Integration of vector function against vector measure

Question

Let $X,Y,Z$ be Banach spaces and let $m\,:\,X\times Y\to Z$ be a bilinear map such that $\|m(x,y)\|\leq C \|x\|\|y\|$ for some fixed constant $C$. Moreover, let $\mu$ be a Borell vector measure on $\mathbb{R}^d$ valued in $Y$ with finite total variation and let $f\,:\,\mathbb{R}^d\to X$ be bounded and measurable (in the sense that an inverse image of a Borel set is Borel). Can one make sense of the following integral $$ \int_{\mathbb{R}^d} m(f(u),\mu(du)) \in Z $$ (possibly under some extra conditions like the condition that the Banach spaces are separable).

Answer 1

$\newcommand\R{\mathbb R}$Suppose that $X$ is separable. Let $$\int_{\R^d}m(x\,1_B,d\mu):=m(x,\mu(B))$$ for all $x\in X$ and Borel $B\subseteq\R^d$. Then, for simple functions $$s_n:=\sum_{j=1}^n x_j\,1_{B_j},$$ with $x_j\in X$ and Borel $B_j\subseteq\R^d$ for all $j$, let $$\int_{\R^d}m\big(s_n,d\mu\big):=\sum_{j=1}^k m(x_j,\mu(B_j)).$$

Then, by approximation, extend this notion of integral to bounded measurable functions $f\colon\R^d\to X$. Here, as in the theory of the Bochner integral, we can use the Pettis theorem implying that a measurable function with values in a separable Banach space is a $\|\mu\|$-almost everywhere limit of a sequence of measurable simple functions, where $\|\mu\|$ is the variation measure for $\mu$ -- cf. e.g. Sections V.4 and V.3 of K. Yosida's Functional Analysis, Sixth Edition.