A map, [ilmath]f:(A,\mathcal{A})\rightarrow(F,\mathcal{F})[/ilmath], is [ilmath]\mathcal{A}/\mathcal{F} [/ilmath] measurable if and only if for some generator [ilmath]\mathcal{F}_0[/ilmath] of [ilmath]\mathcal{F} [/ilmath] we have [ilmath]\forall S\in\mathcal{F}_0[f^{-1}(S)\in\mathcal{A}][/ilmath]
From Maths
(Unknown grade)
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.
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
|