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Disclaimer: This would be too long of a comment, so posting here instead to get some input. This is an attempt to get a handle on Martin Vath's answer. Thanks in advance.

Let $P$ be probability distribution on a compact subset $X$ of $\mathbb R^n$, 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 $$ P_t(f) := P(\{x \in X \mid f(x)>t\}). $$

By the layer-cake representation, one can write

$$ S := \int_X F\,dP = \int_0^bP_t(F)\,dt, $$ where $P_t(F) := \{P_t(f) \mid f \in F\}$.

Thanks to Theorem 4 of Aumann 1965 (the paper cited by Martin Vath, for $S$ to be compact, it suffices that $P_t(F)$ 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 $\mathbb R^k$.

Moreover, to ensure $F$ is uniformly bounded, it suffices to demand $v$ be bounded; we can then take $b = \mathrm{diam}(A)\cdot\sup_{x \in X}\|v(x)\|<\infty$.

Question. For such an $F$, is it true $P_t(F)$ is closed for (almost) any $t \in [0,b]$ ?

Partial solution when $P$ has countable support

Suppose $P = \sum_{i=1}^\infty w_i \delta_{x_i}$, with $(x_i)_i \subseteq X$ and $(w_i)_i \in \ell^1(\mathbb R)$. Then, a direct computation gives $$ \begin{split} P_t(F) &= \left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot \pi(x_i) \,>\, t} \mid \pi \in \Pi\right\}\\ &=\left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot a_i \,>\, t} \mid (a_i)_i \subseteq A\right\}\\ & = \sum_{i=1}^\infty w_i u_i(A)\text{ (Minkowski sum)}, \end{split} $$ where $u_i(a) := 1_{v(x_i)\cdot a \,>\, t} \in \{0,1\}$. Thus, we see that $P_t(F)$ is a subset of values for the subsums of $\sum_{i=1}^\infty w_i$, and so must be closed (since there are only countably many distinct values for these subsums). We thus recover the result established in the original question.

Disclaimer: This would be too long of a comment, so posting here instead to get some input. This is an attempt to get a handle on user Martin Vath's answer. Thanks in advance.

Let $P$ be probability distribution on a compact subset $X$ of $\mathbb R^n$, 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 $$ P_t(f) := P(\{x \in X \mid f(x)>t\}). $$

By the layer-cake representation, one can write

$$ S := \int_X F\,dP = \int_0^bP_t(F)\,dt, $$ where $P_t(F) := \{P_t(f) \mid f \in F\}$.

Thanks to Theorem 4 of Aumann 1965 (the paper cited by Martin Vath, for $S$ to be compact, it suffices that $P_t(F)$ 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 $\mathbb R^k$.

Moreover, to ensure $F$ is uniformly bounded, it suffices to demand $v$ be bounded; we can then take $b = \mathrm{diam}(A)\cdot\sup_{x \in X}\|v(x)\|<\infty$.

Question. For such an $F$, is it true $P_t(F)$ is closed for (almost) any $t \in [0,b]$ ?

Partial solution when $P$ has countable support

Suppose $P = \sum_{i=1}^\infty w_i \delta_{x_i}$, with $(x_i)_i \subseteq X$ and $(w_i)_i \in \ell^1(\mathbb R)$. Then, a direct computation gives $$ \begin{split} P_t(F) &= \left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot \pi(x_i) \,>\, t} \mid \pi \in \Pi\right\}\\ &=\left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot a_i \,>\, t} \mid (a_i)_i \subseteq A\right\}\\ & = \sum_{i=1}^\infty w_i u_i(A)\text{ (Minkowski sum)}, \end{split} $$ where $u_i(a) := 1_{v(x_i)\cdot a \,>\, t} \in \{0,1\}$. Thus, we see that $P_t(F)$ is a subset of values for the subsums of $\sum_{i=1}^\infty w_i$, and so must be closed (since there are only countably many distinct values for these subsums). We thus recover the result established in the original question.

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

Let $P$ be probability distribution on a compact subset $X$ of $\mathbb R^n$, 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 $$ P_t(f) := P(\{x \in X \mid f(x)>t\}). $$

By the layer-cake representation, one can write

$$ S := \int_X F\,dP = \int_0^bP_t(F)\,dt, $$ where $P_t(F) := \{P_t(f) \mid f \in F\}$.

Thanks to Theorem 4 of Aumann 1965 (the paper cited by Martin Vath, for $S$ to be compact, it suffices that $P_t(F)$ 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 $\mathbb R^k$.

Moreovfer, to ensure $F$ is uniformly bounded, it suffices to demand $v$ be bounded; we can then take $b = \mathrm{diam}(A)\cdot\sup_{x \in X}\|v(x)\|<\infty$.

Question. For such an $F$, is it true $P_t(F)$ is closed for (almost) any $t \in [0,b]$ ?

Partial solution when $P$ has countable support

Suppose $P = \sum_{i=1}^\infty w_i \delta_{x_i}$, with $(x_i)_i \subseteq X$ and $(w_i)_i \in \ell^1(\mathbb R)$. Then, a direct computation gives $$ \begin{split} P_t(F) &= \left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot \pi(x_i) \,>\, t} \mid \pi \in \Pi\right\}\\ &=\left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot a_i \,>\, t} \mid (a_i)_i \subseteq A\right\}\\ & = \sum_{i=1}^\infty w_i u_i(A)\text{ (Minkowski sum)}, \end{split} $$ where $u_i(a) := 1_{v(x_i)\cdot a \,>\, t} \in \{0,1\}$. Thus, we see that $P_t(F)$ is a subset of values for the subsums of $\sum_{i=1}^\infty w_i$, and so must be closed (since there are only countably many distinct values for these subsums). We thus recover the result established in the original question.

Disclaimer: This would be too long of a comment, so posting here instead to get some input. This is an attempt to get a handle on Martin Vath's answer. Thanks in advance.

Let $P$ be probability distribution on a compact subset $X$ of $\mathbb R^n$, 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 $$ P_t(f) := P(\{x \in X \mid f(x)>t\}). $$

By the layer-cake representation, one can write

$$ S := \int_X F\,dP = \int_0^bP_t(F)\,dt, $$ where $P_t(F) := \{P_t(f) \mid f \in F\}$.

Thanks to Theorem 4 of Aumann 1965 (the paper cited by Martin Vath, for $S$ to be compact, it suffices that $P_t(F)$ 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 $\mathbb R^k$.

Moreover, to ensure $F$ is uniformly bounded, it suffices to demand $v$ be bounded; we can then take $b = \mathrm{diam}(A)\cdot\sup_{x \in X}\|v(x)\|<\infty$.

Question. For such an $F$, is it true $P_t(F)$ is closed for (almost) any $t \in [0,b]$ ?

Partial solution when $P$ has countable support

Suppose $P = \sum_{i=1}^\infty w_i \delta_{x_i}$, with $(x_i)_i \subseteq X$ and $(w_i)_i \in \ell^1(\mathbb R)$. Then, a direct computation gives $$ \begin{split} P_t(F) &= \left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot \pi(x_i) \,>\, t} \mid \pi \in \Pi\right\}\\ &=\left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot a_i \,>\, t} \mid (a_i)_i \subseteq A\right\}\\ & = \sum_{i=1}^\infty w_i u_i(A)\text{ (Minkowski sum)}, \end{split} $$ where $u_i(a) := 1_{v(x_i)\cdot a \,>\, t} \in \{0,1\}$. Thus, we see that $P_t(F)$ is a subset of values for the subsums of $\sum_{i=1}^\infty w_i$, and so must be closed (since there are only countably many distinct values for these subsums). We thus recover the result established in the original question.

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

Let $P$ be probability distribution on a compact subset $X$ of $\mathbb R^n$, 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 $$ P_t(f) := P(\{x \in X \mid f(x)>t\}). $$

By the layer-cake representation, one can write

$$ S := \int_X F\,dP = \int_0^bP_t(F)\,dt, $$ where $P_t(F) := \{P_t(f) \mid f \in F\}$.

Thanks to Theorem 4 of Aumann 1965 (the paper cited by Martin Vath, for $S$ to be compact, it suffices that $P_t(F)$ 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 $P_t(F)$ is closed for (almost) any $t \in [0,b]$ ?

Partial solution when $P$ has countable support

Suppose $P = \sum_{i=1}^\infty w_i \delta_{x_i}$, with $(x_i)_i \subseteq X$ and $(w_i)_i \in \ell^1(\mathbb R)$. Then, a direct computation gives $$ \begin{split} P_t(F) &= \left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot \pi(x_i) \,>\, t} \mid \pi \in \Pi\right\}\\ &=\left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot a_i \,>\, t} \mid (a_i)_i \subseteq A\right\}\\ & = \sum_{i=1}^\infty w_i u_i(A)\text{ (Minkowski sum)}, \end{split} $$ where $u_i(a) := 1_{v(x_i)\cdot a \,>\, t} \in \{0,1\}$. Thus, we see that $P_t(F)$ is a subset of values for the subsums of $\sum_{i=1}^\infty w_i$, and so must be closed (since there are only countably many distinct values for these subsums). We thus recover the result established in the original question.

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

Let $P$ be probability distribution on a compact subset $X$ of $\mathbb R^n$, 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 $$ P_t(f) := P(\{x \in X \mid f(x)>t\}). $$

By the layer-cake representation, one can write

$$ S := \int_X F\,dP = \int_0^bP_t(F)\,dt, $$ where $P_t(F) := \{P_t(f) \mid f \in F\}$.

Thanks to Theorem 4 of Aumann 1965 (the paper cited by Martin Vath, for $S$ to be compact, it suffices that $P_t(F)$ 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 $\mathbb R^k$.

Moreovfer, to ensure $F$ is uniformly bounded, it suffices to demand $v$ be bounded; we can then take $b = \mathrm{diam}(A)\cdot\sup_{x \in X}\|v(x)\|<\infty$.

Question. For such an $F$, is it true $P_t(F)$ is closed for (almost) any $t \in [0,b]$ ?

Partial solution when $P$ has countable support

Suppose $P = \sum_{i=1}^\infty w_i \delta_{x_i}$, with $(x_i)_i \subseteq X$ and $(w_i)_i \in \ell^1(\mathbb R)$. Then, a direct computation gives $$ \begin{split} P_t(F) &= \left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot \pi(x_i) \,>\, t} \mid \pi \in \Pi\right\}\\ &=\left\{\sum_{i=1}^\infty w_i1_{v(x_i)\cdot a_i \,>\, t} \mid (a_i)_i \subseteq A\right\}\\ & = \sum_{i=1}^\infty w_i u_i(A)\text{ (Minkowski sum)}, \end{split} $$ where $u_i(a) := 1_{v(x_i)\cdot a \,>\, t} \in \{0,1\}$. Thus, we see that $P_t(F)$ is a subset of values for the subsums of $\sum_{i=1}^\infty w_i$, and so must be closed (since there are only countably many distinct values for these subsums). We thus recover the result established in the original question.