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Thanks to your helpful feedback, I have made my claim more precise.

Claim

Given an infinite measure space $\left( X,\mathcal B, \mu\right)$ and an ergodic, invertible, measure preserving and conservative transformation $T$. Let $n_i \in \mathbb Z, i \in \mathbb N$ and $W\in \mathcal B$ be an exhaustive weakly wandering sequence. Then for any measurable set $A$ of finite measure, almost every $x \in A, T^{n_i - n_k}x \in A$ finitely many times (where $k$ is chosen such that $x \in T^{n_k}W$, and $n_i > n_k$).

This is to be contrasted with the recurrence theorem. If $T$ is conservative, then for almost every $x \in A$, $T^{n}x \in A$ for infinitely many $n \in \mathbb N$. I claim that $T^nx \in A$ for finitely many $n \in \{n_i - n_k\}, n_i > n_k$.

Proof

Under these assumptions, an exhaustive weakly wandering set will always exist. Hence there is a set $W$ and integers $n_i, i \in \mathbb N$ such that $T^{n_i}W \cap T^{n_j}W = \emptyset, i \neq j$ and the sets $T^{n_i}W$ cover $X$.

Given any set $A \in \mathcal B$ of finite measure, let $B \subseteq A$ be the set of points which recur infinitely often in $A$. That this set is measurable can be shown by the same method used in the recurrence theorem.

Suppose $x \in B \cap T^{n_k}W$ for some fixed $k$. Without loss of generality, and with some improvement in readability, assume $n_k = 0$, so we need to prove that for $x \in B \cap W$, then only finitely many $T^{n_i}x \in B$ for $n_i > 0$.

Let $I = \{ i : \mu(T^{n_i}W \cap B) > 0\}$. If $|I|<\infty$ and $x \in T^{n_k} W \cap B, k \in I$, then $T^{n_i - n_k}x \in T^{n_i}W$ is disjoint from the cover of $B$ for $i \in \mathbb N - I$. Hence only for $i\in I$ can $T^{n_i - n_k}x$ return to $B$.

Now suppose $|I|=\infty$: that infinitely many of the sets $T^{n_i}W \cap B$ have positive measure. Because $\infty > \mu(A) \geq \mu(B) = \sum_{i=0}^\infty \mu(B \cap T^{n_i}W)$, then for any $\epsilon > 0$ there exists an $N$ such that for all $k > N$

$$\sum_{i=k}^\infty \mu(B \cap T^{n_i}W) < \epsilon$$

Let $\epsilon = \mu(B \cap W) > 0$ and recall for $x \in B \cap W$ that infinitely many $T^{n_i}x \in B$. Then each element of $B\cap W$ will eventually find its way into $B \cap (\cup_{i=N}^\infty T^{n_i}W)$, the measure of the former should be less than or equal to the measure of the latter; yet this is not so. To be precise, for any $x \in B\cap W$. Let $\sigma(x)$ be the smallest return time to $B$ greater than $n_N$. We can subdivide $B\cap W$ into sets according to this return time: $B_{n_i} = \{ x \in B\cap W : \sigma(x) = n_i\}$, where each $B_{n_i}$ has the property that $T^{n_i}B_{n_i} \subset B \cap T^{n_i}W$. Hence

$$ \epsilon = \mu(B\cap W) = \sum_{i=N}^\infty \mu(B_{n_i}) = \sum_{i=N}^\infty \mu(T^{n_i}B_{n_i}) \leq \sum_{i=N}^\infty \mu(B \cap T^{n_i}W) < \epsilon$$

a contradiction. Hence either $\mu(B\cap W) = 0$ or the $T^{n_i}x \in B$ only finitely often.

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The claim doesn't hold. If $T$ is conservative, then for every set of positive measure, almost every point returns infinitely often. Examples of conservative infinite measure-preserving transformations? Boole's transformation (see Aaronson's book); a transformation that I study with Arek Goetz... –  Anthony Quas Jan 31 '12 at 6:00
    
Hi Daniel, Having a bit of a problem understanding something here. If the sets $T^{n_i}W$ form a disjoint cover of $X$, then how can a sub-collection of them also be a cover? –  Anthony Quas Jan 31 '12 at 18:00
    
I meant that a sub collection of them cover $B \subset X$. I've removed that bit now because it was over-complicating the matter. –  Daniel Mansfield Feb 1 '12 at 4:09
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1 Answer

You might enjoy Ergodic theorems by U. Krengel, de Gruyter, 1985. In section 3.1 "The Hopf decomposition", Theorem 1.6 states that dissipativity is characterized by the property that, for all nonnegative, integrable function $f$, $\sum_{n\geq0} f^n\circ T$ is finite a.e. One can replace such functions by characteristic functions of finite measure sets. So the property you gave turns out to be a characterization of dissipativity (of course, one should understand "each element" by "almost each element").

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Thank you Jerome, You are correct, except that I've changed the question now: the dissipative assumption has been dropped, and I claim something more consistent with $T$ being conservative. –  Daniel Mansfield Feb 1 '12 at 4:13
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