Poincaré Recurrence Theorem

Theorem
Let $\struct {X, \Sigma, \mu}$ be a probability space.

Let $T: X \to X$ be a $\mu$-preserving transformation.

For all $A \in \Sigma$:


 * $\ds \map \mu {A \setminus \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A} = 0$

That is, for $\mu$-almost all $x\in A$ there are integers $0 < n_1 < n_2 < \cdots$ such that $\map {T^{n_i} }x \in A$.

Proof
Let:
 * $A _\infty := \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A$.

For $N\in\N$, let:
 * $A_N := \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A$

so that:
 * $A _\infty = \bigcap _{N \mathop =1} ^\infty A_N$

Then, for all $N\in\N$:
 * $A_N = \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A \supseteq \bigcup_{n \mathop = N+1}^\infty T^{-n} \sqbrk A \supseteq A_{N+1}$

On the other hand, for all $N\in\N$:

that implies:
 * $\map \mu {A_N} = \map \mu {T^{-1} \sqbrk {A_N}} = \map \mu {A _{N+1}}$

since $T$ is a $\mu$-preserving transformation.

Therefore: