Hyperfinite equivalence relations are treeable. For the case of uncountable relations, I was wondering if there is a reference to (or simple proof of) the following statement: Let $E$ be a (possibly uncountable) amenable equivalence relation on a standard probability space $(X,\mu)$. Then $E$ is treeable. Or perhaps is this statement incorrect?
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$\begingroup$ To begin with, how does one make treeable an amenable foliation? $\endgroup$– R WCommented Jun 15, 2015 at 19:18
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$\begingroup$ Thanks, R W! I'm not sure I understand your question, or how it relates to mine. Could you elaborate? To clarify, by amenable equivalence relation I mean in the sense of Connes-Feldman-Weiss (e.g., Borel reducible to a hyperfinite equivalence relation, or an orbit equivalence relation of an amenable group). $\endgroup$– VladimirCommented Jun 15, 2015 at 20:58
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$\begingroup$ Greg Hjorth's paper "Non-treeability for product group actions" is a good place to read the definition for a general (uncountable) treeable equivalence relation. Essentially it is the same definition as in the countable case. $\endgroup$– VladimirCommented Jun 16, 2015 at 0:26
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