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

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 $\nu(B_{\nu}^{f^*})=1$.
