MATH41112/61112 Ergodic Theory Lecture 16

16. Ergodicity

§16.1 Ergodicity In this lecture, we introduce what it means to say that a transformation is ergodic with respect to an invariant measure. Ergodicity is an important concept for many reasons, not least because Birkhoff’s Ergodic Theorem holds:

Theorem 16.1 Let T be an ergodic transformation of the probability space (X, B, µ) and let f ∈ L1(X, B, µ). Then

n−1 1 f(T jx) → f dµ n Xj=0 Z for µ-almost every x ∈ X.

§16.2 The definition of ergodicity Definition. Let (X, B, µ) be a probability space and let T : X → X be a measure-preserving transformation. We say that T is an ergodic transfor- mation (or µ is an ergodic measure) if, for B ∈ B,

T −1B = B ⇒ µ(B) = 0 or 1.

Remark. One can view ergodicity as an indecomposability condition. If ergodicity does not hold and we have T −1A = A with 0 < µ(A) < 1, then one can split T : X → X into T : A → A and T : X \ A → X \ A 1 1 with invariant probability measures µ(A) µ(· ∩ A) and 1−µ(A) µ(· ∩ (X \ A)), respectively.

It will sometimes be convenient for us to weaken the condition T −1B = B to µ(T −1B4B) = 0, where 4 denotes the symmetric difference:

A4B = (A \ B) ∪ (B \ A).

The next lemma allows us to do this.

Lemma 16.2 −1 −1 If B ∈ B satisfies µ(T B4B) = 0 then there exists B∞ ∈ B with T B∞ = B∞ and µ(B4B∞) = 0. (In particular, µ(B) = µ(B∞).)

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Proof. For each j ≥ 0, we have the inclusion

j−1 T −jB4B ⊂ T −(i+1)B4T −iB i[=0 j−1 = T −i(T −1B4B) i[=0 and so (since T preserves µ)

µ(T −jB4B) ≤ jµ(T −1B4B) = 0.

Let ∞ ∞ −i B∞ = T B. j\=0 i[=j We have that

∞ ∞ µ B4 T −iB ≤ µ(B4T −iB) = 0.   i[=j Xi=j   ∞ −i Since the sets i=j T B decrease as j increases we hence have µ(B4B∞) = 0. Also, S ∞ ∞ −1 −(i+1) T B∞ = T B j=0 i=j \∞ [∞ −i = T B = B∞, j\=0 i=[j+1 as required. 2

Corollary 16.3 If T is ergodic and µ(T −1B4B) = 0 then µ(B) = 0 or 1.

Remark. Occasionally, instead of saying that µ(A4B) = 0, we will say that A = B a.e. or A = B mod 0.

§16.3 An alternative characterisation of ergodicity The next result characterises ergodicity in a convenient way.

Proposition 16.4 Let T be a measure-preserving transformation of (X, B, µ). The following are equivalent:

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(i) T is ergodic;

(ii) whenever f ∈ L1(X, B, µ) satisfies f ◦ T = f µ-a.e. we have that f is constant µ-a.e.

Remark. We can replace L1 in the statement by measurable or by L2.

Proof. (i) ⇒ (ii): Suppose that T is ergodic and that f ∈ L1(X, B, µ) with f ◦ T = f µ-a.e. For k ∈ Z and n ∈ N, define k k + 1 k k + 1 X(k, n) = x ∈ X | ≤ f(x) < = f −1 , . 2n 2n 2n 2n     Since f is measurable, X(k, n) ∈ B. We have that

T −1X(k, n)4X(k, n) ⊂ {x ∈ X | f(T x) 6= f(x)} so that µ(T −1X(k, n)4X(k, n)) = 0. Hence µ(X(k, n)) = 0 or µ(X(k, n)) = 1.

For each fixed n, the union k∈Z X(k, n) is equal to X up to a set of measure zero, i.e., S µ X4 X(k, n) = 0, ∈Z ! k[ and this union is disjoint. Hence we have

µ(X(k, n)) = µ(X) = 1 ∈Z Xk and so there is a unique kn for which µ(X(kn, n)) = 1. Let ∞

Y = X(kn, n). n=1 \ Then µ(Y ) = 1 and, by construction, f is constant on Y , i.e., f is constant µ-a.e. −1 1 (ii) ⇒ (i): Suppose that B ∈ B with T B = B. Then χB ∈ L (X, B, µ) and χB ◦ T (x) = χB(x) ∀ x ∈ X, so, by hypothesis, χB is constant µ-a.e. Since χB only takes the values 0 and 1, we must have χB = 0 µ-a.e. or χB = 1 µ-a.e. Therefore

µ(B) = χB dµ = 0 or 1, ZX and T is ergodic. 2

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§16.4 Rotations of a circle Fix α ∈ R and define T : R/Z → R/Z by T (x) = x + α mod 1. We have already seen that T preserves Lebesgue measure.

Theorem 16.5 Let T (x) = x + α mod 1.

(i) If α ∈ Q then T is not ergodic.

(ii) If α 6∈ Q then T is ergodic.

Proof. Suppose that α ∈ Q and write α = p/q for p, q ∈ Z with q 6= 0. Define f(x) = e2πiqx ∈ L2(X, B, µ). Then f is not constant but

f(T x) = e2πiq(x+p/q) = e2πi(qx+p) = e2πiqx = f(x).

Hence T is not ergodic. Suppose that α 6∈ Q. Suppose that f ∈ L2(X, B, µ) is such that f ◦T = f a.e. Suppose that f has Fourier series

∞ 2πinx cne . n=−∞ X Then f ◦ T has Fourier series

∞ 2πinα 2πinx cne e . n=−∞ X Comparing Fourier coefficients we see that

2πinα cn = cne ,

2πinα for all n ∈ Z. As α 6∈ Q, e 6= 1 unless n = 0. Hence cn = 0 for n 6= 0. Hence f has Fourier series c0, i.e. f is constant a.e. 2

Exercise 16.1 Show that, when α ∈ Q, the rotation T (x) = x + α mod 1 is not ergodic from the definition, i.e. find an invariant set B = T −1B which has Lebesgue measure 0 < µ(B) < 1.

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