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I'm trying to understand the proof of Lemma 6.2.1 (p.260-261) in Petersen's Ergodic Theory. Specifically, I don't understand why $B_{n}^{A} \in \mathscr{B}(T^{-1}\alpha \vee \dots \vee T^{-n}\alpha)$ holds. Why does it?

Here's the setup:

  • $(X,\mathscr{B}, \mu)$ is a probability space
  • $T\colon X \to X$ is a measure-preserving ergodic transformation
  • $\alpha$ is a countable measurable partition of $X$ with finite entropy
  • $\mathscr{B}(T^{-1}\alpha \vee \dots \vee T^{-k}\alpha)$ denotes the $\sigma$-algebra generated by the common refinement of the partitions $T^{-1}\alpha, \dots, T^{-k}\alpha$.
  • $B_{n}^{A} := \{x \colon f_{1}^{A}(x), \dots, f^{A}_{n-1} \leq \lambda, f_{n}^{A} > \lambda\}$, where $A\in \alpha$ is fixed and $\lambda \geq 0$
  • $f_{k}^{A} = -\log \mu(A|T^{-1}\alpha \vee \dots \vee T^{-k}\alpha)$

So again, my question is: Why is $B_{n}^{A} \in \mathscr{B}(T^{-1}\alpha \vee \dots \vee T^{-n}\alpha)$?

Edit: I really only need to understand this for the case where $T$ is the shift on Cantor space; i.e. $X=\{0,1\}^{\mathbb{N}}$. So an answer in this setting will suffice.

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Is it clear to you that $f^A_k$ is $\mathscr B(T^{-1}\alpha\vee\dots\vee T^{-n}\alpha)|\mathscr B(\mathbb R)$-measurable for any $k\leq n$ and $A$? Note that $B_n^A$ is a pre-image of the function $(f_1^A,\dots,f_n^A)$. –  S.D. Nov 30 '11 at 15:41
    
@S.D. What does $\mathscr{B}(T^{-1}\alpha \vee \dots \vee T^{-n} \alpha)|\mathscr{B}(X)$-measurable mean? (I'm assuming your $\mathbb{R}$ should've been an $X$.) –  Quinn Culver Nov 30 '11 at 16:55
    
Nope, I meant that the function $$f_k^A = -\log \mu (A|T^{-1}\alpha \vee \dots \vee T^{-k}\alpha):(X,\mathscr{B}(T^{-1}\alpha \vee \dots \vee T^{-k}\alpha))\to(\mathbb R,\mathscr{B}(\mathbb R))$$ is measurable, i.e. for any $B\in \mathscr{B}(\mathbb R)$ one has $(f^A_k)^{-1}(B)\in \mathscr{B}(T^{-1}\alpha \vee \dots \vee T^{-k}\alpha)$. $$$$ By the way, I've never seen notation with $\vee$ before. Did you mean that $$T^{-1}\alpha \vee \dots \vee T^{-k}\alpha = \{T^{-1}\alpha_1\cap\dots\cap T^{-k}\alpha_k|\alpha_1,\dots,\alpha_k\in \alpha\}$$ is the common refinement? –  S.D. Nov 30 '11 at 17:03
    
@S.D.Yes, that's what I meant by the symbol $\vee$. That is the notation used for common refinement in Petersen's, Walters', and Billingsley's books. I'm still thinking about the rest of your comments. –  Quinn Culver Nov 30 '11 at 18:40
    
Take a look on the definition of measure $\mu(\cdot|\mathscr F)$ conditional some $\sigma$-algebra $\mathscr F$. –  S.D. Dec 1 '11 at 9:12

1 Answer 1

up vote 2 down vote accepted

Using the definition of $f_k^A$ we see that it is $\mathscr B (T^{-1}\alpha\dots T^{-k}\alpha) |\mathscr B(\mathbb R)$-measurable function which implies the measurability of the map $$ f = (f_1^A,\dots,f_n^A):(X,\mathscr B (T^{-1}\alpha\dots T^{-n}\alpha))\to(\mathbb R^n,\mathscr B(\mathbb R^n)). $$ Finally, the set $C = (-\infty,\lambda]^{n-1}\times(\lambda,\infty)\in \mathscr B(\mathbb R^n)$ and $B_n^A = f^{-1}(C)$ which answers your question

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