Let $T$ be a measureable map on the measureable space $(X,B)$.
It is fairly elementary that the ergodic measures are the extreme points of the space of the $T$-invariant measures, which means that no ergodic measure can be presented as a convex combination of two (different) invariant measures.
It is obvious that this can be generalized by induction to any finite-length convex combinations. My question is if it is true for general integration. It makes sense to me that the following will hold (or something similar), yet I couldn't find any references:
"If $\mu=\int \nu_x dm(x)$ where $\mu$ is ergodic, $\nu_x$ are $T$-invariant, and $m$ is a probability measure on the space of invariant probability measures, then $\nu_x=\mu$ for $m$-a.e. $x$."
Here's an argument I learned from the book 'Ergodic theory with a view towards number theory' by Manfred Einsiedler and Thomas Ward.
We have $(X,\mathscr{B}_X,\mu,T)$, an ergodic system and a decomposition $$ \mu = \int_{Y} \nu_y dm(y) \qquad \qquad (1)$$
where $Y$ is the space of $T$-invariant probability measures. And I make the assumption that the space $(X,\mathscr{B}_X)$ is a Borel space so in particular $C_c(X)$ has a countable dense subset $\lbrace f_i \rbrace$. If both the sets $$\bigcup_i \left\lbrace y\in Y: \int f_i d\mu < \int f_i d\nu_y\right\rbrace,$$ $$\bigcup_i \left\lbrace y\in Y: \int f_i d\mu > \int f_i d\nu_y\right\rbrace$$
have measure zero then we are done. Assume otherwise. Without loss of generality, we have an $f\in C_c(X)$ with $$E:=\left\lbrace y\in Y: \int f d\mu < \int f d\nu_y \right\rbrace$$ having $0<m(E)$. Note, $m(E)$ cannot actually be full measure since that would contradict $(1)$. Hence $0<m(E)<1$. Now we have, for all $g\in C_c(X)$,
$$\int gd\mu = m(E)\cdot\int_Y \frac{1_{E}(y)}{m(E)}\left(\int gd\nu_y\right)dm(y) + m(E^c)\cdot\int_Y \frac{1_{E^c}(y)}{m(E^c)}\left(\int gd\nu_y\right)dm(y)$$ which gives a representation of $\mu$ as a convex sum of two $T$-invariant measures. By ergodicity and hence extremality of $\mu$, each of the measures is in fact $\mu$. This is a contradiction since integrating the first measure in the convex sum against $f$ gives something strictly smaller than $\int fd\mu$.