# Semi-ring of half-closed-half-open 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 half-open-half-closed 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 half-open-half-closed 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 [[semi-ring of sets][. For a proof of this claim see "Proof of claims" below.

## Purpose

Probably the most important use case for this semi-ring is as the domain of a certain kind of pre-measure, namely a pre-measure on a semi-ring 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_i-a_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 (b-a)[/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 pre-measure on a semi-ring may be extended uniquely to a pre-measure on a ring to get a normal pre-measure. Doing this is far easier than trying to define a pre-measure on the ring of sets generated by [ilmath]\mathscr{J}^n[/ilmath].

Once we have a pre-measure we can follow the usual path of extending pre-measures to measures

## Proof of claims

Recall the definition of a semi-ring of sets

A collection of sets, [ilmath]\mathcal{F} [/ilmath][Note 4] is called a semi-ring of sets if[1]:

1. [ilmath]\emptyset\in\mathcal{F}[/ilmath]
2. [ilmath]\forall S,T\in\mathcal{F}[S\cap T\in\mathcal{F}][/ilmath]
3. [ilmath]\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}[/ilmath][ilmath]\text{ pairwise disjoint}[/ilmath][ilmath][S-T=\bigudot_{i=1}^m S_i][/ilmath][Note 5] - this doesn't require [ilmath]S-T\in\mathcal{F} [/ilmath] note, it only requires that their be a finite collection of disjoint elements whose union is [ilmath]S-T[/ilmath].

In order to prove this we will first show that [ilmath]\mathscr{J}^1[/ilmath] (the collection of half-open-half-closed intervals of the form [ilmath][a,b)\subset\mathbb{R} [/ilmath]) is a semi-ring. Then we shall use induction on [ilmath]n[/ilmath] to show it for all [ilmath]\mathscr{J}^n[/ilmath]

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
It's a straightforward and fairly easy proof, but it isn't trivial. As such it is not marked as an easy proof, but nor does it have a high grade. See page 18 in Measures, Integrals and Martingales - René L. Schilling if stuck.

## Conventions

1. For intervals in general we define the following:
1. If [ilmath]\alpha\ge\beta[/ilmath] then [ilmath][\alpha,\beta)=\emptyset[/ilmath]
2. 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)
As such if there is one or more [ilmath]a_i[/ilmath] and [ilmath]b_i[/ilmath] such that [ilmath]a_i\ge b_i[/ilmath] then [ilmath]\[a,b\)=\emptyset[/ilmath]

## Notes

1. Or equivalently, [ilmath]a\in\mathbb{R}^n[/ilmath], either way we get an [ilmath]n[/ilmath]-tuple of real numbers
2. 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]
3. Here [ilmath]\prod[/ilmath] denotes multiplication repeated over a range, in this case multiplication of real numbers
4. An F is a bit like an R with an unfinished loop and the foot at the right. "Semi Ring".
5. 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(S-T=\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(S-T=\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]S-T=\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].