Suppose that $(\Omega,\mathcal{A},\mu)$ is a $\sigma$-finite measure space of infinite measure and $T:\Omega\to\Omega$ a measure-preserving transformation with measurable inverse. Let be $\Omega\_k\in \mathcal{A}$ an increasing sequence such that $\Omega\_k\uparrow\Omega$ and
$\mu(\Omega\_k)<+\infty$ for all $k\in\mathbb{N}$. 


 **Question 1:**  Given a set $A\in\mathcal{A}$, such that $\mu(A)>0$, is it true that the set 
$$E\_k=\{\omega\in A; T^n(w)\notin A\ \forall n\in\mathbb{N}\ \text{and}\ T^{n\_j}(w)\in \Omega\_k \ \text{for some infinite sequence}\ (n\_j(\omega)) \}$$
has zero measure ? 


 **Question 2:**  If $T$ is not invertible is $\mu(E\_k)=0$, in general ? 



Poincare Recurrence Theorem on Infinite Measure Space
If I remember my infinite ergodic theory correctly, any measure-preserving transformation $T$ of a $\sigma$-finite measure space $(\Omega,\mathcal{A},\mu)$ leads to a decomposition of $\Omega$ into a dissipative part $\Omega\_d$ and a conservative part $\Omega\_c$. (This notation is probably non-standard.) The dissipative part is the union of the wandering sets for $T$, where a set $E$ is wandering if it is disjoint from all its preimages $T^{-n}(E)$, $n\geq 0$. The conservative part is thee complement of the dissipative part.


On the dissipative part, everything "goes to infinity" in some sense (eventually leaves $\Omega\_k$, for instance), while on the conservative part, the Poincare recurrence theorem holds. (Conservative measures are exactly those $\sigma$-finite measures for which the Poincare recurrence theorem still works.) I believe this establishes the dichotomy you want.


This may even work for transformations that don't preserve $\mu$, but only preserve the collection of null sets. I'd check a reference on that, though -- details of all of this are in Aaronson's book *An Introduction to Infinite Ergodic Theory*.


