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 $$\mu(E-A_n) \leq \mu(A_0-A_n) = \mu(A_0) - \mu(A_n) = 0.$$ 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 $f^k(x)\notin E$ .