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If for each $t\in I=[0,1]$ I have a measurable space $(X_t,\Sigma_t)$, is there a standard notion which will give a measurable space deserving to be called the integral $\int_I X_t\,\mathrm d t$?

Motivated by this question and curiosity...

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Of course can can replace $I$ and its Lebesgue measure by any other measured space in this. My guess would be that if we take a set $I$ with the $\sigma$-algebra generated by singletons and the counting measure, then $\int_IX_t\,\mathrm d t$ should be the product $\prod_{t\in I}X_i$. –  Mariano Suárez-Alvarez Feb 14 '11 at 4:02
    
I've certainly never seen it in this form & it's getting really late here... First, I think you should allow further degrees of freedom: a measure space structure on $I$ and (at least) a measure class $[\mu_{t}]$ on each $X_{t}$. This should allow under some conditions to produce something deserving the name by playing around with (abelian) von Neumann algebras and direct integrals. I can try to elaborate on that tomorrow (if you're interested) at the moment I suggest to have a look at the wikipedia page en.wikipedia.org/wiki/… –  t.b. Feb 14 '11 at 4:04
    
Sorry, I thought of $I$ as an index set, not the unit interval with Lebesgue measure. –  t.b. Feb 14 '11 at 4:06
    
@Theo, I don't want the $X_t$ to have measures or, rather, if you are going to end up putting a measure on the "integral" then, before that, you are going to need to put a $\sigma$-algebra on it. I just want to know the latter :) In particular, direct integrals of v.N. algebras sort of sidestep this. –  Mariano Suárez-Alvarez Feb 14 '11 at 4:14
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I'm pretty sure there is no sensible notion for such an integral. But it is certainly not a product. Everything that deserves to be called an integral over a probability space integrates constant functions to their value. The product doesn't do this. –  Michael Greinecker Dec 29 '11 at 20:34
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1 Answer 1

An integral is a generalization of a weighted sum, but neither adding measurable spaces nor multiplying them with numbers are meaningful operations. Neither can we build integrals of topological spaces, filters, uniformities, etc.

Since the question is inspired by products of measurable spaces, what one can do is forming direct sums. The booklet Borel spaces by Rao and Rao contains two approaches. Let $(X_\lambda,\mathcal{X}_\lambda)_{\lambda\in\Lambda}$ be a family of measurable spaces. We can assume the underlying spaces to be disjoint and $X=\bigcup_\lambda X_\lambda$. The measurable sets of the direct sum are of the form $\bigcup_\lambda B_\lambda$ for some $(B_\lambda)\in\prod_\lambda \mathcal{X}_\lambda$. The weak direct sum has as its underlying $\sigma$-algebra the family $\sigma\big(\bigcup_\lambda\mathcal{X}_\lambda\big)$. The direct sum and the weak direct sum coincide if and only if $\Lambda$ is countable.

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