Difference between revisions of "Sigma-algebra"
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* if <math>A\in S</math> then <math>A^c\in S</math> | * 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> | * if <math>\{A_n\}_{n=1}^\infty\subset S</math> then <math>\cup^\infty_{n=1}A_n\in S</math> | ||
| − | That is it | + | That is it is closed under [[Complement|complement]] and [[Countable|countable]] [[Union|union]].<br/> |
| + | ===Immediate consequences=== | ||
| + | Among other things immediately we see that: | ||
| + | {{Begin Inline Theorem}} | ||
| + | * {{M|\mathcal{A} }} is closed under [[Set subtraction|set subtraction]] | ||
| + | {{Begin Inline Proof}} | ||
| + | :: 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 | ||
| + | {{End Proof}}{{End Theorem}} | ||
{{Begin Inline Theorem}} | {{Begin Inline Theorem}} | ||
* {{M|\emptyset\in\mathcal{A} }} | * {{M|\emptyset\in\mathcal{A} }} | ||
| Line 22: | Line 29: | ||
{{Begin Inline Proof}} | {{Begin Inline Proof}} | ||
:: To prove this we must check: | :: To prove this we must check: | ||
| − | ::# {{M|\ | + | ::# {{M|\mathcal{A} }} is closed under countable union |
| + | ::#* True by definition of {{Sigma|algebra}} | ||
| + | ::# {{M|\mathcal{A} }} is closed under [[Set subtraction|set subtraction]] | ||
::#* We've already shown this, so this is true too. | ::#* We've already shown this, so this is true too. | ||
:: This completes the proof. | :: This completes the proof. | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
| − | |||
==Important theorems== | ==Important theorems== | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
Revision as of 16:26, 23 August 2015
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
Contents
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]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]\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:
- [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
- Pre-image [ilmath]\sigma[/ilmath]-algebra
See also
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
