First-return times and induced transformations

Let $(X,\mathcal{A},\mu ,T)$ preserve finite measure and let $E\in \mathcal{A}$ satisfy $\mu (E)>0$. For $x\in E$ define the first-return time

$r_E(x)=\min \{n\ge 1:T^{n}x\in E\},\min \mathrm{\varnothing }=\mathrm{\infty }\mathrm{.}$

A nonempty set of return times has a least element by The well-ordering principle. Define the infinitely returning core

$E_{\mathrm{\infty }}=E\cap {\bigcap }_{N\ge 1}{\bigcup }_{n\ge N}{T}^{-n}E\mathrm{.}$

This is the set-limsup convention of Limit superior and limit inferior of a sequence of sets. It is measurable by Sigma-algebras are closed under countable intersections, differences, symmetric differences, and set limits, and Poincare recurrence for finite measure-preserving systems gives $\mu (E\setminus E_{\mathrm{\infty }})=0$. Define the induced transformation and normalized measure by

$T_E:E_{\mathrm{\infty }}⟶E_{\mathrm{\infty }},T_E(x)=T^{r_E(x)}x,{\mu }_E(B)=\frac{\mu (B)}{\mu (E)},B\in \mathcal{A}{\mathrm{\mid }}_{E_{\mathrm{\infty }}}\mathrm{.}$

Here the trace is as in The trace of a sigma-algebra on a subset. Since $E_{\mathrm{\infty }}$ is measurable, the trace equals $\{B\in \mathcal{A}:B\subseteq E_{\mathrm{\infty }}\}$: each intersection with $E_{\mathrm{\infty }}$ is measurable and each such $B$ equals $B\cap E_{\mathrm{\infty }}$. Its closure under relative complement and countable union follows directly from these same operations in $\mathcal{A}$; no sequence of ambient representatives needs to be chosen. This supplies the measurable-subset instance of The trace of a sigma-algebra is a sigma-algebra on the traced subset locally.

For $x\in E_{\mathrm{\infty }}$, $r_E(x)$ is finite, and the infinitely many visits after $r_E(x)$ are exactly positive visits of $T_E(x)$, so $T_E(x)\in E_{\mathrm{\infty }}$. Restricting countable additivity of $\mu$ and dividing by the positive finite number $\mu (E)$ makes ${\mu }_E$ a measure; its total mass is $\mu (E_{\mathrm{\infty }})\mathrm{/}\mu (E)=1$. The value $\mathrm{\infty }$ is allowed for $r_E$ on $E\setminus E_{\mathrm{\infty }}$, but is never used as an iterate in $T_E$. No choice axiom enters this construction.