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 question before on MSE, but have received no answer.

  • $\begingroup$ With “universally measurable” you mean “finite with regard to every finite measure space containing $(\Omega,\Sigma)\times (K,\mathfrak{B}(K))$? Usually this term is only defined for the Borel-σ-algebra (for example in Kechris). $\endgroup$
    – The User
    May 13, 2013 at 11:09
  • $\begingroup$ Yes, with "finite" replaced by "measurable". Dellarcherie and Meyer do this in this generality. $\endgroup$ May 13, 2013 at 11:40
  • $\begingroup$ Yes, sorry, I meant “measurable”. $\endgroup$
    – The User
    May 13, 2013 at 13:54
  • $\begingroup$ @TheUser there is also a generalization of the notion of being "universal", see it used e.g. in this answer: math.stackexchange.com/a/248199/5887 $\endgroup$
    – SBF
    May 21, 2013 at 14:51

1 Answer 1


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.


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