If $T_A$ is ergodic and $\mu\Big(\cup_{n\in\mathbb{N}} T^{-n} A\Big)=1$ then $T$ is also ergodic.

Question

This is exercise 1.8.4 from https://www.staff.science.uu.nl/~kraai101/LectureNotesMM-2.pdf

"Show that if $T_A$ is ergodic and $\mu\Big( \cup_{k\geq 1} T^{-k}(A)\Big)=1$ , then, $T$ is ergodic."

Answer 1

Assume $T$ is not ergodic. Then $\exists$ a not $\mu$-zero $C\subseteq X$ such that $\mu\Big(\cup_{n\in\mathbb{N}} T^{-n}(C)\Big)\neq 1$.

So let $D$ be a set with $\mu(D)>0$ and s.t. no points of $D$ will reach $C$ after $>0$ steps.

Let $A'\subseteq A$ be defined as follows: $\lbrace a\in A| \exists n\in\mathbb{N},d\in D: T^n(d)=a \rbrace$

Now $\cup_{n\in \mathbb{N}} T^{-n}(A') \supseteq D$ a.e. since $\mu\Big(\cup_{n\in\mathbb{N}} T^{-n}(A)\Big) = 1$

hence if $\mu(A')=0$ then $T^{-n}(A')=0$ $\forall n\in\mathbb{N}$ (measure-preserving), hence $0=\mu\Big(\cup_n T^{-n}(A')\Big)\geq \mu(D)>0$

Now that $\mu(A')\neq 0$, almost all points of $C$ (will reach $A$ and hence) will reach $A'$ by ergodicity of $T_A$.

Points of $C$ will come back to $C$ infinitely often (Poincaré), so there are points of $A'$ which will reach $C$ infinitely often. This contradicts the definition of $D$.