# Is an non-singualr invertable ergodic transformation on a measure space isomorphic to its inverse?

A non-singular, invertable, ergodic transformation is the quadriple $(X,\mathcal B, \mu, T)$ where $(X,\mathcal B, \mu)$ is a measure space and $T$ is an invertable, measurable automorphism where $\mu$ and $\mu\circ T$ equivalent measures.

Two such systems $(X,\mathcal B, \mu, T)$ and $(Y,\mathcal C, \nu, S)$ are isomorphic when there exists isomorphism $\phi: X \mapsto Y$ where $$S\phi x = \phi T x$$

It seems it should be obvious that $(X,\mathcal B, \mu, T)$ and $(X,\mathcal B, \mu, T^{-1})$ are isomorphic. If I assume $X$ is a product space and $T$ is the $+1$ odometer action then I can prove it myself. But I would rather quote someone and be done with it.

Do you know of a source to quote for this?

• It isn't true in general, even when the measure is invariant; however, I don't know a reference, except that the construction is bound to be difficult. There even exist AT (approximately transitive) examples (again with invariant measures)—I found one myself, but it is too complicated to give here. May 3 '14 at 13:33
• Why is it obvious that $T$ and $T^{-1}$ should be isomorphic? May 6 '14 at 15:07

Page 108 of Peter Walters' classic textbook An Introduction to Ergodic Theory suggests the reference "An uncountable family of $K$-automorphisms" by Donald Ornstein and Paul Shields [Advances in Mathematics 10 63--88 (1973)] for the fact that there exists an invertible measure-preserving tranformation with the Kolmogorov property which is not isomorphic to its inverse.