Bochner integral over convex sets lies in the convex set?

Question

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 $\bar{\mu}\in C$?

Answer 1

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.

Answer 2

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.