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.