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"
From Maths
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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
[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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