8
$\begingroup$

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.

  • 0
    The reason why one uses these slightly unconventional notions of measurability is to have a sensible theory of integrating vector-valued functions. The only functions we really know how to integrate sensibly are simple functions, or maybe countably-valued functions. The integral for more complex functions is obtained by approximating them with simple functions. But I#m sure there are people more experienced with vector integration who can give you a detailed answer.2012-03-22

2 Answers 2