The ring of sets generated by a semi-ring is the set containing the semi-ring and all finite disjoint unions

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Statement

Given a semi-ring of sets, [ilmath]\mathcal{F} [/ilmath], the ring generated by[Note 1] [ilmath]\mathcal{F} [/ilmath], written [ilmath]R(\mathcal{F})[/ilmath] is exactly the set [ilmath]\mathcal{F} [/ilmath] and all finite unions of elements of [ilmath]\mathcal{F} [/ilmath][1].

Recall the definitions of a ring of sets, a semi-ring of sets and the ring of sets generated by:

A Ring of sets is a non-empty class [ilmath]R[/ilmath][2] of sets such that:

  • [math]\forall A\in R\forall B\in R[A\cup B\in R][/math]
  • [math]\forall A\in R\forall B\in R[A-B\in R][/math]

[math]\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]

A collection of sets, [ilmath]\mathcal{F} [/ilmath][Note 2] 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 3] - 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].

Ring of sets generated by/Definition

Proof that [ilmath]\mathcal{F}_\cup[/ilmath] is a ring

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Mostly done but some trivial results do not have proof. Some of it was rushed, it would benefit from being read over

This page is waiting for a final review, then this notice will be removed.

A ring is defined to be [ilmath]\cup[/ilmath]-closed and [ilmath]\setminus[/ilmath]-closed; however showing it is [ilmath]\setminus[/ilmath]-closed is easier if we show already it is [ilmath]\cap[/ilmath]-closed so we do this instead. Caution:It is not yet said on this website that rings are [ilmath]\cap[/ilmath]-closed, but they are!

  1. [ilmath]\mathcal{F}_\cup [/ilmath] is [ilmath]\cap[/ilmath]-closed
    • Let [ilmath]S,T\in\mathcal{F}_\cup[/ilmath] be given. We want to show that [ilmath]S\cap T\in\mathcal{F}_\cup[/ilmath]
      • First note that this means [ilmath]S=\bigudot_{i=1}^{m_1}S_i[/ilmath] and [ilmath]T=\bigudot_{j=1}^{m_2}T_j[/ilmath] for some [ilmath]m_1,m_2\in\mathbb{N} [/ilmath] and [ilmath]S_i,T_j\in\mathcal{F} [/ilmath]. Now we want to show that [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)\cap\big(\bigudot_{j=1}^{m_2}T_j\big)\in\mathcal{F}_\cup[/ilmath]
      • Secondly, recall the distributivity of intersections across unions, which states:
        • [ilmath]A\cap(\bigcup_{i=1}^n B_i)=\bigcup_{i=1}^n(A\cap B_i)[/ilmath]
      • Furthermore, if the [ilmath]B_i[/ilmath] are pairwise disjoint, then:
        • [ilmath]A\cap(\bigudot_{i=1}^n B_i)=\bigudot_{i=1}^n(A\cap B_i)[/ilmath][Note 4]
      • By immediate application of this proposition we see: [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)\cap\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{j=1}^{m_2}\Big(\big(\bigudot_{i=1}^{m_1}S_i\big)\cap T_j\Big)[/ilmath]
      • By commutivity of intersection we see [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)\cap T_j=T_j\cap\big(\bigudot_{i=1}^{m_1}S_i\big)[/ilmath]
      • By applying distributivity of intersections across unions again we see:
        • [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)\cap T_j=T_j\cap\big(\bigudot_{i=1}^{m_1}S_i\big)=\bigudot_{i=1}^{m_1}(T_j\cap S_i)[/ilmath]
      • Then by substitution: [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)\cap\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{j=1}^{m_2}\Big(\big(\bigudot_{i=1}^{m_1}S_i\big)\cap T_j\Big)=\bigudot_{j=1}^{m_2}\Big(\bigudot_{i=1}^{m_1}(T_j\cap S_i)\Big)[/ilmath]
      • By comutivity of the union and commutivity of intersection we can write this in a slightly neater way:
        • [ilmath]\bigudot_{i=1}^{m_1}\Big(\bigudot_{j=1}^{m_2}(S_i\cap T_j)\Big)[/ilmath]
      • Notice that if [ilmath]A,B\in\mathcal{F} [/ilmath] then as [ilmath]\mathcal{F} [/ilmath] is a semi-ring of sets, [ilmath]A\cap B\in\mathcal{F} [/ilmath] and that we have [ilmath]m_1[/ilmath] unions of [ilmath]m_2[/ilmath] unions each, so there are [ilmath]m_1\times m_2[/ilmath] unions of elements of [ilmath]\mathcal{F} [/ilmath] in total. As [ilmath]m_1,m_2\in\mathbb{N} [/ilmath] so is their product.
    • Thus we have expressed [ilmath]S\cap T[/ilmath] as an element of [ilmath]\mathcal{F}_{\cup} [/ilmath] showing it is [ilmath]\cap[/ilmath]-closed.
  2. [ilmath]\mathcal{F}_\cup[/ilmath] is [ilmath]\setminus[/ilmath]-closed.
    • Let [ilmath]S,T\in\mathcal{F}_\cup[/ilmath] be given, we want to show that [ilmath]S-T\in\mathcal{F}_\cup[/ilmath]
      • First note that this means [ilmath]S=\bigudot_{i=1}^{m_1}S_i[/ilmath] and [ilmath]T=\bigudot_{j=1}^{m_2}T_j[/ilmath] for some [ilmath]m_1,m_2\in\mathbb{N} [/ilmath] and [ilmath]S_i,T_j\in\mathcal{F} [/ilmath]; so we really want to show that [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)\in\mathcal{F}_\cup[/ilmath]
      • Note that [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{i=1}^{m_1}\big(S_i-\bigudot_{j=1}^{m_2}T_j\big)[/ilmath][ilmath]=\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}(S_i-T_j)\big)[/ilmath] Caution:I didn't just pull this out of thin air, I drew a venn diagram for the case [ilmath]m_1=m_2=1[/ilmath] and worked from there, adding some notes on why would be good.
        • As [ilmath]S_i,T_j\in\mathcal{F} [/ilmath] we know there exists a finite sequence of elements of [ilmath]\mathcal{F} [/ilmath], say [ilmath] ({ W_{ijk} })_{ k = 1 }^{ n_{ij} }\subseteq \mathcal{F} [/ilmath] such that [ilmath]S_i-T_j=\bigudot_{k=1}^{n_{ij} }W_{ijk}[/ilmath].
          • Define [ilmath]W_{ij}:=\bigudot_{k=1}^{n_{ij} }W_{ijk} = S_i-T_j[/ilmath], then [ilmath]W_{ij}\in\mathcal{F}_\cup[/ilmath]
      • Thus: [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}(S_i-T_j)\big)[/ilmath][ilmath]=\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}W_{ij}\big)[/ilmath] (really we should say [ilmath]\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}\big(\bigudot_{k=1}^{n_{ij} }W_{ijk}\big)\big)[/ilmath] but it nicer with just [ilmath]W_{ij}\in\mathcal{F}_\cup[/ilmath])
      • We have already shown that [ilmath]\mathcal{F}_\cup[/ilmath] is [ilmath]\cap[/ilmath]-closed, thus [ilmath]\bigcap_{j=1}^{m_2}W_{ij}\in\mathcal{F}_\cup[/ilmath]
        • There exists [ilmath] ({ V_{il} })_{ l = 1 }^{ n_i }\subseteq \mathcal{F} [/ilmath] (notice the [ilmath]V_{il}\in\mathcal{F} [/ilmath] not [ilmath]\mathcal{F}_\cup[/ilmath]) such that [ilmath]\bigudot_{l=1}^{n_i}V_{il}=\bigcap_{j=1}^{m_2}W_{ij}[/ilmath]
      • So: [ilmath]\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{i=1}^{m_1}\big(\bigcap_{j=1}^{m_2}W_{ij}\big)[/ilmath][ilmath]=\bigudot_{i=1}^{m_1}\big(\bigudot_{l=1}^{n_i}V_{il}\big)[/ilmath]
      • This is a finite disjoint union of elements of [ilmath]\mathcal{F} [/ilmath] with [ilmath]\sum_{i=1}^{m_1}n_i[/ilmath] terms, thus [ilmath]S-T=\big(\bigudot_{i=1}^{m_1}S_i\big)-\big(\bigudot_{j=1}^{m_2}T_j\big)=\bigudot_{i=1}^{m_1}\big(\bigudot_{l=1}^{n_i}V_{il}\big)\in\mathcal{F}_\cup[/ilmath]
    • Since [ilmath]S[/ilmath] and [ilmath]T[/ilmath] were arbitrary we have completed the proof that [ilmath]\mathcal{F}_\cup[/ilmath] is [ilmath]\setminus[/ilmath]-closed
  3. [ilmath]\mathcal{F}_\cup[/ilmath] is [ilmath]\cup[/ilmath]-closed
    • Let [ilmath]S,T\in\mathcal{F}_\cup[/ilmath] be given
      • Note [ilmath]S\cup T=(S-T)\udot(S\cap T)\udot(T-S)[/ilmath] and each term of this union is in [ilmath]\mathcal{F}_\cup[/ilmath]
    • This completes the proof.

Proof that [ilmath]\mathcal{F}_\cup[/ilmath] is the ring generated by [ilmath]\mathcal{F} [/ilmath]

Grade: A
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 quite easy, just assume that [ilmath]\mathcal{F}_\cup[/ilmath] isn't the ring generated by [ilmath]\mathcal{F} [/ilmath], that means the generated ring is a proper subset of [ilmath]\mathcal{F}_\cup[/ilmath], so there is an element in [ilmath]\mathcal{F}_\cup[/ilmath] not in[ilmath]R(\mathcal{F})[/ilmath]. That element must be the disjoint union of a finite number of elements from [ilmath]\mathcal{F} [/ilmath], that violates that [ilmath]R(\mathcal{F})[/ilmath] is closed under unions! As the union of these elements is not in the generated ring!

Purpose

The main use of this theorem is to aid with extending a pre-measure on a semi-ring to a pre-measure. Defining a pre-measure on a semi-ring is often very easy, where as defining one on a ring is very tedious.

See also

Notes

  1. At the time of writing the ring generated by page doesn't exist so I write this note to clarify.
    • The ring of sets generated by an arbitrary collection of sets is the smallest ring of sets containing every element in the arbitrary collection
  2. An F is a bit like an R with an unfinished loop and the foot at the right. "Semi Ring".
  3. 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].
  4. The proof of this is easy, as the intersection of sets is a subset of each set we see [ilmath]A\cap B_i\subseteq B_i[/ilmath] for each [ilmath]i[/ilmath]. As the [ilmath]B_i[/ilmath] are pairwise disjoint, we see the [ilmath]A\cap B_i[/ilmath] must be also

References

  1. 1.0 1.1 Measures, Integrals and Martingales - René L. Schilling
  2. Page 19 -Measure Theory - Paul R. Halmos