Checking a separability condition for Bochner measurability

Question

Equip $[0,\infty)$ with the Borel measure $\mu$. I am looking at an integral of the following form:

$$\int_0^\infty R(x)\,dx(\mu),$$

where $R:[0,\infty)\rightarrow B(H,K)$ is a continuous family of bounded linear operators between Hilbert spaces $H$ and $K$. Here $R(x) = (T+x)^{-1}$ is in fact a family of resolvent operators.

I understand that for this integral to make sense, the function $R$ first needs to be strongly $\mu$-measurable in the sense of Bochner integrals. By Pettis' Theorem, this means $R(x)$ needs to be $\mu$-almost-everywhere separably valued.

Question: Is it true that $R$ is $\mu$-almost-everywhere separably valued? That is, is the image of $R$ contained in a closed separable subspace of $B(H,K)$?

(Or maybe there are other tricks one can use to show measurability for functions taking values in $B(H,K)$, or perhaps non-separable $C^*$-algebras?)

Cheers!

Answer 1

This may help; Let $C$ be a measure space, $X$ a Banach space. A countably valued function $x: C\rightarrow X$ is Bochner Integrable if the map $$a - \|x(a)\|$$ is Lebesgue measurable and the inverse image under $x$ of each element in the range of $x$ is a measurable set in $C$.

A general function $x:C\rightarrow X$ is Bochner integrable if and only if there is a sequence $\{x(n)\}$ of countably valued Bochner integrable functions converging pointwise to $x$ in the following way;

1. $\|x(n)(a)->x(a)\|\rightarrow 0$ for almost all $a$ and
2. $\lim \int \|x(a) - xn (a)\|\, dm = 0$

($m = $ Lebesgue measure on $\mathbb{R}$)