Semiring of halfclosedhalfopen intervals
[ilmath]\newcommand{\[}{[\![}\newcommand{\]}{]\!]}\newcommand{\(}{(\!(}\newcommand{\)}{)\!)} [/ilmath]
Definition
Let [ilmath] a:= ({ a_i })_{ i = 1 }^{ n }\subseteq \mathbb{R} [/ilmath]^{[Note 1]}^{[Note 2]} and [ilmath] b:= ({ b_i })_{ i = 1 }^{ n }\subseteq \mathbb{R} [/ilmath] be two finite sequences of the same length (namely [ilmath]n\in\mathbb{N} [/ilmath]), we define [ilmath]\[a,b\)[/ilmath], a halfopenhalfclosed rectangle in [ilmath]\mathbb{R}^n[/ilmath]^{[1]} as follows:
 [ilmath]\[a,b\):=[a_1,b_1)\times\cdots\times[a_n,b_n)\subset\mathbb{R}^n[/ilmath] where [ilmath][\alpha,\beta):=\{x\in\mathbb{R}\ \vert\ \alpha\le x < \beta\}[/ilmath]^{[Convention 1]}
We denote the collection of all such halfopenhalfclosed rectangles by [ilmath]\mathscr{J}^n[/ilmath]^{[1]}, [ilmath]\mathscr{J}(\mathbb{R}^n)[/ilmath]^{[1]} or, provided the context makes the dimensions obvious, simply just [ilmath]\mathscr{J} [/ilmath]^{[1]}. Formally:
 [ilmath]\mathscr{J}^n:=\{\[a,b\)\ \vert\ a,b\in\mathbb{R}^n\}[/ilmath]
Furthermore, we claim [ilmath]\mathscr{J}^n[/ilmath] is a [[semiring of sets][. For a proof of this claim see "Proof of claims" below.
Purpose
Probably the most important use case for this semiring is as the domain of a certain kind of premeasure, namely a premeasure on a semiring that serves as a precursor to the Lebesgue measure, which for the reader's curiosity we include the definition for here:
 [ilmath]\lambda^n:\mathscr{J}^n\rightarrow\overline{\mathbb{R}_{\ge 0} } [/ilmath] with [ilmath]\lambda^n:\[a,b\)\mapsto\prod_{i=1}^n(b_ia_i)[/ilmath]^{[Note 3]}
In the case of [ilmath]\mathscr{J}^1[/ilmath] the Lebesgue measure is just the length of an interval, that is: [ilmath]\lambda^1:[a,b)\mapsto (ba)[/ilmath] and for [ilmath]\mathscr{J}^2[/ilmath] it is the area of a rectangle, for [ilmath]\mathscr{J}^3[/ilmath] volume of a cuboid, and so forth.
We can then use the theorem: a premeasure on a semiring may be extended uniquely to a premeasure on a ring to get a normal premeasure. Doing this is far easier than trying to define a premeasure on the ring of sets generated by [ilmath]\mathscr{J}^n[/ilmath].
Once we have a premeasure we can follow the usual path of extending premeasures to measures
Proof of claims
Recall the definition of a semiring of sets
A collection of sets, [ilmath]\mathcal{F} [/ilmath]^{[Note 4]} is called a semiring of sets if^{[1]}:
 [ilmath]\emptyset\in\mathcal{F}[/ilmath]
 [ilmath]\forall S,T\in\mathcal{F}[S\cap T\in\mathcal{F}][/ilmath]
 [ilmath]\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}[/ilmath][ilmath]\text{ pairwise disjoint}[/ilmath][ilmath][ST=\bigudot_{i=1}^m S_i][/ilmath]^{[Note 5]}  this doesn't require [ilmath]ST\in\mathcal{F} [/ilmath] note, it only requires that their be a finite collection of disjoint elements whose union is [ilmath]ST[/ilmath].
In order to prove this we will first show that [ilmath]\mathscr{J}^1[/ilmath] (the collection of halfopenhalfclosed intervals of the form [ilmath][a,b)\subset\mathbb{R} [/ilmath]) is a semiring. Then we shall use induction on [ilmath]n[/ilmath] to show it for all [ilmath]\mathscr{J}^n[/ilmath]
The message provided is:
Conventions
 ↑ For intervals in general we define the following:
 If [ilmath]\alpha\ge\beta[/ilmath] then [ilmath][\alpha,\beta)=\emptyset[/ilmath]
 If [ilmath]\{X_\alpha\}_{\alpha\in I} [/ilmath] is an arbitrary collection of sets where one or more of the [ilmath]X_\alpha[/ilmath] are the empty set, [ilmath]\emptyset[/ilmath], then:
 [ilmath]\prod_{\alpha\in I}X_\alpha=\emptyset[/ilmath] (here [ilmath]\prod[/ilmath] denotes the Cartesian product)
Notes
 ↑ Or equivalently, [ilmath]a\in\mathbb{R}^n[/ilmath], either way we get an [ilmath]n[/ilmath]tuple of real numbers
 ↑ The symbol [ilmath]\subset[/ilmath] could be used instead of [ilmath]\subseteq[/ilmath] but it doesn't matter, as:
 [ilmath]\big[A\subset B\big]\implies\big[A\subseteq B\big][/ilmath]
 ↑ Here [ilmath]\prod[/ilmath] denotes multiplication repeated over a range, in this case multiplication of real numbers
 ↑ An F is a bit like an R with an unfinished loop and the foot at the right. "Semi Ring".
 ↑ Usually the finite sequence [ilmath] ({ S_i })_{ i = m }^{ \infty }\subseteq \mathcal{F} [/ilmath] being pairwise disjoint is implied by the [ilmath]\bigudot[/ilmath] however here I have been explicit. To be more explicit we could say:
 [ilmath]\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\underbrace{\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)}_{\text{the }S_i\text{ are pairwise disjoint} }\overbrace{\wedge}^\text{and}\left(ST=\bigcup_{i=1}^m S_i\right)\right][/ilmath]
 Caution:The statement: [ilmath]\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)\implies\left(ST=\bigcup_{i=1}^m S_i\right)\right][/ilmath] is entirely different
 In this statement we are only declaring that a finite sequence exists, and if it is NOT pairwise disjoint, then we may or may not have [ilmath]ST=\bigcup_{i=1}^mS_i[/ilmath]. We require that they be pairwise disjoint AND their union be the set difference of [ilmath]S[/ilmath] and [ilmath]T[/ilmath].
 Caution:The statement: [ilmath]\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)\implies\left(ST=\bigcup_{i=1}^m S_i\right)\right][/ilmath] is entirely different
 [ilmath]\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\underbrace{\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)}_{\text{the }S_i\text{ are pairwise disjoint} }\overbrace{\wedge}^\text{and}\left(ST=\bigcup_{i=1}^m S_i\right)\right][/ilmath]
References
 ↑ ^{1.0} ^{1.1} ^{1.2} ^{1.3} ^{1.4} Measures, Integrals and Martingales  René L. Schilling
