$\def\E{\hskip.15ex\mathsf{E}\hskip.10ex}$ Let $B$ be a (maybe nonseparable) Banach space equipped with the Borel $\sigma$-algebra $\mathscr{B}(B)$. Let $R:B\to \mathbb{R}$ be a bounded linear operator.

Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $F$ be a $\mathscr{B}(B)|\mathcal{F}$-measurable mapping $\Omega\to B$. Suppose that $\E \|F\|<\infty$.

Question: is it true without any additional assumptions that $E F$ is well-defined, belongs to $B$ and $$ \E RF= R \E F? $$

Remark: usually (e.g. in the Ledoux-Talagrand book) the separability of the space B is additionally imposed. I wonder whether the statement is true without this assumption. For example, what happens if $B$ is just a space of bounded measurable functions?

$\def\E{\hskip.15ex\mathsf{E}\hskip.10ex}$ Let $B$ be a (maybe nonseparable) Banach space equipped with the Borel $\sigma$-algebra $\mathscr{B}(B)$. Let $R:B\to \mathbb{R}$ be a bounded linear operator.

Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $F$ be a $\mathscr{B}(B)|\mathcal{F}$-measurable mapping $\Omega\to B$. Suppose that $\E \|F\|<\infty$.

Question: is it true without any additional assumptions that $\E F$ is well-defined, belongs to $B$ and $$ \E RF= R \E F? $$

Remark: usually (e.g. in the Ledoux-Talagrand book) the separability of the space B is additionally imposed. I wonder whether the statement is true without this assumption. For example, what happens if $B$ is just a space of bounded measurable functions (in that case one can just define $[\E F](x):=\E[F(x)]$)?