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This is intended to be a really soft question, which might be considered bad for this site. Let me know if that's true. I'll try to make this question as "answerable" as possible.

I've been reading measure and integration theory on Analysis III by H. Amann & J. Escher (Chapter IX and X). This text is famous for its characteristic that everything is presented in the greatest generality. For example, he talks about convergence on Banach spaces right at the beginning.

With this being said, it is not surprising that the text develops the theory of Bochner integrals first, even before Lebesgue integrals. However, what really annoys me is that, at the beginning of each section he says: In the following, let $(X,\mathcal{A},\mu)$ be a complete, $\sigma$-finite measure space. This goes against the spirit of the book! Moreover, I'm utterly confused when the Wikipedia article on Bochner integrals does not assume completeness and $\sigma$-finiteness, and yet it seems that many important results remains true without that assumption.

Now my question is this. What do we lose if we assume all measures are complete and $\sigma$-finite? I know that every measure has a completion, so that probably explains why we assume completeness. But what about $\sigma$-finiteness? Is it just that non-$\sigma$-finite just rarely exist? (This situation is much like topology: We often assume our topological spaces are Hausdorff without regret, because non-Hausdorff spaces are just too pathological.)

To make this question more concrete, let me elaborate:

  1. Do we often encounter non-$\sigma$-finite measures in higher analysis? I fear the theorems here are not general enough for later use. Is that true?
  2. Are there any other books that, like Amann, develops Bochner integrals in detail and presents results without assuming completeness and $\sigma$-finiteness? I've looked into several standard real analysis texts so far, and they either treat it as an exercise (Folland), or in an appendix (Cohn), or omit it altogether. I also found the book Topics in Banach space integration by Schwabik, but unfortunately this book simply assumes $X$ to be a cube in $\mathbb{R}^n$...
  3. What significant results are not true or are much harder to prove if measures are not complete or $\sigma$-compact? I already know Fubini's theorem is one example. (This might be too vague since it's impossible to list all such results. But maybe just name a few?)

Thanks in advance!

Yuxiao Xie
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  • For 1.: The $m$-dimensional Hausdorff measure in $\mathbb{R}^n$ with $m<n$ is not $\sigma$-finite and is used very frequently. Here is a wiki-link for this measure: https://en.wikipedia.org/wiki/Hausdorff_measure – humanStampedist Oct 04 '18 at 14:33
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    As Folland pointed out in his textbook, the usual measurability of a function in real analysis means "Borel"-measurable, not Lebesque measurable, even if we use Lebesque integral for its integration. Here, Borel measurable implies Lebesgue measurable but not vice versa, and the composition of Lebesgue measurable functions is not Lebesgue measurable. – Jae Young Lee Feb 15 '19 at 23:55

1 Answers1

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I asked this question four months ago. Now that I return to it, many doubts that I had then have disappeared. So I'll write a brief answer below, an answer that I wished to get.

  1. $\sigma$-finiteness is important, as many theorems depend on this property. Non-$\sigma$-finite measure may be too pathological in some sense, but they cannot be avoided altogether. The situation is a bit like Hausdorffness in general topology; most topologies that arise in practice are Hausdorff, but still some authors don't impose this condition.

  2. S. Lang's Real and Functional Analysis treats Bochner integrals right from the beginning (without assuming $\sigma$-finiteness or completeness), and his treatment is almost identical to that of Amann. However, once the usual theory of Lebesgue integration of real-valued functions is established, we can derive properties of Bochner integrals therefrom. Actually, vector-valued integration is typically treated in functional analysis, where Bochner integral is the strong kind and there is also a weak kind called Gelfand–Pettis integral. Maybe that should be the right context for this topic.

I hope this is helpful for people who have the same doubts when reading the text by Amann & Escher.

Yuxiao Xie
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    I have posted a complete proof of Fubini's theorem for Banach spaces here. Amann proves Fubini's theorem for only Lebesgue measure on $\mathbb R^n$, while my proof is for general complete $\sigma$-finite measures. – Akira Nov 16 '21 at 10:16
  • The need for $\sigma$-finiteness is already mentioned in your post. I want to emphasize that the completeness of measure is used first in this essential result, which later leads to Fubini's theorem. – Akira Nov 16 '21 at 10:16
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    @Akira Oh another two years have passed. :) Looking back, I feel like after all it was not really worth it worrying about all those technicalities. I for one have forgotten most of those conditions, but it doesn't really matter as long as you're not studying pure measure theory. – Yuxiao Xie Nov 21 '21 at 19:25
  • It's great to see you too :v – Akira Nov 21 '21 at 19:39