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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 $\prod_{t\in I}X_i$.2011-02-14
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    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 http://en.wikipedia.org/wiki/Direct_integral#Direct_integrals_of_von_Neumann_algebras2011-02-14
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    Sorry, I thought of $I$ as an index set, not the unit interval with Lebesgue measure.2011-02-14
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    @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.2011-02-14
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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.2011-12-29

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