Difference between revisions of "Trace sigma-algebra"

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{{Requires references|"Trace", but it's a subspace concept! Find more references}}
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==Definition==
 
==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}}:
 
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}}:

Latest revision as of 12:00, 23 August 2018

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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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Easy - just show definition

References

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