Showing that $\mu(\limsup_{n\to\infty}f^{-n}(A)\setminus A) = 0$ if $\mu(A)=\mu(f^{-1}(A)),\mu(A\setminus f^{-1}(A))=0$ and $\mu$ is finite

Question

Let $\mu$ be a probability measure over $X$, $f:X\to X$ a $\mu$ measurable mapping and suppose that $\mu$ preserves $f$ in the sense that $\forall S\subset X:\mu(S) = \mu(f^{-1}(S))$. Let $A\subset X$ a $\mu$ measurable set such that $\mu(A\setminus f^{-1}(A)) = 0$ and define $B=\limsup_{n\to\infty}f^{-n}(A) = \bigcap_{n=0}^\infty \bigcup_{k\geq n}f^{-k}(A)$. Then $f^{-1}(B) = B$.

I am trying to show that $\mu(B\setminus A) = 0$ but I don't know how to connect the assumption $\mu(A\setminus f^{-1}(A)) = 0$ back to $\mu(B\setminus A)$. Namely, by invariance of $\mu$ and $B$, we can take as high pre-image as we want in the sense that $\mu(B\setminus A) = \mu(B\setminus f^{-N}(A)),\forall N\in\mathbb{N}$. Therefore we can remove any single $f^{-k}(A)$ in the limsup $\bigcap_{n=0}^\infty \bigcup_{k\geq n}f^{-k}(A)$. But this is a finite, single-time, approach to an asymptotic object and I don't see how you could remove all the $f^{-k}(A)$s (or just enough of them) to make $\mu(B\setminus A) = 0$. Something to consider is that for any $N\in\mathbb{N}$, $f^{-N}(A)\setminus A \subset \bigcup_{n=0}^{N-1}f^{-n-1}(A)\setminus f^{-n}(A) = \bigcup_{n=0}^{N-1}f^{-n}(f^{-1}(A)\setminus A)$. But I am not sure how to use this superset to my advantage.

Answer 1

Hints: Verify that $$f^{-n}(A)\setminus A\subseteq \bigcup_k [f^{-k}(A)\setminus f^{-(k-1)}(A)\setminus]$$ by observing that $f^{n}(x) \in A$ and $x \notin A$ implies that there is smallest $k$ such that $f^{k}(x) \in A$]. Next use the hypothesis to see that $\mu ([f^{-k}(A)\setminus f^{-(k-1)}(A)])=0$ for each $k$. It follows that $\mu (\bigcup_n f^{-n}(A)\setminus A)=0$.

[Lemma

$\mu(C)=\mu(D)$ and $\mu (C\setminus D)=0$ implies that $\mu (D\setminus C)=0$

Proof: $\mu(C\cap D)+\mu (C\setminus D)=\mu(C\cap D)+\mu (D\setminus C)$. Cancel the first term.

Corollary

$\mu (f^{-1}(A)\setminus A)=0$

$\mu (f^{-1}(S))=\mu(S)$ for any $S$. Iteration gives $\mu (f^{-j}(S))=\mu (S)$ for any $j$ for any $S$. Also, $\mu (f^{-1}(A)\setminus A)=0$ so $\mu (f^{-j}(A)\setminus f^{-(j-1)}(A))=\mu (f^{-(j-1)} (f^{-1}(A)\setminus A)]=0$ for any $j$ ].