Probabilist often work on Polish spaces. Does somebody know an ("non-exotic") example, for which it is not possible to work on a Polish space, but instead one has to work on a general measurable space? By non-exotic example I mean something like a stochastic process, which is really used in applications, and cannot be defined on a Polish space...(I posted this question also here).

  • $\begingroup$ I'm no probabilist, but shouldn't something like $[0,1]^{\mathbf R_+}$ naturally arise in probabilistic contexts? It's completely regular, but certainly not Polish. $\endgroup$ – tomasz Jul 10 '12 at 12:44
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    $\begingroup$ There is a good link on MO, with examples of what can go wrong when the assumption Polish is being dropped...this might help in constructing an example... $\endgroup$ – Andy Teich Jul 10 '12 at 12:48

There re a number of constructions that do not work for Polish spaces, but a certain class of probability spaces, variously known as super-atomless, saturated, nowhere countably generated and a number of other names. A nice overview can be found here.

A probability space $(\Omega,\Sigma,\mu)$ is saturated if for every two Poilsh spaces $X$ and $Y$, every probability measure $\nu$ on $X\times Y$ and every random variable $f:\Omega\to X$ such that its distribution $\mu f^{-1}$ equals the marginal of $\nu$ on $X$, there is a random variable $g:\Omega\to Y$ such that the joint distribution of $(f,g)$ is $\nu$.

The following definition is conceptually different, but can be shown to be equivalent:

A probability space $(\Omega,\Sigma,\mu)$ is super-atomless if there is no $A\in\Sigma$ satisfying $\mu(A)>0$, such that the pseudo-metric space obtained by endowing the trace $\sigma$-algebra on $A$ with the pseudo-metric $d(A,B)=\mu(A\triangle B)$ is separable.

  • $\begingroup$ Could you explain a bit more? The definition of "saturated probability space" in their article makes no sense, likely because of faulty notations... $\endgroup$ – D. Thomine Jul 10 '12 at 16:46
  • $\begingroup$ @D.Thomine: I hope it is clearer now. $\endgroup$ – Michael Greinecker Jul 10 '12 at 17:16
  • $\begingroup$ Yes, thank you. Now I still have to understand this theory, but that's another problem :) $\endgroup$ – D. Thomine Jul 10 '12 at 17:22
  • $\begingroup$ @MichaelGreinecker: The paper you are referring to seems to give only examples of set-valued functions...are there also examples for single-valued functions or does that already follow from this paper? $\endgroup$ – Andy Teich Jul 11 '12 at 9:04
  • $\begingroup$ @Andy: Section 4 is about the existence of Nash equilibria, which are certain functions, even though the proofs make use of set-valued functions. Here is a paper that uses these kind of spaces to construct extended product spaces which are used in the modeling of ideosyncratic risk in large populations in economics. $\endgroup$ – Michael Greinecker Jul 11 '12 at 9:11

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