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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.

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    $\begingroup$ What is $\pi v$? From what I see here, it seems that somehow I have to assume that each of the functions $\pi$, $v$, and $\pi v$ is $\mathbb R^d$-valued. $\endgroup$
    – Iosif Pinelis
    Apr 28, 2022 at 12:54
  • $\begingroup$ Sorry $v(x) \in R^{d \times k}$ is a matrix and $\pi(x) \in R^k$ is a vector. $v\pi(x) := v(x)\pi(x)$ is a matrix-vector product. $\endgroup$
    – dohmatob
    Apr 28, 2022 at 13:06
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    $\begingroup$ "the Minkowski sum of a finite number of closed subsets in a reflexive Banach space is closed" - this is not true even on the real line (consider the set $\mathbb{N}$ of positive integers and the set $\{-n-1/n|n\in \mathbb{N}\}$) $\endgroup$
    – Fedor Petrov
    Apr 28, 2022 at 13:59
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    $\begingroup$ @dohmatob That is about the sum of closed set and a compact set, not finitely many closed set. The counterexample above does not disappear if you add the compact set $\{0\}$. $\endgroup$
    – Michael Greinecker
    Apr 28, 2022 at 14:09
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    $\begingroup$ @dohmatob It should work if the compact sets are uniformly bounded, otherwise there exists probably a counterexample. I'll take a closer look in a few hours. $\endgroup$
    – Michael Greinecker
    Apr 28, 2022 at 14:31

3 Answers 3

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I guess that if $v$ is $P$-integrable then the answer is positive, and actually the set is compact.

Indeed, what you are looking for in this case is the compactness of the Aumann integral of the measurable multivalued integrably bounded function $v(\cdot)A$ with compact values in a finite-dimensional space.

In Aumann's original paper (Robert J. Aumann, Integrals of set-valued functions, J. Math. Anal. Appl. 12 (1965), 1-12) the compactness of the integral is proved in this case for the Lebesgue measure (Theorem 4), hence the result follows if $P$ has no atoms (Edit: and is separable, see comments below).

I guess that the proof for general measure spaces with atoms is similar, especially since your argument shows that it holds if the measure space is purely atomic. Probably somebody has carried out the proof, though I do not know a reference.

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  • $\begingroup$ Thanks for the nice insight (upvoted). Indeed, it seems Theorem 4 would establish the result for Lebesgue measure. How does one go from non-atomic to general measures on $R^d$ ? Is it via some kind of representation theorem for non-atomic measures (on $R^d$) ? $\endgroup$
    – dohmatob
    Apr 29, 2022 at 9:38
  • $\begingroup$ Also, by $v(\cdot)A$ you meant $v(\cdot )\Pi$, right ? That is, the collection of functions of the form $x \mapsto v(x)\pi(x)$ with $\pi \in \Pi$. $\endgroup$
    – dohmatob
    Apr 29, 2022 at 9:47
  • $\begingroup$ @dohmatob You can decompose $X$ into a countable set on which $P$ is purely atomic and a set on which $P$ is nonatomic. You have a solution for both sets, the Minkowski sum will be again compact and give you your answer. $\endgroup$
    – Michael Greinecker
    Apr 29, 2022 at 11:16
  • $\begingroup$ @MichaelGreinecker Thanks for the insight. My issue is that i don't how to move from Lebesgue (the case treated in the reference paper) to general non-atomic probability measures. But I might be missing something. (BTW, I started a thread on this issue here mathoverflow.net/q/421356/78539) $\endgroup$
    – dohmatob
    Apr 29, 2022 at 11:19
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    $\begingroup$ By $v(\cdot)A$ I mean the multivalued function (which assumes values in the powerset of $\mathbb R^n$); the Aumann integral is defined as the set of integrals over all measurable selections of that multivalued function. Concerning going to Lebesgue measure: I had the theorem in mind that every atomless separable measure space is isomorphic to Lebesgue measure. I forgot to mention "separable". (One might still try to apply Maharam's results for the nonseparable case, but I am not sure whether this will lead to something.) $\endgroup$
    – Martin Väth
    Apr 29, 2022 at 20:26
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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 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.

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  • $\begingroup$ As mentioned in a comment to my reply, I had a direct application of the measure isomorphism theorem in mind (which requires separability). However, I think that simply the proof of the convexity result by Aumann requires only a non-atomaic measure space, that is without loss of generality you can assume in a non-atomic measure space that $A$ is convex, and then again Aumann's proof for compactness should directly hold. Alternatively, once you assume that $A$ is convex, you can apply some of many results which establish equality of the Aumann and the Debreu integral. $\endgroup$
    – Martin Väth
    Apr 30, 2022 at 12:06
  • $\begingroup$ (continuing comment): Note that the Debreu integral is defined by means of approximation of the mutlivalued functions in the hyperspace of nonempty compact (convex?) sets with the Hausdorff metric, hence by definition is compact (and convex). All result establishing this equality are one sense or another based on the weak compactness in $L_1$ of the set of selections of an integrably bounded multivalued map with compact convex values. I am afraid that for this result convexity is crucial. $\endgroup$
    – Martin Väth
    Apr 30, 2022 at 12:08
  • $\begingroup$ Thanks for the input. In fact in my original question $A$ was compact and convex (the unit probability simplex in $\mathbb R^k$). I diidn't know convexity would play in the problem, and so I suppressed it. Concerning the equality of the Aumann and Debreu integrals, is there are clear reference for this result ? (N.B.: I'm only learning all of this set-valued analysis here on the fly based on these interactions). Thanks. $\endgroup$
    – dohmatob
    Apr 30, 2022 at 12:16
  • $\begingroup$ I guess you were referring to something like Theorem 3.11 of this paper citeseerx.ist.psu.edu/viewdoc/…. $\endgroup$
    – dohmatob
    Apr 30, 2022 at 12:44
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Disclaimer. This post is mostly to provide some low-level details for Martin Vath's answer and comments. Note that my previous post https://mathoverflow.net/a/421367/78539 didn't quite correspond to what Martin Vath intended, though it still ends up solving the problem for the special case of countably supported $P$.

We shall work under the following assumption:

Asumption 1. For the function $v:X \to \mathbb R^{d\times k}$ in the question, $\|v\|:x \mapsto \|v(x)\|_{op}$ is $P$-integrable.

Method 1: Via Maraham's theorem

The separable probability measure space $(X,P)$ can be decomposed into the sum of an atomic part (i.e with countable support) and a non-atomic part (i.e containing no atoms) which is isomorphic to $L:=([0,1],\mathcal B([0,1]), dx)$, the standard Lebesgue space. The set $S$ in the question then decomposes as a Minkowski sum of isomorphic images of the versions $S$ corresponding to each part of this decomposition.

Thus, we only need to study the compactness of $S$ in the case where the original probability space $(X,P)$ is atomic (done!) and the case where it is the standard Lebesgue space $L$.

Thus, let $(X,P) = L$, the standard Lebesgue measure space. Consider the set-valued map $F:X \to 2^{\mathbb R^d}$ defined by $$ F(x) := v(x)\cdot A := \{v(x) a \mid a \in A\}, $$ Then, the set $\mathcal S$ in the original question can be written as the Aumann integral of $F$ over $X$, i.e $$ S = \int_X F\,dP := \left\{\int_X f\,dP \,\,\big |\,\, f \in \mathfrak F\right\}, $$ where $\mathfrak F$ be the collection of all integrable functions $f:X \to \mathbb R$ such that $f(x) \in F(x)$ for all $x \in X$.

We observe the following:

  • Since $A$ is closed (compact), $F(x):=v(x)A$ is closed (compact) for every $x \in X$.
  • Since $A\subseteq \mathbb R^k$ is compact, it is clear that $F$ is integrably bounded by the function $h:X \to \mathbb R_+$ defined by $h(x):=\mathrm{diam}(A)\cdot\|v(x)\|_{op}$, meaning that $h$ is $P$-integrable and $$ \|z\| \le h(x),\text{ for every }x \in X, \,z \in F(x). $$

It then follows from Theorem 4 of Robert J. Aumann, Integrals of set-valued functions, J. Math. Anal. Appl. 12 (1965), 1-12) that $S$ is compact.

Method 2: via Debreu-Aumann equivalence

Assumption 2. $A$ is convex.

Under Assumption 1 and Assumption 2, the compactness of the set $S$ defined in the original question follows from Theorem 3.11 of Sambucini (1999). However, it is a bit trickier to verify all the hypotheses: "total measurability of $F$", etc.

Possible issue. Something which appears weird is that Method 2 seems to require convexity of $A$ while Method 1 doesn't.

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    $\begingroup$ Maharam's decomposition you mention holds only if the measure space is separable. Otherwise an uncountable number of copies of intervals is involved, though I do not know the details by heart. (AFAIK, only this non-separable version is due to D. Maharam; the separable version had been known long before.) $\endgroup$
    – Martin Väth
    May 1, 2022 at 10:22
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    $\begingroup$ If you assume that $A$ is convex (and compact), it suffices to apply the earlier mentioned theorem: The set of measurable selections of an integrably bounded measurable multivalued function with compact convex values is weakly compact in $L_1$, Indeed if $a_n=\int f_n$ is a sequence in the image then the weak (sequential) compactness of $f_n$ implies that there is a subsequence with $\int f_{n_k}\to\int f$ weakly for some selection $f$. Because we are in the space $\mathbb R^n$, this implies strong convergence. $\endgroup$
    – Martin Väth
    May 1, 2022 at 10:31
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    $\begingroup$ Maharam's theorem characterizes measure algebras, not measure spaces. This is a coarser classification. For example, the measure algebra of any probability space coincides with the measure algebra of its completion. Nevertheless, a suitable isomorphism theorem exists; one can find it in Royden's real analysis book. $\endgroup$
    – Michael Greinecker
    May 1, 2022 at 12:28
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    $\begingroup$ @dohmatob: I am a bit surprised by the formulation of Theorem 3.1, because it is not defined what a convex correspondence is. I assume that it should be convex-valued and measurable in some previously defined sense, but there is no mentioning of any measurability hypothesis. But Diestel & Uhl have proved various theorems of this kind, maybe their monograph contains clearer formulations. $\endgroup$
    – Martin Väth
    May 1, 2022 at 14:41
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    $\begingroup$ The mapping $I\colon f\mapsto\int_Xf\,dP$ is indeed continuous if image and preimage space are endowed with the weak topology. The essential argument needed to show this is that if $\varphi$ is a bounded functional in the image space then $\varphi\circ I$ is a bounded functional on $L_1$ (with the norm topology). And the final step is of course to use the finite-dimensionality of $\mathbb R^n$ by using that weak and strong topology on the image space coincide. $\endgroup$
    – Martin Väth
    May 1, 2022 at 14:43

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