Let $(0,1)$ the unit interval. An open subset $\mathcal{R}\subseteq(0,1)$ is *regular* if it is the interior of its own closure. The intersection of two regular open sets is regular. Unfortunately, the *union* of two regular open sets is generally *not* regular. (For example, if $0<a<b<c<1$, then the open intervals $(a,b)$ and $(b,c)$ are regular, but their disjoint union $(a,b)\sqcup(b,c)$ isn't.)
Neither is the complement of a regular open set. Thus, if we define $\mathfrak{R}$ to be the family of all regular open subsets of $(0,1)$, then $\mathfrak{R}$ is not a Boolean algebra under the standard set-theoretic operations. However, $\mathfrak{R}$ *is* a Boolean algebra under slightly different operations. If $\mathcal{Q}$ and $\mathcal{R}$ are regular open subsets of $\mathcal{X}$, then define $\mathcal{Q}\vee\mathcal{R}:=\mathrm{int}\left[\mathrm{clos}(\mathcal{Q}\cup\mathcal{R})\right]$. (For example: $(a,b)\vee(b,c)=(a,c)$.) Meanwhile, define $\neg\mathcal{Q}:=\mathrm{int}((0,1)\setminus\mathcal{Q})$. Then $\mathfrak{R}$ is a Boolean algebra under the operations $\vee$, $\cap$, and $\neg$.

We can then define a *finitely additive measure* on $\mathfrak{R}$ in the obvious way: it is a function $\mu:\mathfrak{R}\longrightarrow\mathbb{R}_+$ such that $\mu[\emptyset]=0$ and $\mu[\mathcal{Q}\vee\mathcal{R}]=\mu[\mathcal{Q}]+\mu[\mathcal{R}]$ whenever $\mathcal{Q}$ and $\mathcal{R}$ are disjoint regular open subsets of $(0,1)$. To avoid confusion with the standard notion of measure (defined in terms of disjoint unions), I will sometimes call this a *finitely $\vee$-additive measure* in what follows.

Let $\lambda$ be the Lebesgue measure. Intuitively, it seems that $\lambda$ "should" define a finitely $\vee$-additive measure when restricted to $\mathfrak{R}$. The simplest regular open subsets of $(0,1)$ are finite disjoint unions of open intervals --let us call these *simple* open sets. The simple open sets form a Boolean sub-algebra of the regular open sets, and it is easy to see that $\lambda$ is finitely $\vee$-additive on this sub-algebra. For example

$$ \lambda\left[(a,b)\vee(b,c)\right] \ = \ \lambda\left[(a,c)\right] \ = \ c-a \ = \ c-b+b-a \ = \ \lambda\left[(a,b)\right] + \lambda\left[(b,c)\right]. $$

However, not all regular open sets are simple; in general, a regular open set is a *countable* disjoint union of open intervals, and it is not immediately clear that $\lambda$ will be finitely $\vee$-additive when applied to such sets. Hence my question:

Does the Lebesgue measure induce a finitely $\vee$-additive measure on the Boolean algebra of regular open subsets of $(0,1)$?

This seems like an obvious question to ask, so presumably it was answered a long time ago, and I am just looking in the wrong place. (Oddly, Fremlin's multi-volume encyclopaedic work on measure theory does not seem to address this question.) If the answer is already known, then I would really appreciate a reference to the relevant literature.

Some remarks:

(1) This question arose in the discussion following another question I recently asked about finitely $\vee$-additive measures.

(2) I have focused on the open unit interval $(0,1)$ only for simplicity. Obviously, the same question could be posed for any bounded interval (closed or open). Unbounded intervals might be more complicated, since they contain sets with infinite measure.

(3) More ambitiously, the same question could be posed in higher-dimensional Euclidean spaces. For example, in $\mathbb{R}^2$, it is clear that the Lebesgue measure is finitely $\vee$-additive for regular open sets which are finite disjoint unions of open rectangles. But a general regular open set in $\mathbb{R}^2$ is a complicated beastie (it is not just a countable disjoint union of rectangles), so different strategies may be required.

(4) $\mathfrak{R}$ is in fact a *complete* Boolean algebra. Thus, we can define $\bigvee_{n=1}^\infty \mathcal{R}_n$ for any countable collection $\{\mathcal{R}_n\}_{n=1}^\infty\subseteq\mathfrak{R}$. However, I have focussed on *finite* $\vee$-additivity for a reason: it is easy to show that there $\mathfrak{R}$ cannot support *any* countably $\vee$-additive measure. In particular, the Lebesgue measure *cannot* be countably $\vee$-additive on $\mathfrak{R}$, even if it turns out to be finitely $\vee$-additive. So even though any regular open set is a countable disjoint union of open intervals, this does *not* mean that any finitely $\vee$-additive measure is determined in the obvious way by its behavior on open intervals.