Do the Baire Sets of $\mathbb{R}$ contain the Borel Sets of $\mathbb{R}$? I'm trying to gain some intuitive characterization of the Baire subsets of $\mathbb{R}$.  I've seen different definitions of Baire subsets (which I presume are equivalent) but here is a characterization from Royden:
The Baire subsets of $X$ (denoted $\mathcal{B}$) are equivalent to the smallest $\sigma$-algebra generated by $\{G_\delta\}$ where $G_\delta$ is a countable intersection of open subsets of $X$.
If $X = \mathbb{R}$, wouldn't then $\mathcal{B}$ contain the Borel subsets of $\mathbb{R}$ since any open set $O \subseteq \mathbb{R}$ is also a $G_\delta$ so that since the open sets of $\mathbb{R}$ generate the Borel Sets of $\mathbb{R}$, then $\mathcal{B}$ would at least contain the Borel subsets of $\mathbb{R}$? 
 A: With your definition of the "Baire sets" $\mathcal{B}'$ as the $\sigma$-algebra generated by the $G_\delta$-sets, what we obtain is the $\sigma$-algebra $\mathcal{B}$ of Borel sets. The reason is that $\mathcal{B}'$ contains the open sets, as you observed, so $\mathcal{B}' \supseteq \mathcal{B}$ and since the Borel $\sigma$-algebra $\mathcal{B}$ contains the open sets and is closed under taking countable intersections it follows that $\mathcal{B}$ contains all $G_\delta$-sets, so $\mathcal{B} \supseteq \mathcal{B}'$. Therefore $\mathcal{B} = \mathcal{B}'$ and it would make no sense to distinguish the two concepts.
The usual definition of the Baire $\sigma$-algebra $\mathcal{Ba}$ (in a locally compact Hausdorff space $X$) is the $\sigma$-algebra generated by the compact $G_\delta$-sets, or, equivalently, the $\sigma$-algebra making all continuous functions with compact support measurable. Since the $\sigma$-algebra $\mathcal{B}$ of Borel sets contains all closed sets, it contains all compact subsets of a Hausdorff space, so we always have the inclusion $\mathcal{B} \supseteq \mathcal{Ba}$.
With the first description we can see that in $\mathbb{R}$ the Borel sets and the Baire sets coincide: compact intervals $[a,b]$ are countable intersections of open intervals: $[a,b] = \bigcap_{n=1}^\infty\left(a-\frac{1}{n}, b+\frac{1}{n}\right)$, so every compact interval is a compact $G_\delta$, hence $[a,b] \in \mathcal{Ba}(\mathbb{R})$ whenever $a \leq b$. Furthermore, every open subset of $\mathbb{R}$ is a countable union of open intervals and every open interval $(a,b)$ can be written as $(a,b) = \bigcup_{n=1}^\infty \left[a+ \frac{1}{n}, b-\frac{1}{n}\right]$ which shows that $(a,b) \in \mathcal{Ba}(\mathbb{R})$ and hence $\mathcal{Ba}(\mathbb{R})$ contains all open subsets of $\mathbb{R}$, so $\mathcal{Ba}(\mathbb{R}) \supseteq \mathcal{B}(\mathbb{R})$.
This easily generalizes to $\mathbb{R}^n$ and, more generally, $\mathcal{Ba}(X) = \mathcal{B}(X)$ holds in every second countable locally compact Hausdorff space.
Here are two simple examples you should think about:


*

*For every discrete space $X$ we have $$\mathcal{Ba}(X) = \{A \subseteq X : \text{ either } A \text{ is countable or } X \setminus A \text{ is countable}\}$$
while the $\sigma$-algebra of $X$ is $\mathcal{B}(X) = \mathcal{P}(X)$, the power set of $X$. In particular, $\mathcal{Ba}(X) \subsetneqq \mathcal{B}(X)$ whenever $X$ is uncountable.

*If you take the one-point compactification $X^\ast = X \cup \{\infty\}$ of an uncountable discrete space $X$ then $\{\infty\} \in X^\ast$ is a standard example of a compact set which is not a $G_\delta$-set.
