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Let $T$ be a measure preserving transformation on a measure space $(X, \mathscr{F}, m)$ with infinite measure $m$. Let $A \in \mathscr{F}$ be such that $X = \cup_{k=0}^\infty T^{-k} A \pmod{m}$. Then first hitting time $\tau(x):= \inf\{k \ge 1: T^k x \in A\}$ is finite $m$-a.e. $x \in X$ and the first hitting mapping $\varphi(x):= T^{\tau(x)} x$ is defined a.e. The induced mapping $\varphi_A:=\varphi|_A$ is supposed to be a measure preserving transformation of $(A, \mathscr{F} \cap A, m|_A)$. But is this always true if $m(A)=\infty$?

Proposition 1.5.3 of Aaronson's Introduction to Infinite Ergodic Theory requires that $T$ is conservative and $m(A)<\infty$. The proof can be easily modified to drop either assumption. But I do not see how to drop both of them.

Eventually, everything boils down to showing that $m(T^{-n} B \setminus \cup_{k=0}^{n-1} T^{-k} A) \to 0$ for every measurable $B \subset A$ of finite measure. Is this really true without conservativity or invertibility of $T$? All of this is supposed to be very basic but I cannot figure it out.

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One basic thing: for $X=\mathbb Z$ with counting measure and $T:x\mapsto x+1$, $A$ satisfies the condition iff $\sup A=\infty$, i.e. $A=\{a_1<...<a_n<...\}$. Then $\varphi(a_i)=a_{i+1}$ and of course $\varphi_A^{-1}(a_1)=\emptyset$ so that $\varphi_A$ is not counting measure preserving on $A$ if this means $|\varphi_A^{-1}(B)|=|B|$, but it is if this means $|\varphi_A(B)|=|B|$ ($B\subset A$)..

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  • $\begingroup$ Ahhh, so easy... Thank you! I did not think of shifts, being mistakenly sure that the claim was true for invertible $T$. $\endgroup$ – Vysotsky Oct 18 '17 at 14:50

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