Baire $\sigma$-algebra is generated by closed $G_{\delta}$ sets Let $X$ be a topological space, and let $C(X)$be the set of all real-valued continuous funtions on $X$. Then $B_{o}(X)$, the Baire $\sigma$-algebra on $X$ is the smallest $\sigma$-algebra on $X$ that makes each function in $C(X)$ measurable. I wonder how to prove Baire $\sigma$-algebra is generated by closed $G_{\delta}$ sets.
Essentially one has to show two things: (1) that every closed $G_\delta$ subset of $X$ is an element of $B_0(X)$, and (2) that if any $\sigma$-algebra $A$ on $X$ has every closed $G_\delta$ subset of $X$ as an element, then $B_0(X)\subseteq A$. But I have trouble in showing either of them.
 A: What if your assertion is false?  The Baire sigma-algebra $B_0(X)$ is generated by the zero sets of continuous functions.  We have  
$\{\text{compact }G_\delta\} \subset \{\text{zero set}\} \subset \{\text{closed }G_\delta\}$  
neither reversible in general.  So we should find an example of a (completely regular Hausdorff) space where some closed $G_\delta$ is not a Baire set.  
added, making it a solution
(I looked in $\pi$-Base for "completely regular + not normal" to get some candidates.  I liked this one.)  
Our space $X$ is Thomas' Plank, described in pp. 113-114 of Counterexamples in Topology by Steen & Seebach.
We start with the set $[0,1)$ [but basically what we need is an uncountable set with a distinguished point $0$].  Let $X = \bigcup_{i=0}^\infty L_i$ where $L_0 = \{(x,0): x \in (0,1)\}$ and for $i \ge 1$,
$L_i = \{(x,1/i) : x \in [0,1)\}$.  So $L_0$ has no distinguished point, but the others do.  Description of the topology: If $i > 0$ and $x \ne 0$, then the singleton $\{(x,1/i)\}$ is open.  If $i > 0$ then basis neighborhoods for $(0,1/i)$ are subsets of $L_i$ with finite complements.  Basis neighborhoods for $(x,0)$ are the sets
$$
U_n(x,0) = \{(x,0)\} \cup \{(x,1/i) :  i > n\} .
$$
From Steen & Seebach: $X$ is completely regular, Hausdorff, but not normal.  Our example is adapted from their proof that $X$ is not normal.  
First, some information about continuous functions $f : X \to \mathbb R$.  Fix $i \ge 1$; then by continuity at the point $(0,1/i)$, we see that the sets $\{(x,1/i) : |f(x,1/i) - f(0,1/i)| > 1/n\}$ are cofinite for $n > 0$, and therefore $\{(x,1/i) : f(x,1/i) \ne f(0,1/i)\}$ is cocountable.  Thus, the restriction of $f$ to $L_i$ is constant except for a countable set.  And on $L_0$ the restriction of $f$ is the pointwise limit of the restrictions to $L_i$, so also the restriction of $f$ to $L_0$ is constant except for a countable set.
Now we deduce some information about the zero-set of a continuous function $f$.  The intersection with each $L_i$ is either countable or co-countable.  The set
$$
\{A \subseteq X : \text{ for all }i, A \cap L_i \text{ is either countable or cocountable}\}
$$
is a sigma-algebra, contains all zero sets, therefore contains all Baire sets.    
Now let $E \subset (0,1)$ be uncountable, with uncountable complement.  For example $E = (0,1/2)$.  Let $F = \{(x,0) \in L_0 : x \in E\}$ so that both $F$ and $L_0 \setminus F$ are uncountable.  From our description above, we know $F$ is not a Baire set.
But I claim that $F$ is a closed $G_\delta$ set.  The complement
$$
F^c = \bigcup_{i=1}^\infty L_i \cup \{(x,0) \in L_0 : x \notin E\}
$$
is open; indeed, from the description of the basis for the topology, each point of $F^c$ has a neighborhood contained in $F^c$.  So $F$ is closed.  
Next, fix $n \ge 1$.  The set
$$
F_n = F \cup \bigcup_{i=n}^\infty L_i
$$
is open.  And $F = \bigcap_{n=1}^\infty F_n$ is therefore a $G_\delta$ set.
