A Simple Introduction to Ergodic Theory

Karma Dajani and Sjoerd Dirksin

December 18, 2008 2 Contents

1 Introduction and preliminaries 5 1.1 What is Ergodic Theory? ...... 5 1.2 Measure Preserving Transformations ...... 6 1.3 Basic Examples ...... 9 1.4 Recurrence ...... 15 1.5 Induced and Integral Transformations ...... 15 1.5.1 Induced Transformations ...... 15 1.5.2 Integral Transformations ...... 18 1.6 Ergodicity ...... 20 1.7 Other Characterizations of Ergodicity ...... 22 1.8 Examples of Ergodic Transformations ...... 24

2 The Ergodic Theorem 29 2.1 The Ergodic Theorem and its consequences ...... 29 2.2 Characterization of Irreducible Markov Chains ...... 41 2.3 Mixing ...... 45

3 Measure Preserving Isomorphisms and Factor Maps 47 3.1 Measure Preserving Isomorphisms ...... 47 3.2 Factor Maps ...... 51 3.3 Natural Extensions ...... 52

4 Entropy 55 4.1 Randomness and Information ...... 55 4.2 Deﬁnitions and Properties ...... 56 4.3 Calculation of Entropy and Examples ...... 63 4.4 The Shannon-McMillan-Breiman Theorem ...... 66 4.5 Lochs’ Theorem ...... 72

3 4

5 Hurewicz Ergodic Theorem 79 5.1 Equivalent measures ...... 79 5.2 Non-singular and conservative transformations ...... 80 5.3 Hurewicz Ergodic Theorem ...... 83

6 Invariant Measures for Continuous Transformations 89 6.1 Existence ...... 89 6.2 Unique Ergodicity ...... 96

7 Topological Dynamics 101 7.1 Basic Notions ...... 102 7.2 Topological Entropy ...... 108 7.2.1 Two Deﬁnitions ...... 108 7.2.2 Equivalence of the two Deﬁnitions ...... 114 7.3 Examples ...... 118

8 The Variational Principle 127 8.1 Main Theorem ...... 127 8.2 Measures of Maximal Entropy ...... 136 Chapter 1

Introduction and preliminaries

1.1 What is Ergodic Theory?

It is not easy to give a simple definition of Ergodic Theory because it uses techniques and examples from many fields such as probability theory, statistical mechanics, number theory, vector fields on manifolds, group actions of homogeneous spaces and many more. The word ergodic is a mixture of two Greek words: ergon (work) and odos (path). The word was introduced by Boltzmann (in statistical mechanics) regarding his hypothesis: for large systems of interacting particles in equilib- rium, the time average along a single trajectory equals the space average. The hypothesis as it was stated was false, and the investigation for the conditions under which these two quantities are equal lead to the birth of ergodic theory as is known nowadays. A modern description of what ergodic theory is would be: it is the study of the long term average behavior of systems evolving in time. The collection of all states of the system form a space X, and the evolution is represented by either – a transformation T : X → X, where T x is the state of the system at time t = 1, when the system (i.e., at time t = 0) was initially in state x. (This is analogous to the setup of discrete time stochastic processes). – if the evolution is continuous or if the configurations have spacial structure, then we describe the evolution by looking at a group of transformations G (like Z2, R, R2) acting on X, i.e., every g ∈ G is identified with a transformation Tg : X → X, and Tgg0 = Tg ◦ Tg0 .

5 6 Introduction and preliminaries

The space X usually has a special structure, and we want T to preserve the basic structure on X. For example –if X is a measure space, then T must be measurable. –if X is a topological space, then T must be continuous. –if X has a diﬀerentiable structure, then T is a diﬀeomorphism. In this course our space is a probability space (X, B, µ), and our time is discrete. So the evolution is described by a measurable map T : X → X, so that T −1A ∈ B for all A ∈ B. For each x ∈ X, the orbit of x is the sequence

x, T x, T 2x, . . . .

If T is invertible, then one speaks of the two sided orbit

...,T −1x, x, T x, . . . .

We want also that the evolution is in steady state i.e. stationary. In the language of ergodic theory, we want T to be measure preserving.

1.2 Measure Preserving Transformations

Deﬁnition 1.2.1 Let (X, B, µ) be a probability space, and T : X → X measurable. The map T is said to be measure preserving with respect to µ if µ(T −1A) = µ(A) for all A ∈ B.

This deﬁnition implies that for any measurable function f : X → R, the process

f, f ◦ T, f ◦ T 2,... is stationary. This means that for all Borel sets B1,...,Bn, and all integers r1 < r2 < . . . < rn, one has for any k ≥ 1,

r1 rn µ ({x : f(T x) ∈ B1, . . . f(T x) ∈ Bn}) = r1+k rn+k µ {x : f(T x) ∈ B1, . . . f(T x) ∈ Bn} .

In case T is invertible, then T is measure preserving if and only if µ(TA) = µ(A) for all A ∈ B. We can generalize the definition of measure preserving to the following case. Let T :(X1, B1, µ1) → (X2, B2, µ2) be measurable, then −1 T is measure preserving if µ1(T A) = µ2(A) for all A ∈ B2. The following Measure Preserving Transformations 7 gives a useful tool for verifying that a transformation is measure preserving. For this we need the notions of algebra and semi-algebra. Recall that a collection S of subsets of X is said to be a semi-algebra if (i) ∅ ∈ S, (ii) A ∩ B ∈ S whenever A, B ∈ S, and (iii) if A ∈ S, then n X \A = ∪i=1Ei is a disjoint union of elements of S. For example if X = [0, 1), and S is the collection of all subintervals, then S is a semi-algebra. Or if X = {0, 1}Z, then the collection of all cylinder sets {x : xi = ai, . . . , xj = aj} is a semi-algebra. An algebra A is a collection of subsets of X satisfying:(i) ∅ ∈ A, (ii) if A, B ∈ A, then A ∩ B ∈ A, and finally (iii) if A ∈ A, then X \ A ∈ A. Clearly an algebra is a semi-algebra. Furthermore, given a semi-algebra S one can form an algebra by taking all finite disjoint unions of elements of S. We denote this algebra by A(S), and we call it the algebra generated by S. It is in fact the smallest algebra containing S. Likewise, given a semi-algebra S (or an algebra A), the σ-algebra generated by S (A) is denoted by B(S) (B(A)), and is the smallest σ-algebra containing S (or A). A monotone class C is a collection of subsets of X with the following two properties

∞ – if E1 ⊆ E2 ⊆ ... are elements of C, then ∪i=1Ei ∈ C,

∞ – if F1 ⊇ F2 ⊇ ... are elements of C, then ∩i=1Fi ∈ C. The monotone class generated by a collection S of subsets of X is the smallest monotone class containing S.

Theorem 1.2.1 Let A be an algebra of X, then the σ-algebra B(A) generated by A equals the monotone class generated by A.

Using the above Theorem, one can get an easier criterion for checking that a transformation is measure preserving.

Theorem 1.2.2 Let (Xi, Bi, µi) be probability spaces, i = 1, 2, and T : X1 → X2 a transformation. Suppose S2 is a generating semi-algebra of B2. Then, T is measurable and measure preserving if and only if for each A ∈ S2, we −1 −1 have T A ∈ B1 and µ1(T A) = µ2(A).

Proof. Let

−1 −1 C = {B ∈ B2 : T B ∈ B1, and µ1(T B) = µ2(B)}. 8 Introduction and preliminaries

Then, S2 ⊆ C ⊆ B2, and hence A(S2) ⊂ C. We show that C is a monotone ∞ class. Let E1 ⊆ E2 ⊆ ... be elements of C, and let E = ∪i=1Ei. Then, −1 ∞ −1 T E = ∪i=1T Ei ∈ B1.

−1 ∞ −1 −1 µ1(T E) = µ1(∪n=1T En) = lim µ1(T En) n→∞

= lim µ2(En) n→∞ ∞ = µ2(∪n=1En)

= µ2(E).

Thus, E ∈ C. A similar proof shows that if F1 ⊇ F2 ⊇ ... are elements of C, ∞ then ∩i=1Fi ∈ C. Hence, C is a monotone class containing the algebra A(S2). By the monotone class theorem, B2 is the smallest monotone class containing A(S2), hence B2 ⊆ C. This shows that B2 = C, therefore T is measurable and measure preserving. For example if – X = [0, 1) with the Borel σ-algebra B, and µ a probability measure on B. Then a transformation T : X → X is measurable and measure preserving if and only if T −1[a, b) ∈ B and µ (T −1[a, b)) = µ ([a, b)) for any interval [a, b). – X = {0, 1}N with product σ-algebra and product measure µ. A transformation T : X → X is measurable and measure preserving if and only if −1 T ({x : x0 = a0, . . . , xn = an}) ∈ B, and

−1 µ T {x : x0 = a0, . . . , xn = an} = µ ({x : x0 = a0, . . . , xn = an}) for any cylinder set.

Exercise 1.2.1 Recall that if A and B are measurable sets, then

A∆B = (A ∪ B) \ (A ∩ B) = (A \ B) ∪ (B \ A).

Show that for any measurable sets A, B, C one has

µ(A∆B) ≤ µ(A∆C) + µ(C∆B).

Another useful lemma is the following (see also ([KT]). Basic Examples 9

Lemma 1.2.1 Let (X, B, µ) be a probability space, and A an algebra generating B. Then, for any A ∈ B and any > 0, there exists C ∈ A such that µ(A∆C) < .

Proof. Let

D = {A ∈ B : for any > 0, there exists C ∈ A such that µ(A∆C) < }.

Clearly, A ⊆ D ⊆ B. By the Monotone Class Theorem (Theorem (1.2.1)), we need to show that D is a monotone class. To this end, let A1 ⊆ A2 ⊆ · · · S∞ be a sequence in D, and let A = An, notice that µ(A) = lim µ(An). n=1 n→∞ Let > 0, there exists an N such that µ(A∆AN ) = |µ(A) − µ(AN )| < /2. Since AN ∈ D, then there exists C ∈ A such that µ(AN ∆C) < /2. Then,

µ(A∆C) ≤ µ(A∆AN ) + µ(AN ∆C) < . Hence, A ∈ D. Similarly, one can show that D is closed under decreasing in- tersections so that D is a monotone class containg A, hence by the Monotone class Theorem B ⊆ D. Therefore, B = D, and the theorem is proved.

1.3 Basic Examples

(a) Translations – Let X = [0, 1) with the Lebesgue σ-algebra B, and Lebesgue measure λ. Let 0 < θ < 1, deﬁne T : X → X by

T x = x + θ mod 1 = x + θ − bx + θc.

Then, by considering intervals it is easy to see that T is measurable and measure preserving. (b) Multiplication by 2 modulo 1 – Let (X, B, λ) be as in example (a), and let T : X → X be given by 2x 0 ≤ x < 1/2 T x = 2x mod 1 = 2x − 1 1/2 ≤ x < 1.

For any interval [a, b), a b a + 1 b + 1 T −1[a, b) = [ , ) ∪ [ , ), 2 2 2 2 10 Introduction and preliminaries and λ T −1[a, b) = b − a = λ ([a, b)) . Although this map is very simple, it has in fact many facets. For example, iterations of this map yield the binary expansion of points in [0, 1) i.e., using T one can associate with each point in [0, 1) an inﬁnite sequence of 0’s and 1’s. To do so, we deﬁne the function a1 by ( 0 if 0 ≤ x < 1/2 a1(x) = 1 if 1/2 ≤ x < 1,

n−1 then T x = 2x − a1(x). Now, for n ≥ 1 set an(x) = a1(T x). Fix x ∈ X, for simplicity, we write an instead of an(x), then T x = 2x − a1. Rewriting 2 a1 T x a2 T x we get x = 2 + 2 . Similarly, T x = 2 + 2 . Continuing in this manner, we see that for each n ≥ 1, a a a T nx x = 1 + 2 + ··· + n + . 2 22 2n 2n Since 0 < T nx < 1, we get

n n X ai T x x − = → 0 as n → ∞. 2i 2n i=1

P∞ ai Thus, x = i=1 2i . We shall later see that the sequence of digits a1, a2,... forms an i.i.d. sequence of Bernoulli random variables. (c) Baker’s Transformation – This example is the two-dimensional version of example (b). The underlying probability space is [0, 1)2 with product Lebesgue σ-algebra B × B and product Lebesgue measure λ × λ. Deﬁne T : [0, 1)2 → [0, 1)2 by

( y (2x, 2 ) 0 ≤ x < 1/2 T (x, y) = y+1 (2x − 1, 2 ) 1/2 ≤ x < 1. Exercise 1.3.1 Verify that T is invertible, measurable and measure preserving.

(d) β-transformations√ – Let X = [0, 1) with the Lebesgue σ-algebra B. Let 1+ 5 2 β = 2 , the golden mean. Notice that β = β + 1. Deﬁne a transformation Basic Examples 11

T : X → X by

βx 0 ≤ x < 1/β T x = βx mod 1 = βx − 1 1/β ≤ x < 1.

Then, T is not measure preserving with respect to Lebesgue measure (give a counterexample), but is measure preserving with respect to the measure µ given by Z µ(B) = g(x) dx, B where ( √ 5+3 5 0 ≤ x < 1/β 10√ g(x) = 5+ 5 10 1/β ≤ x < 1.

Exercise 1.3.2 Verify that T is measure preserving with respect to µ, and show that (similar to example (b)) iterations of this map generate expansions for points x ∈ [0, 1) (known as β-expansions) of the form

∞ X bi x = , βi i=1 where bi ∈ {0, 1} and bibi+1 = 0 for all i ≥ 1.

(e) Bernoulli Shifts – Let X = {0, 1, . . . k − 1}Z (or X = {0, 1, . . . k − 1}N), F the σ-algebra generated by the cylinders. Let p = (p0, p1, . . . , pk−1) be a positive probability vector, deﬁne a measure µ on F by specifying it on the cylinder sets as follows

µ ({x : x−n = a−n, . . . , xn = an}) = pa−n . . . pan .

Let T : X → X be deﬁned by T x = y, where yn = xn+1. The map T , called the left shift, is measurable and measure preserving, since

−1 T {x : x−n = a−n, . . . xn = an} = {x : x−n+1 = a−n, . . . , xn+1 = an}, and

µ ({x : x−n+1 = a−n, . . . , xn+1 = an}) = pa−n . . . pan . 12 Introduction and preliminaries

Notice that in case X = {0, 1, . . . k − 1}N, then one should consider cylinder sets of the form {x : x0 = a0, . . . xn = an}. In this case

−1 k−1 T {x : x0 = a0, . . . , xn = an} = ∪j=0 {x : x0 = j, x1 = a0, . . . , xn+1 = an}, and it is easy to see that T is measurable and measure preserving. (f) Markov Shifts – Let (X, F,T ) be as in example (e). We deﬁne a measure ν on F as follows. Let P = (pij) be a stochastic k × k matrix, and q = (q0, q1, . . . , qk−1) a positive probability vector such that qP = q. Deﬁne ν on cylinders by

ν ({x : x−n = a−n, . . . xn = an}) = qa−n pa−na−n+1 . . . pan−1an . Just as in example (e), one sees that T is measurable and measure preserving. (g) Stationary Stochastic Processes– Let (Ω, F, IP) be a probability space, and ...,Y−2,Y−1,Y0,Y1,Y2,... a stationary stochastic process on Ω with values in R. Hence, for each k ∈ Z

IP(Yn1 ∈ B1,...,Ynr ∈ Br) = IP(Yn1+k ∈ B1,...,Ynr+k ∈ Br) for any n1 < n2 < ··· < nr and any Lebesgue sets B1,...,Br. We want to see this process as coming from a measure preserving transformation. Let X = RZ = {x = (. . . , x1, x0, x1,...): xi ∈ R} with the product σ-algebra (i.e. generated by the cylinder sets). Let T : X → X be the left shift i.e. T x = z where zn = xn+1. Deﬁne φ :Ω → X by

φ(ω) = (...,Y−2(ω),Y−1(ω),Y0(ω),Y1(ω),Y2(ω),... ).

Then, φ is measurable since if B1,...,Br are Lebesgue sets in R, then −1 −1 −1 φ ({x ∈ X : xn1 ∈ B1, . . . xnr ∈ Br}) = Yn1 (B1) ∩ ... ∩ Ynr (Br) ∈ F. Deﬁne a measure µ on X by µ(E) = IP φ−1(E) . On cylinder sets µ has the form,

µ ({x ∈ X : xn1 ∈ B1, . . . xnr ∈ Br}) = IP(Yn1 ∈ B1,...,Ynr ∈ Br) . Basic Examples 13

Since −1 T ({x : xn1 ∈ B1, . . . xnr ∈ Br}) = {x : xn1+1 ∈ B1, . . . xnr+1 ∈ Br}, stationarity of the process Yn implies that T is measure preserving. Further- th more, if we let πi : X → R be the natural projection onto the i coordinate, i then Yi(ω) = πi(φ(ω)) = π0 ◦ T (φ(ω)). (h) Random Shifts – Let (X, B, µ) be a probability space, and T : X → X an invertible measure preserving transformation. Then, T −1 is measurable and measure preserving with respect to µ. Suppose now that at each moment instead of moving forward by T (x → T x), we first flip a fair coin to decide whether we will use T or T −1. We can describe this random system by means of a measure preserving transformation in the following way. Let Ω = {−1, 1}Z with product σ-algebra F (i.e. the σ-algebra generated by the cylinder sets), and the uniform product measure IP (see example (e)), and let σ :Ω → Ω be the left shift. As in example (e), the map σ is measure preserving. Now, let Y = Ω × X with the product σ-algebra, and product measure IP × µ. Define S : Y → Y by S(ω, x) = (σω, T ω0 x). Then S is invertible (why?), and measure preserving with respect to IP × µ. To see the latter, for any set C ∈ F, and any A ∈ B, we have (IP × µ) S−1(C × A) = (IP × µ)({(ω, x): S(ω, x) ∈ (C × A))

= (IP × µ)({(ω, x): ω0 = 1, σω ∈ C, T x ∈ A) −1 + (IP × µ) {(ω, x): ω0 = −1, σω ∈ C,T x ∈ A −1 −1 = (IP × µ) {ω0 = 1} ∩ σ C × T A −1 + (IP × µ) {ω0 = −1} ∩ σ C × TA −1 −1 = IP {ω0 = 1} ∩ σ C µ T A −1 + IP {ω0 = −1} ∩ σ C µ (TA) −1 = IP {ω0 = 1} ∩ σ C µ(A) −1 + IP {ω0 = −1} ∩ σ C µ(A) = IP(σ−1C)µ(A) = IP(C)µ(A) = (IP × µ)(C × A).

(h) continued fractions – Consider ([0, 1), B), where B is the Lebesgue σ- algebra. Deﬁne a transformation T : [0, 1) → [0, 1) by T 0 = 0 and for x 6= 0 1 1 T x = − b c. x x 14 Introduction and preliminaries

Exercise 1.3.3 Show that T is not measure preserving with respect to Lebesgue measure, but is measure preserving with respect to the so called Gauss probability measure µ given by

Z 1 1 µ(B) = dx. B log 2 1 + x

An interesting feature of this map is that its iterations generate the continued fraction expansion for points in (0, 1). For if we deﬁne

( 1 1 if x ∈ ( 2 , 1) a1 = a1(x) = 1 1 n if x ∈ ( n+1 , n ], n ≥ 2,

1 1 then, T x = x − a1 and hence x = . For n ≥ 1, let an = an(x) = a1 + T x n−1 a1(T x). Then, after n iterations we see that

1 1 x = = ... = . a + T x 1 1 a + 1 . 1 .. a2 + + n an + T x

pn 1 In fact, if = , then one can show that {qn} are mono- qn 1 a1 + .. 1 a2 + . + an tonically increasing, and

pn 1 |x − | < 2 → 0 as n → ∞. qn qn

The last statement implies that

1 x = . 1 a + 1 1 a + 2 1 a + 3 . .. Recurrence 15 1.4 Recurrence

Let T be a measure preserving transformation on a probability space (X, F, µ), and let B ∈ F. A point x ∈ B is said to be B-recurrent if there exists k ≥ 1 such that T kx ∈ B.

Theorem 1.4.1 (Poincar´eRecurrence Theorem) If µ(B) > 0, then a.e. x ∈ B is B-recurrent.

Proof Let F be the subset of B consisting of all elements that are not B- recurrent. Then, F = {x ∈ B : T kx∈ / B for all k ≥ 1}. We want to show that µ(F ) = 0. First notice that F ∩ T −kF = ∅ for all k ≥ 1, hence T −lF ∩ T −mF = ∅ for all l 6= m. Thus, the sets F,T −1F,... are pairwise disjoint, and µ(T −nF ) = µ(F ) for all n ≥ 1 (T is measure preserving). If µ(F ) > 0, then

−k X 1 = µ(X) ≥ µ ∪k≥0T F = µ(F ) = ∞, k≥0 a contradiction. The proof of the above theorem implies that almost every x ∈ B returns to B infinitely often. In other words, there exist infinitely many integers ni n1 < n2 < . . . such that T x ∈ B. To see this, let D = {x ∈ B : T kx ∈ B for finitely many k ≥ 1}. Then, k ∞ −k D = {x ∈ B : T x ∈ F for some k ≥ 0} ⊆ ∪k=0T F. Thus, µ(D) = 0 since µ(F ) = 0 and T is measure preserving.

1.5 Induced and Integral Transformations

1.5.1 Induced Transformations Let T be a measure preserving transformation on the probability space (X, F, µ). Let A ⊂ X with µ(A) > 0. By Poincar´e’sRecurrence Theo- rem almost every x ∈ A returns to A inﬁnitely often under the action of T . 16 Introduction and preliminaries

For x ∈ A, let n(x) := inf{n ≥ 1 : T nx ∈ A}. We call n(x) the ﬁrst return time of x to A.

Exercise 1.5.1 Show that n is measurable with respect to the σ-algebra F ∩ A on A.

By PoincaréTheorem, n(x) is finite a.e. on A. In the sequel we remove from A the set of measure zero on which n(x) = ∞, and we denote the new set again by A. Consider the σ-algebra F ∩ A on A, which is the restriction of F to A. Furthermore, let µA be the probability measure on A, defined by µ(B) µ (B) = , for B ∈ F ∩ A, A µ(A) so that (A, F ∩A, µA) is a probability space. Finally, define the induced map TA : A → A by n(x) TAx = T x , for x ∈ A.

From the above we see that TA is deﬁned on A. What kind of a transformation is TA?

Exercise 1.5.2 Show that TA is measurable with respect to the σ-algebra F ∩ A.

Proposition 1.5.1 TA is measure preserving with respect to µA.

Proof For k ≥ 1, let

Ak = {x ∈ A : n(x) = k}

k−1 k Bk = {x ∈ X \ A : T x, . . . , T x 6∈ A, T x ∈ A}.

S∞ Notice that A = k=1 Ak, and

−1 −1 T A = A1 ∪ B1 and T Bn = An+1 ∪ Bn+1. (1.1)

Let C ∈ F∩A, since T is measure preserving it follows that µ(C) = µ(T −1C). −1 To show that µA(C) = µA(TA C), we show that

−1 −1 µ(TA C) = µ(T C). Induced and Integral Transformations 17

B1 ...... B1 B2 ...... T 2A \ A ∪ TA ≡ (A, 3) B1 B2 B3 ...... TA \ A ≡ (A, 2) B1 B2 B3 B4 ...... A ≡ (A, 1) A1 A2 A3 A4 A5

Figure 1.1: A tower.

Now, ∞ ∞ −1 [ −1 [ −k TA (C) = Ak ∩ TA C = Ak ∩ T C, k=1 k=1 hence ∞ −1 X −k µ TA (C) = µ Ak ∩ T C . k=1 On the other hand, using repeatedly (1.1) , one gets for any n ≥ 1,

−1 −1 −1 µ T (C) = µ(A1 ∩ T C) + µ(B1 ∩ T C) −1 −1 −1 = µ(A1 ∩ T C) + µ(T (B1 ∩ T C)) −1 −2 −2 = µ(A1 ∩ T C) + µ(A2 ∩ T C) + µ(B2 ∩ T C) . . n X −k −n = µ(Ak ∩ T C) + µ(Bn ∩ T C). k=1 Since ∞ ! ∞ [ −n X −n 1 ≥ µ Bn ∩ T C = µ(Bn ∩ T C), n=1 n=1 it follows that −n lim µ(Bn ∩ T C) = 0. n→∞ 18 Introduction and preliminaries

Thus, ∞ −1 X −k −1 µ(C) = µ(T C) = µ Ak ∩ T C = µ(TA C). k=1 −1 This shows that µA(C) = µA(TA C), which implies that TA is measure preserving with respect to µA.

Exercise 1.5.3 Assume T is invertible. Without using Proposition 1.5.1 show that for all C ∈ F ∩ A,

µA(C) = µA(TAC). √ 1 + 5 Exercise 1.5.4 Let G = , so that G2 = G + 1. Consider the set 2 1 [ 1 1 X = [0, ) × [0, 1) [ , 1) × [0, ), G G G endowed with the product Borel σ-algebra, and the normalized Lebesgue measure λ × λ. Deﬁne the transformation

 y 1  (Gx, ), (x, y) ∈ [0, G ) × [0, 1]  G T (x, y) =  1 + y  (Gx − 1, ), (x, y) ∈ [ 1 , 1) × [0, 1 ). G G G (a) Show that T is measure preserving with respect to λ × λ.

(b) Determine explicitely the induced transformation of T on the set [0, 1)× 1 [0, G ).

1.5.2 Integral Transformations Let S be a measure preserving transformation on a probability space (A, F, ν), and let f ∈ L1(A, ν) be positive and integer valued. We now construct a measure preserving transformation T on a probability space (X, C, µ), such that the original transformation S can be seen as the induced transformation on X with return time f.

(1) X = {(y, i): y ∈ A and 1 ≤ i ≤ f(y), i ∈ N}, Induced and Integral Transformations 19

(2) C is generated by sets of the form

(B, i) = {(y, i): y ∈ B and f(y) ≥ i} ,

where B ⊂ A, B ∈ F and i ∈ N. ν(B) (3) µ(B, i) = Z and then extend µ to all of X. f(y) dν(y) A (4) Deﬁne T : X → X as follows: ( (y, i + 1), if i + 1 ≤ f(y), T (y, i) = (Sy, 1), if i + 1 > f(y).

Now (X, C, µ, T ) is called an integral system of (A, F, ν, S) under f. We now show that T is µ-measure preserving. In fact, it suﬃces to check this on the generators. Let B ⊂ A be F-measurable, and let i ≥ 1. We have to discern the following two cases: (1) If i > 1, then T −1(B, i) = (B, i − 1) and clearly

−1 ν(B) µ(T (B, i)) = µ(B, i − 1) = µ(B, i) = R . A f(y) dν(y)

(2) If i = 1, we write An = {y ∈ A : f(y) = n}, and we have

∞ −1 [ −1 T (B, 1) = (An ∩ S B, n) (disjoint union). n=1 S∞ Since n=1 An = A we therefore ﬁnd that

∞ −1 −1 −1 Xν(An ∩ S B) ν(S B) µ(T (B, 1)) = Z = Z n=1 f(y) dν(y) f(y) dν(y) A A

ν(B) = Z = µ(B, 1) . f(y) dν(y) A 20 Introduction and preliminaries

This shows that T is measure preserving. Moreover, if we consider the induced transformation of T on the set (A, 1), then the ﬁrst return time n(x, 1) = inf{k ≥ 1 : T k(x, 1) ∈ (A, 1)} is given by n(x, 1) = f(x), and T(A,1)(x, 1) = (Sx, 1).

1.6 Ergodicity

Deﬁnition 1.6.1 Let T be a measure preserving transformation on a probability space (X, F, µ). The map T is said to be ergodic if for every measurable set A satisfying T −1A = A, we have µ(A) = 0 or 1.

Theorem 1.6.1 Let (X, F, µ) be a probability space and T : X → X measure preserving. The following are equivalent: (i) T is ergodic.

(ii) If B ∈ F with µ(T −1B∆B) = 0, then µ(B) = 0 or 1.

∞ −n (iii) If A ∈ F with µ(A) > 0, then µ (∪n=1T A) = 1. (iv) If A, B ∈ F with µ(A) > 0 and µ(B) > 0, then there exists n > 0 such that µ(T −nA ∩ B) > 0.

Remark 1.6.1 1. In case T is invertible, then in the above characterization one can replace T −n by T n. 2. Note that if µ(B4T −1B) = 0, then µ(B \ T −1B) = µ(T −1B \ B) = 0. Since B = B \ T −1B ∪ B ∩ T −1B , and T −1B = T −1B \ B ∪ B ∩ T −1B , we see that after removing a set of measure 0 from B and a set of measure 0 from T −1B, the remaining parts are equal. In this case we say that B equals T −1B modulo sets of measure 0. 3. In words, (iii) says that if A is a set of positive measure, almost every x ∈ X eventually (in fact inﬁnitely often) will visit A. 4. (iv) says that elements of B will eventually enter A. Ergodicity 21

Proof of Theorem 1.6.1 (i)⇒(ii) Let B ∈ F be such that µ(B∆T −1B) = 0. We shall deﬁne a measurable set C with C = T −1C and µ(C∆B) = 0. Let

∞ ∞ \ [ C = {x ∈ X : T nx ∈ B i.o. } = T −kB. n=1 k=n

Then, T −1C = C, hence by (i) µ(C) = 0 or 1. Furthermore,

∞ ∞ ! ∞ ∞ ! \ [ [ \ µ(C∆B) = µ T −kB ∩ Bc + µ T −kBc ∩ B n=1 k=n n=1 k=n ∞ ! ∞ ! [ [ ≤ µ T −kB ∩ Bc + µ T −kBc ∩ B k=1 k=1 ∞ X ≤ µ T −kB∆B . k=1

Using induction (and the fact that µ(E∆F ) ≤ µ(E∆G)+µ(G∆F )), one can show that for each k ≥ 1 one has µ T −kB∆B = 0. Hence, µ(C∆B) = 0 which implies that µ(C) = µ(B). Therefore, µ(B) = 0 or 1. S∞ −n −1 (ii)⇒(iii) Let µ(A) > 0 and let B = n=1 T A. Then T B ⊂ B. Since T is measure preserving, then µ(B) > 0 and

µ(T −1B∆B) = µ(B \ T −1B) = µ(B) − µ(T −1B) = 0.

Thus, by (ii) µ(B) = 1. (iii)⇒(iv) Suppose µ(A)µ(B) > 0. By (iii)

∞ ! ∞ ! [ [ µ(B) = µ B ∩ T −nA = µ (B ∩ T −nA) > 0. n=1 n=1

Hence, there exists k ≥ 1 such that µ(B ∩ T −kA) > 0. (iv)⇒(i) Suppose T −1A = A with µ(A) > 0. If µ(Ac) > 0, then by (iv) there exists k ≥ 1 such that µ(Ac ∩ T −kA) > 0. Since T −kA = A, it follows that c µ(A ∩ A) > 0, a contradiction. Hence, µ(A) = 1 and T is ergodic. 22 Introduction and preliminaries 1.7 Other Characterizations of Ergodicity

We denote by L0(X, F, µ) the space of all complex valued measurable functions on the probability space (X, F, µ). Let Z Lp(X, F, µ) = {f ∈ L0(X, F, µ): |f|p dµ(x) < ∞}. X

We use the subscript R whenever we are dealing only with real-valued functions. Let (Xi, Fi, µi), i = 1, 2 be two probability spaces, and T : X1 → X2 a mea- −1 sure preserving transformation i.e., µ2(A) = µ1(T A). Deﬁne the induced 0 0 operator UT : L (X2, F2, µ2) → L (X1, F1, µ1) by

UT f = f ◦ T.

The following properties of UT are easy to prove.

Proposition 1.7.1 The operator UT has the following properties:

(i) UT is linear

(ii) UT (fg) = UT (f)UT (g)

(iii) UT c = c for any constant c.

(iv) UT is a positive linear operator

(v) UT 1B = 1B ◦ T = 1T −1B for all B ∈ F2.

R R 0 (vi) UT f dµ1 = f dµ2 for all f ∈ L (X2, F2, µ2), (where if one side X1 X2 doesn’t exist or is inﬁnite, then the other side has the same property).

p p (vii) Let p ≥ 1. Then, UT L (X2, F2, µ2) ⊂ L (X1, F1, µ1), and ||UT f||p = p ||f||p for all f ∈ L (X2, F2, µ2).

Exercise 1.7.1 Prove Proposition 1.7.1

Using the above properties, we can give the following characterization of ergodicity Other Characterizations of Ergodicity 23

Theorem 1.7.1 Let (X, F, µ) be a probability space, and T : X → X measure preserving. The following are equivalent:

(i) T is ergodic.

(ii) If f ∈ L0(X, F, µ), with f(T x) = f(x) for all x, then f is a constant a.e.

(iii) If f ∈ L0(X, F, µ), with f(T x) = f(x) for a.e. x, then f is a constant a.e.

(iv) If f ∈ L2(X, F, µ), with f(T x) = f(x) for all x, then f is a constant a.e.

(v) If f ∈ L2(X, F, µ), with f(T x) = f(x) for a.e. x, then f is a constant a.e.

Proof The implications (iii)⇒(ii), (ii)⇒(iv), (v)⇒(iv), and (iii)⇒(v) are all clear. It remains to show (i)⇒(iii) and (iv)⇒(i). (i)⇒(iii) Suppose f(T x) = f(x) a.e. and assume without any loss of generality that f is real (otherwise we consider separately the real and imaginary parts of f). For each n ≥ 1 and k ∈ Z, let k k + 1 X = {x ∈ X : ≤ f(x) < }. (k,n) 2n 2n

−1 Then, T X(k,n)∆X(k,n) ⊆ {x : f(T x) 6= f(x)} which implies that

−1 µ T X(k,n)∆X(k,n) = 0.

By ergodicity of T , µ(X(k,n)) = 0 or 1, for each k ∈ Z. On the other hand, for each n ≥ 1, we have [ X = X(k,n) (disjoint union). k∈Z Hence, for each n ≥ 1, there exists a unique integer kn such that µ X(kn,n) = k 1. In fact, X ⊇ X ⊇ ..., and { n } is a bounded increasing sequence, (k1,1) (k2,2) 2n 24 Introduction and preliminaries

k hence lim n exists. Let Y = T X , then µ(Y ) = 1. Now, if n→∞ 2n n≥1 (kn,n) k x ∈ Y , then 0 ≤ |f(x) − k /2n| < 1/2n for all n. Hence, f(x) = lim n , n n→∞ 2n and f is a constant on Y .

(iv)⇒(i) Suppose T −1A = A and µ(A) > 0. We want to show that µ(A) = 1. 2 Consider 1A, the indicator function of A. We have 1A ∈ L (X, F, µ), and 1A ◦ T = 1T −1A = 1A. Hence, by (iv), 1A is a constant a.e., hence 1A = 1 a.e. and therefore µ(A) = 1.

1.8 Examples of Ergodic Transformations

Example 1–Irrational Rotations. Consider ([0, 1), B, λ), where B is the Lebesgue σ-algebra, and λ Lebesgue measure. For θ ∈ (0, 1), consider the transformation Tθ : [0, 1) → [0, 1) deﬁned by Tθx = x + θ (mod 1). We have seen in example (a) that Tθ is measure preserving with respect λ. When is Tθ ergodic?

If θ is rational, then Tθ is not ergodic. Consider for example θ = 1/4, then the set A = [0, 1/8) ∪ [1/4, 3/8) ∪ [1/2, 5/8) ∪ [3/4, 7/8) is Tθ-invariant but µ(A) = 1/2.

p Exercise 1.8.1 Suppose θ = with gcd(p, q) = 1. Find a non-trivial T - q θ invariant set. Conclude that Tθ is not ergodic if θ is a rational.

Claim. Tθ is ergodic if and only if θ is irrational. Proof of Claim.

(⇒) The contrapositive statement is given in Exercise 1.8.1 i.e. if θ is rational, then Tθ is not ergodic.

2 (⇐) Suppose θ is irrational, and let f ∈ L (X, B, λ) be Tθ-invariant. Write f in its Fourier series X 2πinx f(x) = ane . n∈Z Examples of Ergodic Transformations 25

Since f(Tθx) = f(x), then

X 2πin(x+θ) X 2πinθ 2πinx f(Tθx) = ane = ane e n∈Z n∈Z X 2πinx = f(x) = ane . n∈Z P 2πinθ 2πinx Hence, n∈Z an(1 − e )e = 0. By the uniqueness of the Fourier 2πinθ coeﬃcients, we have an(1 − e ) = 0 for all n ∈ Z. If n 6= 0, since θ is 2πinθ irrational we have 1 − e 6= 0. Thus, an = 0 for all n 6= 0, and therefore f(x) = a0 is a constant. By Theorem 1.7.1, Tθ is ergodic. Exercise 1.8.2 Consider the probability space ([0, 1), B × B, λ × λ), where as above B is the Lebesgue σ-algebra on [0, 1), and λ normalized Lebesgue measure. Suppose θ ∈ (0, 1) is irrational, and deﬁne Tθ ×Tθ : [0, 1)×[0, 1) → [0, 1) × [0, 1) by

Tθ × Tθ(x, y) = (x + θ mod (1) , y + θ mod (1) ) .

Show that Tθ × Tθ is measure preserving, but is not ergodic.

Example 2–One (or Two) sided shift. Let X = {0, 1, . . . k − 1}N, F the σ- algebra generated by the cylinders, and µ the product measure deﬁned on cylinder sets by

µ ({x : x0 = a0, . . . xn = an}) = pa0 . . . pan , where p = (p0, p1, . . . , pk−1) is a positive probability vector. Consider the left shift T deﬁned on X by T x = y, where yn = xn+1 (See Example (e) in Subsection 1.3). We show that T is ergodic. Let E be a measurable subset of X which is T -invariant i.e., T −1E = E. For any > 0, by Lemma 1.2.1 (see subsection 1.2), there exists A ∈ F which is a ﬁnite disjoint union of cylinders such that µ(E∆A) < . Then |µ(E) − µ(A)| = |µ(E \ A) − µ(A \ E)| ≤ µ(E \ A) + µ(A \ E) = µ(E∆A) < .

Since A depends on ﬁnitely many coordinates only, there exists n0 > 0 such that T −n0 A depends on diﬀerent coordinates than A. Since µ is a product measure, we have

µ(A ∩ T −n0 A) = µ(A)µ(T −n0 A) = µ(A)2. 26 Introduction and preliminaries

Further,

µ(E∆T −n0 A) = µ(T −n0 E∆T −n0 A) = µ(E∆A) < , and µ E∆(A ∩ T −n0 A) ≤ µ(E∆A) + µ(E∆T −n0 A) < 2. Hence,

|µ(E) − µ((A ∩ T −n0 A))| ≤ µ E∆(A ∩ T −n0 A) < 2.

Thus,

Since > 0 is arbitrary, it follows that µ(E) = µ(E)2, hence µ(E) = 0 or 1. Therefore, T is ergodic.

The following lemma provides, in some cases, a useful tool to verify that a measure preserving transformation deﬁned on ([0, 1), B, µ) is ergodic, where B is the Lebesgue σ-algebra, and µ is a probability measure equivalent to Lebesgue measure λ (i.e., µ(A) = 0 if and only if λ(A) = 0).

Lemma 1.8.1 (Knopp’s Lemma) . If B is a Lebesgue set and C is a class of subintervals of [0, 1) satisfying

(a) every open subinterval of [0, 1) is at most a countable union of disjoint elements from C,

(b) ∀A ∈ C , λ(A ∩ B) ≥ γλ(A), where γ > 0 is independent of A, then λ(B) = 1.

Proof The proof is done by contradiction. Suppose λ(Bc) > 0. Given ε > 0 there exists by Lemma 1.2.1 a set Eε that is a ﬁnite disjoint union of open c intervals such that λ(B 4Eε) < ε. Now by conditions (a) and (b) (that is, writing Eε as a countable union of disjoint elements of C) one gets that λ(B ∩ Eε) ≥ γλ(Eε). Examples of Ergodic Transformations 27

Also from our choice of Eε and the fact that

c c c λ(B 4Eε) ≥ λ(B ∩ Eε) ≥ γλ(Eε) ≥ γλ(B ∩ Eε) > γ(λ(B ) − ε), we have that c c γ(λ(B ) − ε) < λ(B 4Eε) < ε . Hence γλ(Bc) < ε + γε, and since ε > 0 is arbitrary, we get a contradiction.

Example 3–Multiplication by 2 modulo 1–Consider ([0, 1), B, λ) be as in Ex- ample (1) above, and let T : X → X be given by

2x 0 ≤ x < 1/2 T x = 2x mod 1 = 2x − 1 1/2 ≤ x < 1,

(see Example (b), subsection 1.3). We have seen that T is measure preserving. We will use Lemma 1.8.1 to show that T is ergodic. Let C be the collection of all intervals of the form [k/2n, (k + 1)/2n) with n ≥ 1 and 0 ≤ k ≤ 2n − 1. Notice that the the set {k/2n : n ≥ 1, 0 ≤ k < 2n − 1} of dyadic rationals is dense in [0, 1), hence each open interval is at most a countable union of disjoint elements of C. Hence, C satisﬁes the ﬁrst hypothesis of Knopp’s Lemma. Now, T n maps each dyadic interval of the form [k/2n, (k + 1)/2n) linearly onto [0, 1), (we call such an interval dyadic of order n); in fact, T nx = 2nx mod(1). Let B ∈ B be T -invariant, and assume λ(B) > 0. Let A ∈ C, and assume that A is dyadic of order n. Then, T nA = [0, 1) and

1 λ(A ∩ B) = λ(A ∩ T −nB) = λ(T nA ∩ B) λ(A) 1 = λ(B) = λ(A)λ(B). 2n Thus, the second hypothesis of Knopp’s Lemma is satisﬁed with γ = λ(B) > 0. Hence, λ(B) = 1. Therefore T is ergodic.

Exercise 1.8.3 Let β > 1 be a non-integer, and consider the transformation Tβ : [0, 1) → [0, 1) given by Tβx = βx mod(1) = βx − bβxc. Use Lemma 1.8.1 to show that Tβ is ergodic with respect to Lebesgue measure λ, i.e. if −1 Tβ A = A, then λ(A) = 0 or 1. 28 Introduction and preliminaries

Example 4–Induced transformations of ergodic transformations– Let T be an ergodic measure preserving transformation on the probability space (X, F, µ), and A ∈ F with µ(A) > 0. Consider the induced transformation TA on n(x) (A, F ∩ A, µA) of T (see subsection 1.5). Recall that TAx = T x, where n(x) := inf{n ≥ 1 : T nx ∈ A}. Let (as before)

Ak = {x ∈ A : n(x) = k}

k−1 k Bk = {x ∈ X \ A : T x, . . . , T x 6∈ A, T x ∈ A}.

Proposition 1.8.1 If T is ergodic on (X, F, µ), then TA is ergodic on (A, F∩ A, µA).

Proof Let C ∈ F ∩ A be such that T −1C = C. We want to show that A S µA(C) = 0 or 1; equivalently, µ(C) = 0 or µ(C) = µ(A). Since A = k≥1 Ak, −1 S −k S −k we have C = TA C = k≥1 Ak ∩ T C. Let E = k≥1 Bk ∩ T C, and −1 F = E ∪C (disjoint union). Recall that (see subsection 1.5) T A = A1 ∪B1, −1 and T Bk = Ak+1 ∪ Bk+1. Hence, T −1F = T −1E ∪ T −1C [ −(k+1) −1 = (Ak+1 ∪ Bk+1) ∩ T C ∪ (A1 ∪ B1) ∩ T C k≥1 [ −k [ −k = (Ak ∩ T C) ∪ (Bk ∩ T C) k≥1 k≥1 = C ∪ E = F. Hence, F is T -invariant, and by ergodicity of T we have µ(F ) = 0 or 1. –If µ(F ) = 0, then µ(C) = 0, and hence µA(C) = 0. –If µ(F ) = 1, then µ(X \ F ) = 0. Since X \ F = (A \ C) ∪ ((X \ A) \ E) ⊇ A \ C, it follows that µ(A \ C) ≤ µ(X \ F ) = 0.

Since µ(A \ C) = µ(A) − µ(C), we have µ(A) = µ(C), i.e., µA(C) = 1.

S −k Exercise 1.8.4 Show that if TA is ergodic and µ k≥1 T A = 1, then, T is ergodic. Chapter 2

The Ergodic Theorem

2.1 The Ergodic Theorem and its consequences

The Ergodic Theorem is also known as Birkhoﬀ’s Ergodic Theorem or the Individual Ergodic Theorem (1931). This theorem is in fact a generalization of the Strong Law of Large Numbers (SLLN) which states that for a sequence Y1,Y2,... of i.i.d. random variables on a probability space (X, F, µ), with E|Yi| < ∞; one has

n 1 X lim Yi = EY1 (a.e.). n→∞ n i=1

For example consider X = {0, 1}N, F the σ-algebra generated by the cylinder sets, and µ the uniform product measure, i.e.,

n µ ({x : x1 = a1, x2 = a2, . . . , xn = an}) = 1/2 .

Suppose one is interested in finding the frequency of the digit 1. More precisely, for a.e. x we would like to find 1 lim #{1 ≤ i ≤ n : xi = 1}. n→∞ n Using the Strong Law of Large Numbers one can answer this question easily. Define 1, if x = 1, Y (x) := i i 0, otherwise.

29 30 The Ergodic Theorem

Since µ is product measure, it is easy to see that Y1,Y2,... form an i.i.d. Bernoulli process, and EYi = E|Yi| = 1/2. Further, #{1 ≤ i ≤ n : xi = 1} = Pn i=1 Yi(x). Hence, by SLLN one has 1 1 lim #{1 ≤ i ≤ n : xi = 1} = . n→∞ n 2 Suppose now we are interested in the frequency of the block 011, i.e., we would like to find 1 lim #{1 ≤ i ≤ n : xi = 0, xi+1 = 1, xi+2 = 1}. n→∞ n We can start as above by defining random variables 1, if x = 0, x = 1, x = 1, Z (x) := i i+1 i+2 i 0, otherwise. Then, n 1 1 X #{1 ≤ i ≤ n : x = 0, x = 1, x = 1} = Z (x). n i i+1 i+2 n i i=1 It is not hard to see that this sequence is stationary but not independent. So one cannot directly apply the strong law of large numbers. Notice that if T n−1 n−1 is the left shift on X, then Yn = Y1 ◦ T and Zn = Z1 ◦ T . In general, suppose (X, F, µ) is a probability space and T : X → X a measure preserving transformation. For f ∈ L1(X, F, µ), we would like to know under n−1 1 X what conditions does the limit lim f(T ix) exist a.e. If it does exist n→∞ n i=0 what is its value? This is answered by the Ergodic Theorem which was originally proved by G.D. Birkhoff in 1931. Since then, several proofs of this important theorem have been obtained; here we present a recent proof given by T. Kamae and M.S. Keane in [KK]. Theorem 2.1.1 (The Ergodic Theorem) Let (X, F, µ) be a probability space and T : X → X a measure preserving transformation. Then, for any f in L1(µ), n−1 1 X lim f(T i(x)) = f ∗(x) n→∞ n i=0 R R ∗ exists a.e., is T -invariant and X f dµ = X f dµ. If moreover T is ergodic, ∗ ∗ R then f is a constant a.e. and f = X f dµ. The Ergodic Theorem and its consequences 31

For the proof of the above theorem, we need the following simple lemma.

Lemma 2.1.1 Let M > 0 be an integer, and suppose {an}n≥0, {bn}n≥0 are sequences of non-negative real numbers such that for each n = 0, 1, 2,... there exists an integer 1 ≤ m ≤ M with

an + ··· + an+m−1 ≥ bn + ··· + bn+m−1. Then, for each positive integer N > M, one has

a0 + ··· + aN−1 ≥ b0 + ··· + bN−M−1.

Proof of Lemma 2.1.1 Using the hypothesis we recursively ﬁnd integers m0 < m1 < . . . < mk < N with the following properties

m0 ≤ M, mi+1 − mi ≤ M for i = 0, . . . , k − 1, and N − mk < M,

a0 + ... + am0−1 ≥ b0 + ... + bm0−1,

am0 + ... + am1−1 ≥ bm0 + ... + bm1−1, . .

amk−1 + ... + amk−1 ≥ bmk−1 + ... + bmk−1. Then,

a0 + ... + aN−1 ≥ a0 + ... + amk−1

≥ b0 + ... + bmk−1 ≥ b0 + . . . bN−M−1. Proof of Theorem 2.1.1 Assume with no loss of generality that f ≥ 0 (otherwise we write f = f + − f −, and we consider each part separately). f (x) Let f (x) = f(x) + ... + f(T n−1x), f(x) = lim sup n , and f(x) = n n→∞ n f (x) lim inf n . Then f and f are T -invariant. This follows from n→∞ n f (T x) f(T x) = lim sup n n→∞ n f (x) n + 1 f(x) = lim sup n+1 · − n→∞ n + 1 n n f (x) = lim sup n+1 = f(x). n→∞ n + 1 32 The Ergodic Theorem

(Similarly f is T -invariant). Now, to prove that f ∗ exists, is integrable and T -invariant, it is enough to show that Z Z Z f dµ ≥ f dµ ≥ f dµ. X X X For since f − f ≥ 0, this would imply that f = f = f ∗. a.e. R R We ﬁrst prove that X fdµ ≤ X f dµ. Fix any 0 < < 1, and let L > 0 be any real number. By deﬁnition of f, for any x ∈ X, there exists an integer m > 0 such that f (x) m ≥ min(f(x),L)(1 − ). m Now, for any δ > 0 there exists an integer M > 0 such that the set

X0 = {x ∈ X : ∃ 1 ≤ m ≤ M with fm(x) ≥ m min(f(x),L)(1 − )} has measure at least 1 − δ. Deﬁne F on X by f(x) x ∈ X F (x) = 0 L x∈ / X0.

n Notice that f ≤ F (why?). For any x ∈ X, let an = an(x) = F (T x), and bn = bn(x) = min(f(x),L)(1 − ) (so bn is independent of n).We now show that {an} and {bn} satisfy the hypothesis of Lemma 2.1.1 with M > 0 as above. For any n = 0, 1, 2,... n –if T x ∈ X0, then there exists 1 ≤ m ≤ M such that

n n fm(T x) ≥ m min(f(T x),L)(1 − ) = m min(f(x),L)(1 − )

= bn + ... + bn+m−1. Hence,

n n+m−1 an + ... + an+m−1 = F (T x) + ... + F (T x) n n+m−1 n ≥ f(T x) + ... + f(T x) = fm(T x)

≥ bn + ... + bn+m−1.

n –If T x∈ / X0, then take m = 1 since

n an = F (T x) = L ≥ min(f(x),L)(1 − ) = bn. The Ergodic Theorem and its consequences 33

Hence by Lemma 2.1.1 for all integers N > M one has

F (x) + ... + F (T N−1x) ≥ (N − M) min(f(x),L)(1 − ).

Integrating both sides, and using the fact that T is measure preserving one gets Z Z N F (x) dµ(x) ≥ (N − M) min(f(x),L)(1 − ) dµ(x). X X Since Z Z F (x) dµ(x) = f(x) dµ(x) + Lµ(X \ X0), X X0 one has Z Z f(x) dµ(x) ≥ f(x) dµ(x) X X0 Z = F (x) dµ(x) − Lµ(X \ X0) X (N − M) Z ≥ min(f(x),L)(1 − ) dµ(x) − Lδ. N X Now letting ﬁrst N → ∞, then δ → 0, then → 0, and lastly L → ∞ one gets together with the monotone convergence theorem that f is integrable, and Z Z f(x) dµ(x) ≥ f(x) dµ(x). X X

We now prove that Z Z f(x) dµ(x) ≤ f(x) dµ(x). X X Fix > 0, for any x ∈ X there exists an integer m such that

f (x) m ≤ (f(x) + ). m For any δ > 0 there exists an integer M > 0 such that the set

Y0 = {x ∈ X : ∃ 1 ≤ m ≤ M with fm(x) ≤ m (f(x) + )} 34 The Ergodic Theorem has measure at least 1 − δ. Deﬁne G on X by

f(x) x ∈ Y G(x) = 0 0 x∈ / Y0.

n Notice that G ≤ f. Let bn = G(T x), and an = f(x)+ (so an is independent of n). One can easily check that the sequences {an} and {bn} satisfy the hypothesis of Lemma 2.1.1 with M > 0 as above. Hence for any M > N, one has G(x) + ... + G(T N−M−1x) ≤ N(f(x) + ). Integrating both sides yields Z Z (N − M) G(x)dµ(x) ≤ N( f(x)dµ(x) + ). X X R Since f ≥ 0, the measure ν deﬁned by ν(A) = A f(x) dµ(x) is absolutely continuous with respect to the measure µ. Hence, there exists δ0 > 0 such that if µ(A) < δ, then ν(A) < δ0. Since µ(X \ Y0) < δ, then ν(X \ Y0) = R f(x)dµ(x) < δ0. Hence, X\Y0 Z Z Z f(x) dµ(x) = G(x) dµ(x) + f(x) dµ(x) X X X\Y0 N Z ≤ (f(x) + ) dµ(x) + δ0. N − M X

Now, let ﬁrst N → ∞, then δ → 0 (and hence δ0 → 0), and ﬁnally → 0, one gets Z Z f(x) dµ(x) ≤ f(x) dµ(x). X X This shows that Z Z Z f dµ ≥ f dµ ≥ f dµ, X X X hence, f = f = f ∗ a.e., and f ∗ is T -invariant. In case T is ergodic, then the T -invariance of f ∗ implies that f ∗ is a constant a.e. Therefore, Z Z f ∗(x) = f ∗(y)dµ(y) = f(y) dµ(y). X X

The Ergodic Theorem and its consequences 35

Remarks (1) Let us study further the limit f ∗ in the case that T is not ergodic. Let I be the sub-σ-algebra of F consisting of all T -invariant subsets A ∈ F. Notice that if f ∈ L1(µ), then the conditional expectation of f given I (denoted by 1 Eµ(f|I)), is the unique a.e. I-measurable L (µ) function with the property that Z Z f(x) dµ(x) = Eµ(f|I)(x) dµ(x) A A −1 ∗ for all A ∈ I i.e., T A = A. We claim that f = Eµ(f|I). Since the limit function f ∗ is T -invariant, it follows that f ∗ is I-measurable. Furthermore, for any A ∈ I, by the ergodic theorem and the T -invariance of 1A,

n−1 n−1 1 X i 1 X i ∗ lim (f1A)(T x) = 1A(x) lim f(T x) = 1A(x)f (x) a.e. n→∞ n n→∞ n i=0 i=0 and Z Z ∗ f1A(x) dµ(x) = f 1A(x) dµ(x). X X ∗ This shows that f = Eµ(f|I).

(2) Suppose T is ergodic and measure preserving with respect to µ, and let ν be a probability measure which is equivalent to µ (i.e. µ and ν have the same sets of measure zero so µ(A) = 0 if and only if ν(A) = 0), then for every f ∈ L1(µ) one has

n−1 1 X Z lim f(T i(x)) = f dµ n→∞ n i=0 X ν a.e.

Exercise 2.1.1 (Kac’s Lemma) Let T be a measure preserving and ergodic transformation on a probability space (X, F, µ). Let A be a measurable subset of X of positive µ measure, and denote by n the ﬁrst return time map and let TA be the induced transformation of T on A (see section 1.5). Prove that Z n(x) dµ = 1. A 36 The Ergodic Theorem

1 Conclude that n(x) ∈ L (A, µA), and that

n−1 1 X i 1 lim n(TA(x)) = , n→∞ n µ(A) i=0 almost everywhere on A.

√ 1 + 5 Exercise 2.1.2 Let β = , and consider the transformation T : 2 β [0, 1) → [0, 1) given by Tβx = βx mod(1) = βx − bβxc. Deﬁne b1 on [0, 1) by 0 if 0 ≤ x < 1/β b (x) = 1 1 if 1/β ≤ x < 1, Fix k ≥ 0. Find the a.e. value (with respect to Lebesgue measure) of the following limit

1 lim #{1 ≤ i ≤ n : bi = 0, bi+1 = 0, . . . , bi+k = 0}. n→∞ n

Using the Ergodic Theorem, one can give yet another characterization of ergodicity.

Corollary 2.1.1 Let (X, F, µ) be a probability space, and T : X → X a measure preserving transformation. Then, T is ergodic if and only if for all A, B ∈ F, one has

n−1 1 X lim µ(T −iA ∩ B) = µ(A)µ(B). (2.1) n→∞ n i=0

Proof Suppose T is ergodic, and let A, B ∈ F. Since the indicator function 1 1A ∈ L (X, F, µ), by the ergodic theorem one has

n−1 Z 1 X i lim 1A(T x) = 1A(x) dµ(x) = µ(A) a.e. n→∞ n i=0 X The Ergodic Theorem and its consequences 37

Then,

n−1 n−1 1 X 1 X lim 1T −iA∩B(x) = lim 1T −iA(x)1B(x) n→∞ n n→∞ n i=0 i=0 n−1 1 X i = 1B(x) lim 1A(T x) n→∞ n i=0

= 1B(x)µ(A) a.e. 1 Pn−1 Since for each n, the function limn→∞ n i=0 1T −iA∩B is dominated by the constant function 1, it follows by the dominated convergence theorem that n Z n−1 1 X −i 1 X lim µ(T A ∩ B) = lim 1T −iA∩B(x) dµ(x) n→∞ n n→∞ n i=0 X i=0 Z = 1Bµ(A) dµ(x) = µ(A)µ(B). X Conversely, suppose (2.1) holds for every A, B ∈ F. Let E ∈ F be such that T −1E = E and µ(E) > 0. By invariance of E, we have µ(T −iE ∩E) = µ(E), hence n−1 1 X lim µ(T −iE ∩ E) = µ(E). n→∞ n i=0 On the other hand, by (2.1)

n−1 1 X lim µ(T −iE ∩ E) = µ(E)2. n→∞ n i=0 Hence, µ(E) = µ(E)2. Since µ(E) > 0, this implies µ(E) = 1. Therefore, T is ergodic. To show ergodicity one needs to verify equation (2.1) for sets A and B belonging to a generating semi-algebra only as the next proposition shows.

Proposition 2.1.1 Let (X, F, µ) be a probability space, and S a generating semi-algebra of F. Let T : X → X be a measure preserving transformation. Then, T is ergodic if and only if for all A, B ∈ S, one has

n−1 1 X lim µ(T −iA ∩ B) = µ(A)µ(B). (2.2) n→∞ n i=0 38 The Ergodic Theorem

Proof We only need to show that if (2.2) holds for all A, B ∈ S, then it holds for all A, B ∈ F. Let > 0, and A, B ∈ F. Then, by Lemma 1.2.1 (subsection 1.2) there exist sets A0,B0 each of which is a ﬁnite disjoint union of elements of S such that

µ(A∆A0) < , and µ(B∆B0) < .

Since,

−i −i −i −i (T A ∩ B)∆(T A0 ∩ B0) ⊆ (T A∆T A0) ∪ (B∆B0), it follows that

−i −i −i −i |µ(T A ∩ B) − µ(T A0 ∩ B0)| ≤ µ (T A ∩ B)∆(T A0 ∩ B0) −i −i ≤ µ(T A∆T A0) + µ(B∆B0) < 2.

Further,

≤ |µ(B) − µ(B0)| + |µ(A) − µ(A0)|

≤ µ(B∆B0) + µ(A∆A0) < 2.

Hence,

n−1 ! n−1 ! 1 X 1 X µ(T −iA ∩ B) − µ(A)µ(B) − µ(T −iA ∩ B ) − µ(A )µ(B ) n n 0 0 0 0 i=0 i=0 n−1 1 X −i −i ≤ µ(T A ∩ B) + µ(T A0 ∩ B0) − |µ(A)µ(B) − µ(A0)µ(B0)| n i=0 < 4.

Therefore, " n−1 # 1 X lim µ(T −iA ∩ B) − µ(A)µ(B) = 0. n→∞ n i=0 The Ergodic Theorem and its consequences 39

Theorem 2.1.2 Suppose µ1 and µ2 are probability measures on (X, F), and T : X → X is measurable and measure preserving with respect to µ1 and µ2. Then,

(i) if T is ergodic with respect to µ1, and µ2 is absolutely continuous with respect to µ1, then µ1 = µ2,

(ii) if T is ergodic with respect to µ1 and µ2, then either µ1 = µ2 or µ1 and µ2 are singular with respect to each other.

Proof (i) Suppose T is ergodic with respect to µ1 and µ2 is absolutely continuous with respect to µ1. For any A ∈ F, by the ergodic theorem for a.e. x one has n−1 1 X i lim 1A(T x) = µ1(A). n→∞ n i=0 Let n−1 1 X i CA = {x ∈ X : lim 1A(T x) = µ1(A)}, n→∞ n i=0 then µ1(CA) = 1, and by absolute continuity of µ2 one has µ2(CA) = 1. Since T is measure preserving with respect to µ2, for each n ≥ 1 one has

n−1 1 X Z 1 (T ix) dµ (x) = µ (A). n A 2 2 i=0 X

On the other hand, by the dominated convergence theorem one has

Z n−1 Z 1 X i lim 1A(T x)dµ2(x) = µ1(A) dµ2(x). n→∞ n X i=0 X

This implies that µ1(A) = µ2(A). Since A ∈ F is arbitrary, we have µ1 = µ2.

(ii) Suppose T is ergodic with respect to µ1 and µ2. Assume that µ1 6= µ2. Then, there exists a set A ∈ F such that µ1(A) 6= µ2(A). For i = 1, 2 let

n−1 1 X j Ci = {x ∈ X : lim 1A(T x) = µi(A)}. n→∞ n j=0 40 The Ergodic Theorem

By the ergodic theorem µi(Ci) = 1 for i = 1, 2. Since µ1(A) 6= µ2(A), then C1 ∩ C2 = ∅. Thus µ1 and µ2 are supported on disjoint sets, and hence µ1 and µ2 are mutually singular. We end this subsection with a short discussion that the assumption of ergodicity is not very restrictive. Let T be a transformation on the probability space (X, F, µ), and suppose T is measure preserving but not necessarily ergodic. We assume that X is a complete separable metric space, and F the corresponding Borel σ-algebra (in order to make sure that the conditional expectation is well-deﬁned a.e.). Let I be the sub-σ-algebra of T -invariant measurable sets. We can decompose µ into T -invariant ergodic components in the following way. For x ∈ X, deﬁne a measure µx on F by

µx(A) = Eµ(1A|I)(x). Then, for any f ∈ L1(X, F, µ), Z f(y) dµx(y) = Eµ(f|I)(x). X Note that Z Z µ(A) = Eµ(1A|I)(x) dµ(x) = µx(A) dµ(x), X X and that Eµ(1A|I)(x) is T -invariant. We show that µx is T -invariant and ergodic for a.e. x ∈ X. So let A ∈ F, then for a.e. x −1 µx(T A) = Eµ(1A ◦ T |I)(x) = Eµ(IA|I)(T x) = Eµ(IA|I)(x) = µx(A). −1 Now, let A ∈ F be such that T A = A. Then, 1A is T -invariant, and hence I-measurable. Then,

µx(A) = Eµ(1A|I)(x) = 1A(x) a.e. Hence, for a.e. x and for any B ∈ F,

µx(A ∩ B) = Eµ(1A1B|I)(x) = 1A(x)Eµ(1B|I)(x) = µx(A)µx(B). 2 In particular, if A = B, then the latter equality yields µx(A) = µx(A) which implies that for a.e. x, µx(A) = 0 or 1. Therefore, µx is ergodic. (One in fact needs to work a little harder to show that one can ﬁnd a set N of µ-measure zero, such that for any x ∈ X \ N, and any T -invariant set A, one has µx(A) = 0 or 1. In the above analysis the a.e. set depended on the choice of A. Hence, the above analysis is just a rough sketch of the proof of what is called the ergodic decomposition of measure preserving transformations.) Characterization of Irreducible Markov Chains 41 2.2 Characterization of Irreducible Markov Chains

Consider the Markov Chain in Example(f) subsection 1.3. That is X = {0, 1,...N − 1}Z, F the σ-algebra generated by the cylinders, T : X → X the left shift, and µ the Markov measure deﬁned by the stochastic N × N matrix P = (pij), and the positive probability vector π = (π0, π1, . . . , πN−1) satisfying πP = π. That is

µ({x : x0 = i0, x1 = i1, . . . xn = in}) = πi0 pi0i1 pi1i2 . . . pin−1in . We want to find necessary and sufficient conditions for T to be ergodic. To achieve this, we first set

n−1 1 X Q = lim P k, n→∞ n k=0

k (k) th 0 where P = (pij ) is the k power of the matrix P , and P is the k × k identity matrix. More precisely, Q = (qij), where

n−1 1 X (k) qij = lim pij . n→∞ n k=0

1 Pn−1 (k) Lemma 2.2.1 For each i, j ∈ {0, 1,...N −1}, the limit limn→∞ n k=0 pij exists, i.e., qij is well-deﬁned.

Proof For each n,

n−1 n−1 1 X (k) 1 1 X pij = µ({x ∈ X : x0 = i, xk = j}). n πi n k=0 k=0 Since T is measure preserving, by the ergodic theorem,

n−1 n−1 1 X 1 X k ∗ lim 1{x:x =j}(x) = lim 1{x:x =j}(T x) = f (x), n→∞ n k n→∞ n 0 k=0 k=0 where f ∗ is T -invariant and integrable. Then,

n−1 n−1 1 X 1 X k ∗ lim 1{x:x =i,x =j}(x) = 1{x:x =i}(x) lim 1{x:x =j}(T x) = f (x)1{x:x =i}(x). n→∞ n 0 k 0 n→∞ n 0 0 k=0 k=0 42 The Ergodic Theorem

1 Pn−1 Since n k=0 1{x:x0=i,xk=j}(x) ≤ 1 for all n, by the dominated convergence theorem,

n−1 1 1 X qij = lim µ({x ∈ X : x0 = i, xk = j}) πi n→∞ n k=0 n−1 1 Z 1 X = lim 1{x:x0=i,xk=j}(x) dµ(x) πi n→∞ n X k=0 Z 1 ∗ = f (x)1{x:x0=i}(x) dµ(x) πi X 1 Z = f ∗(x) dµ(x) πi {x:x0=i} ∗ which is ﬁnite since f is integrable. Hence qij exists. Exercise 2.2.1 Show that the matrix Q has the following properties: (a) Q is stochastic. (b) Q = QP = PQ = Q2. (c) πQ = π.

We now give a characterization for the ergodicity of T . Recall that the matrix P is said to be irreducible if for every i, j ∈ {0, 1,...N − 1}, there (n) exists n ≥ 1 such that pij > 0. Theorem 2.2.1 The following are equivalent, (i) T is ergodic. (ii) All rows of Q are identical.

(iii) qij > 0 for all i, j. (iv) P is irreducible. (v) 1 is a simple eigenvalue of P .

Proof (i) ⇒ (ii) By the ergodic theorem for each i, j,

n−1 1 X lim 1{x:x =i,x =j}(x) = 1{x:x =i}(x)πj. n→∞ n 0 k 0 k=0 Characterization of Irreducible Markov Chains 43

By the dominated convergence theorem,

n−1 1 X lim µ({x ∈ X : x0 = i, xk = j}) = πiπj. n→∞ n k=0 Hence, n−1 1 1 X qij = lim µ({x ∈ X : x0 = i, xk = j}) = πj, πi n→∞ n k=0 i.e., qij is independent of i. Therefore, all rows of Q are identical. (ii) ⇒ (iii) If all the rows of Q are identical, then all the columns of Q are constants. Thus, for each j there exists a constant cj such that qij = cj for all i. Since πQ = π, it follows that qij = cj = πj > 0 for all i, j. (iii) ⇒ (iv) For any i, j

n−1 1 X (k) lim pij = qij > 0. n→∞ n k=0

(n) Hence, there exists n such that pij > 0, therefore P is irreducible. (iv) ⇒ (iii) Suppose P is irreducible. For any state i ∈ {0, 1,...,N − 1}, let Si = {j : qij > 0}. Since Q is a stochastic matrix, it follows that Si 6= ∅. Let n l ∈ Si, then qil > 0. Since Q = QP = QP for all n, then for any state j

N−1 X (n) (n) qij = qimpmj ≥ qilplj m=0

(n) for any n. Since P is irreducible, there exists n such that plj > 0. Hence, qij > 0 for all i, j.

(iii) ⇒ (ii) Suppose qij > 0 for all j = 0, 1,...,N − 1. Fix any state j, and let qj = max0≤i≤N−1 qij. Suppose that not all the qij’s are the same. Then there exists k ∈ {0, 1,...,N − 1} such that qkj < qj. Since Q is stochastic and Q2 = Q, then for any i ∈ {0, 1,...,N − 1} we have,

N−1 N−1 X X qij = qilqlj < qilqj = qj. l=0 l=0 44 The Ergodic Theorem

This implies that qj = max0≤i≤N−1 qij < qj, a contradiction. Hence, the columns of Q are constants, or all the rows are identical. (ii) ⇒ (i) Suppose all the rows of Q are identical. We have shown above that this implies qij = πj for all i, j ∈ {0, 1,...,N − 1}. Hence πj = 1 Pn−1 (k) limn→∞ n k=0 pij . Let

A = {x : xr = i0, . . . , xr+l = il}, and B = {x : xs = j0, . . . , xs+m = jm} be any two cylinder sets of X. By Proposition 2.1.1 in Section 2, we must show that n−1 1 X lim µ(T −iA ∩ B) = µ(A)µ(B). n→∞ n i=0 Since T is the left shift, for all n sufficiently large, the cylinders T −nA and B depend on different coordinates. Hence, for n sufficiently large,

µ(T −nA ∩ B) = π p . . . p p(n+r−s−m)p . . . p . j0 j0j1 jm−1jm jmi0 i0i1 il−1il Thus,

n−1 1 X lim µ(T −kA ∩ B) n→∞ n k=0 n−1 1 X (k) = πj0 pj0j1 . . . pjm−1jm pi0i1 . . . pil−1il lim pj i n→∞ n m 0 k=0

= (πj0 pj0j1 . . . pjm−1jm )(πi0 pi0i1 . . . pil−1il ) = µ(B)µ(A).

Therefore, T is ergodic.

(ii) ⇒ (v) If all the rows of Q are identical, then qij = πj for all i, j. If PN−1 vP = v, then vQ = v. This implies that for all j, vj = ( i=0 vi)πj. Thus, v is a multiple of π. Therefore, 1 is a simple eigenvalue. ∗ (v) ⇒ (ii) Suppose 1 is a simple eigenvalue. For any i, let qi = (qi0, . . . , qi(N−1)) th ∗ denote the i row of Q then, qi is a probability vector. From Q = QP , we get ∗ ∗ qi = qi P . By hypothesis π is the only probability vector satisfying πP = P , ∗ hence π = qi , and all the rows of Q are identical. Characterization of Irreducible Markov Chains 45 2.3 Mixing

As a corollary to the ergodic theorem we found a new deﬁnition of ergodicity; namely, asymptotic average independence. Based on the same idea, we now deﬁne other notions of weak independence that are stronger than ergodicity.

Deﬁnition 2.3.1 Let (X, F, µ) be a probability space, and T : X → X a measure preserving transformation. Then, (i) T is weakly mixing if for all A, B ∈ F, one has

n−1 1 X −i lim µ(T A ∩ B) − µ(A)µ(B) = 0. (2.3) n→∞ n i=0

(ii) T is strongly mixing if for all A, B ∈ F, one has

lim µ(T −iA ∩ B) = µ(A)µ(B). (2.4) n→∞

Notice that strongly mixing implies weakly mixing, and weakly mixing implies ergodicity. This follows from the simple fact that if {an} is a sequence n−1 1 X of real numbers such that lim a = 0, then lim |a | = 0, and n→∞ n n→∞ n i i=0 n−1 1 X hence lim a = 0. Furthermore, if {a } is a bounded sequence, n→∞ n i n i=0 then the following are equivalent:

n−1 1 X (i) lim |a | = 0 n→∞ n i i=0

n−1 1 X (ii) lim |a |2 = 0 n→∞ n i i=0 (iii) there exists a subset J of the integers of density zero, i.e. 1 lim #({0, 1, . . . , n − 1} ∩ J) = 0, n→∞ n

such that limn→∞,n/∈J an = 0. 46 The Ergodic Theorem

Using this one can give three equivalent characterizations of weakly mixing transformations, can you state them?

Exercise 2.3.1 Let (X, F, µ) be a probability space, and T : X → X a measure preserving transformation. Let S be a generating semi-algebra of F.

(a) Show that if equation (2.3) holds for all A, B ∈ S, then T is weakly mixing.

(b) Show that if equation (2.4) holds for all A, B ∈ S, then T is strongly mixing.

Exercise 2.3.2 Consider the one or two-sided Bernoulli shift T as given in Example (e) in subsection 1.3, and Example (2) in subsection 1.8. Show that T is strongly mixing.

Exercise 2.3.3 Let (X, F, µ) be a probability space, and T : X → X a measure preserving transformation. Consider the transformation T × T de- ﬁned on (X × X, F × F, µ × µ) by T × T (x, y) = (T x, T y).

(a) Show that T × T is measure preserving with respect to µ × µ.

(b) Show that T × T is ergodic, if and only if T is weakly mixing. Chapter 3

Measure Preserving Isomorphisms and Factor Maps

3.1 Measure Preserving Isomorphisms

Given a measure preserving transformation T on a probability space (X, F, µ), we call the quadruple (X, F, µ, T ) a dynamical system . Now, given two dynamical systems (X, F, µ, T ) and (Y, C, ν, S), what should we mean by: these systems are the same? On each space there are two important structures:

(1) The measure structure given by the σ-algebra and the probability measure. Note, that in this context, sets of measure zero can be ignored.

(2) The dynamical structure, given by a measure preserving transformation.

So our notion of being the same must mean that we have a map

ψ :(X, F, µ, T ) → (Y, C, ν, S) satisfying

(i) ψ is one-to-one and onto a.e. By this we mean, that if we remove a (suitable) set NX of measure 0 in X, and a (suitable) set NY of measure 0 in Y , the map ψ : X \ NX → Y \ NY is a bijection.

(ii) ψ is measurable, i.e., ψ−1(C) ∈ F, for all C ∈ C.

47 48 Measure Preserving Isomorphism and Factor Maps

(iii) ψ preserves the measures: ν = µ ◦ ψ−1, i.e., ν(C) = µ (ψ−1(C)) for all C ∈ C.

Finally, we should have that

(iv) ψ preserves the dynamics of T and S, i.e., ψ ◦ T = S ◦ ψ, which is the same as saying that the following diagram commutes.

T ...... NN...... ψ. . ψ ...... 0...... S ...... 0 N ..... N

This means that T -orbits are mapped to S-orbits:

x → T x → T 2x → · · · → T nx → ↓ ↓ ↓ ↓ ↓ ψ(x) → S (ψ(x)) → S2 (ψ(x)) → · · · → Sn (ψ(x)) →

Definition 3.1.1 Two dynamical systems (X, F, µ, T ) and (Y, C, ν, S) are isomorphic if there exist measurable sets N ⊂ X and M ⊂ Y with µ(X\N) = ν(Y \ M) = 0 and T (N) ⊂ N,S(M) ⊂ M, and finally if there exists a measurable map ψ : N → M such that (i)–(iv) are satisfied.

Exercise 3.1.1 Suppose (X, F, µ, T ) and (Y, C, ν, S) are two isomorphic dynamical systems. Show that

(a) T is ergodic if and only if S is ergodic.

(b) T is weakly mixing if and only if S is weakly mixing.

Examples (1) Let K = {z ∈ C : |z| = 1} be equipped with the Borel σ-algebra B on K, and Haar measure (i.e., normalized Lebeque measure on the unit circle). Measure Preserving Isomorphisms 49

Define S : K → K by Sz = z2; equivalently Se2πiθ = e2πi(2θ).. One can easily check that S is measure preserving. In fact, the map S is isomorphic to the map T on ([0, 1), B, λ) given by T x = 2x (mod 1) (see Example (b) in subsection 1.3, and Example (3) in subsection 1.8). Define a map φ : [0, 1) → K by φ(x) = e2πix. We leave it to the reader to check that φ is a measurable isomorphism, i.e., φ is a measurable and measure preserving bijection such that Sφ(x) = φ(T x) for all x ∈ [0, 1). (2) Consider ([0, 1), B, λ), the unit interval with the Lebesgue σ-algebra, and Lebesgue measure. Let T : [0, 1) → [0, 1) be given by T x = Nx − bNxc. Iterations of T generate the N-adic expansion of points in the unit interval. Let Y := {0, 1,...,N − 1}N, the set of all sequences (yn)n≥1, with yn ∈ {0, 1,...,N − 1} for n ≥ 1. We now construct an isomorphism between ([0, 1), B, λ, T ) and (Y, F, µ, S), where F is the σ-algebra generated by the cylinders, and µ the uniform product measure defined on cylinders by 1 µ({(y ) ∈ Y : y = a , y = a , . . . , y = a }) = , i i≥1 1 1 2 2 n n N n for any (a1, a2, a3,...) ∈ Y , and where S is the left shift. Define ψ : [0, 1) → Y = {0, 1,...,N − 1}N by

∞ X ak ψ : x = 7→ (a ) , N k k k≥1 k=1

P∞ k where k=1 ak/N is the N-adic expansion of x (for example if N = 2 we get the binary expansion, and if N = 10 we get the decimal expansion). Let

C(i1, . . . , in) = {(yi)i≥1 ∈ Y : y1 = i1, . . . , yn = in} . In order to see that ψ is an isomorphism one needs to verify measurability and measure preservingness on cylinders: i i i i i i + 1 ψ−1 (C(i , . . . , i )) = 1 + 2 + ··· + n , 1 + 2 + ··· + n 1 n N N 2 N n N N 2 N n and 1 λ ψ−1(C(i , . . . , i )) = = µ (C(i , . . . , i ))) . 1 n N n 1 n Note that

N = {(yi)i≥1 ∈ Y : there exists a k ≥ 1 such that yi = N −1 for all i ≥ k} 50 Measure Preserving Isomorphism and Factor Maps is a subset of Y of measure 0. Setting Y˜ = Y \N , then ψ : [0, 1) → Y˜ is a bijection, since every x ∈ [0, 1) has a unique N-adic expansion generated by T . Finally, it is easy to see that ψ ◦ T = S ◦ ψ. Exercise 3.1.2 Consider ([0, 1)2, B × B, λ × λ), where B × B is the product Lebesgue σ-algebra, and λ × λ is the product Lebesgue measure Let T : [0, 1)2 → [0, 1)2 be given by  1  (2x, y), 0 ≤ x < 1  2 2 T (x, y) =  1  (2x − 1, (y + 1)), 1 ≤ x < 1.  2 2 Show that T is isomorphic to the two-sided Bernoulli shift S on {0, 1}Z, F, µ, where F is the σ-algebra generated by cylinders of the form

∆ = {x−k = a−k, . . . , x` = a` : ai ∈ {0, 1}, i = −k, . . . , `}, k, ` ≥ 0,

1 1 1 k+`+1 and µ the product measure with weights ( 2 , 2 ) (so µ(∆) = ( 2 ) ). √ 1 + 5 Exercise 3.1.3 Let G = , so that G2 = G + 1. Consider the set 2 1 [ 1 1 X = [0, ) × [0, 1) [ , 1) × [0, ), G G G endowed with the product Borel σ-algebra. Deﬁne the transformation

 y 1  (Gx, ), (x, y) ∈ [0, G ) × [0, 1]  G T (x, y) =  1 + y  (Gx − 1, ), (x, y) ∈ [ 1 , 1) × [0, 1 ). G G G (a) Show that T is measure preserving with respect to normalized Lebesgue measure on X. (b) Now let S : [0, 1) × [0, 1) → [0, 1) × [0, 1) be given by

 y 1  (Gx, ), (x, y) ∈ [0, G ) × [0, 1]  G S(x, y) =  G + y  (G2x − G, ), (x, y) ∈ [ 1 , 1) × [0, 1). G2 G Factor Maps 51

Show that S is measure preserving with respect to normalized Lebesgue measure on [0, 1) × [0, 1).

1 (c) Let Y = [0, 1) × [0, G ), and let U be the induced transformation of T on Y , i.e., for (x, y) ∈ Y , U(x, y) = T n(x,y), where n(x, y) = inf{n ≥ 1 : T n(x, y) ∈ Y }. Show that the map φ : [0, 1) × [0, 1) → Y given by y φ(x, y) = (x, ) G deﬁnes an isomorphism from S to U, where Y has the induced measure structure (see Section 1.5).

3.2 Factor Maps

In the above section, we discussed the notion of isomorphism which describes when two dynamical systems are considered the same. Now, we give a precise deﬁnition of what it means for a dynamical system to be a subsystem of another one.

Definition 3.2.1 Let (X, F, µ, T ) and (Y, C, ν, S) be two dynamical systems. We say that S is a factor of T if there exist measurable sets M1 ∈ F and M2 ∈ C, such that µ(M1) = ν(M2) = 1 and T (M1) ⊂ M1,S(M2) ⊂ M2, and finally if there exists a measurable and measure preserving map ψ : M1 → M2 which is surjective, and satisfies ψ(T (x)) = S(ψ(x)) for all x ∈ M1. We call ψ a factor map.

Remark Notice that if ψ is a factor map, then G = ψ−1C is a T -invariant sub-σ-algebra of F, since

T −1G = T −1ψ−1C = ψ−1S−1C ⊆ ψ−1C = G.

Examples Let T be the Baker’s transformation on ([0, 1)2, B × B, λ × λ), given by  (2x, 1 y), 0 ≤ x < 1  2 2 T (x, y) =  1 1 (2x − 1, 2 (y + 1)), 2 ≤ x < 1, 52 Measure Preserving Isomorphism and Factor Maps and let S be the left shift on X = {0, 1}N with the σ-algebra F generated by the cylinders, and the uniform product measure µ. Deﬁne ψ : [0, 1)×[0, 1) → X by ψ(x, y) = (a1, a2,...), a where x = P∞ n is the binary expansion of x. It is easy to check that ψ n=1 2n is a factor map.

Exercise 3.2.1 Let T be the left shift on X = {0, 1, 2}N which is endowed with the σ-algebra F, generated by the cylinder sets, and the uniform product measure µ giving each symbol probability 1/3, i.e., 1 µ ({x ∈ X : x = i , x = i , . . . , x = i }) = , 1 1 2 2 n n 3n where i1, i2, . . . , in ∈ {0, 1, 2}. Let S be the left shift on Y = {0, 1}N which is endowed with the σ-algebra G, generated by the cylinder sets, and the product measure ν giving the symbol 0 probability 1/3 and the symbol 1 probability 2/3, i.e., 2 1 µ ({y ∈ Y : y = j , y = j , . . . , y = j }) = ( )j1+j2+...+jn ( )n−(j1+j2+...+jn), 1 1 2 2 n n 3 3 where j1, j2, . . . , jn ∈ {0, 1}. Show that S is a factor of T .

Exercise 3.2.2 Show that a factor of an ergodic (weakly mixing/strongly mixing) transformation is also ergodic (weakly mixing/strongly mixing).

3.3 Natural Extensions

Suppose (Y, G, ν, S) is a non-invertible measure-preserving dynamical system. An invertible measure-preserving dynamical system (X, F, µ, T ) is called a natural extension of (Y, G, ν, S) if S is a factor of T and the factor map ψ ∞ m −1 satisﬁes ∨m=0T ψ G = F, where

∞ _ T kψ−1G k=0 is the smallest σ-algebra containing the σ-algebras T kψ−1G for all k ≥ 0. Natural Extensions 53

Example Let T on {0, 1}Z, F, µ be the two-sided Bernoulli shift, and S on {0, 1}N∪{0}, G, ν be the one-sided Bernoulli shift, both spaces are endowed with the uniform product measure. Notice that T is invertible, while S is not. Set X = {0, 1}Z, Y = {0, 1}N∪{0}, and deﬁne ψ : X → Y by

ψ (. . . , x−1, x0, x1,...) = (x0, x1,...) .

Then, ψ is a factor map. We claim that

∞ _ T kψ−1G = F. k=0

W∞ k −1 To prove this, we show that k=0 T ψ G contains all cylinders generating F. Let ∆ = {x ∈ X : x−k = a−k, . . . , x` = a`} be an arbitrary cylinder in F, and let D = {y ∈ Y : y0 = a−k, . . . , yk+` = a`} which is a cylinder in G. Then,

−1 k −1 ψ D = {x ∈ X : x0 = a−k, . . . , xk+` = a`} and T ψ D = ∆.

This shows that ∞ _ T kψ−1G = F. k=0 Thus, T is the natural extension of S. 54 Measure Preserving Isomorphism and Factor Maps Chapter 4

Entropy

4.1 Randomness and Information

Given a measure preserving transformation T on a probability space (X, F, µ), we want to define a nonnegative quantity h(T ) which measures the average uncertainty about where T moves the points of X. That is, the value of h(T ) reflects the amount of ‘randomness’ generated by T . We want to define h(T ) in such a way, that (i) the amount of information gained by an application of T is proportional to the amount of uncertainty removed, and (ii) that h(T ) is isomorphism invariant, so that isomorphic transformations have equal entropy. The connection between entropy (that is randomness, uncertainty) and the transmission of information was first studied by Claude Shannon in 1948. As a motivation let us look at the following simple example. Consider a source (for example a ticker-tape) that produces a string of symbols ··· x−1x0x1 ··· from the alphabet {a1, a2, . . . , an}. Suppose that the probability of receiving symbol ai at any given time is pi, and that each symbol is transmitted independently of what has been transmitted earlier. Of course we must have here P that each pi ≥ 0 and that i pi = 1. In ergodic theory we view this process as the dynamical system (X, F, B, µ, T ), where X = {a1, a2, . . . , an}N, B the σ-algebra generated by cylinder sets of the form

∆n(ai1 , ai2 , . . . , ain ) := {x ∈ X : xi1 = ai1 , . . . , xin = ain }

µ the product measure assigning to each coordinate probability pi of seeing

55 56 Entropy

the symbol ai, and T the left shift. We deﬁne the entropy of this system by n X H(p1, . . . , pn) = h(T ) := − pi log2 pi . (4.1) i=1

If we define log pi as the amount of uncertainty in transmitting the symbol ai, then H is the average amount of information gained (or uncertainty removed) per symbol (notice that H is in fact an expected value). To see why this is an appropriate definition, notice that if the source is degenerate, that is, pi = 1 for some i (i.e., the source only transmits the symbol ai), then H = 0. In this case we indeed have no randomness. Another reason to see why this 1 definition is appropriate, is that H is maximal if pi = n for all i, and this agrees with the fact that the source is most random when all the symbols are equiprobable. To see this maximum, consider the function f : [0, 1] → R+ defined by ( 0 if t = 0, f(t) = −t log2 t if 0 < t ≤ 1. Then f is continuous and concave downward, and Jensen’s Inequality implies that for any p1, . . . , pn with pi ≥ 0 and p1 + ... + pn = 1, n n 1 1 X 1 X 1 1 H(p , . . . , p ) = f(p ) ≤ f( p ) = f( ) = log n , n 1 n n i n i n n 2 i=1 i=1 so H(p1, . . . , pn) ≤ log2 n for all probability vectors (p1, . . . , pn). But 1 1 H( ,..., ) = log n, n n 2 1 1 so the maximum value is attained at ( n ,..., n ).

4.2 Deﬁnitions and Properties

So far H is defined as the average information per symbol. The above definition can be extended to define the information transmitted by the occurrence of an event E as − log2 P (E). This definition has the property that the information transmitted by E ∩ F for independent events E and F is the sum of the information transmitted by each one individually, i.e.,

− log2 P (E ∩ F ) = − log2 P (E) − log2 P (F ). Deﬁnitions and Properties 57

The only function with this property is the logarithm function to any base. We choose base 2 because information is usually measured in bits. In the above example of the ticker-tape the symbols were transmitted independently. In general, the symbol generated might depend on what has been received before. In fact these dependencies are often ‘built-in’ to be able to check the transmitted sequence of symbols on errors (think here of the Morse sequence, sequences on compact discs etc.). Such dependencies must be taken into consideration in the calculation of the average information per symbol. This can be achieved if one replaces the symbols ai by blocks of symbols of particular size. More precisely, for every n, let Cn be the collection of all possible n-blocks (or cylinder sets) of length n, and deﬁne

X Hn := − P (C) log P (C) .

C∈Cn

1 Then n Hn can be seen as the average information per symbol when a block of length n is transmitted. The entropy of the source is now deﬁned by

H h := lim n . (4.2) n→∞ n

The existence of the limit in (4.2) follows from the fact that Hn is a subadditive sequence, i.e., Hn+m ≤ Hn + Hm, and proposition (4.2.2) (see proposition (4.2.3) below). Now replace the source by a measure preserving system (X, B, µ, T ). How can one define the entropy of this system similar to the case of a source? The symbols {a1, a2, . . . , an} can now be viewed as a partition A = {A1,A2,...,An} of X, so that X is the disjoint union (up to sets of measure zero) of A1,A2,...,An. The source can be seen as follows: with each point x ∈ X, we associate an infinite sequence ··· x−1, x0, x1, ··· , where i xi is aj if and only if T x ∈ Aj. We define the entropy of the partition α by

n X H(α) = Hµ(α) := − µ(Ai) log µ(Ai) . i=1

Our aim is to define the entropy of the transformation T which is independent of the partition we choose. In fact h(T ) must be the maximal entropy over all possible finite partitions. But first we need few facts about partitions. 58 Entropy

Exercise 4.2.1 Let α = {A1,...,An} and β = {B1,...,Bm} be two partitions of X. Show that

−1 −1 −1 T α := {T A1,...,T An} and α ∨ β := {Ai ∩ Bj : Ai ∈ α, Bj ∈ β} are both partitions of X.

The members of a partition are called the atoms of the partition. We say that the partition β = {B1,...,Bm} is a reﬁnement of the partition α = {A1,...,An}, and write α ≤ β, if for every 1 ≤ j ≤ m there exists an 1 ≤ i ≤ n such that Bj ⊂ Ai (up to sets of measure zero). The partition α ∨ β is called the common reﬁnement of α and β.

Exercise 4.2.2 Show that if β is a reﬁnement of α, each atom of α is a ﬁnite (disjoint) union of atoms of β.

Given two partitions α = {A1,...An} and β = {B1,...,Bm} of X, we deﬁne the conditional entropy of α given β by

X X µ(A ∩ B) H(α|β) := − log µ(A ∩ B) . µ(B) A∈α B∈β

(Under the convention that 0 log 0 := 0.) The above quantity H(α|β) is interpreted as the average uncertainty about which element of the partition α the point x will enter (under T ) if we already know which element of β the point x will enter.

Proposition 4.2.1 Let α, β and γ be partitions of X. Then,

(a) H(T −1α) = H(α) ;

(b) H(α ∨ β) = H(α) + H(β|α);

(d) H(α ∨ β) ≤ H(α) + H(β);

(e) If α ≤ β, then H(α) ≤ H(β); Deﬁnitions and Properties 59

(f) H(α ∨ β|γ) = H(α|γ) + H(β|α ∨ γ); (g) If β ≤ α, then H(γ|α) ≤ H(γ|β); (h) If β ≤ α, then H(β|α) = 0. (i) We call two partitions α and β independent if µ(A ∩ B) = µ(A)µ(B) for all A ∈ α, B ∈ β . If α and β are independent partitions, one has that H(α ∨ β) = H(α) + H(β) .

Proof We prove properties (b) and (c), the rest are left as an exercise.

X X H(α ∨ β) = − µ(A ∩ B) log µ(A ∩ B) A∈α B∈β X X µ(A ∩ B) = − µ(A ∩ B) log µ(A) A∈α B∈β X X + − µ(A ∩ B) log µ(A) A∈α B∈β = H(β|α) + H(α). We now show that H(β|α) ≤ H(β). Recall that the function f(t) = −t log t for 0 < t ≤ 1 is concave down. Thus, X X µ(A ∩ B) H(β|α) = − µ(A ∩ B) log µ(A) B∈β A∈α X X µ(A ∩ B) µ(A ∩ B) = − µ(A) log µ(A) µ(A) B∈β A∈α X X µ(A ∩ B) = µ(A)f µ(A) B∈β A∈α ! X X µ(A ∩ B) ≤ f µ(A) µ(A) B∈β A∈α X = f(µ(B)) = H(β). B∈β 60 Entropy

Exercise 4.2.3 Prove the rest of the properties of Proposition 4.2.1

Wn−1 −i Now consider the partition i=0 T α, whose atoms are of the form Ai0 ∩ −1 −(n−1) T Ai1 ∩ ... ∩ T Ain−1 , consisting of all points x ∈ X with the property n−1 that x ∈ Ai0 , T x ∈ Ai1 ,...,T x ∈ Ain−1 .

Exercise 4.2.4 Show that if α is a ﬁnite partition of (X, F, µ, T ), then

n−1 n−1 j _ X _ H( T −iα) = H(α) + H(α| T −iα). i=0 j=1 i=1

To deﬁne the notion of the entropy of a transformation with respect to a partition, we need the following two propositions.

Proposition 4.2.2 If {an} is a subadditive sequence of real numbers i.e., an+p ≤ an + ap for all n, p, then a lim n n→∞ n exists.

Proof Fix any m > 0. For any n ≥ 0 one has n = km + i for some i between 0 ≤ i ≤ m − 1. By subadditivity it follows that a a a a a a n = km+i ≤ km + i ≤ k m + i . n km + i km km km km

Note that if n → ∞, k → ∞ and so lim supn→∞ an/n ≤ am/m. Since m is arbitrary one has a a a lim sup n ≤ inf m ≤ lim inf n . n→∞ n m n→∞ n

Therefore limn→∞ an/n exists, and equals inf an/n.

Proposition 4.2.3 Let α be a ﬁnite partitions of (X, B, µ, T ), where T is a 1 Wn−1 −i measure preserving transformation. Then, limn→∞ n H( i=0 T α) exists. Deﬁnitions and Properties 61

Wn−1 −i Proof Let an = H( i=0 T α) ≥ 0. Then, by Proposition 4.2.1, we have

n+p−1 _ −i an+p = H( T α) i=0 n−1 n+p−1 _ _ ≤ H( T −iα) + H( T −iα) i=0 i=n p−1 _ −i = an + H( T α) i=0 = an + ap.

Hence, by Proposition 4.2.2

n−1 an 1 _ lim = lim H( T −iα) n→∞ n n→∞ n i=0 exists. We are now in position to give the deﬁnition of the entropy of the transformation T .

Deﬁnition 4.2.1 The entropy of the measure preserving transformation T with respect to the partition α is given by

n−1 1 _ −i h(α, T ) = hµ(α, T ) := lim H( T α) , n→∞ n i=0 where n−1 _ X H( T −iα) = − µ(D) log(µ(D)) .

i=0 Wn−1 −i D∈ i=0 T α Finally, the entropy of the transformation T is given by

h(T ) = hµ(T ) := sup h(α, T ) . α

The following theorem gives an equivalent deﬁnition of entropy.. 62 Entropy

Theorem 4.2.1 The entropy of the measure preserving transformation T with respect to the partition α is also given by

n−1 _ h(α, T ) = lim H(α| T −iα) . n→∞ i=1

Wn −i Proof Notice that the sequence {H(α| i=1 T α) is bounded from below, Wn −i and is non-increasing, hence limn→∞ H(α| i=1 T α) exists. Furthermore,

n n j _ 1 X _ lim H(α| T −iα) = lim H(α| T −iα). n→∞ n→∞ n i=1 j=1 i=1 From exercise 4.2.4, we have

n−1 n−1 j _ X _ H( T −iα) = H(α) + H(α| T −iα). i=0 j=1 i=1 Now, dividing by n, and taking the limit as n → ∞, one gets the desired result Theorem 4.2.2 Entropy is an isomorphism invariant.

Proof Let (X, B, µ, T ) and (Y, C, ν, S) be two isomorphic measure preserving systems (see Deﬁnition 1.2.3, for a deﬁnition), with ψ : X → Y the corresponding isomorphism. We need to show that hµ(T ) = hν(S). −1 −1 −1 Let β = {B1,...,Bn} be any partition of Y , then ψ β = {ψ B1, . . . , ψ Bn} −1 is a partition of X. Set Ai = ψ Bi, for 1 ≤ i ≤ n. Since ψ : X → Y is an isomorphism, we have that ν = µψ−1 and ψT = Sψ, so that for any n ≥ 0 and Bi0 ,...,Bin−1 ∈ β

−1 −(n−1) ν Bi0 ∩ S Bi1 ∩ ... ∩ S Bin−1 −1 −1 −1 −1 −(n−1) = µ ψ Bi0 ∩ ψ S Bi1 ∩ ... ∩ ψ S Bin−1 −1 −1 −1 −(n−1) −1 = µ ψ Bi0 ∩ T ψ Bi1 ∩ ... ∩ T ψ Bin−1 −1 −(n−1) = µ Ai0 ∩ T Ai1 ∩ ... ∩ T Ain−1 . Setting

−(n−1) −(n−1) A(n) = Ai0 ∩ ... ∩ T Ain−1 and B(n) = Bi0 ∩ ... ∩ S Bin−1 , Calculation of Entropy and Examples 63 we thus ﬁnd that

n−1 1 _ −i hν(S) = sup hν(β, S) = sup lim Hν( S β) β β n→∞ n i=0 1 X = sup lim − ν(B(n)) log ν(B(n)) β n→∞ n Wn−1 −i B(n)∈ i=0 S β 1 X = sup lim − µ(A(n)) log µ(A(n)) −1 n→∞ n ψ β Wn−1 −i −1 A(n)∈ i=0 T ψ β −1 = sup hµ(ψ β, T ) ψ−1β

≤ sup hµ(α, T ) = hµ(T ) , α where in the last inequality the supremum is taken over all possible ﬁnite partitions α of X. Thus hν(S) ≤ hµ(T ). The proof of hµ(T ) ≤ hν(S) is done similarly. Therefore hν(S) = hµ(T ), and the proof is complete.

4.3 Calculation of Entropy and Examples

Calculating the entropy of a transformation directly from the definition does not seem feasible, for one needs to take the supremum over all finite partitions, which is practically impossible. However, the entropy of a partition is relatively easier to calculate if one has full information about the partition under consideration. So the question is whether it is possible to find a partition α of X where h(α, T ) = h(T ). Naturally, such a partition contains all the information ‘transmitted’ by T . To answer this question we need some notations and definitions. For α = {A1,...,AN } and all m, n ≥ 0, let

m ! −n ! _ _ σ T −iα and σ T −iα i=n i=−m

Wm −i W−n −i be the smallest σ-algebras containing the partitions i=n T α and i=−m T α W−∞ −i respectively. Furthermore, let σ i=−∞ T α be the smallest σ-algebra con- Wm −i W−n −i taining all the partitions i=n T α and i=−m T α for all n and m. We W∞ −i call a partition α a generator with respect to T if σ i=−∞ T α = F, 64 Entropy where F is the σ-algebra on X. If T is non-invertible, then α is said to be W∞ −i a generator if σ ( i=0 T α) = F. Naturally, this equality is modulo sets of measure zero. One has also the following characterization of generators, saying basically, that each measurable set in X can be approximated by a ﬁnite disjoint union of cylinder sets. See also [W] for more details and proofs.

Proposition 4.3.1 The partition α is a generator of F if for each A ∈ F m and for each ε > 0 there exists a ﬁnite disjoint union C of elements of {αn }, such that µ(A4C) < ε.

We now state (without proofs) two famous theorems known as Kolmogorov- Sinai’s Theorem and Krieger’s Generator Theorem. For the proofs, we refer the interested reader to the book of Karl Petersen or Peter Walter.

Theorem 4.3.1 (Kolmogorov and Sinai, 1958) If α is a generator with respect to T and H(α) < ∞, then h(T ) = h(α, T ).

Theorem 4.3.2 (Krieger, 1970) If T is an ergodic measure preserving transformation with h(T ) < ∞, then T has a ﬁnite generator.

We will use these two theorems to calculate the entropy of a Bernoulli shift. Example (Entropy of a Bernoulli shift)–Let T be the left shift on X = {1, 2, ··· , n}Z endowed with the σ-algebra F generated by the cylinder sets, and product measure µ giving symbol i probability pi, where p1 + p2 + ... + pn = 1. Our aim is to calculate h(T ). To this end we need to ﬁnd a partition α which generates the σ-algebra F under the action of T . The natural choice of α is what is known as the time-zero partition α = {A1,...,An}, where

Ai := {x ∈ X : x0 = i} , i = 1, . . . , n .

Notice that for all m ∈ Z,

−m T Ai = {x ∈ X : xm = i} , and

−1 −m Ai0 ∩ T Ai1 ∩ · · · ∩ T Aim = {x ∈ X : x0 = i0, . . . , xm = im} . Calculation of Entropy and Examples 65

Wm −i In other words, i=0 T α is precisely the collection of cylinders of length m (i.e., the collection of all m-blocks), and these by deﬁnition generate F. Hence, α is a generating partition, so that

m−1 ! 1 _ h(T ) = h(α, T ) = lim H T −iα . m→∞ m i=0

First notice that − since µ is product measure here − the partitions

α, T −1α, ··· ,T −(m−1)α are all independent since each speciﬁes a diﬀerent coordinate, and so

H(α ∨ T −1α ∨ · · · ∨ T −(m−1)α) = H(α) + H(T −1α) + ··· + H(T −(m−1)α) n X = mH(α) = −m pi log pi. i=1

Thus, n n 1 X X h(T ) = lim (−m) pi log pi = − pi log pi . m→∞ m i=1 i=1

Exercise 4.3.1 Let T be the left shift on X = {1, 2, ··· , n}Z endowed with the σ-algebra F generated by the cylinder sets, and the Markov measure µ given by the stochastic matrix P = (pij), and the probability vector π = (π1, . . . , πn) with πP = π. Show that

n n X X h(T ) = − πipij log pij j=1 i=1

Exercise 4.3.2 Suppose (X1, B1, µ1,T1) and (X2, B2, µ2,T2) are two dynamical systems. Show that

hµ1×µ2 (T1 × T2) = hµ1 (T1) + hµ2 (T2). 66 Entropy 4.4 The Shannon-McMillan-Breiman Theorem

In the previous sections we have considered only finite partitions on X, however all the definitions and results hold if we were to consider countable partitions of finite entropy. Before we state and prove the Shannon-McMillan- Breiman Theorem, we need to introduce the information function associated with a partition. Let (X, F, µ) be a probability space, and α = {A1,A2,...} be a finite or a countable partition of X into measurable sets. For each x ∈ X, let α(x) be the element of α to which x belongs. Then, the information function associated to α is defined to be X Iα(x) = − log µ(α(x)) = − 1A(x) log µ(A). A∈α

For two ﬁnite or countable partitions α and β of X, we deﬁne the conditional information function of α given β by

X X µ(A ∩ B) I (x) = − 1 (x) log . α|β (A∩B) µ(B) B∈β A∈α

We claim that X Iα|β(x) = − log Eµ(1α(x)|σ(β)) = − 1A(x) log E(1A|σ(β)), (4.3) A∈α where σ(β) is the σ-algebra generated by the finite or countable partition β, (see the remark following the proof of Theorem (2.1.1)). This follows from the fact (which is easy to prove using the definition of conditional expectations) that if β is finite or countable, then for any f ∈ L1(µ), one has

X 1 Z E (f|σ(β)) = 1 fdµ. µ B µ(B) B∈β B R Clearly, H(α|β) = X Iα|β(x) dµ(x).

Exercise 4.4.1 Let α and β be ﬁnite or countable partitions of X. Show that Iα W β = Iα + Iβ|α. The Shannon-McMillan-Breiman Theorem 67

Now suppose T : X → X is a measure preserving transformation on −1 (X, F, µ), and let α = {A1,A2,...} be any countable partition. Then T = −1 −1 {T A1,T A2,...} is also a countable partition. Since T is measure preserving one has,

X −1 X −1 −1 IT α(x) = − 1T Ai (x) log µ(T Ai) = − 1Ai (T x) log µ(Ai) = Iα(T x).

Ai∈α Ai∈α Furthermore, n Z 1 _ −i 1 lim H( T α) = lim IWn T −iα(x) dµ(x) = h(α, T ). n→∞ n + 1 n→∞ n + 1 i=0 i=0 X The Shannon-McMillan-Breiman theorem says if T is ergodic and if α has 1 ﬁnite entropy, then in fact the integrand IWn T −iα(x) converges a.e. to n + 1 i=0 h(α, T ). Notice that the integrand can be written as

n ! 1 1 _ −i IWn T −iα(x) = − log µ ( T α)(x) , n + 1 i=0 n + 1 i=0

Wn −i Wn −i where ( i=0 T α)(x) is the element of i=0 T α containing x (often referred to as the α-cylinder of order n containing x). Before we proceed we need the following proposition.

Proposition 4.4.1 Let α = {A1,A2,...} be a countable partition with ﬁnite ∗ entropy. For each n = 1, 2, 3,..., let f (x) = I Wn −i (x), and let f = n α| i=1 T α supn≥1 fn. Then, for each λ ≥ 0 and for each A ∈ α,

µ ({x ∈ A : f ∗(x) > λ}) ≤ 2−λ.

Furthermore, f ∗ ∈ L1(X, F, µ).

Proof Let t ≥ 0 and A ∈ α. For n ≥ 1, let

n ! A _ −i fn (x) = − log Eµ 1A| T α (x), i=1 and A A A Bn = {x ∈ X : f1 (x) ≤ t, . . . , fn−1(x) ≤ t, fn (x) > t}. 68 Entropy

A Notice that for x ∈ A one has fn(x) = fn (x), and for x ∈ Bn one has Wn −i −t Wn −i Eµ (1A| i=1 T α)(x) < 2 . Since Bn ∈ σ ( i=1 T α), then

Z µ(Bn ∩ A) = 1A(x) dµ(x) Bn Z n ! _ −i = Eµ 1A| T α (x) dµ(x) Bn i=1 Z −t −t ≤ 2 dµ(x) = 2 µ(Bn). Bn

Thus,

∗ µ ({x ∈ A : f (x) > t}) = µ ({x ∈ A : fn(x) > t, for some n}) A = µ {x ∈ A : fn (x) > t, for some n} ∞ = µ (∪n=1A ∩ Bn) ∞ X = µ (A ∩ Bn) n=1 ∞ −t X −t ≤ 2 µ(Bn) ≤ 2 . n=1

We now show that f ∗ ∈ L1(X, F, µ). First notice that

µ ({x ∈ A : f ∗(x) > t}) ≤ µ(A), hence,

µ ({x ∈ A : f ∗(x) > t}) ≤ min(µ(A), 2−t). The Shannon-McMillan-Breiman Theorem 69

Using Fubini’s Theorem, and the fact that f ∗ ≥ 0 one has Z Z ∞ f ∗(x) dµ(x) = µ ({x ∈ X : f ∗(x) > t}) dt X 0 Z ∞ X = µ ({x ∈ A : f ∗(x) > t}) dt 0 A∈α X Z ∞ = µ ({x ∈ A : f ∗(x) > t}) dt A∈α 0 X Z ∞ ≤ min(µ(A), 2−t) dt A∈α 0 X Z − log µ(A) X Z ∞ = µ(A) dt + 2−t dt A∈α 0 A∈α − log µ(A) X X µ(A) = − µ(A) log µ(A) + log 2 A∈α A∈α e 1 = Hµ(α) + < ∞. loge 2

So far we have deﬁned the notion of the conditional entropy Iα|β when α and β are countable partitions. We can generalize the deﬁnition to the case α is a countable partition and G is a σ-algebra as follows (see equation (4.3)),

Iα|G(x) = − log Eµ(1α(x)|G). W∞ −i Wn −i If we denote by i=1 T α = σ(∪n i=1 T α), then

Iα| W∞ T −iα(x) = lim Iα| Wn T −iα(x). (4.4) i=1 n→∞ i=1 Exercise 4.4.2 Give a proof of equation (4.4) using the following important theorem, known as the Martingale Convergence Theorem (and is stated to our setting)

Theorem 4.4.1 (Martingale Convergence Theorem) Let C1 ⊆ C2 ⊆ · · · be a 1 sequence of increasing σalgebras, and let C = σ(∪nCn). If f ∈ L (µ), then

Eµ(f|C) = lim Eµ(f|Cn) n→∞ µ a.e. and in L1(µ). 70 Entropy

Exercise 4.4.3 Show that if T is measure preserving on the probability space (X, F, µ) and f ∈ L1(µ), then

f(T nx) lim = 0, µ a.e. n→∞ n

Theorem 4.4.2 (The Shannon-McMillan-Breiman Theorem) Suppose T is an ergodic measure preserving transformation on a probability space (X, F, µ), and let α be a countable partition with H(α) < ∞. Then,

1 lim IWn T −iα(x) = h(α, T ) a.e. n→∞ n + 1 i=0

Proof For each n = 1, 2, 3,..., let f (x) = I Wn −i (x). Then, n α| i=1 T α

IWn −i (x) = IWn −i (x) + I Wn −i (x) i=0 T α i=1 T α α| i=1 T α

= IWn−1 −i (T x) + fn(x) i=0 T α

= IWn−1 −i (T x) + I Wn−1 −i (T x) + fn(x) i=1 T α α| i=1 T α 2 = IWn−2 −i (T x) + fn−1(T x) + fn(x) i=0 T α . . n n−1 = Iα(T x) + f1(T x) + ... + fn−1(T x) + fn(x).

1 Let f(x) = Iα| W∞ T −iα(x) = limn→∞ fn(x). Notice that f ∈ L (X, F, µ) R i=1 since X f(x) dµ(x) = h(α, T ). Now letting f0 = Iα, we have

n 1 1 X k IWn T −iα(x) = fn−k(T x) n + 1 i=0 n + 1 k=0 n n 1 X 1 X = f(T kx) + (f − f)(T kx). n + 1 n + 1 n−k k=0 k=0

By the ergodic theorem,

n X Z lim f(T kx) = f(x) dµ(x) = h(α, T ) a.e. n→∞ k=0 X The Shannon-McMillan-Breiman Theorem 71

n 1 X We now study the sequence { (f − f)(T kx)}. Let n + 1 n−k k=0 ∗ FN = sup |fk − f|, and f = sup fn. k≥N n≥1

n n−N 1 X 1 X |f − f|(T kx) = |f − f|(T kx) n + 1 n−k n + 1 n−k k=0 k=0 n 1 X + |f − f|(T kx) n + 1 n−k k=n−N+1 n−N 1 X ≤ F (T kx) n + 1 N k=0 N−1 1 X + |f − f|(T n−kx). n + 1 k k=0 If we take the limit as n → ∞, then by exercise (4.4.3) the second term tends to 0 a.e., and by the ergodic theorem as well as the dominated convergence theorem, the first term tends to zero a.e. Hence, 1 lim IWn T −iα(x) = h(α, T ) a.e. n→∞ n + 1 i=0 The above theorem can be interpreted as providing an estimate of the size Wn −i of the atoms of i=0 T α. For n sufficiently large, a typical element A ∈ Wn −i i=0 T α satisfies 1 − log µ(A) ≈ h(α, T ) n + 1 or −(n+1)h(α,T ) µ(An) ≈ 2 . W∞ −i Furthermore, if α is a generating partition (i.e. i=0 T α = F, then in the conclusion of Shannon-McMillan-Breiman Theorem one can replace h(α, T ) by h(T ). 72 Entropy 4.5 Lochs’ Theorem

In 1964, G. Lochs compared the decimal and the continued fraction expansions. Let x ∈ [0, 1) be an irrational number, and suppose x = .d1d2 ··· is the decimal expansion of x (which is generated by iterating the map Sx = 10x (mod 1)). Suppose further that

1 x = = [0; a , a , ··· ] (4.5) 1 1 2 a + 1 1 a + 2 1 a3 + ... is its regular continued fraction (RCF) expansion (generated by the map 1 1 T x = x − b x c). Let y = .d1d2 ··· dn be the rational number determined by the ﬁrst n decimal digits of x, and let z = y + 10−n. Then, [y, z) is the decimal cylinder of order n containing x, which we also denote by Bn(x). Now let 1 y = 1 b1 + .. 1 b2 + . + bl and 1 z = 1 c1 + .. 1 c2 + . + ck be the continued fraction expansion of y and z. Let

m (n, x) = max {i ≤ min (l, k) : for all j ≤ i, bj = cj} . (4.6)

In other words, if Bn(x) denotes the decimal cylinder consisting of all points y in [0, 1) such that the ﬁrst n decimal digits of y agree with those of x, and if Cj(x) denotes the continued fraction cylinder of order j containing x, i.e., Cj(x) is the set of all points in [0, 1) such that the ﬁrst j digits in their continued fraction expansion is the same as that of x, then m(n, x) is the largest integer such that Bn(x) ⊂ Cm(n,x)(x). Lochs proved the following theorem: Lochs’ Theorem 73

Theorem 4.5.1 Let λ denote Lebesgue measure on [0, 1). Then for a.e. x ∈ [0, 1) m(n, x) 6 log 2 log 10 lim = . n→∞ n π2

In this section, we will prove a generalization of Lochs’ theorem that allows one to compare any two known expansions of numbers. We show that Lochs’ theorem is true for any two sequences of interval partitions on [0, 1) satisfying the conclusion of Shannon-McMillan-Breiman theorem. The content of this section as well as the proofs can be found in [DF]. We begin with few deﬁnitions that will be used in the arguments to follow.

Deﬁnition 4.5.1 By an interval partition we mean a ﬁnite or countable partition of [0, 1) into subintervals. If P is an interval partition and x ∈ [0, 1), we let P (x) denote the interval of P containing x.

∞ Let P = {Pn}n=1 be a sequence of interval partitions. Let λ denote Lebesgue probability measure on [0, 1).

Deﬁnition 4.5.2 Let c ≥ 0. We say that P has entropy c a.e. with respect to λ if log λ (P (x)) − n → c a.e. n

Note that we do not assume that each Pn is reﬁned by Pn+1. ∞ ∞ Suppose that P = {Pn}n=1 and Q = {Qn}n=1 are sequences of interval partitions. For each n ∈ N and x ∈ [0, 1), deﬁne

mP,Q (n, x) = sup {m | Pn (x) ⊂ Qm (x)} .

∞ ∞ Theorem 4.5.2 Let P = {Pn}n=1 and Q = {Qn}n=1 be sequences of interval partitions and λ Lebesgue probability measure on [0, 1). Suppose that for some constants c > 0 and d > 0, P has entropy c a.e with respect to λ and Q has entropy d a.e. with respect to λ. Then

m (n, x) c P,Q → a.e. n d 74 Entropy

Proof First we show that m (n, x) c lim sup P,Q ≤ a.e. n→∞ n d Fix ε > 0. Let x ∈ [0, 1) be a point at which the convergence conditions of c + η the hypotheses are met. Fix η > 0 so that c < 1 + ε. Choose N so c − η d that for all n ≥ N −n(c+η) λ (Pn (x)) > 2 and −n(d−η) λ (Qn (x)) < 2 . n c o Fix n so that min n, n ≥ N, and let m0 denote any integer greater than d c (1 + ε) n. By the choice of η, d

λ (Pn (x)) > λ (Qm0 (x)) so that Pn (x) is not contained in Qm0 (x). Therefore c m (n, x) ≤ (1 + ε) n P,Q d and so m (n, x) c lim sup P,Q ≤ (1 + ε) a.e. n→∞ n d Since ε > 0 was arbitrary, we have the desired result. Now we show that m (n, x) c lim inf P,Q ≥ a.e. n→∞ n d c Fix ε ∈ (0, 1). Choose η > 0 so that ζ =: εc − η 1 + (1 − ε) > 0. For d j c k each n ∈ letm ¯ (n) = (1 − ε) n . For brevity, for each n ∈ we call an N d N element of Pn (respectively Qn)(n, η) −good if

−n(c−η) λ (Pn (x)) < 2

(respectively −n(d+η) λ (Qn (x)) > 2 .) Lochs’ Theorem 75

For each n ∈ N, let Pn (x) is (n, η) − good and Qm¯ (n) (x) is (m ¯ (n) , η) − good Dn (η) = x : . and Pn (x) " Qm¯ (n) (x)

If x ∈ Dn (η), then Pn (x) contains an endpoint of the (m ¯ (n) , η) −good interval Qm¯ (n) (x). By the deﬁnition of Dn (η) andm ¯ (n),

λ (Pn (x)) −nζ < 2 . λ Qm¯ (n) (x)

Since no more than one atom of Pn can contain a particular endpoint of an −nζ atom of Qm¯ (n), we see that λ (Dn (η)) < 2 · 2 and so

∞ X λ (Dn (η)) < ∞, n=1 which implies that λ {x | x ∈ Dn (η) i.o.} = 0. Sincem ¯ (n) goes to inﬁnity as n does, we have shown that for almost every x ∈ [0, 1), there exists N ∈ N, so that for all n ≥ N, Pn (x) is (n, η) −good and Qm¯ (n) (x) is (m ¯ (n) , η) −good and x∈ / Dn (η). In other words, for almost every x ∈ [0, 1), there exists N ∈ N, so that for all n ≥ N, Pn (x) is (n, η) −good and Qm¯ (n) (x) is (m ¯ (n) , η) −good and Pn (x) ⊂ Qm¯ (n) (x). Thus, for almost every x ∈ [0, 1), there exists N ∈ N, so that for all n ≥ N, mP,Q (n, x) ≥ m¯ (n), so that m (n, x) c P,Q ≥ b(1 − ε) c. n d This proves that m (n, x) c lim inf P,Q ≥ (1 − ε) a.e. n→∞ n d Since ε > 0 was arbitrary, we have established the theorem. The above result allows us to compare any two well-known expansions of numbers. Since the commonly used expansions are usually performed for points in the unit interval, our underlying space will be ([0, 1), B, λ), where B is the Lebesgue σ-algebra, and λ the Lebesgue measure. The expansions we have in mind share the following two properties. 76 Entropy

Definition 4.5.3 A surjective map T : [0, 1) → [0, 1) is called a number theoretic fibered map (NTFM) if it satisfies the following conditions:

(a) there exists a ﬁnite or countable partition of intervals α = {Ai; i ∈ D} such that T restricted to each atom of α (cylinder set of order 0) is monotone, continuous and injective. Furthermore, α is a generating partition.

(b) T is ergodic with respect to Lebesgue measure λ, and there exists a T invariant probability measure µ equivalent to λ with bounded density. dµ dλ (Both and are bounded, and µ(A) = 0 if and only if λ(A) = 0 dλ dµ for all Lebesgue sets A.).

Let T be an NTFM with corresponding partition α, and T -invariant measure µ equivalent to λ. Let L, M > 0 be such that

Lλ(A) ≤ µ(A) < Mλ(A)

Wn−1 −i for all Lebesgue sets A (property (b)). For n ≥ 1, let Pn = i=0 T α, then by property (a), Pn is an interval partition. If Hµ(α) < ∞, then Shannon- McMillan-Breiman Theorem gives

log µ(Pn(x)) lim − = hµ(T ) a.e. with respect to µ. n→∞ n Exercise 4.5.1 Show that the conclusion of the Shannon-McMillan-Breiman Theorem holds if we replace µ by λ, i.e.

log λ(Pn(x)) lim − = hµ(T ) a.e. with respect to λ. n→∞ n Iterations of T generate expansions of points x ∈ [0, 1) with digits in D. We refer to the resulting expansion as the T -expansion of x. Almost all known expansions on [0, 1) are generated by a NTFM. Among them are the n-adic expansions (T x = nx (mod 1), where n is a positive integer), β expansions (T x = βx (mod 1), where β > 1 is a real number), 1 1 continued fraction expansions (T x = x − b x c), and many others (see the book Ergodic Theory of Numbers). Lochs’ Theorem 77

Exercise 4.5.2 Prove Theorem (4.5.1) using Theorem (4.5.2). Use the fact that the continued fraction map T is ergodic with respect to Gauss measure µ, given by Z 1 1 µ(B) = dx, B log 2 1 + x π2 and has entropy equal to h (T ) = . µ 6 log 2

Exercise 4.5.3 Reformulate and prove Lochs’ Theorem for any two NTFM maps S and T on [0, 1). 78 Entropy Chapter 5

Hurewicz Ergodic Theorem

In this chapter we consider a class of non-measure preserving transformations. In particular, we study invertible, non-singular and conservative transformations on a probability space. We first start with a quick review of equivalent measures, we then define non-singular and conservative transformations, and state some of their properties. We end this section by giving a new proof of Hurewicz Ergodic Theorem, which is a generalization of Birkhoff Ergodic Theorem to non-singular conservative transformations.

5.1 Equivalent measures

Recall that two measures µ and ν on a measure space (Y, F) are equivalent if µ and ν have the same null-sets, i.e.,

µ(A) = 0 ⇔ ν(A) = 0,A ∈ F.

The theorem of Radon-Nikodym says that if µ, ν are σ-ﬁnite and equivalent, then there exist measurable functions f, g ≥ 0, such that Z Z µ(A) = f dν and ν(A) = g dµ. A A

Furthermore, for all h ∈ L1(µ) (or L1(ν)), Z Z Z Z h dµ = hf dν and h dν = hg dµ.

79 80 Hurewicz Ergodic Theorem

dµ dν Usually the function f is denoted by and the function g by . dν dµ Now suppose that (X, B, µ) is a probability space, and T : X → X a measurable transformation. One can deﬁne a new measure µ ◦ T −1 on (X, B) by µ ◦ T −1(A) = µ(T −1A) for A ∈ B. It is not hard to prove that for f ∈ L1(µ), Z Z f d(µ ◦ T −1) = f ◦ T dµ (5.1)

Exercise 5.1.1 Starting with indicator functions, give a proof of (5.1).

Note that if T is invertible, then one has that Z Z f d(µ ◦ T ) = f ◦ T −1 dµ (5.2)

5.2 Non-singular and conservative transformations

Deﬁnition 5.2.1 Let (X, B, µ) be a probability space and T : X → X an invertible measurable function. T is said to be non-singular if for any A ∈ B,

µ(A) = 0 if and only if µ(T −1A) = 0.

Since T is invertible, non-singularity implies that

µ(A) = 0 if and only if µ(T nA) = 0, n 6= 0.

This implies that the measures µ ◦ T n deﬁned by µ ◦ T n(A) = µ(T nA) is equivalent to µ (and hence equivalent to each other). By the theorem of Radon-Nikodym, there exists for each n 6= 0, a non-negative measurable dµ ◦ T n function ω (x) = (x) such that n dµ Z n µ(T A) = ωn(x) dµ(x). A We have the following propositions. Non-singular and conservative transformations 81

Proposition 5.2.1 Suppose (X, B, µ) is a probability space, and T : X → X is invertible and non-singular. Then for every f ∈ L1(µ), Z Z Z n f(x) dµ(x) = f(T x)ω1(x) dµ(x) = f(T x)ωn(x) dµ(x). X X X Proof We show the result for indicator functions only, the rest of the proof is left to the reader. Z −1 1A(x) dµ(x) = µ(A) = µ(T (T A)) X Z = ω1(x) dµ(x) T −1A Z = 1A(T x)ω1(x) dµ(x). X Proposition 5.2.2 Under the assumptions of Proposition 5.2.1, one has for all n, m ≥ 1, that

n ωn+m(x) = ωn(x)ωm(T x), µ a.e.

Proof For any A ∈ B, Z Z n n n ωn(x)ωm(T x)dµ(x) = 1A(x)ωm(T x)d(µ ◦ T )(x) A X Z −n = 1A(T x)ωm(x)dµ(x) X Z m = 1T nA(x)d(µ ◦ T )(x) X Z m n m+n = µ ◦ T (T A) = µ(T A) = ωn+m(x)dµ(x). A n Hence, ωn+m(x) = ωn(x)ωm(T x), µ a.e. Exercise 5.2.1 Let (X, B, µ) be a probability space, and T : X → X an invertible non-singular transformation. For any measurable function f, set Pn−1 i fn(x) = i=0 f(T x)ωi(x), n ≥ 1, where ω0(x) = 1. Show that for all n, m ≥ 1, n fn+m(x) = fn(x) + ωn(x)fm(T x). 82 Hurewicz Ergodic Theorem

Deﬁnition 5.2.2 Let (X, B, µ) be a probability space, and T : X → X a measurable transformation. We say that T is conservative if for any A ∈ B with µ(A) > 0, there exists an n ≥ 1 such that µ(A ∩ T −nA) > 0.

Note that if T is invertible, non-singular and conservative , then T −1 is also non-singular and conservative. In this case, for any A ∈ B with µ(A) > 0, there exists an n 6= 0 such that µ(A ∩ T nA) > 0.

Proposition 5.2.3 Suppose T is invertible, non-singular and conservative on the probability space (X, B, µ), and let A ∈ B with µ(A) > 0. Then for µ a.e. x ∈ A there exist inﬁnitely many positive and negative integers n, such that T nx ∈ A.

Proof Let B = {x ∈ A : T nx∈ / A for all n ≥ 1}. Note that for any n ≥ 1, B ∩ T −nB = ∅. If µ(B) > 0, then by conservativity there exists an n ≥ 1, such that µ(B∩T −nB) is positive, which is a contradiction. Hence, µ(B) = 0, and by non-singularity we have µ(T −nB) = 0 for all n ≥ 1. Now, let C = {x ∈ A; T nx ∈ A for only ﬁnitely many n ≥ 1}, then S∞ −n C ⊂ n=1 T B, implying that ∞ X µ(C) ≤ µ(T −nB) = 0. n=1 Therefore, for almost every x ∈ A there exist inﬁnitely many n ≥ 1 such that n −1 T x ∈ A. Replacing T by T yields the result for n ≤ −1. Proposition 5.2.4 Suppose T is invertible, non-singular and conservative, then ∞ X ωn(x) = ∞, µ a.e. n=1 P∞ Proof Let A = {x ∈ X : n=1 ωn(x) < ∞}. Note that ∞ ∞ [ X A = {x ∈ X : ωn(x) < M}. M=1 n=1 If µ(A) > 0, then there exists an M ≥ 1 such that the set

∞ X B = {x ∈ X : ωn(x) < M} n=1 Hurewicz Ergodic Theorem 83

R P∞ has positive measure. Then, B n=1 ωn(x) dµ(x) < Mµ(B) < ∞. However, ∞ ∞ Z X X Z ωn(x) dµ(x) = ωn(x) dµ(x) B n=1 n=1 B ∞ X = µ(T nB) n=1 ∞ X Z = 1T nB(x) dµ(x) n=1 X Z ∞ X −n = 1B(T x) dµ(x). X n=1 R P∞ −n Hence, X n=1 1B(T x) dµ(x) < ∞, which implies that ∞ X −n 1B(T x) < ∞ µ a.e. n=1 Therefore, for µ a.e. x one has T −nx ∈ B for only ﬁnitely many n ≥ 1, contradicting Proposition 5.2.3. Thus µ(A) = 0, and ∞ X ωn(x) = ∞, µ a.e. n=1

5.3 Hurewicz Ergodic Theorem

The following theorem by Hurewicz is a generalization of Birkhoﬀ’s Ergodic Theorem to our setting; see also Hurewicz’ original paper [H]. We give a new prove, similar to the proof for Birkhoﬀ’s Theorem; see Section 2.1 and [KK].

Theorem 5.3.1 Let (X, B, µ) be a probability space, and T : X → X an invertible, non-singular and conservative transformation. If f ∈ L1(µ), then n−1 X i f(T x)ωi(x) lim i=0 = f (x) n−1 ∗ n→∞ X ωi(x) i=0 84 Hurewicz Ergodic Theorem

exists µ a.e. Furthermore, f∗ is T -invariant and Z Z f(x) dµ(x) = f∗(x) dµ(x). X X Proof Assume with no loss of generality that f ≥ 0 (otherwise we write f = f + − f −, and we consider each part separately). Let

n−1 fn(x) = f(x) + f(T x)ω1(x) + ··· + f(T x)ωn−1(x),

gn(x) = ω0(x) + ω1(x) + ··· + ωn−1(x), ω0(x) = g0(x) = 1,

f (x) f (x) f(x) = lim sup n = lim sup n , n−1 n→∞ X n→∞ gn(x) ωi(x) i=0 and f (x) f (x) f(x) = lim inf n = lim inf n . n−1 n→∞ X n→∞ gn(x) ωi(x) i=0 n By Proposition (5.2.2), one has gn+m(x) = gn(x) + gm(T x). Using Exercise (5.2.1) and Proposition (5.2.4), we will show that f and f are T -invariant. To this end,

fn(T x) f(T x) = lim sup n n→∞ gn(T x)

fn+1(x) − f(x) ω (x) = lim sup 1 n→∞ gn+1(x) − g(x) ω1(x) f (x) − f(x) = lim sup n+1 n→∞ gn+1(x) − g(x) f (x) g (x) f(x) = lim sup n+1 · n+1 − n→∞ gn+1(x) gn+1(x) − g(x) gn+1(x) − g(x) f (x) = lim sup n+1 n→∞ gn+1(x) = f(x). Hurewicz Ergodic Theorem 85

(Similarly f is T -invariant). Now, to prove that f∗ exists, is integrable and T -invariant, it is enough to show that Z Z Z f dµ ≥ f dµ ≥ f dµ. X X X

For since f − f ≥ 0, this would imply that f = f = f∗. a.e. R R We ﬁrst prove that X fdµ ≤ X f dµ. Fix any 0 < < 1, and let L > 0 be any real number. By deﬁnition of f, for any x ∈ X, there exists an integer m > 0 such that f (x) m ≥ min(f(x),L)(1 − ). gm(x) Now, for any δ > 0 there exists an integer M > 0 such that the set

X0 = {x ∈ X : ∃ 1 ≤ m ≤ M with fm(x) ≥ gm(x) min(f(x),L)(1 − )} has measure at least 1 − δ. Deﬁne F on X by

f(x) x ∈ X F (x) = 0 L x∈ / X0.

n Notice that f ≤ F (why?). For any x ∈ X, let an = an(x) = F (T x)ωn(x), and bn = bn(x) = min(f(x),L)(1 − )ωn(x). We now show that {an} and {bn} satisfy the hypothesis of Lemma 2.1.1 with M > 0 as above. For any n = 0, 1, 2,... n –if T x ∈ X0, then there exists 1 ≤ m ≤ M such that

n n fm(T x) ≥ min(f(x),L)(1 − )gm(T x).

Hence, n n ωn(x)fm(T x) ≥ min(f(x),L)(1 − )gm(T x)ωn(x). Now,

n bn + ... + bn+m−1 = min(f(x),L)(1 − )gm(T x)ωn(x) n ≤ ωn(x)fm(T x) n n+1 n+m−1 = f(T x)ωn(x) + f(T x)ωn+1(x) + ··· + f(T x)ωn+m−1(x) n n+1 n+m−1 ≤ F (T x)ωn(x) + F (T x)ωn+1(x) + ··· + F (T x)ωn+m−1(x)

= an + an+1 + ··· + an+m−1. 86 Hurewicz Ergodic Theorem

n –If T x∈ / X0, then take m = 1 since

n an = F (T x)ωn(x) = Lωn(x) ≥ min(f(x),L)(1 − )ωn(x) = bn.

Hence by T -invariance of f, and Lemma 2.1.1 for all integers N > M one has

N−1 F (x)+F (T x)+ω1(x)+···+ωN−1(x)F (T x) ≥ min(f(x),L)(1−)gN−M (x).

Integrating both sides, and using Proposition (5.2.1) together with the T - invariance of f one gets Z Z N F (x) dµ(x) ≥ min(f(x),L)(1 − )gN−M (x) dµ(x) X X Z = (N − M) min(f(x),L)(1 − ) dµ(x). X Since Z Z F (x) dµ(x) = f(x) dµ(x) + Lµ(X \ X0), X X0 one has Z Z f(x) dµ(x) ≥ f(x) dµ(x) X X0 Z = F (x) dµ(x) − Lµ(X \ X0) X (N − M) Z ≥ min(f(x),L)(1 − ) dµ(x) − Lδ. N X Now letting ﬁrst N → ∞, then δ → 0, then → 0, and lastly L → ∞ one gets together with the monotone convergence theorem that f is integrable, and Z Z f(x) dµ(x) ≥ f(x) dµ(x). X X

We now prove that Z Z f(x) dµ(x) ≤ f(x) dµ(x). X X Hurewicz Ergodic Theorem 87

Fix > 0, for any x ∈ X there exists an integer m such that f (x) m ≤ (f(x) + ). gm(x) For any δ > 0 there exists an integer M > 0 such that the set

Y0 = {x ∈ X : ∃ 1 ≤ m ≤ M with fm(x) ≤ (f(x) + )gm(x)} has measure at least 1 − δ. Deﬁne G on X by

f(x) x ∈ Y G(x) = 0 0 x∈ / Y0.

n Notice that G ≤ f. Let bn = G(T x)ωn(x), and an = (f(x) + )ωn(x). We now check that the sequences {an} and {bn} satisfy the hypothesis of Lemma 2.1.1 with M > 0 as above. n –if T x ∈ Y0, then there exists 1 ≤ m ≤ M such that

n n fm(T x) ≤ (f(x) + )gm(T x).

Hence,

n n ωn(x)fm(T x) ≤ (f(x)+)gm(T x)ωn(x) = (f(x)+)(ωn(x)+···+ωn+m−1(x).

By Proposition (5.2.2), and the fact that f ≥ G, one gets

n n+m−1 bn + ... + bn+m−1 = G(T x)ωn(x) + ··· + G(T x)ωn+m−1(x) n n+m−1 ≤ f(T x)ωn(x) + ··· + f(T x)ωn+m−1(x) n = ωn(x)fm(T x)

≤ (f(x) + )(ωn(x) + ··· + ωn+m+1(x))

= an + ··· an+m−1.

n –If T x∈ / Y0, then take m = 1 since

n bn = G(T x)ωn(x) = 0 ≤ (f(x) + )(ωn(x)) = an.

Hence by Lemma 2.1.1 one has for all integers N > M

N−M−1 G(x) + G(T x)ω1(x) + ... + G(T x)ωN−M−1(x) ≤ (f(x) + )gN (x). 88 Hurewicz Ergodic Theorem

Integrating both sides yields Z Z (N − M) G(x)dµ(x) ≤ N( f(x)dµ(x) + ). X X R Since f ≥ 0, the measure ν deﬁned by ν(A) = A f(x) dµ(x) is absolutely continuous with respect to the measure µ. Hence, there exists δ0 > 0 such that if µ(A) < δ, then ν(A) < δ0. Since µ(X \ Y0) < δ, then ν(X \ Y0) = R f(x)dµ(x) < δ0. Hence, X\Y0 Z Z Z f(x) dµ(x) = G(x) dµ(x) + f(x) dµ(x) X X X\Y0 N Z ≤ (f(x) + ) dµ(x) + δ0. N − M X

Now, let ﬁrst N → ∞, then δ → 0 (and hence δ0 → 0), and ﬁnally → 0, one gets Z Z f(x) dµ(x) ≤ f(x) dµ(x). X X This shows that Z Z Z f dµ ≥ f dµ ≥ f dµ, X X X hence, f = f = f∗ a.e., and f∗ is T -invariant.

Remark We can extend the notion of ergodicity to our setting. If T is non- singular and conservative, we say that T is ergodic if for any measurable set A satisfying µ(A∆T −1A) = 0, one has µ(A) = 0 or 1. It is easy to check that the proof of Proposition (1.7.1) holds in this case, so that T ergodic implies that each T -invariant function is a constant µ a.e. Hence, if T is invertible, non-singular, conservative and ergodic, then by Hurewicz Ergodic Theorem one has for any f ∈ L1(µ),

n−1 X f(T ix)ω (x) i Z lim i=0 = fdµ µ a.e. n−1 n→∞ X X ωi(x) i=0 Chapter 6

Invariant Measures for Continuous Transformations

6.1 Existence

Suppose X is a compact metric space, and let B be the Borel σ-algebra i.e., the σ-algebra generated by the open sets. Let M(X) be the collection of all Borel probability measures on X. There is natural embedding of the space X in M(X) via the map x → δx, where δX is the Dirac measure concentrated at x (δx(A) = 1 if x ∈ A, and is zero otherwise). Furthermore, M(X) is a convex set, i.e., pµ+(1−p)ν ∈ M(X) whenever µ, ν ∈ M(X) and 0 ≤ p ≤ 1. Theorem 6.1.2 below shows that a member of M(X) is determined by how it integrates continuous functions. We denote by C(X) the Banach space of all complex valued continuous functions on X under the supremum norm. Theorem 6.1.1 Every member of M(X) is regular, i.e., for all B ∈ B and every > 0 there exist an open set U and a closed sed C such that C ⊆ B ⊆ U such that µ(U \ C) < .

Idea of proof Call a set B ∈ B with the above property a regular set. Let R = {B ∈ B : B is regular }. Show that R is a σ-algebra containing all the closed sets. Corollary 6.1.1 For any B ∈ B, and any µ ∈ M(X), µ(B) = sup µ(C) = inf µ(U). B⊆U:U open C⊆B:C closed

89 90 Invariant measures for continuous transformations

Theorem 6.1.2 Let µ, m ∈ M(X). If Z Z f dµ = f dm X X for all f ∈ C(X), then µ = m.

Proof From the above corollary, it suﬃces to show that µ(C) = m(C) for all closed subsets C of X. Let > 0, by regularity of the measure m there exists an open set U such that C ⊆ U and m(U \ C) < . Deﬁne f ∈ C(X) as follows ( 0 x∈ / U f(x) = d(x,X\U) x ∈ U . d(x,X\U)+d(x,C)

Notice that 1C ≤ f ≤ 1U , thus Z Z µ(C) ≤ f dµ = f dm ≤ m(U) ≤ m(C) + . X X Using a similar argument, one can show that m(C) ≤ µ(C) + . Therefore, µ(C) = m(C) for all closed sets, and hence for all Borel sets. This allows us to deﬁne a metric structure on M(X) as follows. A sequence {µn} in M(X) converges to µ ∈ M(X) if and only if Z Z lim f(x) dµn(x) = f(x) dµ(x) n→∞ X X for all f ∈ C(X). We will show that under this notion of convergence the space M(X) is compact, but ﬁrst we need The Riesz Representation Representation Theorem.

Theorem 6.1.3 (The Riesz Representation Theorem) Let X be a compact metric space and J : C(X) → C a continuous linear map such that J is a positive operator and J(1) = 1. Then there exists a µ ∈ M(X) such that R J(f) = X f(x) dµ(x).

Theorem 6.1.4 The space M(X) is compact. Existence 91

Idea of proof Let {µn} be a sequence in M(X). Choose a countable dense R subset of {fn} of C(X). The sequence { X f1 dµn} is a bounded sequence of R (1) complex numbers, hence has a convergent subsequence { X f1 dµn }. Now, R (1) the sequence { X f2 dµn } is bounded, and hence has a convergent sub- R (2) R (2) sequence { X f2 dµn }. Notice that { X f1 dµn } is also convergent. We (i) continue in this manner, to get for each i a subsequence {µn } of {µn} (i) (j) R (i) such that for all j ≤ i, {µn } is a subsequence of {µn } and { X fj dµn } (n) R (n) converges. Consider the diagonal sequence {µn }, then { X fj dµn } con- R (n) verges for all j, and hence { X f dµn } converges for all f ∈ C(X). Now R (n) deﬁne J : C(X) → C by J(f) = limn→∞{ X f dµn }. Then, J is linear, continuous (|J(f)| ≤ supx∈X |f(x)|), positive and J(1) = 1. Thus, by Riesz Representation Theorem, there exists a µ ∈ M(X) such that R (n) R (n) J(f) = limn→∞{ X f dµn } = X f dµ. Therefore, limn→∞ µn = µ, and M(X) is compact. Let T : X → X be a continuous transformation. Since B is generated by the open sets, then T is measurable with respect to B. Furthermore, T induces in a natural way, an operator T : M(X) → M(X) given by (T µ)(A) = µ(T −1A)

i for all A ∈ B. Then T µ(A) = µ(T −iA). Using a standard argument, one can easily show that Z Z f(x) d(T µ)(x) = f(T x)dµ(x) X X for all continuous functions f on X. Note that T is measure preserving with respect to µ ∈ M(X) if and only if T µ = µ. Equivalently, µ is measure preserving if and only if Z Z f(x) dµ(x) = f(T x) dµ(x) X X for all continuous functions f on X. Let M(X,T ) = {µ ∈ M(X): T µ = µ} be the collection of all probability measures under which T is measure preserving. 92 Invariant measures for continuous transformations

Theorem 6.1.5 Let T : X → X be continuous, and {σn} a sequence in M(X). Deﬁne a sequence {µn} in M(X) by

n−1 1 X i µ = T σ . n n n i=0

Then, any limit point µ of {µn} is a member of M(X,T ).

Proof We need to show that for any continuous function f on X, one R R has X f(x) dµ(x) = X f(T x) dµ. Since M(X) is compact there exists a

µ ∈ M(X) and a subsequence {µni } such that µni → µ in M(X). Now for any f continuous, we have Z Z Z Z

f(T x) dµ(x) − f(x) dµ(x) = lim f(T x) dµnj (x) − f(x) dµnj (x) X X j→∞ X X Z nj −1 1 X i+1 i = lim f(T x) − f(T x) dσn (x) j→∞ n j j X i=0 1 Z nj = lim (f(T x) − f(x)) dσnj (x) j→∞ nj X 2 sup |f(x)| ≤ lim x∈X = 0. j→∞ nj

Therefore µ ∈ M(X,T ).

Theorem 6.1.6 Let T be a continuous transformation on a compact metric space. Then,

(i) M(X,T ) is a compact convex subset of M(X).

(ii) µ ∈ M(X,T ) is an extreme point (i.e. µ cannot be written in a non- trivial way as a convex combination of elements of M(X,T )) if and only if T is ergodic with respect to µ.

Proof (i) Clearly M(X,T ) is convex. Now let {µn} be a sequence in M(X,T ) converging to µ in M(X). We need to show that µ ∈ M(X,T ). Existence 93

Since T is continuous, then for any continuous function f on X, the function f ◦ T is also continuous. Hence, Z Z f(T x) dµ(x) = lim f(T x) dµn(x) X n→∞ X Z = lim f(x) dµn(x) n→∞ X Z = f(x) dµ(x). X Therefore, T is measure preserving with respect to µ, and µ ∈ M(X,T ). (ii) Suppose T is ergodic with respect to µ, and assume that

µ = pµ1 + (1 − p)µ2 for some µ1, µ2 ∈ M(X,T ), and some 0 < p ≤ 1. We will show that µ = µ1. Notice that the measure µ1 is absolutely continuous with respect to µ, and T is ergodic with respect to µ, hence by Theorem (2.1.2) we have µ1 = µ. Conversely, (we prove the contrapositive) suppose that T is not ergodic with respect to µ. Then there exists a measurable set E such that T −1E = E, and 0 < µ(E) < 1. Deﬁne measures µ1, µ2 on X by µ(B ∩ E) µ (B ∩ (X \ E)) µ (B) = and µ (B) = . 1 µ(E) 1 µ(X \ E)

Since E and X \ E are T -invariant sets, then µ1, µ2 ∈ M(X,T ), and µ1 6= µ2 since µ1(E) = 1 while µ2(E) = 0. Furthermore, for any measurable set B,

µ(B) = µ(E)µ1(B) + (1 − µ(E))µ2(B), i.e. µ1 is a non-trivial convex combination of elements of M(X,T ). Thus, µ is not an extreme point of M(X,T ). Since the Banach space C(X) of all continuous functions on X (under the sup norm) is separable i.e. C(X) has a countable dense subset, one gets the following strengthening of the Ergodic Theorem. Theorem 6.1.7 If T : X → X is continuous and µ ∈ M(X,T ) is ergodic, then there exists a measurable set Y such that µ(Y ) = 1, and

n−1 1 X Z lim f(T ix) = f(x) dµ(x) n→∞ n i=0 X for all x ∈ Y , and f ∈ C(X). 94 Invariant measures for continuous transformations

Proof Choose a countable dense subset {fk} in C(X). By the Ergodic Theorem, for each k there exists a subset Xk with µ(Xk) = 1 and

n−1 Z 1 X i lim fk(T x) = fk(x) dµ(x) n→∞ n i=0 X

T∞ for all x ∈ Xk. Let Y = k=1 Xk, then µ(Y ) = 1, and

n−1 Z 1 X i lim fk(T x) = fk(x)dµ(x) n→∞ n i=0 X for all k and all x ∈ Y . Now, let f ∈ C(X), then there exists a subsequence {fkj } converging to f in the supremum norm, and hence is uniformly convergent. For any x ∈ Y , using uniform convergence and the dominated convergence theorem, one gets

n−1 n−1 1 X i 1 X i lim f(T x) = lim lim fk (T x) n→∞ n n→∞ j→∞ n j i=0 i=0 n−1 1 X i = lim lim fk (T x) j→∞ n→∞ n j i=0 Z Z

= lim fkj dµ = f dµ. j→∞ X X

Theorem 6.1.8 Let T : X → X be continuous, and µ ∈ M(X,T ). Then T is ergodic with respect to µ if and only if

n−1 1 X δ i → µ a.e. n T x i=0

Proof Suppose T is ergodic with respect to µ. Notice that for any f ∈ C(X), Z i f(y) d(δT ix)(y) = f(T x), X Existence 95

Hence by theorem 6.1.7, there exists a measurable set Y with µ(Y ) = 1 such that

n−1 Z n−1 Z 1 X 1 X i lim f(y) d(δT ix)(y) = lim f(T x) = f(y) dµ(y) n→∞ n n→∞ n i=0 X i=0 X

1 Pn−1 for all x ∈ Y , and f ∈ C(X). Thus, n i=0 δT ix → µ for all x ∈ Y .

1 Pn−1 Conversely, suppose n i=0 δT ix → µ for all x ∈ Y , where µ(Y ) = 1. Then for any f ∈ C(X) and any g ∈ L1(X, B, µ) one has

n−1 1 X Z lim f(T ix)g(x) = g(x) f(y) dµ(y). n→∞ n i=0 X

By the dominated convergence theorem

n−1 1 X Z Z Z lim f(T ix)g(x) dµ(x) = g(x)dµ(x) f(y) dµ(y). n→∞ n i=0 X X X

Now, let F,G ∈ L2(X, B, µ). Then, G ∈ L1(X, B, µ) so that

n−1 1 X Z Z Z lim f(T ix)G(x) dµ(x) = G(x)dµ(x) f(y)dµ(y) n→∞ n i=0 X X X

for all f ∈ C(X). Let > 0, there exists f ∈ C(X) such that ||F − f||2 < which implies that | R F dµ − R fdµ| < . Furthermore, there exists N so that for n ≥ N one has

n−1 Z 1 X Z Z f(T ix)G(x) dµ(x) − Gdµ f dµ < . n X i=0 X X 96 Invariant measures for continuous transformations

Thus, for n ≥ N one has

n−1 Z 1 X Z Z F (T ix)G(x) dµ(x) − Gdµ F dµ n X i=0 X X Z n−1 1 X i i ≤ F (T x) − f(T x) |G(x)| dµ(x) n X i=0 n−1 Z 1 X Z Z + f(T ix)G(x)dµ(x) − Gdµ fdµ n X i=0 X X Z Z Z Z

+ fdµ Gdµ − F dµ G dµ X X X X < ||G||2 + + ||G||2. Thus,

n−1 1 X Z Z Z lim F (T ix)G(x) dµ(x) = G(x) dµ(x) F (y) dµ(y) n→∞ n i=0 X X X for all F,G ∈ L2(X, B, µ) and x ∈ Y . Taking F and G to be indicator functions, one gets that T is ergodic. Exercise 6.1.1 Let X be a compact metric space and T : X → X be a continuous homeomorphism. Let x ∈ X be periodic point of T of period n, i.e. T nx = x and T jx 6= x for j < i. Show that if µ ∈ M(X,T ) is ergodic n−1 1 X and µ({x}) > 0, then µ = δ i . n T x i=0 6.2 Unique Ergodicity

A continuous transformation T : X → X on a compact metric space is uniquely ergodic if there is only one T -invariant probabilty measure µ on X. In this case, M(X,T ) = {µ}, and µ is necessarily ergodic, since µ is an extreme point of M(X,T ). Recall that if ν ∈ M(X,T ) is ergodic, then there exists a measurable subset Y such that ν(Y ) = 1 and

n−1 1 X Z lim f(T ix) = f(y) dν(y) n→∞ n i=0 X Unique Ergodicity 97 for all x ∈ Y and all f ∈ C(X). When T is uniquely ergodic we will see that we have a much stronger result.

Theorem 6.2.1 Let T : X → X be a continuous transformation on a compact metric space X. Then the following are equivalent:

n−1 1 X (i) For every f ∈ C(X), the sequence { f(T jx)} converges uniformly n j=0 to a constant.

n−1 1 X (ii) For every f ∈ C(X), the sequence { f(T jx)} converges pointwise n j=0 to a constant.

(iii) There exists a µ ∈ M(X,T ) such that for every f ∈ C(X) and all x ∈ X. n−1 1 X Z lim f(T ix) = f(y) dµ(y). n→∞ n i=0 X (iv) T is uniquely ergodic.

Proof (i) ⇒ (ii) immediate. (ii) ⇒ (iii) Deﬁne L : C(X) → C by

n−1 1 X L(f) = lim f(T ix). n→∞ n i=0 By assumption L(f) is independent of x, hence L is well deﬁned. It is easy to see that L is linear, continuous (|L(f)| ≤ supx∈X |f(x)|), positive and L(1) = 1. Thus, by Riesz Representation Theorem there exists a probability measure µ ∈ M(X) such that Z L(f) = f(x) dµ(x) X for all f ∈ C(x). But

n−1 1 X L(f ◦ T ) = lim f(T i+1x) = L(f). n→∞ n i=0 98 Invariant measures for continuous transformations

Hence, Z Z f(T x) dµ(x) = f(x) dµ(x). X X Thus, µ ∈ M(X,T ), and for every f ∈ C(X),

n−1 1 X Z lim f(T ix) = f(x) dµ(x) n→∞ n i=0 X for all x ∈ X. (iii) ⇒ (iv) Suppose µ ∈ M(X,T ) is such that for every f ∈ C(X),

n−1 1 X Z lim f(T ix) = f(x) dµ(x) n→∞ n i=0 X for all x ∈ X. Assume ν ∈ M(X,T ), we will show that µ = ν. For any f ∈ n−1 1 X C(X), since the sequence { f(T jx)} converges pointwise to the constant n j=0 R function X f(x) dµ(x), and since each term of the sequence is bounded in absolute value by the constant supx∈X |f(x)|, it follows by the Dominated Convergence Theorem that

n−1 Z 1 X Z Z Z lim f(T ix) dν(x) = f(x) dµ(x)dν(y) = f(x) dµ(x). n→∞ n X i=0 X X X But for each n,

n−1 Z 1 X Z f(T ix) dν(x) = f(x) dν(x). n X i=0 X R R Thus, X f(x) dµ(x) = X f(x) dν(x), and µ = ν. (iv) ⇒ (i) The proof is done by contradiction. Assume M(X,T ) = {µ} and n−1 1 X suppose g ∈ C(X) is such that the sequence { g ◦T j} does not converge n j=0 uniformly on X. Then there exists > 0 such that for each N there exists n > N and there exists xn ∈ X such that n−1 Z 1 X j g(T xn) − g dµ ≥ . n j=0 X Unique Ergodicity 99

Let n−1 n−1 1 X 1 X j µ = δ j = T δ . n n T xn n xn j=0 j=0 Then, Z Z | gdµn − g dµ| ≥ . X X

Since M(X) is compact. there exists a subsequence µni converging to ν ∈ M(X). Hence, Z Z | g dν − g dµ| ≥ . X X By Theorem (6.1.5), ν ∈ M(X,T ) and by unique ergodicity µ = ν, which is a contradiction.

Example If Tθ is an irrational rotation, then Tθ is uniquely ergodic. This is a consequence of the above theorem and Weyl’s Theorem on uniform distri- bution: for any Riemann integrable function f on [0, 1), and any x ∈ [0, 1), one has n−1 1 X Z lim f(x + iθ − bx + iθc) = f(y) dy. n→∞ n i=0 X As an application of this, let us consider the following question. Consider the sequence of ﬁrst digits

{1, 2, 4, 8, 1, 3, 6, 1,...} obtained by writing the ﬁrst decimal digit of each term in the sequence

{2n : n ≥ 0} = {1, 2, 4, 8, 16, 32, 64, 128,...}.

For each k = 1, 2,..., 9, let pk(n) be the number of times the digit k appears in the ﬁrst n terms of the ﬁrst digit sequence. The asymptotic relative fre- p (n) quency of the digit k is then lim k . We want to identify this limit n→∞ n for each k ∈ {1, 2,... 9}. To do this, let θ = log10 2, then θ is irrational. For k = 1, 2,..., 9, let Jk = [log10 k, log10(k + 1)). By unique ergodicity of Tθ, we have for each k = 1, 2,..., 9, n−1 1 X j k + 1 lim 1Jk (T (0)) = λ(Jk) = log10 . n→∞ n θ k i=0 100 Invariant measures for continuous transformations

Returning to our original problem, notice that the ﬁrst digit of 2i is k if and only if k · 10r ≤ 2i < (k + 1) · 10r for some r ≥ 0. In this case,

r + log10 k ≤ i log10 2 = iθ < r + log10(k + 1).

This shows that r = biθc, and

log10 k ≤ iθ − biθc < log10(k + 1).

i i But Tθ (0) = iθ − biθc, so that Tθ (0) ∈ Jk. Summarizing, we see that the ﬁrst i i digit of 2 is k if and only if Tθ (0) ∈ Jk. Thus, n−1 pk(n) 1 X i k + 1 lim = lim 1Jk (Tθ (0)) = log10 . n→∞ n n→∞ n k i=0 Chapter 7

Topological Dynamics

In this chapter we will shift our attention from the theory of measure preserving transformations and ergodic theory to the study of topological dynamics. The field of topological dynamics also studies dynamical systems, but in a different setting: where ergodic theory is set in a measure space with a measure preserving transformation, topological dynamics focuses on a topological space with a continuous transformation. The two fields bear great similarity and seem to co-evolve. Whenever a new notion has proven its value in one field, analogies will be sought in the other. The two fields are, however, also complementary. Indeed, we shall encounter a most striking example of a map which exhibits its most interesting topological dynamics in a set of measure zero, right in the blind spot of the measure theoretic eye. In order to illuminate this interplay between the two fields we will be merely interested in compact metric spaces. The compactness assumption will play the role of the assumption of a finite measure space. The assumption of the existence of a metric is sufficient for applications and will greatly simplify all of our proofs. The chapter is structured as follows. We will first introduce several basic notions from topological dynamics and discuss how they are related to each other and to analogous notions from ergodic theory. We will then devote a separate section to topological entropy. To conclude the chapter, we will discuss some examples and applications.

101 102 Basic Notions 7.1 Basic Notions

Throughout this chapter X will denote a compact metric space. Unless stated otherwise, we will always denote the metric by d and occasionally write (X, d) to denote a metric space if there is any risk of confusion. We will assume that the reader has some familiarity with basic (analytic) topology. To jog the reader’s memory, a brief outline of these basics is included as an appendix. Here we will only summarize the properties of the space X, for future reference.

Theorem 7.1.1 Let X be a compact metric space, Y a topological space and f : X −→ Y continuous. Then,

• X is a Hausdorﬀ space

• Every closed subspace of X is compact

• Every compact subspace of X is closed

• X has a countable basis for its topology

• X is normal, i.e. for any pair of disjoint closed sets A and B of X there are disjoint open sets U, V containing A and B, respectively

• If O is an open cover of X, then there is a δ > 0 such that every subset A of X with diam(A) < δ is contained in an element of O. We call δ > 0 a Lebesgue number for O.

• X is a Baire space

• f is uniformly continuous

• If Y is an ordered space then f attains a maximum and minimum on X

• If A and B are closed sets in X and [a, b] ⊂ R, then there exists a continuous map g : X −→ [a, b] such that g(x) = a for all x ∈ A and g(x) = b for all x ∈ b 103

The reader may recognize that this theorem contains some of the most important results in analytic topology, most notably the Lebesgue number lemma, Baire’s category theorem, the extreme value theorem and Urysohn’s lemma. Let us now introduce some concepts of topological dynamics: topological transitivity, topological conjugacy, periodicity and expansiveness.

Deﬁnition 7.1.1 Let T : X −→ X be a continuous transformation. For n x ∈ X, the forward orbit of x is the set FOT (x) = {T (x)|n ∈ Z≥0}. T is called one-sided topologically transitive if for some x ∈ X, FOT (x) is dense in X. n If T is invertible, the orbit of x is deﬁned as OT (x) = {T (x)|n ∈ Z}. If T is a homeomorphism, T is called topologically transitive if for some x ∈ X OT (x) is dense in X.

In the literature sometimes a stronger version of topological transitivity is introduced, called minimality.

Deﬁnition 7.1.2 A homeomorphism T : X −→ X is called minimal if every x ∈ X has a dense orbit in X.

Example 7.1.1 Let X be the topological space X = {1, 2,..., 1000} with the discrete topology. X is a ﬁnite space, hence compact, and the discrete metric ( 0 if x = y ddisc(x, y) = 1 if x 6= y induces the discrete topology on X. Thus X is a compact metric space. If we now deﬁne T : X −→ X by the permutation (12 ··· 1000), i.e. T (1) = 2, T (2) = 3,...,T (1000) = 1, then it is easy to see that T is a homeomorphism. Moreover, OT (x) = X for all x ∈ X, so T is minimal.

Example 7.1.2 Fix n ∈ Z>0. Let X be the topological space X = {0} ∪ 1 { nk |k ∈ Z>0} equipped with the subspace topology inherited from R. Then X is a compact metric space. Indeed, if O is an open cover of X, then O contains an open set U which contains 0. U contains all but finitely many points of X, so by adding an open set Uy to U from O with y ∈ Uy, for each x y not in U, we obtain a finite subcover. Define T : X −→ X by T (x) = n . 104 Basic Notions

Note that T is continuous, but not surjective as T −1({1}) = ∅. Observe that FOT (1) = X − {0} and X − {0} = X, so T is one-sided topologically transitive as x = 1 has a dense forward orbit.

The following theorem summarizes some equivalent deﬁnitions of topological transitivity. Theorem 7.1.2 Let T : X −→ X be a homeomorphism of a compact metric space. Then the following are equivalent: 1. T is topologically transitive 2. If A is a closed set in X satisfying TA = A, then either A = X or A has empty interior (i.e. X − A is dense in X)

3. For any pair of non empty open sets U, V in X there exists an n ∈ Z such that T nU ∩ V 6= ∅ Proof

Recall that OT (x) is dense in X if and only if for every U open in X, U ∩ OT (x) 6= ∅. (1)⇒(2): Suppose that OT (x) = X and let A be as stated. Suppose that A does not have an empty interior, then there exists an open set U such that U is open, non-empty and U ⊂ A. Now, we can ﬁnd a p ∈ Z such that p n T (x) ∈ U ⊂ A. Since T A = A for all n ∈ Z, OT (x) ⊂ A and taking closures we obtain X ⊂ A. Hence, either A has empty interior, or A = X. ∞ n (2)⇒(3): Suppose U, V are open and non-empty. Then W = ∪n=−∞T U is open, non-empty and invariant under T . Hence, A = X − W is closed, TA = A and A 6= X. By (2), A has empty interior and therefore W is dense in X. Thus W ∩ V 6= ∅, which implies (3). (3)⇒(1): For an arbitrary y ∈ X we have: y ∈ OT (x) if and only if every m open neighborhood V of y intersects OT (x), i.e. T (x) ∈ V for some m ∈ Z. ∞ By theorem 7.1.1 there exists a countable basis U = {Un}n=1 for the topology of X. Since every open neighborhood of y contains a basis element Ui such that y ∈ Ui, we see that OT (x) = X if and only if for every n ∈ Z>0 there is m some m ∈ Z such that T (x) ∈ Un. Hence, ∞ ∞ \ [ m {x ∈ X|OT (x) = X} = T Un n=1 m=−∞ 105

∞ m But ∪m=−∞T Un is T -invariant and therefore intersects every open set in ∞ m X by (3). Thus, ∪m=−∞T Un is dense in X for every n. Now, since X is a Baire space (c.f. theorem 7.1.1), we see that {x ∈ X|OT (x) = X} is itself dense in X and thus certainly non-empty (as X 6= ∅).

An analogue of this theorem exists for one-sided topological transitivity, see [W]. Note that theorem 7.1.2 clearly resembles theorem (1.6.1) and (2) implies that we can view topological transitivity as an analogue of ergodicity (in some sense). This is also reﬂected in the following (partial) analogue of theorem (1.7.1).

Theorem 7.1.3 Let T : X −→ X be continuous and one-sided topologically transitive or a topologically transitive homeomorphism. Then every continuous T -invariant function is constant.

Proof

Let f : X −→ Y be continuous on X and suppose f ◦ T = f. Then n f ◦ T = f, so f is constant on (forward) orbits of points. Let x0 ∈ X have a dense (forward) orbit in X and suppose f(x0) = c for some c ∈ Y . Fix ε > 0 and let x ∈ X be arbitrary. By continuity of f at x, we can ﬁnd a δ > 0 such that d(f(x), f(˜x)) < ε for anyx ˜ ∈ X with d(x, x˜) < δ. But then, since x0 has a dense (forward) orbit in X, there is some n ∈ Z (n ∈ Z≥0) such that n d(x, T (x0)) < δ. Hence,

n d(f(x), c) = d(f(x), f(T (x0))) < ε

Since ε > 0 was arbitrary, our proof is complete.

Exercise 7.1.1 Let X be a compact metric space with metric d and let T : X −→ X is a topologically transitive homeomorphism. Show that if T is an isometry (i.e. d(T (x),T (y)) = d(x, y), for all x, y ∈ X) then T is minimal.

Deﬁnition 7.1.3 Let X be a topological space, T : X −→ X a transformation and x ∈ X. Then x is called a periodic point of T of period n (n ∈ Z>0) if T n(x) = x and T m(x) 6= x for m < n. A periodic point of period 1 is called a ﬁxed point. 106 Basic Notions

Example 7.1.3 The map T : R −→ R, deﬁned by T (x) = −x has one ﬁxed point, namely x = 0. All other points in the domain are periodic points of period 2.

Deﬁnition 7.1.4 Let X be a compact metric space and T : X −→ X a homeomorphism. T is said to be expansive if there exist a δ > 0 such that: if x 6= y then there exists an n ∈ Z such that d(T n(x),T n(y)) > δ. δ is called an expansive constant for T .

Example 7.1.4 Let X be a ﬁnite space with the discrete topology. X is a compact metric space, since the discrete metric ddisc induces the topology on X. Now, any bijective map T : X −→ X is an expansive homeomorphism and any 0 < δ < 1 is an expansive constant for T .

For a non-trivial example of an expansive homeomorphism, see exercise 7.3.1. Expansiveness is closely related to the concept of a generator. Deﬁnition 7.1.5 Let X be a compact metric space and T : X −→ X a homeomorphism. A ﬁnite open cover α of X is called a generator for T if ∞ −n the set ∩n=−∞T An contains at most one point of X, for any collection ∞ {An}n=−∞, Ai ∈ α. Theorem 7.1.4 Let X be a compact metric space and T : X −→ X a homeomorphism. Then T is expansive if and only if T has a generator.

Proof

Suppose that T is expansive and let δ > 0 be an expansive constant for T . δ Let α be the open cover of X deﬁned by α = {B(a, 2 )|a ∈ X}. By compact- ∞ −n ness, there exists a ﬁnite subcover β of α. Suppose that x, y ∈ ∩n=−∞T Bn, m m Bi ∈ β for all i. Then d(T (x),T (y)) ≤ δ, for all m ∈ Z. But by expansiveness, there exists an l ∈ Z such that d(T l(x),T l(y)) > δ, a contradiction. Hence, x = y. We conclude that β is a generator for T . Conversely, suppose that α is a generator for T . By theorem 7.1.1, there exists a Lebesgue number δ > 0 for the open cover α. We claim that δ/4 is an expansive constant for T . Indeed, if x, y ∈ X are such that d(T m(x),T m(y)) ≤ δ/4, for all m ∈ Z. Then, for all m ∈ Z, the closed ball B(T m(x), δ/4) contains T m(y) and has diameter δ/2 < δ. Hence,

m B(T (x), δ/4) ⊂ Am, 107

∞ −m for some Am ∈ α. It follows that x, y ∈ ∩m=−∞T Am.But α is a generator, ∞ −m so ∩m=−∞T Am contains at most one point. We conclude that x = y, T is expansive.

Exercise 7.1.2 In the literature, our definition of a generator is sometimes ∞ −n called a weak generator. A generator is then defined by replacing ∩n=−∞T An ∞ −n by ∩n=−∞T An in definition 7.1.5. Show that in a compact metric space both concepts are in fact equivalent.

Exercise 7.1.3 Prove the following basic properties of an expansive homeomorphism T : X −→ X: a. T is expansive if and only if T n is expansive (n 6= 0) b. If A is a closed subset of X and T (A) = A, then the restriction of T to A, T |A, is expansive c. Suppose S : Y −→ Y is an expansive homeomorphism. Then the product map T × S : X × Y −→ X × Y is expansive with respect to the metric d = max{dX , dY }.

Deﬁnition 7.1.6 Let X, Y be compact spaces and let T : X −→ X, S : Y −→ Y be homeomorphisms. T is said to be topologically conjugate to S if there exists a homeomorphism φ : X −→ Y such that φ ◦ T = S ◦ φ. φ is called a (topological) conjugacy.

Note that topological conjugacy deﬁnes an equivalence relation on the space of all homeomorphisms. The term ‘topological conjugacy’ is, in a sense, a misnomer. The following theorem shows that topological conjugacy can be considered as the counter- part of a measure preserving isomorphism. Theorem 7.1.5 Let X, Y be compact spaces and let T : X −→ X, S : Y −→ Y be topologically conjugate homeomorphisms. Then, 1. T is topologically transitive if and only if S is topologically transitive

2. T is minimal if and only if S is minimal

3. T is expansive if and only if S is expansive 108 Two Deﬁnitions

Proof

The proof of the ﬁrst two statements is trivial. For the third statement we note that ∞ ∞ ∞ \ −n −1 \ −1 −n −1 \ −n T (φ An) = φ ◦ S (An) = φ ( S An) n=−∞ n=−∞ n=−∞ The lefthandside of the equation contains at most one point if and only if the righthandside does, so a ﬁnite open cover α is a generator for S if and only if φ−1α = {φ−1A|A ∈ α} is a generator for T . Hence the result follows from theorem 7.1.4.

7.2 Topological Entropy

Topological entropy was first introduced as an analogue of the succesful concept of measure theoretic entropy. It will turn out to be a conjugacy invariant and is therefore a useful tool for distinguishing between topological dynamical systems. The definition of topological entropy comes in two flavors. The first is in terms of open covers and is very similar to the definition of measure theoretic entropy in terms of partitions. The second (and chronologically later) definition uses (n,ε) separating and spanning sets. Interestingly, the latter definition was a topological dynamical discovery, which was later engineered into a similar definition of measure theoretic entropy.

7.2.1 Two Definitions We will start with a definition of topological entropy similar to the definition of measure theoretic entropy introduced earlier. To spark the reader’s recognition of the similarities between the two, we use analogous notation and terminology. Before kicking off, let us first make a remark on the assumptions about our space X. The first definition presented will only require X to be compact. The second, on the other hand, only requires X to be a metric space. We will obscure from these subtleties and simply stick to the context of a compact metric space, in which the two definitions will turn out to be equivalent. Nevertheless, the reader should be aware that both definitions represent generalizations of topological entropy in different directions 109 and are therefore interesting in their own right. Let X be a compact metric space and let α, β be open covers of X. We say that β is a refinement of α, and write α ≤ β, if for every B ∈ β there is an A ∈ α such that B ⊂ A. The common refinement of α and β is defined to n be α ∨ β = {A ∩ B|A ∈ α, B ∈ β}. For a finite collection {αi}i=1 of open Wn n covers we may define i=1 αi = {∩i=1Aji |Aji ∈ αi}. For a continuous transformation T : X −→ X we define T −1α = {T −1A|A ∈ α}. Note that these are all again open covers of X. Finally, we define the diameter of an open cover α as diam(α) := supA∈α diam(A), where diam(A) = supx,y∈A d(x, y).

Exercise 7.2.1 Show that T −1(α ∨ β) = T −1(α) ∨ T −1(β) and show that α ≤ β implies T −1α ≤ T −1β.

Definition 7.2.1 Let α be an open cover of X and let N(α) be the number of sets in a finite subcover of α of minimal cardinality. We define the entropy of α to be H(α) = log(N(α)).

The following proposition summarizes some easy properties of H(α). The proof is left as an exercise. Proposition 7.2.1 Let α be an open cover of X and let H(α) be the entropy of α. Then 1. H(α) ≥ 0 2. H(α) = 0 if and only if N(α) = 1 if and only if X ∈ α 3. If α ≤ β, then H(α) ≤ H(β) 4. H(α ∨ β) ≤ H(α) + H(β) 5. For T : X −→ X continuous we have H(T −1α) ≤ H(α). If T is surjective then H(T −1α) = H(α).

Exercise 7.2.2 Prove proposition 7.2.1 (Hint: If T is surjective then T (T −1A) = A).

We will now move on to the definition of topological entropy for a continuous transformation with respect to an open cover and subsequently make this definition independent of open covers. 110 Two Definitions

Deﬁnition 7.2.2 Let α be an open cover of X and let T : X −→ X be continuous. We deﬁne the topological entropy of T with respect to α to be:

n−1 1 _ h(T, α) = lim H( T −iα) n→∞ n i=0 We must show that h(T, α) is well-deﬁned, i.e. that the right hand side exists.

1 Wn−1 −i Theorem 7.2.1 limn→∞ n H( i=1 T α) exists. Proof

Deﬁne n−1 _ −i an = H( T α) i=0

Then, by proposition (4.2.2),, it suﬃces to show that {an} is subadditive. Now, by (4) of proposition 7.2.1 and exercise 7.2.1

n+p−1 _ −i an+p = H( T α) i=0 n−1 p−1 _ _ ≤ H( T −iα) + H(T −n T −jα) i=0 j=0

≤ an + ap

This completes our proof.

Proposition 7.2.2 h(T, α) satisﬁes the following:

1. h(T, α) ≥ 0

2. If α ≤ β, then h(T, α) ≤ h(T, β)

3. h(T, α) ≤ H(α) 111

Proof

These are easy consequences of proposition 7.2.1. We will only prove the third statement. We have:

n−1 n−1 _ X H( T −iα) ≤ H(T −iα) i=0 i=0 ≤ nH(α)

Here we have subsequently used (4) and (5) of the proposition.

Finally, we arrive at our sought deﬁnition:

Deﬁnition 7.2.3 (I) Let T : X −→ X be continuous. We deﬁne the topological entropy of T to be:

h1(T ) = sup h(T, α) α where the supremum is taken over all open covers of X.

We will defer a discussion of the properties of topological entropy till the end of this chapter.

Let us now turn to the second approach to defining topological entropy. This approach was first taken by E.I. Dinaburg. We shall first define the main ingredients. Let d be the metric on the compact metric space X and for each n ∈ Z≥0 define a new metric by:

i i dn(x, y) = max d(T (x),T (y)) 0≤i≤n−1

Definition 7.2.4 Let n ∈ Z≥0, ε > 0 and A ⊂ X. A is called (n, ε)- spanning for X if for all x ∈ X there exists a y ∈ A such that dn(x, y) < ε. Define span(n, ε, T ) to be the minimum cardinality of an (n, ε)-spanning set. A is called (n, ε)-separated if for any x, y ∈ A dn(x, y) > ε. We define sep(n, ε, T ) to be the maximum cardinality of an (n, ε)-separated set. 112 Two Definitions

Figure 7.1: An (n, ε)-separated set (left) and (n, ε)-spanning set (right). The ε dotted circles represent open balls of dn-radius 2 (left) and of dn-radius ε (right), respectively.

Note that we can extend this deﬁnition by letting A ⊂ K, where K is a compact subset of X. This generalization to the context of non-compact metric spaces was devised by R.E. Bowen. See [W] for an outline of this extended framework. We can also formulate the above in terms of open balls. If B(x, r) = {y ∈ X|d(x, y) < r} is an open ball in the metric d, then the open ball with centre x and radius r in the metric dn is given by:

n−1 \ −i i Bn(x, r) := T B(T (x), r) i=0 Hence, A is (n, ε)-spanning for X if:

[ X = Bn(a, ε) a∈A and A is (n, ε)-separated if:

(A − {a}) ∩ Bn(a, ε) = ∅ for all a ∈ A

Deﬁnition 7.2.5 For n ∈ Z≥0 and ε > 0 we deﬁne cov(n, ε, T ) to be the minimum cardinality of a covering of X by open sets of dn-diameter less than ε.

The following theorem shows that the above notions are actually two sides of the same coin. 113

Theorem 7.2.2 cov(n, 3ε, T ) ≤ span(n, ε, T ) ≤ sep(n, ε, T ) ≤ cov(n, ε, T ). Furthermore, sep(n, ε, T ), span(n, ε, T ) and cov(n, ε, T ) are ﬁnite, for all n, ε > 0 and continuous T . Proof

We will only prove the last two inequalities. The ﬁrst is left as an exercise to the reader. Let A be an (n, ε)-separated set of cardinality sep(n, ε, T ). Suppose A is not (n, ε)-spanning for X. Then there is some x ∈ X such that dn(x, a) ≥ ε, for all a ∈ A. But then A ∪ {x} is an (n, ε)-separated set of cardinality larger than sep(n, ε, T ). This contradiction shows that A is an (n, ε)-spanning set for X. The second inequality now follows since the cardinality of A is at least as large as span(n, ε, T ). To prove the third inequality, let A be an (n, ε)-separated set of cardinality sep(n, ε, T ). Note that if α is an open cover of dn-diameter less than ε, then no element of α can contain more than 1 element of A. This holds in particular for an open cover of minimal cardinality, so the third inequality is proved. The ﬁnal statement follows for span(n, ε, T ) and cov(n, ε, T ) from the compactness of X and subsequently for sep(n, ε, T ) by the last inequality.

Exercise 7.2.3 Finish the proof of the above theorem by proving the ﬁrst inequality.

1 Lemma 7.2.1 The limit cov(ε, T ) := limn→∞ n log(cov(n, ε, T )) exists and is ﬁnite. Proof

Fix ε > 0. Let α, β be open covers of X consisting of sets with dm- diameter and dn-diameter, smaller than ε and with cardinalities cov(m, ε, T ) and cov(n, ε, T ), respectively. Pick any A ∈ α and B ∈ β. Then, for x, y ∈ A ∩ T −mB,

i i dm+n(x, y) = max d(T (x),T (y)) 0≤i≤m+n−1 ≤ max{ max d(T i(x),T i(y)), max d(T j(x),T j(y))} 0≤i≤m−1 m≤j≤m+n−1 < ε 114 Equivalence of the two Deﬁnitions

−m −m Thus, A∩T B has dm+n-diameter less than ε and the sets {A∩T B|A ∈ α, B ∈ β} form a cover of X. Hence,

cov(m + n, ε, T ) ≤ cov(m, ε, T )· cov(n, ε, T )

∞ Now, since log is an increasing function, the sequence {an}n=1 deﬁned by an = log cov(n, ε, T ) is subadditive, so the result follows from proposition 5.

Motivated by the above lemma and theorem, we have the following deﬁnition

Deﬁnition 7.2.6 (II) The topological entropy of a continuous transformation T : X −→ X is given by:

1 h2(T ) = lim lim log(cov(n, ε, T )) ε↓0 n→∞ n 1 = lim lim log(span(n, ε, T )) ε↓0 n→∞ n 1 = lim lim log(sep(n, ε, T )) ε↓0 n→∞ n

Note that there seems to be a concealed ambiguity in this deﬁnition, since h2(T ) still depends on the chosen metric d. As we will see later on, in compact metric spaces h2(T ) is independent of d, as long as it induces the topology of X. However, there is deﬁnitely an issue at this point if we drop our assumption of compactness of the space X.

7.2.2 Equivalence of the two Definitions We will now show that definitions (I) and (II) of topological entropy coincide on compact metric spaces. This will justify the use of the notation h(T ) for both definitions of topological entropy. In the meantime, we will continue to write h1(T ) for topological entropy in the definition using open covers and h2(T ) for the second definition of entropy. After all the work done in the last section, our only missing ingredient is the following lemma: 115

∞ Lemma 7.2.2 Let X be a compact metric space. Suppose {αn}n=1 is a sequence of open covers of X such that limn→∞ diam(αn) = 0, where diam is the diameter under the metric d. Then limn→∞ h1(T, αn) = h1(T ). Proof

We will only prove the lemma for the case h1(T ) < ∞. Let ε > 0 be arbitrary and let β be an open cover of X such that h1(T, β) > h1(T ) − ε. By theorem 7.1.1 there exists a Lebesgue number δ > 0 for β. By assumption, there exists an N > 0 such that for n ≥ N we have diam(αn) < δ. So, for such n, we ﬁnd that for any A ∈ αn there exists a B ∈ β such that A ⊂ B, i.e. β ≤ αn. Hence, by proposition 7.2.2 and the above,

h1(T ) − ε < h1(T, β) ≤ h1(T, αn) ≤ h1(T ) for all n ≥ N.

Exercise 7.2.4 Finish the proof of lemma 7.2.2. That is, show that if h1(T ) = ∞, then limn→∞ h1(T, αn) = ∞. Theorem 7.2.3 For a continuous transformation T : X −→ X of a compact metric space X, deﬁnitions (I) and (II) of topological entropy coincide. Proof

Step 1: h2(T ) ≤ h1(T ). ∞ For any n, let {αk}k=1 be the sequence of open covers of X, defined by 1 αk = {Bn(x, 3k )|x ∈ X}. Then, clearly, the dn-diameter of αk is smaller than 1 Wn−1 −i k . Now, i=0 T αk is an open cover of X by sets of dn-diameter smaller 1 1 Wn−1 −i than k , hence cov(n, k ,T ) ≤ N( i=0 T αk). Since limk→∞ diam(αk) = 0, by lemma 7.2.2, we can subsequently take the log, divide by n, take the limit for n → ∞ and the limit for k → ∞ to obtain the desired result. Step 2: h1(T ) ≤ h2(T ). ∞ 1 2 Let {βk}k=1 be defined by βk = {B(x, k )|x ∈ X}. Note that δk = k is a δk Lebesgue number for βk, for each k. Let A be an (n, 2 )-spanning set for X δk i δk of cardinality span(n, 2 ,T ). Then, for each a ∈ A the ball B(T (a), 2 ) is i δk contained in a member of βk (for all 0 ≤ i < n), hence Bn(T (a), 2 ) is con- Wn−1 −i Wn−1 −i 1 tained in a member of i=0 T βk. Thus, N( i=0 T βk) ≤ span(n, 4k ,T ). Proceeding in the same way as in step 1, we obtain the desired inequality. 116 Equivalence of the two Definitions

Properties of Entropy We will now list some elementary properties of topological entropy. Theorem 7.2.4 Let X be a compact metric space and let T : X −→ X be continuous. Then h(T ) satisﬁes the following properties:

1. If the metrics d and d˜ both generate the topology of X, then h(T ) is the same under both metrics

2. h(T ) is a conjugacy invariant

n 3. h(T ) = n· h(T ), for all n ∈ Z>0 4. If T is a homeomorphism, then h(T −1) = h(T ). In this case, h(T n) = |n|· h(T ), for all n ∈ Z Proof

(1): Let (X, d) and (X, d˜) denote the two metric spaces. Since both induce the same topology, the identity maps i :(X, d) −→ (X, d˜) and j :(X, d˜) −→ (X, d) are continuous and hence by theorem 7.1.1 uniformly continuous. Fix ε1 > 0. Then, by uniform continuity, we can subsequently ﬁnd an ε2 > 0 and ε3 > 0 such that for all x, y ∈ X:

˜ d(x, y) < ε1 ⇒ d(x, y) < ε2 ˜ d(x, y) < ε2 ⇒ d(x, y) < ε3

Let A be an (n, ε1)-spanning set for T under d. Then A is also (n, ε2)- ˜ spanning for T under d. Analogously, every (n, ε2)-spanning set for T under ˜ d is (n, ε3)-spanning for T under d. Therefore, ˜ span(n, ε3, T, d) ≤ span(n, ε2,T, d) ≤ span(n, ε1, T, d) where the fourth argument emphasizes the metric. By subsequently taking the log, dividing by n, taking the limit for n → ∞ and the limit for ε1 ↓ 0 (so that ε2, ε3 ↓ 0) in these inequalities, we obtain the desired result. (2): Suppose that S : Y −→ Y is topologically conjugate to T , where Y is a 117

compact metric space, with conjugacy ψ : Y −→ X. Let dX be the metric on ˜ ˜ X. Then the map dY deﬁned by dY (x, y) = dX (ψ(x), ψ(y)) deﬁnes a metric on Y which induces the topology of Y . Indeed, let BdY (y0, η) be a basis element of the topology of Y . Then ψ(BdY (y0, η)) is open in X, as ψ is an open map. Hence, there is an open ball BdX (ψ(y0), η˜) ⊂ ψ(BdY (y0, η)), since −1 the open balls are a basis for the topology of X. Now, ψ (BdX (ψ(y0), η˜)) ⊂

BdY (y0, η) and −1 ˜ y ∈ ψ (BdX (ψ(y0), η˜)) ⇔ dY (y0, y) = dX (ψ(y), ψ(y0)) < η˜ hence, B ˜ (y , η˜) ⊂ B (y , η). Analogously, for any y ∈ Y and δ˜ > 0, there dY 0 dY 0 0 is a δ > 0 such that B (y , δ) ⊂ B ˜ (y , δ˜) dY 0 dY 0 ˜ Thus, dY induces the topology of Y . Now, for x1, x2 ∈ X, we have:

dX (T (x1),T (x2)) = dX (T ◦ ψ(y1),T ◦ ψ(y2))

= dX (ψ ◦ S(y1), ψ ◦ S(y2)) ˜ = dY (S(y1),S(y2))

Where we have used that x1 = ψ(y1), x2 = ψ(y2) for some y1, y2 ∈ Y , since ψ is a bijection. Thus, we see that the n−diameters (in the metrics dX and ˜ ˜ dY ) remain the same under ψ. Hence, cov(n, ε, T, dX ) = cov(n, ε, S, dY ), for ˜ all ε > 0 and n ∈ Z≥0. By (1), h(S) is the same under dY and dY , so it now easily follows that h(S) = h(T ). (3): Observe that max d(T ni(x),T ni(y)) ≤ max d(T j(x),T j(y)) 1≤i≤m−1 1≤j≤nm−1 n 1 n therefore, span(m, ε, T ) ≤ span(nm, ε, T ). This implies m span(m, ε, T ) ≤ n n nm span(nm, ε, T ), thus h(T ) ≤ nh(T ). Fix ε > 0. Then, by uniform continuity of T i on X (c.f. theorem 7.1.1), we can find a δ > 0 such that d(T i(x),T i(y)) < ε for all x, y ∈ X satisfying d(x, y) < δ and i = 0, . . . , n − 1. Hence, for x, y ∈ X satisfying d(x, y) < δ, n we find dn(x, y) < ε. Let A be an (m, δ)-spanning set for T . Then, for all x ∈ X there exists some a ∈ A such that max d(T ni(x),T ni(a)) < δ 0≤i≤m−1 118 Equivalence of the two Definitions

But then, by the above,

max max d(T ni+k(x),T ni+k(a)) = max d(T j(x),T j(a)) < ε 0≤k≤n−1 0≤i≤m−1 0≤j≤nm−1

Thus, span(mn, ε, T ) ≤ span(m, δ, T n) and it follows that h(T n) ≥ nh(T ) (4): Let A be an (n, ε)-separated set for T . Then, for any x, y ∈ A dn(x, y) > ε. But then,

max d(T −i(T n−1(x)),T −i(T n−1(y))) = max d(T i(x),T i(y)) > ε 0≤i≤n−1 0≤i≤n−1 so T n−1A is an (n, ε)-separated set for T −1 of the same cardinality. Con- versely, every (n, ε)-separated set B for T −1 gives the (n, ε)-separated set T −(n−1)B for T . Thus, sep(n, ε, T ) = sep(n, ε, T −1) and the ﬁrst statement readily follows. The second statement is a consequence of (3).

Exercise 7.2.5 Show that assertion (4) of theorem 7.2.4, i.e. h(T −1) = h(T ), still holds if X is merely assumed to be compact.

Exercise 7.2.6 Let (X, dX ), (Y, dY ) be compact metric spaces and T : X −→ X, S : Y −→ Y continuous. Show that h(T × S) = h(T ) + h(S). (Hint: ﬁrst show that the metric d = max{dX , dY } induces the product topology on X × Y ).

7.3 Examples

In this section we provide a few detailed examples to get some familiarity with the material discussed in this chapter. We derive some simple properties of the dynamical systems presented below, but leave most of the good stuﬀ for you to discover in the exercises. After all, the proof of the pudding is in the eating!

Example. (The Quadratic Map) Consider the following biological population model. Suppose we are studying a colony of penguins and wish to model the evolution of the population through time. We presume that there is a certain limiting population level, P ∗, which cannot be exceeded. When- ever the population level is low, it is supposed to increase, since there is 119 plenty of ﬁsh to go around. If, on the other hand, population is near the up- per limit level P ∗, it is expected to decrease as a result of overcrowding. We arrive at the following simple model for the population at time n, denoted by Pn:

∗ Pn+1 = λPn(P − Pn)

∗ Where λ is a ‘speed’ parameter. If we now set P = 1, we can think of Pn as the population at time n as a percentage of the limiting population level. Then, writing x = P0, the population dynamics are described by iteration of the following map:

Deﬁnition 7.3.1 The Quadratic Map with parameter λ, Qλ : R −→ R, is deﬁned to be

Qλ(x) = λx(1 − x)

Of course, in the context of the biological model formulated above, our attention is restricted to the interval [0, 1] since the population percentage cannot lie outside this interval. The quadratic map is deceivingly simple, as it in fact displays a wide range of interesting dynamics, such as periodicity, topological transitivity, bifurca- tions and chaos.

Let us take a closer look at the case λ > 4. For x ∈ (−∞, 0) ∪ (1, ∞) n it can be shown that Qλ(x) → −∞ as n → ∞. The interesting dynamics occur in the interval [0, 1]. Figure 7.2 shows a phase portrait for Qλ on [0, 1], which is nothing more than the graph of Qλ together with the graph of the line y = x. In a phase portrait the trajectory of a point under iteration of the map Qλ is easily visualized by iteratively project- ing from the line y = x to the graph and vice versa. From our phase portrait for Qλ it is immediately visible that there is a ‘leak’ in our interval, i.e. the subset A1 = {x ∈ [0, 1]|Qλ(x) ∈/ [0, 1]} is non-empty. Deﬁne i n An = {x ∈ [0, 1]|Qλ(x) ∈ [0, 1] for 0 ≤ i ≤ n − 1,Qλ(x) ∈/ [0, 1]} and set ∞ C = [0, 1]−∪n=1An. This procedure is reminiscent of the construction of the ‘classical’ Cantor set, by removing middle-thirds in subsequent steps. In the exercises below it is shown that C is indeed a Cantor set. 120 Equivalence of the two Deﬁnitions

Figure 7.2: Phase portrait of the quadratic map Qλ for λ > 4 on [0, 1]. The arrows indicate the (partial) trajectory of a point in [0, 1]. The dotted lines divide [0, 1] in three parts: I0, A1 and I1 (from left to right).

Example 7.3.1 . (Circle Rotations) Let S1 denote the unit circle in C. We 1 1 iθ deﬁne the rotation map Rθ : S −→ S with parameter θ by Rθ(z) = e z = ei(θ+ω), where θ ∈ [0, 2π) and z = eiω for some ω ∈ [0, 2π). It is easily seen that Rθ is a homeomorphism for every θ. The rotation map is very similar to the earlier introduced shift map Tθ. This is not surprising, considering the fact that the interval [0, 1] is homeomorphic to S1 after identiﬁcation of the endpoints 0 and 1.

1 1 Proposition 7.3.1 Let θ ∈ [0, 2π) and Rθ : S −→ S be the rotation map. a 1 If θ is rational, say θ = b , then every z ∈ S is periodic with period b. Rθ is minimal if and only if θ is irrational.

Proof

The ﬁrst statement follows trivially from the fact that e2niπ = 1 for n ∈ Z. 121

Suppose that θ is irrational. Fix ε > 0 and z ∈ S1, so z = eiω for some ω ∈ [0, 2π). Then

n m i(ω+nθ) i(ω+mθ) Rθ (z) = Rθ (z) ⇔ e = e ⇔ ei(n−m)θ = 1 ⇔ (n − m)θ ∈ Z

n n ∞ Thus, the points {Rθ (z)|n ∈ Z} are all distinct and it follows that {Rθ (z)}n=1 is an inﬁnite sequence. By compactness, this sequence has a convergent subsequence, which is Cauchy. Therefore, we can ﬁnd integers n > m such that n m d(Rθ (z),Rθ (z)) < ε where d is the arc length distance function. Now, since Rθ is distance preserv- l ing with respect to this metric, we can set l = n−m to obtain d(Rθ(z), z) < ε. l l l Also, by continuity of Rθ, Rθ maps the connected, closed arc from z to Rθ(z) l 2l onto the connected closed arc from Rθ(z) to Rθ (z) and this one onto the arc 2l 3l connecting Rθ (z) to Rθ (z), etc. Since these arcs have positive and equal length, they cover S1. The result now follows, since the arcs have length smaller than ε, and ε > 0 was arbitrary.

As mentioned in the proof, Rθ is an isometry with respect to the arc length distance function and this metric induces the topology on S1. Hence, we see that span(n, ε, Rθ) = span(1, ε, Rθ) for all n ∈ Z≥0 and it follows that h(Rθ) = 0, for any θ ∈ [0, 2π).

Example 7.3.2 . (Bernoulli Shift Revisited) In this example we will rein- troduce the Bernoulli shift in a topological context. Let Xk = {0, 1, . , k−1}Z + N∪{0} (or Xk = {0, . , k − 1} ) and recall that the left shift on Xk is defined by L : Xk −→ Xk, L(x) = y, yn = xn+1. We can define a topology on Xk (or + Xk ) which makes the shift map continuous (in fact, a homeomorphism in the case of Xk) as follows: we equip {0, . . . , k − 1} with the discrete topology and subsequently give Xk the corresponding product topology. It is easy to see that the cylinder sets form a basis for this topology. By the Tychonoff theorem (see e.g. [Mu]), which asserts that an arbitrary product of compact 122 Equivalence of the two Definitions

spaces is compact in the product topology, our space Xk is compact. In the exercises you are asked to show that the metric −l d(x, y) = 2 if l = min{|j| : xj 6= yj} induces the product topology, i.e. the open balls B(x, r) with respect to this metric form a basis for the product topology. Here we set d(x, y) = 0 if x = y, this corresponds to the case l = ∞. We will end this example by calculating the topological entropy of the full shift L. Proposition 7.3.2 The topological entropy of the full shift map is equal to h(L) = log(k). Proof

+ We will prove the proposition for the left shift on Xk , the case for the left shift ∞ ∞ + on Xk is similar. Fix 0 < ε < 1 and pick any x = {xi}i=0, y = {yi}i=0 ∈ Xk . Notice that if at least one of the first n symbols in the itineraries of x and y differ, then i i dn(x, y) = max d(T (x),T (y)) = 1 > ε 0≤i≤n−1 + where we have used the metric d on Xk defined above. Thus, the set An + consisting of sequences a ∈ Xk such that ai ∈ {0, . . . , k−1} for 0 ≤ i ≤ n−1 and aj = 0 for j ≥ n is (n, ε)-separated. Explicitly, a ∈ An is of the form

a0a1a2 . . . an−1000 ... Since there are kn possibilities to choose the ﬁrst n symbols of an itinerary, we obtain sep(n, ε, L) ≥ kn. Hence, 1 h(T ) = lim lim log(sep(n, ε, L)) ≥ log(k) ε↓0 n→∞ n −l To prove the reverse inequality, take l ∈ Z>0 such that 2 < ε. Then An+l + is an (n, ε, L)-spanning set, since for every x ∈ Xk there is some a ∈ An+l for which the ﬁrst n + l symbols coincide. In other words, d(x, a) < 2−l < ε. Therefore, 1 n + l h(T ) = lim lim log(span(n, ε, L)) ≤ lim lim log(k) = log(k) ε↓0 n→∞ n ε↓0 n→∞ n and our proof is complete. 123

Figure 7.3: Phase portrait of the teepee map T on [0, 1].

Example 7.3.3 . (Symbolic Dynamics) The Bernoulli shift on the bise- quence space Xk is a topological dynamical system whose dynamics are quite well understood. The subject of symbolic dynamics is devoted to using this knowledge to unravel the dynamical properties of more difficult systems. The scheme works as follows. Let T : X −→ X be a topological dynamical system and let α = {A0,...,Ak−1} be a partition of X such that Ai ∩ Aj = ∅ for i 6= j (Note the difference with the earlier introduced notion of a partition). We assume that T is invertible. Now, for x ∈ X, we let φi(x) be the index of the element of α which contains T i(x), i ∈ Z. This defines a map φ : X −→ Xk, called the Itinerary map generated by T on α,

∞ φ(x) = {φi(x)}i=−∞ The image of x under φ is called the itinerary of x. Note that φ satisfies φ ◦ T = L ◦ φ. Of course, if T is not invertible, we may apply the same + procedure by replacing Xk by Xk . We will now apply the above procedure to determine the number of periodic points of the teepee (or tent) map T : [0, 1] −→ [0, 1], defined by (see figure 7.3) ( 1 2x if 0 ≤ x ≤ 2 T (x) = 1 2(1 − x) if 2 ≤ x ≤ 1 1 1 Let I0 = [0, 2 ], I1 = [ 2 , 1]. We would like to define an itinerary map 124 Equivalence of the two Definitions

+ ψ : [0, 1] −→ X2 generated on {I0,I1}. Unfortunately, the fact that I0 ∩ I1 6= ∅ causes some complications. Indeed, we have an ambiguous choice of 1 the itinerary for x = 2 , since it can be described by either (01000 ...) or 1 (11000 ...), or any other point in [0, 1] which will end up in x = 2 upon iteration of T . We will therefore identify any pair of sequences of the form (a0a1 . . . al01000 ...) and (a0a1 . . . al11000 ...) and replace the two sequences ˜ + by a single equivalence class. The resulting space will be denoted by X2 and ˜ + the resulting itinerary map again by ψ : [0, 1] −→ X2 . We will ﬁrst show that ψ is injective. Suppose that there are x, y ∈ [0, 1] n n 1 with ψ(x) = ψ(y). Then T (x) and T (y) lie on the same side of x = 2 for n all n ∈ Z≥0. Suppose that x 6= y. Then, for all n ∈ Z≥0, T ([x, y]) does 1 p not contain the point x = 2 . But then no rational numbers of the form 2n , n p ∈ {1, 2, 3,..., 2 − 1}, are in [x, y] for all n ∈ Z>0. Since these form a dense set in [0, 1], we obtain a contradiction. We conclude that x = y, ψ is one-to-one. ∞ ˜ + Now let us show that ψ is also surjective. Let a = {ai}i=0 ∈ X2 . If a is one of the aforementioned equivalence classes, we use the ‘0’ element to represent a. Now, deﬁne

n An = {x ∈ [0, 1]|x ∈ Ia0 ,T (x) ∈ Ia1 ,...,T (x) ∈ Ian } −1 −n = Ia0 ∩ T Ia1 ∩ ... ∩ T Ian

−i Since T is continuous, T Iai is closed for all i and therefore An is closed for n ∈ Z≥0. As [0, 1] is a compact metric space, it follows from theorem 7.1.1 ∞ that An is compact. Moreover, {An}n=0 forms a decreasing sequence, in the ∞ sense that An+1 ⊂ An. Therefore, the intersection ∩n=0An is not empty and ∞ a is precisely the itinerary of x ∈ ∩n=0An. We conclude that ψ is onto. ˜ + ˜ + Now, since ψ ◦ T = L ◦ ψ, with L : X2 −→ X2 the (induced) left shift, we ˜ + have found a bijection between the periodic points of L on X2 and those of T on [0, 1]. The periodic points for T are quite difficult to find, but the periodic points of L are easily identified. First observe that there are no ˜ + periodic points in the earlier defined equivalence classes in X2 since these points are eventually mapped to 0. Now, any periodic point a of L of period + n in X2 must have an itinerary of the form

(a0a1 . . . an−1a0a1 . . . an−1a0a1 ...) Hence, T has 2n periodic points of period n in [0, 1]. 125

Exercise 7.3.1 In this exercise you are asked to prove some properties of Z + N∪{0} the shift map. Let Xk = {0, 1, . , k − 1} and Xk = {0, 1, . , k − 1} . a. Show that the map d : Xk × Xk −→ Xk, deﬁned by

−l d(x, y) = 2 if l = min{|j| : xj 6= yj}

is a metric on Xk that induces the product topology on Xk.

∗ b. Show that the map d : Xk × Xk −→ Xk, deﬁned by

∞ X |xn − yn| d∗(x, y) = 2|n| n=−∞

is a metric on Xk that induces the product topology on Xk.

+ c. Show that the one-sided shift on Xk is continuous. Show that the two- sided shift on Xk is a homeomorphism. d. Show that the two-sided shift L : Xk −→ Xk is expansive.

Exercise 7.3.2 Let Qλ be the quadratic map and let C,A1 be as in the ﬁrst example. Set [0, 1] − A1 = I0 ∪ I1, where I0 is the√ interval left of A1 and I1 0 the interval to the right. We assume that λ > 2+ 5, so that |Qλ(x)| > 1 for + N + all x ∈ I0 ∪ I1 (check this!). Let X2 = {0, 1} and let φ : C −→ X2 be the itinerary map generated by Qλ on {I0,I1}. This exercise serves as another demonstration of the usefulness of symbolic dynamics. a. Show that φ is a homeomorphism. b. Show that φ is a topological conjugacy between Qλ and the one-sided + shift map on X2 .

n c. Prove that Qλ has exactly 2 periodic points of period n in C. Show that the periodic points of Qλ are dense in C. Finally, prove that Qλ is topologically transitive on C.

Exercise 7.3.3 Let Qλ be the quadratic map and let C be as in the ﬁrst example. This exercise shows that C is a Cantor set, i.e. a closed, totally disconnected and perfect√ subset of [0, 1]. As in the previous exercise, we will assume that λ > 2 + 5. 126 Equivalence of the two Deﬁnitions a. Prove that C is closed. b. Show that C is totally disconnected, i.e. the only connected subspaces of C are one-point sets. c. Demonstrate that C is perfect, i.e. every point is a limit point of C. Chapter 8

The Variational Principle

In this chapter we will establish a powerful relationship between measure theoretic and topological entropy, known as the Variational Principle. It asserts that for a continuous transformation T of a compact metric space X the topological entropy is given by h(T ) = sup{hµ(T )|µ ∈ M(X,T )}. To prove this statement, we will proceed along the shortest and most popular route to victory, provided by [M]1. The proof is at times quite technical and is therefore divided into several digestable pieces.

8.1 Main Theorem

For the ﬁrst part of the proof of the Variational Principle we will only use some properties of measure theoretic entropy. All the necessary ingredients are listed in the following lemma:

Lemma 8.1.1 Let α, β, γ be ﬁnite partitions of X and T a measure preserving transformation of the probability space (X, F, µ). Then,

1. Hµ(α) ≤ log(N(α)), where N(α) is the number of sets in the partition 1 of non-zero measure. Equality holds only if µ(A) = N(α) , for all A ∈ α with µ(A) > 0

2. If α ≤ β, then hµ(T, α) ≤ hµ(T, β)

1Our exposition of Misiurewicz’s proof follows [W] to a certain extent

127 128 Equivalence of the two Deﬁnitions

3. hµ(T, α) ≤ hµ(T, γ) + Hµ(α|γ)

n 4. hµ(T ) = n · hµ(T ), for all n ∈ Z>0 Proofs

(1): This is a straighforward consequence of the concavity of the function f : [0, ∞) −→ R, deﬁned by f(t) = −t log(t)(t > 0), f(0) = 0. In this P notation, Hµ(α) = A∈α f(µ(A)). (2): α ≤ β implies that for every B ∈ β, there some A ∈ α such that B ⊂ A −i −i and hence also T B ⊂ T A, i ∈ Z≥0. Therefore, for n ≥ 1

n−1 n−1 _ _ T −iα ≤ T −iβ i=0 i=0

The result now follows from proposition . (3): By proposition (4.2.1)(e) and (b), respectively,

n−1 n−1 n−1 _ _ _ H( T −iα) ≤ H( T −iγ ∨ T −iα) i=0 i=0 i=0 n−1 n−1 n−1 _ _ _ = H( T −iγ) + H( T −iα| T −iγ) i=0 i=0 i=0

Now, by successively applying proposition (4.2.1) (d), (g) and ﬁnally using the fact that T is measure preserving, we get

n−1 n−1 n−1 n−1 _ _ X _ H( T −iα| T −iγ) ≤ H(T −iα| T −iγ) i=0 i=0 i=0 i=0 n−1 X ≤ H(T −iα|T −iγ) i=0 = nH(α|γ)

Combining the inequalities, dividing both sides by n and taking the limit for n → ∞ gives the desired result. Main Theorem 129

(4): Note that

n−1 k−1 n−1 _ 1 _ _ h(T n, T −iα) = lim H( T −nj( T −iα)) k→∞ k i=0 j=0 i=0 kn−1 n _ = lim H( T −iα) k→∞ kn i=0 m−1 n _ = lim H( T −iα) m→∞ m i=0 = nh(T, α)

Wn−1 −i Notice that { i=0 T α | α a partition of X} ⊂ {β | β a partition of X}. Hence,

nh(T ) = n sup h(T, α) α n−1 _ = sup h(T n, T −iα) α i=0 ≤ sup h(T n, α) α = h(T n)

Wn−1 −i Finally, since α ≤ i=0 T α, we obtain by (2),

n−1 _ h(T n, α) ≤ h(T n, T −iα) = nh(T, α) i=0

n So h(T ) ≤ nh(T ). This concludes our proof.

Exercise 8.1.1 Use lemma 8.1.1 and proposition (4.2.1) to show that for an invertible measure preserving transformation T on (X, F, µ):

n h(T ) = |n|h(T ), for all n ∈ Z We are now ready to prove the ﬁrst part of the theorem. 130 Equivalence of the two Deﬁnitions

Theorem 8.1.1 Let T : X −→ X be a continuous transformation of a compact metric space. Then h(T ) ≥ sup{hµ(T )|µ ∈ M(X,T )}. Proof

Fix µ ∈ M(X,T ). Let α = {A1,...,An} be a finite partition of X. Pick 1 ε > 0 such that ε < n log n . By theorem 16, µ is regular, so we can find closed n sets Bi ⊂ Ai such that µ(Ai − Bi) < ε, i = 1, . . . , n. Define B0 = X − ∪i=1Bi and let β be the partition β = {B0,...,Bn}. Then, writing f(t) = −t log(t),

n n X X µ(Bi ∩ Aj) H (α|β) = µ(B )f( ) µ i µ(B ) i=0 j=1 i n X µ(B0 ∩ Aj) = µ(B ) f( ) 0 µ(B ) j=1 0 in the ﬁnal step we used the fact that for i ≥ 1,

Bi ∩ Ai = Bi

Bi ∩ Aj = ∅ if i 6= j

We can deﬁne a σ-algebra on B0 by F ∩ B0 = {F ∩ B0|F ∈ F} and deﬁne the conditional measure µ(·|B0): F ∩ B0 −→ [0, 1] by

µ(A ∩ B0) µ(A|B0) = µ(B0)

It is easy to check that (B0, F ∩ B0, µ(·|B0)) is a probability space and

µ(·|B0) ∈ M(B0,T |B0 ). Now, noting that α0 := {A∩B0|A ∈ α} is a partition of B0 under µ(·|B0), we get

n X µ(B0 ∩ Aj) H (α|β) = µ(B ) f( ) = µ(B )H (α ) µ 0 µ(B ) 0 µ(·|B0) 0 j=1 0

Hence, we can apply (1) of lemma 8.1.1 and use the fact that

n n n µ(B0) = µ(X − ∪i=1Bi) = µ(∪i=1Ai − ∪i=1Bi) = µ(∪i=1(Ai − Bi)) < nε Main Theorem 131 to obtain

Hµ(α|β) ≤ µ(B0) log(n) < nε log(n) < 1

Deﬁne for each i, i = 1, . . . , n, the open set Ci by Ci = B0 ∪ Bi = X − ∪j6=iBj. Then we can deﬁne an open cover of X by γ = {C1,...,Cn}. By (1) of lemma 8.1.1 we have for m ≥ 1, in the notation of the lemma, Wm−1 −i Wm−1 −i Hµ( i=0 T β) ≤ log(N( i=0 T β)). Note that γ is not necessarily a par- Wm−1 −i tition, but by the proof of (1) of lemma 8.1.1, we still have Hµ( i=0 T β) ≤ m Wm−1 −i log(2 N( i=0 T γ)), since β contains (at most) one more set of non-zero measure. Thus, it follows that

hµ(T, β) ≤ h(T, γ) + log 2 ≤ h(T ) + log 2 and by (3) of lemma 8.1.1 we get obtain

hµ(T, α) ≤ hµ(T, β) + Hµ(α|β) ≤ h(T ) + log 2 + 1

Note that if µ ∈ M(X,T ), then also µ ∈ M(X,T m), so the above inequality holds for T m as well. Applying (4) of lemma 8.1.1 and (3) of theorem 7.2.4 leads us to mhµ(T ) ≤ mh(T ) + log 2 + 1. By dividing by m, taking the limit for m → ∞ and taking the supremum over all µ ∈ M(X,T ), we obtain the desired result. We will now ﬁnish the proof of the Variational Principle by proving the opposite inequality. This is the hard part of the proof, since it does not only require more machinery, but also a clever trick.

Lemma 8.1.2 Let X be a compact metric space and α a ﬁnite partition of X. Then, for any µ, ν ∈ M(X,T ) and p ∈ [0, 1] we have Hpµ+(1−p)ν(α) ≥ pHµ(α) + (1 − p)Hν(α)

Proof

Deﬁne the function f : [0, ∞) −→ R by f(t) = −t log(t)(t > 0), f(0) = 0. Then f is concave and so, for A measurable, we have:

0 ≤ f(pµ(A) + (1 − p)ν(A)) − pf(µ(A)) − (1 − p)f(ν(A))

The result now follows easily. 132 Equivalence of the two Deﬁnitions

Exercise 8.1.2 Suppose X is a compact metric space and T : X −→ X a continuous transformation. Use (the proof of) lemma 8.1.2 to show that for any µ, ν ∈ M(X,T ) and p ∈ [0, 1] we have hpµ+(1−p)ν(T ) ≥ phµ(T ) + (1 − p)hν(T ).

Exercise 8.1.3 Improve the result in the exercise above by showing that we can replace the inequality by an equality sign, i.e. for any µ, ν ∈ M(X,T ) and p ∈ [0, 1] we have hpµ+(1−p)ν(T ) = phµ(T ) + (1 − p)hν(T ).

Recall that the boundary of a set A is deﬁned by ∂A = A − A.

Lemma 8.1.3 Let X be a compact metric space and µ ∈ M(X). Then,

1. For any x ∈ X and δ > 0, there is a 0 < η < δ such that µ(∂B(x, η)) = 0

2. For any δ > 0, there is a ﬁnite partition α = {A1,...,An} of X such that diam(Aj) < δ and µ(∂Aj) = 0, for all j

3. If T : X −→ X is continuous, µ ∈ M(X,T ) and Aj ⊂ X measurable n−1 −j such that µ(∂Aj) = 0, j = 0, . . . , n − 1, then µ(∂(∩j=0 T Aj)) = 0

4. If µn → µ in M(X) and A is a measurable set such that µ(∂A) = 0, then limn→∞ µn(A) = µ(A) Proof

(1): Suppose the statement is false. Then there exists an x ∈ X and δ > 0 ∞ such that for all 0 < η < δ µ(∂B(x, η)) > 0. Let {ηi}i=1 be a sequence of distinct real numbers satisfying 0 < ηi < δ and ηi ↑ δ for i → ∞. Then, ∞ P∞ µ(∪i=1∂B(x, ηi)) = i=1 µ(∂B(x, ηi)) = ∞, contradicting the fact that µ is a probability measure on X. (2): Fix δ > 0. By (1), for each x ∈ X, we can find an 0 < ηx < δ/2 such that µ(∂B(x, ηx)) = 0. The collection {B(x, ηx)|x ∈ X} forms an open cover of X, so by compactness there exists a finite subcover which we denote by β = {B1,...,Bn}. Define α by letting A1 = B1 and for 0 < j ≤ n let j−1 Aj = Bj − (∪k=1Bk). Then α is a partition of X, diam(Aj) < diam(Bj) < δ n n and µ(∂Aj) ≤ µ(∪i=1∂Bi) = 0, since ∂Aj ⊂ ∪i=1∂Bi. n−1 −j n−1 −j n−1 −j (3): Let x ∈ ∂(∩j=0 T Aj). Then x ∈ ∩j=0 T Aj, but x∈ / ∩j=0 T Aj. −j −k That is, every open neighborhood of x intersects every T Aj, but x∈ / T Ak Main Theorem 133

−k −k for some 0 ≤ k ≤ n − 1. Hence, x ∈ T Ak − T Ak and by continuity of T k, k k −k T (x) ∈ T (T Ak) − Ak ⊂ Ak − Ak = ∂Ak n−1 −j n−1 −j Thus, ∂(∩j=0 T Aj) ⊂ ∪j=0 T ∂Aj and the statement readily follows. (4): Recall that µn → µ in M(X) for n → ∞ if and only if: Z Z lim f(x)dµn(x) = f(x)dµ(x) n→∞ X X 1 for all f ∈ C(X). Let A be as stated and deﬁne Ak = {x ∈ X|d(x, A) > k }, k ∈ Z>0. Then, since A and X − Ak are closed, it follows from theorem 7.1.1 that for every k there exists a function fk ∈ C(X) such that 0 ≤ fk ≤ 1, fk(x) = 1 for all x ∈ A and fk(x) = 0 for all x ∈ X − Ak. Now, for each k, Z Z Z lim µn(A) = lim 1A(x)dµn(x) ≤ lim fk(x)dµn(x) = fk(x)dµ(x) n→∞ n→∞ X n→∞ X X

Note that fk ↓ 1A in µ-measure for k → ∞, so by the Dominated Convergence Theorem, Z lim µn(A) ≤ lim fk(x)dµ(x) = µ(A) = µ(A) n→∞ k→∞ X where the ﬁnal inequality follows from the assumption µ(∂A) = 0. The proof of the opposite inequality is similar. Lemma 8.1.4 Let q, n be integers such that 1 < q < n. Deﬁne, for 0 ≤ j ≤ n−j q − 1, a(j) = b q c, where bc means taking the integer part. Then we have the following

1. a(0) ≥ a(1) ≥ · · · ≥ a(q − 1)

2. Fix 0 ≤ j ≤ q − 1. Deﬁne

Sj = {0, 1, . . . , j − 1, j + a(j)q, j + a(j)q + 1, . . . , n − 1}

Then

{0, 1, . . . , n − 1} = {j + rq + i|0 ≤ r ≤ a(j) − 1, 0 ≤ i ≤ q − 1} ∪ Sj

and card(Sj) ≤ 2q. 134 Equivalence of the two Deﬁnitions

n−j 3. For each 0 ≤ j ≤ q − 1, (a(j) − 1)q + j ≤ b q − 1cq + j ≤ n − q. The numbers {j + rq|0 ≤ j ≤ q − 1, 0 ≤ r ≤ a(j) − 1} are distinct and no greater than n − q.

The three statements are clear after a little thought.

Example 8.1.1 Let us work out what is going on in lemma 8.1.4 for n = 10−0 10, q = 4. Then a(0) = b 4 c = 2 and similarly, a(1) = 2, a(2) = 2 and a(3) = 1. This gives us S0 = {8, 9}, S1 = {0, 9}, S2 = {0, 1} and S3 = {0, 1, 2, 7, 8, 9}. For example, S3 is obtained by taking all nonnegative integers smaller than 3 and adding the numbers 3+a(3)q = 7, 3+a(3)q+1 = 8 and 3 + a(3)q + 2 = 9. Note that card(Sj) ≤ 8. One can readily check all 10−3 properties stated in the lemma, e.g. (a(3) − 1)q + 3 ≤ b 4 − 1c · 4 + 3 = 3 ≤ 6 = n − q.

We shall now finish the proof of our main theorem. The proof is presented in a logical order, which is in this case not quite the same as thinking order. Therefore, before plunging into a formal proof, we will briefly discuss the main ideas. We would like to construct a Borel probability measure µ with measure the- 1 oretic entropy hµ(T ) ≥ sep(ε, T ) := limn→∞ n log(sep(n, ε, T )). To do this, Wn−1 −i we first find a measure σn with Hσn ( i=0 T α) equal to log(sep(n, ε, T )), where α is a suitably chosen partition. This is not too difficult. The problem is to find an element of M(X,T ) with this property. Theorem (6.1.5) sug- gest a suitable measure µ which can be obtained from the σn. The trick in lemma 8.1.4 plays a crucial role in getting from an estimate for Hσn to one Wn−1 −i for Hµ. The idea is to first remove the ‘tails’ Sj of the partition i=0 T α, make a crude estimate for these tails and later add them back on again. Lemma 8.1.3 fills in the remaining technicalities.

Theorem 8.1.2 Let T : X −→ X be a continuous transformation of a compact metric space. Then h(T ) ≤ sup{hµ(T )|µ ∈ M(X,T )}.

Proof

Fix ε > 0. For each n, let En be an (n, ε)-separated set of cardinality sep(n, ε, T ). Deﬁne σ ∈ M(X) by σ = (1/sep(n, ε, T )) P δ , n n x∈En x Main Theorem 135

where δx is the Dirac measure concentrated at x. Define µn ∈ M(X) by 1 Pn−1 −i µn = n i=0 σn ◦T . By theorem 19, M(X) is compact, hence there exists a ∞ subsequence {nj}j=1 such that {µnj } converges in M(X) to some µ ∈ M(X). By theorem (6.1.5), µ ∈ M(X,T ). We will show that hµ(T ) ≥ sep(ε, T ), from which the result clearly follows. By (2) of lemma 8.1.3, we can find a µ-measurable partition α = {A1,...,Ak} of X such that diam(Aj) < ε and µ(∂Aj) = 0, j = 1, . . . , k. We may as- Wn−1 −i sume that every x ∈ En is contained in some Aj. Now, if A ∈ i=0 T α, then A cannot contain more than one element of En. Hence, σn(A) = 0 or k Wn−1 −i 1/sep(n, ε, T ). Since ∪j=1Aj contains all of En, we see that Hσn ( i=0 T α) = log(sep(n, ε, T )). Fix integers q, n with 1 < q < n and define for each 0 ≤ j ≤ q − 1 a(j) as in lemma 8.1.4. Fix 0 ≤ j ≤ q − 1. Since

n−1 a(j)−1 q−1 _ _ _ _ T −iα = ( T −(rq+j)( T −iα)) ∨ ( T −iα)

i=0 r=0 i=0 i∈Sj we ﬁnd that

n−1 _ −i log(sep(n, ε, T )) = Hσn ( T α) i=0 a(j)−1 q−1 X −rq−j _ −i X −i ≤ Hσn (T ( T α)) + Hσn (T α)

r=0 i=0 i∈Sj a(j)−1 q−1 X _ −i −(rq+j) ≤ Hσn◦T ( T α) + 2q log k r=0 i=0

Here we used proposition (4.2.1)(d) and lemma 8.1.1. Now if we sum the above inequality over j and divide both sides by n, we obtain by (3) and (1) of lemma 8.1.4

n−1 q−1 2 q 1 X _ −i 2q log(sep(n, ε, T )) ≤ H −l ( T α) + log k n n σn◦T n l=0 i=0 q−1 _ 2q2 ≤ H ( T −iα) + log k µn n i=0 136 Measures of Maximal Entropy

Wq−1 −i By (3) of lemma 8.1.3, each atom A of i=0 T α has boudary of µ-measure zero, so by (4) of the same lemma, limj→∞ µnj (A) = µ(A). Hence, if we replace n by nj in the above inequality and take the limit for j → ∞ we Wq−1 −i obtain qsep(ε, T ) ≤ Hµ i=0 T α. If we now divide both sides by q and take the limit for q → ∞, we get the desired inequality.

Corollary 8.1.1 (The Variational Principle) The topological entropy of a continuous transformation T : X −→ X of a compact metric space X is given by h(T ) = sup{hµ(T )|µ ∈ M(X,T )} To get a taste of the power of this statement, let us recast our proof of the invariance of topological entropy under conjugacy ((2) of theorem 7.2.4). Let φ : X1 → X2 denote the conjugacy. We note that µ ∈ M(X1,T1) if and only −1 if µ ◦ φ ∈ M(X2,T2) and we observe that hµ(T1) = hµ◦φ−1 (T2). The result now follows from the Variational Principle. It is that simple.

8.2 Measures of Maximal Entropy

The Variational Principle suggests an educated way of choosing a Borel probability measure on X, namely one that maximizes the entropy of T .

Deﬁnition 8.2.1 Let X be a compact metric space and T : X −→ X be continuous. A measure µ ∈ M(X,T ) is called a measure of maximal entropy if hµ(T ) = h(T ). Let Mmax(X,T ) = {µ ∈ M(X,T )|hµ(T ) = h(T )}. If Mmax(X,T ) = {µ} then µ is called a unique measure of maximal entropy.

Example. Recall that the topological entropy of the circle rotation Rθ is given by h(Rθ) = 0. Since hµ(Rθ) ≥ 0 for all µ ∈ M(X,Rθ), we see that every µ ∈ M(X,Rθ) is a measure of maximal entropy, i.e. Mmax(X,Rθ) = M(X,Rθ). More generally, we have h(T ) = 0 and hence Mmax(X,T ) = M(X,T ) for any continuous isometry T : X −→ X.

Measures of maximal entropy are closely connected to (uniquely) ergodic measures, as will become apparent from the following theorem.

Theorem 8.2.1 Let X be a compact metric space and T : X −→ X be continuous. Then 137

• Mmax(X,T ) is a convex set.

• If h(T ) < ∞ then the extreme points of Mmax(X,T ) are precisely the ergodic members of Mmax(X,T ).

• If h(T ) = ∞ then Mmax(X,T ) 6= ∅. If, moreover, T has a unique measure of maximal entropy, then T is uniquely ergodic.

• A unique measure of maximal entropy is ergodic. Conversely, if T is uniquely ergodic, then T has a measure of maximal entropy. Proofs

(1): Let p ∈ [0, 1] and µ, ν ∈ Mmax(X,T ). Then, by exercise 8.1.3,

hpµ+(1−p)ν(T ) = phµ(T ) + (1 − p)hν(T ) = ph(T ) + (1 − p)h(T ) = h(T )

Hence, pµ + (1 − p)ν ∈ Mmax(X,T ). (2): Suppose µ ∈ Mmax(X,T ) is ergodic. Then, by theorem 6.1.6, µ cannot be written as a non-trivial convex combination of elements of M(X,T ). Since Mmax(X,T ) ⊂ M(X,T ), µ is an extreme point of Mmax(X,T ). Conversely, suppose µ is an extreme point of M(X,T ) and suppose there is a p ∈ (0, 1) and ν1, ν2 ∈ M(X,T ) such that µ = pν1 + (1 − p)ν2. By exercise 8.1.3, h(T ) = hµ(T ) = phν1 (T ) + (1 − p)hν2 (T ). But by the Variational Principle, h(T ) = sup{hµ(T )|µ ∈ M(X,T )}, so h(T ) = hν1 (T ) = hν2 (T ). In other words, ν1, ν2 ∈ Mmax(X,T ), thus µ = ν1 = ν2. Therefore, µ is also an extreme point of M(X,T ) and we conclude that µ is ergodic. For the second part, suppose that Mmax(X,T ) 6= ∅. (3): By the Variational Principle, for any n ∈ Z>0, we can ﬁnd a µn ∈ n M(X,T ) such that hµn (T ) > 2 . Deﬁne µ ∈ M(X,T ) by

∞ N ∞ X µn X µn X µn µ = = + 2n 2n 2n n=1 n=1 n=N But then, N X µn h (T ) ≥ > N µ 2n n=1

This holds for arbitrary N ∈ Z>0, so hµ(T ) = h(T ) = ∞ and µ is a measure of maximal entropy. 138 Measures of Maximal Entropy

Now suppose that T has a unique measure of maximal entropy. Then, for 1 1 any ν ∈ M(X,T ), hµ/2+ν/2(T ) = 2 hµ(T ) + 2 hν(T ) = ∞. Hence, µ = ν, M(X,T ) = {µ}. (4): The ﬁrst statement follows from (2), for h(T ) < ∞, and (3), for h(T ) = ∞. If T is uniquely ergodic, then M(X,T ) = {µ} for some µ. By the Variational Principle, hµ(T ) = h(T ).

Example 8.2.1 For θ irrational Rθ is uniquely ergodic with respect to Lebesgue measure. By the theorem above, it follows that Lebesgue measure is also a unique measure of maximal entropy for Rθ.

Exercise 8.2.1 Let X = {0, 1, . . . , k − 1}Z and let T : X −→ X be the full two-sided shift. Use proposition 7.3.2 to show that the uniform product measure is a unique measure of maximal entropy for T .

We end this section with a generalization of the above exercise.

Proposition 8.2.1 Every expansive homeomorphism of a compact metric space has a measure of maximal entropy.

Proof

Let T : X −→ X be an expansive homeomorphism and let δ > 0 be an expansive constant for T . Fix 0 < ε < δ. Deﬁne µ ∈ M(X,T ) as in the proof of theorem 8.1.2. Then, by the proof, hµ(T ) ≥ sep(ε, T ). We will show that h(T ) = sep(ε, T ). It then immediately follows that µ ∈ Mmax(X,T ). Pick any 0 < η < ε and let A be an (n, η)-separated set of cardinality sep(n, η, T ). By expansiveness, we can ﬁnd for any x, y ∈ A some k = kx,y ∈ Z such that d(T k(x),T k(y)) > ε

By theorem 7.2.2, A is ﬁnite, so l := max{|kx,y| : x, y ∈ A} is in Z>0. Now, by our choice of l, for x, y ∈ T −lA we have

max d(T i(x),T i(y)) > ε 0≤i≤2l+n−1 139

So T −lA is an (2l + n, ε, T )-separated set of cardinality sep(n, η, T ). Hence, 1 sep(η, T ) = lim log(sep(n, η, T )) n→∞ n 1 ≤ lim log(sep(2l + n, ε, T )) n→∞ n 1 ≤ lim log(sep(2l + n, ε, T )) n→∞ 2l + n = sep(ε, T )

Conversely, if A is (n, ε)-separated then A is also (n, η)-separated, so

sep(ε, T ) ≤ sep(η, T ).

We conclude that sep(ε, T ) = sep(η, T ). Since 0 < η < ε was arbitrary, this shows that h(T ) = limη↓0 sep(η, T ) = sep(ε, T ). This completes our proof.

From the proof we extract the following corollary.

Corollary 8.2.1 Let T : X −→ X be an expansive homeomorphism and let δ be an expansive constant for T . Then h(T ) = sep(ε, T ), for any 0 < ε < δ.

At this point we would like to make a remark on the proof of the above proposition. One might be inclined to believe that the measure µ constructed in the proof of theorem 8.1.2 is a measure of maximal entropy for any continuous transformation T . The reader should note, though, that µ still depends on ε and it is therefore only because of the above corollary that µ is a measure of maximal entropy for expansive homeomorphisms. 140 Measures of Maximal Entropy Bibliography

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(X, d), 102 sep(n, ε, T ), 111 (n, ε)-separated, 111 span(n, ε, T ), 111 (n, ε)-spanning, 111 B(x, r), 112 Stationary Stochastic Processes, 12 B (x, r), 112 n algebra, 7 FO (x), 103 T algebra generated, 7 H(α), 109 atoms of a partition, 58 Mmax(X,T ), 136 N(α), 109 Baker’s Transformation, 10 OT (x), 103 Bernoulli Shifts, 11 Qλ, 119 binary expansion, 10 Rθ, 120 Birkhoﬀ’s Ergodic Theorem, 29 T −1α, 109 X, 102 Cantor set, 125 α ∨ β, 109 circle rotation, 120 β-transformations, 10 common reﬁnement, 58 Wn of open cover, 109 i=1 αi, 109 ∂A, 132 conditional expectation, 35 cov(ε, T ), 113 conditional information function, 66 cov(n, ε, T ), 112 conservative, 82 d, 102 Continued Fractions, 13 ddisc, 103 diameter dn, 111 of a set, 109 diam(A), 109 of open cover, 109 diam(α), 109 dynamical system, 47 h(T ), 114 h(T, α), 110 entropy of the partition, 57 h1(T ), 111 entropy of the transformation, 61 h2(T ), 114 equivalent measures, 79 sep(ε, T ), 134 ergodic, 20

143 144 Measures of Maximal Entropy ergodic decomposition, 40 monotone class, 7 expansive, 106 constant, 106 natural extension, 52 extreme point, 92 non-singular transformation, 80 factor map, 51 orbit, 103 first return time, 16 forward, 103 fixed point, 105 periodic point, 105 generator, 64, 106 phase portrait, 119 weak, 107 PoincaréRecurrence Theorem, 15 homeomorphism quadratic map, 118 conjugate, 107 Radon-Nikodym Theorem, 79 expansive, 106, 107 random shifts, 13 generator for, 106 refinement minimal, 103 of open cover, 109 topologically transitive, 103 regular measure, 89 weak generator for, 107 Riesz Representation Representation Hurewicz Ergodic Theorem, 83 Theorem, 90 induced map, 16 semi-algebra, 7 induced operator, 22 Shannon-McMillan-Breiman Theo- information function, 66 rem, 70 integral system, 19 shift map irreducible Markov chain, 42 full, 121 isometry, 105 strongly mixing, 45 isomorphic, 47 subadditive sequence, 60 Kac’s Lemma, 35 symbolic dynamics, 123 Knopp’s Lemma, 26 topological conjugacy, 107 Lebesgue number, 102 topological entropy Lochs’ Theorem, 72 definition (I), 111 definition (II), 114 Markov measure, 41 of open cover, 109 Markov Shifts, 12 properties, 116 measure of maximal entropy, 136 wrt open cover, 110 unique, 136 topologically transitive, 103 measure preserving, 6 homeomorphism, 103 145

one-sided, 103 translations, 9 uniquely ergodic, 96

Variational Principle, 136 weakly mixing, 45