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The following question about disintegration of measures on product spaces has been asked on this website:

Given two measurable spaces $X_1$ and $X_2$, and a measure on the product measurable space $X_1\times X_2$, what are some necessary and/or sufficient conditions for the measure on $X_1\times X_2$ to be the composition of some measure on $X_1$ and some transition measure from $X_1$ to $X_2$? (See the question here.)

The answer given to that question mentions an example by Andersen and Jessen that shows that one cannot hope for such a disintegration to hold in general.

I would like to ask whether, if $X_1$ and $X_2$ are both standard probability spaces (also known as Lebesgue-Rokhlin spaces), with measures $m_1,m_2$ respectively, and the probability measure $m$ on $X_1\times X_2$ satisfies $m\circ \pi_i^{-1}=m_i$ for $i=1,2$ (where $\pi_i$ is projection to the $i$-th coordinate), then the above disintegration result holds, more precisely we can disintegrate $m$ relative to $m_2$, where almost every measure $m_{x}$ in the disintegration is a probability measure on $X_1$.

(Perhaps the Andersen-Jessen example rules this out too, but I could not check this because the link provided in the answer mentioned above does not work anymore, apparently.)

Peter
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  • How do you define standard probability spaces? The usual definition I'm familiar with (measurably isomorphic to a Polish space with the Borel $\sigma$-algebra) differs from the spaces Rohlin was working with. In the former case, disintegrations certainly exist, and I think the latter case is treated somewhere in Rohlin's paper. – Michael Greinecker Dec 15 '17 at 05:51
  • I use any of the equivalent definitions of standard probability space, namely Definition 9.4.6 in Bogachev's Measure Theory´ Vol. 2 (where they are called Lebesgue-Rohlin spaces) or thetopological' definition in De La Rue's paper Espaces de Lebesgue´, or, equivalently, Definition 6.8 in the book `Operator Theoretic Aspects of Ergodic Theory´, which is close to the definition you are mentioning. Surely all these definitions are equivalent, right? Can you point me to a precise place in the literature where the disintegration result I ask for in my question appears? Thank you for any help. – Peter Dec 15 '17 at 11:44
  • These definitions are not all equivalent; Lebesgue measure on $[0,1]$ is Lebesgue-Rohlin, but there are more Lebesgue measurable sets ($2^\mathfrak{c}$) than Borel sets ($\mathfrak{c}$). For the measurably-isomorphic-to-Polish space definition, you can combine Theorem 10.4.5. in Volume 2 of Bogachev's book together with the inner regularity of probability measures on Polish spaces. In that case, the compact sets form the desired compact class. For more general Lebesgue-Rohlin spaces, I have no answer. – Michael Greinecker Dec 15 '17 at 15:09
  • Thank you very much for the useful reference. Something remains unclear to me though: for the three definitions of standard probability space I mentioned earlier (Bogachev's, De la Rue's, and the one in the book by Eisner&al), I don't see how they fail to be equivalent. Both Bogachev and De la Rue prove that a standard probability space in their definition is mod 0 isomorphic to an interval with Lebesgue measure and some atoms, and the third definition also involves isomorphism mod 0 with a standard Borel space with probability measure... (tbc) – Peter Dec 15 '17 at 16:57
  • ... So it seems to me that these three definitions are indeed equivalent, but that they are not equivalent to the definition with measurable isomorphism that you mention, because measurable isomorphism is much stronger than mod 0 isomorphism (if I understand correctly, measurable isomorphism is a bi-measurable measure-preserving bijection, whereas in a mod 0 isomophism you allow for null sets to be removed from each of the two measure spaces). Do you agree? – Peter Dec 15 '17 at 17:00
  • I'm not sure what you mean. Lebesgue-Rohlin spaces are complete probability spaces and therefore must contain many more measurable sets than the Borel sets (unless they are countable). – Michael Greinecker Dec 16 '17 at 01:15
  • Apologies for my lack of clarity. Please let me ask my main question in my last comments again: are you saying that the three definitions of standard probability space that I mentioned (Definition 9.4.6 in Bogachev (called Lebesgue-Rohlin space), Definition 1-1 in De la Rue's paper (called Lebesgue space), and Definition 6.8 in the book by Eisner et al (called standard probability space)) are not equivalent? If you are, which definition differs from which? – Peter Dec 16 '17 at 13:02
  • There are as many Borel sets in an uncountable Polish space as there are real numbers. If you take the completion of a Borel measure on an uncountable Polish space, you get as many measurable sets as there are subsets of real numbers. – Michael Greinecker Dec 16 '17 at 14:18
  • So there are a lot of measure zero sets in the completion that are not Borel sets. Taking away such a set and taking the trace $\sigma$-algebra and restricted measure gives you a space isomorphic mod 0 to the space you started with and the resulting space is Lebesgue-Rohlin according to the definition in Bogachev. The canonical injection identifies it with a non-Borel subset of the original space. It follows from Bogachev Theorem 6.8.6. that the smaller space cannot be standard Borel in the sense of Eisner et al. There are more Lebesgue-Rohlin spaces than standard probability spaces. – Michael Greinecker Dec 16 '17 at 14:29
  • Thank you for your reply, but either I am missing something or there is a contradiction between what you wrote and the literature. In the book by Eisner et al, Remark 7.22 on page 120 ends with the following sentence:

    "The theorem shows that standard probability spaces in our definition are the same as Rokhlin’s Lebesgue spaces."

     – Peter Dec 16 '17 at 16:56
  • Maybe, you are using a different edition of Eisner et al, I only see it on page 129. Anyways, one might read the remark as that these spaces are equivalent under the relevant notion of equivalence, isomorphism mod 0. The problem I see, related to your initial problem, is that it is not clear how Lebesgu-Rohlin spaces relates to taking products. – Michael Greinecker Dec 16 '17 at 18:14
  • The product of two measurable spaces isomorphic (related by bimeasurable bijections) to Polish spaces is again Polish, so the product space in your question is a standard probability space. I don't know whether a Probability space on a product set is Lebesgu-Rohlin if the marginals are. – Michael Greinecker Dec 16 '17 at 18:17
  • There is also a sceptical remark (Remark 40) in "An Outline of Ergodic Theory" by Kalikow and McCutcheon: "A classic reference in the theory of Lebesgue spaces, axiomatically defined, is Rohlin (1952), though this and most of the literature on axiomatic treatments is fraught with vagueness and ambiguity (if not confusion), due in part to a cavalier attitude toward sets of measure zero." – Michael Greinecker Dec 16 '17 at 18:21

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