Difference between revisions of "Notes:Halmos measure theory skeleton"

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* [[additive set function]]
 
* [[additive set function]]
 
* ''measure'', {{M|\mu}} - [[extended real valued]], non negative, [[countably additive set function]] defined on a [[ring of sets]]
 
* ''measure'', {{M|\mu}} - [[extended real valued]], non negative, [[countably additive set function]] defined on a [[ring of sets]]
 +
** ''complete measure'' - {{M|\mu}} is complete if:
 +
*** {{M|1=\forall A\in\mathcal{R}\forall B[B\subseteq A\wedge\mu(A)=0\implies B\in\mathcal{R} ]}}
 
* ''hereditary'' system - a system of sets, {{M|\mathcal{E} }} such that if {{M|E\in\mathcal{E} }} then {{M|\forall F\in\mathcal{P}(E)[F\in\mathcal{E}]}}
 
* ''hereditary'' system - a system of sets, {{M|\mathcal{E} }} such that if {{M|E\in\mathcal{E} }} then {{M|\forall F\in\mathcal{P}(E)[F\in\mathcal{E}]}}
 
** ''hereditary ring generated by''
 
** ''hereditary ring generated by''
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* {{M|\mu^*}}-measurable - given an ''outer measure'' {{M|\mu^*}} on a hereditary {{sigma|ring}} {{M|\mathcal{H} }} a set {{M|A\in\mathcal{H} }} is ''{{M|\mu^*}}-measurable'' if:
 
* {{M|\mu^*}}-measurable - given an ''outer measure'' {{M|\mu^*}} on a hereditary {{sigma|ring}} {{M|\mathcal{H} }} a set {{M|A\in\mathcal{H} }} is ''{{M|\mu^*}}-measurable'' if:
 
** {{M|1=\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B\cap A')]}}
 
** {{M|1=\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B\cap A')]}}
−
*** '''PROBLEM: How can we do [[complementation]] in a ring?'''
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*** '''PROBLEM: How can we do [[complementation]] in a ring?'''<ref group="Solution">We only have to deal with {{M|A\cap B'}} which could just be Halmos abusing notation. If we take it literally ("{{M|1=B':=\{x(\in?)\vert x\notin B\} }}") we may be able to work through this. Note that [[the intersection of sets is a subset of each set]] so {{M|A\cap B'\subseteq A}} and is in fact {{M|1==A-B}}, so what Halmos is really saying is:
 +
* {{M|1=A\in\mathcal{H} }} is {{M|\mu^*}}-measurable if:
 +
** {{M|1=\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)]}}</ref>
 +
*** '''Solution: ''' Note that {{M|1=S\cap T'=S-T}}, so Halmos is really saying:
 +
**** {{M|1=A\in\mathcal{H} }} is {{M|\mu^*}}''-measurable'' if:
 +
***** {{M|1=\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B - A)]}}
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* ''Theorem: '' if {{M|\mu^*}} is an ''outer measure'' on a hereditary {{sigma|ring}} {{M|\mathcal{H} }} and if {{M|\mathcal{S} }} is the class of all {{M|\mu^*}}-measurable sets, ''then {{M|\mathcal{S} }} is a [[ring of sets]]''
 +
* ''Theorem: '' (p46) - {{M|\mathcal{S} }} is a [[sigma-ring]]
 +
* ''Theorem: '' - Every set of {{M|1=\mu^*=0}} belongs to {{M|\mathcal{S} }} and the set function {{M|\bar{\mu}:\mathcal{S}\rightarrow\bar{\mathbb{R} }_{\ge 0} }} is a ''complete'' measure ({{AKA}}: {{M|\bar{\mu} }} is the ''measure induced by'' {{M|\mu^*}}).
 +
* ''Theorem'' - Every set in {{M|\sigma_R(\mathcal{R})}} is {{M|\mu^*}}''-measurable'' (the {{sigma|ring}} generated by {{M|\mathcal{R} }})
 +
** Alternatively: {{M|\sigma_R(\mathcal{R})\subseteq\mathcal{S} }}
 +
* ''Theorem'' - {{M|1=\forall A\in\mathbf{H}(\mathcal{R})[\underbrace{\mu^*(A)}_{\text{Outer} }=\underbrace{\text{inf}\{\bar{\mu}(B)\ \vert\ A\subseteq B\in\mathcal{S}\} }_{\text{Induced by } \bar{\mu} }=\underbrace{\{\bar{\mu}(B)\ \vert\ A\subseteq B\in\sigma_R(\mathcal{R})\} }_{\text{Induced by } \bar{\mu} }]}}
 +
** That is to say that:
 +
*** Outer measure induced by {{M|\bar{\mu} }} on {{M|\sigma_R(\mathcal{R})}} AND
 +
*** the outer measure induced by {{M|\bar{\mu} }} on {{M|\mathcal{S} }}
 +
*** agree with {{M|\mu^*}}
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* ''completion of a measure'' - p55
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* ''Theorem: '' if {{M|\mu}} is a {{sigma|finite}} measure on a {{sigma|ring}} {{M|\mathbb{R} }} and if {{M|\mu^*}} is the outer measure it induces, then:
 +
** The ''completion'' of the extension of {{M|\mu}} to {{M|\sigma_R(\mathcal{R})}} is identical with {{M|\mu^*}} on the class of all {{M|\mu^*}}-measurable sets
 +
==Solutions==
 +
<references group="Solution"/>

Revision as of 20:05, 22 March 2016

Skeleton

  • Ring of sets
  • Sigma-ring
  • additive set function
  • measure, [ilmath]\mu[/ilmath] - extended real valued, non negative, countably additive set function defined on a ring of sets
    • complete measure - [ilmath]\mu[/ilmath] is complete if:
      • [ilmath]\forall A\in\mathcal{R}\forall B[B\subseteq A\wedge\mu(A)=0\implies B\in\mathcal{R} ][/ilmath]
  • hereditary system - a system of sets, [ilmath]\mathcal{E} [/ilmath] such that if [ilmath]E\in\mathcal{E} [/ilmath] then [ilmath]\forall F\in\mathcal{P}(E)[F\in\mathcal{E}][/ilmath]
    • hereditary ring generated by
  • subadditivity
  • outer measure, [ilmath]\mu^*[/ilmath] (p42) - extended real valued, non-negative, monotone and countably subadditive set function on an hereditary [ilmath]\sigma[/ilmath]-ring with [ilmath]\mu^*(\emptyset)=0[/ilmath]
    • Theorem: If [ilmath]\mu[/ilmath] is a measure on a ring [ilmath]\mathcal{R} [/ilmath] and if:
      • [ilmath]\forall A\in\mathbf{H}(\mathcal{R})[\mu^*(A)=\text{inf}\{\sum^\infty_{n=1}\mu(A_n)\ \vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R} \wedge A\subseteq \bigcup^\infty_{n=1}A_n\}][/ilmath] then [ilmath]\mu^*[/ilmath] is an extension of [ilmath]\mu[/ilmath] to an outer measure on [ilmath]\mathbf{H}(\mathcal{R})[/ilmath]
    • [ilmath]\mu^*[/ilmath] is the outer measure induced by the measure [ilmath]\mu[/ilmath]
  • [ilmath]\mu^*[/ilmath]-measurable - given an outer measure [ilmath]\mu^*[/ilmath] on a hereditary [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{H} [/ilmath] a set [ilmath]A\in\mathcal{H} [/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
    • [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B\cap A')][/ilmath]
      • PROBLEM: How can we do complementation in a ring?[Solution 1]
      • Solution: Note that [ilmath]S\cap T'=S-T[/ilmath], so Halmos is really saying:
        • [ilmath]A\in\mathcal{H}[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
          • [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B - A)][/ilmath]
  • Theorem: if [ilmath]\mu^*[/ilmath] is an outer measure on a hereditary [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{H} [/ilmath] and if [ilmath]\mathcal{S} [/ilmath] is the class of all [ilmath]\mu^*[/ilmath]-measurable sets, then [ilmath]\mathcal{S} [/ilmath] is a ring of sets
  • Theorem: (p46) - [ilmath]\mathcal{S} [/ilmath] is a sigma-ring
  • Theorem: - Every set of [ilmath]\mu^*=0[/ilmath] belongs to [ilmath]\mathcal{S} [/ilmath] and the set function [ilmath]\bar{\mu}:\mathcal{S}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] is a complete measure (AKA: [ilmath]\bar{\mu} [/ilmath] is the measure induced by [ilmath]\mu^*[/ilmath]).
  • Theorem - Every set in [ilmath]\sigma_R(\mathcal{R})[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable (the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]\mathcal{R} [/ilmath])
    • Alternatively: [ilmath]\sigma_R(\mathcal{R})\subseteq\mathcal{S} [/ilmath]
  • Theorem - [ilmath]\forall A\in\mathbf{H}(\mathcal{R})[\underbrace{\mu^*(A)}_{\text{Outer} }=\underbrace{\text{inf}\{\bar{\mu}(B)\ \vert\ A\subseteq B\in\mathcal{S}\} }_{\text{Induced by } \bar{\mu} }=\underbrace{\{\bar{\mu}(B)\ \vert\ A\subseteq B\in\sigma_R(\mathcal{R})\} }_{\text{Induced by } \bar{\mu} }][/ilmath]
    • That is to say that:
      • Outer measure induced by [ilmath]\bar{\mu} [/ilmath] on [ilmath]\sigma_R(\mathcal{R})[/ilmath] AND
      • the outer measure induced by [ilmath]\bar{\mu} [/ilmath] on [ilmath]\mathcal{S} [/ilmath]
      • agree with [ilmath]\mu^*[/ilmath]
  • completion of a measure - p55
  • Theorem: if [ilmath]\mu[/ilmath] is a [ilmath]\sigma[/ilmath]-finite measure on a [ilmath]\sigma[/ilmath]-ring [ilmath]\mathbb{R} [/ilmath] and if [ilmath]\mu^*[/ilmath] is the outer measure it induces, then:
    • The completion of the extension of [ilmath]\mu[/ilmath] to [ilmath]\sigma_R(\mathcal{R})[/ilmath] is identical with [ilmath]\mu^*[/ilmath] on the class of all [ilmath]\mu^*[/ilmath]-measurable sets

Solutions

  1. ↑ We only have to deal with [ilmath]A\cap B'[/ilmath] which could just be Halmos abusing notation. If we take it literally ("[ilmath]B':=\{x(\in?)\vert x\notin B\}[/ilmath]") we may be able to work through this. Note that the intersection of sets is a subset of each set so [ilmath]A\cap B'\subseteq A[/ilmath] and is in fact [ilmath]=A-B[/ilmath], so what Halmos is really saying is:
    • [ilmath]A\in\mathcal{H}[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
      • [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)][/ilmath]
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