Let $A_{n}=\cup_{{k=n}}^{\infty}f^{{-k}}E$. Clearly, $E\subset A_{0}$ and $A_{i}\subset A_{j}$ when $j\leq i$. Also, $A_{i}=f^{{j-i}}A_{j}$, so that $\mu(A_{i})=\mu(A_{j})$ for all $i,j\geq 0$, by the $f$-invariance of $\mu$. Now for any $n>0$ we have $E-A_{n}\subset A_{0}-A_{n}$, so that

Hence $\mu(E-A_{n})=0$ for all $n>0$, so that $\mu(E-\cap_{{n=1}}^{\infty}A_{n})=\mu(\cup_{{n=1}}^{\infty}E-A_{n})=0$. But $E-\cap_{{n=1}}^{\infty}A_{n}$ is precisely the set of those $x\in E$ such that for some $n$ and for all $k>n$ we have $f^{k}(x)\notin E$. $\Box$