Borel-Cantelli Lemma in Terms of Transformation

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I am studying dynamical system, and by accident I found a paper here https://arxiv.org/pdf/math/9912178.pdf

In the introduction part, the authors firstly state the Borel-Cantelli lemma, then they said that in terms of a measure preserving transformation, the Borel-Cantelli lemma can be re-stated as follow:

Let $T$ be a transformation preserving a probability measure $\mu$, and let $A_{n}\subset X$ be a sequence of measurable sets, such that $\sum_{n}\mu(A_{n})<\infty$, then for almost every point $x\in X$, there are only finitely many $n$ such that $T^{n}(x)\in A_{n}$.

The authors did not provide the proof but this lemma really interested me so I tried to prove it, but stuck in the end.

Here is my attempt:

Set $B_{n}:\{x\in X: T^{n}(x)\in A_{n}\}$, and then define $C_{N}:=\bigcup_{n\geq N}B_{n}$ and $C:=\bigcap_{N=1}^{\infty}\bigcup_{n\geq N}B_{n},$ then $C_{N}\searrow C$, and thus $$\lim_{N\rightarrow\infty}\mu(C_{N})=\mu(C).$$

Notice that $C$ is also $T^{n}(x)\in A_{n}$ for infinitely many $n$, so we want to show $\mu(C)=0$.

Use the monotonicity, we have $$\mu(C_{N})=\mu\Big(\bigcup_{n\geq N}B_{n}\Big)\leq\sum_{n=N}^{\infty}m(B_{n}).$$

Then, I tried to use $\sum_{n}\mu(A_{n})<\infty$, but it don't know how to connect $B_{n}$ to $A_{n}$.

What should I do?

Thank you!

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Okay I figured out the answer, I forgot to use the fact that $T$ preserves measure.

Note that $B_{n}=T^{-n}(A_{n})$, and thus $$\mu(B_{n})=\mu\Big(T^{-n}(A_{n})\Big)=\mu(A_{n})$$

Therefore, since $\sum_{n}\mu(A_{n})<\infty$, $\sum_{n\geq N}\mu(A_{n})<\epsilon$ and thus $\mu(C_{N})<\epsilon$ for all $N$.

Therefore, $\mu(C)=\lim_{N\rightarrow\infty}\mu(C_{N})=0$.