TITLE: Definitions of measurability for operator-valued functions
QUESTION [8 upvotes]: If $\mathfrak{X}$ is a Banach space, a function $T: \mathbb{R} \to \mathcal{L}(\mathfrak{X})$ is defined to be 

uniformly measurable if it is an a.e. norm limit of a sequence of countably valued functions from $\mathbb{R}$ to $\mathcal{L}(\mathfrak{X})$
strongly measurable if, for each $x \in \mathfrak{X}$, $t \mapsto T(t)x$ is an a.e. norm limit of a sequence of countably valued functions from $\mathbb{R}$ to $\mathfrak{X}$
weakly measurable if, for each $x \in \mathfrak{X}$ and $\ell \in \mathfrak{X}^*$, $t \mapsto \ell(T(t) x)$ is a measurable function from $\mathbb{R}$ to $\mathbb{C}$

If I had to guess from the terminology, I would have said that uniform, strong, and weak measurability are just measurability with respect to the Borel $\sigma$-algebras generated by the uniform, strong, and weak topologies on $\mathcal{L}(\mathfrak{X})$.  Would that be equivalent?  Or are all those Borel $\sigma$-algebras the same for some reason that I'm not seeing?  Or is the resulting integration theory not as satisfactory?  The books I'm reading (Hille/Phillips and Dunford/Schwartz) don't have any discussion of why we don't define measurability in (what seems to me) the natural way.

REPLY [3 votes]: In the context of Bochner integration, strong measurability is the natural definition for measurability, because the Bochner integral is defined as the limit of simple/countably valued functions. I think weak measurability is more natural in the context of the Pettis integral, but is also a handy tool in proving strong measurability. Weak and strong measurability are related by Pettis' measurability theorem.
I think what you would call the natural definition is Borel measurability with respect to the norm topology. For what I know, this relates to strong measurability the same way as weak measurability does:
A function $f: \Omega \to X$ is strongly measurable if and only if $f$ is seperably valued and for all $B \in \mathcal B(\Omega)$ we have that $f^{-1} (B)$ is measurable.
Here $X$ is a Banach space and $\Omega$ is a measurable space. Ryan gives a nice overview in the book "Introduction to Tensor Products of Banach spaces", but the proof is erroneous. These lecture notes are the only reference I could find where this proposition is proved. I did not check this proof thoroughly yet.
So strong measurability is not Borel measurabilty with respect to the strong topology.