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There are two concurrent theories of measure/integration on a locally compact topological spaces: either as positive linear forms on the space of continuous functions with compact support, or as Borel measures. The theorem of Riesz establishes the relation between these two point of views, which are essentially equivalent: if one imposes the Borel measure to be inner-regular (the measure of any Borel set is the supremum of the measures of its compact subsets), then there is indeed a bijection.

However, this result is hard to find in the litterature (I found it in Fremlin's book, as well as in a 1975 paper of Pollard and Topsoe). Rudin's Real and complex analysis proves a variant where he requires outer-regularity. Of course, the main source of examples for a book like Rudin's consists in open subsets of $\mathbf R^n$, for which all Borel measures are automatically inner-regular. On the other hand, having a bijection is particularly nice in some contexts, such as locally compact groups. (Actually, in my own research, I have to construct measures on some non-metrizable locally compact spaces, and we do it by working on each compact set at a time, hence the inner-regularity point of view is both natural and necessary.)

After this presentation, here are my questions :

  1. What are the good reasons to favor at first the outer-regularity instead of having a nice Riesz theorem, and proving the outer-regularity when needed?

  2. What is a standard reference for the form of the theorem of Riesz that gives a bijection between positive linear forms and inner-regular Borel measures on a locally compact space?

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    $\begingroup$ Rudin proves uniqueness, see Thm 2.14 in his book $\endgroup$ – Asaf Oct 27 '14 at 18:26
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    $\begingroup$ And as for a reference, I would go for Bogachev's Measure Theory books (Riesz' theorem is covered in the second volume). His writing is meticulous, although very technical and I wouldn't recommend it as a textbook. $\endgroup$ – Asaf Oct 27 '14 at 18:49
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    $\begingroup$ @Asaf: You're absolutely right! I was so obsessed with the problem with inner-regularity that I did not remember that Rudin mentioned uniqueness. By the way, the proof of uniqueness is simple since the measure is determined by its values on compact sets... $\endgroup$ – ACL Oct 27 '14 at 21:43
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1 - There is no reason to favor outer regularity in the context of locally compact spaces.

Note however that all finite Borel measures on metric spaces are outer regular (but not necessarily inner regular, due to a shortening of compact subsets). This is used e.g. to show that Lipschitz bounded integrable functions are dense in the space of all integrable functions, an important result in analysis on metric spaces.

2 - Bogachev, measure theory, T II, Theorem 7.11.3.

Finally note that without separability assumption (say $\sigma$-finiteness), there may be an inner regular measure and a outer-regular measure associated to the linear form that are distinct (but coincide on compact sets). See the discussion after Theorem 9.11.2 in Bogachev. The measure obtained in the book of Rudin is the outer-regular one, it does not satisfy the full inner-regularity property (inner-regularity holds only for Borel sets of finite measure). When it comes to Haar measure, Hewitt and Ross build the outer regular one in their book whereas Bogachev builds the inner-regular (Radon) one (cf 9.11.1, 9.11.4).

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Rudin's definition of regular measure (outer regular on all measurable sets, compact-inner-regular on open sets) is better for generalizing all of the classical theorems about Lebesgue measure to arbitrary locally compact Hausdorff spaces. It gives a generalization of the Vitali-Carathéodory Theorem, which says that real-valued integrable functions can be arbitrarily approximated in $L^1$ norm above/below by upper/lower-semicontinuous functions.

This is mildly annoying, because inner regularity properties are definitely the right approach for more general topological spaces.

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  • $\begingroup$ Out of curiosity, what are the right $\endgroup$ – François G. Dorais Nov 10 '17 at 23:05
  • $\begingroup$ @FrançoisG.Dorais It appears your question got cut off. $\endgroup$ – Cameron Zwarich Nov 10 '17 at 23:06

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