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I had a look on the proof of the following Recurrence Theorem of Poincaré:

Let $(\Omega,\Sigma,T,m)$ be a conservative dynamical system in measure theory for which the function $T^{-1}$ preserves null sets. If $f\colon\Omega\to\mathbb{R}$ is measurable, it follows that $$ \liminf_{n\to\infty}\lvert f(x)-f(T^n(x))\rvert=0~\text{a.s.} $$

Proof

Let $B\subset\mathbb{R}$ be a measurable set with $m(f^{-1}(B))>0$ and $\text{diam}(B)<\varepsilon$. With the Theorem of Halmos (cited below) it follows that $$ S_{\infty}1_{f^{-1}(B)}=1_{f^{-1}(B)}+1_{f^{-1}(B)}\circ T+1_{f^{-1}(B)}\circ T^2+\ldots+1_{f^{-1}(B)}\circ T^n+\ldots=\infty $$ alomost surely on $f^{-1}(B)$.

So it follows that $f(T^n(z))\in B$ for almost all $n\geqslant 0$. So with Halmos it is $$ \lvert f(z)-f(T^n(z))\rvert\leqslant\text{diam}(B)=\sup_{x,y\in B}\lvert x-y\rvert<\varepsilon. $$ From this it follows that $$ \liminf_{n\to\infty}\lvert f(z)-f(T^n(z))\rvert<\varepsilon~~~~~(*) $$ almost surely on $f^{-1}(B)$.

Now if one covers $\Omega$ by sets $f^{-1}(B)$ with diam smaller than $\varepsilon$ one has (*) almost surely and the Theorem follows with $\varepsilon\to 0$.


Is that okay?

Why is it possible to cover $\Omega$ by set of the form $f^{-1}(B)$ with $\text{diam}<\varepsilon$? Is $\Omega$ $\sigma$-finite?


Last but not least

here is the Theorem of Halmos that is used:

Let $(\Omega,\Sigma,T,m)$ a dynamical System in measure theory so that $T^{-1}$ preserves null sets. If $A\in\Sigma$ has positive measure then $$ A\in\mathcal{C}(T)\Leftrightarrow S_{\infty}1_B=\infty \text{ a.s. on } B \text{ for each }B\subset A\cap\Sigma. $$ (Here with $\mathcal{C}(T)$ the conservative part of the system is meant.)

With Greetings and best regards

math12

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It doens't matter whether $\Omega$ is $\sigma$-finite or not: it will always be the case that $$\Omega = \bigcup_{n \in \mathbb{Z}} f^{-1} (\epsilon n, \epsilon (n + 1)].$$

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  • $\begingroup$ Sorry, but: Why? $\endgroup$ – math12 Jul 16 '14 at 5:58
  • $\begingroup$ @math12 If $f : \Omega \to \mathbb{R}$, then every point of $\Omega$ is assigned a value in $\mathbb{R}$, and $\mathbb{R} = \cup_{n \in \mathbb{Z}} (\epsilon n, \epsilon (n + 1)]$. $\endgroup$ – A Blumenthal Jul 21 '14 at 17:44
  • $\begingroup$ If $f$ is a function, then the inverse image $f^{-1}$ commutes with unions and intersections. $\endgroup$ – A Blumenthal Jul 21 '14 at 17:46

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