# Integral of measurable spaces

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...

• 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 $\prod_{t\in I}X_i$. 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. Feb 14 '11 at 4:14
• 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. Dec 29 '11 at 20:34

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.