Poincare Recurrence Part 2
In this post I discuss a variation on Poincare Recurrence.
Problem
Set Up
We follow Einsiedler and Ward Exercise 2.2.1. Let \((X, \mathcal{M}, \mu, T)\) be a measure-preserving system (meaning \(\mu(T^{-1}(A)) = \mu(A)\) for every \(A\) measurable). Suppose we only know that \(\mu\) is a finitely additive measure, meaning that if \(\{A_i\}_{i=1}^n\) is a finite collection of disjoint measurable sets, then
\[\mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{i=1}^n \mu(A_i).\]Notice we really only require disjoint almost everywhere, meaning that pairwise the measure of their intersection is \(0\). We’re also implicitly assuming that \(\mu\) is a probability measure as well, so \(\mu(X) = 1\).
Under this set up, our goal is to prove that if \(\mu(A) > 0\), then there exists \(n \leq 1/\mu(A)\) so that \(\mu(A \cap T^{-n}(A)) > 0\). In other words, we have a bound on return time for a measurable set.
Solution
Assume that \(\mu(A \cap T^{-n}(A)) = 0\) for every \(1 \leq n \leq 1/\mu(A)\). To help with notation, let \(N := 1/\mu(A)\). Let
\[B_i := \bigcup_{j=0}^{i} T^{-j}(A).\]Observe that
\[\mu(B_1) = \mu(A \cup T^{-1}(A)) = \mu(A) + \mu(T^{-1}(A)) = 2 \mu(A)\]since we have these sets are almost everywhere disjoint. We claim by induction that for \(i < N\) if
\[\mu(B_i) = \sum_{j=0}^i \mu(T^{-j}(A))\]then
\[\mu(B_{i+1}) = \sum_{j=0}^{i+1} \mu(T^{-j}(A)).\]It suffices to check that \(T^{-i-1}(A)\) is almost everywhere disjoint from \(T^{-j}(A)\) for \(0 \leq j \leq i\). Here we can use the fact that \(T\) is measure preserving, so
\[\mu(T^{-j}(A) \cap T^{-i-1}(A)) = \mu(A \cap T^{j-i-1}(A)).\]Since \(0 \leq j < i+1\) we see \(-(i+1) \leq j-i-1 < 0\), so by assumption this is \(0\). Thus
\[0 \leq \mu(B_i \cap T^{-i-1}(A)) \leq \sum_{j=0}^i \mu(T^{-j}(A) \cap T^{-i-1}(A)) = 0,\]and hence
\[\mu(B_{i+1}) = \mu(B_i) + \mu(T^{-i-1}(A)) = \sum_{j=0}^{i+1} \mu(T^{-j}(A)).\]Using induction and the fact that \(T\) is measure preserving, we see that
\[\mu\left( \bigcup_{i=0}^N T^{-i}(A) \right) = (N+1) \cdot \mu(A) > 1,\]which is impossible. This gives us that for some \(1 \leq n \leq 1/\mu(A)\) we must have \(\mu(A \cap T^{-n}(A)) > 0\).