**Disclaimer:** This would be too long of a comment, so posting here instead to get some input. Thanks.

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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 $\mathbb R^k$.

>**Question.** *For such an $F$, is it true $\mu_t(F)$ is closed for (almost) any $t \in F$ ?*