# Bochner integral over convex sets lies in the convex set?

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

• Aren't you satisfied with the intergal being in the closure of $C$? Commented Mar 19, 2022 at 16:27
• 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?)
– ABIM
Commented Mar 19, 2022 at 16:29
• See the notion of "measure-convex set" ... mathoverflow.net/a/152951/454 Commented Mar 20, 2022 at 1:07

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.
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.