Difference between revisions of "Trace sigma-algebra"
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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}}: | ||
Revision as of 11:59, 23 August 2018
Stub grade: B
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More results would be good. Relation to pullback too
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:
The message provided is:
Easy - just show definition
References
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