Is it true that any sub-$\sigma$-algebra of a Rokhlin-Lebesgue space is induced (up to completion) by a measurable map into another Rokhlin-Lebesgue space? In other words, is it true that conditional expectation with respect to any sub-$\sigma$-algebra is the same as conditional expectation with respect to some random variable (valued in a R.-L. space)?
1 Answer
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Yes - this is one of the key results of the Rokhlin theory. Namely, any complete sub-$\sigma$-algebra of a Lebesgue space can be realized as the preimage $\sigma$-algebra of a quotient map.
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$\begingroup$ What can be said about the target space of this quotient map? One could always take an identity map and the sub-algebra as the $\sigma$-algebra on the target. Could you please give a reference to the precise statement. $\endgroup$ Jun 24, 2015 at 10:08
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$\begingroup$ What do you mean? In the same way as there exists only one purely non-atomic Lebesgue space isomorphic to the unit interval, there exists only one quotient map with purely non-atomic conditional measures, which is the projection of the unit square endowed with the Lebesgue measure onto to the unit interval. $\endgroup$– R WJun 24, 2015 at 10:14
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$\begingroup$ It was not clear from your original post, what you mean by "quotient map". So the target is also a Lebesgue space. Thanks. How about a reference? $\endgroup$ Jun 24, 2015 at 10:20
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$\begingroup$ What is the reason then for using "sub-$\sigma$-algebra conditioned expectation" language instead of conditioning wrt a partition or a random variable? $\endgroup$ Jun 24, 2015 at 10:21
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2$\begingroup$ Unfortunately, there is no accessible presentation of Rokhlin's theory. His original paper is quite difficult to read, and the language is definitely out of date nowadays. However, there is a short synopsis as an appendix to Kornfeld-Fomin-Sinai's "Ergodic theory". On the other hand, if one wants to define just the conditional expectations (rather than conditional measures), this can be done in a pretty straightforward way. This fact (along with other "cultural differences") is the reason why Rokhlin's theory is virtually unknown to probabilists. $\endgroup$– R WJun 24, 2015 at 10:32