I have the following problem. Given a function $f: S \times A \to [0, \infty]$ where $S$ is a state space of the form $S = (L \times \mathbb{R}^V)$, for finite sets L and V. $A$ is a Borel subset of $\mathbb{R}$ and indicates the allowed actions. Assume $f$ is universally measurable in each component, that is, the functions $f(s, \cdot)$ and $f(\cdot, a$) are universally measurable. I also have that the infimization $f^*(s) = \inf_{a\in A} f(s, a)$ is universally measurable.
What I want to know is, whether for any arbitrary $\epsilon > 0$, there is a an $\epsilon$-optimal selection $\sigma: S \to A$ such that $f(s, \sigma(s)) \leq f^*(s) + \epsilon$ and $\sigma$ is universally measurable.
To my knowledge, there exists such selections if $f$ is assumed to be lower semianalytic. This result is based on Von Neumann's selection/choice theorem. However, in that case an analytically measurable selection exists. I am wondering if by weakening the measurability condition on the selection, we can also weaken the conditions on $f$ in some way.