
> 
> 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](https://math.stackexchange.com/questions/383488/from-universal-measurability-to-measurability) this question before on MSE, but have received no answer.



From universal measurability to measurability
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](http://www.jstor.org/stable/2297861) (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. 


