Sigmaalgebra
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Sigma algebra  
[math]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/math] For a [ilmath]\sigma[/ilmath]algebra [ilmath](X,\mathcal{A}\subseteq\mathcal{P}(X))[/ilmath]
[math]\forall\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}\left[\bigcup_{n=1}^\infty A_n\in\mathcal{A}\right][/math] 
 Note: A Sigmaalgebra of sets, or [ilmath]\sigma[/ilmath]algebra is very similar to a [ilmath]\sigma[/ilmath]ring of sets.
 A ring of sets is to an algebra of sets as a [ilmath]\sigma[/ilmath]ring is to a [ilmath]\sigma[/ilmath]algebra
Contents
Definition
Given a set [ilmath]X[/ilmath] a [ilmath]\sigma[/ilmath]algebra on [ilmath]X[/ilmath] is a family of subsets of [ilmath]X[/ilmath], [ilmath]\mathcal{A} [/ilmath]^{[Note 1]}, such that^{[1]}:
 [ilmath]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/ilmath]  Stable under complements
 [ilmath]\forall\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}\left[\bigcup_{n=1}^\infty A_n\in\mathcal{A}\right][/ilmath]  Stable under countable union
Note on Alternative Definitions
Many books have slightly different definitions of a [ilmath]\sigma[/ilmath]algebra, the definition above is actually equivalent to the longer definitions one might see around.
The two properties above give rise to all the others.
Usually on this project differing definitions are listed, but due to the trivial proofs that they share properties, this was omitted.
To show this yourself, do the following:
 Show a [ilmath]\sigma[/ilmath]algebra is closed under setsubtraction, [ilmath]\forall A,B\in\mathcal{A}[AB\in\mathcal{A}][/ilmath]
 Use this to show [ilmath]\emptyset\in\mathcal{A} [/ilmath], notice [ilmath]\emptyset=AA\in\mathcal{A}[/ilmath] for any [ilmath]A\in\mathcal{A} [/ilmath]
 [ilmath]X\in\mathcal{A} [/ilmath] as [ilmath]\emptyset^C\in\mathcal{A} [/ilmath]
This resolves most ambiguities.
Immediate consequences
Among other things immediately we see that:
 [ilmath]\mathcal{A} [/ilmath] is closed under set subtraction
 As [ilmath]AB=(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
One might argue that we only know [ilmath]\mathcal{A} [/ilmath] is closed under countable union, not finite (2) union. so we cannot know [ilmath]A^C\cup B\in\mathcal{A} [/ilmath], thus this proof is invalid. It is not invalid.
The person who mentioned this argued that we require setsubtraction in order to show [ilmath]\emptyset\in\mathcal{A} [/ilmath]  which we do not yet know, so we cannot construct the family:
 [ilmath]\{A_n\}_{n=1}^\infty[/ilmath] with [ilmath]A_1=A^C[/ilmath], [ilmath]A_2=B[/ilmath], [ilmath]A_i=\emptyset[/ilmath] for [ilmath]i> 2[/ilmath]
And they are not wrong, however:
 [ilmath]A^C\cup B=A^C\cup B\cup B\cup B\ldots[/ilmath]
So
 [ilmath]\{A_n\}_{n=1}^\infty[/ilmath] with [ilmath]A_1=A^C[/ilmath] and [ilmath]A_i=B[/ilmath] for [ilmath]i> 1[/ilmath] is something we can construct.
 [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)
The message provided is:
 [ilmath]\emptyset\in\mathcal{A} [/ilmath]
 [ilmath]\forall A\in\mathcal{A} [/ilmath] we have [ilmath]AA\in\mathcal{A} [/ilmath] (by closure under set subtraction), as [ilmath]AA=\emptyset[/ilmath], [ilmath]\emptyset\in\mathcal{A} [/ilmath]
 [ilmath]X\in\mathcal{A} [/ilmath]^{[Note 2]}
 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
 To prove this we must check:
 [ilmath]\mathcal{A} [/ilmath] is closed under countable union
 True by definition of [ilmath]\sigma[/ilmath]algebra
 [ilmath]\mathcal{A} [/ilmath] is closed under set subtraction
 We've already shown this, so this is true too.
 [ilmath]\mathcal{A} [/ilmath] is closed under countable union
 This completes the proof.
 To prove this we must check:
Important theorems
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
 Preimage [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
Notes
 ↑ So [ilmath]\mathcal{A}\subseteq\mathcal{P}(X)[/ilmath]
 ↑ Measures, Integrals and Martingales puts this in the definition of [ilmath]\sigma[/ilmath]algebras
References

OLD PAGE
A Sigmaalgebra 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^{[Note 1]} if^{[1]}^{[2]}
 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
 As [ilmath]AB=(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)
 [ilmath]\emptyset\in\mathcal{A} [/ilmath]
 [ilmath]\forall A\in\mathcal{A} [/ilmath] we have [ilmath]AA\in\mathcal{A} [/ilmath] (by closure under set subtraction), as [ilmath]AA=\emptyset[/ilmath], [ilmath]\emptyset\in\mathcal{A} [/ilmath]
 [ilmath]X\in\mathcal{A} [/ilmath]^{[Note 2]}
 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
 To prove this we must check:
 [ilmath]\mathcal{A} [/ilmath] is closed under countable union
 True by definition of [ilmath]\sigma[/ilmath]algebra
 [ilmath]\mathcal{A} [/ilmath] is closed under set subtraction
 We've already shown this, so this is true too.
 [ilmath]\mathcal{A} [/ilmath] is closed under countable union
 This completes the proof.
 To prove this we must check:
Important theorems
The intersection of [ilmath]\sigma[/ilmath]algebras is a [ilmath]\sigma[/ilmath]algebra
TODO: Proof  see PTACC page 5, also in Halmos AND in that other book
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
 Preimage [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
Notes
 ↑ Some books (notably Measures, Integrals and Martingales) give [ilmath]X\in\mathcal{A} [/ilmath] as a defining property of [ilmath]\sigma[/ilmath]algebras, however the two listed are sufficient to show this (see the immediate consequences section)
 ↑ Measures, Integrals and Martingales puts this in the definition of [ilmath]\sigma[/ilmath]algebras
References
 ↑ Halmos  Measure Theory  page 28  Springer  Graduate Texts in Mathematics  18
 ↑ Measures, Integrals and Martingales  Rene L. Schilling
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