step 2, i.e., g E Lq(µ). Assume first that1 i of nonnegative simple measurablefunctions increasing to IgIq . Let

A :_ {x E Ig(x) >01 and B := {<01.Define

Sn W :- (XA - XB) (On W) 11PI x E X.Then is a sequence of simple measurable functions, and

1/P

Jj'SnjIP - On

Further, since(x)I,9W 'snW = 19W I (On W) 1/P and (5())1/ 9 < l9

we have

Hence

0

Thus

(X) Sn ('T) J (q(x))u//9= On (T)

J On(x)dµ(x) < f Sn(x)9(x)dµ(x) T(Sn)J

On(x)dµ(x) IuTiiusniiP

SIP

= I(f Ondit)

Ondtz) : JIT11.

Thus

f Ond/-L : I I Tl1q.

Since this holds b n, by the monotone convergence theorem, we have

I: J

This proves that g E Lq(µ) and I:5 I I in the case 1 < p < oo. For thecase p = 1, since d E E S,

dµT(XE) = IE 9

we have

E9dµ I < I <

I

Thus it follows from theorem 8.1.9 that

I9(x)I JITIJ for a.e. x(µ).

Hence g E and II9II 11Th. Thus V 1 < p < oo, the functiong E L9 (X, S, µ) with II9119 <_ ITIf. This proves the claim of step 2.

To prove step 3, first note that from (10.26) we have, for every simplemeasurable function s,

T(s) = f sgdjc.

Using theorem 8.6.1, given f E Lp(X, 8,,c.c) we choose a sequence {s}>i ofsimple functions in Lp (X, s, u) such that lim ((f - sn I I p = 0. Then, by the

0ocontinuity of T, T (f) = lim T(s). Also, using Holder's inequality, in case

n-*oo1 < p < oo, we have

ffgd_ f Sn9dA

In case p = 1, clearly

fgdµ - f Sn9dµ

If - SnI I9I dµ < Ilf SnIIp H<

fIfsnIIgfdµ :5 Ilf SnIli hqlloo-

Thus in either case we will have

lim fsmgdi=ffgdi.

Hence d f E LP '(X, S, µ),

T (f) - limo T (Sn) - limo f Sngdµ - ffd.Further, from proposition 10.4.1, we have I= ITo prove the unique-ness of g, let there exist some gl E LP(µ) such that

J fgdµ = ffid.Then g - gl E Lq(X, S, µ) and

(Tg_g1)(f) = f fgd - f fgidµ = 0, d f E L' (A).

10.4 Bounded linear functionals on LP '(X, s, ,L). 381

Thus

0 = I II = 119 - 91 IIq,

i.e., g = g1. This completes the proof in the case ,a(X) < +oo.To prove the theorem when ,a is a-finite, let {X}> 1 be an increasing

sequence of sets in s such that U=1 Xn = X and ,a(Xn) < +oo for every n.Consider the continuous linear map Tn : Lp (Xn, s n Xn, ,a) --- f R, definedby

Tn(f) = T (f ), V fEL(X,XnS,,a),where each f is treated as a function on X, with f 0 on Xn. Thenby the earlier case, there is a function gn c Lq (Xn, Xn n s, ,a) such thatV IEL(X,S,/i),

. dµ'T(XXn f ) = 1Xn f 9

If we treat gn as an element of LP (X, S, µ), with gn =_ 0 on Xnc, then

Since {X}>1 is an increasing sequence and every gn is uniquely determined(except for a set of µ-measure zero), we may assume that gn+l(x) = gn(x)for x E Xn for every n. Define g on X by

9W := gn(x), if x E Xn-

Then g is a well-defined measurable function on (X, S), and the sequence{increases to IgI, d x E X. Hence , by the monotone conver-gence theorem and continuity of T, we have, in case 1 < p < oo,

f l9l4 dµ = lim f Ignj9dµ c IITII.

Thus g E Lq(X, S, µ) when 1 

f and IxXn f I < f,by the dominated convergence theorem XXn f

__>f in LP (X, S, µ). Also,

XX" fg --> fg and IX fgl IfI E Lp(X,S,µ). Thus, again by the domi-nated convergence theorem b f E LP (X, S, µ),

f fgdµ lim f fgdµ - moo f fgdl-L = nmoo T .f) - T (.f)n n

The claim that 11ThI = IlMq now follows from theorem 10.4.3.

10.4.4. Remarks:

(i) The claim of proposition 10.4.1 for the case 1 c p < oo holds even whenµ is not a-finite (analyze the proof carefully).

(ii) When 1 < p < oo, theorem 10.4.3 can be extended to cases when p isnot necessarily a-finite. We show that there exists a set W E S such that µrestricted to W is a-finite and T restricted to W' is the zero map. For this,let E E S with p being a-finite on E. Let gE denote the unique element ofLQ (E, En s, p) such that V f c LP (X, s, µ)

T(xEf) =E

fgd/L.

If we define gE = 0 outside E, then gE E L9(X, S, µ). Also if A C E, thenit follows from the uniqueness of gE that gA = gE a.e. (ii) on A. For everyE E S such that µ is Q-finite on E, let

1/ (E) .- = f 19Ejqdp.

Then d A C E with A c S,

v(A) = fIgAIdµ <

since

f 19E I

f I9EIq

dµ = II 9E II q =II7'E IIqj

where TE is the restriction of T to LP (E, E fl s, p), we have JITEII < 11Th.Hence the set

S :_ {v(E) S E E S, µ being Q-finite on E}

is bounded above. Note that S is a nonempty set. Let a := sup S, and let{E,, },>1 be a sequence of sets from S such that µ is Q-finite on each E,,with

lim v(En) = a.Ti-oo

Let00

.W.=UEn

Then µ is Q-finite on W. Further, since

v(En) < v(W) < a d n,we have a = v(W). Let f c LP' (X, S, µ). Then the set

A:= fx c X I If (x)I :,/-- 01 c S.

Further, µ is Q-finite on A, because A = Uk 1 where An _ {x E X I n <If(x)l < n + 1} and µ(An) < +oo V n. Thus µ is Q-finite on A U W also,and

v (A U W) < a - v (W).

10.4 Bounded linear functionals on LP (X, s, p). 383

As v(W) < v(A U W), we have v(W) = v(A U W), i.e.,

f l7l 4dp = fIgwId.

Thus gAuw (x) = 0 for a.e. x on A \ W. Since gA (x) = 9Auw (x) for a.e.(µ)x E A, we have gA(x) = 0 for a. e. (µ)x c A \ W. Thus

T(fXwc) = T(fXAnwc) = fgd/ = 0.

Hence T restricted to W° is the zero map. In fact, if we define g(x) := gw (x)on W, and equal to 0 elsewhere, then for f E LP (X, S,

T(f) = 7'(fXW) = f fgWdµ = f f 9dµ

Clearly, g E Lq(X, S, µ), and, as before, lIgIIy = 11Th.

(iii) When p = 1, the condition that µ is a-finite in theorem 10.4.3 is neces-sary. For example, let X = [0,1], S = {A C X I A or A' is countable} andµ the counting measure on S. Then f E L1 (X, S, µ) if f vanishes on all butcountably many points al, a2i ... in X with E°_° 1 If (ai)l < +oo Considerthe linear map T : L1(X, S, µ) I[8, defined by

T(f) E xf (x)

SinceIT(f)l < L If (x)l = Ilf Ill,

T is a continuous linear functional on L1(X, S, µ). Clearly, if T (f) = f fgdµfor some g E L,,O (X, S, µ), then we should have g(x) = x d x. But thefunction g(x) = x is not a measurable function on (X, S).

(iv) For p = 1, theorem 10.4.3 can be extended for a class of measure spaceswhich are more general than the space of a-finite measures. These are calleddecomposable measure spaces. For this, we refer the reader to Hewitt andStromberg [18].

(v) Theorem 10.4.3 remains valid for Lp(X, S, µ), the class of complex-valuedmeasurable function on X such that f If V'd/2 < -}-oo. For this, one can con-sider (Re T)(f) := Re (T(f)) and (Tm T)(f) := Im (T(f)) and apply theorem10.4.3.

Appendix A

Extended real numbers

A.1. Extended real numbers

Let ][8 denote the set of all real numbers and let

R* := R U 1+001 U I-001,

where +oo and -oo are two symbols read as plus infinity and minusinfinity. We extend the algebraic operations and the order relation of l[8 toIl8* as follows:

1. Forevery xeR, -oo<x<--boo.2. For every x c R,

(-oo) + x = -00 and (+oo) + x = +oo;(+oo) + (+oo) = +oo and (-oo) + (-oo) = -00.

3. For every x c IR,

x(-boo) = (+oo)x = +o0X (- 00) = (- 00) X = - 00

X(+00) = (+00)X = -00X(-00) = (-00)X = +00

ifx>0,

ifx<0.

Further,

(+00)0= (-00)0=0, (±oo)(+oo) = (+00) and (±oo)(-oo) = (TOO).

Note that the relations -oo + (+oo) and (+oo) + (-oo) are not defined.

385

386 Appendix A

The set R*, also denoted as [-oo, +oo], with the above properties iscalled the set of extended real numbers. The symbol +oo is also denotedby oo, when no confusion arises.

A.2. sup(A) and inf(A)

For a nonempty set A C ][8*, we write sup(A) :_ +oo if Af11E8 is not boundedabove, and inf(A) -oo if A n 1[8 is not bounded below. Thus

sup(A) and inf(A) always exist for every nonempty subset A of ]E8

A.3. Limits of sequences in IR*

For {x}> l any monotonically increasing sequence in IR* which is notbounded above, we say {x}> l is convergent to +oo and write

lim xn=+00.

Similarly, if {Xn}n>i is a monotonically decreasing sequence which is notbounded below, we say {xn}n>1 is convergent to -oo and write

lien xn = -00.

Hence

every monotone sequence in lR* is convergent.

Thus for any sequence {Xn}n>i in R ,the sequences xk} j> 1 andf inf k> j xk } j > 1 always converge. We write

lien sup xn lien (sup xk )3`00 k > j

and

lien inf xn : = lien (inf Xk) .

jlilnn,o.o sup xn is called the limit superior of the sequence {xn}n>1 andlien infn,oo xn is called the limit inferior of the sequence {x}>. Notethat

lien inf xn < lien sup xn .

We say a sequence {x}>I is convergent to x E JR* if lien infn,o-o xn =lim supn,. xn =: x, say. In that case we write limn,oo xn := x.

A.4. Series in JR*

Let be a sequence in R* such that for every n E N , sn :_ Ek=1 xk iswell-defined. We say that the series EO) 1 xk is convergent to x if {s}>is convergent. We write this as x = E 1 Xk, x being called the sum of

Extended real numbers 387

00 OOthe series i1x/ . For example, if each Xk > 0, then the series i=i x/

is always convergent.

Let be a sequence in R*, x,z > 0 for every n. Let a : N -- N beany bijective map. Then the series E' 1 x,(k) is called an arrangementof the series >I=

1xk.

For every rearrangement a, the series E001 x() and i=i x con-verge to the same sum.

Clearly, both the series are convergent in R*. Let a 1 x/ and QEOkO=1 x To prove a - in view of the symmetry, it is enough to prove

that 3 < a. Nown m

k=1 k=1

00 00

where in the the middle term m := max{ a (1) , , a(k)j. Hence ,8 < a,proving the required claim.

Similar arguments apply to any double-indexed sequence ofnonnegative elements of ]R* . For every fixed n, the series EM=1 xn,m is con-vergent in R*. Let yn := EOO

°1 xn,,-n . Further, E°° 1 yn is also convergent inR*. Let y := E00 1 yn . We can also define for all fixed m, zm := I=i xn,mand z := E'=1 z,-n in We show that both these processes lead to thesame sum, i.e., y = z, which is written as

00 00 00

xn,m=1 m=1

0 = E X,(k) _< 1: Xk _< E Xk = Ce)

00

nz=1 \n=1

To see this, we note that V r, s c N,r s s 00

E E xn,m xn,,-n_< 1: 1:

Hencen=1 m=1

y

00

n=1 m=1

00

E

m=1 n=1

xn,m

00

k=1

00

E00

xn,m .

rn=1 n=

E E xn,m =: Z.m=1 n=1

The reverse inequality follows from symmetry.

A similar result holds for rearrangement of double-indexed series of non-negative terms in IR* :

Let a : N -* N xN be any bijective map, and let {Xn,m}n>i,m>.i be adouble-indexed sequence of nonnegative elements of R*. Then

00 00 00

E xa(r) = E E xn,mr=1 n=1 m=0

00 00

E E >Xn,m.m=1 n=1

Appendix B

Axiom of choice

Let A, B be two sets. One defines A x B, the Cartesian product of A andB, to be the empty set if either A or B or both are empty, and to be the setof all ordered pairs (a, b), a E A, b E B, when both A and B are nonempty.Similarly, for a finite family of nonempty sets Al , ... , A7z, we define theirCartesian product to be the set

Al x...xAn:={(xl,x2,...xn) I xiEAi,i=1,2,...,n}.

Obviously, Al x x An is a nonempty set. We can think of Al x ... x An asthe set of all functions f : {1, 2,... , n} - U=1 A i with f (i) = xi E Ai foreach i. One can copy this to define the Cartesian product of any arbitraryfamily of sets, say {Aa}aEI, to be the set

fj A« = faEI

: I -* U AaaEI

f(a)EA,IVaEI -

However, there is no surety that the set fJaEI A. is nonempty, i.e., we do notknow that there always exists at least one function f :I --p UaEj Aa suchthat f (a) E Aa V a E I, although intuitively it seems obvious that such afunction should always exist. However, this cannot be proved with the usualaxioms of set theory. (For a short introduction to axiomatic set theory, seeRana [30]. For detailed account of axiomatic set theory, the axiom of choice,and its history, see Halmos [15] and Fraenkel [12]). The way out of the abovedilemma is to treat this an axiom itself, called the axiom of choice:

If {Aty}ci is a nonempty family of sets such that each A« is nonempty,then there exists a function f I -- UaE I A« such that f (a) EA«`daEl.

389

390 Appendix B

Such a function is called a choice function.The axiom of choice has many equivalent formulations; a useful one is

the following:

If {Aa I a E I} is a nonempty family of pairwise disjoint sets suchthat Aa =/- 0 for every a c I, then there exists a set E C UaEI Aasuch that E fl A« consists of precisely one element for each a E I.

The axiom of choice finds applications in many diverse branches of math-ematics. (We have used it in section 4.6 to construct nonmeasurable subsetsof ][8.)

Appendix C

Continuum hypothesis

Let X and Y be two sets. We say X and Y are equipotent if there existsa bijection between them. We write this as X ti Y. In a sense, equipotentsets have same `number' of elements.

We say a set A is finite if A ti {1, 2, ... , n} for some n c N, and we sayA is countable if A ti N, the set of natural numbers. A set which is notcountable is called uncountable. For example, N, Z, Q are all countablesets while R is uncountable. (For a detailed discussion, see Rana [30].)

The statement X ti Y means X and Y have the same `number of ele-ments', and can be made precise as follows (see Halmos [15] for details). LetC be a collection of sets such that any two members of C are equipotent toeach other. Then one can assign a symbol, called its cardinal number,to each A E C, denoted by card(A). Thus

card(A) = card(B) if A B.

The cardinal number of a set A is also called the cardinality of A. For exam-ple, for any set A which is equipotent to {1, 2)... , n}, we write card(A) = n.For any set A ti N, we write card(A) = No , called aleph-nought. For anyset A ti R, we write card(A) = c, called cardinality of the continuum.For finite sets, it is easy to see that if card(A) = n, then card (P(A)) = 2n,where P(A) is the set of all subsets of A. We define, for any nonempty setX

card(P(X)) := 2card(X)

For example,

card(P(N)) := and card(P(R)) := 2`.

391

392 Appendix C

For a finite set A, we know that

card(P(A)) = 2card(A) > card(A).

Can the same be said about arbitrary sets which are not necessarily finite?For two sets A and B, we say card(A) -_- card(B) if there exists a one-onemap from A into B. We write card(A) < card(B) if card(A) card(B) butcard(A) -7 card(B). Now we ask the question:

Is card(A) < card(P(A)) for any nonempty set A?

The answer is in the affirmative and is due to George Cantor (see Rana[30] for a proof). It can be shown that c, i.e., P(N) R. Thus wehave the following:

This raises the following natural question:

. Does there exist a cardinal number a such that ?Zo < a < c ?

That is, does there exist a set A C ][8 such that A 96 ][8 and N C A butA 96 N? The answer to this question is not known. The statement that theanswer to the above question is in the negative is called the continuumhypothesis:

There does not exist any cardinal number between o and 20 = c.

An equivalent formulation of this is the following:

The set JR can be well-ordered in such a way that each element of JRis preceded by only countably many elements.

We used this in section 3.4, to prove Ulam's theorem. It is known thatthe continuum hypothesis is independent of the Zermelo-Fraenkel axioms ofset theory.

Appendix D

Urysohn's lemma

In theorem 8.6.2, we needed the following result:

Let K be a compact subset of Il8' and U an open subset of Il8n withK C U. Then there exists a continuous function g : ][8n [0, 1] suchthat g(x) = 1 Vx E K and g(x) = 0 V x V U.

This follows from the following result, known as Urysohn's lemma:

D.1. Theorem: Let (X, d) be any metric space and A, B be closed subsetsof X such that AnB = 0. Then there exists a continuous function g : X --j[0, 1] such that g(x) = 1 b x E A and g(x) = 0 b x E B.

Proof: For any nonempty set Y C X, and x E X, let

d(x, Y) := inf{d(x, y) I y E Y}.

Then it is easy to see that for every x, z E X

I d(x, Y) - d(z, Y) < d(x, z).

Thus x --- d(x, Y) is a uniformly continuous function. Further, if Y is aclosed set, then d(x, Y) = 0 if x E Y.

Now, given disjoint closed sets A and B, define b x E X,

d(x, B)g(X) :=

d(x, A) + d(x, B)

Clearly g is a continuous function with the required properties.

393

394 Appendix D

D.2. Corollary: Let A be a closed subset of X and U an open subset of Xsuch that A C U. Then there exists a continuous function g : X - [0, 1]such that g(x) = 1 d x E A and g(am) = 0 d X VU.

Proof: This follows from theorem D.1 with B = Uc.

D.3. Corollary: Let (X, d) be a metric space and x, y E X, x 4 y. Thenthere exists a continuous function g : X -* [0, 1] such that g(x) 4 g(y).

Proof: This follows from theorem D.1 with A = {x}, B = {y}.

The second corollary has a great significance. It ensures that metricspaces have a rich supply of real-valued continuous functions: any two dis-tinct points can be separated by a real-valued continuous function. Urysohn'slemma can be extended to some topological spaces. For details see Hewittand Stromberg [18], Munkres [26].

Appendix E

Singular valuedecomposition of amatrix

We consider matrices with real entries only. For an m x n matrix A = [aJ,its transpose is the n x m matrix [bJ := A', where bid = aji, for 1 < i < nand for 1 < j < m. We write the elements of Rn as 1 x n matrices.

E.1. Theorem: Let A be an m x n real matrix of rank r. Then there existmatrices P, Q and D with the following properties:

(i) P is an m x r matrix with PtP = Id.

(ii) Q is an n x r matrix with QtQ = Id.

(iii) D is an r x r diagonal matrix with nonzero real entries.

(iv) A = PDQt.

The representation A = PDQt is called the singular value decompo-sition of A.

Proof: Consider the m x m real matrix B := AAt. Since B is symmetric,all its eigenvalues are real. In fact, it is easy to see that all the eigenvaluesof B are nonnegative and there exists a complete set of orthonormal eigen-vectors of B. Since rank(B) = rank(A), B has r positive eigenvalues, sayA1, ... , A, Let xl,... , x,,... , x,n E R'n be the complete set of orthonormaleigenvectors of B, xi being the eigenvector corresponding to the eigenvalue

395

396 Appendix E

Ai ,1 < i < r. ThenAjxt if 1 < i < r,

AAtxi =Ot if r+1 <i <m.

Thus for r + 1 <i <m,xiAAtx? = 0, i.e., IxA!2 = 0, i.e., xiA = 0.

Letxi A

yi 1 <i<r.Ai

Clearly y1, ... , y,, are orthonormal vectors in W1. Let P be the m x r matrixwhose column vectors are 4,... , xt,, and let Q be the n x r matrix whosecolumn vectors are y1, ... , yt,Y. Then

A = IA(+ ... + 4Xm)Ax1x1A + ... + xt,xrA

x1( A1y1) + ... +xtr( Aryr)

PDQt,

where D is the r x r diagonal matrix with entries X1, ... , A,..

E.2. Corollary: Let A be an n x n real nonsingular matrix. Then thereexist n x n orthogonal matrices P1 and P2 and an n x n diagonal matrix Dhaving positive entries such that A = P1 DP2 .

Proof: Take Pl = P and P2 = Qt, where P, Q are the matrices as given bytheorem E.1.

E.3. Note: The positive real numbers X1, ... , VA-r, obtained in the proofof theorem E.1 are called the singular values of the matrix A.

Appendix F

Functions of boundedvariation

F.1. Definition: Let f [a, b] ) R, and let P :_ {a = xO < xl < ... <xn = b} be a partition of [a, b]. Let

n

Vb(pl f) If (Xk) - f (Xk-1)1-

k=1

We call ab(P, f) the variation on f over [a, b] with respect to the partitionP. Let

ab(f) := sup{ ab(P, f) I P a partition of [a, b] }.

The extended real number ab (f) is called the total variation of f over[a, b]. The function f is said to be of bounded variation if ab(f) < +oo.

F.2. Example: Suppose f [a, b] ) R is a monotonically increasing ormonotonically decreasing function. Then for every partition P of [a, b],

ab(P,f) = I

Vab(f) = If (b) - f (a) I < +oo.

Thus every monotone function is of bounded variation.

F.3. Proposition: For functions f and g on [a, b], the following hold:

(i) If f is of bounded variation, then f is a bounded function.

397

398 Appendix F

(ii) If f and g are of bounded variation, then so are the functions f +g, f- g, f g and o f for every a c R.

Proof: Exercise.

F.4. Proposition: Let f [a, b] ----) ][8 be of bounded variation. Forx, y E [a, b] with x < y, let Vy(f) denote the variation of f in the interval[x, y] if x < y, and 0 if x = y. Then the following hold:

(1) (f)+V(f)=V(f), V a < c < b.(ii) ax (f ), x E [a, b], is an increasing function.

(iii) I< Vy(f)-Va(f), b a<x<y<b.(iv) The function VQ (f) - f (x), a < x < b, is an increasing function.

Proof: (i) It is easy to see that for every partition P of [a, b],

Va(f+Vb(f)Hence Q6(f) < Vac Y+ Vb(f ). To prove the reverse inequality, let c > 0 begiven. Choose partitions Pl and P2 of [a, c] and [c, b], respectively, such that

Vac(f) -e/2 < Va(P1j) and Vb(f)-E/2 < Vb(P21 f)Let P := P1 U P2. Then P is a partition of [a, b], and

Va(f)+Vb(.f)-E < ab(fSince c > 0 is arbitrary, we get the required reverse inequality.

(ii) Since Va (f) > 0 V x E [a, b], it follows from (i) that Va (f) is anincreasing function of x.

(iii) Let P be any partition of [a, x]. Then Pl := P U {y} is a partitionof [a, b], and

Vay(f) >Since this holds for every partition P of [a, x], we get

Vay(f) >This proves (iii).

(iv) Let a < x < y < b. Then it follows from (iii) that

f(y)-f(x) C I C Va (f)- ax(f)Thus

(V(f)-f(y))-(V(f)-f(x)) >- 0 .

This proves (iv).

F.5. Theorem (Jordan): Let f [a, b] ) R. Then f is of boundedvariation iff f is the difference of two monotonically increasing functions.

Functions of bounded variation 399

Proof: If f = g - h, where g and h are monotonically increasing functions,then it follows from example F.2 and proposition F.3 that f is of boundedvariation. Conversely, let f be of bounded variation. Let g(x) := V(f)and h(x) := Va (f) - f (x). Then g and h are monotonically increasing, byproposition F.4, and f = g - h.

F.6. Exercise:

(i) Let f : [a, b] ) R be a differentiable function such that its derivativeis a bounded function. Show that f has bounded variation.(Hint: Use Lagrange's mean value theorem.)

(ii) Let f : [0, 1] - R be defined by

f(x):= x2 sin if x 0,

0 ifx=0.Show that f is a continuous function of bounded variation.

(iii) Let f : [0, 1] R be defined by

f(x):=< x sin if x 0,

0 ifx=0.Show that f is uniformly continuous but is not of bounded variation.

Appendix G

Differentiabletransformations

For x E TR , x := (xl,... , x.,,), let IxI := (E 1x01/2 For a E Tn and r > 0,let B(a, r) {x E IRn Ix - al < r}. We denote by {e',e2,... , en} thestandard basis of Wn .

G.1. Definition: Let U C IRn be an open set and T : U -* IRm be amapping. We say that T is differentiable at a E U if there exists a linearmapping A : Rn -p TRt such that

limh--+O

T (a + h) - T (a) - A(h)h

= 0.

Equivalently, for each e > 0 there exists S > 0 such that B(a, S) C U andb y E B(a, S),

I T(x) - T(a) + A(x - a) I < clx - al.

Whenever such an A exists, it is unique and is denoted by dT(a), called thedifferential of T at a. We say that T is a differentiable mapping on Uif T is differentiable at every point a E U.

For T : U , TRt, let T (x) = (Ti(x),T2(x),... , Tm(x)). Then thefunctions T2,1 < i < m, are called the coordinate functions of T, and wewrite this as T = (Ti,... , Each TZ is a map from U -* R. We denoteby (DT)(x), 1 <j < n, the jth partial derivative of TZ at x, whenever itexists.

401

402 Appendix G

G.2. Theorem: Let T U C ][8n ) ][8'n be differentiable at a c U.Then T as continuous at a and all the partial derivatives (DT)(a) exist,1 < i, j < n, with DjTi(a) being the iti' coordinate of (dT(a))(eJ).

Proof: By definition, given E > 0 there exists b > 0 such that V x EB(a,6)CU,

T(x) - T(a) - dT(a)(x - a) I < Ix-aI.Thus

T(x) - T(a) I < dT(a)(x - a) I + elx - al. (G.1)

Let dT(a) have the matrix [a] with respect to some bases on ][8n and R.Then for x = (x1i ... , xn), a = (al,... , an),

I dT(a)(x - a) I =n i n

\k=1 V=1

By the Cauchy-Schwarz inequality,

Hence

kj(xj - aj) aki12

n

71

kj2dT(a)(x - a) I <

Let us write

Then

IIdT(a)II =n

k=1 j=1

1/2 1/2

(Xj - aj )2

j=1

1/2 1/2

E(xj - ai)2

j=1j=1

kj I2

I dT(a)(x - a) I c IIx - al. (G.2)

From (G.1) and (G.2), we have

I T(x) - T(a) I < (c + IldT(a)ll)lx - al,

which clearly implies the continuity of T at a. Next we show that (DT)(a)exists for every i, j. Let Ai(ei) denote the ith coordinate of (dT(a))(e), forevery 1 < j < n and 1 < i < m. Then b h E I[8, h =h0,

Ti(a + hey) - Ti (a) - 1tAi(ei) I I T(a + hey) - T(a) - dT(a)(hei)h h

This is because, for x = (xi,... , xn) E Rn we have lxii < x , 1 < i < n. Itfollows from the above inequality and the differentiability of T at a that the

Differentiable transformations 403

partial derivative (DT)(a) exists and is the ith coordinate of (dT(a))(ei).

G.3. Theorem: Let U be an open subset of ][8n and T : U --) ][8m be amapping with coordinate functions Tl, ... ,T,,,. Then the following hold:

(i) T is differentiable at a (E U iff each Ti : U -) ][8 is differentiable ata. Further, in that case the itl' row of the matrix of (dT)(a) is thematrix vector of (dT2)(a).

(ii) If all the mxn partial derivatives (DjTi)(a) exist d a E U and arecontinuous (in that case, we say that T is a Cl-mapping on U),then T is differentiable.

Proof: (i) Let T be differentiable at a E U and Ti be its ith component.Let Ai denote the ith coordinate map of the linear map (dT)(a). Then forhE1[8"') 0,

Ti(a + h) - Ti(a) - Ai(h)<

T(a + h) - T(a) - dT(a)(h)h h

Hence Ti is differentiable at a with (dTi)(a) the ith coordinate map of(dT)(a). Thus the ith row in the matrix of (dT)(a) is the 1 x n matrixof (dT)(a).

Conversely, suppose each Ti is differentiable at a. Consider the linearmap A whose ith coordinate function is (dTi)(a). Then

I T(a + h) - T(a) - A(h) I

I T(a+h) -T,(a) - (dT)(a)(h) z

Tj (a + h) - Tj (a) - (dTj) (a) (h) I -

i=1

From this it follows that T is differentiable at a and (dT)(a) has ith coordi-nate map (dT)(a). This proves (i).

(ii) To show that T is differentiable, it is enough to show that each Ti isdifferentiable. Thus we may assume without loss of generality that m = 1.We are given that (D3T)(a) exists d a E U, 1 < j < n, and is a continuousfunction. We shall show that T is differentiable. In view of theorem G.2,the natural candidate for (dT)(a) is given by

n

(dT)(a)(x) E(DjT)(a)xj, x = (xi,... , xn) E U.j=1

404 Appendix G

We show that (dT)(a) is indeed as defined above. We prove it by inductionon n. For n = 1, it is obvious. So assume this for n - 1. To prove itfor n, for x = (Xi,... , x,,,) E ][8n let mi(x) :_ (xi,... ) xn_1). Note that-x(x + y) = -x(x) + ma(y) and the set 7r(U) = {y E I[8n-lly = -F(x) for somex E U} is an open subset of I[8n-1. Now, for any a = (al, ... , an) E U andh = (hi,... , hn) E ll8n such that a + h E U, we have

T(a + h) - T(a) - E(DjT)(a)hjj=1

= T(7r(a + h), ate, + ham,) - T(7r(a + h), ate) - (DT)(a)hn

+T(7r(a + h), ate) - T(7r(a), ate,) - E (Dj T)(a)hj (G.3)j=1

Consider the function T : 7r(U) ll8 defined by

T(t) := T(t, ate), t E 7r(U).

It is easy to check that the (DST) (t) exist for all 1 < j < n - 1. In fact,(DST) (t) = (DT)(t, an), and hence t has all partial derivatives in 7r (U), andthey are continuous. Thus, by the induction hypothesis, t is differentiableat every t E 7r(U). In particular, T is differentiable at er(a), and by theoremG.2 we have

n-1

(d)((a))(t) _ (Dj T)(a)tj,j=1

for any t = (t1,... , tn_1) E Il8n-1. Thus given e > 0, we can choose 0 < b' < Esuch that for all s E ][8n-1 with Isl < 8' < 1 we have (ir(a) + s) E U and

n-1

(a) + s) -T(e(a)) - E(DjT)(a)sj < els1/2. (G.4)1

Next, by the continuity of (DT)(a), we can choose b > 0 such that 0 < 6J' and d x E II8n with Ix-al < J, we have x E U and

I (DnT) (x) - (DnT) (a) I < E/2. (G.5)

Now, fix h := (h1,... , ham) E ll8n with I hI < J. Then 1,7r (h) I < J', and by themean-value theorem for one variable, we have 0 < 0 < 1 such that

T(7r(a + h), an+ hn) - T(,7r(a + h), an) = hn(DnT)(-x(a + h), a + 9hn).

Then

(7r (a + h), an + ehn) - a I < (7r (h), ehn) I < h < 6. (G.6)

Thus, using (G.5) and (G.6), we get

T(-x (a + h), an + ham,) - T(-x (a + h), ate,) - (DT)(a)hn I < EI hnI/ 2. (G.7)

Combining (G.3), (G.4) and (G.7), we have, d IhI < S,

T(a + h) - T(a) - E(DjT)(a)h I C El7r(h)1/2 + ejhnj/2 < e.j=1

Hence T is differentiable at a.

G.4. Definition: Let U be an open subset of Rn and T : U ) Rn bedifferentiable at a E U. Let the matrix of (dT)(a) with respect to a basis ofRn be denoted by [(dT)(a)], i.e., the ijth element of [is (DjTi)(a),where Ti is the ith coordinate function of T. The Jacobian of T at a isdefined to be the determinant of the matrix [(dT)(a)] and is denoted byJT(a). Note that if T is a Cl-mapping on U, then JT(a) is a continuousfunction on U.

G.5. Theorem (Inverse function): Let U be an open subset of Rn andT : U ) Rn be a C1-mapping. Suppose that JT(a) = 0 V a E U. Then forevery a E U, there exists an open neighborhood V of a with V C U such that

(i) T is one-one on V.(ii) T(V) is in an open neighborhood of T(a).(iii) T-1 : T(V) --) V is a C' -mapping and

(d7'-1)(y) _ ((d7')(7'-1(y)))-1, d y c V.

Proof (S. Kumaresan): Since T is a Cl-map, by theorem G.3, T isdifferentiable on U. Let a E U be fixed arbitrarily. Without loss of generality,we may assume that (dT) (a) = Id, the identity transformation on ]I8n, forotherwise, we can consider the transformation S := ((dT)(a))-1T. Since inthe matrix representation of (dT)(x), the ijth entry is (DjTi)(x), it is easyto show that d t E Rn ,

1/2

(dT) (x) (t) < E E I DjTi (x)(i=1 j=1

Let

Then VtETRT,

11(dT)(x)ll EE(j=j j=1

Dj 7'i (x) I2

I (dT)(x)(t) I < IItI.

Since each (DjTi) (x) is continuous, it follows that x (dT) (x) 11 is also acontinuous map. Thus we can choose 6 > 0 such that for every x E B(a, 6)

I I (dT) (x) - (dT) (a) I I < 1/ 2. (G.8)

406 Appendix G

Next, let X1, x2 E B(a, 6). Then by the mean-value theorem for vector-valuedfunctions, we have some 0 < 6 < 1 such that

T(xi) - T(x2) - dT(a)(xi- x2) I< (dT)(xi + B(x2 - xi))(xi - X2)- (dT)(a)(xi - X2)

< (sup IB(x2 - xl) - (dT)(a)) J- x11. (G.9)

Since for every 0 < B < 1,

xl+ B(x2 - xl) - a = (1 - B)(xl - a) + B(x2 - a),

we have

Jxl + B(x2 - xl) - al < (1 - 9)Ixl - al + 9x2 - al < S.

Hence using (G.8) we have for every 0 < B < 1,

I (dT)(xi + B(x2 - xl)) - (dT)(a) I < 1/2. (G.10)

From (G.9) and (G.10) it follows that for every X1, x2 E B(a, S),

T(xi) - T(xa) - (dT(a))(xi - x2) I C Jxi - x2I/2. (G.11)

Since (dT) (a) = Id, from (G.11) it follows that T is one-one on B(a, 6). Thisproves W.

We show next that there exists Sl > 0 such that B(T(a), J1) C T(B(a, S)).For this, let Sl > 0 be arbitrary and let y E B(T(a), 81). We construct asequence recursively as follows. Take xO = a and for k > 1 let

Xk := Xk-l+ y - T(Xk-l)-

Clearly, xO E B(xo, S). Assume xO, xl,... , xn_1 E B(xo, 8). Since

xk - xk-1 = xk-1 - xk-2 - (T(xk_1) - 7'(xk-2)),

by (G.11) we have

Ixk - xk-11 I- xk-21/2 < Ixl - xol/2k-i < 51/2k-1

Thus if we choose Sl = J/2, then

k

jXk - XOJ < E(1/2)j-l Ixi - x0l < J.j=1

Thus is a Cauchy sequence in W1 and hence is convergent. Letx := lim Xk. Then

k-+oo

Ix - X01 lim IX, - x0n-- oo

n

k=1

< limn--oo

< limn-, o0

n

k=

k - xk-i

-k6 < 6.

Hence x c B(a, 6), and clearly

X = lim Xk = lim (xk_1 + y - 7'(Xk-1)) _ x + y -T(om).k->oo k-+oo

Thus T (x) = y. This proves that

B(T(a), 6/2) C T(B(a, S)).

Let V := B(a, 6) and y c T(B(a, 6)). Let x c B(a, 6) be such that T(x) = y,and let r > 0 be such that B(x, ,q) C B(a, 6). Then

B(y,,q/2) C T(B(x,,q)) C T(B(a,6)).

Hence y is an interior point of T(B(a, 6)). Thus T(V) is an open neighbor-hood of T(a). This proves (ii).

To prove (iii), let y E B(T(a), 6/2) and let x := T-1(y), x E B(a, S). Letk be such that y + k E B(T(a), S), and let x + h E B(a, S) be such thatT(x + h) = y + k. Then by the differentiability of T at x,

k = T(x + h) - y = T(x + h) -T(om) = (dT)(x)(h) + o(IhI), (G.12)

where o(lhl)/lhl ---> 0 as IhI ---> 0. Also, using (G.11) we haveI< IhI/2.Hence IhI < 31kl/2. Using this and (G.12), we get

T-1(y +k) -T-1(y) = h = ((dT)(x))'(k) +o(jkl)where o(jkj)/jkj ---> 0 as IkI ---> 0. Thus T-1 is differentiable at y and(dT-1)(y) = ((dT)(x))'. This proves (iii).

Further, since for every y

(dT(T'(y))) o ((dT')(y)) = Id,it follows that for every 1 < i, j < n,

m

(Dj Tj) (T- 1 (y)) (DkT, 1) (Y) = 6ij -(G.13)

j=1

Note that T-1 is continuous on T(B(a, 6)) and each DjTi is continuous onB(a,8). Equation (G.13), by Cramer's rule, expresses each (Dj (Ti 1)(y)) as

408 Appendix G

the quotient of continuous functions (the denominator being nonzero) andhence is a continuous function. Thus T-1 is also a Cl-map.

G.6. Theorem: Let U be an open subset of Il8' and T : U -> W C Il8' bea one-one Cl-mapping. Further, let JT(x) 0 V x E U. Then the followingstatements hold:

(i) W is an open set.(ii) T-1 : W -)U is a Cl-mapping.(iii) T-1 : W ) U is a homeomorphism.

Proof: Let x E W and T(x) = y, for y E U. Since U is open, we can findb > 0 such that B(a, J) C U. Then, as shown in theorem G.5,

B(y, J/2) C T(B(a, J)) C W.Hence W is also open. Since T is a C1-mapping and JT(x) 0 V x E U,by the inverse function theorem, T-1 is also a C'-mapping. In particular,T-1 is also continuous.

References

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Index

Rio, 112A®Z3, 210a.e., 125absolutely continuous

- complex measure, 366- function, 176- measure, 311- signed measure, 356, 359

aleph-nought, 391algebra, 55

- generated, 57almost everywhere, 125almost uniformly convergent, 250analytic set, 112antiderivative, 30approximate identity, 291Archemedes, 30arrangement, 387Arzela's theorem, 40, 157

BR, 95x312 , 211

x3X , 82Baire, Rene, 44Banach

- algebra with identity, 290- algebra, commutative, 290- lattice, 365- spaces, 266

Bernoulli, Daniel, 31Bessel's inequality, 298, 308binomial distribution, 69Borel, Emile, 44Borel

- measurable function, 227- subsets, 95, 102

- subsets of R2, 211bounded

- convergence theorem, 151- linear functional, 302, 375- variation, 397

C, 243C(R), 162C[a,b], 41, 160COO [a, b], 167COO-function, 286COO (1R), 164

COO (U), 286CEO (U), 286C1-mapping, 403Co (RTh ), 295

CS(R), 162Cc (RTh ), 280, 295CnE,56c, 112

XA(x), 30Cantor sets, 23, 24Cantor's ternary set, 25Caratheodory, Constantin, 44cardinal number, 391cardinality

- of a set, 391- of the continuum, 112, 391

Cartesian product, 389Cauchy, Augustein-Louis, 31Cauchy in measure, 259Cauchy-Schwartz inequality, 264, 296chain rule, 204change of variable for Riemann integration,

200, 207change of variable formula

413

414

- abstract, 332- linear, 334- nonlinear, 338

charactristic function, 30Chebyshev's inequality, 146, 270choice function, 390closed subspace, 298complete measure space, 92completion of a measure space, 92complex measure, 365complex numbers, 243conditional expectation, 305, 321conjugate, 374continuity from above or below, 66continuum hypothesis, 392convergence,

- almost everywhere, 248- almost uniform, 250- pointwise, 248- uniform, 248

convergence in- Lp, 267- measure, 255- pth mean, 267- probability, 261

convergent to- +00, 386- -CO) 386

convolution, 283coordinate functions, 401countable,

Index

Dini derivatives, 180Dini, Ulisse, 37direct substitution, 36Dirichlet function, 32Dirichlet, Peter Gustev Lejeune, 32discrete,

- measure, 68- probability measure, 68

distribution,- binomial, 68- discrete probability, 68- function, 64, 115- function, probability, 115- of the measurable function, 142- Poisson, 68- uniform, 68

dominated convergence theorem, an exten-sion, 274

Ex, 215Ey, 215Egoroff's theorem, 249equicontinuous, 271equipotent, 391equivalent, 158essential supremum, 291essentially bounded, 291Euclid, 30Euler's identity, 310Euler, Leonhard, 30example,

- set, 391 - Riemann's, 32- subadditivity, 4 - Vitali's, 113

countably - Volterra's, 37- additive, 59 extended,- subadditive, 59 - integration by parts, 200

counting measure, 225, 312critical values, 98cylinder set, 58cylindrical coordinates transformation, 344

(Dic)(x), (Dic)(x), 327(D+ f) (c) , (D f) (c), 180(D+ f) (c), (D- f) (c), 180dv (x) , 320dicD i(x), 324d'Alembert, Jean, 31Darboux, Gaston, 34Darboux's theorem, 18Denjoy integral, 172derivative of a measure, 324Descartes, Rene, 31det(T), 334determinant, 334differentiable, 324, 401differential, 401

- real numbers, 1, 386extension,

- of a measure, 60, 83, 88- of the dominated convergence theorem,

274

(f * g)) 283(f) g), 296f g, 43, 158f -1(C), 57f,,-') f, 255fn - f, 248fn -- f, 248f n a ) f, 248fn a. u f, 250

.F (C), 57Fatou's lemma, 140finite

- additivity property, 1- set, 391

Index

- signed measure, 346finitely additive, 59Fourier

- coefficients, 32, 307- series, 32, 307

Fourier, Joseph, 31Fubini's theorem, 189, 221, 222function,

- absolutely continuous, 176- C°° , 286- characteristic, 30- choice, 390- Dirichlet's, 32- generalized step, 48- imaginary part of, 244- indefinite integral of, 175- indicator, 30- infinitely differentiable, 163- integrable, 143, 244, 358- Lebesgue singular, 181- length, 1- Lipschitz, 177- locally integrable, 286- measurable, 135- negative part of , 49- nonnegative measurable, 118, 123- nonnegative simple measurable, 118- of bounded variation, 397- popcorn, 26- positive part of, 49, 135- real part of, 244- Riemann integrable, 10- Riemann integral of, 10- simple, 48- simple measurable, 135- step, 47- support of, 162- vanishing at infinity, 283- with compact support, 280

fundamental theorem of calculus, 34, 191,195, 197

gamma function, 173gauge integral, 172generalized Riemann integral, 172generated,

- algebra, 57- monotone class, 86

Hilbert space, 297Holder's inequality, 263

415

I, 1, 55Z, 551o) 96Id, 82Ir, 82Im (f ), 243f f dµ, 123, 244fsdµ,119fab f (x)dx, 10fE f dµ, 125, 145fbf(x)dx, 10

fab f (X)dx , 10improper Riemann integral, 168, 169indefinite integral, 175indicator function, 30infinitely differentiable functions, 164inner product, 296inner regular, 103integrable, 143, 244, 358integral, 123, 144, 244

- lower, 10- of nonnegative simple measurable func-

tion, 119- over E , 125- upper, 10

integration- by parts, 36, 199- by substitution, 206- of radial functions, 234

intervals,- with dyadic endpoints, 82- with rational endpoints, 82

inverse- function theorem, 405- substitution, 36

Jacobian, 405Jordan, Camille, 44Jordan decomposition theorem, 350Jordan's theorem, 398

kernel, 303Kurzweil-Henstock integral, 172

- o-algebra, 81 L, 135graph of the function, 219 L+) 123

4) 48Haar measure, 108 Lo+ 118Hahn L1 (X) S, a), 244

- decomposition, 350, 373 L 1 c (IIBn) , 286- theorem, 349 Li(X) S,µ), 144

Hankel, Hermann, 36 L 1 [a, b], 154Heine-Borel theorem, 108 L1-metric, 159

416

L1(E)7 154Li(X,S,L+), 358Li(X,S,tt-), 358Lp(X,S,/c), 261Lp(/2), 2611CF) 88,CR, 95lim infhic (D (x), 180lim infhTc (D (x), 180lim suphtc (D(x), 180lim suphTc (D(x), 180

Lebesgue- decomposition theorem, 318- dominated convergence, 246- integrable functions, 154- integral, 154- measurable sets, 95, 229, 239- measure, 95, 239- measure space, 95, 229- outer measure, 95- points, 194, 331- singular function, 181

Lebesgue, Henri, 44Lebesgue-Stieltjes measure, 115Lebesgue-Young theorem, 179left-open, right-closed intervals, 55Leibniz rule, 36Leibniz, Gottfried Wilhem, 30length function, 1

- countable additivity of, 4- countable subadditivity of, 4- finite additivity of, 1- monotonicity property of, 1- translation invariance of, 4

limit inferior and superior, 386Lipschitz function, 177locally integrable, 286lower,

- derivative, 327- integral, 10- left derivative, 180- left limit, 180- right derivative, 180- right limit, 180- sum, 7- variation, 351

Luzin theorem, 254

M(X)7 260M(X, S), 371.M (C), 86Mb (X, S), 362[t Iv,319/2 V v, 365/2 A v, 365/2T-1, 332/2-null set, 347

Index

A+, 351351

c* 71ILF, 62mean value property, 156measurable,

- cover, 90- function, 49, 135, 260- kernel, 90- nonnegative function, 123- partition, 129, 352- rectangle, 209, 212- set, 76- space, 92- transformation, 332

measure, 59- absolutely continuous, 311- complex, 365- counting, 225- discrete, 68- Haar, 108- induced by a transformation, 332- inner regular, 103- Lebesgue, 95- Lebesgue-Stieltjes, 89, 115- lower and upper variation of, 351- outer, 73- outer regular, 102- signed, 345- singular, 319- total variation of a, 351- total variation of, 368

measure space, 92- complete, 92- completeness of, 92- completion of, 92

metric, L1, 41Minkowski's inequality, 264monotone,

- class, 86- class generated, 86- convergence theorem, 127

monotonicity property, 1

v « /c, 311, 356, 366negative part of a

- function, 16, 49, 135- signed measure, 351

negative set, 347Neumann, John von, 315Newton, Issac, 30nonnegative simple measurable function, 118norm, 159, 266

- induced by the inner product, 297- of a bounded linear functional, 376- of a complex measure, 371- of a partition, 6

Index

- of a signed measure, 363normed linear space, 266null

- set, 22- subset, 92

01111

w(f, J), 20w(f,x), 20open intervals, 82, 96orthogonal, 297

- complement, 298oscillation of a function,

- at a point, 20- in an interval, 20

outer measure, 73- induced, 71- Lebesgue, 95

outer regular, 102outer regularity of L A, 102

yth norm of f, 262parallelogram identity, 297Parseval's identity, 309partial sum of the Fourier series, 307partition,

-6- measurable, 129- norm of, 6- refinement of, 7- regular, 14

Perron integral, 172pointwise, 248Poisson distribution, 69polar,

- coordinate, 341- coordinate transformation, 340- representation, 372

popcorn function, 26positive part,

- of a function, 16, 49, 135- of a signed measure, 351

positive set, 347power set, 55probability, 94

- distribution function, 115- measure, discrete, 68- space, 94

product,- measure, 238- measure space, 213, 238- of measures ti and v, 213- a-algebra, 210, 212

projection theorem, 301pseudo-metric, 43Pythagoras identity, 298

(', BF, ILF), 314R1 1Re (f), 243R-integrable, 171Z [a, b], 18Radon, Johann, 44Radon-Nikodym derivative, 320Radon-Nikodym theorem,

- for complex measures, 367- for finite measures, 354- for measures, 319- for signed measures, 357

random variable, 261real part, 244refinement of a partition, 7regular

- measure, 324- partition, 14, 15

regularity of AR2 , 229

regularization of a function, 287representation, standard, 118Riemann, Bernhard, 32Riemann

- integrable, 10- integral, 10- sum, 17

Riemann-Lebesgue lemma, 163Riemann's example, 32Riesz representation, 303Riesz representation theorem, 377Riesz theorem, 257Riesz, Friedrich, 44Riesz-Fischer theorem, 265, 308

S(P, f), 17S1, 298sl V s2, 120si A s2, 120S(C), 81S*1 76Saks' theorem, 97section of E at x or y, 215semi-algebra, 54set function, 59

- countably additive, 59- countably subadditive, 59- finite, 84- finitely additive, 59- induced, 62- monotone, 59- a-finite, 84

sigma algebra, 80a-algebra,

- generated, 81- monotone class technique, 88- monotone class theorem, 87- of Borel subsets of III, 95

417

418

- of Borel subsets of X, 82- product, 210- technique, 82

a-finite,- set function, 84- signed measure, 346

a-set, 57signed measure, 345simple

- function, 48, 279- function technique, 152- measurable function, 135- nonnegative measurable function, 118

singular- measure, 319- value decomposition, 395- values, 396

smaller, 365smoothing of a function, 281space,

- Banach, 266- Hilbert, 297- inner product, 297- Lebesgue measure, 95- measurable, 92- measure, 92- normed, 266- probability, 94- product measure, 213

spherical coordinates transformation, 344standard representation, 118Steinhaus theorem, 104step function, 47, 50, 251subspace, 298

- closed, 298sum,

- lower or upper, 7- Riemann, 17

supp (f)) 162support, 162symmetric moving average, 281

Theorem,- Arzela's, 40, 157- bounded convergence, 151- Darboux's, 18- Egoroff's, 249- Fubini, 189, 221, 222, 239- Fundamental theorem of calculus, 34- Hahn decomposition, 349- Heine-Borel, 108- inverse function, 405- Jordan, 398- Jordan decomposition, 350- Lebesgue - Young, 179- Lebesgue's dominated convergence, 148- Luzin's, 254

- monotone convergence, 127- Riesz representation, 377- Riesz-Fischer, 159, 308- Saks', 97- a-algebra monotone class, 87- Steinhaus, 104- Ulam's, 61- von Neumann, 315- Vitali covering, 108

topological,- group, 108- vector space, 261

total variation, 397- of a complex measure, 368- of a signed measure, 351

totally finite, 84transition,

- measure, 229- probability, 229

translation invariance, 4, 230triangle inequality, 266truncation sequence, 141

Ulam's theorem, 61ultrafilter, 70uncountable, 391uniform distribution, 69uniformly,

- absolutely continuous, 271- integrable, 275

upper,- derivative, 327- integral, 10- left derivative, 180- left limit, 180- right derivative, 180- right limit, 180- sum, 7- variation, 351

Urysohn's lemma, 280, 393

Va(f), 397Va (P, f ), 397V j (F), 19800variation, 397

- on]R, 198Vitali

- cover, 108- covering theorem, 108

Vitali's example, 113Volterra's example, 37Volterra, Vito, 37von Neumann theorem, 315

Weirstrass, Karl, 36

(X, S, TI), 92

Index

Index of notations

Prologue

R : Set of real numbers, 1

R* : Extended real numbers, 1

Z : The collection of all intervals, 1

[0, +oo] : The set {x c R* x > 0}, 1

0 : Empty set, 1

Chapter 1

11P11 Norm of a partition P, 6

L(P, f) Lower sum of f with respect to P, 6

U(P, f) Upper sum of f with respect to P, 7

P1 U P2 Common refinement of Pl and P2, 7

L1dx : Lower integral, 10

fb

f (x)dx : Riemann integral of f over [a, b], 10

ina.

b

P*f+

1-

f(x)dx Upper integral, 10

Regular partition of an interval 14

Positive part of a function, 16

Negative part of a function, 16

419

420 Index of notations

S(P, f) Riemann sum of f with respect to P, 17

R[a, b] Set of Riemann integrable functions on [a, b], 18

W (.f, J)w(f, x)XA

Chapter 2

Oscillation of f in the interval J, 20: Oscillation of f at x, 20: Characteristic function of the set A, 30

Lo : Collection of simple functions on III, 48

Chapter 3

ZThe algebra generated by intervals, 52

The collection of left-open, right-closed intervals, 55P(X) Power set of X; the collection of all subsets of X, 55C fl Ef -1(E)T(C)µFµ*

Subsets of E which are elements of C, 56The set {xjf(x) E E}, 57

: The algebra generated by C, 57Set function induced by the function F, 62Outer measure induced by µ, 71

S*S(C)

Collection of µ*-measurable sets, 76Sigma-algebra (cr-algebra) generated by C, 81

,t3X The Q-algebra of Borel subsets of a topological space X, 82

BR

ZrZd

A4 (C)GF

AF(X, S)

: The Q-algebra of Borel subsets of ][8 82: Open intervals with rational end points, 82Subintervals of [0, 1] with dyadic end points, 82

: Monotone class generated by C, 86The Q-algebra of µF-measurable sets, 88Lebesgue-Stieltjes measure induced by F, 89Measurable space, 92

(X, S, p) : Measure space, 92(X, s, A) : The completion of a measure space (X, S, p), 92

Chapter 4

A* : Lebesgue outer measure, 95£ : a-algebra of Lebesgue measurable subsets of III, 95A The Lebesgue measure on III, 95ZO : The collection of all open intervals in III,diameter(E) : The diameter of a subset E of III, 96

f'(x)A+xxEc

The derivative of f at x, 96The set {y + xjy E A}, 101The set {xyy E E}, 103Cardinality of the continuum, 110

96


ci

No

Chapter 5

The first uncountable ordinal, 111Aleph nought, the cardinality of the set N, 112

ILo : The collection of nonnegative simple measurablefunctions, 118

f sdtc : Integral of a function s E L with respect to ,cc, 119

Si V 82 : Maximum of the functions sl and s2, 120sl A 82 : Minimum of the functions sl and s2, 120L+ : The class of nonnegative measurable functions, 123

fdtc Integral of f E IL+, 123

P a.e. x(µ) on YP a.e. (µ)x E Y

JEf dµ

LIL0

L, (x, S, µ)Ll (X ), Li (µ)

P holds for almost every x E Y

with respect to µ, 125

Integral of f over E with respect to µ, 125

The class of measurable functions, 135The class of simple measurable functions, 135

} : The space of µ-integrable functions, 1441

Ll (E) The space of integrable functions on E, 154Ll [a, b] The space of integrable functions on [a, b], 154

IIiiC[a, b]

The Ll-norm of f, 158The space of continuous functions on [a, b], 160

supp(f)Cc(Il8)

COO(R)

C°°[a, b]

Cc` (R)

00

Support of a function, 162The space of continuous functions on R withcompact support, 162The space of infinitely differentiablefunctions on R, 164The space of infinitely differentiablefunctions on [a, b], 167The space of functions in C°°(R) withcompact support, 168

f(x)dxJ Improper Riemann integral of f over [a, oo), 168a


Chapter 6

Q6(f) Variation of f over the interval [a, b], 176lih cnf (P (x) L ower right limit of I at c, 180

lim sup (P (x) U pper right limit of (P at c, 180hJc

nf (P (x)lim Lower left limit of (P at c 180h c,

lim sup (P (x) U pper left limit of (P at c, 180hTc

L ower right derivative of f at c, 180(Df)(c) U pper right derivative of f at c, 180(D_f)(c) L ower left derivative of f at c, 180(Df)(c)V±Z (F)

Chapter 7

UV

pper left derivative of f at c,ariation of F on R, 198

180

A. ®B : Product of the a-algebra A with 8, 2118R2 : The a-algebra of Borel subsets of R2, 211(X x Y, A ® B, µ x v) : The product measure space, 213

Ex Section of E at x, 215Ey Section of E at y, 215,CR2 Lebesgue measurable subsets of R 2, 229

AR2 Lebesgue measure on J 2, 229Z2 The set {I x JJI, J E Z}, 229det T Determinant of T, 232

n n n

XZ, ,A I µ2 : Product of a finite number2=1 2=1 2=1

of measure spaces, 237

,CRn Lebesgue measurable subsets of Rn, 239ARn Lebesgue measure on Rn, 239BRn The o--algebra of Borel subsets of Rn, 239B(x, r) The open ball in Rn with center at x

and radius r, 240

Chapter 8

CRe(f)Im (f)Li (X s µ)Li (X s µ)A p ) fIn ) f

Ufn) f.fn) f.fn) f.M(X)

Lp(X S, µ)LP(IL)

fhf*gL110c(Rn)

C°°(U)C°° (U)B(0,1)Lc,O (X, S, µ)

lif llC*

Field of complex numbers, 243Real part of a complex-valued function f, 243Imaginary part of a complex-valued function f, 244Real-valued /c-integrable functions, 244Complex-valued /C-integrable functions, 244

fn converges to f pointwise, 248

fn converges to f almost everywhere, 248

fn converges to f uniformly, 248

fn converges to f almost uniformly, 250

fn converges to f in measure, 255The set of all measurable functions on X, 260

The space of pth-power integrable functions of f, 262

pth norm of f , 262The function fh(x) := f (x + h) V x, 281Convolution of f with g, 283Space of locally integrable functions on IlBn, 286The set of infinitely differentiable functions on U, 287The set f f E C°°(U) I supp(f) is compact}, 287

Closure of the ball B(0,1), 287Space of essentially bounded functions, 291Essential supremum of f, 292

Co(Il8n) The set of continuous functions on Il8nvanishing at infinity, 295

L2 (X, S, µ) Space of square integrable functions, 296

111112 L2-norm of f, 296(f, g)S1

Inner product of f, g E L2, 296

The set ff E L2 (X, S, A) l (f , g) = 0 d g E S}, 298

Chapter 9

v<Cµ v is absolutely continuous with respect to µ, 311

(IR, 13F, µF) Completion of the measure space (R, X3R, µF), 314µlv µ is singular with respect to v, 319

µ (x) Radon-Nikodym derivative of v with respect to µ, 320

(D/2)(x)(Dµ) (x)

Derivative of µ at x, 324Upper derivative of µ at x, 327

(D/2)(x) Lower derivative of µ at x, 327

µT-1 The measure induced by T, 332JT(X) Jacobian of T at x, 336

Chapter 10

µ+ Upper variation of a signed measure µ, 351µ- Lower variation of a signed measure µ, 351

W I Total variation of a signed measure µ, 351V « Absolute continuity of a signed measure v

with respect to a measure µ, 356

J fdµ Integral of f with respect to a signed measure µ, 358

V <C µ v absolutely continuous withrespect to a signed measure µ, 359

.A46 (X, S) Space of all finite signed measures on (X, S), 362

11till Norm of a signed measure µ, 363V < v is smaller than µ, 365µ V V Maximum of µ and v, 365µ A v Minimum of µ and v, 365V « µ Absolute continuity of a complex measure v with respect

to a measure µ, 366

1v1 Total variation of a complex measure v, 368Nl (X, S) Space of all complex measures on (X, S), 371

llmll Norm of a complex measure µ, 371

IITM Norm of a bounded linear functional, 376

Appendices

X ti Y :

card(A) :

X equipotent to Y, 391Cardinality, cardinal number of A, 391

N :

c

2 card(x)

Cardinality of N, 391Cardinality of the continuum, 391

Cardinality of the power set of X, 391At Transpose of a matrix, 395Q6(P, f) Variation of f over [a, b] with respect to

ab(f)a partition P, 397Variation of f over [a, b], 397

(dT)(a) Differential of a differentiable mapping T, 401(DT)(x)JT(X)

jth partial derivative of Ti at x, 401Jacobian of T at x, 405

Titles in This Series

45 Inder K. Rana, An introduction to measure and integration, second edition, 200244 Jim Agler and John E. McCarthy, Pick interpolation and Hilbert function spaces, 200243 N. V. Krylov, Introduction to the theory of random processes, 200242 Jin Hong and Seok-Jin Kang, Introduction to quantum groups and crystal bases, 200241 Georgi V. Smirnov, Introduction to the theory of differential inclusions, 200240 Robert E. Greene and Steven G. Krantz, Function theory of one complex variable,

2002

39 Larry C. Grove, Classical groups and geometric algebra, 200238 Elton P. Hsu, Stochastic analysis on manifolds, 200237 Hershel M. Farkas and Irwin Kra, Theta constants, Riemann surfaces and the modular

group, 2001

36 Martin Schechter, Principles of functional analysis, second edition, 200235 James F. Davis and Paul Kirk, Lecture notes in algebraic topology, 200134 Sigurdur Helgason, Differential geometry, Lie groups, and symmetric spaces, 200133 Dmitri Burago, Yuri Burago, and Sergei Ivanov, A course in metric geometry, 200132 Robert G. Bartle, A modern theory of integration, 200131 Ralf Korn and Elke Korn, Option pricing and portfolio optimization: Modern methods

of financial mathematics, 200130 J. C. McConnell and J. C. Robson, Noncommutative Noetherian rings, 200129 Javier Duoandikoetxea, Fourier analysis, 200128 Liviu I. Nicolaescu, Notes on Seiberg-Witten theory, 200027 Thierry Aubin, A course in differential geometry, 200126 Rolf Berndt, An introduction to symplectic geometry, 200125 Thomas Friedrich, Dirac operators in Riemannian geometry, 200024 Helmut Koch, Number theory: Algebraic numbers and functions, 200023 Alberto Candel and Lawrence Conlon, Foliations I, 200022 Gi nter R. Krause and Thomas H. Lenagan, Growth of algebras and Gelfand-Kirillov

dimension, 200021 John B. Conway, A course in operator theory, 200020 Robert E. Gompf and Andras I. Stipsicz, 4-manifolds and Kirby calculus, 199919 Lawrence C. Evans, Partial differential equations, 199818 Winfried Just and Martin Weese, Discovering modern set theory. II: Set-theoretic

tools for every mathematician, 199717 Henryk Iwaniec, Topics in classical automorphic forms, 199716 Richard V. Kadison and John R. Ringrose, Fundamentals of the theory of operator

algebras. Volume II: Advanced theory, 199715 Richard V. Kadison and John R. Ringrose, Fundamentals of the theory of operator

algebras. Volume I: Elementary theory, 199714 Elliott H. Lieb and Michael Loss, Analysis, 199713 Paul C. Shields, The ergodic theory of discrete sample paths, 199612 N. V. Krylov, Lectures on elliptic and parabolic equations in Holder spaces, 199611 Jacques Dixmier, Enveloping algebras, 1996 Printing10 Barry Simon, Representations of finite and compact groups, 19969 Dino Lorenzini, An invitation to arithmetic geometry, 19968 Winfried Just and Martin Weese, Discovering modern set theory. I: The basics, 19967 Gerald J. Janusz, Algebraic number fields, second edition, 19966 Jens Carsten Jantzen, Lectures on quantum groups, 1996

TITLES IN THIS SERIES

5 Rick Miranda, Algebraic curves and Riemann surfaces, 19954 Russell A. Gordon, The integrals of Lebesgue, Denjoy, Perron, and Henstock, 19943 William W. Adams and Philippe Loustaunau, An introduction to Grobner bases,

1994

2 Jack Graver, Brigitte Servatius, and Herman Servatius, Combinatorial rigidity,1993

1 Ethan Akin, The general topology of dynamical systems, 1993

From a review of the first edition:

The material is presented with generous details and helpful examples at a level suit-able for an introductory course or for self study.

-Zentralblatt MATH

Integration is one of the two cornerstones of analysis. Since the fundamentalwork of Lebesgue, integration has been presented in terms of measure theory.This introductory text starts with the historical development of the notion ofthe integral and a review of the Riemann integral. From here, the reader is nat-urally led to the consideration of the Lebesgue integral, where abstract integra-tion is developed via measure theory.The important basic topics are all covered:the Fundamental Theorem of Calculus, Fubini's Theorem, L,, spaces, the Radon-Nikodym Theorem, change of variables formulas, and so on.

The book is written in an informal style to make the subject matter easily acces-sible. Concepts are developed with the help of motivating examples, probingquestions, and many exercises. It would be suitable as a textbook for an intro-ductory course on the topic or for self-study.

AMS on the Webwww.ams.org

[inder k. rana] an introduction to measure and int(bookzz.org)

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