Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In probability theory, there are many results which are valid in purely measurable settings, usually beginning with the assumption, "let $(\Omega, \mathcal F, \mathbb P)$ be an abstract probability space." On the other hand, many results require a more "locally measurable" structure, often taking the form of a Borel $\sigma$-algebra on a topological space. In an effort to compromise between the two concepts, I am wondering if there is a good notion of a measurable fibration.

Let $E$ be a measurable space, let $B$ be a topological space, and let $\pi : E \to B$ be a Borel-measurable function. I would like to define a "measurable homotopy lifting property" with respect to an abstract probability space $(\Omega, \mathcal F, \mathbb P)$.

Let $I = [0,1]$ be the unit interval, equipped with its Borel $\sigma$-algebra $\mathcal B(I)$ and Lebesgue measure $\lambda$. A "measurable homotopy" should be a measurable map $f : \Omega \times I \to B$, where the product $\Omega \times I$ is equipped with the tensor product $\sigma$-algebra $\mathcal F \otimes \mathcal B(I)$. Measures push forward, so this naturally endows $B$ with a probability measure $P_f = f_*(\mathbb P \otimes \lambda)$.

Normally, one next considers a lift $\tilde f_0 : \Omega \to E$ of the map $f_0 = f|_{\Omega \times \{0\}}$. This makes sense in topology, because by lifting at a point one can "tug" the rest of the homotopy up to $E$. However, this doesn't make sense in this context: the set $\Omega \times \{0\}$ has measure zero, hence is meaningless from the point of view of measure theory. Hence:

Question: is there a generalization of the homotopy lifting property to this measurable setting?

Even thought $0 \in I$ has no special meaning probabilistically, the concept of a random number $\iota \in I$ with distribution $\lambda$ does make sense. In fact, the product measure $\mathbb P \otimes \lambda$ represents choosing a random $\omega \in \Omega$ and random $\iota \in I$ independently from one another. Consequently, a random point $(\omega, \iota)$ picks out a particular function $f_{\iota} := f|_{\Omega \times \{\iota\}}$. Diagrammatically this results in a mess of arrows, so I'll stop the speculation and leave the question to the community.

share|improve this question
    
You could consider $[0,\epsilon] \hookrightarrow [0,1]$ instead of $\lbrace0\rbrace \hookrightarrow [0,1]$. Homotopically it's the same, and possibly more what you want: the map from the small interval seems to be a better choice of 'initial data' for the lift. –  David Roberts Jan 30 '13 at 23:59
    
Or perhaps it's better to think about what you want trivial cofibrations to be first, and then define fibrations to be those with the right lifting property against those... –  David Roberts Jan 31 '13 at 0:00
2  
I'm ill-equipped to understand the topological part of the picture, but in Dan Rudolph's book, Fundamentals of Measurable Dynamics, he spends quite a lot of time doing a fibre construction in Lebesgue spaces. –  Anthony Quas Jan 31 '13 at 0:09
add comment

2 Answers

This seems to be similar to the concept of measurable partition. One related result is the Rokhlin theorem, see http://w3.impa.br/~viana/out/rokhlin.pdf This notion is extensively used in ergodic theory, see books by Sinai (and coauthors).

share|improve this answer
add comment

It seems you are looking for something similar to the notion of "measurable foliation" (lamination) - it is defined by requiring the transverse structure to be just measurable (rather than smooth or continuous which is the case for foliations or laminations, respectively) and endowed with a holonomy (quasi-)invariant measure. Have a look, for instance at the book "Global analysis on foliated spaces" by Moore and Schochet.

share|improve this answer
1  
@R W thanks! I just bought a copy of Moore & Schochet. –  Tom LaGatta Jan 31 '13 at 0:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.