$\mu,\nu$ 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.
 A: 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 $\nu(B_{\nu}^{f^*})=1$.
A: I came up with a "elementary" solution.


*

*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, $\mu=\nu$.

*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 $\mu=w=\nu$.
