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. 



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



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



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




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**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](https://mathoverflow.net/questions/243782/does-the-lebesgue-measure-induce-a-finitely-additive-measure-on-the-boolean-alge), 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.




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**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](https://mathoverflow.net/questions/243782/does-the-lebesgue-measure-induce-a-finitely-additive-measure-on-the-boolean-alge).



Finitely additive measures on Boolean algebras of regular open subsets: Is there a relationship with Borel measures? A theory of integration?
The fact you are probably looking for is that, for any Baire space $X$ (e.g. a completely metrizable space or a compact Hausdorff space) the inclusion map $\mathfrak{R}(X) \rightarrow \mathfrak{B}o(X)/\mathfrak{M}(X)$ is an isomorphism of complete Boolean algebras. $\mathfrak{B}o(X)$ is the $\sigma$-algebra of Borel sets and $\mathfrak{M}(X)$ the $\sigma$-ideal of meagre Borel sets. In fact, this quotient Boolean algebra can also be taken to be "sets with the Baire property" modulo meagre sets. A proof can be found in Halmos's "Lectures on Boolean Algebras" as Theorem 4 in section 13, or in section 514I in Fremlin's "Measure Theory". However, once the definitions are understood it is not difficult to rediscover the proof.


For any Boolean algebra $A$, we can define $M(A)$ to be the set of finitely additive probability measures on $A$. For any homomorphism of Boolean algebras $f : A \rightarrow B$, we can define  

$$
M(f)(\mu)(a) = \mu(f(a))
$$
and $M(f)(\mu)$ can be easily verified to be a finitely additive probability measure on $A$ if $\mu \in M(B)$, i.e. $M(f) : M(B) \rightarrow M(A)$. In technical terms, this is a contravariant functor from the category of Boolean algebras (with Boolean homomorphisms) to the category of sets (with functions). As described above, we have an isomorphism $i : \mathfrak{R}(X) \rightarrow \mathfrak{B}o(X)/\mathfrak{M}(X)$, and we have the quotient morphism $p : \mathfrak{B}o(X) \twoheadrightarrow \mathfrak{B}o(X)/\mathfrak{M}(X)$, whose kernel is $\mathfrak{M}(X)$. Therefore if $\mu \in M(\mathfrak{R}(X))$, $M(i^{-1} \circ p)(\mu) \in M(\mathfrak{B}o(X))$, and vanishes on $\mathfrak{M}(X)$. Therefore, if $X$ is a Baire space, any finitely additive measure on $\mathfrak{R}(X)$ defines a finitely additive measure on $\mathfrak{B}o(X)$ vanishing on meagre sets. The well-definedness argument used to prove the other direction (any finitely additive Borel measure on $X$ vanishing on meagre sets defines a measure on the regular open sets) should now also be obvious. 


Note that if $X$ is a Polish space in which each point has empty interior, one can adapt the usual proof that there is a comeagre set of measure zero (based on enumerating neighbourhoods of the rationals) to show that there are no *countably additive* probability measures on $\mathfrak{R}(X)$. 


For question 2, we can observe that, as every clopen set is regular open, and the Boolean operations agree, so is an injective Boolean homomorphism $j : \mathfrak{B}(X) \rightarrow \mathfrak{R}(X)$. This is an isomorphism iff $X$ is extremally disconnected, \emph{i.e.} the closure of any open set is open. In such a case, e.g. any stonean space, $M(j)$ is an isomorphism between the sets of finitely additive measures. Furthermore, if $j$ is not an isomorphism, this can only be if it is not surjective, and then we can use the Stone duality to obtain two 2-valued measures $\mu \neq \nu$ on $\mathfrak{R}(X)$ such that $M(j)(\mu) = M(j)(\nu)$. So the answer to question 2 is that they are isomorphic iff $X$ is extremally disconnected.


For question 3, if $X$ is a Baire space you can use a the finitely additive version of the usual integration theory. Here is my original answer to keep the later comments making sense:
As for question 3, even if $X$ is not a Baire space, one can take the Stone space of $\mathfrak{R}(X)$, which is a stonean space $Y$ whose clopen sets and regular open sets are both isomorphic to $\mathfrak{R}(X)$. As it is stonean, it is a compact Hausdorff space, and therefore a Baire space. I suspect you will have already encountered this construction in Fremlin's book. 


However, it is not clear how continuous functions on the stone space relate to functions on the original space, except in the case of finite linear combinations of $\chi\_U$ for $U$ a regular open set.


