# Algebra of sets

(Redirected from Algebra (measure theory))
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Defining properties: Algebra of sets [ilmath]\mathcal{A}\subseteq\mathcal{P}(X)[/ilmath]For an algebra of sets, [ilmath]\mathcal{A} [/ilmath] on [ilmath]X[/ilmath] [ilmath]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/ilmath] [ilmath]\forall A,B\in\mathcal{A}[A\cup B\in\mathcal{A} ][/ilmath]
Note: Every algebra of sets is a ring of sets (see below)

## Definition

An algebra of sets is a collection of sets, [ilmath]\mathcal{A} [/ilmath] such that:

• [ilmath]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/ilmath][Note 1]
• In words: For all [ilmath]A[/ilmath] in [ilmath]\mathcal{A} [/ilmath] the complement of [ilmath]A[/ilmath] (with respect to [ilmath]X[/ilmath]) is also in [ilmath]\mathcal{A} [/ilmath]
• [ilmath]\forall A,B\in\mathcal{A}[A\cup B\in\mathcal{A}][/ilmath]
• In words: For all [ilmath]A[/ilmath] and [ilmath]B[/ilmath] in [ilmath]\mathcal{A} [/ilmath] their union is also in [ilmath]\mathcal{A} [/ilmath]

Claim 1: Every algebra of sets is also a ring of sets

## Immediate properties

TODO: Do this as a list of inline theorem boxes

• [ilmath]\mathcal{A} [/ilmath] is [ilmath]\setminus[/ilmath]-closed
• [ilmath]\emptyset\in\mathcal{A} [/ilmath]
• [ilmath]X\in\mathcal{A} [/ilmath]
• [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]-closed

## Proof of claims

Claim 1: Every algebra of sets is also a ring of sets

This is trivial. In order to show [ilmath]\mathcal{A} [/ilmath] is a ring of sets we require two properties:

1. [ilmath]\forall A,B\in\mathcal{A}[A\cup B\in\mathcal{A}][/ilmath] - this is clearly satisfied by definition of an algebra of sets
2. [ilmath]\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}][/ilmath] - that is [ilmath]\mathcal{A} [/ilmath] must be [ilmath]\setminus[/ilmath]-closed

This completes the proof