Collection of finite unions of half-open intervals form an algebra

[psie]

TL;DR Summary

I'm confused over a claim made in my lecture notes, namely that the collection of finite unions of half-open intervals in ##\mathbb R## forms an algebra.

I'm reading in these notes the following passage (I only have a question about the last two sentences):

We briefly consider a generalization of one-dimensional Lebesgue measure, called Lebesgue-Stieltjes measures on ##\mathbb{R}##. These measures are obtained from an increasing, right-continuous function ##F: \mathbb{R} \rightarrow \mathbb{R}##, and assign to a half-open interval ##(a, b]## the measure $$\mu_{F}((a, b])=F(b)-F(a) .$$ The use of half-open intervals is significant here because a Lebesgue-Stieltjes measure may assign nonzero measure to a single point. Thus, unlike Lebesgue measure, we need not have ##\mu_{F}([a, b])=\mu_{F}((a, b])##. Half-open intervals are also convenient because the complement of a half-open interval is a finite union of (possibly infinite) half-open intervals of the same type. Thus, the collection of finite unions of half-open intervals forms an algebra.

The last two sentences confuse me. Which sets does the author have in mind? I know what an algebra is (basically a sigma algebra, but not closed under countable infinite operations, but finite operations), but I'm unsure which sets the author has in mind. Specifically, is it true that the collection of finite unions of half-open intervals forms an algebra?

Looking at Folland's book, he claims that the collection of finite disjoint unions of half open intervals of the form ##\emptyset## or ##(a,b]## or ##(a,\infty)##, where ##-\infty\leq a <b<\infty##, forms an algebra, but my notes claim that the union doesn't have to be disjoint to form an algebra. Grateful for any clarification.

 

[pasmith]

Which of the axioms of an algebra of sets do you not believe to hold for the collection of finite unions of half-open intervals?
- Closed under complementation: [itex]\mathbb{R} \setminus (a, b] = (-\infty, a] \cup (b ,\infty)[/itex].
- The empty set [itex](a, a][/itex] or [itex](\infty, \infty)[/itex] is in the collection.
- Closed under binary unions, given that the union of two half-open intervals is either two disjoint half-open intervals or a single half-open interval.

 

[psie]

Ok, I think I understand @pasmith, thank you. It doesn't matter here to specify whether the finite union is disjoint or not, i.e. Folland writes that a collection of finite disjoint unions of half-open intervals is an algebra, but I don't think the word disjoint is necessary, since we'll get an algebra nevertheless.

 

[pasmith]

More to the point, if [itex](a,b][/itex] and [itex](c,d][/itex] are not disjoint then their union is [itex](\min(a,c), \max(b,d)][/itex]. Thus a finite union of arbitrary half-open intervals is a finite union of disjoint half-open intervals.