Compactness of the integral of a set-valued function

Question

Let $X$ be a compact space (e.g. a compact subset of $\mathbb R^n$) and $P$ be a probability measure on $X$. Let $A$ be a compact subset of some $\mathbb R^d$. Finally, let $F$ be the collection of $P$-measurable functions from $X$ to $A$ and let $v$ be an $P$-integrable function from $X$ to $\mathbb R^d$. Define $S \subseteq \mathbb R$ by

$$ S := \int_X \langle F, v\rangle\,dP = \left\{\int_X \langle f(x),v(x)\rangle\,dP \mid f \in F\right\}, $$

where $\langle a,b\rangle$ is the inner-product of vectors $a$ and $b$.

Question. Under what minimal conditions on $P$ is $S$ compact ?

Note: The case where $P$ has countable support has been solved here (see end of question).

---

Related: Under what general conditions is the set $S := \left\{\int_{X}v(x)\pi(x)\,\mathrm{d}P(x) \mid \pi: X \to A\right\}$ closed?

Answer 1

Clearly, the integral of this correspondence coincides with the integral of correspondence $\phi:X\to 2^\mathbb{R}$ given by $\phi(x)=\big\{\langle a,v(x)\rangle \mid a\in A\big\}$. Moreover, $\phi$ is clearly compact-valued and is integrably bounded, since $$\max_{a\in A} \big|\langle a,v(x)\rangle\big|\leq \max_{a\in A} \|a\| \|v(x)\|$$ by the Cauchy-Schwarz inequality and the norm of an integrable function is integrable. By Proposition 7 on page 73 of the 1974 book "Core and Equilibria of a Large Economy" by Werner Hildenbrand shows that the integral of a closed valued integrable correspondence from a probability space to a Euclidean space is compact.

The only subtlety one needs to take care of is that Hildenbrand defines the integral of a correspondence with almost everywhere selections instead of actual measurable selections. This is not an issue if $\phi$ admits a measurable selection. We can apply the Kuratwki - Ryll-Nardzewski measurable selection theorem if we can show that the set $\{x\in X\mid \phi(x)\cap O\neq\emptyset\}$ is measurable for each open set of reals $O$. Let $V\subseteq\mathbb{R}^d$ be given by $V=\{v\in\mathbb{R}^d \mid \langle a,v\rangle\in O \text{ for some } a\in A\}$ and for each $a\in A$ let $V_a=\{v\in\mathbb{R}^d \mid \langle a,v\rangle\in O\}$, an open set. Then $V=\bigcup_{a\in A} V_a$ is open too and $$\{x\in X\mid \phi(x)\cap (b,c)\neq\emptyset\}=v^{-1}(V).$$ So if $v$ is Borel measurable, we can take the integal with respect to Borel selections. Otherwise, we can define it with respect to measurable selections for the $P$-completion.