Basic properties of expectation in non-separable Banach spaces

Question

$\def\E{\hskip.15ex\mathsf{E}\hskip.10ex}$ 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 $[\E F](x):=\E[F(x)]$)?

Answer 1

$\newcommand{\E}{\operatorname{\mathsf{E}}}$ You do not need the separability of $B$ to define $\E F$ for a random vector $F\colon\Omega\to B$; however, you need to assume that $F$ is strongly measurable, in the sense that there is a sequence of finitely-valued random vectors $F_n$ in $B$ such that $\|F_n(\omega)-F(\omega)\|\to0$ for almost all $\omega\in\Omega$.

By Bochner's theorem, if $F$ is strongly measurable, then $\E\|F\|<\infty$ iff $F$ is Bochner-integrable, in the sense that for some sequence of finitely-valued random vectors $F_n$ in $B$ we have $\|F_n(\omega)-F(\omega)\|\to0$ for almost all $\omega$ and $\E\|F_n-F\|\to0$; then $\E F:=\lim_n\E F_n$, with naturally defined $\E F_n$.

It is then known and easy to check that $\E RF=R\E F$ for any strongly measurable random vector $F$ with $\E\|F\|<\infty$ and any bounded linear operator $R\colon B\to\mathbb R$; that is, the Bochner integrability implies the Pettis integrability.

See e.g. Yosida, Sections V.4 and V.5 for further details.

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Concerning your remark that usually the Banach space is assumed to be separable: this is done to ensure the measurability in a number of instances, including the measurability of the sum of random vectors.

Answer 2

A counterexample for random variable with nonseparable range.

Let $\omega_1$ be the smallest uncountable ordinal. Let $\Omega = [0,\omega_1)$, the set of countable ordinals, with the order topology. Then the Banach space $B = C[0,\omega_1)$ will be used. Note: every real-avalued continuous function on $[0,\omega_1)$ is bounded. Moreover, $$ \lim_{\alpha \to \omega_1} f(\alpha) $$ exists for every $f \in B$ and $\phi : B \to \mathbb R$ defined by $$ \phi(f) = \lim_{\alpha \to \omega_1} f(\alpha) $$ is a bounded linear functional. [The Stone-Cech compactification of $\Omega$ is the one-point compactification of $\Omega$.]

Our measure space is $(\Omega,\mathcal F, \mathbb P)$, where $\mathcal F$ is the countable-cocountable sigma-algebra, and $\mathbb P$ is $0$ on countable sets and $1$ on cocountable sets. Note: if $f \in C[0,\omega_1)$, then the integral is $$ \int f(\alpha)\;\mathbb P(d\alpha) = \lim_{\alpha \to \omega_1} f(\alpha) = \phi(f) . $$

Define $F : \Omega \to B$ by $F(\alpha) = \mathbf1_{(\alpha,\omega_1)} $, the indicator function of the interval $(\alpha,\omega_1)$, which is a clopen set.

We will show that that there is no $\mathbb E[F] \in B$ with the property $\mathbb E[R\circ F] = R\big(\mathbb E[F]\big)$ for all bounded linear functionals $R : B \to \mathbb R$. Suppose it does exist.

Fix $\xi \in [0,\omega_1)$. Then $f \mapsto f(\xi)$ is a bounded linear functional, and $$ F(\alpha)(\xi) = \begin{cases} 1,\quad \alpha < \xi \\ 0,\quad \alpha \ge \xi \end{cases} $$ and thus $$ 0 = \lim_{\alpha \to \omega_1} F(\alpha)(\xi) = \int F(\alpha)(\xi)\;\mathbb P(d\alpha) = \left(\mathbb E[F]\right)(\xi) $$ This holds for all $\xi$ so $\mathbb E[F] = \mathbf 0$, the zero element of $B$.

On the other hand $\phi$ defined above is a bounded linear functional, and $$ \phi(F(\alpha)) = \lim_{\xi \to \omega_1}F(\alpha)(\xi) = 1 $$ for all $\alpha$.

So $$ \mathbb E[\phi\circ F] =\int\phi(F(\alpha))\;\mathbb P(d\alpha) =\int 1\;\mathbb P(d\alpha) = 1 . $$ This is not equal to $\phi\big(\mathbb E[F]\big) = \phi(\mathbf 0) = 0$.

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Now I'm wondering if $F$ is Borel.