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.