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


$\newcommand{\E}{\operatorname{\mathsf{E}}}$ 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.

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.

  • $\begingroup$ Iosif, thanks a lot for a detailed answer. Do you know if the same would be true for conditional expectations: if $F$ is strongly measurable with $E\|F\|<\infty$, $\mathcal{G}$ is a sub-$\sigma$-algebra, then $E(F|\mathcal{G})$ is well-defined and strongly measurable? $\endgroup$ – Oleg Aug 12 at 21:00
  • $\begingroup$ Also I wonder about the case when $B$ is the space of bounded functions on, say, $[0,1]$. If $F$ is not strongly measurable (just measurable), can it happen that the function $x\to EF(x)$ is not measurable? $\endgroup$ – Oleg Aug 12 at 21:02
  • $\begingroup$ Also I am a bit puzzled about your last comment: if $F$, $G$ are measurable, isn't it always true that $F+G$ is measurable? Is there a counterexample? $\endgroup$ – Oleg Aug 12 at 21:48
  • $\begingroup$ $F$, $G$ measruable need not imply $F+G$ measurable in the nonseparable case. See mathoverflow.net/a/313792/454 $\endgroup$ – Gerald Edgar Aug 12 at 21:51
  • $\begingroup$ @Oleg : I don't know answers, right away, to your additional questions, on the conditional expectation and $EF(x)$, but can try to think about them. In this case and in general, I believe it's best, in more aspects than one, to post the additional questions separately. $\endgroup$ – Iosif Pinelis Aug 12 at 23:15

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 $\phi\big(\mathbb E[F]\big) = \phi(\mathbf 0) = 0$.


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

  • $\begingroup$ Gerald, thanks for this interesting counter-example. Do you think something similar can happen if $B$ is a space of bounded Borel measurable functions? $\endgroup$ – Oleg Aug 13 at 9:29

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