Let $\mu$ be a probability measure on measurable space $X=\mathbb R^n$ (euclidean), and let $F$ be a family of $\mu$-measurable functions $X \mapsto \mathbb R$ which are uniformly bounded, i.e $b:=\sup_{f \in F}\|f\|_\infty < \infty$. For any $t \in \mathbb R$, let $\mu_t(f) := \mu(\{x \in X \mid f(x) > t\})$.
Assume that $\mu$ is non-atomic, in case such an assumption makes any difference.
Question. What minimal assumptions on $F$ ensure that the set $\mu_t(F) := \{\mu_t(f) \mid f \in F\} \subseteq \mathbb R$ closed for (Lebesgue-almost) any $t \in [0,b]$ ?
Context
My goal is to ultimately show that the set-valued integral $S:=\int_X F\,d\mu$ is compact. I figured out that (using Layer-cake representation), $$ S:=\int_X F\,d\mu = \int_0^b \mu_t(F)\,dt. $$
Thanks to Theorem 4 of Aumann 1965, compactness of $S$ would follow immediate if one can show that $\mu_t(F)$ is closed for (almost) any $t \in [0, b]$.
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