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In Paul Halmos' Measure Theory book, section 53, he defines a content on a locally compact Hausdorff space to be a set function, $\lambda$ that is additive, subadditive, monotone, and $0\le\lambda(C)<\infty$ for all $C$ compact. The "Borel sets" he considers(section 52) in this book are the smallest $\sigma$-ring generated by the class of compact sets. The difference between $\sigma$-ring and $\sigma$-field is that the superset need not be contained. It can be shown, and is given as an exercise by Halmos, that the smallest $\sigma$-ring generated by compact sets is the same as the smallest $\sigma$-ring generated by open bounded sets. A bounded set is one that is contained in a compact set. Halmos then shows that a content induces a Borel Regular measure. He introduces the inner content and an outer measure in order to do so. Regular is the important word for this question. In fact, later in this chapter he proves the Riesz Representation Theorem, proving the existence of a Regular measure induced from positive linear functional of the continuous functions of compact support. In the next chapter, Halmos provides an existence proof of the Haar Measure by creating a left-invariant content, which then induces a Regular measure.

The issue I am having is that in almost every other source(books, wikipedia), the Haar Measure --and the Reisz-Rep measure which is related because one can prove the existence via a left or right invariant functional--is a Radon measure. A Radon measure is outer regular on Borel sets, finite on compact sets, and inner regular on open sets and Borel sets with finite measure. I do realize that in the other books, the Borel sets are the smallest $\sigma$-algebra generated by open sets.

The questions I have are:

1) Is the Borel $\sigma$-ring generated by compact sets in a locally compact Hausdorff space the same as the Borel $\sigma$-field generated by the open sets. I think the answer is no. One reason is a $\sigma$-ring and and the other is a $\sigma$-field. I don't even think that the $\sigma$-ring generated by compact sets is the same as the $\sigma$-ring generated by open sets. I think the best you can say is bounded open sets(mentioned above). Perhaps they are the same when the space is separable. Is this true? Is the $\sigma$-algebra generated by compact sets the same as the $\sigma$-algebra generated by the open sets the same.

2)Although this questions is more about basic measure theory than the Haar measure, in the end I want to show that the measure Halmos created is the same as the one constructed in most other books. In order to do this, I need to go from a measure defined on the $\sigma$-ring generated on compact sets to the $\sigma$-field generated by open sets. How do I do this?

Thank you in advance for all the help.

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up vote 3 down vote accepted

The situation is indeed a delicate one and one needs to carefully check the conventions before transferring a result from one context to another. The situation is summarized in Royden's Real Analysis, though with just a few examples.

For a given space $X$, the main players are:

  • $\mathcal{B}a$ — the σ-algebra generated by the compact Gδ sets. (The Baire algebra.)

  • $\mathcal{B}c$ — the smallest σ-algebra with respect to which all continuous real-valued functions are measurable.

  • $\mathcal{B}k$ — the σ-algebra generated by the compact sets.

  • $\mathcal{B}o$ — the σ-algebra generated by the closed sets. (The Borel algebra.)

  • $\mathcal{S}$ — the smallest σ-ring containing the compact sets.

  • $\mathcal{R}$ — the smallest σ-ring containing the compact Gδ sets.

In general, we have the inclusions $$\mathcal{B}a \subseteq \mathcal{B}k \subseteq \mathcal{B}o\quad\mbox{and}\quad\mathcal{B}a \subseteq \mathcal{B}c \subseteq \mathcal{B}o,$$ but $\mathcal{B}c$ and $\mathcal{B}k$ are not necessarily related. Moreover, $\mathcal{B}a = \mathcal{B}c \cap \mathcal{B}k$ and $\mathcal{B}o$ is generated by $\mathcal{B}c \cup \mathcal{B}k$. The σ-rings $\mathcal{S}$ and $\mathcal{R}$ consist of the σ-bounded elements of $\mathcal{B}a$ and $\mathcal{B}o$, respectively.

  • When $X$ is σ-compact and locally compact, then $$\mathcal{R} = \mathcal{B}a = \mathcal{B}c \quad\mbox{and}\quad\mathcal{S} = \mathcal{B}k = \mathcal{B}o.$$ A compact example showing the strict inequality is $\beta\mathbb{N}$.

  • When $X$ is metrizable, more generally when closed sets are Gδ, we have $$\mathcal{R} = \mathcal{S} \subseteq \mathcal{B}a = \mathcal{B}k \subseteq \mathcal{B}c = \mathcal{B}o.$$ An example showing strict inequality is the space of irrational numbers.

  • When $X$ is locally compact and separable, then we have $$\mathcal{R} = \mathcal{S} = \mathcal{B}a = \mathcal{B}k = \mathcal{B}c = \mathcal{B}o.$$

That said, the uniqueness (up to normalization) of Haar measure and it's (inner) regularity guarantee that all constructions will agree on their common domain of definition. Proving this from scratch does require some work depending on where you start and end.

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If your space X is sigma-compact (can be written as the union of a countable family of compact sets) then everything is okay, I think. This follows as X is in the Borel sigma-ring, and hence the Borel sigma-ring is a sigma-field (or sigma-algebra). If U is open, then $X\setminus U$ is closed, and so can be written as the union of a countable family of compacts (intersect each member of your initial countable family with $X\setminus U$). Thus $X\setminus U$, and so U, is in the Borel sigma-ring/field. So the two possible definitions of the Borel sigma-algebra agree.

Now, groups are rather nice in this regard. If G is a locally compact group, then pick some compact neighbourhood of the identity, say A. We may suppose that $A=A^{-1}$. Then let H be the union $\bigcup_{n\geq 1} A^n$ where $A^2 = \{ab:a,b\in A\}$ and so forth. Then H is an open subgroup of G, so H is also closed. Each $A^n$ is compact, so H is also sigma-compact. Thus the above applies to H. What's really nice is that the coset space $G/H$ carries the discrete topology, and so the Haar measure on G can be reconstructed from that of H and the counting measure on $G/H$. So, for example, $L^1(G) \cong \ell^1(L^1(H),G/H)$. In essential, for measure theory issues, you need only worry about H.

Note: Francois posted an answer as I was typing: he makes some similar points to my first paragraph.

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+1: Groups are much nicer than the general case. I wonder if the same argument generalizes to a useful class of uniform spaces. – François G. Dorais Jun 20 '10 at 18:56
Nice explanation. Minor quibble/question: in penultimate sentence fo you mean something like $\ell^1(G/H, L^1(H))$ - or have I got my notation mixed up? – Yemon Choi Jun 20 '10 at 19:51
@Yemon: Yeah: I can never remember is the indexing set comes first or second... I think your notation is more standard. – Matthew Daws Jun 20 '10 at 20:53

The answer to your first question is no. A not very interesting counterexample is given by the interval [0,1] together with the discrete topology. All sets are open, hence the Borel $\sigma$-algebra contains all the subsets of [0,1]. On the other hand, compact sets have a finite number of elements. So the $\sigma$-algebra they generate contains only sets that are countable, or with countable complements.

If you assume that your space is metric separable, then it's ok, as explained by F.Dorais in his answer. Let me add a few words with respect to regularity. First recall that a Polish space is a topological space which is homeomorphic to a complete separable metric space. Locally compact separable metric spaces are Polish. We have the following result due to Oxtoby and Ulam.

Theorem let X be a Polish space and $\mu$ a finite Borel measure on X. Then $\mu$ is inner regular: all Borel set can be written as a countable union of compact sets together with a set of zero measure.

As a result, a finite measure on a Polish space is uniquely determined by its values on compact sets.

The book by Halmos is very good but a bit outdated in my opinion. I would suggest instead the book of Donald L. Cohn, "measure theory", who deals, among other things, with the Haar measure.

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