Suppose $(X,\mathcal{B}, \mu, T)$ and $(X,\mathcal{B}, \nu, T)$ are both ergodic ppt. I'm a bit confused how the B Ergodic Theroem works since the LHS of the equation doesn't involve $\mu$ or $\nu$, except that $f\in L^1(\mu)$ or $L^1(\nu)$. It seems that given any $E\in \mathcal{B}$ if $f=1_E$ then $f \in L^1(\mu)\cap L^1(\nu)$ so $$\frac{1}{n}\sum_{k=0}^{n-1} f\circ T^k \to \nu(E) \text{ and } \mu(E)$$
Therefore, all ergodic measures on $(X,\mathcal{B}, T)$ are equal..... This is obviously false, but I can't spot the fallacy.
The convergence is for almost all points with respect to some measure. In particular, $$ \frac{1}{n}\sum_{k=0}^{n-1} f\circ T^k \to \nu(E) $$ $\nu$-almost everywhere, while $$ \frac{1}{n}\sum_{k=0}^{n-1} f\circ T^k \to \mu(E) $$ $\mu$-almost everywhere. For example, if the two measures are singular, no information about $\mu$ is given by the first limit and no information about $\nu$ is given by the second limit.