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I have been trying to get an understandable, self-contained proof about the regularity of Baire measures in locally compact spaces. Failing to follow Royden's proof - which required the lemma I asked about in this question - I found this article on JSTOR which was promising and slightly more general.

In all that follows, the Baire $\sigma$-algebra is the smallest $\sigma$-algebra in which the compactly supported continuous real valued functions on a set are measurable. However, since I am asking only about locally compact Hausdorff (LCH) spaces, we may just consider the Baire algebra as that generated by all compact $G_\delta$ sets - $G_\delta$ meaning the countable intersection of open sets - as it can be shown they are equal in this instance. A Baire measure is a measure on the Baire algebra which is finite on all the compact sets.

The JSTOR article proves (at first) that:

For any paracompact LCH space $X$ and a Baire measure $\mu$ thereon, $\mu$ is outer regular.

Reading their proof, I fall at the first hurdle:

Let $E$ be a Baire subset of $X$ [...] outer regularity is well-known if $E$ is $G_\delta$ or if $\mu(E)=\infty$. Hence we may assume that $\mu(E)\lt\infty$ and that $A:=X\setminus E$ is $\sigma$-bounded.

$\sigma$-bounded means "contained in a countable union of compact sets". Naturally if $X$ is itself $\sigma$-compact this is obvious, but why should $A$ be $\sigma$-bounded in general?

I tried to reach a contradiction - if $A$ is not $\sigma$-bounded, then $\mu(E)=\infty$, or perhaps that if $A$ is not $\sigma$-bounded, $E$ is evidently outer regular.

$A$ not being $\sigma$-bounded implies that any countable union of compact sets will not cover $A$. In particular, for any collection of compact sets $\{K_n:n\in\Bbb N\}$ for which $\bigcup_{n\in\Bbb N}K_n\subseteq A$, $E\subseteq\bigcap_{n\in\Bbb N}(X\setminus K_n)$. This does not seem to imply outer regularity however, since each of the sets $X\setminus K_n$ may well have infinite $\mu$-measure and we don't have equality, so I'm not sure if we can say that $E$ is $G_\delta$, only that $E$ is a subset of a $G_\delta$ set.

What is the obvious "without loss of generality" principle that I am missing?

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  • $\begingroup$ In fact they say it’s well-known for $E\in \mathcal{R}_a$; they even mention that right at the beginning of the paper quoting Halmos. $\endgroup$ Mar 5, 2022 at 13:27

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Lemma 5 in Royden real analysis (3rd ed) on page 333 says: if $E$ is Baire then $E$ or $X\setminus E$ is $\sigma$-bounded. If $E$ were $E$ would be in $\mathcal{R}_a$ (lemma 6 same page) and that case was already discarded as known (with an appeal to an earlier edition of Royden).

FYI the paper is freely available here (no JSTOR needed). Free archives FTW!

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  • $\begingroup$ I am on the $4$ edition, and sadly there is no mention of this that I've seen (I've read pretty much every page before this - this is chapter $20$ I am on, page $480$). Side-note: are compact Baire sets outer-regular? I know that they form a regular subspace, but are they themselves regular in the parent space? $\endgroup$
    – FShrike
    Mar 5, 2022 at 13:11
  • $\begingroup$ @FShrike yes compact Baire sets are $G_\delta$ so quite trivially outer regular. $\endgroup$ Mar 5, 2022 at 13:12
  • $\begingroup$ compact + Baire implies $G_\delta$? This has not been mentioned in the text thus far. Yes, the compact $G_\delta$ sets generate the Baire algebra but in the converse this is not clear in the absence of a metric. Am I supposed to show that if it is not $G_\delta$, it cannot be Baire? $\endgroup$
    – FShrike
    Mar 5, 2022 at 13:18
  • $\begingroup$ @FShrike yes it’s in both Royden and Halmos. The proofs are not from contradiction but rather clever. Look them up. Your library should have them. $\endgroup$ Mar 5, 2022 at 13:22
  • $\begingroup$ Well, thank you for noting Halmos’ text. I found an online copy and was able to get the necessary details from that. Amusingly, in my failed attempts to google a self-contained answer online, the top searches were always comments from you yourself on “ask a topologist”! $\endgroup$
    – FShrike
    Mar 5, 2022 at 15:36

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