Semi-ring of sets/Definition

Definition

$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$

$\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$

$\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }$

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

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

Notes

Jump up ↑ An F is a bit like an R with an unfinished loop and the foot at the right. "Semi Ring".
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]
  - 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

Jump up ↑ Measures, Integrals and Martingales - René L. Schilling

[1] Jump up ↑ An F is a bit like an R with an unfinished loop and the foot at the right. "Semi Ring".

[3] 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:
$\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\underbrace{\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)}_{\text{the }S_i\text{ are pairwise disjoint} }\overbrace{\wedge}^\text{and}\left(S-T=\bigcup_{i=1}^m S_i\right)\right]$
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 $T$ .

[3] $\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\underbrace{\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)}_{\text{the }S_i\text{ are pairwise disjoint} }\overbrace{\wedge}^\text{and}\left(S-T=\bigcup_{i=1}^m S_i\right)\right]$
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 $T$ .

[4] 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 $T$ .

[5] 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 $T$ .

[MIAMRLS-2] Jump up ↑ Measures, Integrals and Martingales - René L. Schilling

[Note 1]

[1]

[Note 2]

Semi-ring of sets/Definition

Definition

Notes

References

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