Sigma-algebra

A Sigma-algebra of sets, or [ilmath]\sigma[/ilmath]-algebra is very similar to a [ilmath]\sigma[/ilmath]-ring of sets.

Like how ring of sets and algebra of sets differ, the same applies to [ilmath]\sigma[/ilmath]-ring compared to [ilmath]\sigma[/ilmath]-algebra

Definition

A non empty class of sets [ilmath]S[/ilmath] is a [ilmath]\sigma[/ilmath]-algebra if

• if [math]A\in S[/math] then [math]A^c\in S[/math]
• if [math]\{A_n\}_{n=1}^\infty\subset S[/math] then [math]\cup^\infty_{n=1}A_n\in S[/math]

That is it is closed under complement and countable union.

Immediate consequences

Among other things immediately we see that:

• [ilmath]\mathcal{A} [/ilmath] is closed under set subtraction

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As [ilmath]A-B=(A^c\cup B)^c[/ilmath] and a [ilmath]\sigma[/ilmath]-algebra is closed under complements and unions, this shows it is closed under set subtraction too

• [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]-closed (furthermore, that [ilmath]\mathcal{A} [/ilmath] is in fact [ilmath]\sigma[/ilmath]-[ilmath]\cap[/ilmath]-closed - that is closed under countable intersections)

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See Properties of a class of sets closed under set subtraction

• [ilmath]\emptyset\in\mathcal{A} [/ilmath]

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[ilmath]\forall A\in\mathcal{A} [/ilmath] we have [ilmath]A-A\in\mathcal{A} [/ilmath] (by closure under set subtraction), as [ilmath]A-A=\emptyset[/ilmath], [ilmath]\emptyset\in\mathcal{A} [/ilmath]

• [ilmath]X\in\mathcal{A} [/ilmath]

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As [ilmath]\emptyset\in\mathcal{A} [/ilmath] and it is closed under complement we see that [ilmath]\emptyset^c\in\mathcal{A} [/ilmath] (by closure under complement) and [ilmath]\emptyset^c=X[/ilmath] - the claim follows.

• [ilmath]\mathcal{A} [/ilmath] is a [ilmath]\sigma[/ilmath]-algebra [ilmath]\implies[/ilmath] [ilmath]\mathcal{A} [/ilmath] is a [ilmath]\sigma[/ilmath]-ring

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To prove this we must check:

1. [ilmath]\mathcal{A} [/ilmath] is closed under countable union
   • True by definition of [ilmath]\sigma[/ilmath]-algebra
2. [ilmath]\mathcal{A} [/ilmath] is closed under set subtraction
   • We've already shown this, so this is true too.

This completes the proof.

Important theorems

The intersection of [ilmath]\sigma[/ilmath]-algebras is a [ilmath]\sigma[/ilmath]-algebra

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TODO: Proof - see PTACC page 5, also in Halmos AND in that other book

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Common [ilmath]\sigma[/ilmath]-algebras

See also: Index of common [ilmath]\sigma[/ilmath]-algebras

• [ilmath]\sigma[/ilmath]-algebra generated by
• Trace [ilmath]\sigma[/ilmath]-algebra
• Pre-image [ilmath]\sigma[/ilmath]-algebra

See also

• Types of set algebras
• [ilmath]\sigma[/ilmath]-algebra generated by
• [ilmath]\sigma[/ilmath]-ring
• Properties of a class of sets closed under set subtraction