Difference between revisions of "A map from two sigma-algebras, A and B, is measurable if and only if for some generator of B (call it G) we have the inverse image of S is in A for every S in G"
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* {{M|f^{-1}(\mathcal{G})\subseteq\mathcal{A} }} | * {{M|f^{-1}(\mathcal{G})\subseteq\mathcal{A} }} | ||
==Proof== | ==Proof== | ||
| + | ==={{M|\implies}}: {{M|f:A\rightarrow B}} is {{M|\mathcal{A}/\mathcal{F} }}-measurable {{M|\implies}} for some generator {{M|\mathcal{G} }} of {{M|\mathcal{F} }} we have {{M|1=\forall S\in\mathcal{G}[f^{-1}(S)\in\mathcal{A}]}}=== | ||
| + | * Let {{M|S\in\mathcal{G} }} be given | ||
| + | ** Note that {{M|\mathcal{G}\subseteq\sigma(\mathcal{G})}}, so by the [[implies-subset relation]] we see {{M|S\in\mathcal{G}\implies S\in\sigma(\mathcal{G})}} | ||
| + | ** By [[measurable map|the definition of {{M|\mathcal{A}/\mathcal{F} }}-measurable]]: | ||
| + | *** {{M|\forall S\in F[f^{-1}(S)\in\mathcal{A}]}} | ||
| + | ** Thus {{M|1=S\in\mathcal{G}\implies S\in\sigma(\mathcal{G})=\mathcal{F} }} | ||
| + | *** But as we've just seen, if {{M|S\in\mathcal{F} }} then {{M|f^{-1}(S)\in\mathcal{A} }} | ||
| + | * So {{M|f^{-1}(S)\in\mathcal{A} }} | ||
| + | This completes the proof | ||
| + | ==={{M|\impliedby}}:=== | ||
{{Todo|See ref<ref name="PAS"/> page 6, also lemma 7.2 in<ref name="MIAMRLS"/>}} | {{Todo|See ref<ref name="PAS"/> page 6, also lemma 7.2 in<ref name="MIAMRLS"/>}} | ||
==Notes== | ==Notes== | ||
Latest revision as of 13:23, 18 March 2016
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Contents
- 1 Statement
- 2 Proof
- 2.1 [ilmath]\implies[/ilmath]: [ilmath]f:A\rightarrow B[/ilmath] is [ilmath]\mathcal{A}/\mathcal{F} [/ilmath]-measurable [ilmath]\implies[/ilmath] for some generator [ilmath]\mathcal{G} [/ilmath] of [ilmath]\mathcal{F} [/ilmath] we have [ilmath]\forall S\in\mathcal{G}[f^{-1}(S)\in\mathcal{A}][/ilmath]
- 2.2 [ilmath]\impliedby[/ilmath]:
- 3 Notes
- 4 References
Statement
A map from a [ilmath]\sigma[/ilmath]-algebra [ilmath](A,\mathcal{A})[/ilmath] to another [ilmath]\sigma[/ilmath]-algebra [ilmath](F,\mathcal{F})[/ilmath], [ilmath]f:A\rightarrow F[/ilmath], is [ilmath]\mathcal{A}/\mathcal{F} [/ilmath] measurable if and only if for some generator, [ilmath]\mathcal{G} [/ilmath], of [ilmath]\mathcal{F} [/ilmath][Note 1] we have[1][2]:
- [ilmath]\forall S\in\mathcal{G}[f^{-1}(S)\in\mathcal{A}][/ilmath]
Which we may alternatively write (for brevity, see: abuses of the implies-subset relation) as:
- [ilmath]f^{-1}(\mathcal{G})\subseteq\mathcal{A} [/ilmath]
Proof
[ilmath]\implies[/ilmath]: [ilmath]f:A\rightarrow B[/ilmath] is [ilmath]\mathcal{A}/\mathcal{F} [/ilmath]-measurable [ilmath]\implies[/ilmath] for some generator [ilmath]\mathcal{G} [/ilmath] of [ilmath]\mathcal{F} [/ilmath] we have [ilmath]\forall S\in\mathcal{G}[f^{-1}(S)\in\mathcal{A}][/ilmath]
- Let [ilmath]S\in\mathcal{G} [/ilmath] be given
- Note that [ilmath]\mathcal{G}\subseteq\sigma(\mathcal{G})[/ilmath], so by the implies-subset relation we see [ilmath]S\in\mathcal{G}\implies S\in\sigma(\mathcal{G})[/ilmath]
- By the definition of [ilmath]\mathcal{A}/\mathcal{F} [/ilmath]-measurable:
- [ilmath]\forall S\in F[f^{-1}(S)\in\mathcal{A}][/ilmath]
- Thus [ilmath]S\in\mathcal{G}\implies S\in\sigma(\mathcal{G})=\mathcal{F}[/ilmath]
- But as we've just seen, if [ilmath]S\in\mathcal{F} [/ilmath] then [ilmath]f^{-1}(S)\in\mathcal{A} [/ilmath]
- So [ilmath]f^{-1}(S)\in\mathcal{A} [/ilmath]
This completes the proof
[ilmath]\impliedby[/ilmath]:
TODO: See ref[2] page 6, also lemma 7.2 in[1]
Notes
- ↑ Thus [ilmath]\mathcal{F}=\sigma(\mathcal{G})[/ilmath]
References
- ↑ 1.0 1.1 Measures, Integrals and Martingales - René L. Schilling
- ↑ 2.0 2.1 Probability and Stochastics - Erhan Cinlar
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