Non-intersecting splitting algebras in a non-Bernoulli transformation

TL;DR

This paper demonstrates that a measure-preserving transformation can have two non-intersecting splitting invariant sigma algebras without being Bernoulli, answering a question from 1975.

Contribution

It constructs an example of a transformation with two non-intersecting splitting sigma algebras, showing such a transformation need not be Bernoulli.

Findings

Existence of non-Bernoulli transformations with two non-intersecting splitting sigma algebras

Counterexample to a longstanding question by Thouvenot (1975)

Clarification of the relationship between splitting sigma algebras and Bernoulli property

Abstract

Given a measure preserving transformation $T$ on a Lebesgue $\sigma$ algebra, a complete $T$ invariant sub $\sigma$ algebra is said to split if there is another complete $T$ invariant sub $\sigma$ algebra on which $T$ is Bernoulli which is completely independent of the given sub $\sigma$ algebra and such that the two sub $\sigma$ algebras together generate the entire $\sigma$ algebra. It is easily shown that two splitting sub $\sigma$ algebras with nothing in common imply $T$ to be K. Here it is shown that $T$ does not have to be Bernoulli by exhibiting two such non-intersecting $\sigma$ algebras for the $T,T^{-1}$ transformation, negatively answering a question posed by Thouvenot in 1975.