Let $\{U_{n}:n\in\mathbb{N}\}$ be a basis of open sets for $X$, and for each $n$ define

From theorem 1 we know that $\mu(U_{n}^{{\prime}})=0$. Let $N=\bigcup_{{n\in\mathbb{N}}}U_{n}^{{\prime}}.$ 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 $x\in U_{n}-U_{n}^{{\prime}}$ which by definition of $U_{n}^{{\prime}}$ means that there exists $n\geq 1$ such that $f^{n}(x)\in U_{n}\subset U$; thus $x$ is recurrent. $\Box$