Let $T: X \to X$ be an invertible measure preserving ergodic transformation on a probability space.
- Why is it that $X=\bigcup_{i,k<i}T^{-k}A_i$, where $A_i=\{n(x)=i\}$, $n(x)$ the first return map and $A$ is a set of positive measure?
From ergodicity we know that $X=\bigcup_i T^{-i}(A)$ a.e. (Which tells us that actually $n(x)$ is defined on all of $X$ mod $0$). and poincare recurrence theorem tells us that $A=\bigcup_k A_k$ a.e. The problem is that I don't know why we have the stronger condition $X=\bigcup_{i,k<i}T^{-k}A_i$ with $i<k$ instead of just $X=\bigcup_{i,k}T^{-k}A_i$.
This is False. Take a simple dynamical system made of five points $\{p_1,p_2,\cdots, p_5 \}$, with $T(p_i)= p_{i+1}$ if $i<5$ and $T(p_5)=p_1$.
We take uniform measure on this five points, of course this is an ergodic system.
Now take $A= \{p_1, p_3\}$, we have $n(p_1)=2$ and $n(p_3)=3$. Then $$ \begin{matrix} T^{-0}(\{p_1\})=\{p_1 \}\\ T^{-0}(\{p_3\})=\{p_3\} \\ T^{-1}(\{p_1\})=\{p_5\} \\ T^{-1}(\{p_3\})=\{p_2\} \\ T^{-2}(\{p_3\})=\{p_1\} \\ \end{matrix} $$ Making the union of all that miss $p_4$.