# A pre-measure on a semi-ring may be extended uniquely to a pre-measure on a ring

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## Statement

Given a pre-measure on a semi-ring, [ilmath]\tilde{\mu}:\mathcal{F}\rightarrow\overline{\mathbb{R}_{\ge 0} } [/ilmath] (that is a function whose domain is a semi-ring of sets that is countably additive with [ilmath]\tilde{\mu}(\emptyset)=0[/ilmath]) then we may extend [ilmath]\srmu[/ilmath] to a pre-measure, [ilmath]\rmu:R(\mathcal{F})\rightarrow\overline{\mathbb{R}_{\ge 0} } [/ilmath][Note 1]; furthermore this extension is unique[1]. The details follow:

• The ring generated by a semi-ring is exactly the set of all finite disjoint unions of elements from that semiring.
• That is to say, [ilmath]R(\mathcal{F})=\left\{\left.\bigudot_{i=1}^nA_i\ \right\vert\ (A_i)_{i=1}^n\subseteq\mathcal{F}\right\}[/ilmath]
• so any [ilmath]A\in R(\mathcal{F}) [/ilmath] can be written as [ilmath]A=\bigudot_{i=1}^n A_i[/ilmath] for some finite sequence of pariwise disjoint sets, [ilmath] ({ A_i })_{ i = 1 }^{ n }\subseteq \mathcal{F} [/ilmath][Note 2]
• We define the induced pre-measure, [ilmath]\rmu:R(\mathcal{F})\rightarrow\overline{\mathbb{R}_{\ge 0} } [/ilmath] as follows:
• [ilmath]\rmu:\bigudot_{i=1}^nA_i\mapsto\sum_{i=1}^n\srmu(A_i)[/ilmath], and we claim this map is well-defined

## Prerequisites

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

## Proof

Grade: A
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## Notes

1. Here [ilmath]R(X)[/ilmath] denotes the ring of sets generated by a collection of sets, [ilmath]X[/ilmath].
2. I've mentioned it a few times but in case it isn't clear:
• For [ilmath]A\in R(\mathcal{F})[/ilmath] we have [ilmath]A=\bigudot_{i=1}^nA_i[/ilmath] for some finite sequence, [ilmath] ({ A_i })_{ i = 1 }^{ n }\subseteq \mathcal{F} [/ilmath], note the elements of the sequence are in [ilmath]\mathcal{F} [/ilmath]