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

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].

Ring of sets generated by/Definition

Proof

The proof will have two distinct steps:

Showing that [ilmath]\mathcal{F}_\cup:=\{(A_i)_{i=1}^m\ \vert\ m\in\mathbb{N}\ \wedge\ (A_i)_{i=1}^m\subseteq R\}[/ilmath] is indeed a ring
Showing that [ilmath]\mathcal{F}_\cup=R(\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:

Do it on paper first, consult page 39 of Books:Measures, Integrals and Martingales - René L. Schilling

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.

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].

References

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

[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] The ring of sets generated by an arbitrary collection of sets is the smallest ring of sets containing every element in the arbitrary collection

[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].

[5] [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].

[6] 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].

[7] 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].

[MIAMRLS-2] 1.0 ^1.1 Measures, Integrals and Martingales - René L. Schilling

[3] Page 19 -Measure Theory - Paul R. Halmos

[Note 1]

[1]

[2]

[Note 2]

[Note 3]

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

Proof

Purpose

See also

Notes

References

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