Is the boundary of an open set in a $\sigma$-space empty? Recall that a Boolean space is a $\sigma$-space in case the closure of every open Borel set is open.
Let $\{B_i\}$ be a denumerable family of open-closed sets in a $\sigma$-space $X$. Then $\bigcup_i B_i$ is open but not closed (hence, not open-closed). Since $X$ is a $\sigma$-space, however, the closure of $\bigcup_i B_i$ is open and, consequently, the closure of $\bigcup_i B_i$ is open-closed.
Now, $\bigcup_i B_i$ and ${\rm cl}\ \bigcup_i B_i$ symmetrically differ by a meagre set, namely, the boundary of $\bigcup_i B_i$. This boundary is meager, but is it empty? (I think not, because the boundary of set is empty if and only if the set is open-closed, and $\bigcup_i B_i$ is open. But I am not sure...)
 A: In effect, you are asking if $\bigvee\limits_{i \in \omega}B_i = \bigcup\limits_{i \in \omega}B_i$, where the left hand side is the closure of the union, which is the join/supremum of the family $\{B_i\}_{i \in \omega}$ in the Boolean algebra of clopen sets of $X$, which we will write as $\mathrm{Clopen}(X)$. It is possible to characterize exactly when this happens.
We first consider the case when the family is not "genuinely infinite", in that there exists a finite set $I \subset \omega$ such that $\bigcup\limits_{i \in \omega}B_i = \bigcup\limits_{i \in I} B_i$. It follows that $\bigcup\limits_{i \in I} B_i$ is clopen, so the boundary of $\bigcup\limits_{i \in \omega}B_i$ is empty.
In the other case, suppose the family is "genuinely infinite", which is to say that for any finite set $I \subset \omega$, $\bigcup_\limits{i \in I}B_i$ is a proper subset of $\bigcup\limits_{i \in \omega}B_i$. We will show that the boundary of $\bigcup\limits_{i \in \omega}B_i$ is not empty. 
We do this using Stone duality and the Boolean prime ideal theorem. Let's define $B = \mathrm{cl}\left(\bigcup_{i \in \omega}B_i\right) = \bigvee_{i \in \omega}B_i$. We are looking for a point $x \in X$ such that $x \in B$ but $x \not\in \bigcup_{i \in \omega}B_i$. By Stone duality, this is equivalent to finding an ultrafilter $u$ on $\mathrm{Clopen}(X)$ such that $B \in u$ but $B_i \not\in u$ for all $i \in I$. Let $\mathcal{F}$ be the principal filter on $B$ in $\mathrm{Clopen}(X)$, i.e. the set of clopen sets containing $B$, and let $\mathcal{I}$ be the ideal generated by the set $\{B_i\}_{i \in \omega}$ in $\mathrm{Clopen}(X)$, i.e. the set of clopen sets $G$ such that there exists a finite set $I \subset \omega$ such that $G \subseteq \bigcup_{i \in I}G_i$. Our assumption that the join is "genuinely infinite" ensures that $\mathcal{F} \cap \mathcal{I} = \emptyset$. So by the Boolean prime ideal theorem there exists an ultrafilter $u$ such that $\mathcal{F} \subseteq u$ and $u \cap \mathcal{I} = \emptyset$. Therefore $B \in u$ and $B_i \not\in u$ for all $i \in \omega$, as required.
