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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.

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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. –  Michael Greinecker Mar 22 '12 at 1:56

2 Answers 2

"Uniformly measurable" is the same as "Bochner measurable" when $L(X)$ is considered to be a Banach space. But since $L(X)$ is (usually) non-separable in the norm, this is not "measurability" with respect to the norm sigma-algebra. Similar remarks for strongly measurable. For weakly measurable: this definition says the inverse images of certain subbasic (and basic) open sets are measurable, but of course there is no reason to think the inverse images of uncountable unions of basic open sets are measurable. And (since you don't need them to be measurable anyway to do integration) why would you want to add that requirement?

If you try doing things using your so-called "natural" way of defining measurability, I think you will be in big trouble very soon!

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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.

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How is Ryan's proof of the Pettis measurability theorem erroneous? Are you talking about a minor slip or a major goof (I couldn't find either at a quick glance)? –  t.b. Aug 1 '12 at 11:09
    
It was discovered in a seminar I attended about Ryan's book that there is a small error in the proof of (ii) implies (iii) which makes the reasoning brake down. To be honest, I cannot remember exactly what it was. I will take a look again and try to contact the one who discovered it. –  Ron Aug 1 '12 at 14:01
    
Thanks. I agree that there's a leap in the argument where he claims that $f^{-1}(U) \in \Sigma$ for all weakly open sets. Note that what he really needs for his argument to work is that pre-images of closed balls are measurable and that is easy to prove: Since $Y$ is separable, there is a countable family $\varphi_n$ of functionals such that $\|y\| = \sup_{n} |\varphi_n(y)|$ for all $y$ and this yields that claim easily and the end of the argument goes through. Thanks for pointing that out. –  t.b. Aug 1 '12 at 15:11
    
By the way: a great reference for vector-valued integration is Diestel-Uhl's Vector measures and the Pettis measurability theorem is proved on page 42. The original reference is B.J. Pettis, On integration in vector spaces, Trans. AMS 44 (1938), 277-304. –  t.b. Aug 1 '12 at 15:45
    
Thanks for the fix of the proof and the tip for the book! I noted that Diestel uses a finite measure space, where Hille-Phillips does not make this requirement. –  Ron Aug 9 '12 at 13:46

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