# Understanding Proof of Poincare Recurrence Theorem

Theorem Let $$(X,\Sigma,\mu$$ be a finite measure space, $$f:X\to X$$ be a measurable map, and $$\mu$$ be $$f$$-invariant measure. For each set $$A\in\Sigma$$, we have $$\mu(\{x\in A: f^n(x)\in A$$ for infinitely many n $$\})=\mu(A)$$.

Proof: Let $$B=\{x\in A: f^n(x)\in A$$ for infinitely many n $$\}$$. We have \begin{align} B=A\cap\bigcap_{n=1}^{\infty}A_n=A\setminus\bigcup_{n=1}^{\infty}(A\setminus A_n) \end{align} where $$A_n=\bigcup_{k=n}^{\infty}f^{-k}(A)$$. We note that $$A\setminus A_n\subset A_0\setminus A_n=A_0\setminus f^{-n}(A_0)$$. Since $$f^{-n}(A_0)=A_n\subset A_0$$, and the measure $$\mu$$ is finite, it follows that: \begin{align} 0&\leq\mu(A\setminus A_n)\\ &\leq\mu(A_0\setminus f^{-n}(A_0))\\ &=\mu(A_0)-\mu(f^{-n}(A_0))\\ &=0 \end{align}

(because $$\mu$$ is $$f$$-invariant). It follows that $$\mu(B)=\mu(A)$$. QED

Question 1: The assumption that $$\mu$$ is finite is only relevant for the very last step $$\mu(A_0)-\mu(f^{-n}(A_0))=0$$, so we don't get $$\infty-\infty$$, correct?

Question 2: I'm trying to see why $$B=A\cap\bigcap_{n=1}^{\infty}A_n$$: If $$x\in B$$, then $$x\in A$$ and $$f^m(x)\in A$$ for infinitely many $$m$$. Let $$I=\{m_1, m_2,...\}$$ be the sequence of such $$m$$. Then $$x\in A\cap\bigcap_IA_I\subseteq A\cap\bigcap_{n=1}^{\infty}A_n$$. So $$B\subseteq A\cap\bigcap_{n=1}^{\infty}A_n$$. Conversely, let $$x\in A\cap\bigcap_{n=1}^{\infty}A_n$$. Suppose $$f^k(x)\in A$$ for only finitely many $$k$$. Let $$M$$ be the largest such $$k$$. Then since $$x\in A\cap A_{M+1}$$, $$f^N(x)\in A$$ for some $$N>M$$. Contradiction, hence, there are infinitely many $$k$$'s and so $$A\cap\bigcap_{n=1}^{\infty}A_n\subseteq B$$. Is this right?

Question 3: Why, in the first two equalities of the proof, does the index begin at $$1$$ instead of $$0$$?

Q1: Heuristically speaking, PRT is a measure-theoretic pigeonhole principle of sorts, iterations of a measure-preserving map exhaust the whole space eventually, which is why there is recurrence (consider a translation on $\Bbb{R}$). For practical purposes finiteness of the measure is required for the so-called "excision property": if $A\subseteq B,$ then $\mu(B\setminus A)=\mu(B)-\mu(A)$ (the standard proof of this goes like this: $\mu(B)=\mu(B\setminus A)+\mu(A) \implies \mu(B)-\mu(A)=\mu(B\setminus A)$, where cancellation makes sense provided that $\mu(A)$ is a number).
Q3: It could, but since we are taking the intersection with $A$, $n=0$ is redundant: