# Measurable isomorphism between two non-totally ergodic systems

Suppose $(X,\mathcal A,\mu,T)$ is a finite measure-preserving system. Then we define a new measure system $(X^{(K)},\mathcal A^{(K)},\mu^{(K)},T^{(K)})$ defined by $X^{(K)}=X\times \{1,2,...,K\}$ for any positive integer $K$, $\mu^{(K)}(A\times \{k\})=\dfrac{\mu(A)}{K}$ for $1\leq k\leq K$ and $A\in\mathcal A$, and define $T^{(K)}:X^{(K)}\to X^{(K)}$ by $T^{(K)}(x,n)=(x,n+1)$ if $n<K$ and $(Tx,0)$ if $n=K$.

Then one can check that even if $T$ is ergodic, $T^{(K)}$ is not totally ergodic.

Now suppose we have a finite measure preserving system $(Y,\mathcal C,\nu,S)$ which is not totally ergodic.

Question: Find if possible a measure system $(X,\mathcal A,\mu,T)$ such that $S$ is measurably isomorphic to the system $(X^{(K)},\mathcal A^{(K)},\mu^{(K)},T^{(K)})$ for some positive integer $K$.

I get it that my $K$ will be the least positive integer $n$ for which $S^n$ is NOT ergodic. But I cannot define my $X$. Is it true that we will always get such an $X$?

Assume $(Y,\nu,S)$ is ergodic (but not totally ergodic), let $T = S^K$, where $K$ is the least positive integer $n$ for which $S^n$ is not ergodic, and let $Z$ be an ergodic component of $(Y,\nu,T)$. Since $S$ is ergodic, the ergodic decomposition of $(Y,\nu,T)$ is $Y = Z \cup S Z \cup \cdots \cup S^{K-1} Z$. So it is easy to see that $(Y,\nu,S)$ is isomorphic to $(Z^{(K)}, \mu^{(K)}, T^{(K)})$.
• It's worth adding that this is not unique. There may be larger $n$ values for which the decomposition becomes finer. – Anthony Quas Jun 30 '16 at 19:48