Does product measure being ergodic imply component ergodic?

Question

Suppose $(X,B,\mu,T)$ is measure preserving dynamical system.

Consider a singleton set $\{0\}$ with counting measure, and suppose $S:X\times \{0\}\to X\times \{0\}$ defined by $S(x,0)=(Tx,0)$ is ergodic wrt the product measure.

Then can we say $T$ is ergodic?

I think the answer should be yes, but there seem to be some gaps: if $f:X\to \mathbb{R}$ is measurable wrt $\mu$ and $f\circ T= f$, then I suppose the idea would be to consider $g:X\times \{0\}\to \mathbb{R}$ defined by $g(x,0)=f(x)$. This clearly satisfies $g\circ S= g$, and so $g$ is constant almost everywhere with respect to the product measure.

However, I am unsure on how to pass to the original measure (I am not even sure if my $g$ as defined is measurable. The counting measure is throwing me off!). So I suppose those are the two questions:

1) Is my $g$ measurable? 2) How can I "clinch" the proof?

Thanks in advance!

Answer 1

$g$ is surely measurable. If $(\mu \times \nu )(A)=0%=$ the there exists $y$ such that the 'section' $A^{y}=\{x: (x,y) \in A\}$ satisfies $\mu (A^{y})=0$ (by Fubini's Theorem). This proves that $f$ is a constant a.e. $[\mu]$.