Disclaimer: I didn't include the following claim to the title as I don't know whether it has any well known name.
Consider the following claim:
1: Let $(X,\mathcal{F}\mu)$ be a finite measure space and $f:X\to X$ be a $\mu$-measurable function such that $\forall S\in \mathcal{F}:\mu(S) = \mu(f^{-1}(S))$. Let $A\subset \mathcal{F}$ be any set element such that $\mu(A) > 0$. Then there exists a $\mu$-measurable set $B\subset A$ and an integer $M\in\mathbb{N}$ such that $0 < \mu(B)$ and $f^{M}(B)\subset A$.
I am trying to show that the prior claim is equivalent with the Poincaré recurrence theorem in finite measure spaces (or at least with one version of the recurrence theorem)
(Poincaré) Let $(X,\mathcal{F},\mu)$ be a finite measure space and $f:X\to X$ be a $\mu$-measurable function such that $\forall S\in \mathcal{F}:\mu(S) = \mu(f^{-1}(S))$. Then if $A\in \mathcal{F}:\mu(A)> 0$, for $\mu$-almost every point $x\in A$ there exists infinitely many indices $n\in\mathbb{N}: f^n(x)\in A$.
It is not too hard to show that the prior version of Poincaré implies 1. But I am quite stuck at trying to show the converse. If we define $N_n := \{x\in A: x\not\in f^{-n}(A)\}$ then the set of points of $A$ that return to $A$ at most finitely many times is given by $N := \liminf_{n\to\infty}N_n = \bigcup_{n\in\mathbb{N}}\bigcap_{k\geq n}N_k$. Observe that $N_n=A\setminus f^{-n}(A)$.
So we would like to show that this set has measure zero. To my understanding our only immediate connection to our assumption is that we can take a sequence of $B_n$s such that $A\supset B_1\supset B_2\supset \cdots$, $\forall n\in\mathbb{N}:\mu(B_n) > 0$ and $\exists M_n\in\mathbb{N}:f^{M_n}(B_n)\subset A$. Since the sets $B_n$ have positive measure one idea could be to establish a contradiction by bounding $B_n$s above by a set of measure zero. But I don't know how to do this nor do I know how to really utilize the aforementioned sequence of $B_n$s.
You are close to the finish line if you modify one detail in your argument. Namely, you can take your $B_n$ and $M_n$ such that $f^{M_1}(B_1) \subset A$ and $f^{M_n}(B_n) \subset B_{n-1}$ as opposed to the weaker condition $f^{M_n}(B_n) \subset A$. Then, $f^{M_1+\dots+M_n}(B_n) \subset A$.
In words, we proved that for any set $A$ of positive measure and any $N \in \mathbb{N}$ there exists a subset $B \subset A$ of positive measure and an $M \geq N$ with $f^M(B) \subset A$.
Since $N^n := \bigcap_{k \geq n} N_k$ contains all points $x \in A$ with $f^k(x) \notin A$ for all $k \geq n$, we have in particular that $f^k(x) \notin N^n$ for all $k \geq n$. With the above, we conclude that $N^n$ has measure zero. Finally, $N = \bigcup_n N^n$ has measure zero as a countable union of measure zero sets.