Difference between revisions of "Trace sigma-algebra"
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(Created page with "Let {{M|E\subseteq X}}, then {{M|1=\mathcal{A}_E:=\mathcal{A}\cap E:=\{A\cap E\vert A\in\mathcal{A}\} }} {{Todo|Measures Integrals and Martingales - page 16}} {{Definition|Me...") |
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| − | Let {{M| | + | {{Stub page|grade=B|More results would be good. Relation to pullback too}} |
| − | {{ | + | __TOC__ |
| − | + | ==Definition== | |
| + | Let {{M|(X,\mathcal{A})}} be a [[sigma-algebra|{{sigma|algebra}}]] and let {{M|Y\subseteq X}} be any [[subset]] of {{M|X}}, then we may construct a {{sigma|algebra}} on {{M|Y}} called the ''trace {{sigma|algebra}}'', {{M|\mathcal{A}_Y}} given by{{rMIAMRLS}}: | ||
| + | * {{M|1=\mathcal{A}_Y:=\left\{Y\cap A\ \vert A\in\mathcal{A}\right\} }} | ||
| + | '''Claim: ''' {{M|(Y,\mathcal{A}_Y)}} is a {{sigma|algebra}} | ||
| + | ==Proof of claims== | ||
| + | {{Begin Inline Theorem}} | ||
| + | '''[[Trace sigma-algebra/Proof of claim that it actually is a sigma-algebra|Claim 1]]: ''' that {{M|(Y,\mathcal{A}_Y)}} is indeed a [[sigma-algebra|{{sigma|algebra}}]] | ||
| + | {{Begin Inline Proof}} | ||
| + | {{:Trace sigma-algebra/Proof of claim that it actually is a sigma-algebra}} | ||
| + | {{End Proof}}{{End Theorem}} | ||
| + | ==References== | ||
| + | <references/> | ||
| + | {{Measure theory navbox|plain}} | ||
{{Definition|Measure Theory}} | {{Definition|Measure Theory}} | ||
Latest revision as of 12:00, 23 August 2018
Stub grade: B
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More results would be good. Relation to pullback too
Contents
Definition
Let [ilmath](X,\mathcal{A})[/ilmath] be a [ilmath]\sigma[/ilmath]-algebra and let [ilmath]Y\subseteq X[/ilmath] be any subset of [ilmath]X[/ilmath], then we may construct a [ilmath]\sigma[/ilmath]-algebra on [ilmath]Y[/ilmath] called the trace [ilmath]\sigma[/ilmath]-algebra, [ilmath]\mathcal{A}_Y[/ilmath] given by[1]:
- [ilmath]\mathcal{A}_Y:=\left\{Y\cap A\ \vert A\in\mathcal{A}\right\}[/ilmath]
Claim: [ilmath](Y,\mathcal{A}_Y)[/ilmath] is a [ilmath]\sigma[/ilmath]-algebra
Proof of claims
Claim 1: that [ilmath](Y,\mathcal{A}_Y)[/ilmath] is indeed a [ilmath]\sigma[/ilmath]-algebra
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The message provided is:
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Easy - just show definition
References
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