Poincaré's Recurrent Theorem

For almost every point $x$, if the system is at state $x$ at time zero, then it will return arbitrarily close to $x$ infinitely many times in the arbitrarily far future. Proof: Fix some $\epsilon>0$ and define $W:=\{x:d(T^n x,x) > \epsilon\}$. We can partition $W$ into finitely many disjoint sets $W_i$ of diameter at most $\epsilon$. For every fixed $i$ we have $T^{-n}(W_i)$ are mutually disjoint for all $n$. To see this assume the contrary, i.e. for some $n$ and $k$ we have $T^{-n}(W_i) \cap T^{-(n+k)}(W_i) \neq \emptyset$. Therefore there exists some $x \in T^{-k}(W_i) \cap W_i$, hence $T^{-k}(W_i) \cap W_i \neq \emptyset$. On the one hand, $x \in T^{-k}(W_i)$ implies that $d(x,T^k(x))<\epsilon$. On the other hand, by the definition of $W$, $x \in W_i \subset W$ implies that $d(x,T^k(x))>\epsilon$. ok so we agree that $T^{-n}(W_i)$ are pairwise disjoint for all $n$. Therefore $m(X) \geq \sum_i m(T^{-n}(W_i))$. Now here's where I'm stuck, because somehow, by Louiseville's theorem, these sets are supposed to have the same measure! On the other hand $m(X)<\infty$. Therefore, we must have $m(W_i) = 0$ why? Therefore, almost every $x$ has the property that $d(T^n(x),x)<\epsilon$ for some $n \geq 1$. The ergodic hypothesis, that for certain invariant measures $\mu$, many functions $f:X\rightarrow \mathbf R$, and many states $x$, the time average $\lim_{T\infty}\frac{1}{T}\int_{0}^T f(T_t(x))dt$ exists and equals the space average $\frac{1}{\mu(X)}\int f d\mu$ of $f$ is a quantitative version of Poincaré's recurrence theorem. Imagine $f$ is the indicator function of the $\epsilon$-ball around a state $x$. Then, the time average of $f$ is the frequency of times when $T_t(x)$ is $\epsilon$=away from $x$. The ergodic hypothesis states that this frequency converges to the measure of the $\epsilon$-ball around $x$.