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$\newcommand{\E}{\operatorname{\mathsf{E}}}$ 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, Section V.4 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.

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

$\newcommand{\E}{\operatorname{\mathsf{E}}}$ 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 and $\E\|F\|<\infty$, then $F$ 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 $\E F:=\lim_n EF_n$, with naturally defined $\E F_n$.

It is then a known and easy to check fact 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, Section V.4 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.

$\newcommand{\E}{\operatorname{\mathsf{E}}}$ 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, Section V.4 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.

You do not need the separability of $B$ to define $EF$ 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 and $E\|F\|<\infty$, then $F$ 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 $EF:=\lim_n EF_n$, with naturally defined $EF_n$.

$\newcommand{\E}{\operatorname{\mathsf{E}}}$ 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 and $\E\|F\|<\infty$, then $F$ 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 $\E F:=\lim_n EF_n$, with naturally defined $\E F_n$.

It is then a known and easy to check fact 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, Section V.4 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.