If $T\times T$ is ergodic, then so is $T\times T\times T$

81 Views Asked by At

Given a measure preserving system $(X, \mathcal{F}, \mu, T)$, show that $T \times T$ is ergodic with respect to $\mu\times \mu$ if and only if $T \times T \times T$ is ergodic with respect to $\mu\times \mu\times \mu$.

I have already shown that the reverse implication holds. For "$\Rightarrow$", maybe it suffices to show that for all $A\times B\times C\in \mathcal{F}\times \mathcal{F}\times \mathcal{F}$ there is a set of full measure, $M$ say, such that all $(a,b,c)\in M$ have an $n\in\mathbb{N}: (T^{3})^n(a,b,c)\in A\times B\times C$. It surely suffices to show that a.e. all $(a,b,c)$ will enter arbitary sets in $\sigma(\mathcal{F}^3)$, but I am not sure whether we can do it for only a generating $\pi$-system.

Since $T\times T$ is ergodic, then there is $n: (T^2)^n\in A\times B$ and there is $m: (T^2)^m \in B\times C$. But then we can't conclude that $T^3(a,b,c)$ eventually enters $A\times B\times C$?

Is there anyone who has a further idea? Thanks in advance.