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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| − | + | {{DISPLAYTITLE:A map, {{M|f:(A,\mathcal{A})\rightarrow(F,\mathcal{F})}}, is {{M|\mathcal{A}/\mathcal{F} }} measurable ''if and only if'' for some generator {{M|\mathcal{F}_0}} of {{M|\mathcal{F} }} we have {{M|\forall S\in\mathcal{F}_0[f^{-1}(S)\in\mathcal{A}]}}}} | |
| + | {{Stub page}} | ||
| + | __TOC__ | ||
==Statement== | ==Statement== | ||
| − | A map | + | A [[map]] from a [[sigma-algebra|{{sigma|algebra}}]] {{M|(A,\mathcal{A})}} to another {{sigma|algebra}} {{M|(F,\mathcal{F})}}, {{M|f:A\rightarrow F}}, is [[measurable map|{{M|\mathcal{A}/\mathcal{F} }} measurable]] {{iff}} for some [[Generator (sigma-algebra)|generator]], {{M|\mathcal{G} }}, of {{M|\mathcal{F} }}<ref group="Note">Thus {{M|1=\mathcal{F}=\sigma(\mathcal{G})}}</ref> we have{{rMIAMRLS}}<ref name="PAS">Probability and Stochastics - Erhan Cinlar</ref>: |
| + | * {{M|\forall S\in\mathcal{G}[f^{-1}(S)\in\mathcal{A}]}} | ||
| + | Which we may alternatively write (for brevity, see: [[abuses of the implies-subset relation]]) as: | ||
| + | * {{M|f^{-1}(\mathcal{G})\subseteq\mathcal{A} }} | ||
==Proof== | ==Proof== | ||
| − | {{Todo|See ref<ref name="PAS"/> page 6}} | + | ==={{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"/>}} | ||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
==References== | ==References== | ||
<references/> | <references/> | ||
| + | {{Measure theory navbox|plain}} | ||
{{Theorem Of|Measure Theory}} | {{Theorem Of|Measure Theory}} | ||
Latest revision as of 13:23, 18 March 2016
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Contents
[hide]Statement
A map from a σ-algebra (A,A) to another σ-algebra (F,F), f:A→F, is A/F measurable if and only if for some generator, G, of F[Note 1] we have[1][2]:
- ∀S∈G[f−1(S)∈A]
Which we may alternatively write (for brevity, see: abuses of the implies-subset relation) as:
- f−1(G)⊆A
Proof
⟹: f:A→B is A/F-measurable ⟹ for some generator G of F we have ∀S∈G[f−1(S)∈A]
- Let S∈G be given
- Note that G⊆σ(G), so by the implies-subset relation we see S∈G⟹S∈σ(G)
- By the definition of A/F-measurable:
- ∀S∈F[f−1(S)∈A]
- Thus S∈G⟹S∈σ(G)=F
- But as we've just seen, if S∈F then f−1(S)∈A
- So f−1(S)∈A
This completes the proof
⟸:
TODO: See ref[2] page 6, also lemma 7.2 in[1]
Notes
- Jump up ↑ Thus F=σ(G)
References
- ↑ Jump up to: 1.0 1.1 Measures, Integrals and Martingales - René L. Schilling
- ↑ Jump up to: 2.0 2.1 Probability and Stochastics - Erhan Cinlar
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