A Truly Trivial Use of the Recurrence Theorem

Let $X$ be a measure space and let $T:X\to X$ be a measure-preserving measurable transformation. If $E\subseteq X$ and $x\in E$ then $x$ is said to be recurrent with respect to $E$ and $T$ if there exists a positive integer $n$ such that $T^n(x) \in E$.

Theorem. If $X$ has finite measure then almost every point of $E$ is recurrent.

The proof of this comes down to showing that for all nonnegative $n$ the sets $T^{-n}(F)$ are pairwise disjoint where $F$ is the set of nonrecurrent points of $E$. Since $T$ is measure preserving and $X$ has finite measure, this means that $F$ has measure zero.

It's a short argument, but nonetheless encapsulates certain arguments under the umbrella of iterated measurable transformations. For example, consider the following: if $G$ is a finite group and $g\in G$ then there exists a positive integer $n$ such that $g^n = 1$ where $1\in G$ is the identity. The usual proof: since $g$ is finite, $g^n = g^m$ for some $n$ and $m$ distinct with $n > m$ and hence $g^{n-m} = 1$.

We can prove this with the recurrence theorem. We define the transformation $T:G\to G$ via $T(h) = gh$, and put the counting measure on $G$: that is, $\mu(\{g\}) = 1$ and the corresponding $\sigma$-algebra is the power set of $G$. The transformation $T$ is measure preserving since it is bijective, and the resulting measure space has finite measure since $G$ is finite.

Since every nonempty subset of $G$ has positive measure, every point of every set is recurrent with respect to that set by the recurrence theorem. This applies to the set $\{1\}$. Hence, there exists a positive integer $n$ such that $T^n(1) = 1$, or in other words, $g^n = 1$.

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