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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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    @StéphaneLaurent: I'm aware of that, as mush of its "disjunction" meaning, whereas here the "conjunction" $\wedge$ seems to be more appropriate. – 2013-02-20

1 Answers 1

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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