The ring of sets generated by a semi-ring is the set containing the semi-ring and all finite disjoint unions
Contents
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]
A collection of sets, [ilmath]\mathcal{F} [/ilmath][Note 2] is called a semi-ring 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][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].
Proof that [ilmath]\mathcal{F}_\cup[/ilmath] is a ring
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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!
- [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.
- 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]
- [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]
- 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].
- 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
- Let [ilmath]S,T\in\mathcal{F}_\cup[/ilmath] be given, we want to show that [ilmath]S-T\in\mathcal{F}_\cup[/ilmath]
- [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.
- Let [ilmath]S,T\in\mathcal{F}_\cup[/ilmath] be given
Proof that [ilmath]\mathcal{F}_\cup[/ilmath] is the ring generated by [ilmath]\mathcal{F} [/ilmath]
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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
- ↑ 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
- ↑ 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(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].
- 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
- [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]
- ↑ 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
- ↑ 1.0 1.1 Measures, Integrals and Martingales - René L. Schilling
- ↑ Page 19 -Measure Theory - Paul R. Halmos
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