Let $\underline{X}=(X,\mathcal{B},\mu,T)$ and $\underline{Y}=(Y,\mathcal{B},\mu,S)$ be ergodic measure preserving systems on Borel probability spaces. A joining of $\underline{X}$ and $\underline{Y}$ is a Borel probability measure $\rho$ on $X \times Y$, which is invariant under $T \times S$, such that the pushforward measures of $\rho$ onto $X$ and $Y$ are $\mu$ and $\nu$ respectively, that is $(\pi_X)_*\rho = \mu$ and $(\pi_Y)_*=\nu$.
Question: How to show that in this case, there exists a joining that is ergodic for $T \times S$?
I don't quite understand the standard proofs in standard texts: The general strategy is to take some joining $\rho$ (the product measure is a joining, so this can be done) and to consider the ergodic decomposition $\rho= \int \rho_z d\tau(z)$. One then shows (this I do understand) quite easily that we have a decomposition $$\mu=\int (\pi_X)_*\rho_z d\tau (z).$$
How does it follow from here that $\mu=(\pi_X)_*\rho_z$ for almost all $z \in X \times Y$? It is just claimed in a text I was reading that it follows from the result that extreme invariant measures are the ergodic ones, but I can't see how that works.
If anyone knows an alternative proof to the question, that would be also appreciated.
I think I can answer my question.
Philosophy: The equation $$\mu= \int (\pi_X)_*\rho_z d\tau(z)$$ is, in some sense, a convex combination of the $(\pi_X)_* \rho_z$, thus $\mu$ being ergodic means that $\mu$ is extreme and thus almost all the $(\pi_x)_* \rho_z$ must be the same, thus equal to $\mu$.
Rigorous Proof: It is enough to show that for all continuous $f:X \to \mathbb{R}$ we have that $\int f d\mu= \int f d (\pi_X)_{*} \rho_z$ for almost all $z$ (Since $C(X)$ has a countable dense set, we may then find a conull set of $z$ for each such continuous function in our countable dense set, and then take a countable intersection on which the functionals $\int d\mu$ and $\int d(\pi_X)_*\rho_z$ agree). Thus suppose for contradiction that this wasn't the case; then $\int f d (\pi_X)_{*} \rho_z$ is not an almost constant function of $z$ and so we may find a non-trivial partition $Z=Z_1 \sqcup Z_2$ such that $Z_1$ and $Z_2$ are both not $\tau$-null and $$\{ \int f d(\pi_X)_*\rho_z | z \in Z_1 \}< \{ \int f d(\pi_X)_*\rho_z | z \in Z_2 \}$$ in the sense that any element on the left side is strictly smaller than any element on the right side (this is a standard measure theoretic exercise which hinges on the fact that the mapping $$z \mapsto \int f d(\pi_X)_*\rho_z$$ is a measurable real valued function). Then we may express $$\mu= \mu(Z_1)\frac{1}{\mu(Z_1)} \int_{Z_1} (\pi_X)_*\rho_z d\tau(z) + \mu(Z_2) \frac{1}{\mu(Z_2)} \int_{Z_2} (\pi_X)_*\rho_z d\tau(z)$$ which is a genuine convex combination. It is non-trivial since those functionals disagree at $f$, by construction.