Does ergodicity of product dynamical system imply ergodicity of components?

Question

Let $\mathrm{X}=(X,\mathcal{F},\mu,T)$, $\mathrm{Y}=(Y,\mathcal{G},\nu,S)$ be measure preserving dynamical systems. Assume that the product dynamical system $$\mathrm{X}\times\mathrm{Y}~=~(X\times Y,\mathcal{F}\otimes\mathcal{G},\mu\times\nu,T\times S)$$ is ergodic. Does this imply $X$ and $Y$ are each by itself ergodic?

I know that if $\mathrm{X},\mathrm{Y}$ are ergodic then $\mathrm{X}\times\mathrm{Y}$ is not necessarily ergodic too, but can't find anything about the converse.

Edit: I realized that an easy way to see this is just to look at $T\times S(A\times Y)$ and $T\times S(X\times B)$, $A\in\mathcal{F},B\in\mathcal{G}$, which coincide with $T(A),S(B)$.

Answer 1

Proving the contrapositive, suppose that $\mathrm{X}$ is not ergodic. Let $A \subset X$ be an invariant subset so that $A$ and $X-A$ have nonzero measure. It follows that $A \times Y \subset X \times Y$ is an invariant subset, its measure is equal to that of $A$ which is nonzero, and the measure of $(X \times Y) - (A \times Y) = (X - A) \times Y$ is equal to the measure of $X-A$ which is nonzero (Fubini's Theorem is being applied here). So $\mathrm{X} \times \mathrm{Y}$ is not ergodic.

Answer 2

More generally if $\pi:(\mu,f\curvearrowright X)\to (\nu,g\curvearrowright Y)$ is a factor map, then for any $g$-invariant measurable function $\phi:Y\to \mathbb{R}$ the pullback $\phi\circ \pi:X\to\mathbb{R}$ is $f$-invariant and measurable. If in addition $(\mu,f)$ is ergodic then $\phi\circ \pi$ is constant $\mu$ almost everywhere, whence $\phi$ is constant $\nu$ almost everywhere.