absolute continuity vs total singularity - george washington university
TRANSCRIPT
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Absolute continuity vs total singularity
Absolute continuity vs totalsingularity
E. Arthur (Robbie) Robinson
The George Washington University
February 16, 2012
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Absolute continuity vs total singularity
Outline
1 Introduction
2 preliminaries
3 The Banach Lemma
4 Villani’s lemma
5 Proof of Kakeya’s lemma
6 Next. . .
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Absolute continuity vs total singularity
Introduction
1 Introduction
2 preliminaries
3 The Banach Lemma
4 Villani’s lemma
5 Proof of Kakeya’s lemma
6 Next. . .
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Absolute continuity vs total singularity
Introduction
Part 1
In the first lecture I want to prove the following theorem (andsome related results).
Lemma (Kakeya’s Lemma)
If f : [0, 1]→ R is continuous, strictly monotone and satisfies|f ′(x)| ≥ a > 0 a.e., then f−1(x) is absolutely continuous, and|(f−1)′(x)| ≤ 1/a a.e..
This was stated without proof by Soichi Kakeya in his 1924 paperon representing real numbers. The necessity of this theorem wasmissed in Fritz Schweiger’s 1995 account of Kakeya’s theorem(which he thus states incorrectly).
I was surprised by how difficult it turned out to be!
We will begin by reviewing some of the relevant background.
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Absolute continuity vs total singularity
Introduction
Part 2
In the second lecture, I will discuss examples of strictly increasingcontinuous functions f : [0, 1]→ [0, 1] with f ′(x) = 0 a.e. I callthese totally singular (TS) functions.
The most famous TS function is Minkowski’s “question mark”function ?(x). The inverse ?−1(x), which is also TS, is known asJohn H. Conway’s “box” function. Both implement a strictlymonotone bijection between the rationals and dyadic rationals in[0, 1].
We will also look at some examples of TS functions due to RaphelSalem, which are related to the Koch snowflake.
This will be in 2 weeks.
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Absolute continuity vs total singularity
Introduction
Math Classes
Some results and definitions quoted in this talk come from specificmath classes. The GW numbering for them is:
M 1231: Calculus I
M 1232: Calculus II
M 4239/40: Undergraduate Analysis I & II
M 6214: Measure theory (Graduate Analysis I)
M 6215: Functional Analysis (Graduate Analysis II)
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Absolute continuity vs total singularity
preliminaries
1 Introduction
2 preliminaries
3 The Banach Lemma
4 Villani’s lemma
5 Proof of Kakeya’s lemma
6 Next. . .
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Absolute continuity vs total singularity
preliminaries
Monotonicity
A function f : [0, 1]→ R is increasing if x < y, x, y ∈ [0, 1]implies f(x) ≤ f(y).A function f : [0, 1]→ R is strictly increasing if x < y,x, y ∈ [0, 1] implies f(x) < f(y).
The definitions of decreasing and strictly decreasing areanalogous.
A function is monotonic if it is either increasing or decreasingand strictly monotonic if it is either strictly increasing orstrictly decreasing.
(M 1231)
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Absolute continuity vs total singularity
preliminaries
properties of monotonic functions
A monotonic function is continuous except, at most, on acountable set. The discontinuities are all jumps. By changingcountably many values, we can assume it is right continuous(M 4839/40 or M 6214).
A monotonic function is differentiable almost everywhere (theset of x so that
f ′(x) = limh→0
f(x+ h)− f(x)h
either does not exist or is ±∞ has measure zero (M 6214)).
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Absolute continuity vs total singularity
preliminaries
The Riesz representation theorem
Any regular finite Borel measure µ on [0, 1] is defined by a rightcontinuous increasing function g : [0, 1]→ R.
In particular,µ([a, b]) = g−(b)− g(a),
andµ((a, b)) = g(b)− g+(a),
where [a, b] ⊆ [0, 1]. (g+ and g− are right and left limits.)
Conversely, any such g defines a measure µg. (M6215)
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Absolute continuity vs total singularity
preliminaries
Bounded variation
A function has bounded variation (BV) if
V (f) = supP
n∑k=1
|f(xk)− f(xk−1)| <∞,
where P is a partition of [0, 1] of the form
P : 0 = x0 < x1 < · · · < xn = 1.
Using Riesz representation, µf is a signed measure.
This is an element of the ball B of radius V (f) in the dualC([0, 1])∗ via f 7→
∫fdµg. This ball is weak-* compact
(M 6215).
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Absolute continuity vs total singularity
preliminaries
Properties of bounded variation
Clearly, any monotonic f has V (f) = |f(1)− f(0)| <∞ (i.e.,BV).
If V (f) <∞ (i.e., BV) then there exist increasing g anddecreasing h so that f(x) = g(x) + h(x). Almost unique(M 6214).
BV functions are continuous except on an at most countableset.
BV functions are differentiable almost everywhere.
If f ∈ C1([0, 1]), then f has BV (M 4839/40).
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Absolute continuity vs total singularity
preliminaries
Continuity
A function f : [0, 1]→ R is continouos (C) if for all ε > 0 andx ∈ [0, 1], there exists δ > 0 so that if |x− y| < δ, y ∈ [0, 1]then |f(x)− f(y)| < ε (M 1231).
A function f : [0, 1]→ R is uniformly continouos (UC) if forall ε > 0, there exists δ > 0 so that if |x− y| < δ, x, y ∈ [0, 1]then |f(x)− f(y)| < ε (M 4239).
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Absolute continuity vs total singularity
preliminaries
Absolute continuity
Let I = {(xi, yi) : i = 1, . . . , n} denote a finite collection ofdisjoint open intervals in [0, 1].We say f : [0, 1]→ R is absolutely continuous (AC) if for all ε > 0there exists δ > 0 so that for any I with
n∑i=1
(xi − yi) < δ
we haven∑i=1
|f(xi)− f(yi)| < ε.
(M 6214).
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Absolute continuity vs total singularity
preliminaries
Continuity properties
AC =⇒ UC =⇒ C.
C =⇒ UC (since [0, 1] is compact. M 4239/40)
AC =⇒ BV (M 6214).
Thus AC =⇒ f ′(x) exists a.e.
UC (or C) 6 =⇒ BV (M 6214).
Thus UC 6 =⇒ AC.
Also UC + BV 6 =⇒ AC (we will see this and much more).
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Absolute continuity vs total singularity
preliminaries
Fundamental theorem of calculus
If f : [0, 1]→ R is increasing then
f(x) ≥ f(0) +∫ x
0f ′(t)dt.
f : [0, 1]→ R is AC if and only if
f(x) = f(0) +
∫ x
0f ′(t)dt.
(both M 6214)
This holds for f ∈ C1 (M 1231) so C1 =⇒ AC.
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Absolute continuity vs total singularity
preliminaries
Examples & Facts
The Cantor function is an increasing continuous surjectionf : [0, 1]→ [0, 1] so that f ′(x) = 0 a.e. (f ′(x) = 0 on a denseopen set). It is not strictly increasing (not 1:1). (SC=“singularcontinuous”)
(Source: http://mathworld.wolfram.com/CantorFunction.html)
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Absolute continuity vs total singularity
preliminaries
Examples & Facts
There exist strictly increasing continuous functionsg : [0, 1]→ [0, 1] so that g′(x) = 0 a.e. These correspond tosingular Borel measures that are positive on open sets (!) Theseare the topic of the second lecture. (TS=“totally singular”)
(Source:http://en.wikipedia.org/wiki/Minkowski%27s question mark function)
The Cantor and ? functions both show UC + BV 6 =⇒ AC.
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Absolute continuity vs total singularity
preliminaries
Examples & Facts
(Lebesgue decomposition theorem) An increasing f can bewritten as a sum of increasing functions
f(x) = fac(x) + fsc(x) + fd(x)
where fac ∈ AC, fsc is continuous with f ′sc(x) = 0 a.e. (SC),and fd is a step function (M 6214).
If f is strictly a increasing bijection, then f is continuous, andf−1 is also strictly increasing, bijective and thus continuous.
If A,B ⊆ [0, 1] are dense open sets, and f : A→ B is astrictly increasing bijection, then f extends to a strictlyincreasing bijection f : [0, 1]→ [0, 1].
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Absolute continuity vs total singularity
The Banach Lemma
1 Introduction
2 preliminaries
3 The Banach Lemma
4 Villani’s lemma
5 Proof of Kakeya’s lemma
6 Next. . .
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Absolute continuity vs total singularity
The Banach Lemma
Statement
Let λ denote Lebesgue measure on I = [0, 1]. A functionf : I → R is said to satisfy Lusin’s property-N if λ(f(E)) = 0whenever E ⊆ I has λ(E) = 0 (see Natanson).
Lemma (Banach)
A continuous function f : I → R of bounded variation is absolutelycontinuous if and only if it satisfies Lusin’s property-N.
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma
First we prove property-N implies absolute continuity.
For simplicity, we assume f is strictly increasing.
The case of strictly decreasing is the same.
The case of bounded variation is true but harder and we skip ithere.
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma. . .
Let I denote a finite set of disjoint open intervals in Ii ⊆ I.Let |I| = ∪Ii∈IIi.
Let J = f(I) = {f(Ii)} be the collection of their images.
Recall that ϕ is absolutely continuous if for all ε > 0 there existsδ > 0 so that λ(|I|) < δ implies λ(|f(I)|) < ε (here we use that fis monotone).
Suppose f satisfies property-N but is not absolutely continuous.
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma. . .
Let I denote a finite set of disjoint open intervals in Ii ⊆ I.Let |I| = ∪Ii∈IIi.
Let J = f(I) = {f(Ii)} be the collection of their images.
Recall that ϕ is absolutely continuous if for all ε > 0 there existsδ > 0 so that λ(|I|) < δ implies λ(|f(I)|) < ε (here we use that fis monotone).
Suppose f satisfies property-N but is not absolutely continuous.
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma. . .
Since f is not absolutely continuous, there exists ε0 > 0 and
I1, I2, I3, . . .
withλ(|In|) ≥ ε0 (1)
for all n = 1, 2, 3, . . . , and
∞∑n=1
λ(|f(In)|) <∞. (2)
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma. . .
Let Nn(y) = χ|f(In)|(y) (the characteristic function of |f(In)|).
Then by (1) ∫ϕ(I)
Nn(y) ≥ ε0 > 0 (3)
for all n.
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma. . .
LetB = lim sup |In|.
By (2), the Borel-Cantelli lemma (see two slides ahead) impliesλ(B) = 0.
LetA = lim sup |f(In)|.
Note that A = f(B).
Then λ(A) = 0 since f satisfies property-N.
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma. . .
Now, y ∈ A iff Nn(y) 6= 0 infinitely often.
It follows that limn→∞Nn(y) = 0 for y ∈ f(I)\A (that is, a.e. y).
By the Dominated Convergence Theorem (M 6214)
limn→∞
∫f(I)
Nn(y)dy = 0,
contradicting (3).
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Absolute continuity vs total singularity
The Banach Lemma
Aside: The Borel-Cantelli Lemma
Let In be a sequence of measurable sets in [0, 1]. Recall that
J := lim supn→∞
In =
∞⋂n=1
∞⋃k=n
In.
Then x ∈ J if and only if x ∈ In for infinitely many n.
The sets In are independent if
λ(⋂k=1
Ink) =
∏k=1
λ(Ink),
for any n1 < n2 < · · · < n` where ` <∞.
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Absolute continuity vs total singularity
The Banach Lemma
Aside: The Borel-Cantelli Lemma
Lemma (Borel-Cantelli)
Let In ⊆ [0, 1] be a sequence of measurable sets, and let λ beLebesgue measure (λ[0, 1] = 1).
If∑∞
n=1 λ(In) <∞ then λ(lim supn→∞ In) = 0.∑∞n=1 λ(In) =∞ and the sets In are independent then
λ(lim supn→∞ In) = 1.
We used only the first part (sometimes called the “easy half”).The proof can be an exercise in (M 6214). The second part (calledthe “hard half”) is considered to be part of probability theory.
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma: Converse. . .
Suppose f : [0, 1]→ R is absolutely continuous and increasing (theproof works for BV, but is harder).
Let E ⊆ [0, 1] with λ(E) = 0.
Let ε > 0 be given, and choose δ > 0 for f ∈ AC.
Choose a (possibly infinite) set I of disjoint open intervalsIi ⊆ [0, 1] with
E ⊆ |I|,
andλ(|I|) < δ.
(possible since λ(E) = 0).
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Absolute continuity vs total singularity
The Banach Lemma
Proof of Banach’s lemma: Converse. . .
Let |I| = ∪Ii∈IIi (an open set).
Let J = f(I) = {f(Ii)} be the collection of their images. This isalso a collection of intervals (here we use that f is strictlymonotone).
For any finite I ′ ⊆ I we have that J ′ = f(J ) satisfies λ(|J ′|) < ε(because f ∈ AC).
Thus λ(|J |) ≤ ε, so λ(|J |) = 0 since ε > 0 was arbitrary.
But f(E) ⊆ |J | = f(|I|), so λ(E) = 0. It follows that f satisfisproperty-N.
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Absolute continuity vs total singularity
Villani’s lemma
1 Introduction
2 preliminaries
3 The Banach Lemma
4 Villani’s lemma
5 Proof of Kakeya’s lemma
6 Next. . .
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Absolute continuity vs total singularity
Villani’s lemma
Villani’s lemma
Lemma (Villani)
Let ϕ : [a, b]→ R be continuous, strictly monotone. Then ϕ−1(x)is absolutely continuous if and only if |ϕ′(x)| > 0 a.e..
This lemma appears in a paper by Villani (1984) who says he“believes it is known” but cannot find it in the literature.
We often use this in the case where ϕ({a, b}) = {0, 1} so thatf(x) = ϕ−1(x) satisfies f : [0, 1]→ R is continuous andmonotone. The lemma shows f ∈ AC.
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Absolute continuity vs total singularity
Villani’s lemma
An application
Let f : [0, 1]→ [0, 1] be a piecewise continuous, piecewisemonotone interval map (regarded as a dynamical system).
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
Figure: The Renyi-Bolyai map: T (x) = x2 + 2x mod 1.
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Absolute continuity vs total singularity
Villani’s lemma
An application. . .
The map f is called non-singular if λ(f−1(E)) = 0 for any E withλ(E) = 0.
We see that f is nonsingular iff |f ′(x)| > 0 a.e. True forRenyi-Bolyai.
Let ρ(x) ≥ 0,∫ 10 ρ(x)dx = 1, and let µρ(E) =
∫E ρ(x)dx
(µρ = ρλ). We call ρ an invariant density if
µρ(f−1(E)) = µρ(E).
Clearly non-singularity is necessary for the existence of an invariantdensity ρ.
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Absolute continuity vs total singularity
Villani’s lemma
Proof of Villani’s lemma
Let
[c, d] = ϕ([a, b]),
λ=Lebeague measure on [a, b], and
λ′=Lebeague measure on [c, d].
By the Lebesgue decomposition (M 6214) we have
λ′ ◦ ϕ = µ+ σ where µ ⊥ σ.
Here µ << λ is the absolutely continuous part.
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Absolute continuity vs total singularity
Villani’s lemma
Proof of Villani’s lemma
Now we use the following fact (see e.g. Rudin).
|ϕ′(x)| = dµ
dλ(x),
that relates the ordinary derivative ϕ′ to the Radon-Nikodymderivative dµ
dλ (M 6214). Thus
λ′ ◦ ϕ = |ϕ′(x)|λ+ σ where λ ⊥ σ.
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Absolute continuity vs total singularity
Villani’s lemma
Proof of Villani’s lemma
By the Banach lemma, it suffices to show ϕ−1 has property-N(i.e., λ ◦ ϕ−1 << λ).
We show λ(A) > 0 implies λ′(ϕ(A)) > 0.
But
(λ′ ◦ ϕ)(A) =∫A|ϕ′(x)| dλ(x) + σ(A) > 0,
since |ϕ′(x)| > 0 a.e.
We skip the converse since we don’t use it.
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
1 Introduction
2 preliminaries
3 The Banach Lemma
4 Villani’s lemma
5 Proof of Kakeya’s lemma
6 Next. . .
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
Kakeya’s Lemma
Lemma (Kakeya)
Let ϕ : [a, b]→ R be continuous, strictly monotone and satisfying|ϕ′(x)| > a > 0 a.e., then |ϕ−1(y)| < 1/a a.e..
Proof:For a continuous monotonic function ϕ, the inverse functiontheorem
(ϕ−1)′(ϕ(x)) =1
ϕ′(x),
holds whenever ϕ′(x) exists and is nonzero (see Wade(M 4239/40)).
Note that ϕ′(x) exists λ a.e.. Let E = {x : ϕ′(x) ≤ a}
By assumption, λ(E) = 0.
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
Proof. . .
Now ϕ(E) = {y : ϕ′(ϕ−1(y)) ≤ a}
Let F be the set of y so that (ϕ−1)′(y) does not exist.
Since ϕ−1 is monotone, λ(F ) = 0.
It suffices to show λ(ϕ(E) ∪ F ) = λ(ϕ(E)) = 0, because thenλ((ϕ(E) ∪ F )c) = 1, and the inverse function theorem implies
(ϕ−1)′(y) =1
ϕ′(ϕ−1(y))< 1/a for λ a.e. y.
This follows from the next lemma.
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
Natanson’s lemma
Lemma (Natanson)
Let ϕ : [0, 1]→ R and let 0 ≤ ϕ′(x) ≤ a for x ∈ E. Then
λ∗(ϕ(E)) ≤ aλ∗(E),
where λ∗ denotes Lebesgue outer measure.
Comment: If λ∗(E) = 0 then E is measurable, and this impliesλ(E) = 0. This is what happens in our case.
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
Proof of Natanson’s Lemma
Fix ε > 0. Choose G open with E ⊆ G and
λ(G) < λ∗(E) + ε.
Also fix a0 > a.
For each x ∈ E there is a sequence hn > 0, hn → 0 so that
limn→∞
ϕ(x+ hn)− ϕ(x)hn
= ϕ′(x) ≤ a.
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
Proof of Natanson’s Lemma
By choosing n large enough we can assume:
ϕ(x+ hn)− ϕ(x)hn
< a0,
andIn(x) := [x, x+ hn] ⊆ G.
Assume this WOLOG for all n.
Let Jn(x) = ϕ(In(x)) = [ϕ(x), ϕ(x+ hn)].
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
Proof of Natanson’s Lemma
Now {In(x) : x ∈ E} is a Vitali cover of E (this means theintervals that cover each point x ∈ E are arbitrarily small).
The Vitali Covering Lemma (see Natanson or Rudin, M 6214) saysthere is a sequence of disjoint intervals:
En1(x1), En2(x2), En3(x3), . . .
such that
λ∗
(E\
∞⋃k=0
Ink(xk)
)= 0.
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
Proof of Natanson’s Lemma
Thus
λ∗(ϕ(E)) ≤∞∑k=0
λ(Jnk(xk)) < a0
∞∑k=0
λ(Ink(xk)).
Since Ink(xk) are pairwise disjoint, we have
∞∑k=0
λ(Ink(xk)) = λ
( ∞⋃k=0
Ink(xk)
),
and⋃∞k=0 Ink
(xk) ⊆ G.
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Absolute continuity vs total singularity
Proof of Kakeya’s lemma
Proof of Natanson’s Lemma
Finally,λ ∗ (ϕ(E)) < a0λ(G) < a0(λ
∗(E) + ε).
The proof is finished by letting a0 → a and ε→ 0.
λ∗(ϕ(E)) ≤ aλ∗(E).
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Absolute continuity vs total singularity
Next. . .
1 Introduction
2 preliminaries
3 The Banach Lemma
4 Villani’s lemma
5 Proof of Kakeya’s lemma
6 Next. . .
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Absolute continuity vs total singularity
Next. . .
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Absolute continuity vs total singularity
Next. . .
References
Kakeya, S., On the generalized scale of notation, Japan J.Math, 1, (1926), 95-108.
Natanson, I. P., Theory of Functions of a Real Variable,Frederick Unger, New York, (1955).
Rudin, W., Real and Complex Analysis, McGraw-HillScience/Engineering/Math, 3rd ed, (1986)
Schweiger, F., Ergodic Theory of Fibred Systems and MetricNumber Theory, Oxford University Press, (1995)
Villani, A. On Lusin’s condition for the inverse function,Rendiconti del Circolo Matematico di Palermo, Serie II, 33,331-335, (1984).
Wade, W. R., Introduction to Analysis, 4th ed, Pearson,(2010)