Semi-ring of sets

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 $S-T$ .

Purpose

The main motivation for semi-rings (in Measure Theory at least) is to let us provide a pre-measure on a semi-ring (a kind of pre-measure^{[Note 3]}) and then use a theorem to prove this can be extended to a normal pre-measure (a similar structure define on a ring of sets instead). Then we can apply extending pre-measures to outer-measures to obtain an outer-measure, all without going through the tedious task of defining a pre-measure on a ring and doing only the basics by defining it on a semi-ring.

Examples

Lebesgue pre-measure on a semi-ring - in one dimension the semi-ring, \mathscr{J}^1, here is the collection of all half-open-half-closed intervals on the real line, [a,b)\subset\mathbb{R} (with [a,b):=\{x\in\mathbb{R}\ \vert a\le x < r\}) for a,b\in\mathbb{R} with the convention that if then [a,b)=\emptyset.
1. Clearly, $\emptyset\in\mathscr{J}^1$
2. Let $a,b,c,d\in\mathbb{R}$ , Suppose $a<b$ and $c<d$ (as if either interval is the empty set the result is trivial). Suppose they partially intersect with $a<c$ and $b<d$ , then clearly $[a,b)\cap[c,d)=[c,b)$ , this is the most difficult case.
3. Using the same variables, the "hardest" case is that of $a<c<d<b$ so we have $[a,b)-[c,d)$ with $[c,d)$ being inside $[a,b)$ , then: $[a,b)-[c,d)=[a,c)\cup[d,b)$ . The other cases are easier still.
- Showing even $\mathscr{J}^1$ is a ring of sets is very tedious. If the reader cannot see this, he should try it. Where as defining pre-measure on a semi-ring instead is something we've already done most of the work for!

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 $T$ .
Jump up ↑ Many authors consider a pre-measure to be something we can extend to a measure somehow. We do not use this. Instead we define a pre-measure as being a function with certain properties on a ring of sets. This is useful because a pre-measure, under this definition, is almost a measure. A ring of sets is closed under all the elementary set operations.
We also adopt the convention of calling anything that can be extended to either a pre-measure (and thus an outer-measure and later a measure) a where $X$ is say a semi-ring or something.
All we need to do is show the pre-measure on $X$ extends uniquely to a pre-measure to allow the theorems (extending pre-measures to measures) to yield us a measure.

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 $T$ .

[4] Jump up ↑ Many authors consider a pre-measure to be something we can extend to a measure somehow. We do not use this. Instead we define a pre-measure as being a function with certain properties on a ring of sets. This is useful because a pre-measure, under this definition, is almost a measure. A ring of sets is closed under all the elementary set operations.
We also adopt the convention of calling anything that can be extended to either a pre-measure (and thus an outer-measure and later a measure) a where $X$ is say a semi-ring or something.
All we need to do is show the pre-measure on $X$ extends uniquely to a pre-measure to allow the theorems (extending pre-measures to measures) to yield us a measure.

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

[Note 1]

[1]

[Note 2]

[Note 3]

Semi-ring of sets

Contents

Definition

Purpose

Examples

See also

Notes

References

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