Regularity of Dirac measure on Baire sets

Question

Suppose $X$ is a locally compact Hausdorff space. Define the Baire sets in $X$, denoted by $\mathcal Ba(X)$, to be the smallest $\sigma$-algebra that contains all compact $G_\delta$ subsets of $X$. Let $x \in X$. Define the Dirac measure $\delta_x$ on $\mathcal Ba(X)$ by $\delta_x(E)$ = 1 if $x \in E$ and 0 otherwise. Must Dirac measure be regular on the Baire sets?

Answer 1

Let $\mathcal A_1$ be the collection of all sets $E \subseteq X$ such that there exists a compact $G_\delta$ set $K$ with $x \in K \subseteq E$. Let $\mathcal A_2$ consist of the complements $E^c = X \setminus E$ for $E \in \mathcal A_1$. Let $\mathcal A = \mathcal A_1 \cup \mathcal A_2$.

claim $\mathcal A$ is a sigma-algebra. Proof:

$X \in \mathcal A$ since $X \in \mathcal A_1$

countable unions: let $E_n \in \mathcal A$ for $n=1,2,3,\dots$. If $A_m \in \mathcal A_1$ for some $m$, then there is compact $G_\delta$ set $K$ with $x \in K \subseteq A_m \subseteq \bigcup A_n$, so $\bigcup A_n \in \mathcal A_1$ and $\bigcup A_n \in \mathcal A$. On the other hand, if $A_m \in \mathcal A_2$ for all $m$, then there exist compact $G_\delta$ sets $K_m$ with $x \in K_m \subseteq E_m^c$ for all $m$, and then $K=\bigcap K_n$ is a compact $G_\delta$ set with $x \in K \subseteq \bigcap A_n^c = (\bigcup A_n)^c$, so $\bigcup A_n \in \mathcal A_2$ and $\bigcup A_n \in \mathcal A$.

complements: $E \in \mathcal A_1 \Longleftrightarrow E^c \in \mathcal A_2$.

claim $\mathcal A$ contains all compact $G_\delta$ sets.
Let $L$ be a compact $G_\delta$ set. If $x \in L$, then $L \in \mathcal A_1$ so $L \in \mathcal A$. On the other hand, suppose $x \not\in L$. By a Lemma*, there exists a compact $G_\delta$ set $K$ with $x \in K \subseteq L^c$. So $L \in \mathcal A_2$ and $L \in \mathcal A$.

conclude $\mathcal{Ba}(X) \subseteq \mathcal A$.

Now let $E$ be any Baire set. So $E \in \mathcal A$. First assume $x \not\in E$. Then $E \in \mathcal A_2$ and $E^c \in \mathcal A$, so there is a compact $G_\delta$ set $K$ with $x \in K \subseteq E^c$. But $K^c$ is an open Baire set, $\varnothing \subseteq E \subseteq K^c$ with $0=\delta_x(\varnothing)\le \delta_x(E)\le\delta_x(K^c)=0$. On the other hand, assume $x \in E$. Then $E \in \mathcal A_1$, so there is a compact $G_\delta$ set $K$ with $x \in K \subseteq E$. This time, $X$ is an open Baire set, and $1 = \delta_x(K) \le \delta_x(E) \le \delta_x(X) = 1$.

*${}$ Lemma If $U$ is open and $x \in U$, then there is a compact $G_\delta$ set $K$ with $x \in K \subseteq U$.

Answer 2

I have developed another approach using well-known results:

Every Baire measure on a $\sigma$-ring is regular (see Halmos, $\textit {Measure Theory}$, p. 228). The $\sigma$-algebra generated by a $\sigma$-ring consists of the sets in that $\sigma$-ring and their complements (standard result). It follows that if $E \in \mathcal Ba(X)$ then either $E$ or $X \setminus E$ is regular. From this we see that any finite measure on $X$ is outer regular. To prove that Dirac measure is inner regular, let $E$ be a Baire set. If $x \notin E$, the empty set fills the bill. Otherwise, use the outer regularity of $X \setminus E$ to obtain a closed Baire set $F$ with $x \in F \subset E$. Use the Lemma in the answer above to obtain a compact $G_\delta$ set $K \ni x$. Then $H := F \cap K$ is a compact Baire set with $x \in H \subset E$.