An outline of ergodic theory by Steven Kalikow

By Steven Kalikow

An advent to ergodic thought for graduate scholars, and an invaluable reference for the pro mathematician.

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Xn . Let Wn = n be the set of words of length n. Suppose k < m, w is a word of length k and u is a word of length m. We write r f (w; u) = # i : 1 ≤ i ≤ m − k + 1, u i u i+1 · · · u i+k−1 = w m−k+1 for the relative frequency with which w occurs as a subword of u. If B ⊂ Wk we write r f (B; u) = # i : 1 ≤ i ≤ m − k + 1, u i u i+1 · · · u i+k−1 ∈ B m−k+1 for the relative frequency with which the members of B occur collectively as subwords of u. If v ∈ we write r f (w; v) = limn→∞ r f w; wn (v) , should this limit exist, and similarly for r f (B; v).

And μ(X \ i=0 Idea of proof. Start with a tiny set C, and let S be the set of all points T n x such that x ∈ C, n is a non-negative multiple of N , and {T x, T 2 x, . . , T n+N x} is entirely outside of C. Then {S, T S, T 2 S, . . , T N S} is a disjoint cover of all of the space except C ∪ T −1 C ∪ T −2 C ∪ · · · ∪ T −N C, because by ergodicity, the translates of C cover the whole space. Sketch of proof. Let C be a set of measure less than N and let S = T k N ω : k ∈ Z, k ≥ 0, ω ∈ C, C∩{T ω, T 2 ω, T 3 ω, .

43 That is, the measure ν constructed in Theorem 226. Notice that it is unique, by Comment 223. 48 Measure-preserving systems 236. Exercise. Recall the discussion in 217. e. Conclude that μ(C) = μx (C) dμ(x). then n1 i=1 Use this fact to show that the conclusion to the theorem holds whenever f is a finite linear combination of cylinder set indicator functions. • In order to get the result in full from the foregoing exercise, we have to be able to deal with the error function that arises when you approximate a general f by a finite linear combination of cylinder set indicator functions.

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