Difference between revisions of "Sigma-algebra"

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{{:Sigma-algebra/Infobox}}
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: '''Note: ''' A ''Sigma-algebra'' of sets, or {{sigma|algebra}} is very similar to a [[Sigma-ring|{{sigma|ring}}]] of sets.
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:: A [[ring of sets]] is to an [[algebra of sets]] as a [[sigma-ring|{{sigma|ring}}]] is to a ''{{sigma|algebra}}''
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__TOC__
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==Definition==
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{{:Sigma-algebra/Definition}}
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===Note on Alternative Definitions===
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{{Begin Notebox}}
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Many books have slightly different definitions of a {{sigma|algebra}}, the definition above is actually equivalent to the longer definitions one might see around.<br/> The two properties above give rise to all the others.
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{{Begin Notebox Content}}
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Usually on this project differing definitions are listed, but due to the trivial proofs that they share properties, this was omitted.
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To show this yourself, do the following:
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# Show a {{sigma|algebra}} is closed under [[set-subtraction]], {{M|\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}]}}
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# Use this to show {{M|\emptyset\in\mathcal{A} }}, notice {{M|1=\emptyset=A-A\in\mathcal{A} }} for any {{M|A\in\mathcal{A} }}
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# {{M|X\in\mathcal{A} }} as {{M|\emptyset^C\in\mathcal{A} }}
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This resolves most ambiguities.
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{{End Notebox Content}}{{End Notebox}}
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==Immediate consequences==
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Among other things immediately we see that:
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{{Begin Inline Theorem}}
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* {{M|\mathcal{A} }} is closed under [[Set subtraction|set subtraction]]
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{{Begin Inline Proof}}
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:: As {{M|1=A-B=(A^c\cup B)^c}} and a {{sigma|algebra}} is closed under complements and unions, this shows it is closed under set subtraction too
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{{Begin Notebox}}
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One might argue that we only know {{M|\mathcal{A} }} is closed under ''countable'' union, not ''finite'' (2) union. so we cannot know {{M|A^C\cup B\in\mathcal{A} }}, thus this proof is invalid. It is not invalid.
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{{Begin Notebox Content}}
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The person who mentioned this argued that we ''require'' [[set-subtraction]] in order to show {{M|\emptyset\in\mathcal{A} }} - which we do not yet know, so we cannot construct the family:
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* {{M|1=\{A_n\}_{n=1}^\infty}} with {{M|1=A_1=A^C}}, {{M|1=A_2=B}}, {{M|1=A_i=\emptyset}} for {{M|i> 2}}
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And they are not wrong, however:
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* {{M|1=A^C\cup B=A^C\cup B\cup B\cup B\ldots}}
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So
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* {{M|1=\{A_n\}_{n=1}^\infty}} with {{M|1=A_1=A^C}} and {{M|1=A_i=B}} for {{M|i> 1}} ''is'' something we can construct.
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{{End Notebox Content}}{{End Notebox}}
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{{End Proof}}{{End Theorem}}
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{{Begin Inline Theorem}}
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* {{M|\mathcal{A} }} is {{M|\cap}}-closed (furthermore, that {{M|\mathcal{A} }} is in fact {{M|\sigma}}-{{M|\cap}}-closed - that is closed under countable intersections)
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{{Begin Inline Proof}}
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{{Requires proof|We should do this here}}
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: See [[Class of sets closed under set-subtraction properties|Properties of a class of sets closed under set subtraction]]
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{{End Proof}}{{End Theorem}}
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{{Begin Inline Theorem}}
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* {{M|\emptyset\in\mathcal{A} }}
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{{Begin Inline Proof}}
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:: {{M|\forall A\in\mathcal{A} }} we have {{M|A-A\in\mathcal{A} }} (by closure under set subtraction), as {{M|1=A-A=\emptyset}}, {{M|\emptyset\in\mathcal{A} }}
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{{End Proof}}{{End Theorem}}
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{{Begin Inline Theorem}}
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* {{M|X\in\mathcal{A} }}<ref group="Note">''Measures, Integrals and Martingales'' puts this in the definition of {{sigma|algebras}}</ref>
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{{Begin Inline Proof}}
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:: As {{M|\emptyset\in\mathcal{A} }} and it is closed under complement we see that {{M|\emptyset^c\in\mathcal{A} }} (by closure under complement) and {{M|1=\emptyset^c=X}} - the claim follows.
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{{End Proof}}{{End Theorem}}
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{{Begin Inline Theorem}}
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* {{M|\mathcal{A} }} is a {{sigma|algebra}} {{M|\implies}} {{M|\mathcal{A} }} is a [[Sigma-ring|{{Sigma|ring}}]]
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{{Begin Inline Proof}}
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:: To prove this we must check:
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::# {{M|\mathcal{A} }} is closed under countable union
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::#* True by definition of {{Sigma|algebra}}
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::# {{M|\mathcal{A} }} is closed under [[Set subtraction|set subtraction]]
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::#* We've already shown this, so this is true too.
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:: This completes the proof.
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{{End Proof}}{{End Theorem}}
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==Important theorems==
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* [[The intersection of sigma-algebras is a sigma-algebra|The intersection of {{sigma|algebras}} is a {{sigma|algebra}}]]
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==Common {{Sigma|algebras}}==
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'''See also: [[Index of common sigma-algebras|Index of common {{Sigma|algebras}}]]'''
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* [[Sigma-algebra generated by|{{Sigma|algebra}} generated by]]
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* [[Trace sigma-algebra|Trace {{Sigma|algebra}}]]
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* [[Pre-image sigma-algebra|Pre-image {{Sigma|algebra}}]]
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==See also==
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* [[Types of set algebras]]
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* [[Sigma-algebra generated by|{{Sigma|algebra}} generated by]]
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* [[Sigma-ring|{{Sigma|ring}}]]
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* [[Class of sets closed under set-subtraction properties|Properties of a class of sets closed under set subtraction]]
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==Notes==
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<references group="Note"/>
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==References==
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<references/>
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{{Measure theory navbox|plain}}
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{{Definition|Measure Theory}}
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{{Theorem Of|Measure Theory}}
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=OLD PAGE=
 
A '''Sigma-algebra''' of sets, or {{sigma|algebra}} is very similar to a [[Sigma-ring|{{sigma|ring}}]] of sets.  
 
A '''Sigma-algebra''' of sets, or {{sigma|algebra}} is very similar to a [[Sigma-ring|{{sigma|ring}}]] of sets.  
  

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Sigma algebra
[math]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/math]
[math]\forall\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}\left[\bigcup_{n=1}^\infty A_n\in\mathcal{A}\right][/math]
For a [ilmath]\sigma[/ilmath]-algebra [ilmath](X,\mathcal{A}\subseteq\mathcal{P}(X))[/ilmath]
Note: A Sigma-algebra 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

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:

  1. Show a [ilmath]\sigma[/ilmath]-algebra is closed under set-subtraction, [ilmath]\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}][/ilmath]
  2. Use this to show [ilmath]\emptyset\in\mathcal{A} [/ilmath], notice [ilmath]\emptyset=A-A\in\mathcal{A}[/ilmath] for any [ilmath]A\in\mathcal{A} [/ilmath]
  3. [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:

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

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 set-subtraction 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)

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See Properties of a class of sets closed under set subtraction

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

[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][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:
  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

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

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

See also

Notes

  1. ↑ So [ilmath]\mathcal{A}\subseteq\mathcal{P}(X)[/ilmath]
  2. ↑ Measures, Integrals and Martingales puts this in the definition of [ilmath]\sigma[/ilmath]-algebras

References

  1. ↑ Measures, Integrals and Martingales - René L. Schilling

v • d • e
 
Measure Theory
Overview of the basic and important objects studied in measure theory
Set-structures
Measures
Pre-measure [ilmath]\leadsto[/ilmath] outer-measure [ilmath]\leadsto[/ilmath] measure (see Extending pre-measures to measures (via: Extending pre-measures to outer-measures [ilmath]\rightarrow[/ilmath] "Restricting outer-measures to measures" maybe?). Alternatively: inner-measure can be used.
measure space
Functions
Spaces
Integrals




OLD PAGE

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[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:

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)

See Properties of a class of sets closed under set subtraction

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

[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][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:
  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


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

See also

Notes

  1. ↑ 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)
  2. ↑ Measures, Integrals and Martingales puts this in the definition of [ilmath]\sigma[/ilmath]-algebras

References

  1. ↑ Halmos - Measure Theory - page 28 - Springer - Graduate Texts in Mathematics - 18
  2. ↑ Measures, Integrals and Martingales - Rene L. Schilling



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