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.


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 $|\varphi_A(B)|=|B|$ ($B\subset A$)..

  • $\begingroup$ 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

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.