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In a previous discussion (von neumann algebras and measurable spaces), the connexion between von Neumann algebras and localized measured spaces was clarified. I would like to have a category theory formulation of this correspondance. The morphisms associated to von Neumann algebras are simply $*$-homomorphisms. What are the morphisms associated with (localized) measured spaces?

I guess it is some set of equivalence classes of measurable functions, but I can't find any rigorous formulation for this.

Thanks for your help!

[EDIT:] I don't think this question is a duplicate since it is focused on the morphisms associated to measured spaces (those provided with measurable sets and sets of measure 0), not hyperstonian morphisms. I found an explicit answer here (Is there an introduction to probability theory from a structuralist/categorical perspective?): morphisms correspond to maps $(X,M,N)\rightarrow (Y,P,Q)$ such that the preimage of every element of $P$ is a union of an element of $M$ and a subset of an element of $N$ and the preimage of every element of $Q$ is a subset of an element of $N$

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Issam Ibnouhsein
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  • If I understand the question correctly, then the paper "Two closed categories of filters" by Andreas Blass http://matwbn.icm.edu.pl/ksiazki/fm/fm94/fm94115.pdf discusses the category of filters which is a special case of the category which you want. – Joseph Van Name Sep 06 '14 at 19:37
  • Also, in my new paper "Representations of algebras in varieties generated by infinite primal algebras", I consider a generalization of the category of filters mentioned in Andreas Blass' paper. This category can be used to characterize up-to-equivalence the pro-completion of the category of sets and more general categories. – Joseph Van Name Sep 06 '14 at 19:43
  • Of course, I do not believe that we obtain an equivalence of categories between the categories mentioned above unless one restricts the category of measured spaces to an appropriate full subcategory. – Joseph Van Name Sep 06 '14 at 19:49
  • Thanks for the comments. In my understanding, there is an equivalence with localized measured spaces. However, I only need to make a rigorous formulation of such a correspondance in a math section of a PhD in physics. Therefore a full understanding of the papers you pointed out goes way beyond my needs. Do you have an idea about what the explicit morphisms associated to localized measured spaces are? – Issam Ibnouhsein Sep 06 '14 at 19:52
  • The morphisms between measured spaces $(X,\mathcal{M},\mathcal{I}),(Y,\mathcal{N},\mathcal{J})$ where $(X,\mathcal{M})$ is a $\sigma$-algebra and $\mathcal{I}$ is a $\sigma$-complete ideal should simply be the functions $f:X\rightarrow Y$ where $f^{-1}[R]\in\mathcal{M}$ for each $R\in\mathcal{N}$ and where $f^{-1}[R]\in\mathcal{I}$ whenever $R\in\mathcal{I}$. – Joseph Van Name Sep 06 '14 at 20:04
  • One should obtain an equivalence of categories if we restrict this category to just the $\sigma$-algebras of the form $(2^{X},\mathcal{M})$ where $2^{X}$ is the Cantor cube and $\mathcal{M}$ is the Baire $\sigma$-algebra on $2^{X}$. This should follow from the fact that $\mathcal{M}$ is freely generated as a $\sigma$-complete Boolean algebra by all half cubes ${f\in 2^{X}|f(x)=1}$ where $x$ ranges over all elements in $x$. – Joseph Van Name Sep 06 '14 at 20:06
  • Thanks a lot! Concerning your second comment, I thought the proof p.46 in Sakai's book was enough to proclaim category equivalence. Why do you feel there is a need for restriction on the $\sigma$-algebra type? – Issam Ibnouhsein Sep 06 '14 at 21:58
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    This question was asked before by me, you can find a statement there: http://mathoverflow.net/questions/23408/reference-for-the-gelfand-neumark-theorem-for-commutative-von-neumann-algebras. See also the link there, which explains morphisms of measurable spaces in more detail. – Dmitri Pavlov Sep 07 '14 at 11:35
  • Thanks for the comment Dmitri. However I don't see my question as a duplicate since it is focused on the morphisms associated to measured spaces (those provided with measurable sets and sets of measure 0), not hyperstonian morphisms. – Issam Ibnouhsein Sep 07 '14 at 14:53
  • I found an explicit answer to my question here (http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical): morphisms correspond to maps $(X,M,N)\rightarrow (Y,P,Q)$ such that the preimage of every element of $P$ is a union of an element of $M$ and a subset of an element of $N$ and the preimage of every element of $Q$ is a subset of an element of $N$. – Issam Ibnouhsein Sep 07 '14 at 15:11

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