Q:

Product of uniquely ergodic continuous disjoint maps is uniquely ergodic

Let $g: X \rightarrow X$ and $h: Y \rightarrow Y$ be continuous uniquely ergodic maps between compact metric spaces. Assume $g$ and $h$ are disjoint. Show that $g \times h$ is uniquely ergodic.
Attempt Let $\mu$ be the unique ergodic measure on $X$ and $\nu$ be the unique ergodic measure on $Y$. Then $g \times h$ is ergodic with respect to the product measure $\mu \times \nu$. I think the way to go is now to show that every other ergodic measure $m$ on $X \times Y$ yields a contradiction to the disjointness property of $g$ and $h$. Sadly I have no idea how to start this argument. Can somebody give a hint? Or am I on the totally wrong path?
Definition Two measure preserving maps $r: (A,\mu) \rightarrow (A,\mu)$ and $s: (B,\nu) \rightarrow (B,\nu)$ between probability spaces are disjoint if the product measure $\mu \times \nu$ is the only joining.

A:

I think it goes like this:
Let $\mu$ be the unique ergodic measure on $X$ and $\nu$ be the unique ergodic measure on $Y$.
Now suppose $\eta$ is a finite $(g \times h)$-invariant Borel measure. The objective is to show there is only one such measure. Let $\pi_X: X \times Y \to X$ and $\pi_X: X \times Y \to Y$ be the projection maps. Try showing the following:

Compute $(\pi_X)_*\eta$ and show it is equal to $\mu$ by unique ergodicity.
Do the same thing for $(\pi_Y)_*\eta$ and $\nu$.
Then $\eta$ is a joining by definition. Now by the disjointness hypothesis and your definition, $\eta$ must be $\mu \times \nu$.

(So you don't need to assume $\mu \times \nu$ is ergodic to start with.)

