$\newcommand{\E}{\mathsf{E}}$ $\newcommand{\P}{\mathsf{P}}$ The following additional question was asked in a comment by user Oleg:

Suppose that $(\Omega,\mathcal F,\P)$ is a probability space, $B$ is a (not necessarily separable) Banach space, and $F\colon\Omega\to B$ is a strongly measurable random vector in $B$, with $\E\|F\|<\infty$. If now $\mathcal G$ is a sub-σ-algebra of $\mathcal F$, is then there a well-defined and strongly measurable conditional expectation $\E(F|\mathcal G)$?

The strong measurability of a random vector $F$ means that there is a sequence of finitely-valued random vectors $F_n$ in $B$ such that $\|F_n(\omega)-F(\omega)\|\to0$ for $\P$-almost all $\omega\in\Omega$.

I thought it makes sense to post this question separately, which is what is being done here.


1 Answer 1


$\newcommand{\E}{\mathsf{E}}$ $\newcommand{\P}{\mathsf{P}}$ The affirmative answer to this question is provided by Scalora, Theorem 2.1, page 354, which can be stated as follows, using the setting and notation in the question:

If $F\colon\Omega\to B$ is a Bochner-integrable random vector, then there is a $\P$-almost surely unique Bochner-integrable strongly $\mathcal G$-measurable random vector $\E(F|\mathcal G)=:G$ in $B$ such that $\E G1_A=\E F1_A$ for all $A\in \mathcal G$.

Note that in Scalora's paper the strong measurability is part of the definition of the Bochner integrability. Note also that then, by Bochner's theorem, $F$ is Bochner-integrabile iff $F$ is strongly measurable and the condition $\E\|F\|<\infty$ holds.

It is not a big advantage here to allow the Banach space to be non-separable. Indeed, the strong measurability of a random vector implies that it is almost separably valued; see e.g. the note in the middle of page 348 of Scalora's paper.

  • $\begingroup$ Thank you very much! $\endgroup$
    – Oleg
    Aug 23, 2019 at 14:51

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