Let $(\Omega,\mathcal{A},\mathbb{P})$ and $(\Omega',\mathcal{A}',\lambda)$ be two measure spaces. If we have an ergodic map $f\colon \Omega \rightarrow\Omega$ and a bijection $h\colon \Omega'\rightarrow \Omega$ I would like to be able to make the conclusion that $$T:=h^{-1}\circ f\circ h$$ is ergodic if $T$ is measure preserving.
Is this even possible? I can only show ergodicity modulo the bijective transformation, e.g. $A\in\mathcal{A}$ is $f$ invariant (wrt. $\mathbb{P}$) iff $h(A)$ is $T$ invariant (wrt. $\lambda$). But this is a trivial result.
Best Answer
Lemma: Suppose $(\Omega,\mathcal{A},\mathbb{P},f)$ is ergodic and let $(\Omega^\prime,\mathcal{A}^\prime)$ be measurable space. If $h:\Omega\to\Omega^\prime$ is measurable and such that for some measure preserving $g:\Omega^\prime\to\Omega^\prime$ $$ g\circ h=h\circ f, $$ Then $(\Omega^\prime,\mathcal{A}^\prime,h^*\mathbb{P},g)$ is ergodic.
From this your question is immediate.
Proof: Let $A^\prime\in\mathcal{A}^\prime$ be invariant under $g$. Then $f^{-1}(h^{-1}(A^\prime))=h^{-1}(g^{-1}(A^\prime))=h^{-1}(A^\prime)$. Hence, by ergodicity of $\mathbb{P}$, $h^*\mathbb{P}(A^\prime)\in\{0,1\}$.