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Suppose we are given a measure space $(X,\mathfrak{M},\mu)$, but we know nothing more than this information. (Assume that $\mu$ is a positive, extended real-valued function.) Is there any "nice" way to tell whether or not there exists a topological space $Y$ and a measure $\nu$ on the Borel sets $B(Y)$ of $Y$ such that there is a bimeasurable bijection between $X$ and $Y$ i.e. a measure space isomorphism of $(X,\mathfrak{M},\mu)$ with $(Y,B(Y),\nu)$?

Either this question is ridiculous as asked, or there is probably some kind of set-theoretic business connected to it...sorry if it's the former!

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In this answer: math.stackexchange.com/a/31967/8157 Byron Schmuland spoke of something vaguely similar, namely the possibility of choosing a "universal space" for the Skorokhod representation theorem. Maybe it can be useful, I don't know. –  Giuseppe Negro Dec 14 '12 at 15:50
    
Interesting...it makes me wonder if it is possible to realize Borel spaces somehow as dense in the probability spaces. Kind of a weird idea...(of course this is just a reflex response to your suggestion, though). –  Jon Bannon Dec 14 '12 at 15:52
    
I'm sorry I cannot help you, it's just that your question vaguely reminded me of the Skorokhod representation theorem, but the problem you pose is different. I'm just guessing that it could be a starter, nothing more than that. –  Giuseppe Negro Dec 14 '12 at 15:56
    
You certainly can't have a general isomorphism: assume you have a Dirac measure. It's not going to be isomorphic to a Borel measure. The imbedding above (Skorohod) might work, but I'm too rusty to tell. –  gnometorule Dec 14 '12 at 16:14

1 Answer 1

up vote 4 down vote accepted

Not every $\sigma$-algebra is the Borel $\sigma$-algebra of a topological space, so the answer is no.

But there is a sense in which the answer is almost yes for probability spaces. If we identify measurable sets $A$ and $B$ if $\mu(A\Delta B)=0$, we get a measure agebra $(A,\mu)$ where $A$ contains the equivalence classes and $\mu$ is defined in the obvious way. If the underlying measure space is atomless, we get by a representation theorem of Maharam that $(A,\mu)$ coincides with the measure algebra of a countable convex combination of coin-flipping probability spaces $\{0,1\}^\kappa$.

The result can be generalized to certain measure spaces that are not atomless and might be infinite. Volume 3 of Fremlin's treatise discusses these things in great detail.

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Thanks, Michael. I didn't expect that every sigma algebra is the Borel sigma algebra of a topological space, but wanted to know if there was an easy way to tell if this happens for a given sigma algebra. Your "almost yes" part is precisely the kind of thing I'm wondering about! –  Jon Bannon Dec 14 '12 at 16:25
    
I guess this answer has reduced the question to something like "Does Maharam's representation theorem and its known extensions finish the job?" In light of this, I'm accepting this answer. –  Jon Bannon Dec 14 '12 at 16:29
    
If you are willing to enter the realm of measure algebras then there's no need of invoking Maharam's classification theorem. The much more elementary Stone representation theorem gives you an honest topological measure space whose measure algebra is (canonically isomorphic to) the one you started with. Since Jon is an operator algebraist it might be worth pointing out that the resulting space is the same as the Gelfand spectrum of $L^\infty(\mu)$. In this context variants of this result also go by the name of "Mackey's point realization theorem". –  t.b. Dec 16 '12 at 13:29

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