I'm trying to follow a proof in my book of the Poincare Recurrence Theorem, but I have three questions about this proof:

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).

Q2: Your argument is correct, but perhaps there is a more straightforward argument:

\begin{align} B&\stackrel{\tiny\mbox{def}}{=}\{x\in A\mid f^n(x)\in A\mbox{ FIM } n\} =\{x\in A\mid \exists n_k\subseteq n: f^{n_k}(x)\in A\}\\ &=\{x\in A\mid \forall n,\exists k\geq n: f^k(x)\in A\}=\bigcap_{n\geq1}\bigcup_{k\geq n}A\cap f^{-k}A= A\cap \bigcap_{n\geq1}\bigcup_{k\geq n}f^{-k}A. \end{align}

Q3: It could, but since we are taking the intersection with $A$, $n=0$ is redundant:

\begin{align} A\cap\bigcap_{n\geq0}A_n&= A\cap A_0 \cap \bigcap_{n\geq1}A_n \\ &= A\cap \left(f^{-0}A\cup\bigcup_{n\geq1}f^{-n}A\right)\cap\bigcap_{n\geq1}A_n \\ &\quad\quad\quad\quad= A\cap (A\cup A_1)\cap\bigcap_{n\geq1}A_n = A\cap\bigcap_{n\geq1}A_n. \end{align}

P.S.: For future reference I believe the book you are following is Barreira & Valls' Dynamical Systems: An Introduction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.