Disclaimer: This would be too long of a comment, so posting here instead to get some input. Thanks.
Let $P$ be probability distribution on the measurable space $X=\mathbb R^n$ (euclidean), and let $F$ be a collection of $P$-measurable functions. We are interested in the compactness of the set-valued integral $$ S:=\int_X F\,dP = \left\{\int_X f\,dP \mid f \in F\right\}. $$
Suppose $F$ is uniformly bounded, i.e., $b := \sup_{f \in F}\|f\|_\infty < \infty$, and for any For any $t \in [0,b]$ and $f \in F$, define $$ \mu_t(f) := P(\{x \in X \mid f(x)>t\}). $$
By the Layer-cake representation, one can write
$$ S := \int_X F\,dP = \int_0^b\mu_t(F)\,dt, $$ where $\mu_t(F) := \{\mu_t(f) \mid f \in F\}$.
Thanks to Theorem 4 of Aumann 1965 (the paper cited by Martin Vath, for $S$ to be compact, it suffices that $\mu_t(F)$ be closed for (almost) any $t \in [0,b]$.
In my specific question (and taking $d=1$ for simplicity), $F$ is collection of functions of the form $x \mapsto v(x)\cdot \pi(x)$, where $v:X \mapsto \mathbb R^k$ is a bounded $P$-measurable function function and $\pi$ runs over $P$-measurable functions $X \mapsto A$, with $A$ being a fixed compact subset of $\mathbb R^k$. It is easy to see that in this case,
$$ \mu_t(F) = \{P(v(x)\cdot a > t) \mid a \in A\}. $$
Question. For such an $F$, is it true $\mu_t(F)$ is closed for (almost) any $t \in F$ ?