1

الورحلة الثالثة/كلية العلىم/ قسن الرياضيات

Real Analysis التحليل الرياضي

اعداد :أ.م.د. نهاد سالن الوظفر

2012-2012هدرسىا الواده

أ.م.د. نهاد سالن الوظفر

أ.د. وسن خالد

Real Analysis

Third Class

2

Chapter (1)

Real and rational numbers

The axiom of real numbers -:

Let be a triple consist of a non-empty set with the operation of

addition and multiplication. We say the triple is a field if it satisfies

the following properties:-

1) and (Commutative la)

2) and (Associative law)

3) (Distributive law)

4) There is distinct real number 0 and 1 s.t and

5) For each there's a real number such that and if

there is a real number

such that

Example-:

The real numbers from a field and the rational numbers (which are the real

number that can be written as

, where and integers and )

The order relation-:

The real numbers ordered by the relation , which has the following

properties:-

6) For each pair of real numbers and exactly one of the following is

true

7) If , and , then ( (transitive)

8) If , then for any and if , then .

3

A field with an order relation satisfying (6( ,)7( ,)8 ) is an order field. Thus

the real numbers form ordered field. The rational numbers also forms an

ordered field .

Supremum of a set :-

A set of real numbers is bounded above if there is a real number such

that for each . In this case, is an upper bound of . If is an

upper bound of , then so is any larger number, because of property (7)

If is an upper bound of , but no number less than , then is a

supremum of , and we write .

Example:-

If is the set of negative numbers, then any non-negative number

is an upper bound of , and .

If is the set of negative integers, then any number such that

is an upper bound of , and

The example shows that a supremum of a set may or may not be in the

set since contains it's supremum but dose not

Infimum of a set :-

A set of real numbers is bounded below if there is a real number such

that, for each . In this case is a lower bound of so is any

smaller number because of property (7). If is a lower bound of but no

number greater than , then is an infimum of , and we write

i f .

4

Remark :-

If is a non-empty set of real numbers, we write to indicate

that is unbounded above and i f to indicate that is unbound-

ed below.

Example:-

Let , { } then and . i f

Example:-

Let , { } then and i f .

If is the set of all integers, then and i f

H.W): Find and i f , state whether they are in .

1- { }

2- { }

3- { | | }

The relation between the field of rational of numbers and real number:

Proposition (1-1):-

Every orderd field contains a subfield similar to field of rational numbers.

Proof:- Let be an orderd field (1 is the identity element with

respect to operation) ( , is the identity of )

Claim (1)

5

Proof (1) Suppose the result is not true i.e there exists a positive integer

.and It's clear that and

) since (

The result is not true.

Trivial.

Claim (2)

Proof : if clearly .

If .

Then by (1) . Thus ( Contains a copy of Z).

( .is a group), such that , hence

( Contains a copy of )

( is a field),

such that (

) .

(

)

(binary operation).

( Contains a copy of Q).

Corollary (1-2):-

orderd field,

.

Q/ Is .

To answer this question, we beginning by this proposition:

6

Proposition (1-3):-

The equation has no solution in .

Proof: Suppose the result is not true i.e the equation has a root in

say

and the greatest common divisor

.

If and are odd, then .

If is odd and is even, , then

.

If is even and is odd, (, then

.

If and are even, (,then

..

So that there is no rational number satisfy this equation.

H.W:-

The equation has no solution in .

Proposition (1-4):-

The equation has only one real positive root.

Proof: Let { } . is bounded above (2,3,

upper bound of ). , since , ( and ,

Since is complete orderd field, then by (completeness property: Every non

empty subset of has an upper bound, then it has ), then has

a least upper bound say .

7

Claim:- , (i.e the least upper bound of is a root of equation ).

If not, then either or .

1. If , take ,

Choose satisfies:

Hence . Thus

2. If , take

Choose satisfies:

Hence , since

Uniquness:

Let such that and . Then either or

Thus .

8

Corollary (1-5):-

. (The field of rational numbers is proper subfield of the

field of real numbers ).

Proof: √ , from (1.4).

√ , from (1.3).

Corollary (1-6):-

is not complete orderd field.

Proof: Let { } .

√ .

Thus is not complete orderd field.

Remark (1-7):-

Let denote the set of irrational numbers, . is

complete orderd field (√ ) .

Now, we study the set and how we distribute the elements of and

the element of in . We start by the following theorem:

Theorem (1-8) : (Archimedean property):-

For each real numbers and there exists a positive integer

such that

Proof: Suppose the result is not true

9

Consider the set { } , then is bounded

above by . Since , then by the completeness of real numbers has a

least upper bound in say

Since , then , hence is not upper bound of , then

such that , ,

.

Thus the result is true.

Corollary (1.9):- , there exists a positive integer such that

.

Proof: Take . By (1.8) such that ,

hence

.

Theorem (1.10)-: (The density of rational numbers)

For each real numbers and with , there exists a rational number

between and

Proof:

(1 ) If

Define { }

Choose be the smallest positive integer satisfies

From (1 ) and (2)

10

is the rational

number between and .

If , then such that ,

hence by the previous result, such that

,

is the rational number between and .

(2)

is the rational number between and .

(3)

And by (1) there exists a rational number

is the rational number

Corollary (1-11) :-

For each real numbers and there exists an infinite countable set of

rational numbers between and

Proof: by (1.10 ( .

by (1.10 (

And

Generally between and and between and .

Thus we have infinite countable set between and

11

Theorem(1.12):- The density of irrational number

For each real numbers and with , there exists an irrational

number between and .

Proof: Suppose the result is not true i.e between and there is only

rational number by (1.10 ) (

√ √ √ √

√ √ √

√ , If , hence a contradiction

Corollary (1.13) :-

For any real numbers and there exists an infinite countable set of

irrational numbers between and .

Proof : by (1.12 ( .

by (1.12 (

And

Generally between and and between and .

we have infinite countable set { } between and

Example:- .

, by Arch., then

. The number is

12

Chapter (2)

The sequences of real numbers

Definition(2.1) :-

Let be a function, then , is called a

sequence of real numbers which will be denoted by ⟨ ⟩ or { }.

⟨ ⟩

Examples:-

1. ⟨

⟩

2. ⟨

⟩

3. ⟨ ⟩

4. ⟨ ⟩

5. ⟨

⟩

6. ⟨

⟩

Converging sequences:

-:Definition(2.2)

Let ⟨ ⟩ be a sequence of real numbers ,we say that ⟨ ⟩ is converging

sequence if there exists a real number satisfies for all

there exist a positive integer (depend on ) such that | | i.e

if , then i .

13

Otherwise the sequence is divergence.

Proposition (2-3):-

If the sequence ⟨ ⟩ is convergence sequence, then the limit point is

unique.

Proof: Suppose that and such that , then

| |. Since

, in particular take

(

) such that | |

.

Since

(

) such that | |

.

| | | | .

. | | | |.

{ } .

Examples:-

1) Is ⟨

⟩ converge to

⟨

⟩

Let , to find such that: |

|

Proof: |

|

, since .

14

By Archimedean such that

Thus |

|

.

2) Is ⟨ ⟩ ⟨ ⟩ converge to 3,

⟨ ⟩

| | .

3) Let ⟨ ⟩ be define by:

{

⟨ ⟩

This sequence convergence to .

| | .

4) Let ⟨ ⟩ ⟨ ⟩ be a divergence sequence.

⟨ ⟩

If , then for all , contain all odd terms but

doesn't contain any even term and since the even terms are infinite, then

.

15

If , then for all contain all even terms but

doesn't contain any odd term and since the odd terms are infinite, then

.

If or

| | | | .

If we choose { } , then any open interval

doesn't contain any term of the sequence and hence .

Thus ⟨ ⟩ is a divergence sequence.

H.W: Which of the following sequence convergence or divergence.

1. ⟨

⟩.

2. ⟨

⟩.

3. ⟨ ⟩.

Bounded sequences:

Definition(2.4) :-

A sequence ⟨ ⟩ of real numbers is said to be a bounded sequence ,if

there exists a real number such that | | .

. Examples-:

1. ⟨ ⟩ ⟨

⟩ is bounded sequence since

.

2. ⟨ ⟩ ⟨ ⟩is bounded sequence since .

3. ⟨ ⟩ {

⟨ ⟩

16

This sequence is bounded since .

4. ⟨ ⟩ ⟨ ⟩ is bounded sequence since

.

4. ⟨ ⟩ ⟨ ⟩ is not bounded sequence

since . (bounded below but not bounded above).

Proposition (2-5):-

Every convergence sequence is a bounded sequence.

Proof: Let ⟨ ⟩ be a convergence sequence, that convergence to

i.e

, such that | | .

| | | | | | .

Then | | | |

Hence | | | | .

| | | | | | | | | | | |

Take {| | | | | | | | | | | | }.

| | .

Example:-

⟨ ⟩ is not bounded sequence and by this

theorem is divergence.

Remark(2.6):-

17

The converse of proposition (2.5) is not true in general, as the following

example shows.

Example:-

⟨ ⟩ is bounded sequence which is a divergence sequence.

Monotonic sequences:

Definition(2.7) :-

Le ⟨ ⟩ be a sequence, we say that ⟨ ⟩ is a non- decreasing sequence ,if

.

⟨ ⟩ is an increasing sequence, if .

⟨ ⟩ is a non- increasing sequence, if .

And ⟨ ⟩ is a decreasing sequence ,if .

And we say that ⟨ ⟩ is a monotonic sequence ,if ⟨ ⟩ satisfies one of the

above conditions .

Examples-:

1) ⟨

⟩

is decreasing sequence.

2) ⟨

⟩

is an increasing sequence.

3) ⟨ ⟩ is a non- increasing sequence and a non-

decreasing sequence.

4) ⟨ ⟩ is not monotonic sequence.

Proposition (2-8):-

18

Every bounded monotonic sequence is convergence sequence.

Proof: Let ⟨ ⟩ be a sequence in , since ⟨ ⟩ is bounded sequence. ,

such that | | .

{ } bounded ( above and below ).

1) Suppose ⟨ ⟩ is a non- decreasing sequence,

Since is bounded above, then by completeness of real number has a least

upper bound say .

.

Claim:

then

is not an upper bound.

such that

| |

.

(2) Suppose ⟨ ⟩ is a non- increasing sequence,

i.e , such that | | .

Since is bounded below, where { }, then by

completeness of real number has greatest lower bound, say .

Claim: h h | |

.

19

i f … (1).

is not a lower bound (since ).

such that …(2)

Since ⟨ ⟩ is a non- increasing sequence, then …(3).

From (1), (2), (3) .

.

Then | |

.

Thus ⟨ ⟩ is converges.

Examples:-

1. ⟨ ⟩ ⟨

⟩

..

{ } ,

- .

This sequence is decreasing and bounded (below, above).

{ }.

2. Converges monotonic.

5. Let ⟨ ⟩ {⟨ ⟩

}

⟨ ⟩

It is converges but not monotonic sequence.

Cauchy sequences:

20

Definition(2-10) :-

A sequence ⟨ ⟩ is called a Cauchy sequence if there exist a

positive integer such that | | .

Proposition )2-11(

Every convergence sequence in or is a Cauchy sequence.

Proof: Let ⟨ ⟩ be a convergence sequence, that convergence to

i.e

such that | |

.

| | | |.

| | | |.

.

Thus | |

Remark)2-12( :

The converse of Proposition (2-11) is not true in general in the field of

rational number.

We need the following lemma:

Lemma (2-13):-

For any real number , there exists a sequence of rational number

converge to .

Proof: Let

21

By the density of rational numbers such that

, then

And by the density of rational numbers such that

.

Continue in this way we get a sequence of rational numbers ⟨ ⟩

… (*)

Claim : from (*) | |

.

(Arch.) such that

| |

(

Thus | |

i.e

Remark (2.12) :-

The converge of proposition (2.11) in general is not true in .

Proof: Let √

then by lemma (2.13) a sequence of rational numbers ⟨ ⟩ such

that √ , since √ , then by proposition (2.11) ⟨ ⟩ is a

Cauchy sequence, but ⟨ ⟩ is not converges in

H.W:

(1) ⟨

⟩ { }

22

(2) For any real number there exists a sequence of irrational numbers

converge to .

Theorem (2.14):- (The nested intervals theorem)

Let ⟨ ⟩ be a sequence of closed intervals such that . Then

.

Moreover if ⟨| |⟩ converges to zero, then consists of only one

point.

Proof: Let ⟨ ⟩ [ [ [

Let { } { }

.

. Then .

So that each element in is an upper bound of . Thus is

bounded above.

By completeness of real numbers has a least upper bound say .

and

. Hence . Thus .

-If ⟨| |⟩

Suppose, there exists another point , such that and

| |

23

Since ⟨| |⟩ then such that | |

| | | |

Thus

Remark (2.15):

In general theorem (2.14) is not true if the interval is not closed. As

the following example show:

Example: (

)

If { }

such that

?

i.e (

) thus

.

Completeness of real numbers

Every Cauchy sequence in is converging in .

Proposition (2.17):

Every Cauchy sequence is a bounded sequence.

Proof: let ⟨ ⟩ be a Cauchy sequence, i.e such

that | |

In particular take

| | | | | |

24

| | | |

Take { | | | | | | | | }

Thus | | .

Proposition (2.18):

Let ⟨ ⟩ and ⟨ ⟩ be two convergence sequences such that

and , then:

1. .

2. .

3.

4.

.

Proof: (4)

Since then

such that | |

| |

Since then

such that | |

| |

Since ⟨ ⟩ is converge, then s.t | | .

Since ⟨ ⟩ is converge, then s.t | | .

|

| |

|

| | | |

| || |

| || |

| || | .

25

| |

| |

| |

| |

{ }.

Countable sets is countable set.

Proposition (2.19):

is not countable set.

Proof: Let { } be a countable set

Let be a closed interval in such that | | and .

Let be a closed interval in such that | |

and and .

Let be a closed interval in such that | |

and and

| | and |

| , by nested

theorem { }

and . Then . Thus

Corollary (2.20):

The set of irrational number is uncountable set. (The union of two

countable set is countable)

Proof: If not, then , then countable C!

26

Thus is not countable.

Chapter (4)

The metric spaces:

Definition(4.1):

An order pair is called a metric space if is a non-empty set and

is a function

Satisfies:

1)

2)

3)

4)

is called the distance function, and the elements of are called the

element of the space.

27

Examples(4.2):

1) ; the set of real numbers and

is defined by | |

| |

| |

| | | | | | | |

| | | | | | | |

is a metric space.

2) If such that

{ } .

If .

Defined: by:

√

√∑

‖ ‖

√∑

√∑

∑

√∑

√∑

28

To prove (4) we need the following:

Lemma (4.3): The Cauchy - Schwarz inequality

For each real numbers we have:

| |

√

√

Lemma (4.4):

For each real numbers we have:

|

|

√

√

Proof:

By lemma (4.3)

√

√

√

√

√

29

√∑

Let

√∑

√∑

√∑

[By lemma (4.4)]

.

3) Let is a non-empty set define:

By:

{

}

1)

2)

3)

4)

4) If such that

{ }

Defined by:

30

| | | |

1) | | | |

2) | | | |

| | | |

3) | | | |

| | | |

4) | | | |

Let

| | | |

| | | | | | | |

H.W: If . Defined by:

| | | |

Is a metric space?

Remarks(4.5):

Let be a metric space, then

1) For any , we have .

| |

31

2) For any , we have

Proof(1):

From (1) we get:

From (2) we get:

| |

Proof(2):

By induction on the element of .

, then

Suppose the result is true for any

i.e

To prove is true for any

by (3)

32

Basic principles of topology:

Definition(4.6):

Let be a metric space, and , then:

{ }

Is called a ball of radius and center .

{ }

Is called a disk of radius and center .

Examples:

1) is a metric space.

{ | | }

{ }

( | )

{ | | }

[

2) is a metric space

, √

-; is a usual

distance

𝑥

𝑥 𝑟 𝑥 𝑟

𝑥

33

{

}

is a metric space; is a usual distance

,

√

-

3) is a metric space

Where defined by | | | |

{ | | | | }

{ | | | | }

We have the following cases:

Definition(4.7):

𝑟

34

Let be a metric space and, , is called an open set if for

each there exists , , such that:

Examples:

1) Every ball in any metric space is an open set.

{ }

Proof: let

Take , to proof

Let

, to prove

In particular every open interval in R is an open set,

are open sets.

| | ( ( | )

𝑎

𝑏 |𝑏 𝑎|

𝑥

𝑟 𝑑

35

[ is not an open set.

[

2) { }

is not open subset in .

Since the ball with center is not contain in .

{ }

is open subset in

Since the ball with center is contain in .

3) The set of rational (irrational) number is not open set.

Since any interval in with center

doe ’ o i io o y

(by density of irrational).

Also any interval in doe ’ contain irrational only because of the

density of rational number) not open.

Proposition(4.8):

Let be a metric space, and be a collection of all open subset of X,

then satisfies the following:

1) .

2) The union of any number of open sets is open. (i.e The union of any

element of is again in .

𝐻

𝑥

𝐻

36

3) The intersection of a finite number of element of is again in .

Proof:2) Let { } be any number of open sets in .

To prove (i.e is open in ).

Let .

is open

3) Let be a finite number of open sets in .

To prove is open in .

Let .

Take { }

is open

Remark(4.9):

The i e e io of i fi i e e of o e e eed ’ e o e A

the following example shows:

37

Example:

, let (

) { }

( ( ( | ) ) )

If

(

).

If

(

)

is only zero.

Note:

{ } is not open, since. { }

( | )

Remark:

If is a metric space, then we can define a topological space from

this metric space by taking the set of all open subsets of and by

proposition (4.8) we easily seen that is a topological space

But if is a topological space, then in general we couldn't get a

metric space from this topological space as the following example

shows:-

Example:

Let { } { }.

-1 1 0

𝜖 0 𝜖

38

is a topological space

But we cannot define a distance between the elements of .

Proposition(4,12):-

Let be a metric space and , then is open iff is a union of

balls

Proof:- let be an open set

Then such that

.

are balls

every ball is an open set is open (by proposition (4.8)).

Definition (4.13):-

Let be a metric space (topological space) and , then is

closed in if is open in .

Examples:-

1- [ [ is closed

Since [ is open. [ ]

The union of open set in a metric space is open.

𝑎 𝑏

𝑋 𝐷𝑟 𝑥

39

In general any disk is a closed set.

{ }

{ } is an open set

2- Every finite subset of a metric space is a closed set.

Proof:- let { }

T.P is open.

Let

Take i { }

is open

is closed.

3- { }

is closed not open subset in .

Since the ball with center is not contain in .

{ }

is open not closed subset in 𝐻

𝑥

𝑥

𝑥 𝑥5

𝑥

𝐻

𝑋 𝐻 o e

𝑐𝑙𝑜𝑠𝑒𝑑 𝑛𝑜𝑡 𝑜𝑝𝑒𝑛

𝑋 𝐻 𝑜𝑝𝑒𝑛 𝑛𝑜𝑡 𝑐𝑙𝑜𝑠𝑒𝑑

40

Since the ball with center is contain in .

4- is not closed is not open

is not closed

5- (Integers number) is closed

Balls open | ( ) | ( ) | ( ) | ( )|

is closed

6- are closed sets

.is open is closed, is open, is closed.

Proposition (4.14):-

Let be a metric space (topological space) and let be the

collection of all closed subsets of . Then satisfies the followings:

1) (i.e , and are closed )

2) The union of finite numbers of elements in is an element is (i.e,

the union of finite numbers of closed set is again a closed set)

o o e

41

3) The intersection of finite or infinite numbers of elements of is an

element in

(i.e, the intersection of finite or infinite numbers of closed set is

closed)

Proof: (H.w)

Remark:

Let then

.

Definition (4.15):

Let be a metric space and and , we say.

that is a cluster point for , if every open set contain contains

another element in and

i.e for any open set { }

Note:-

We will denote the set of all cluster points of by .

is called the closer of )

{ e oi of }

Example:-

1) , find

𝑝

42

( ( | ) ) (1)

( ) (2)

( | ) ( ) (3)

[ [

a) If , then any open interval , we have:- { }

.

b) If , then any open interval contain satisfies :-

c) For any [ , then ,

and | |

2) Let ,

-

a) If

, take (

)

( ,

-)

b) If

{ } , if (

)

( | | )

if

(

) and

c) If and

𝑎 𝑏

P=a

𝑏 𝑎

𝑎 𝑏

𝑝

𝑝

43

d) If { } , then any open set (interval) contain , (

)

and .

Since

(

)

{ } only zero

( | )

{ } ,

-

{ }

not closed.

3) Let be a metric space and be any finite subset of , then

.

Sol: let { } , let , if , then

.

Then

i { }

{ }

Now,

Let i { }

-1 P=0 1

44

is not a cluster point

.

4) Let be the set of rational numbers in with the usual distance.

a) If , then any open set (open interval) , we

have:- { } . (By the density rational number)

b) If , then any open set (open interval) such that

, we have . (By the density irrational number)

.

( | ) ( | )

(H.W)

Find .

Proposition (4.16):

Let be a metric space and , then is closed iff

contains all its cluster points (i.e )

Proof: suppose the result is not true i.e a cluster point for such

that .

is closed, then is open, hence ( is a

cluster point for ).

𝑅

𝑟 𝑄 𝑠 𝑄

45

let ,T.P is closed i.e is open.

Let is not a cluster point.

open set and .

In particular a ball is open

is closed.

Example:

,

- is not closed.

| | | | |

is not open, any ball

Definition (4.17):

Let be a metric space and and , define.

i f{ }

is called the distant between the point and the set

Remark (4.18):

If be a metric space and , then

i f { }

46

If , then i f { } i f{ }

The converse of remark (4.18) is not true in general as the following

example show:

Example:

Let .

( )

i f { }

i f { }

i f { }

Proposition (4.19):

Let be a metric space and , then iff

or is a cluster point of .

Proof: suppose that T.P is a cluster point for .

If is not a cluster point for .

a ball such that

(since )

is a cluster point for .

If by remark (4.18) .

If is a cluster point for , then for any open set

𝑝 𝑎

𝑎 𝑏

47

{ }

In particular a ball

i f { }

Corollary (4.20):

Let be a metric space and , then

{ } .

Proof: by proposition (4.19) ( iff or is a

cluster point for ).

Definition (4.21):

Let be a metric space and ⟨ ⟩ be a sequence in , we say that

⟨ ⟩ is a convergence sequence if there exists such that

satisfise:

i.e any ball with center and radius contain most of the terms of the

sequence.

48

Proposition (4.22):

If ⟨ ⟩ is a convergence sequence in that converges to , then is

unique.

Proof: Suppose there exists another limit point for ⟨ ⟩

i.e and .

take

and

such that

and , then each of balls

and

contain most of the term of the sequence but

a

contradiction.

Definition (4.23):

Let be a metric space and ⟨ ⟩ be a sequence in , we say that

⟨ ⟩ is a Cauchy sequence if such that:

Proposition (4.24):

Every convergence sequence in a metric space is a Cauchy sequence.

49

Proof: Let ⟨ ⟩ be a convergence sequence that converge to i.e

.

Let , then

such that

Remark (4.25):

The converse of proposition (4.24) in general is not true.

Proof: Let { } | | { }

⟨

⟩ in { }

{ }

⟨

⟩ is not a convergence sequence

By proposition (4.24) is a Cauchy sequence but not converges in

{ }.

Definition (4.26):

A metric space is called a complete metric space if every Cauchy

sequence in is a convergence sequence in .

50

Theorem (4.27):

is called a complete metric space .

Proof: let ⟨ ⟩ be a Cauchy sequence in .

such that

( ) √

And

| |

And | |

⟨ ⟩ is a Cauchy sequence in and ⟨ ⟩ is a Cauchy sequence in .

is complete

and

such that | |

such that | |

.

Claim: .

( ( ))

51

{ }

H.W: In

Definition (4.28):

Let be a metric space and , then .is a subspace

of a metric space , where |

Proposition (4.29):

Let be a metric space and if ⟨ ⟩ is a sequence in

such that ⟨ ⟩ converge to , iff either or is a cluster point for

.

Proof: if that T.P is a cluster point for .

, then any ball contain most of the terms of the

sequence, hence

is a cluster point for .

If , then ⟨ ⟩ .

If is a cluster point for , then for any ball

, we have:

( { })

52

Then (

{ })

⟨ ⟩ is a sequence in .

Claim: ⟨ ⟩ converge to .

.

Proposition (4.30):

Let be a complete metric space and if is a closed

set, then is a complete metric space

Proof: let ⟨ ⟩ be a Cauchy sequence in . T.P ⟨ ⟩ converge to .

⟨ ⟩ is a Cauchy sequence in .

is complete

⟨ ⟩ .

By proposition (4.29) either or is a cluster point for .

If , then we are done.

If is a cluster point for

Since is closed, then by proposition (4.16) .

53

Definition (4.31):

Let be a metric space and and let

{ }

is bounded below since

If is bounded above, then we say that is a bounded set and in this

case (R is complete) we write

( ) i

Examples:

1. .

{ }

( )

is a bounded set. ( )

2. [ .

{ }

( )

is a bounded set. [ ]

3. .

{ }

Is not bounded above, hence is not bounded set.

𝑎 𝑏

𝑎 𝑏

54

Proposition (4.32):

Let be a metric space and , is bounded if and only if

, there exists such that

Proof: let be a bounded.

Then { }

is bounded set (above) such that .

In particular

let , then such that .

(upper bound).

Cantor nested sets theorem(4.33):

Let be a metric space and ⟨ ⟩ be a sequence of bounded sets

such that:

1) .

2) is a non-empty closed sets.

3) The sequence ⟨di ⟩ converges to zero.

If is a complete metric space, then consist of only one

point.

Proof: , let since .

Since di , then such that di .

.

55

from(1)

di , then ⟨ ⟩ is a Cauchy

sequence.

Since is a complete metric space, hence ⟨ ⟩ is converge to

.

Claim: { }.

.

most of the term of the sequence in .

most of the term of the sequence in .

by proposition (4.29) either is a cluster point for

(intersection of closed sets is closed).

is closed, hence .(proposition4.16)

Uniqueness: Suppose .

di

such that di .

In particular when

di .

Contracting mapping principle theorem(4.34):

Let be a metric space and be a mapping satisfies

there exists such that:

( )

56

(in this case is called a contracting mapping), if is complete, then

there exists only one point such that ( is called a fixed point.

Proof: Since

let .

Let .

Claim: ⟨ ⟩ is a Cauchy sequence.

if

(

)

(

)

(

) (

)

.

{

}

{ (

) (

)

(

)}

{

}

57

{ }

By using mathematical induction

Such that

Claim: .

( )

Claim: uniqueness

Suppose

( )

58

Example:

Let [ [ be a mapping satisfies such that

| | | | [

[ is closed subset of a complete metric space , then by propostion

(4.30) [ is complete.

By theorem (4.34) has only one fixed point.

Remark(4.35):

If is a differentiable mapping satisfies | |

such that , then is a contracting mapping.

|

|

| | | | [

And hence has exactly only one fixed point.

Example:

Let [ [ defined by:

5

i [ such that

5

o

| o | | i |

59

| | |

5

o |

5 | |

| o |

5

5

the mapping

5

i has only one fixed point.

i.e the equation has only one root.

Compact space:

Defnition(4.36):

Let be a metric space and and let { } be a family of

open sets in , we say that { } is an open covering for if

.

Note:

Every set has at least one open covering , since .

Defnition(4.37):

is called a compact subset of , if for any open covering for , there

exists a finite open subcovering for .

60

i.e

if any open covering { } for , there exists

such that

.

In this case if , then is compact.

Defnition(4.38):

If { } is an open covering for , we say that { } , is an open sub

covering from { } , if { } .

Examples:

1) Every finite set in any metric space is compact.

Let { }

Let { } is an open covering for .

i.e

,

, then such that .

then such that

then such that

61

{

} is an open subcovering for { } .

2) Let ,

- is a compact subset of .

Let { } is an open covering for . i.e ,

, then such that .

{

} is an open subcovering for { } .

is a compact subset of .

3) is not compact subset of .

, let (

) (

)

Claim: { } has no finite subcovering for if there exists a finite

subcovering from { } for , then:

(

) (

) (

) (

)

62

(

) a contraction since .

( | | ) )

(

)

H.W:

[ are not compact.

Proposition (4.40):

Let be a compact metric space, if is a closed subset of , then

is compact

Proof: let , is a closed.

Let { } is an open covering for . i.e ,

Since is compact, then

.

(

)

(

) (

)

(

)

ϵ

𝑚

63

Examples:

1) Let ,

- [

[ is a compact subset of .

is closed

2) is not compact subset of .

[

[ is a compact subset of

Note:

If is not compact, then is not closed.

Theorem: (4.41):

Let be a metric space, if is a compact subset of , then is

closed.

Proof: Suppose that is not closed.

i.e a cluster point for such that .

is a cluster point for , then any open set such that , we have

In particular any ball of form

.

is closed set.

Let

is an open set .

64

Claim:

{ }

Suppose

a contraction since

.

.

/ is open

is an open covering for

Since is compact, then

Since , then , then .

Proposition: (4.42):

Let be a metric space and be a compact subset of , then is

bounded.

Proof: let .

{ }

.

( ) , { } is an open covering for

65

Since is compact, then

Then , is bounded.

Examples:

1) is not compact subset of since not bounded and not closed.

2) is not compact subset of since not bounded and not closed.

3) is not compact since not bounded.

Hein Borel theorem(4.43):

Any bounded closed subset of is compact.

Proof: let be a bounded closed subset of

Since is bounded, then there exists an open interval (ball) such that

, and hence , where .

Let { } is an open covering for . i.e , and suppose that

can't be covered by a finite subcovering from { } .

Divide into two equal closed intervals { } at least one of the sets

or can't be covered by a finite subcovering from { }

for otherwise, we get covered by a finite

subcovering from { } a contradiction with .

Let be the set which can't be covered by a finite subcovering from

{ } .

Divide into two equal closed intervals { } at least one of the sets

or can't be covered by a finite subcovering from { }

say .

66

continue in this way, hence we get a sequence of closed intervals ⟨ ⟩ a

satisfies:

1) .

2) is a non-empty closed sets.

3) The sequence ⟨| |⟩ ⟨

⟩

And can't be covered by a finite subcovering from { } by

the nested intervals theorem we get { }

Claim: a cluster point for .

Let be an open set such that

Since | | , then such that by Archimedean

, but is an infinite set, { } .

a cluster point for

Since is closed, then .

Since , then such that

.

Then such that , hence

a contradiction

since can't be covered by a finite subcovering from { } .

Corollary(4.44):

Let , then is compact iff is closed and bounded.

Proof: by proposition (4.42) every compact set is bounded and by

proposition (4.41) every compact set is closed.

by Heine-Borel theorem (every bounded and closed subset of is

compact).

67

Examples:

1) ,

- is compact.

2) ,

- is not compact.

3) is not compact.

4) is not compact.

5) [ is compact.

6) { } is compact (every finite set is closed and every finite

set is bounded).

Chapter (5)

Continuity:

Definition(5.1):

Let and be metric spaces and let be a function, is said to

be continuous at if such that for any , if

, then ( ) .

𝑓

𝑥 𝑓 𝑥 𝑋

𝑋 𝐵

𝐵

68

i.e is continuous at , if for any ball in with center and radius

( ), there exists a ball in with center and radius such

that .

Note:- if is continuous at each , we say that is continuous.

Proposition (5.2):-

Let be a function, then is continuous at iff for any open set in with is open in , where { }.

Proof:- let be an open in such that . To prove is open in

. Since , then aball ( ) but is continuous so a ball in

such that . Hence

let ( ) be a ball in with center and radius

. To show a ball in such that .

Since is open in , then by assumption is open in ,

clearly (since ), then a ball in such that

(by definition of open set), hence ( ) ( ). Thus

is continuous at .

Proposition (5.3):-

Let be a function, is continuous at iff for any closed set in

with is closed in .

Proof:- (H.W)

Hint: . is closed. To prove is closed, we have to

show is open.

69

Proposition (5.4):-

Let and be two metric spaces and let be a mapping, is

continuous at iff for each sequence ⟨ ⟩ converge to , the sequence

⟨ ⟩ converge to .

Proof:- :- Suppose is continuous at and let ⟨ ⟩ be a sequence in . To

prove ⟨ ⟩ converge to .

let be an open set such that , since is continuous at , then

is open in , clearly . Since , then contain most of the

terms of the sequence ⟨ ⟩.i.e contain most of the terms of the sequence ⟨ ⟩.

Thus .

Suppose the result is not true i.e such that

such that ,if

, then ( ) i.e a sequence ⟨ ⟩ in

such that .(by Archimedes such that

, then

).but a contradiction, thus the

result is true and is continuous at .

Examples(5-5):

5) Let is defined by . Is continuous?

Let . To prove is continuous at . Let ⟨ ⟩ be a sequence in such that

, we have to show . and

is continuous everywhere since .

6) Let is defined by f . Is f continuous?

70

Let . To prove is continuous at . Let ⟨ ⟩ be a sequence in such that

, we have to show . and ,

. Thus and is continuous at .

7) Let is defined by f

. Is continuous?

Let and let To prove such that if | | , we have

to show | | .

| | |

| |

|

| |

If we take , then | | , but | | | | | | , then | |

| | , which implies that | | | | .

So that | |

| |

, then choose i { } | | .

Now, it is easy to show that satisfies this relation. In fact if | | , then

| | |

| |

|

| |

| |

. Thus is

continuous.

Real value mapping:-

Let be a metric space, the mapping is called a real valued mapping.

Proposition (5.6):-

Let be a metric space and be a real valued mapping, if are

continuous at , then:-

1. is continuous at such that

2. is continuous at such that

3.

is continuous at such that

=

.

4. is continuous at such that .

5. | | is continuous at such that | | | |.

71

Example:

If [ and , then

Proof:- (3)

Let ⟨ ⟩ be a sequence in such that , we have to show

.

, since continuous at , then. and ,

hence

, then

. Thus

is continuous at .

Proposition :

Let , and be metric spaces and be a continuous

mapping at and be a continuous mapping at , then

is a continuous mapping at .

Proof:- (H.W)

Definition (5.7):-

Let be a real valued mapping, we say that is bounded if there exists

such that | | .

, { }.

Proposition (5.8):-

Let be a continuous mapping, if is compact, then is compact, hence

is bounded and closed.

72

Proof:

Let { } be an open covering for , is open in

, since is a continuous, then is open in , since

.

Claim: .

Let , then from (1), hence

, then , hence and

. Thus . So that { } is an open covering

for .

Since is compact, then such that

, then

{

} and

,. Thus is compact. Also

since is compact, then by Hein Boral theorem is bounded and closed.

Remark:-

Let be a real valued mapping defined by

, then.

1) is continuous mapping.

2) is not compact.

3) is not bounded.

In fact such that

.

Definition (5.9):-

Let be a mapping, if there exists such that ,

then is called a minimum point, if there exists such that

, then is called a maximum point.

Proposition (5.10):-

Let be a continuous mapping, if is compact, then there exists

such that .

73

has minimum and maximum point).

Proof:

By proposition (5.8) is compact and hence is closed and bounded, since is

bounded (below, above).

Below: such that , if { } ,

then such that , then . Thus is the

minimum point.

If , then is a cluster point for ,

(Since , hence is a cluster point for ). Thus

(since is closed). Then such that , then

. Thus is the minimum point.

Above: (H.W).

Uniform Continuity

Definition(5.11):

Let and be metric spaces and let be a mapping, we say that

is uniformly continuous if such that, if , then

( ) .

Clearly every uniformly continuous mapping is continuous, but the convers is

not true as the following example show:

𝑓

𝑋 𝑦 𝑓 𝑥 𝑓 𝑦 𝑋

𝐵

𝐵

74

Example:

Let is defined by . Is continuous?

Let . To prove is continuous at . Let ⟨ ⟩ be a sequence in such that

, we have to show .

. Thus and is continuous.

The proof of continuity by using definition:

let , if | | , then | | | | | |

, hence | | | | and | | | | | | | |

| | | | | | | | | | | | | | | |

| | . Take i ,

| | -

Thus | | and is continuous.

But is not uniformly continuous.

Take

| |

, by Archimedean

such that

, | |

.

| | |

| |(

) |

|

|

. Thus is not uniformly continuous.

Notice that is uniformly continuous on .

let , such that , if | | | |

| | | | | | | | | | | | | | | |

| |

Take

.

Example:

Let is defined by

. is continuous but not

uniformly continuous on . Take i { | | }

75

Let we must show that there exists such that

| | but | | . By Archimedean there exist a positive integer

such that

.

Let

.

| | |

| |

|

, such that

, hence

| |

| | |

| |

|

By Archimedean there exist a positive integer such that

.

Or let

, then | |

but | | |

|

.

Thus is not uniformly continuous.

Notice that is uniformly continuous on [ .

let , such that , if | | | |

| | |

| |

|

| |

| |

.

So that if , we can find . In this case if | | , then

| | |

|

| |

| |

. Thus is not uniformly

continuous.

H.W:

Let is defined by

. Prove that is continuous

but not uniformly continuous on .

Theorem (5.12):-

Let be a continuous mapping, if is compact, then is uniformly

continuous.

76

Proof:

let , such that , if | | ,

let , let

be an open interval, since is

continuous, then is open in and ( ) and .

Let

be a ball with center and radius

, hence {

}

is an

open covering for

, since is compact, then

such that

.

Choose ,

-.

Claim:- is satisfies the condition of uniformly continuous.

, if | | .

Since

, then such that i.e

.

, since is

continuous, then

| | | | | | |

|

.

Corollary (5.13):-

If [ is a continuous mapping, then is uniformly continuous.

Examples:-

1. Let [ be defined by .

Let ⟨ ⟩ be a sequence such that .

77

Since , then .

2. Let [ be defined by

[ .

Let ⟨ ⟩ be a sequence such that .

Since , then , hence is uniformly continuous.

3. Let [ be defined by i [ .

Let ⟨ ⟩ be a sequence such that . i

i

Since , then , hence is uniformly continuous.

Definition (5.14):-The intermediate value property

Let [ be a mapping, is said to be satisfies the intermediate value

property, if for all [ and for each between and , then there exists

between and such that .

Theorem (5.15):- The intermediate value - theorem

Let [ be a continuous mapping and between and , there

exists in [ such that

Proof:

Let [ , since between and , then either or

1) If , let

.

78

If , then we are done.

If not , then either , then or ,

then .

Let [ [ or [ [ , then either or

Let [ , let

.

If , then we are done.

If either , then or , then

.

Let [ [ or [ [ , then either or

.

Continuo in this way we get a sequence of closed intervals ⟨ ⟩ such that

.and | | sinc

, hence by nested intervals

theorem of closed intervals { }

Claim:-

and , Since | | , then , such that | | .

| | | | and | | | | , since is

continuous, then and and by (2)

. Thus .

If , then either or .

If .

If .

2) (H.W).

79

Examples:-

1) Let [ be a continuous mapping, if [ ,

then has at least one real root.

Proof: If

Thus By intermediate value theorem such that where

If

Then

Then such that

2) If odd and continuous.

by satisficing theorem (5.15)

If

such that

If

Hence such that

3) If is even, then may have no real root.

Example:-

Prewar Theorem (5.16):-

Let be a continuous mapping, then has at least one fixed point where

disk in .

Example: Let [ [ be a continuous mapping, then has at least one fixed point.

Sol:

Let is continuous mapping on [ .

80

.

.

By theorem (5.15) [ such that

has at least one fixed point.

Chapter (6)

Sequences and series of functions:

Definition (6.1):

Let . Define { i i g } , the sequence ⟨ ⟩ is

called a sequence of function where .

Definition (6.2):(Point wise convergence and uniform convergence)

Let ⟨ ⟩ be a sequence of function on , we say that ⟨ ⟩ converges to a function on

, if such that | | .

.

In this case, we say that ⟨ ⟩ converges point wise to a function for short, we write

→

i

And ⟨ ⟩ converges uniformly to a function on , if

such that | | , for short we write

81

Examples (6-3):

8) , let be defined by

. Is ⟨ ⟩ converges point wise

to ?

⟨ ⟩ ⟨

⟩ . ⟨ ⟩

.

1) i ⟨ ⟩ i

, then a sequence .

2) Let

| | |

| |

|

| |

, by Archimedean property | | ,

then | |

| | |

|

| |

| |

.

Thus

→

But

does not converge to uniformly.

Since if | |

, then | | .

Which is contradiction, since is not bounded.

To show the sequence ⟨

⟩ is converges uniformly to a function

.

| | |

| |

|

| |

, by Archimedean property on ,a

, then

|

|

. Thus

on .

9) , let [ be defined by . Is ⟨ ⟩

converges point wise to ?

⟨ ⟩ is decreasing sequence and bounded below by zero, so it is converges sequence. In

fact if , which implies that , therefore If

, then .

If , then ,then .

82

Thus → ,

But does not converge uniformly.

Is | | [ ?

Specially is | | [ ?

if

, then | | (

)

.

Thus ⟨ ⟩ does not converge uniformly on [ .

To show the sequence ⟨ ⟩ is converges uniformly to a function [

[ [ .

| | | | | | , by Archimedean property on ,

, then

| | [ .

Thus on [ .

10) , let be defined by

. Is ⟨ ⟩ converges

point wise to ?

⟨ ⟩ ⟨

⟩ .

Let

| | |

|

| |

| |, by Archimedean property on | |

| | , then

| |

| | |

|

| |

| | . Thus

→

But

does not converge to uniformly.

Since if if

, then | | |

(

) |

.

Thus ⟨ ⟩ does not converge uniformly on [ .

To show the sequence ⟨

⟩ is converges uniformly to a function

.

83

| | |

| |

| |

|

| |

, by

Archimedean property on , , then

|

|

.

Thus ⟨ ⟩ converges uniformly on .

Proposition (6.4):-

Let ⟨ ⟩ be a sequence of function such that ⟨ ⟩ convergence point wise to a function on

, and ⟨ ⟩ | | , then ⟨ ⟩ converge uniformly to ⟨ ⟩ .

| |

| |

| |

Example:

, let [ be defined by

. Show that whether

⟨ ⟩ convergence uniformly or not.?

| | |

| [ |

| and by proposition

(5.10) [ |

| [ |

|

.

.

Then is increasing function, hence ⟨ ⟩ ⟨

⟩ so that . Thus

. By ( 6.4).

The following propositions give some properties of uniformly converges.

84

Proposition (6.5):-

Let ⟨ ⟩ be a sequence of mapping on , if is bounded on and ⟨ ⟩

converges uniformly to on , then is bounded on .

Proof:

To proof | | .

Since i.e such that | |

.

Since is bounded , then such that | | , hence

| | .

| | | |

| | | | .

Thus | | .

Proposition (6.6):-

Let ⟨ ⟩ be a sequence of continuous mapping on such that ⟨ ⟩ converges uniformly to

on , then is continuous.

Proof:

To proof is continuous at a point , it's enough to show that for each sequence ⟨ ⟩

converge to on , the sequence ⟨ ⟩ converge to .

Let and ⟨ ⟩ be a sequence on such that T.P. .

Since is continuous and , then , such that

| |

.

Take , since , such that | |

.

85

| | | |

| | | | | |

,

take { }. Thus .

Remark:

The above Proposition is not true if → for example

,then .

Thus → ,

But is not continuous.

Theorem (6.7):-

Let ⟨ ⟩ be a sequence of continuous mapping on that converges to on , if either

or and is

compact, then ⟨ ⟩ converges uniformly to on .

Proof:

Case (1):- When , to proof .

Let we will prove that

.

is continuous on .

, hence .

Since → , then

→ .

Thus s.t | | .

is continuous | |

whenever | | .

( i.e ball in , a ball with center in such that and

.

So that { } is an open covering for , ( .

86

Since is compact, such that , take

{ } | | .

Thus

.

Case (2):- When (increasing) (H.W).

Definition (6.8):

Let ⟨ ⟩ be a sequence of mapping on , we say that ⟨ ⟩ is uniformly bounded sequence

if there exists a real number such that | | .

i.e | |

| |

Example:

, let [ be defined by

[ . Show that ⟨ ⟩

uniformly bounded

⟨ ⟩

| | [ .

Definition (6.9):

Let ⟨ ⟩ be a sequence of mapping on , we say that ⟨ ⟩ is a bounded converging to on

if:-

1) ⟨ ⟩ converges to on . →

2) ⟨ ⟩ uniformly bounded sequence.

Example:

87

, let [ be defined by [ . Show that

⟨ ⟩ uniformly bounded

1) → where ,

2) | | | | [ .

⟨ ⟩ is uniformly bounded sequence. Thus → .

Theorem (6.10):

Let ⟨ ⟩ be a sequence of mapping on that converges uniformly to on , if ,

is bounded, then ⟨ ⟩ is a bounded converges to on .

Proof:

To proof | | .

Since by proposition (6.5) is bounded on i.e | |

.

Also i e such that | |

.

| | | | | | | | .

Take {| | | | | | | | }.

Uniformly converges bounded converges point wise converges.

But the converse in general is not true.

Example:

, let [ be defined by [ . Is ⟨ ⟩

uniformly converges?

1) → where ,

and →

2) uniformly .

88

Thus ⟨ ⟩ is not uniformly converges sequence.

Example:

, let be defined by

. Is ⟨ ⟩

uniformly bounded

→ ?

|

| |

|

s.t

.

Then | |

.

⟨

⟩

is not uniformly bounded

If ⟨

⟩ i ifo y o ded , then | | |

|

.

By Archimedes on

If , then

, then

with .

If ?

Hence ⟨ ⟩ is not uniformly bounded

Proposition (6.11):

Let ⟨ ⟩ be a bounded convergence sequence to on , then ⟨ ⟩ is bounded.

Proof:

To proof | | .

Since →

1) →

2) ⟨ ⟩ is uniformly bounded.

89

i e such that | | . In

particular | | . Since ⟨ ⟩ is bounded, then

| | .

| | | | | | | |

.

| | .

Seris of mapping (6.12)

Let ⟨ ⟩ be a sequence of real valued mapping where on , the sum

∑ is called the series of mappings.

∑

∑ .

⟨ ⟩ is called the sequence of partial sums of ∑

If ⟨ ⟩ converges uniformly to a function on , then ∑ and the

convergence is uniformly on

If ⟨ ⟩ converges point wise to a function In this case, ∑ and the

convergence point wise on

Example:

90

∑ . Geometric series.

When , then ∑ .

, then

.

If | | , then . thus

If | | , then is diverges series since ⟨ ⟩ not bounded, hence diverges.

Thus ∑

only on .

Power seris (6.13)

The power series is of the form:-

∑

When , then

∑

When , then

∑

Thus ∑ converges when .

Example:

∑ .

Since ⟨ ⟩ is not bounded , then ⟨ ⟩ is diverges sequence,

hence ∑ is diverges series.

Thus the series ∑ is converges only when .

91

Theorem (6.14):

Let ∑

be a power series, if ∑

converges at , then

∑

converges at each such that | | | | .

Proof:

∑

∑ | |

∑ | |

|

|

Since ∑ | |

is a converges series, then by proposition (3.5) the sequence

⟨| |⟩ convergence to zero, hence ⟨|

|⟩ is bounded sequence i.e

| | .

∑ | |

∑ | |

|

|

∑ |

|

is a geometric

series, hence ∑ |

|

converges when |

| but | | | | . Hence |

| .

Remark (6.15):

Let ∑

be a power series

1) ∑

converges only at .

2) ∑

absolutely converges on .

3) There exists such that ∑

absolutely converges for each with

| | in this case is called the radius of convergence of the series and (– ) is called

an interval of convergence.

Theorem (6.16):

Let ∑

be a power series with .

1) If the sequence ⟨|

|⟩ converges to , then

when , and if ,

then . Thus (– ) is a convergence interval.

2) If the sequence ⟨|

|⟩ is not bounded, then ∑

converges only when

.

92

Examples (6.17):

1. ∑

,

⟨|

|⟩ ⟨|

|⟩ ⟨

⟩ ⟨

⟩

Hence , , then ∑

Converges . Thus (– ) is a

convergence interval.

2. ∑

,

⟨|

|⟩ ⟨|

|⟩ ⟨

⟩

Hence

, , then ∑

Converges . Thus (– ) is a convergence

interval.

3. ∑ ,

⟨|

|⟩ ⟨|

|⟩ ⟨ ⟩ not bounded.

Hence ∑ Converges only at .

(H.W): ∑ (

)

.

Chapter (7)

Riemann integration:

93

Definition (7.1):

Let [ e o ded i g , and { }

is called a Riemann partition, put [ , since is bounded, then has and

f. Let { } f { }

{ } f { } .

Clearly:- .

∑ | | is called Riemann upper sum, and ∑ | |

is called

Riemann lower sum.

Clearly:- , (since ).

Definition (7.2):

A partition on [ is called refirement for if every in is in .

Example:- [ [ . [ | ]

Proposition (7.3):-

If is a refirement for , then , and

Proof: (H.W)

Proposition (7.4):-

For any partitions on [ we have .

Proof:

Let , clearly is a refirement for and . Thus

.

94

Let { i io o } bound below.

{ i io o } bound above.

By proposition (7.4), we have each element in is a lower bound for and each

element in is an upper bound for .

And by completeness of , has a greatest lower bound and has a least upper

bound.

Now, let i f( ) which is called Riemann upper integral and

( ) which is called Riemann lower integral.

Clearly that:- .

If , then is called Riemann integrable. O.W we say that is not Riemann

integrable.

Examples (7.5):-

1) Let [ be defined by . is Riemann integrable?

Let { } be a partition on [ .

{ } f { }

∑ | | | | | | | |

| | | | | | | |

∑ | | | | | | | |

| | | | | | | |

{ i io o } .

{ i io o }

i f( )

( )

. Thus is Riemann integrable.

95

2) - Let [ be defined by [ . is Riemann integrable?

Let ,

- be a partition on [ .

{ } f { }

∑ | | | | | | | | | |

*

+ *

+ *

+ *

+

(

)

∑ | | | | | | | | | |

*

+ *

+ *

+ *

+

[

(

)

,

- .

,

-

i f( )

( )

. Thus is Riemann integrable.

96

3) [ such that

. is Riemann integrable? (H.W).

4) Let [ be defined by { [

[ . is Riemann

integrable?

Let { } be a partition on [ .

{ } f { }

∑ | | | | | | | |

| | | | | | | |

∑ | | | | | | | |

| | | | | | | |

{ i io o } .

{ i io o }

i f( )

( )

. Thus is not Riemann integrable

Q1/ Is there exists a discontinuous mapping in finite infinite of point and Riemann integrable?

Q2/ Is every continuous mapping and Riemann integrable?

Q3/ Is there exists a relation between points of continuity and Riemann integrable?

There exist discontinuous mappings in a point and Riemann integrable.

5) Let [ be defined by ,

. Is Riemann integrable? is

not continuous at .

Let *

+ *

+ *

+ be a partition on [ .

*

+

*

+

*

+

97

{ } , *

+- { } .

{ } , *

+- { } .

{ } , *

+- { } .

∑ | |

| | |

| | |

{ } , *

+- { } .

{ } , *

+- { }

{ } , *

+- { }

∑ | |

| | |

| | |

{ i io o } ,

- .

{ i io o } ,

-

i f( )

( )

. Thus is Riemann integrable.

98

6) Let [ be defined by {

. is Riemann integrable? (H.W)

*

+ *

+ *

+ *

+ *

+

To answer question tow

Lemma (7.6):

Let [ be a bounded function is Riemann integrable iff for each , there

exists a partition on [ such that .

Proof:

Let , since is Riemann integrable, then

( ) { i io o }.

i.e there exists a partition o such that

.

( ) { i io o }.

i.e there exists a partition o such that

.

Let clearly is a refirement to each and .

By proposition (7.3)

(since is integrable)

Let and there exists a partition on such that .

99

. Thus is Riemann integrable

Proposition (7.7):

Let [ be a bounded function, if is continuous, then is Riemann

integrable.

Proof:

By previous proposition (5.10) has a minimum and a maximum points and also by

proposition (5.12) is uniformly continuous.

Divided into equal closed intervals each of length

, [

is uniformly continuous on .

Let if | | , then | |

.

∑ | |

∑ | |

∑ | |

{ } { }

∑| || |

∑

| |

Thus is Riemann integrable.by (7.6)

Monotonic function and Riemann integrable:

Definition (7.8):

Let [ be a function is called a non-decreasing (increasing) if

[ if , then and is said to be a non-increasing

(decreasing) if [ if , then .

Examples:

100

1. i , is continuous function but not monotonic.

2. | | [ is monotonic function but not continuous.

Remarks (7.9):

1) Let [ be a monotonic function, then is bounded.

If is non-decreasing, then [ .

If is non-increasing, then [ .

2) Let [ be a monotonic function and is non-decreasing, then is non

increasing and if is non-increasing, then is non-decreasing.

Theorem (7.10):

Let [ be a monotonic function, then is Riemann integrable.

Proof: (H.W)

Definition (7.11):

Let is called a negligible set (zero set) if for each , there exists a countable

collection of open intervals { } such that.

1.

2. ∑ | | .

Remarks examples (7.12):

1) Every finite set is a negligible set.

Let { } .

Let , (

)

1.

2. ∑ | | ∑

.

2) In general every countable (finite or infinite) set is a negligible set.

Let { } .

101

Let , (

)

1)

2) ∑ | | ∑

∑

∑

.

In particular (the set of rational numbers) is a zero set.

3) If is a negligible set and , then is a negligible set.

Proof:

Since is a negligible set, then { } of open intervals such that

1.

2. ∑ | | .

Since , and ∑ | | | | , hence we are done.

4) The union of a countable number of negligible sets is again negligible set.

Proof:

Let { } .be a countable collection number of a negligible set. T.P. is a negligible set.

,

- a countable collection of open intervals such that 1)

,

2) ∑ | |

.

1)

2) ∑ |

| ∑ ∑ |

| ∑

.

5) Every interval in is not a negligible set.

Proof:

Every open covering for | | , any intervals is of length equal or greater than | | and

hence when

| | .

The condition (2) is not hold.

6) is not a negligible set.

.

102

is not a negligible set by (5) and is a negligible set by (2)

If is a zero set, then . is a zero set C!

Theorem (7.13): (Lebesgue theorem in Riemann integration)

Let [ be a bounded function, then is Riemann integrable if and only if the

set of discontinuous points [ of on [ is a negligible set.

Example: Every empty set is a zero set.

Let (

)

1)

2) | |

.

Corollary (7.14):

Let [ be a monotonic function, then the set of discontinuous points of on

[ , is a zero set.

Proof:

By remark (7.9) is bounded and by theorem (7.10) is Riemann integrable, then by

(7.13) is a zero set.

Corollary (7.15):

Let [ be a bounded Riemann integrable function and let [ be a

bounded function, if [ [ , then is Riemann integrable.

Proof:

Since [ is bounded and Riemann integrable, then by (7.13)the set of on

[ is a zero set and every subset of a zero set is also is a zero set, hence [ is a zero set

and again by (7.13), then is Riemann integrable

103

Proposition (7.16):

Let [ be a bounded function and let [ , if is Riemann integrable on

[ , then is Riemann integrable on [ and [ . Moreover.

∫

∫

∫

Proof:

Let and be partitions on [ and [ respectively.

.

.

Notice that is Riemann integrable on [ and [ .

By corollary (7.15)

∫

∫

∫

∫

And

∫

From and we get

(

)

Thus

Remark (7.17):

Let [ { [ o ded ie i eg e f io } , then

104

[ is a vector space.

Let [ [ [ [ ,.

then [

, is a zero set.

, let [ , is a zero set. .

is a zero set, then is a zero set.

Thus is a Riemann integrable, then [ .

Now, define [

(Number)

1)

2) [

Proposition (7.18):

Let [ be a bounded function, if is Riemann integrable and

[ , then

.

Proof:

Let be a partition on [ .

{ }.

[ , { } { }

∑ | | , since

∑ | | | | | | | |

{ i io o [ }

( )

Since is Riemann integrable

Corollary (7.19):

105

Let [ be bounded functions, if and are Riemann integrable and

[ , then

.

Proof:

Let [

Since is bounded, then | | [ .

Since is bounded, then | | [ .

| | | | | | | | | | | | [

Then is bounded and is Riemann integrable [ [ and by proposition

(7.18)

, then

. Thus

.

Corollary (7.20):

If [ , then | | [ and |

| | |

.

Proof:

1. Since | | , { }

Since [ , then by Lebesgue theorem [ is a negligible set. Hence [ | | is

a negligible set and then by Lebesgue theorem | | [ .

2. Since | | | | [ , then by corollary (7.19) | |

| |

, thus |

| | |

.

Remark :

The convers in general is not true i.e if | | [ , then needn't be [ .

Example:

{ [

[ .

| | [ is Riemann integrable but is not Riemann integrable

Remark :

Clearly that

, but if

is ?

106

In general No

Examples:

1) Let [ . be defined by [

but

2) Let [ . be defined by ,

.

but

Proposition (7.21):

Let [ and is continuous function, [ and

,

then .

Proof:

Suppose that the result is not true (i.e) [ .

Let (

) be a ball in , since is continuous on [ , then a ball

in [ such that

Let closed interval, since , then .

is closed and bounded, then by Hien-Borel theorem is compact, hence is

continuous on a compact space, hence has minimum and maximum points.

i { } from (*) since

∫ ∫ | |

| | | | | |

| | | | | |

| |

Definition (7.22):

𝐸

107

Let ⟨ ⟩ be a sequence of real valued functions on [ , we say that ⟨ ⟩ converges (point

wise) to on [ if [ such that | |

.

And we say that ⟨ ⟩ converges uniformly to on [ if such that

| | [ .

: If ⟨ ⟩ is a sequence of real valued bounded function on [ that converges point wise

to on [ and the sequence ⟨ ⟩ is Riemann integrable on [ . Is Riemann

integrable?

Answer: No in general as the following example show:

Example:

Let [ , let { } be the set of rational numbers in [

[ be defined by { { }

{ } .

{ { }

{ }

{ { }

{ }

{ { }

{ }

[ { } is a negligible set . Hence [ by Lebesgue.

Claim: → , where {

[

[

,

108

,

,

If { } , then [

Thus → and [ .

H.W:

2

converges point wise to .

: If ⟨ ⟩ is a sequence of real valued bounded functions on [ that converges point wise

to on [ , if the sequence ⟨ ⟩ is Riemann integrable on [ and Riemann

integrable. Is i i ?

Answer: No. in general as the following example show:

Example:

[ be defined by

{

.

109

{

{

{

. Hence [

→

, hence in general i i

Note: If the converges is uniformly the answer for two questions are yes.as in the following

theorem:

Theorem (7.23):

Let ⟨ ⟩ be a sequence of bounded functions on [ that converges uniformly to on [

and if [ , then [ .

110

Moreover ⟨ ⟩ converges to i.e i i .

Proof:

Since and ⟨ ⟩ is bounded , then is bounded by proposition (6.5).

Let the set of discontinuous points of on [ .

Since [ , then is a negligible set.

Let is a negligible set, then is continuous on [ .

Since and ⟨ ⟩ is continuous on [ , then by proposition (6.6) is continuous

on [ , then where the set of discontinuous points of , then

is a negligible set and hence [ .

|∫ ∫ | |∫ | ∫| |

Then such that | |

.

Chapter (8)

Differentiation:

Definition (8.1):

Let be a function, we say that is differentiable at if for any

sequence ⟨ ⟩ in such that and , there exists a real number

such that the sequence

i.e:- ⟨ ⟩ in such that , such that

111

is called the derivative of at , also is denoted by

|

i

Otherwise is not differentiable at .

Remark (8.2):

If is differentiable at each , then we say that is differentiable.

Theorem (8.3):-

Let be a function then is differentiable at iff there exists a

real number and a continuous function with satisfies

[

Proof:

Since is differentiable at , then ⟨ ⟩ in such that

, such that

Define as follows

{

Claim: is continuous

Let T.P .

112

by

Hence is continuous.

.

Thus

[

T.P is differentiable at we have continuous a

[

Let ⟨ ⟩ be a sequence in such that and ,

Since and is continuous at , then

,

from

Thus

. And is differentiable at

Proposition (8.4):-

Let be a function if is differentiable at , then is a continuous

at .

Proof:

113

Since is differentiable at , then a real number and a continuous function

with satisfies

Since and are continuous functions.

Then is a continuous at .

Remark :-

The convers of the above proposition in general is not true as the following example shows.

Examples:-

7) Let be defined by | | { [

.

is a continuous at .

Let , | |

But is not differentiable at i.e ⟨ ⟩ in such that ,

such that

(

)

(

)

Thus is not unique and is not differentiable at

Now, we have some examples about differentiation:

8) Let be defined by .

is differentiable

Let and let ⟨ ⟩ ,

114

9) - Let be defined by .

is differentiable

Let and let ⟨ ⟩ ,

10) Let be defined by . is differentiable?

Let and let ⟨ ⟩ ,

Then is differentiable

11) Let be defined by {

.

is not differentiable (H.W).

Proposition (8.5):

Let be differentiable functions at , then:

1. is differentiable at and .

2. is differentiable at and .

3. is differentiable at and .

4.

is differentiable at and (

)

.

Proof:(4)

Let ⟨ ⟩ ,

115

( ) (

)

0

1 0

1

Since , are differentiable function at , then

and such that.

And continuous

0

1 0

1

( )

Proposition (8.6): (Chain Rule)

Let be a differentiable function at and be a differentiable function

at , then is differentiable at and ( ) ; are

open intervals.

Proof:

Since is differentiable at and a continuous function

with satisfies:

116

Since is differentiable at , ( ) and a

continuous function with satisfies:

( ) ( ) ( )[

( ) ( ) [ [

( ) ( )

[ [ ( )]

( ) .

Examples:-

1) Let be a function defined by ,.

. (H.W)

2)

( )

Definition (8.7):

Let be a function, let , we say that is increasing at , if there

exists an open interval such that if , then , and

if , then .

And is decreasing at , if there exists an open interval such that if

, then , and if , then .

117

If is increasing at , then is increasing function and if is decreasing at

, then is decreasing.

Theorem (8.8):

Let be a differentiable function at , if , then is increasing at

and if , then is decreasing at . Hence if , then there exists an open

interval such that is 1-1 and on .

Proof:

The inverse function theorem (8.9):

Let be a differentiable function at , if , then there exists an open

interval and an inverse function of where and is differentiable at .

Moreover

; is open intervals.

Proof:

Since , then by theorem (8.8) there exists an open interval , and

is 1-1.

is 1-1 and onto since if . Hence has inverse say

.

, .

, , chain rule ( ) at , then

( ) , then

, where [given].

Since is differentiable at , then continuous and satisfies:

[

( ) ( ) ( ) [

( )]

( ) [

( )]

( ) *

( )+

118

( ) *

( )+

( ) *

( )+

( )

Thus ( ) is continuous on . And

( )

i e

Definition (8.10):

Let be a function , we say that is a local maximum point, if there

exists an open interval , such that , , and we say that is a

local minimum point , if there exists an open interval and .

Proposition (8.11):

Let be a differentiable function, if is either a local minimum point or a local

maximum point, then

Proof:

If , then either , then is decreasing at , or , then is

increasing at in each case is not local minimum and not local maximum a contradiction,

hence

Remark (8.12):-

In general the convers of the above proposition is not true as the following example shows:-

Example:-

Let be defined by .

119

Since is increasing at , then is not local minimum and not local maximum

Roll's theorem (8.13):

Let [ be a differentiable function on and continuous on [ , if

, then there exists such that .

Proof:

If is constant, then

If is not constant

Since is continuous on [ (compact set), then has maximum and minimum values say

.

i.e [ such that [

Clearly since is not constant.

maximum and minimum points, then is local minimum, then by (8.11),

put

Clearly and , since if , then

Or , then

Then is constant C!

Say , then , put .

And , then , put .

Mean Value Theorem (8.14):

Let [ be a differentiable function on and continuous on [ , then

there exists such that

.

Proof:

Define [ by:-

is differentiable function on and continuous on [

120

So that , then by Roll's theorem such that .

Then

. Thus

Chapter (9)

Measure Theory:

The length of open bounded intervals:

Step 1. If { } is an open bounded interval, then the length of

is denoted by ( ) and defined by:

{

Where is the empty set

1. ( )

{ }

2. If are two open intervals with , then .

3. If are two open intervals, then and

, if , then .

In general if are open intervals, then ∑

If are disjoint, then ∑

.

4. If { } is a countable number of open intervals, then ∑ and

if { } are disjoint , then ∑ .

121

5. . Where { }.

{| | } {| | }

The length of open bounded sets:

Step 2. If is any bounded open subset of

Lemma (9.1):-

Every open bounded subset of can be written as a union of a countable number of

disjoint unique open intervals and this representation is unique.

{ } are open intervals , . Hence by lemma

(9.1) if is an open subset (interval) of , then ,

Let ∑ (disjoin).

Is ∑ exits?

Let

∑

Let ⟨ ⟩ be the sequence of partial sum of ∑ .

Then . Thus ⟨ ⟩ is an increasing sequence

Since , is bounded, then is bounded, hence an open interval (ball)

, then .

{| | } ∑

.

∑ , then is bounded, hence ⟨ ⟩ is bounded and monotonic

sequence.

122

Thus ∑ converges ∑ and ∑ exists.

Note :-

1 ∑ ,

2 If are open bounded subset of with , then .

By (9.1) [ , then ∑ ∑ . Thus ].

3 If are open bounded subset of , then .

∑ ( ) ∑ ∑

In general if are open bounded intervals, then ∑

If are disjoint, then ∑

.

4 If { } is a countable collection of open bounded subset of , then

∑ and if { } are disjoint , then

∑ .

5 If is open bounded set, then

, where { }.

∑ ∑

Step 3. If is any bounded subset of , let

{ }

is bounded, then a ball (open interval) open and bounded such that .

{ } bounded below, since is complete,

Let i f { o e d o ded}

is called the outer measure of . (for short we write by ).

Examples:-

12) If

i f { o e d o ded}.

123

i f , (

) -

i f { }

13) - If { }

i f { { } o e d o ded}.

i f , { } (

) - , is o e d o ded.

i f { }

14) If { } (

)

i f { }

i f , ∑ ∑

- .

15) If { } is an infinite countable subset of , (

) ,

i f { ∑ }

i f { ∑ ∑

∑

(

)

} i f 2

3 i f ,

- .

16) If [

i f ,

[ (

) o e d o ded-

124

17) If [ (H.W)

18) If is a bounded zero set , then and conversely if , then is a

zero set

Proof:

Let be a bounded zero set

i.e { } of open interval such that:

1. .

2. ∑ | | .

T.P i f { o e d o ded}

Take .

i f { ∑ }

Conversely, let . T.P is a zero set.

Let , then open and bounded subset of such that .

i f { o e d o ded}

Since is open, then by lemma is a union of open balls (intervals in ).

Thus { } of open intervals in .

1. .

2. Since , then ∑ | | .

Bounded measurable sets:

Definition (9.2):

Let be a bounded subset of , we say that is a measurable set, if an

open bounded subset of such that and .

Note: If is a measurable set, we put

Examples:

1) If . is measurable set?

125

Let ,

take (

) (

)

(

)

((

)) (

)

.

Hence is measurable.

2) If { } . is measurable set?

Let (

) .

{ }

((

) (

))

. ((

) (

))/ .

∑. (

) (

)/

∑

∑

3) If [ bounded. is measurable set?

[ i f , [ (

) o e d o ded-

i f , ( (

)) - i f { } .

4) If [ . is measurable set? (H.W)

Proposition (9.3):

Let be a bounded measurable set, then

126

1 .

2 If are measurable sets with , then .

3 If are measurable sets, then , then

,

If are measurable disjoint sets, then .

4 If { } is a collection of measurable bounded sets, if is bounded, then is

measurable set and ∑ .

If { } disjoint, then ∑

5 If is bounded measurable set and , then .

Proposition (9.4):

Let be a bounded subset of . is a measurable set iff an open

bounded subset of , such that .

Examples:

1) Every open bounded set is a measurable set.

Let , take .

2) Every bounded interval is a measurable set.

[ , let , take (

)

[ ( (

) [ ) ( (

) (

))

((

)) ((

))

.

By Arch median

3) Every bounded countable (finite or infinite) subset of R is a measurable set

Let { } , let (

) ,

( { } ) ∑

127

Theorem (9.5):

let be a bounded measurable subset of , then.

1) .

2) If are bounded measurable sets such that , if is measurable, then

is measurable and

3) If are bounded measurable sets, with is bounded, then

is

measurable and ∑

.

If , then ∑

4) If are bounded measurable subsets of with is bounded, then

is measurable sets and ∑ .

If { } disjoint, then ∑

5) If is bounded measurable subsets of , then for any , is a measurable

set and .

Chapter (10)

Lebesgue Theory of Integration:

Definition (10.1): Lebesgue Partition

Let be a bounded measurable subset of and { } be a finite collection of subset of

. { } is a Lebesgue partition on if satisfies:

1)

2) are measurable sets

3) is a zero set.

Notes:

1) If { } and { }

, we say is a refinement to if

128

2) If { } and { }

are Lebesgue partitions on , then {

} is a Lebesgue partition on .

Definition (10.2): Lebesgue Integral

Let be a bounded measurable set and be a bounded function. ) If { }

is a partition on , put

{ } f { } .

And put { } f { }

Clearly:- .

Now define

∑ is called Lebesgue upper sum for Lebesgue partition .

And

∑ is called Lebesgue lower sum Lebesgue partition .

Clearly:- .

Remarks :

(1) If is a refinement to , then.

(2) For any two partitions and on .

{ i io o }

{ i io o } .

From (1) is bound below and is bound above, by completeness of

129

Put i f( ) which is called Lebesgue upper integral and

( ) which is called Lebesgue lower integral.

Clearly that:- . If , then is called Lebesgue integrable and we

write

.

Remarks (10.3):

Every Riemann partition is a Lebesgue partition.

Proof:

Let [ be a closed interval. It's clear that [ measurable.

Let { } be a partition on [

1. [

2. is a measurable sets

3. or only one element which is a zero set.

Hence we have the following result.

Proposition (10.4):-

If [ is a bounded function and is Riemann integrable, then is Lebesgue

integrable

Proof:

By remark (10.3) every Riemann partition is a Lebesgue partition

{ i io o [ }

{ i io o [ }

{ i io o [ }

{ i io o [ }

It's clear that and

130

Then

( ) ( )

( ) ( )

This means that

Since is Riemann integrable, then and hence .

Remark :

The converse of remark (10.4) is not true in general as the following example shows.

Example :

Let [ be defined by { [

[

[ is not a zero set, then by Lebesgue theorem [ . Thus is not

Riemann integrable

Let [ { } where [ and [ .

Claim: is a Lebesgue partition

1) [

2) [ di joi

[ .

.

3) is a zero set.

{ } f { }

131

∑

∑

{ i io } .

{ i io }

i f( )

( )

. Thus is Lebesgue integrable.

Proposition (10.5):-

Let be a measurable bounded set and be a bounded function, then is

Lebesgue integrable iff for each , there exists a Lebesgue partition such that

.

Proof:

Compare with Lemma (7.6) in chapter (7).

Some properties of lebesgue integral:

Remarks :

19) Let be a measurable bounded set and be a function defined by

, then is Lebesgue integrable and

Proof:

Let { } be any partition on .

{ } f { }

132

∑

( )

and zero set, then

∑

∑

Thus .

∑

( )

{ i io } .

{ i io }

i f( )

( )

.

Thus is Lebesgue integrable.

20) Let is a bounded measurable set and be a bounded Lebesgue integrable

function, if , then

.

Proof:

Let { } be a Lebesgue partition on , { }

∑

( )

∫

i f{ { i io }}

133

21) If is a zero set and is a bounded function, then is Lebesgue integrable and

.

Proof:

Let { } be a Lebesgue partition on .

1) .

2) are measurable sets

3) is a zero set.

Since is a zero set, then each is a zero set and .

{ } f { } , then

∑

Similarly .

Then , hence is Lebesgue integrable and

.

22) If is a bounded measurable set and is a bounded Lebesgue integrable function,

, then

.

Proof:

From (2) , then

. Thus

Proposition (10.6):-

Let be bounded function, be a measurable bounded set if are subsets of

such that and and is Lebesgue integrable, then

∫

∫

∫

134

Proposition (10.7):

Let be a measurable bounded set and be bounded Lebesgue integrable

functions, then.

3)

4)

Corollary (10.8):

If is a measurable bounded set and are bounded Lebesgue integrable

functions such that , then

.

Proof:

Let

By (4)

.

Then

.

Corollary (10.9):

If is a measurable bounded set and is a bounded Lebesgue integrable function ,

then | | is Lebesgue integrable and |

| | |

.

Proof: (H.W)

Definition (10.10):

Let be functions, if there exists a zero set such that

[ , then we say that almost everywhere

Proposition (10.11):

135

Let is be bounded measurable set and be bounded functions, if is

Lebesgue integrable and , then is Lebesgue integrable and

.

Example:

Let [ be defined by {

[ { }

[ √

{ √ }

.

[ , [ √

[ { √ }

Measurable functions and integrable functions

Definition(Measurable functions) (10.12):

Let , be bounded function, is said to be a measurable function if for each

open set in , is a measurable set.

Remarks:

1) If is a measurable function, then is a measurable set.

Since is open, then is a measurable set.

2) If is a measurable set and is a continuous function, then is a measurable function.

Proof:

Let be any open set, since is a continuous function, then , is

open set and is a measurable set, hence is a measurable set.

Proposition (10.13):

If and is a function, then the following are equivalent:

1) is a measurable function.

136

2) For each closed set , then is a measurable set.

3) For each [ [ , then ([ ) ([ ) ( ) are

measurable sets

3) For each , then ( ) ( ) are measurable sets.

4) For each , then ([ ) ( ) are measurable sets.

Corollary(10.14):

Let is a function:

1) is a measurable function iff for each , then ( ) is a measurable set.

2) is a measurable function iff for each , then ([ ) is a measurable set.

3) is a measurable function iff for each , then ( ) is a measurable set.

4) is a measurable function iff for each , then ( ) is a measurable set.

Example:

Let [ be defined by { [

[ .

Sol: let , is open

{

[ [

[

{ [ }

[ is bounded countable set, the [

[ [ [ [disjoint]

[ [ [

[

In each case is a measurable set, hence is a measurable function.

Remark:

If [ is a monotonic function, then is a measurable function? (why)

Proposition (10.15):

137

If and is a measurable function and are continuous function,

then is a measurable function.

Proof:

Let be any open set in , since is a continuous function, then is open set in .

( ) , since is open set in and is a measurable

function, hence ( ) is a measurable set.

Bounded variation functions

Definition(10.16):

Let [ be a function and Let { } be a

partition on [ , [

Let

∑| |

Let { i io o [ } , then is bound below.

If is bound above, then has least upper bound.

Put ( ) .

is called the variation of on [

is called the bounded variation function.

Otherwise if is bound above, then is not bounded variation function.

Remark (10.17):

If [ is a bounded variation function, then is bounded.

Proof:

To proof | | [ .

Let { } be a partition on [ , [

138

∑| |

{ i io o [ } .

( ) exists, then | |∑ | | [

Take .

Thus | | [

Remark (10.18):

If [ is a bounded monotonic function, then is a bounded variation

function.

Remark (10.19):

If [ are bounded variation functions, then is a bounded variation

function.

Proof:

Let { } be a partition on [ , [

∑ | | .

∑ | |

∑ | | ∑ | |

{ i io o [ }

{ i io o [ }

{ i io o [ } { i io o [ }

( ) ( ) ( )

Thus

139

Remark (10.20):

If [ is a bounded variation function, then is a bounded variation function.

Proof: (H.W)