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.