I am given the following information: Let $(X,\mathcal{F},\mu)$ be a probability space and let $T:X \rightarrow X$ be measurable and strongly mixing. Consider the probability space $(Y,\mathcal{G},\nu)$ where
$ Y = (X \times \left \{ 0 \right \}) \cup (X \times \left \{ 1 \right \}) \cup (X \times \left \{ 2 \right \}), $
$\mathcal{G}$ is the $\sigma$-algebra generated by the sets $A \times i$ for $A \in \mathcal{F}$ and $i \in \left \{0,1,2 \right \}$ and $\nu$ is the measure defined by $\nu(A \times i) = \frac{1}{3}\mu(A)$. Define the map $S:Y \rightarrow Y$ by $S(x,0) = (x,1), S(x,1) = (x,2)$ and $S(x,2) = (Tx,0)$.
I have to show that the map $S$ is ergodic. I can show it is measure preserving, but I am not sure how to proceed from this? Can someone give a hint?
Thanks in advance.
Suppose $E$ is an $S$-invariant set: $S^{-1}E = E$. Write $E = E_0\sqcup E_1\sqcup E_2$ where $E_i \in X\times\{i\}$. Then $E_0 = E_1, E_1 = E_2$, and $\pi(E_2) = T^{-1}\pi(E_0)$ (where $\pi$ is the projection map onto $X$). So, $\pi(E_0) = T^{-1}\pi(E_0)$, which means $\pi(E_0)$ has measure $0$ or $1$ since $T$ is ergodic. This means $E_0$ has measure $0$ or $\frac{1}{3}$, which means $E$ has measure $0$ or $1$, as desired.