A regular open set is an open set which equals the interior of its closure: $U=\overline U^\circ$. A regular closed set is a closed set which equals the closure of its interior: $U=\overline{U^\circ}$.
Let $R(\Bbb R^n)$ refer to the Boolean algebra of regular open sets, where $U\vee V:=(\overline{U\cup V})^\circ$ and $U\wedge V:=U\cap V$. (In fact, I believe this is isomorphic to the Boolean algebra of regular closed sets, where $U\vee V:=U\cup V$ and $U\cap V:=\overline{(U\cap V)^\circ}$ , where the isomorphism is given by the closure operator. So it doesn't matter if we choose the regular open sets or the regular closed sets.)
Is there a countably additive measure on this algebra? Naïvely, we might want to just use the Lebesgue measure (on either the open interior or the closed closure, depending on which isomorphic copy of $R$ we take), but unfortunately there exist regular open sets whose boundaries are Osgood curves (curves with positive area) which can ruin things. EDIT: Even in dimension 1 we can have regular open sets whose boundaries are fat Cantor sets.
I'm wondering about potentially averaging these: if $U\in R(\Bbb R^n)$, defining the measure $m(U)=\frac12\mu(U^\circ)+\frac12\mu(\overline U)$. Does that work?
If $\mu$ is a (countably additive) measure on $R(\mathbb{R}^n)$ (for $n>0$), then for all $U\in R(\mathbb{R}^n)$, $\mu(U)=0$ or $\mu(U)=\infty$. Indeed, suppose $0<\mu(U)<\infty$ for some $U$. Pick a countable dense subset $\{q_n:n\in\mathbb{N}\}\subset U$. Since $\mu$ is finite on $U$, $\mu(B_\epsilon(q_n))\to 0$ as $\epsilon\to 0$ for each $n$, so you can pick a sequence of $\epsilon_n>0$ such that $\mu(B_{\epsilon_n}(q_n))<\mu(U)/2^{n+1}$ for each $n$. Now observe that $\bigcup_n B_{\epsilon_n}(q_n)$ is dense in $U$ so $U$ is the join of the sets $B_{\epsilon_n}(q_n)$ in $R(\mathbb{R}^n)$. But $\sum_n B_{\epsilon_n}(q_n))<\sum_n \mu(U)/2^{n+1}<\mu(U)$, so this contradicts countable additivity.