MATH41112/61112 Ergodic Theory Lecture 16 16. Ergodicity x16.1 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 2 L1(X; B; µ). Then n−1 1 f(T jx) ! f dµ n Xj=0 Z for µ-almost every x 2 X. x16.2 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 2 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 n A ! X n A 1 1 with invariant probability measures µ(A) µ(· \ A) and 1−µ(A) µ(· \ (X n 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 n B) [ (B n A): The next lemma allows us to do this. Lemma 16.2 −1 −1 If B 2 B satisfies µ(T B4B) = 0 then there exists B1 2 B with T B1 = B1 and µ(B4B1) = 0. (In particular, µ(B) = µ(B1).) 1 MATH41112/61112 Ergodic Theory Lecture 16 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 1 1 −i B1 = T B: j\=0 i[=j We have that 1 1 µ B4 T −iB ≤ µ(B4T −iB) = 0: 0 1 i[=j Xi=j @ A 1 −i Since the sets i=j T B decrease as j increases we hence have µ(B4B1) = 0. Also, S 1 1 −1 −(i+1) T B1 = T B j=0 i=j \1 [1 −i = T B = B1; 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. x16.3 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: 2 MATH41112/61112 Ergodic Theory Lecture 16 (i) T is ergodic; (ii) whenever f 2 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 2 L1(X; B; µ) with f ◦ T = f µ-a.e. For k 2 Z and n 2 N, define k k + 1 k k + 1 X(k; n) = x 2 X j ≤ f(x) < = f −1 ; : 2n 2n 2n 2n Since f is measurable, X(k; n) 2 B. We have that T −1X(k; n)4X(k; n) ⊂ fx 2 X j f(T x) 6= f(x)g 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 k2Z X(k; n) is equal to X up to a set of measure zero, i.e., S µ X4 X(k; n) = 0; 2Z ! k[ and this union is disjoint. Hence we have µ(X(k; n)) = µ(X) = 1 2Z Xk and so there is a unique kn for which µ(X(kn; n)) = 1. Let 1 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 2 B with T B = B. Then χB 2 L (X; B; µ) and χB ◦ T (x) = χB(x) 8 x 2 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 3 MATH41112/61112 Ergodic Theory Lecture 16 x16.4 Rotations of a circle Fix α 2 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 2 Q then T is not ergodic. (ii) If α 62 Q then T is ergodic. Proof. Suppose that α 2 Q and write α = p=q for p; q 2 Z with q 6= 0. Define f(x) = e2πiqx 2 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 62 Q. Suppose that f 2 L2(X; B; µ) is such that f ◦T = f a.e. Suppose that f has Fourier series 1 2πinx cne : n=−∞ X Then f ◦ T has Fourier series 1 2πinα 2πinx cne e : n=−∞ X Comparing Fourier coefficients we see that 2πinα cn = cne ; 2πinα for all n 2 Z. As α 62 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 2 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. 4.