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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.

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  • $\begingroup$ One should mention that the induced measure is definitely not countably additive, because one can have an open dense set in the unit interval of arbitrarily small Lebesgue measure --- place an interval of size $\epsilon/2^n$ about the $n^{th}$ rational. The join of these intervals is the whole interval, since it is dense, but the measures of the finite joins is bounded by $\epsilon$. $\endgroup$ – Joel David Hamkins Jul 6 '16 at 11:07
  • $\begingroup$ Hi Joel. I agree ---this was the reason for "Remark (4)" above. $\endgroup$ – Marcus Pivato Jul 6 '16 at 11:16
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This is a great question! But unfortunately, the answer is no, the Lebesgue measure on the unit interval is not finitely $\vee$-additive.

Theorem. There are two disjoint regular open sets $L$ and $R$ in the unit interval, with Lebesgue measure as small as desired, but whose union is dense, and so $L\vee R$ has full measure.

Proof. Consider the construction of a fat Cantor set, obtained by successively omitting much less than the middle third of each of the remaining intervals. By omitting less, you can arrange that the resulting Cantor set has measure as close to $1$ as desired.

Let $U$ be the union of those omitted intervals, the complement of the fat Cantor set. This is an open dense set of some measure $\epsilon$, as small as desired. (The set $U$, being open dense, is not itself regular.)

Let $L$ be the union of the open left-halves of the omitted intervals, and let $R$ be the union of the open right-halves of those intervals. So $L$ and $R$ form a disjoint open partition of $U$, minus the countably many center points of the omitted intervals, and the measure of $L$ and $R$ are each $\epsilon/2$.

I claim that each of $L$ and $R$ are regular open sets. To see this, notice first that between any two of the intervals used to construct $L$, there is an interval of $R$, and vice versa. Suppose that $u$ is an open interval contained in the closure of $L$. Since $R$ is open and disjoint from $L$ and hence also from the closure of $L$, it must be that $u$ contains no points from $R$. It follows that $u$ can contain points from at most one interval of $L$, and from this it follows that $u\subset L$. So the interior of the closure of $L$ is $L$ itself and therefore $L$ is regular open. A similar argument shows that $R$ is regular open.

Meanwhile, since the union $U=L\cup R$ is open dense in the unit interval, it follows that $L\vee R$ is the whole interval, and so the measure of $L\vee R$ is $1$.

So we have $\lambda(L)+\lambda(R)=\epsilon<1=\lambda(L\vee R)$, which violates finite $\vee$-additivity. QED

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    $\begingroup$ Thank you very much, Joel. This is a beautiful argument. But I think it would help to add a bit more detail on why $L$ and $R$ are regular. I suggest something like the following. Every number in $[0,1]$ has a unique quaternary (i.e. base-4) expansion. (That is, "unique" once we exclude expansions ending in "3333333...".) Let us say that a number is tetradic rational if it is a rational number whose denominator is a power of 4 ---equivalently, its quaternary expansion ends in an infinite sequence of zeros. [Continued in next comment...] $\endgroup$ – Marcus Pivato Jul 7 '16 at 5:29
  • $\begingroup$ Consider the Cantor set $K$ obtained by deleting all numbers which contain a "1" or a "2" anywhere in their quaternary expansion, except for tetradic rationals which have only a single "1" and end in "10000...". (Yes, I know this is not a "fat" Cantor set, but I'm only interested in the topological properties here, not the measure.) In this case, $U$ is the set of all points in $[0,1]$ containing a "1" or a "2" anywhere in their quaternary expansions (except the aforementoned tetradics). [continued] $\endgroup$ – Marcus Pivato Jul 7 '16 at 5:31
  • $\begingroup$ In your notation, $L$ is the set of all points in $U$ such that the first "1" appears before the first "2", while $R$ is the set of all points in $U$ such that that the first "2" appears before the first "1". Clearly, $U=R\sqcup L$, and $L$ and $R$ are open (because a small enough perturbation will not change the first "1" into a "2" or vice versa...) Let $R'$ be the set of left endpoints of $R$-intervals; equivalently, $R'$ is the set of all tetradic rationals whose quaternary expansion contains no "1" and ends with "20000..." [continued...] $\endgroup$ – Marcus Pivato Jul 7 '16 at 5:33
  • $\begingroup$ Then the closure of $L$ is the set $K \sqcup L \sqcup R'$. (Proof: Any sequence in $L$ either converges to a point in $L$, or it converges to a point in $K$ (by pushing the "initial 1" further and further down the quaternary expansion until it vanishes altogether), or it converges from the left to a point in $R'$.) Since $K$ and $R'$ are both nowhere dense, the interior of $K \sqcup L \sqcup R'$ is simply $L$, which shows that $L$ is regular. Finally, there is a self-homeomorphism of [0,1] transforming my sets $L$ and $R$ into your $L$ and $R$, so they are regular also, as you claimed. $\endgroup$ – Marcus Pivato Jul 7 '16 at 5:35

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