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I have a reference request concerning Proposition 1.6 in the following article

The setting: Let $G$ be a locally compact, second countable group. Let $S = (S, \mu)$ be a Polish space. Assume we have a Borel measurable action $G \times S \rightarrow S$. Assume that $\mu$ is quasi invariant.

The statement: There is a standard measure space $E = (E, \nu)$ with G invariant measurable map $\phi :S \rightarrow E$ such that $\phi_*(\mu)=\nu$ and $\mu = \int^\oplus \mu_y d \nu(y)$, where $\mu_y$ is supported on $\phi^{-1}(y)$, $\mu_y$ quasi invariant and ergodic for almost all $y$.

I have tried to recover this result via Choquet theory, but I am not sure what topology to put on the quasi invariant measures, since measure classes are not closed in the $*$ topology. What is the right topology on quasi invariant Radon measures, such that they form a locally compact convex subset of a topological vectorspace?

Additional question: If the action is topological, say $E$ is a Polish space, and smooth in the sense that $G \backslash X$ is $T_0$ or equivalently almost Hausdorff, how can we relate $E$ and $G \backslash X$? Since $E$ is Hausdorff and $G \backslash X$ only almost Hausdorff, I am not sure how to relate the topologies.

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up vote 5 down vote accepted

This is a bit too long for a comment, hence I post it as an answer.

I honestly don't know where you can find a group theoretic version of ergodic decomposition proved via Choquet theory (and I'm not convinced that it exists in the setting you're interested in).

However, the exact result you quote from Zimmer is proved as Theorem 1.1 in the carefully written paper

G. Greschonig, K. Schmidt, Ergodic Decomposition of quasi-invariant probability measures, Colloq. Math. 84/85 (2000), part 2, 495–514, MR1784210.

You'll find many interesting references in there.

For a lot of extremely helpful results that are used in and around Zimmer's work, I recommend Section 2 of

David Fisher, Dave Witte Morris, and Kevin Whyte, Nonergodic actions, cocycles and superrigidity, New York Journal of Mathematics, Volume 10 (2004) 249–269, MR2114789.

I can't say anything on your final question.

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Sorry, I was confusing the above reference with something else dealing with countable groups only. It perfectly answers the first (main) question in the generality, I was asking for. The statement for Radon measures can be reduced to that of probability measures, since $X$ as a Polish space is paracompact. Thanks a lot. I deleted my previous comment and accepted your answer. – Marc Palm Aug 25 '11 at 7:27
But it seems that the theorem was only folklore for about twenty years, since the survey of Zimmer is from the eighties and your article from 2000. – Marc Palm Aug 25 '11 at 7:29
Thanks for clarifying. Yes, as I stated on there are many things in this topic that are folklore and known to certain groups of people but tremendously difficult to locate in the literature. There are results of Mackey and Ramsay (70's) that give the theorem more or less directly --- after a certain amount of translation between different settings. I was going through similar pains when I was learning about the topic (I still am). Fisher-Witte Morris-Whyte did a tremendous job in cleaning up many things in the second article I'm linking to. – Theo Buehler Aug 25 '11 at 7:46
Thanks for the encouraging words. So I am not the only one troubling with the different languages, measurabiliy issues and such. But actually the first article is exactly using Choquet theory. I still fail to show uniqueness of the measure, but that's probably an issue of my ignorance towards Choquet theory so far. I sldo had a look at your PhD thesis... nice appendix;) – Marc Palm Aug 25 '11 at 22:24
Thanks for that info. That appendix was probably the most painful result of the entire thesis... (and all that only to get rid of a hypothesis that is always satisfied in the applications). Yes, you're right Choquet theory enters (always in some form) but the reduction to the countable case and equivalence relations is what I perceive as the hard step and I don't see a way of circumventing that reduction and argue directly from Choquet. That's what I meant by saying I doubt that it exists but maybe I overstated my case... For Choquet theory Phelps's book is unsurpassed. – Theo Buehler Aug 26 '11 at 1:53

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