Let $\mathbb P$ and $\mathbb Q $ different probability measures on $(\Omega, \cal F)$ and $T:\Omega \to \Omega$ a measurable function, which is $\mathbb P$ and $\mathbb Q$ ergodic.
Show $A \in \cal F$ exists, such that $\mathbb P(A)=1$ and $\mathbb Q(A)=0$.
Since $T$ is $\mathbb P$ and $\mathbb Q$ ergodic $\forall A \in \cal F$ with $A=T^{-1}(A)$, $\mathbb P(A) \in \{0,1\}$ and $\mathbb Q(A) \in \{0,1\}$. Furthermore I know since T is ergodic $\lim_{n\to\infty}\frac 1n \sum_{k=0}^{n-1}\mathbb P(A \cap T^{-k}(B))=\mathbb P(A)\mathbb P(B)$. I was wondering if I can use this lemma here.
It exists some $ B \in \cal F$ such that $P (B) \ne Q (B)$. And I considered $A$ to be T-invariant, but still I can not lead this in the right direction.. Some help is highly appreciated!
Since $\mathbb P\neq\mathbb Q$, there exists some event $B\in\mathcal F$ such that $\mathbb P(B)\neq\mathbb Q(B)$. Let $f_n(\omega)$ be the cardinality of the set $\{T^i(\omega):0\le i\le n-1\}\cap B$, or written another way, $$f_n(\omega)=\sum_{i=0}^{n-1}\mathbf1_B(T^i(\omega)).$$By the ergodic theorem, $f_n/n\to\mathbb P(B)$ $\mathbb P$-almost surely and $f_n/n\to\mathbb Q(B)$ $\mathbb Q$-almost surely. In particular, with $A=\{\omega:f_n(\omega)/n\to\mathbb P(B)\}$, we have that $\mathbb P(A)=1$ and $\mathbb Q(A)=0$.