Pre-image [ilmath]\sigma[/ilmath]-algebra

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Pre-image [ilmath]\sigma[/ilmath]-algebra
[math]\{f^{-1}(A')\ \vert\ A'\in\mathcal{A}'\}[/math]

is a [ilmath]\sigma[/ilmath]-algebra on [ilmath]X[/ilmath] given a [ilmath]\sigma[/ilmath]-algebra [ilmath](X',\mathcal{A}')[/ilmath] and a map [ilmath]f:X\rightarrow X'[/ilmath].

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Add to sigma-algebra index, link to other pages, general expansion. Needs to be exemplary as a lot of search traffic enters here.
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Let [ilmath]\mathcal{A}'[/ilmath] be a [ilmath]\sigma[/ilmath]-algebra on [ilmath]X'[/ilmath] and let [ilmath]f:X\rightarrow X'[/ilmath] be a map. The pre-image [ilmath]\sigma[/ilmath]-algebra on [ilmath]X[/ilmath][1] is the [ilmath]\sigma[/ilmath]-algebra, [ilmath]\mathcal{A} [/ilmath] (on [ilmath]X[/ilmath]) given by:

  • [math]\mathcal{A}:=\left\{f^{-1}(A')\ \vert\ A'\in\mathcal{A}'\right\}[/math]

We can write this (for brevity) alternatively as:

Claim: [ilmath](X,\mathcal{A})[/ilmath] is indeed a [ilmath]\sigma[/ilmath]-algebra

Proof of claims

Claim 1: [ilmath](X,\mathcal{A})[/ilmath] is indeed a [ilmath]\sigma[/ilmath]-algebra

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Should be pretty easy, it's just showing the definitions

See also


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


Let [ilmath]f:X\rightarrow X'[/ilmath] and let [ilmath]\mathcal{A}'[/ilmath] be a [ilmath]\sigma[/ilmath]-algebra on [ilmath]X'[/ilmath], we can define a sigma algebra on [ilmath]X[/ilmath], called [ilmath]\mathcal{A} [/ilmath], by:

  • [ilmath]\mathcal{A}:=f^{-1}(\mathcal{A}'):=\left\{f^{-1}(A')\vert\ A'\in\mathcal{A}'\right\}[/ilmath]

TODO: Measures Integrals and Martingales - page 16