# Does locally compact and paracompact imply $X\setminus E$ is contained in a countable union of compact sets, when $E$ has finite (Baire) measure?

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?

• 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. Mar 5, 2022 at 13:27

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).
• 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? Mar 5, 2022 at 13:11
• @FShrike yes compact Baire sets are $G_\delta$ so quite trivially outer regular. Mar 5, 2022 at 13:12
• 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? Mar 5, 2022 at 13:18