Baire sets in locally compact Hausdorff spaces

Question

(This is a follow-up to Compact $G_\delta$ subsets of locally compact Hausdorff spaces.)

Suppose $X$ is a locally compact Hausdorff space. The Baire sets in $X$, denoted by $\mathcal Ba(X)$, comprise the smallest $\sigma$-algebra that contains all compact $G_\delta$ subsets of $X$. Let $F$ be a closed subset of $X$. Then of course $F$ is also locally compact and Hausdorff. Is $\mathcal Ba(F) = \{B \cap F \; | \; B \in \mathcal Ba(X)\}$?

Answer 1

Definition. $\mathcal A = \{B \cap F \; | \; B \in \mathcal Ba(X)\}$. $\mathcal B = \{B \subset X \; | \; B \cap F \in \mathcal Ba(F)\}$.

Theorem 1. $\mathcal A \subset \mathcal Ba(F)$.

Proof: It is easy to see that $\mathcal B$ is a $\sigma$-algebra that contains all compact $G_\delta$ subsets of $X$. Therefore, $\mathcal Ba(X) \subset \mathcal B$, which from the definitions implies the desired result.

Before proving the reverse inclusion, we establish some lemmas.

Definition. Let $f$ be a real-valued function on $X$. The support of $f$, denoted by supp $f$, is the closure of $\{x \in X \; | \; f(x) \neq 0\}$.

Lemma 1. Let $K$ be a compact subset of $X$. Then there is a continuous function $f: X \to [0,1]$ for which $f = 1$ on $K$ and such that $f$ has compact support.

I will not prove this, as it is a standard result.

Lemma 2. Let $K$ be a compact subset of $X$, and let $O$ be an open subset of $X$ with $K \subset O$. Then there is a continuous function $f: X \to [0,1]$ for which $f = 1$ on $K$, $f = 0$ on $X \setminus O$, and such that $f$ has compact support contained in $O$.

Proof: $O$ is a locally compact Hausdorff space, so by Lemma 1, we obtain $f: O \to [0,1]$ for which $f = 1$ on $K$ and such that $f$ has compact support. Define $g = f$ on $O$ and $g = 0$ on $X \setminus O$. Then $g$ has the required properties.

Lemma 3. Let $K$ be a compact subset of $X$, and let $O$ be an open subset of $X$ with $K \subset O$. Then there is a compact $G_\delta$ subset $H$ of $X$ with $K \subset H \subset O$.

Proof: Obtain $f$ satisfying Lemma 2. Define $H = \{x \; | \; f(x) = 1\}$. Clearly $H$ is a closed subset of the compact set supp $f$, so is itself compact. It is also clear that $K \subset H \subset O$. $H = \cap_1 ^ \infty \{x \; | \; f(x) > 1 - 1/n \}$, so it is a $G_\delta$.

Theorem 2. $\mathcal Ba(F) \subset \mathcal A$.

Proof: It is easy to see that $\mathcal A$ is a $\sigma$-algebra of subsets of $F$.

Let $K$ be a compact $G_\delta$ subset of $F$. Then $K = \cap_1 ^ \infty O_n \cap F$, where each $O_n$ is open in $X$. Use Lemma 3 to obtain compact $G_\delta$ subsets $H_n$ of $X$ with $K \subset H_n \subset O_n$. We have $K \subset \cap_1 ^ \infty H_n \cap F \subset \cap_1 ^ \infty O_n \cap F = K$, whence $K = \cap_1 ^ \infty H_n \cap F$. Now each $H_n$, being a compact $G_\delta$ in $X$, is in $\mathcal Ba(X)$, so $\cap_1 ^ \infty H_n$ is also in $\mathcal Ba(X)$. From the definition, $K \in \mathcal A$.

Thus, $\mathcal A$ contains all compact $G_\delta$ subsets of $F$. The result follows.

Answer 2

As noted it suffices to show: every compact $G_\delta$ set $K \subseteq F$ belongs to $\{B \cap F\;|\;B \in \mathcal{Ba}(X)\}$ ...

How about this lemma:
Suppose $K \subseteq U \subseteq X$, where $K$ compact, $U$ open. Then there exist $K_1$ compact, $U_1$ open with $K \subseteq U_1 \subseteq K_1 \subseteq U$.