Absolute continuity vs total singularity - George Washington University

transcript

Absolute continuity vs total singularity

Absolute continuity vs totalsingularity

E. Arthur (Robbie) Robinson

The George Washington University

February 16, 2012

Outline

1 Introduction

2 preliminaries

3 The Banach Lemma

4 Villani’s lemma

5 Proof of Kakeya’s lemma

6 Next. . .

Introduction

1 Introduction

2 preliminaries

3 The Banach Lemma

4 Villani’s lemma

6 Next. . .

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.

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.

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)

preliminaries

1 Introduction

2 preliminaries

3 The Banach Lemma

4 Villani’s lemma

6 Next. . .

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)

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)).

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)

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).

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).

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).

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).

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).

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) +

0f ′(t)dt.

(both M 6214)

This holds for f ∈ C1 (M 1231) so C1 =⇒ AC.

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)

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.

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].

The Banach Lemma

1 Introduction

2 preliminaries

3 The Banach Lemma

4 Villani’s lemma

6 Next. . .

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.

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.

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.

The Banach 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.

The Banach 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)

The Banach 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.

The Banach 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.

The Banach 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).

The Banach Lemma

Aside: The Borel-Cantelli Lemma

Let In be a sequence of measurable sets in [0, 1]. Recall that

J := lim supn→∞

∞⋂n=1

∞⋃k=n

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 ` <∞.

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.

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).

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.

Villani’s lemma

1 Introduction

2 preliminaries

3 The Banach Lemma

4 Villani’s lemma

6 Next. . .

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.

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

Figure: The Renyi-Bolyai map: T (x) = x2 + 2x mod 1.

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.

Villani’s lemma

Proof of Villani’s lemma

[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.

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.

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.

(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.

Proof of Kakeya’s lemma

1 Introduction

2 preliminaries

3 The Banach Lemma

4 Villani’s lemma

6 Next. . .

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.

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.

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.

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.

By choosing n large enough we can assume:

ϕ(x+ hn)− ϕ(x)hn

andIn(x) := [x, x+ hn] ⊆ G.

Assume this WOLOG for all n.

Let Jn(x) = ϕ(In(x)) = [ϕ(x), ϕ(x+ hn)].

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

∞⋃k=0

Ink(xk)

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.

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).

Next. . .

1 Introduction

2 preliminaries

3 The Banach Lemma

4 Villani’s lemma

6 Next. . .

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)

Absolute continuity vs total singularity - George Washington University

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