Semi-ring of half-closed-half-open intervals

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Definition

Let a:= ({ a_i })_{ i = 1 }^{ n }\subseteq \mathbb{R} [Note 1][Note 2] and b:= ({ b_i })_{ i = 1 }^{ n }\subseteq \mathbb{R} be two finite sequences of the same length (namely n\in\mathbb{N} ), we define \[a,b\), a half-open-half-closed rectangle in \mathbb{R}^n[1] as follows:

  • \[a,b\):=[a_1,b_1)\times\cdots\times[a_n,b_n)\subset\mathbb{R}^n where [\alpha,\beta):=\{x\in\mathbb{R}\ \vert\ \alpha\le x < \beta\}[Convention 1]

We denote the collection of all such half-open-half-closed rectangles by \mathscr{J}^n[1], \mathscr{J}(\mathbb{R}^n)[1] or, provided the context makes the dimensions obvious, simply just \mathscr{J} [1]. Formally:

  • \mathscr{J}^n:=\{\[a,b\)\ \vert\ a,b\in\mathbb{R}^n\}

Furthermore, we claim \mathscr{J}^n 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:

  • \lambda^n:\mathscr{J}^n\rightarrow\overline{\mathbb{R}_{\ge 0} } with \lambda^n:\[a,b\)\mapsto\prod_{i=1}^n(b_i-a_i)[Note 3]

In the case of \mathscr{J}^1 the Lebesgue measure is just the length of an interval, that is: \lambda^1:[a,b)\mapsto (b-a) and for \mathscr{J}^2 it is the area of a rectangle, for \mathscr{J}^3 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 \mathscr{J}^n.


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

Proof of claims

[Expand]

Recall the definition of a semi-ring of sets

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

Grade: B
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. Jump up For intervals in general we define the following:
    1. If \alpha\ge\beta then [\alpha,\beta)=\emptyset
    2. If \{X_\alpha\}_{\alpha\in I} is an arbitrary collection of sets where one or more of the X_\alpha are the empty set, \emptyset, then:
      • \prod_{\alpha\in I}X_\alpha=\emptyset (here \prod denotes the Cartesian product)
    As such if there is one or more a_i and b_i such that a_i\ge b_i then \[a,b\)=\emptyset

Notes

  1. Jump up Or equivalently, a\in\mathbb{R}^n, either way we get an n-tuple of real numbers
  2. Jump up The symbol \subset could be used instead of \subseteq but it doesn't matter, as:
    • \big[A\subset B\big]\implies\big[A\subseteq B\big]
  3. Jump up Here \prod denotes multiplication repeated over a range, in this case multiplication of real numbers
  4. Jump up An F is a bit like an R with an unfinished loop and the foot at the right. "Semi Ring".
  5. Jump up Usually the finite sequence ({ S_i })_{ i = m }^{ \infty }\subseteq \mathcal{F} being pairwise disjoint is implied by the \bigudot however here I have been explicit. To be more explicit we could say:
    • \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]
      • Caution:The statement: \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] 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 S-T=\bigcup_{i=1}^mS_i. We require that they be pairwise disjoint AND their union be the set difference of S and T.

References

  1. Jump up to: 1.0 1.1 1.2 1.3 1.4 Measures, Integrals and Martingales - René L. Schilling