Poincaré recurrence but infinite measure I'm trying prove the following problem:
Let $f : M → M$ be an invertible transformation and suppose that $µ$ is an invariant infinite measure. Let $B ⊂ M$ be a set with finite measure. Prove that, given any measurable set $A ⊂ M$ with positive measure, $µ$-almost every point $p ∈ A$ either returns to $A$ an infinite number of times or has only a finite number of iterates in $B$.
Obs1: µ infinite measure means µ(M)=$\infty$;
Obs2: f is a function preserving µ;
Obs3: $f^{-1}$ isn't requered be measurable.
Obs4: $µ(B)<\infty$, $µ(A)>0$
Obs5: $n$ iterates means $f^{n}(p)$ for a point $p$
What I tried:
Separate the set A in 9 parts, combining iterates that never return to B, return a finite times to B, and never return to B with the analogous to A, where the iterates are done over points of the set A. But I didn't get success in prove that the appropriate intersections have null measures. 
I hope any suggestions to proceed ahead, thank you! 
 A: First, note that $M$ can be partitioned into $M_1 :=\limsup_n f^{-n} B$ (the set of points which come back infinitely often in $B$) and its complement $M_2$, and both are $f$-invariant sets.
Points in $M_2 \cap A$ are points in $A$ which go finitely often in $B$. They are good. If $M_1 \cap A$ has zero measure, then we are done; let us assume that $\mu (M_1 \cap A) >0$, and let us look at the measure-preserving system $(M_1, f, \mu_{|M_1})$. Since $M_1 = \limsup_n (f^{-n} (B \cap M_1))$ has positive measure, $B \cap M_1$ has positive and finite measure. All is left to prove is that almost any point in $M_1 \cap A$ comes back infinitely often in $M_1 \cap A$.
Hence, without loss of generality, by focusing on $M_1$, we can assume that any point comes back infinitely often in $B$, but that $\mu$ may be finite or infinite. Then we want to prove that almost any point in $A$ comes back infinitely often in $A$. Without loss of generality, we may also assume that $\mu (B)=1$.
For $x \in M$, let $g (x) := |\{n \geq 0: f^n (x) \in A\}|$. Write $M_0 := \{g<+\infty\}$, which is $f$-invariant.
Let $n(x) := \inf \{n >0: f^n (x) \in B \} < +\infty$. For $x \in B$, let $f_B (x) := f^{n(x)} (x)$ be the induced map in $B$. Then, by general nonsense, $(B, f_B, \mu_{|B})$ is a measure-preserving dynamical system, and $B \cap M_0$ is $f_B$-invariant.
Then $(g \circ f_B^n)_{n \geq 0}$ converges almost surely to $0$ on $B \cap M_0$, but the distribution of $(g \circ f_B^n)_{n \geq 0}$ is stationary on $B$, so $g \equiv 0$ almost everywhere on $B \cap M_0$.
Now, do the same induction process with $A \cup B$. We get a measure-preserving system $(A \cup B, f_{A \cup B}, \mu_{|A \cup B})$. Note that $M_0$ is $f_{A \cup B}$-invariant, so that $((A \cup B)\cap M_0, f_{A \cup B}, \mu_{|(A \cup B)\cap M_0})$ is still measure-preserving. Hence, for all $n \geq 0$:
$$\mu_{|(A \cup B)\cap M_0} (f_{A \cup B}^{-n} (A) \cap B) = \mu_{|(A \cup B)\cap M_0} (f_{A \cup B}^{-n} (B) \cap A).$$
But $g \equiv 0$ on $B \cap M_0$, so $\mu_{|(A \cup B)\cap M_0} (f_{A \cup B}^{-n} (A) \cap B)=0$. Hence, $\mu_{|(A \cup B)\cap M_0} (f_{A \cup B}^{-n} (B) \cap A)=0$.
Hence, for almost every $x \in A \cap M_0$, we have $f_{A \cup B} (x) \notin B$ for all $n \geq 0$, that is, the trajectory $f^n (x)$ never go to $B$. If $M_0 \cap A$ has positive measure, this contradicts the hypothesis that every trajectory goes infinitely often through $B$.
Hence, $\mu (A \cap M_0) = 0$: almost any point in $A$ comes back infinitely often in $A$.
