This probel is problem $1.1.3$ of Fundation of ergodic theory by Marcelo Viana and Krerley Oliveira.
Suppose that $f : M → M$ preserves a probability measure $\mu$. Let $B\subset M$
be a measurable set satisfying any one of the following conditions:$\ \ \ 1.\ \ \mu (B-f^{-1}(B))=0\\$, $\ \ \ \ \ 2.\ \ \mu (f^{-1}(B)-B)=0 $, $\ \ \ \ \ 3. \ \ \mu (B\Delta f^{-1}(B))=0 \ \ $ and $\ \ 4. \ \ f(B) \subset B$. Show that there exist $C\subset M$ such that $f^{-1}(C)= C $ and $\mu (B\Delta C)=0 $.
I know Since $\mu$ is a finite invariant measure we have
$$\mu (B-f^{-1}(B)) = \mu (B)-\mu (f^{-1}(B)\cap B) =\mu (f^{-1}(B))-\mu (f^{-1}(B)\cap B) = \mu (f^{-1}(B)-B)$$
Therefore the three first conditions are the same and every one implies the otheres. I tried to set $C = f^{-1}(B)\cap B$ and i stock to prove $f^{-1}(C)= C $. Once again I set $C = \bigcap_{n=0}^{\infty} f^{-n}(B) $ and failed to show that $\mu (B\Delta C)=0 $.
Best Answer
Your answer $C = \bigcap_{n=0}^{\infty} f^{-n}(B)$ is correct. Remark that $C \subset B$ so $C \Delta B=B-C$. Let's call $D=B-C$ and show that $\mu(D)=0$.
We have that $D= \{x \in B, \exists k \in \mathbb{N}-0, f^k(x)\notin B \}$. We will decompose $D$ into a countable union of set of measure zero. Let's call $D^k=\{x \in B,f^i(x)\in B, 0\leq i <k, f^k(x)\notin B \}$, we have $D= \bigcup_{k\geq 1} D^k$.
$D^1=\{ x\in B, f(x) \notin B \}=B-f^{-1}(B)$, so by hypothesis $\mu(D^1)=0$. Then for $k>1$, $D^k= f^{-1}(D^{k-1})\cap B \subset f^{-1}(D^{k-1})$. As $f$ is measure preserving and by induction $\mu(D^k)=0$. Then $\mu(D)=0$ as a countable union of measure zero set.