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Suppose we have a measure space $(X,a)$, a measurable space $Y$ and for every $x\in X$ we have a measure $b_x$ on $Y$. Suppose that $(Z,c)$ is a measure space such that as a measurable space $Z=X\times Y$ and the measure $c$ is the integral of measures $b_x$ with respect to the measure $a$.

I have a vague intuition that $(Z,c)$ is a coproduct of the family $(Y,b_x)$ along $(X, a)$

Question: Is there a category of measure spaces and a categorical construction in it which captures this intuition?

One could ask a similar question about a category of Hilbert spaces and integrals of Hilbert spaces over measure spaces, and preferably the "categorical construction" in question should also answer this problem.

The categorical construction in question should preferably be similar to coproduct, and it would be very nice if it actually was a coproduct in some category.

On the other hand, maybe my vague intuition is wrong, and comments on that would also be appreciated.

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It looks like the measure on $Z$ is specified in a very asymmetric way, so I can't see how it could be a coproduct. –  S. Carnahan Jan 8 '11 at 18:44
    
@Scott: if (X,a) is a finite set with the counting measure then (Z,c) is a coproduct of (Y,b_x) in the category of measure spaces with measure preserving maps. If the measure is other than counting - more "assymetric", as you could maybe put it - then the answer should be more complicated and involve some additional structure, but imho there should be a satisfactory answer. –  Łukasz Grabowski Jan 8 '11 at 20:43
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Dependent sum? See ncatlab.org/nlab/show/dependent+product –  David Roberts Jan 8 '11 at 23:44
    
I've heard words along the lines of "direct integral" --- something that generalizes direct sums. I think the context is slightly different, but maybe it's close enough that you can adapt the definition? –  Theo Johnson-Freyd Jan 9 '11 at 5:02
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1 Answer

Yes, there is such a category, namely the category of measurable spaces and morphisms of measurable spaces equipped with fiberwise measures (also known as operator valued weights). See this answer for an introduction and this answer for a list of further references.

Intuitively, a morphism f: X→Y in this category is a morphism of measurable spaces together with a measure on each fiber of f that varies measurably with respect to Y. In particular, if Y is the point, then a morphism f: X→pt is simply a measurable space equipped with a measure, i.e., a measured space.

As David Roberts has already pointed out, the categorical construction that captures the intuition of the question is called the dependent sum.

In the notation of the nLab artcle we have B=Z, A=X, I=pt, the morphism f: A→I is the measure a, the morphism g: B→A is the projection map Z=X×Y→X equipped with the family of measures b_x.

The dependent sum of g indexed by f exists and is the composition fg, which in our case is the measure c.

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Thanks. I have to think it through. –  Łukasz Grabowski Jan 15 '11 at 20:08
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