Let $\Omega$ be a Polish space equipped with a Borel probability measure $\mathbb{P}$, and let $\theta \colon \Omega \to \Omega$ be a measurably invertible map that is mixing with respect to $\mathbb{P}$. Let $X$ be a Polish space. Let $\varphi \colon \Omega \times X \to X$ be a measurable map, define $\Theta \colon \Omega \times X \to \Omega \times X$ by $$ \Theta(\omega,x) \ = \ (\theta\omega,\varphi(\omega,x)), $$ and define $\Theta_2 \colon \Omega \times X^2 \to \Omega \times X^2$ by $$ \Theta_2(\omega,x,y) \ = \ (\theta\omega,\varphi(\omega,x),\varphi(\omega,y)). $$ Let $\mu$ be a $\Theta$-invariant probability measure that can be written as $$ \mu(A) \ = \ \int_\Omega \mu_\omega(A_\omega) \, \mathbb{P}(d\omega) \hspace{5mm} \forall \, A \in \mathcal{B}(\Omega \times X) $$ [where $A_\omega$ is the $\omega$-section of $A$], and define the probability measure $\mu_2$ on $\Omega \times X^2$ by $$ \mu_2(A) \ = \ \int_\Omega \mu_\omega \!\otimes\! \mu_\omega(A_\omega) \, \mathbb{P}(d\omega) \hspace{5mm} \forall \, A \in \mathcal{B}(\Omega \times X^2).$$ (Note that since $\theta$ is measurably invertible, $\mu_2$ is invariant under $\Theta_2$; this is because invariance of measures on a product space with invertible base is equivalent to equivariance of the disintegration.)
If $\mu_2$ is ergodic with respect to $\Theta_2$, does it follow that the probability measure $\mu \otimes \mu$ on $\Omega \times X \times \Omega \times X$ is ergodic with respect to the product map $\Theta \!\times\! \Theta$?
If not, are there natural additional conditions under which it does follow (e.g. $\varphi(\omega,\cdot)$ is continuous for all $\omega$)?
My intuition is that the answer is yes, but I've found it difficult to prove.
