$\newcommand{\ba}{\mathcal{Ba}}$I understand that there are different used definitions, so here are mine:
The Baire $\sigma$-algebra $\ba(X)$ on a topological space, $X$, is the smallest $\sigma$-algebra on $X$ for which all the compactly supported continuous real valued functions on $X$, $C_c(X)$ are measurable. If $A\subseteq X$ has $A\in\ba(X)$, then $A$ is a "Baire set".
The motivation for this lemma is that I need it in order to show that every Baire measure on a $\sigma$-compact space is regular, which is the culmination of a chapter of topological measure.
If $X$ is a locally compact Hausdorff space (LCH) it can be shown that $\ba(X)$ is the smallest $\sigma$-algebra containing the compact $G_\delta$ sets of $X$. I know further that LCH spaces possess the property that you can always extend the characteristic function of a compact set to get a function in $C_c(X)$; you can even do this with a "partition of unity". Further, for every compact $G_\delta$ set $K$ in a LCH space $X$ there is an $f\in C_c(X)$ with $K=f^{-1}\{1\}$. I do not have any more knowledge than this, from the text I am following - Royden and Fitzpatrick, $4$th edition - so I presume that the above is sufficient to complete the exercise.
I need to show the following as an exercise:
Let $X$ be an LCH space, and $F\subseteq X$ a closed Baire set. Show that: $$\forall A\subseteq F:A\in\ba(F)\iff A\in\ba(X)$$
So I know that $F$ forms a LCH subspace as it is closed. I have shown further that: $$\mathcal{F}=\{A\subseteq F:A\in\ba(X)\}$$Forms a $\sigma$-algebra on $F$, where it is closed under relative complementation with $F$ since $F\in\ba(X)$. The result is the same as saying $\mathcal{F}=\ba(F)$. The problem:
- Since $F$ is closed, we cannot say that a set is open in $F$ iff. it is open in $X$, so the $G_\delta$ sets of $F$ are not necessarily the $G_\delta$ sets of $X$.
It is very unclear to me how to make the argument. I have confused myself heavily on whether or not it is sufficient to show that every $G_\delta$ subset of $F$, in the $F$-topology, is a $G_\delta$ subset of $F$ in the $X$-topology; or whether or not it is sufficient to show that every $f\in C_c(F)$ is $\mathcal{F}$-measurable.
I think that, for an $f\in C_c(F)$, since its support is entirely contained in $F$ the continuous extension to the $0$ function outside of $F$ will yield an $f_0\in C_c(X)$ and $f_F^{-1}(\mathcal{O})=f_0^{-1}(\mathcal{O})$ for any open $\mathcal{O}\subset\Bbb R\setminus\{0\}$ and this will be an open set in $X$ and in $F$ as it will be a pure subset of $F$; likewise will $f^{-1}\{0\}$ be closed in $X$, but I am not convinced that in general $f$ will be continuous in the $X$-topology for those problem sets $\mathcal{O}\ni0$. Then I cannot confirm that $f\in C_c(F)$ will be $\mathcal{F}$-measurable without further assistance.
Any help would be appreciated.
$\newcommand{\o}{\mathcal{O}}\newcommand{\f}{\mathcal{F}}\newcommand{\ba}{\mathcal{Ba}}$The bullet point I marked as a "problem" is not actually a problem:
Let $K$ be compact and $G_\delta$ in $X$, $K=\bigcap_{n\in\Bbb N}\o_n$ for $\o_n$ open in $X$. Then:
$$K\cap F=\bigcap_{n\in\Bbb N}(F\cap\o_n)$$And althought $F\cap\o_n$ is not open in $X$, it is open in $F$ for each $n$ by definition of subspace topology. Then $K\cap F$ is a $G_\delta$ set in $F$ - it is compact since $F$ is closed. Likewise, now let $K$ be compact and $G_\delta$ in $F$. We can express:
$$K=\bigcap_{n\in\Bbb N}(F\cap\o_n)$$For some $\o_n$ open in $X$, by definition of subspace topology. We know $K$ is compact in $X$ as well, since $F$ is closed. If $K=F$, then $K=F=\bigcap_{n\in\Bbb N}\o_n$ and $K=F$ is $G_\delta$ in $X$. IF $K\subset F$ strictly, then there will be an $N\in\Bbb N$ for which $n\gt N$ has $\o_n\subset F$ and $F\cap\o_n=\o_n$. Then $K$ is $G_\delta$ in $X$ with $K=\bigcap_{n\gt N}\o_n$ (we may wlog take the sequence $\{\o_n\}$ as descending).
$\ba(F)$ is generated by the compact $G_\delta$ sets in $F$. From the above, we know every such set is the intersection $K\cap F$ for a compact $G_\delta$ set in $X$. Then the collection of all such $(K\cap F)$ will generate the $\sigma$-algebra of $\ba(F)$, but we know that $\f$ contains all such sets $K\cap F$ and is a $\sigma$-algebra (on $F$) itself, so by minimality $\f=\ba(F)$, since $F\in\ba(X)$. $\blacksquare$