Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(\Omega,\Sigma)$ be a measurable space and $K$ be a compact metrizable space endowed with its Borel $\sigma$-algebra $\mathcal{B}(K)$. Let $A\subseteq\Omega\times K$ be universally measurable and such that $$C_\omega=\{x\in K:(\omega,x)\in A\}$$ is closed for all $\omega\in\Omega$. Let $\Sigma_u$ the universal completion of $\Sigma$. Is it then the case that $A$ is in $\Sigma_u\otimes \mathcal{B}(K)$?

I face the problem when I want to obtain a certain set-valued function that satisfies a strong measurability condition. It is enough for my purposes to get a closed-valued set-valued function with a graph measurable with respect to the completion on my underlying probability space. I can obtain the desired set-valued function as a projection, but this only gives me universal measurability of the graph.

As a first step, it might be intersting to know whether the conjecture holds in the case that $A(\omega)$ contains a single element for all $\omega$, so that it really is a function.

If $f:\Omega\to K$ has a universally measurable graph, is $f$ then $\Sigma_u$-$\mathcal{B}(K)$-measurable?

I have asked this now also on Mathoverflow.

share|cite|improve this question
Is $\Sigma_u = \Sigma \cup$ all sets that are measurable w.r.t. any complete measure on $\Omega$ whose domain contains $\Sigma$? – hot_queen May 6 '13 at 20:04
@hot_queen Yes, it is. – Michael Greinecker May 6 '13 at 20:05
Is the following true: For every open set U in K, the set $\{\omega \in \Omega: A_{\omega} \cap U = \phi \}$ is in $\Sigma_u$? If not, then can you give an example where this fails? – hot_queen May 6 '13 at 20:33
@hot_queen That is actually the kind of measurability condition I want to end up with. In general, your condition implies the measurability of the graph. If $\Omega$ is standard Borel, I would get the desired measurability. For the general case, I'm not sure. – Michael Greinecker May 6 '13 at 20:54
up vote 3 down vote accepted

I found a solution that suffices for what I do. It is based on strengthening the assumption that the graph $A$ is universally measurable to it being analytic. The notion of analyticity being used is that a subset $S$ of a measurable space $(M,\mathcal{M})$ is analytic if there is a compact metric space $T$ with Borel $\sigma$-algebra $\mathcal{B}(T)$ and a product measurable set $X\in\mathcal{M}\otimes\mathcal{B}(T)$ such that $S=\pi_M(X)$, where $\pi_M$ is the projection onto $M$. Analytic sets are always universally measurable. This notion of analyticity is extensively developed in the book Probability and Potential by Dellacherie and Meyer. A nice guide to the essentials can be found in this paper (JSTOR required).

The countable intersection or union of analytic sets is again analytic. If an analytic set is analytic in the product of some measurable space and a compact metric space, then the projection on the first coordinate is again analytic. An important fact from the theory of set-valued functions on a measurable space is that if the values are closed subsets of separable metric space, then the following condition is sufficient for the graph to be product measurable: The set $$\{\omega:C_\omega\cap O\neq\emptyset\}$$ is measurable for each open set $O$.

So assume that $A$ is analytic and $O$ is open. Then $$\{\omega:C_\omega\cap O\neq\emptyset\}=\pi_\Omega \big(A\cap(\Omega\times O)\big)$$ and hence analytic by the facts above and therefore universally measurable.

share|cite|improve this answer

Your Answer


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.