Let $(X,B)$ be a measure space, $T:X\to X$ be a measurable transformation and $\mu,\nu$ be two ergodic probability measures on $(X,B)$. I am trying to prove that either $\mu=\nu$ or $\mu\perp\nu$ without the help of the Birkhoff Ergodic Theorem. Currently I am stuck at concluding that if we assume that $\mu(A)=\nu(A)$ for all such $A\in B$ that $A = T^{-1}(A)$, then either $\mu=\nu$ or there exists some $A\in B$ such that $A\neq T^{-1}(A)$ and $\mu(A)=1,\nu(A)=0$ or the other way around.
To be precise, by defining $\eta(A)=\frac{1}{2}(\mu+\nu)$, if $\eta(A)=\frac{1}{2}$ for some $T$-invariant $A\in B$, then $\mu$ and $\nu$ are singular. Currently I am dealing with the case that for all $T$-invariant sets $A\in B$, $\eta(A)=0$ or $\eta(A)=1$. But as of writing I am not seeing how to either construct such a concentrated set $A$ or to extend the equality between $\mu$ and $\nu$ on the $T$-invariant sets $A\in B$.
You can use the fact that ergodic measures are extreme points in the space of probability measures.
Then, $\eta$ is not ergodic as non trivial combination of extreme points. Then there must be some invariant $A_0$ such that $0<\eta(A)<1$. Since $\mu$ and $\nu$ are ergodic, either $\mu(A_0)=0$ and $\nu(A_0)=1$, or $\mu(A_0)=1$ and $\nu(A_0)=0$.