proof of Poincaré recurrence theorem 2

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

$$U_n^{\prime }=\{x\in U_n:\forall n\ge 1,f^n(x)\notin U_n\}.$$

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 $n\ge 1$ such that $f^n(x)\in U_n\subset U$; thus $x$ is recurrent. $\mathrm{\square }$