The set of all [ilmath]\mu^*[/ilmath]-measurable sets is a ring

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Statement

[ilmath]\mathcal{S} [/ilmath], the set of all [ilmath]\mu^*[/ilmath] measurable sets, is a ring of sets.

Recall that given an outer-measure, [ilmath]\mu^*:H\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath], where [ilmath]H[/ilmath] is a hereditary [ilmath]\sigma[/ilmath]-ring that we call a set, [ilmath]A\in H[/ilmath] [ilmath]\mu^*[/ilmath]-measurable if:

• [ilmath]\forall B\in H[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)][/ilmath].
• See the page [ilmath]\mu^*[/ilmath]-measurable set for more information.

Proof

Recall what we must show in order for [ilmath]\mathcal{S} [/ilmath] to be a ring of sets:

It is sufficient to show only:

• [ilmath]\forall A,B\in \mathcal{S}[A\cup B\in \mathcal{S}][/ilmath]
• [ilmath]\forall A,B\in \mathcal{S}[A-B\in \mathcal{S}][/ilmath]

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Extremely well documented in Halmos

See also

• The set of all [ilmath]\mu^*[/ilmath]-measurable sets is a [ilmath]\sigma[/ilmath]-ring
• The restriction of an outer-measure to the set of all [ilmath]\mu^*[/ilmath]-measurable sets is a measure