Let $\{U_n : n\in \N\}$ be a basis of open sets for $X$ , and for each $n$ define $$U_n' = \{x\in U_n : \forall n\geq 1,\, f^n(x) \notin U_n \}.$$ From theorem 1 we know that $\mu(U_n')=0$ . Let $N=\bigcup_{n\in \N} U_n'.$ Then $\mu(N)=0$ . We assert that if $x\in X-N$ then $x$ is recurrent. In fact, given a neighborhood $U$ of $x$ , there is a basic neighborhood $U_n$ such that $x\subset U_n\subset U$ , and since $x\notin N$ we have that $x\in U_n-U_n'$ which by definition of $U_n'$ means that there exists $n\geq 1$ such that $f^n(x)\in U_n\subset U$ ; thus $x$ is recurrent.