Non-ergodic measure

Is there an easy way to see that if $\mu$ and $\nu$ are $T$-invariant measures on the same space $X$, and $\mu \neq \nu$, then $\frac{1}{2}(\mu+\nu)$ is NOT ergodic?

I know that ergodic measures are the extreme points of the convex set of invariant measures, but the proof of this fact requires the Radon-Nikodym Theorem. I'm just wondering if the above case has an elementary proof. So far, I don't see it.

Solution 1:

I've a proof which relies on Birkhoff ergodic theorem, so I'm not sure it fits the term "elementary". Anyway, we don't need to know the fact that ergodic measures are the extreme points of the convex set of invariant measures.

Assume that $\lambda:=\frac{\mu+\nu}2$ is ergodic. Then by von Neumann's or Birkhoff's ergodic theorem, we get that given $A\in\mathcal A$, $$\frac 1n\sum_{j=0}^{n-1}\chi_{T^jA}\to \frac{\mu(A)+\nu(A)}2\quad\lambda-\mbox{ almost everywhere}.$$ Hence it also hold $\mu$ almost everywhere, and integrating with respect to $\mu$, we get that $$2\mu(A)=\mu(A)+\nu(A),$$ hence the measures $\mu$ and $\nu$ are equal.

Note that it could also work if we consider $\alpha \mu+(1-\alpha)\nu$ where $\alpha\in (0,1)$.