$\mu,\nu$ ergodic implies $\mu\perp\nu$

Question

Let $T:\Omega\to\Omega$ a measurable function and $\mu,\nu$ $T$-ergodic measures on $\Omega$. I am trying to prove that $\mu\perp\nu$ (this is, they concentrate in disjoint sets).

My attempt was define $w=\mu+\nu$ and use Radon-Nikodým to obtain $f,g\in L^1(\Omega)$ such that $\mu\sim fdw,\ \nu\sim gdw$. Then I only have to show that $w(\{f,g>0\})=0$, but I couldn't progress much.

Note: for $\mu$ being ergodic I mean "$\mu$ is $T$-invariant and, for every measurable $A$, $\mu(A\triangle T^{-1}A)=0$ implies $\mu(A)\in\{0,1\}$.

Note2: I am not able to use the ergodic theorem.

Answer 1

I came up with a "elementary" solution.

1. If $\nu$ a probability $T$-invariant measure and $\mu$ an ergodic measure such that $\nu\ll\mu$, then $\nu=\mu$. In fact, let $f\in L^1(\Omega)$ such that $d\nu=fd\mu$, write $A=\{f<1\}$ and suppose $\mu(A\backslash T^{-1}A)>0$. Then, since $\nu(A\backslash T^{-1}A)=\nu(T^{-1}A\backslash A)$, we have the bounds $$\mu(A\backslash T^{-1}A) > \int_{A\backslash T^{-1}A}fd\mu = \nu(A\backslash T^{-1}A) = \nu(T^{-1}A\backslash A)=\int_{T^{-1}A\backslash A}fd\mu\geq \mu(T^{-1}A\backslash A)$$ This contradictions shows that $\mu(A\triangle T^{-1}A)=0$ and, since $\mu(\Omega)=1$, $\mu(A)=0$. Again, $\mu(\Omega)=1$ implies $\mu(\{f>1\})=0$. Then, $\mu=\nu$.

2. Now, let $\mu,\nu$ $T$-ergodic measures, take any $p\in(0,1)$ and write $w=p\mu+(1-p)\nu$. Clearly, $w$ is a $T$-invariant probability and $\mu,\nu\ll w$. Take $f,g$ in $L^1$ such that $d\mu=fdw,\ d\nu=gdw$ and put $F=\{f>0\},\ G=\{g>0\}$. If we prove $w(F\cap G)=0$ we are done. Suppose $w(F\cap G)>0$ and let's prove that $w$ is ergodic: if $A$ is measurable such that $w(A\triangle T^{-1} A)=0$, then, since $\mu,\nu$ are ergodic, $\mu(A),\nu(A)\in\{0,1\}$. The non trivial case is (without loss of generality) $\mu(A)=1,\nu(A)=0$, but this never happen: if it happens, we would have $$0 = \nu(A) \geq \nu(A\cap F\cap G) = \int_{A\cap F\cap G}gdw$$ so, $w(A\cap F\cap G)=0$. Then, since $\mu(A)=1$, $$\int_{F\cap G}fdw = \mu(F\cap G) = \mu(A\cap F\cap G) \leq w(A\cap F\cap G)=0$$ wich implies $w(F\cap G)=0$, wich is a contradiction. We deduce then that $w$ is ergodic and, therefore, from 1., that $\mu=w=\nu$.

Answer 2

For a measure $\lambda$ and a measurable function, $f$, let \begin{align*} B_{\lambda}^{f} &= \left\{x: \lim_{n} n^{-1}\sum_{i=1}^{n}{f(T^{i}(x)) =E_{\lambda}[f]}\right\} \end{align*} Since $\mu \neq \nu$, there exists a measurable $f^*$ such that $E_{\nu}[f^*] \neq E_{\mu}[f^*]$. Therefore, $B_{\mu}^{f^*} \cap B_{\nu}^{f^*}=\emptyset$. Also conclude from the Ergodic theorem that $\mu(B_{\mu}^{f^*})=1$ and $\nu(B_{\nu}^{f^*})=1$.