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Roughly speaking, this question asks whether there is a measure-conjugacy between two transformations if there are measure-conjugacies between their ergodic components.

Suppose $(X,\mu)$ is a standard probability space and $T$ and $S$ are measure-preserving transformations of $(X,\mu)$. By the ergodic decomposition theorem, there are standard probability spaces $(Y,\nu), (Z,\zeta)$ and Borel maps $\phi:Y \to M_1(X), \psi:Z\to M_1(X)$ (where $M_1(X)$ is the space of Borel probability measures on $X$) such that for a.e. $y \in Y$, $\phi(y)$ is ergodic, $T$-invariant and $\int \phi(y)~d\nu(y)=\mu$. Similarly, for a.e. $z\in Z$, $\psi(z)$ is ergodic and $S$-invariant and $\int \psi(z)~d\zeta(z)=\mu$.

Now suppose there is a measure-space isomorphism $\Omega:(Y,\nu) \to (Z,\zeta)$ such that for a.e. $y \in Y$, $(T,X,\phi(y))$ is measurably conjugate to $(S,X,\psi(\Omega(y)))$. Then is $T$ measurably conjugate to $S$?

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This is probably related to the following question (and, most likely, can be obtained from it or from a modification of the argument): let M and N be two finite von Neumann algebras with centers $Z(M)$ and $Z(N)$ and faithful traces $\tau_M$, $\tau_N$. Assume that there is a (trace-preserving) isomorphism $(Z(M),\tau_M)\cong L^\infty(X,\mu)\stackrel{\alpha}{\to} L^\infty(Y,\nu)\cong (Z(N),\tau_N)$ so that the central components $M_x$ and $N_{\alpha(x)}$ are a.e. isomorphic. Does it follow that $M\cong N$?

This was proved in this form by Effros [Trans. Amer. Math. Soc. 121 (1966), 434--454; MR0192360 (33 #585)] with a later (shorter) proof by Elliott [MR0310659 (46 #9757) An extension of some results of Takesaki in the reduction theory of von Neumann algebras. Pacific J. Math. 39 (1971), 145–148.]

The proofs rely on existence of Borel structure on von Neumann algebras and Borel selection theorems (so, I would guess, they should go through in your context as well?).

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Thanks! It seems reasonable that the same ideas should work in my context; I'll look into it. – Lewis Bowen Jul 30 '11 at 18:27

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