Absolute Continuity Vs Total Singularity

Absolute continuity vs total singularity

E. Arthur (Robbie) Robinson

The George Washington University

February 16, 2012 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. . . 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. . . Absolute continuity vs total singularity Introduction Part 1

In the ﬁrst lecture I want to prove the following theorem (and some related results).

Lemma (Kakeya’s Lemma) If f : [0, 1] → R is continuous, strictly monotone and satisﬁes |f 0(x)| ≥ a > 0 a.e., then f −1(x) is absolutely continuous, and |(f −1)0(x)| ≤ 1/a a.e..

This was stated without proof by Soichi Kakeya in his 1924 paper on representing real numbers. The necessity of this theorem was missed 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. Absolute continuity vs total singularity Introduction Part 2

In the second lecture, I will discuss examples of strictly increasing continuous functions f : [0, 1] → [0, 1] with f 0(x) = 0 a.e. I call these 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 as John H. Conway’s “box” function. Both implement a strictly monotone bijection between the rationals and dyadic rationals in [0, 1]. We will also look at some examples of TS functions due to Raphel Salem, which are related to the Koch snowﬂake. This will be in 2 weeks. Absolute continuity vs total singularity Introduction Math Classes

Some results and deﬁnitions quoted in this talk come from speciﬁc math 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) 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. . . 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 are analogous. A function is monotonic if it is either increasing or decreasing and strictly monotonic if it is either strictly increasing or strictly decreasing. (M 1231) Absolute continuity vs total singularity preliminaries properties of monotonic functions

A monotonic function is continuous except, at most, on a countable set. The discontinuities are all jumps. By changing countably many values, we can assume it is right continuous (M 4839/40 or M 6214). A monotonic function is diﬀerentiable almost everywhere (the set of x so that f(x + h) − f(x) f 0(x) = lim h→0 h either does not exist or is ±∞ has measure zero (M 6214)). Absolute continuity vs total singularity preliminaries The Riesz representation theorem

Any regular ﬁnite Borel measure µ on [0, 1] is deﬁned by a right continuous increasing function g : [0, 1] → R.

In particular, µ([a, b]) = g−(b) − g(a), and µ((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) Absolute continuity vs total singularity preliminaries Bounded variation

A function has bounded variation (BV) if

n X V (f) = sup |f(xk) − f(xk−1)| < ∞, P k=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 dual ∗ R C([0, 1]) via f 7→ fdµg. This ball is weak-* compact (M 6215). 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 and decreasing h so that f(x) = g(x) + h(x). Almost unique (M 6214). BV functions are continuous except on an at most countable set. BV functions are diﬀerentiable almost everywhere. If f ∈ C1([0, 1]), then f has BV (M 4839/40). Absolute continuity vs total singularity preliminaries Continuity

A function f : [0, 1] → R is continouos (C) if for all > 0 and x ∈ [0, 1], there exists δ > 0 so that if |x − y| < δ, y ∈ [0, 1] then |f(x) − f(y)| < (M 1231).

A function f : [0, 1] → R is uniformly continouos (UC) if for all > 0, there exists δ > 0 so that if |x − y| < δ, x, y ∈ [0, 1] then |f(x) − f(y)| < (M 4239). Absolute continuity vs total singularity preliminaries Absolute continuity

Let I = {(xi, yi): i = 1, . . . , n} denote a ﬁnite collection of disjoint open intervals in [0, 1]. We say f : [0, 1] → R is absolutely continuous (AC) if for all > 0 there exists δ > 0 so that for any I with

n X (xi − yi) < δ i=1 we have n X |f(xi) − f(yi)| < . i=1 (M 6214). 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 0(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). Absolute continuity vs total singularity preliminaries Fundamental theorem of calculus

If f : [0, 1] → R is increasing then Z x f(x) ≥ f(0) + f 0(t)dt. 0

f : [0, 1] → R is AC if and only if Z x f(x) = f(0) + f 0(t)dt. 0 (both M 6214) This holds for f ∈ C1 (M 1231) so C1 =⇒ AC. Absolute continuity vs total singularity preliminaries Examples & Facts

The Cantor function is an increasing continuous surjection f : [0, 1] → [0, 1] so that f 0(x) = 0 a.e. (f 0(x) = 0 on a dense open set). It is not strictly increasing (not 1:1). (SC=“singular continuous”)

(Source: http://mathworld.wolfram.com/CantorFunction.html) Absolute continuity vs total singularity preliminaries Examples & Facts

There exist strictly increasing continuous functions g : [0, 1] → [0, 1] so that g0(x) = 0 a.e. These correspond to singular Borel measures that are positive on open sets (!) These are 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. Absolute continuity vs total singularity preliminaries Examples & Facts

(Lebesgue decomposition theorem) An increasing f can be written as a sum of increasing functions

f(x) = fac(x) + fsc(x) + fd(x)

0 where fac ∈ AC, fsc is continuous with fsc(x) = 0 a.e. (SC), and fd is a step function (M 6214). If f is strictly a increasing bijection, then f is continuous, and f −1 is also strictly increasing, bijective and thus continuous. If A, B ⊆ [0, 1] are dense open sets, and f : A → B is a strictly increasing bijection, then f extends to a strictly increasing bijection f : [0, 1] → [0, 1]. 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. . . Absolute continuity vs total singularity The Banach Lemma Statement

Let λ denote Lebesgue measure on I = [0, 1]. A function f : I → R is said to satisfy Lusin’s property-N if λ(f(E)) = 0 whenever E ⊆ I has λ(E) = 0 (see Natanson).

Lemma (Banach) A continuous function f : I → R of bounded variation is absolutely continuous if and only if it satisﬁes Lusin’s property-N. 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 it here. 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∈I Ii.

Let J = f(I) = {f(Ii)} be the collection of their images. Recall that ϕ is absolutely continuous if for all > 0 there exists δ > 0 so that λ(|I|) < δ implies λ(|f(I)|) < (here we use that f is monotone). Suppose f satisﬁes property-N but is not absolutely continuous. 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∈I Ii.

Since f is not absolutely continuous, there exists 0 > 0 and

I1, I2, I3,...

with λ(|In|) ≥ 0 (1) for all n = 1, 2, 3,... , and

∞ X λ(|f(In)|) < ∞. (2) n=1 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) Z Nn(y) ≥ 0 > 0 (3) ϕ(I) for all n. Absolute continuity vs total singularity The Banach Lemma Proof of Banach’s lemma. . .

Let B = lim sup |In|. By (2), the Borel-Cantelli lemma (see two slides ahead) implies λ(B) = 0. Let A = lim sup |f(In)|.

Note that A = f(B). Then λ(A) = 0 since f satisﬁes property-N. 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) Z lim Nn(y)dy = 0, n→∞ f(I)

contradicting (3). 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 sup In = In. n→∞ n=1 k=n

Then x ∈ J if and only if x ∈ In for inﬁnitely many n.

The sets In are independent if

` ` \ Y λ( Ink ) = λ(Ink ), k=1 k=1

for any n1 < n2 < ··· < n` where ` < ∞. 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 be Lebesgue measure (λ[0, 1] = 1). P∞ If n=1 λ(In) < ∞ then λ(lim supn→∞ In) = 0. P∞ 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 (called the “hard half”) is considered to be part of probability theory. Absolute continuity vs total singularity The Banach Lemma Proof of Banach’s lemma: Converse. . .

Suppose f : [0, 1] → R is absolutely continuous and increasing (the proof works for BV, but is harder). Let E ⊆ [0, 1] with λ(E) = 0. Let > 0 be given, and choose δ > 0 for f ∈ AC. Choose a (possibly inﬁnite) set I of disjoint open intervals Ii ⊆ [0, 1] with E ⊆ |I|, and λ(|I|) < δ. (possible since λ(E) = 0). Absolute continuity vs total singularity The Banach Lemma Proof of Banach’s lemma: Converse. . .

Let |I| = ∪Ii∈I Ii (an open set).

Let J = f(I) = {f(Ii)} be the collection of their images. This is also a collection of intervals (here we use that f is strictly monotone). For any finite I0 ⊆ I we have that J 0 = f(J ) satisfies λ(|J 0|) < (because f ∈ AC). Thus λ(|J |) ≤ , so λ(|J |) = 0 since > 0 was arbitrary. But f(E) ⊆ |J | = f(|I|), so λ(E) = 0. It follows that f satisfis property-N. 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. . . Absolute continuity vs total singularity Villani’s lemma Villani’s lemma

Lemma (Villani) −1 Let ϕ :[a, b] → R be continuous, strictly monotone. Then ϕ (x) is absolutely continuous if and only if |ϕ0(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 that −1 f(x) = ϕ (x) satisﬁes f : [0, 1] → R is continuous and monotone. The lemma shows f ∈ AC. Absolute continuity vs total singularity Villani’s lemma An application

Let f : [0, 1] → [0, 1] be a piecewise continuous, piecewise monotone interval map (regarded as a dynamical system).

1.0

0.8

0.6

0.4

0.2

0.0 0.0 0.2 0.4 0.6 0.8 1.0

Figure: The R´enyi-Bolyai map: T (x) = x2 + 2x mod 1. 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 with λ(E) = 0. We see that f is nonsingular iﬀ |f 0(x)| > 0 a.e. True for R´enyi-Bolyai.

R 1 R Let ρ(x) ≥ 0, 0 ρ(x)dx = 1, and let µρ(E) = E ρ(x)dx (µρ = ρλ). We call ρ an invariant density if

−1 µρ(f (E)) = µρ(E).

Clearly non-singularity is necessary for the existence of an invariant density ρ. 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 λ0=Lebeague measure on [c, d]. By the Lebesgue decomposition (M 6214) we have

λ0 ◦ ϕ = µ + σ where µ ⊥ σ.

Here µ << λ is the absolutely continuous part. Absolute continuity vs total singularity Villani’s lemma Proof of Villani’s lemma

Now we use the following fact (see e.g. Rudin).

dµ |ϕ0(x)| = (x), dλ that relates the ordinary derivative ϕ0 to the Radon-Nikodym dµ derivative dλ (M 6214). Thus λ0 ◦ ϕ = |ϕ0(x)|λ + σ where λ ⊥ σ. 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 λ0(ϕ(A)) > 0.

But Z (λ0 ◦ ϕ)(A) = |ϕ0(x)| dλ(x) + σ(A) > 0, A since |ϕ0(x)| > 0 a.e. We skip the converse since we don’t use it. 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. . . 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 |ϕ0(x)| > a > 0 a.e., then |ϕ−1(y)| < 1/a a.e..

Proof: For a continuous monotonic function ϕ, the inverse function theorem 1 (ϕ−1)0(ϕ(x)) = , ϕ0(x) holds whenever ϕ0(x) exists and is nonzero (see Wade (M 4239/40)). Note that ϕ0(x) exists λ a.e.. Let E = {x : ϕ0(x) ≤ a} By assumption, λ(E) = 0. Absolute continuity vs total singularity Proof of Kakeya’s lemma Proof. . .

Now ϕ(E) = {y : ϕ0(ϕ−1(y)) ≤ a} Let F be the set of y so that (ϕ−1)0(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 then λ((ϕ(E) ∪ F )c) = 1, and the inverse function theorem implies 1 (ϕ−1)0(y) = < 1/a for λ a.e. y. ϕ0(ϕ−1(y))

This follows from the next lemma. Absolute continuity vs total singularity Proof of Kakeya’s lemma Natanson’s lemma

Lemma (Natanson) 0 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 implies λ(E) = 0. This is what happens in our case. Absolute continuity vs total singularity Proof of Kakeya’s lemma Proof of Natanson’s Lemma

Fix > 0. Choose G open with E ⊆ G and

λ(G) < λ∗(E) + .

Also ﬁx a0 > a.

For each x ∈ E there is a sequence hn > 0, hn → 0 so that ϕ(x + h ) − ϕ(x) lim n = ϕ0(x) ≤ a. n→∞ hn 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) < a0, hn and In(x) := [x, x + hn] ⊆ G.

Assume this WOLOG for all n.

Let Jn(x) = ϕ(In(x)) = [ϕ(x), ϕ(x + hn)]. 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 the intervals that cover each point x ∈ E are arbitrarily small).

The Vitali Covering Lemma (see Natanson or Rudin, M 6214) says there is a sequence of disjoint intervals:

En1 (x1),En2 (x2),En3 (x3),...

such that ∞ ! ∗ [ λ E\ Ink (xk) = 0. k=0 Absolute continuity vs total singularity Proof of Kakeya’s lemma Proof of Natanson’s Lemma

Thus ∞ ∞ ∗ X X λ (ϕ(E)) ≤ λ(Jnk (xk)) < a0 λ(Ink (xk)). k=0 k=0

Since Ink (xk) are pairwise disjoint, we have

∞ ∞ ! X [ λ(Ink (xk)) = λ Ink (xk) , k=0 k=0 S∞ and k=0 Ink (xk) ⊆ G. Absolute continuity vs total singularity Proof of Kakeya’s lemma Proof of Natanson’s Lemma

Finally, ∗ λ ∗ (ϕ(E)) < a0λ(G) < a0(λ (E) + ).

The proof is ﬁnished by letting a0 → a and → 0.

λ∗(ϕ(E)) ≤ aλ∗(E). 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. . . Absolute continuity vs total singularity Next. . . 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-Hill Science/Engineering/Math, 3rd ed, (1986) Schweiger, F., Ergodic Theory of Fibred Systems and Metric Number 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)