Let $(\Omega,\Sigma,\mu)$ be a probability space, $E$ be a separable Banach space, $f:\Omega\rightarrow E$ be Borelmeasurable and consider the Bochnerintegral $$ \bar{\mu}:=\int_{\omega\in \Omega}\, f(\omega)\, \mu(d\omega). $$ If there is a convex set $C\subseteq E$, such that $f(\Omega)$ is a compact subset of $C$ then, is $\bar{\mu}\in C$?

$\begingroup$ Aren't you satisfied with the intergal being in the closure of $C$? $\endgroup$– Jochen WengenrothCommented Mar 19, 2022 at 16:27

$\begingroup$ Ah, that's precisely my problem; in my setting, I need it to genuinely lie in C. The most I can assume is that $f(\Omega)$ is a compact subset of $C$ (instead of assuming that it is only contained in $C$; would this be enough?) $\endgroup$– ABIMCommented Mar 19, 2022 at 16:29

$\begingroup$ See the notion of "measureconvex set" ... mathoverflow.net/a/152951/454 $\endgroup$– Gerald EdgarCommented Mar 20, 2022 at 1:07
2 Answers
Let $\Omega=\mathbb N=\{1,2,3,\cdots\}$ and $\mu(\{n\}) = 2^{n}$ with $\Sigma = 2^\Omega$. Consider $E=c_0$ with $f:\Omega\rightarrow E$ given by $f(n+1) = 2^{n} e_n$ where $e_n$ is the $n$th standard unit vector basis element, and $f(1)=0$. Then the image of $f$ is $\{0\} \cup \{ 2^{n}e_n\}$ which is compact in $c_0$. The integral is $$ \int_\Omega f = \sum_n 2^{n} f(n) = \frac12 \sum_n 4^{n} e_n \in c_0. $$ If we take $C$ to be the convex hull of $\{0\} \cup \{ 2^{n} e_n\}$ (no closure!) then $f$ maps into $C$ but the integral of $C$ is only in the closure of $C$, not $C$ itself.
I had the same problem many years ago (if I remember correctly), and the answer was negative. I think that I had found a counterexample in the monograph of Diestel and Uhl.
If $E$ has finite dimension, the answer is positive, because the convex hull of a compact subset is compact in this case and thus automatically closed.

4$\begingroup$ I think that this would be a great answer with the counterexample (or a precise reference), but, as is, this is probably more of a comment. $\endgroup$– LSpiceCommented Mar 19, 2022 at 19:23

$\begingroup$ I do not have the references here, so I cannot provide them. Anyway, Matthew's answer is very similar to the example I have seen at that time. $\endgroup$ Commented Mar 20, 2022 at 8:46