Let $X$ be a compact subset of $\mathbb R^n$ and let $A$ be a compact subset of $\mathbb R^k$. Let $P$ be a probability distribution on $X$ and $v$ be a $P$-measurable function from $X$ to $\mathbb R^{d \times k}$.

Assumption.$v$ is bounded on $X$, that is, there exists $R>0$ such that $\sup_{x \in X}\lVert v(x)\rVert_\text{op} \le R$.

Consider the set $S \subseteq \mathbb R^d$ defined by

$$ S := \left\{\int_{X}v\pi\,\mathrm{d}P \mid \pi \in \Pi\right\}, $$

where $\Pi$ is the set of $P$-measurable functions from $X$ to $A$.

Question 1.Under what general conditions is $S$ a closed subset of $\mathbb R^d$ ?

Perhaps even more generally,

Question 2.What is the closure $\overline S$ of $S$ in $\mathbb R^d$ ?

## Partial solution when $P$ has countable support

Suppose $P = \sum_{i=1}^\infty w_i\delta_{x_i}$, for some $x_1,x_2,\dotsc \in X$, and $0\le w_1,w_2,\dotsc$, with $\sum_{i=1}^\infty w_i = 1$. Let $M_i := w_iv(x_i) \in \mathbb R^{d \times k}$ for all $i$. Then, one computes $$ \begin{split} S = \left\{\sum_{i=1}^\infty w_iv(x_i)\pi(x_i) \mid \pi \in \Pi\right\} &= \left\{\sum_{i=1}^\infty M_i a_i \mid a_1,a_2,\ldots \in A\right\}\\ & = B_1 + B_2 + \ldots, \end{split} $$

where $B_i := \{M_i a \mid a \in A\}$. It is clear that each $B_i$ is compact in $\mathbb R^d$. Because $C := B_1 \times B_2 \times \dotsb$ is compact and the funciton $f:C \to S$ defined by $f(b_1,b_2,\dotsc) := \sum_{i=1}^\infty w_i b_i$ is continuous, we deduce that $S$ is compact, and therefore closed.

3more comments