I somehow got stuck on the following line in Oxtoby: "The class of Borel sets that have an $F_\sigma$ subset and a $G_\delta$ superset of equal measure is a $\sigma$-algebra that includes all closed sets." We work here with a Borel measure (=measure defined for all Borel sets) on a metric space.
I have problems with showing that
$\mathcal S=\{A\subseteq X; (\exists F\subseteq A\subseteq G) \mu(G\setminus F)=0, F\text{ is }F_\sigma, G\text{ is }G_\delta\}$
is a $\sigma$-algebra.
As Oxtoby devotes only one grammatical sentence to this claim, I guess it should be relatively easy and I am overlooking some more straightforward approach.
What I tried so far:
Complements: $F\subseteq A\subseteq G$ $\Rightarrow$ $X\setminus G\subseteq X\setminus A\subseteq X\setminus F$.
Finite intersections: Let $F_1\subseteq A\subseteq G_1$, $F_2\subseteq B\subseteq G_2$ and $\mu(G_1\setminus F_1)=\mu(G_2\setminus F_2)=0$. Then $F_1\cap F_2\subseteq A\cap B\subseteq G_1\cap G_2$ and
$G_1\cap G_2\setminus F_1\cap F_2 = (G_1\cap G_2\setminus F_1)\cup (G_1\cap G_2\setminus F_2)\subseteq$
$\subseteq (G_1\setminus F_1)\cup(G_2\setminus F_2),$
hence $\mu(G_1\cap G_2\setminus F_1\cap F_2)=0$. (Also, $G_1\cap G_2$ is $G_\delta$ and $F_1\cap F_2$ is $F_\sigma$.)
Using finite intersections and complements, we can show that $\mathcal S$ is closed under differences of pairs of sets.
Disjoint countable unions: Let $F_i\subseteq A_i \subseteq G_i$, where $A_i$'s are disjoint measurable sets.
Then also A_i\subseteq G_i\setminus \bigcup\limits_{n\ne i}F_n=: G_i' and $G_i'$'s are disjoint $G_\delta$-sets.
Thus we get
\bigcup_{i=1}^\infty F_i \subseteq \bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty G'_i
and \mu(\bigcup_{i=1}^\infty F_i)=\mu(\bigcup_{i=1}^\infty G'_i).
But I do not know whether disjoint union of $G_\delta$ sets is again a $G_\delta$-set.
EDIT 2: As shown by Theo in the answer bellow, it is not true in general.
Using differences and disjoint countable unions, I could get arbitrary countable unions.
I also tried to prove that $\mathcal S$ is closed under countable intersections directly from the definition $\mathcal S$, but in that proof I obtained a countable intersection of $F_\sigma$-sets, which is not necessarily $F_\sigma$.
EDIT:
Based on Theo's suggestion: There are several equivalent characterizations of $\mathcal S$:
$\mathcal S=\{A\subseteq X; (\forall \varepsilon>0)(\exists V\subseteq A\subseteq U) \mu(U\setminus V)<\varepsilon, V\text{ is close}, U\text{ is open}\}$
$\mathcal S=\{A\subseteq X; (\forall \varepsilon>0)(\exists F\subseteq A\subseteq U) \mu(U\setminus F)<\varepsilon, F\text{ is }F_\sigma, U\text{ is open}\}$
Now I can modify the above construction to work for any countable union. For a system $\{A_i; i\in\mathbb N\}$ of sets from $\mathcal S$ I choose $F_i\subseteq A_i \subseteq U_i$ such that $\mu(U_i\setminus F_i)<\varepsilon.2^{-i}$, $U_i$ is open, $F_i$ is $F_\sigma$. Let $U:=\bigcup_{i=1}^\infty U_i$ and $F:=\bigcup_{i=1}^\infty F_i$. Then $U$ is open, $F$ is $F_\sigma$ and
$U\setminus F= (\bigcup_{i=1}^\infty U_i) \setminus (\bigcup_{i=1}^\infty F_i) \subseteq \bigcup_{i=1}^\infty (U_i\setminus F_i),$
hence $\mu(U\setminus F)<\varepsilon$. We also have
$F=\bigcup_{i=1}^\infty F_i \subseteq \bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty U_i=U,$
which implies $\bigcup_{i=1}^\infty A_i\in \mathcal S$.