The following problem looks simple, but it seems there is more needed than just using the definition of ergodicity.
Let $(X, \mathcal{F}, \mu, T)$ and $(X, \mathcal{F}, \nu, T)$ be two measure preserving dynamical systems. Show that if $\mu\neq \nu$, then the measure $\rho := \frac{1}{2}(\mu+\nu)$ is not ergodic for $T$.
These are my ideas: Suppose $T$ is ergodic. Since $\mu\neq \nu$, there is a set $A\in \mathcal{F}: \mu(A)\neq \nu(A)$, so wlog we can assume $\mu(A)<\rho(A)<\nu(A)$, whence it follows that $\rho(A)\notin\lbrace 0,1\rbrace$ and hence can't be $T$-invariant. It seems like we must use that $T$-invariant sets form a $\sigma$-algebra.
Is there anyone with further ideas?
Below is an argument based on the pointwise ergodic theorem. There is also a direct argument without using the ergodic theorem, which you can find (if I remember correctly) in the book of Walters, but I find the following to be more conceptual.
Suppose that $\rho$ is ergodic for $T$. This implies that both $\mu$ and $\nu$ are ergodic as well. Let $B\subseteq X$ be a measurable set, and consider its characteristic function $1_B(\cdot)$. By the ergodic theorem, there exists a measurable set $Y\subseteq X$ with $\rho(Y)=1$ such that, for each $x\in Y$, the ergodic average \begin{align} \overline{1_B}(x) &:= \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}1_B(T^k x) \end{align} exists and satisfies $\overline{1_B}(x)=\int1_B\,\mathrm{d}\rho=\rho(B)$. Similarly, there exists a measurable set $Y_\mu\subseteq X$ with $\mu(Y_\mu)=1$ such that $\overline{1_B}(x)=\int1_B\,\mathrm{d}\mu=\mu(B)$ for each $x\in Y_\mu$.
But $\rho(Y)=1$ implies $\mu(Y)=1$, thus $\mu(Y\cap Y_\mu)=1$. In particular, $Y\cap Y_\mu\neq\varnothing$. Take $x\in Y\cap Y_\mu$. We have $\rho(B)=\overline{1_B}(x)=\mu(B)$. It follows that $\rho=\mu=\nu$.