Let $\mathcal{X}$ be a topological space. An open subset $\mathcal{R}\subseteq\mathcal{X}$ 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. 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 $\mathcal{X}$, 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 $\mathcal{Q}\vee\mathcal{R}:=\mathrm{int}\left[\mathrm{clos}(\mathcal{Q}\cup\mathcal{R})\right]$, and define $\neg\mathcal{Q}:=\mathrm{int}(\mathcal{X}\setminus\mathcal{Q})$. 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 $\mathcal{X}$. 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.

So far, this is all standard material: the Boolean algebra structure of regular open sets is well-known, and the idea of defining a finitely additive measure on an arbitrary Boolean algebra has been around for a long time. (See, e.g. volume III of Fremlin's books on measure theory for discussions of both.) But I am interested in three rather specific questions.

(1) What is the relationship (if any) between finitely $\vee$-additive measures on $\mathfrak{R}$ and Borel measures on $\mathcal{X}$?

In some simple cases, a Borel measure on $\mathcal{X}$ "induces" a finitely $\vee$-additive measure on $\mathfrak{R}$. For example, let $\mathcal{X}=[0,1]$ (the unit interval) with the usual topology; then the Lebesgue measure induces a finitely $\vee$-additive measure on the regular open subsets of $[0,1]$ in the obvious way. However, not every Borel measure on $\mathcal{X}$ induces a finitely $\vee$-additive measure on $\mathfrak{R}$ in this way (for example, "atoms" generally create problems). Conversely, not every finitely $\vee$-additive measure on $\mathfrak{R}$ seems to arise from a Borel probability measure.

You might think that the issue here is the disconnect between finite additivity and countable additivity. To avoid this, let $\mathfrak{B}$ be the Boolean algebra generated by all open and closed subsets of $\mathcal{X}$ under the standard set-theoretic operations. We can define *finitely additive measures* on $\mathfrak{B}$ in the standard way (in terms of disjoint unions). Any Borel measure obviously induces a finitely additive measure on $\mathfrak{B}$ (but not conversely). So we could weaken question (1) to the following:

(2) What is the relationship (if any) between finitely $\vee$-additive measures on $\mathfrak{R}$ and finitely additive measures on $\mathfrak{B}$?

Another question has to do with integration. There is a well-developed theory of integration for any finitely additive or countably additive measure defined on any Boolean algebra of subsets with the standard set-theoretic operations. But this doesn't obviously extend to $\vee$-additive measures.

(3) Is there a well-behaved integration theory for finitely $\vee$-additive measures on $\mathfrak{R}$?

Here, by "well-behaved", I mean that the integration operator is defined for some reasonable domain $\mathcal{F}$ of real-valued functions on $\mathcal{X}$ (e.g. all bounded continuous real-valued functions on $\mathcal{X}$), it is linear on $\mathcal{F}$, it is continuous with respect to some reasonable topology on $\mathcal{F}$, and it is increasing relative to the pointwise ordering of $\mathcal{F}$.

Question (3) is closely related to (1) and (2) because clearly, if we could represent a finitely $\vee$-additive measure in terms of a Borel measure (for example), then we could just invoke the standard integration theory for Borel measures to obtain a positive answer to (3). On the other hand, a positive answer to (3) might lead to a positive answer to (1) and/or (2) via some form of the Riesz Representation Theorem.

I have some ideas about how to answer (1), (2) and (3), but I am worried that I am "reinventing the wheel". These seem to be obvious questions, so I would be surprised if someone hadn't already answered them a long time ago. However, I have looked in the obvious places (e.g. I have searched through Fremlin's encyclopaedic texts on measure theory, done keyword searches on MathSciNet, etc.) and I haven't found anything. But this question lies a bit outside my area of expertise, so perhaps I just looked in the wrong place. So I would be very grateful for any pointers to any literature. Also, I have stated the questions when $\mathcal{X}$ is "any" topological space, but it is likely that we need to impose additional hypotheses on $\mathcal{X}$ (e.g. compact Hausdorff) to get a useful answer.

**Update:** I have removed my earlier "example", because (1) it was unnecessarily complicated, (2) it used the Lebesgue measure, which is not obviously finitely $\vee$-additive, and (3) it was actually not well-defined. Here is a much simpler (and hopefully correct) example.

Let $\mathcal{X}:=[-1,1]$ with the usual topology. Let $\mathfrak{F}$ be the collection of all regular open subsets of $\mathcal{X}$ that contain 0. Then $\mathfrak{F}$ is a filter in the Boolean algebra $\mathfrak{R}$. Use the Ultrafilter Lemma to extend $\mathfrak{F}$ to an ultrafilter $\mathfrak{U}\subset\mathfrak{R}$. Now, for all $\mathcal{R}\in\mathfrak{R}$, define $\mu[\mathcal{R}]:=1$ if $\mathcal{R}\in\mathfrak{U}$, whereas $\mu[\mathcal{R}]:=0$ if $\mathcal{R}\not\in\mathfrak{U}$.

Clearly, $\mu$ is a finitely $\vee$-additive measure. Heuristically, $\mu$ is like a "point mass" at zero, but with an additional feature: if the point 0 lies on the boundary between a regular set $\mathcal{R}$ and its negation $\neg\mathcal{R}$, then exactly *one* of $\mathcal{R}$ or $\neg\mathcal{R}$ gets to "claim ownership" of 0; this decision is made by the ultrafilter $\mathfrak{U}$. For example, exactly *one* of the following two statements is true:

- For all $\epsilon>0$, $\mu[(0,\epsilon)]=1$ while $\mu[(-\epsilon,0]=0$.
- For all $\epsilon>0$, $\mu[(0,\epsilon)]=0$ while $\mu[(-\epsilon,0]=1$.

The ultrafilter $\mathfrak{U}$ also decides "ownership" in more complicated cases. For example, let

$$\mathcal{E}_+ \ := \ \bigsqcup_{n=1}^\infty \left(\frac{1}{2n+1},\frac{1}{2n}\right) \quad\mbox{and}\quad \mathcal{O}_+ \ := \ \bigsqcup_{n=1}^\infty \left(\frac{1}{2n},\frac{1}{2n-1}\right) $$ while

$$\mathcal{E}_- \ := \ \bigsqcup_{n=1}^\infty \left(\frac{-1}{2n},\frac{-1}{2n+1}\right) \quad\mbox{and}\quad \mathcal{O}_- \ := \ \bigsqcup_{n=1}^\infty \left(\frac{-1}{2n-1},\frac{-1}{2n}\right). $$

These are four disjoint regular open sets, and clearly, $\mathcal{X} = \mathcal{E}_+\vee\mathcal{O}_+\vee\mathcal{E}_-\vee\mathcal{O}_-$. Thus, precisely *one* of the four sets $\mathcal{E}_+$, $\mathcal{O}_+$, $\mathcal{E}_-$, and $\mathcal{O}_-$ gets $\mu$-measure 1 (i.e. claims "ownership" of 0), while the other three get $\mu$-measure 0 ---the ultrafilter $\mathfrak{U}$ decides which one.

This example is important for the following reason. In his answer to question (1) (below), Robert Furber argued that a finitely $\vee$-additive measure on $\mathfrak{R}$ can be seen as an ordinary (finitely additive) measure on the algebra of sets with the Baire property which vanishes on meagre sets. As I understand it, the argument works like this:

Given any set with the Baire property $\mathcal{B}\subset\mathcal{X}$, there is a (unique) regular open set $\mathcal{R}\subset\mathcal{X}$ and a meagre set $\mathcal{M}\subset\mathcal{X}$ such that $\mathcal{B}=\mathcal{R}\triangle \mathcal{M}$. In this case, define $\mu^*[\mathcal{B}]:=\mu[\mathcal{R}]$. (In particular, this means $\mu^*[\mathcal{M}]=0$ for all meagre sets $\mathcal{M}\subset\mathcal{X}$.) If $\mathfrak{B}$ is the $\sigma$-algebra of sets with the Baire property, then we thereby obtain a finitely additive measure function $\mu^*$ on $\mathfrak{B}$.

I believe this argument is correct. Yet the above example seems to contradict this statement, since it seems to have a "point mass" at 0, and the set $\{0\}$ is obviously meagre. However, on closer inspection, there is no contradiction: we obtain $\mu^*[\{0\}]=0$, whereas $\mu^*[\mathcal{B}]$ for many Baire sets $\mathcal{B}$ which "touch" 0.

**Further remark.** As Robert pointed out, it is not obvious that the Lebesgue measure induces a finitely $\vee$-additive measure on the Boolean algebra of regular open sets. I have opened this as a separate question.