I am reading a book on Ergodic Theory and feeling confusion in some cases :
Let $(X, \mathscr{B}, \mu)$ be a normalized measure space and let $\tau : X \rightarrow X$ be measure preserving such that $\tau(A) \in \mathscr{B}$ for each $A \in \mathscr{B}$. If $$ \lim_{n \to \infty} \mu (\tau^n A) = 1$$ for every $A \in \mathscr{B}$, $\mu(A) > 0$, then $\tau$ is exact.
Theorem : Let $(X, \mathscr{B}, \mu)$ be a normalized measure space and let $\tau : (X, \mathscr{B}, \mu) \rightarrow (X, \mathscr{B}, \mu)$ be measure preserving. Then $\tau$ is exact if and only if $$ \mathscr{B}^T = \bigcap_{n=0}^{\infty} \tau^{-n}(\mathscr{B}) $$ consists of the sets of $\mu$-measure 0 or 1. \end{theorem}
Proof : Let us assume that $A \in \mathscr{B}^T$, $0<\mu(A)<1$ and let $B_n \in \mathscr{B}$ be such that $A = \tau^{-n}B_n$, $n = 1, 2, \dots.$ Since $\tau$ preserves $\mu$, we have $\mu(B_n) = \mu(A)$, $n = 1, 2, \dots$. We also have $\tau^n (A) = \tau^n (\tau^{-n} B_n) \subset B_n$. Hence, $\mu(\tau^n (A)) \leq \mu(A) < 1 $ for $n = 1, 2, \dots,$ which contradicts the exactness of $\tau$. Let $A \in \mathscr{B}$ and $\mu(A) > 0$. If $ \lim_{n \to +\infty} \mu (\tau^n A) < 1$, we may assume that for some $a < 1$, $\mu(\tau^n(A)) \leq a < 1$, $n = 1, 2, \dots$. For any $n \geq 0$ we have $\tau^{-(n+1)}(\tau^{n+1}A) \supset \tau^{-n} (\tau^n A)$. Thus, the set $B = \bigcup_{n=0}^{\infty} \tau^{-n}(\tau^n A)$ belongs to $\mathscr{B}^T$. Since $B \supset A$ and $\mu (B) \geq \mu(A) >0$, $\mu(B) = 1$. On the other hand, $$ \mu(B) = \lim_{n \to + \infty} \mu (\tau^{-n}(\tau^n A)) = \lim_{n \to + \infty} \mu (\tau^n (A)) \leq a <1.$$ My questions : 1. How can I define the statement of the theorem more clearly, specially the intersection part. 2. In line 2 of the proof, $A = \tau^{-n}B_n$, $n = 1, 2, \dots.$ , is it correct ? Should it not be $A = \cap_{n}\tau^{-n}B_n$, $n = 1, 2, \dots.$ ? If yes, Then how will it change the rest of the calculation ? If no, then why ?