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

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==Skeleton==
 
==Skeleton==
* [[Ring of sets]]
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If we were to cut out all the "extra" (and useful) theorems into just the core that let us get from [[pre-measure|pre-measures]] to [[measure|measures]] we would be left with what I call the "skeleton".
* [[Sigma-ring]]
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The "core" of Halmos' measure theory book is the following:
* [[additive set function]]
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# [[Ring of sets]], {{M|\mathcal{R} }} - '''DONE''' - [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 20:31, 3 April 2016 (UTC)
* ''measure'', {{M|\mu}} - [[extended real valued]], non negative, [[countably additive set function]] defined on a [[ring of sets]]
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# [[Sigma-ring|{{sigma|ring}}]], {{M|\mathcal{R} }}
** ''complete measure'' - {{M|\mu}} is complete if:
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# [[Measure]], {{M|\mu}}, countably additive extended real valued set function on a {{sigma|ring}}, {{M|\mathcal{R} }}, {{M|\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }}
*** {{M|1=\forall A\in\mathcal{R}\forall B[B\subseteq A\wedge\mu(A)=0\implies B\in\mathcal{R} ]}}
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# [[Pre-measure]], {{M|\bar{\mu} }}, countably additive extended real valued set function defined on a ring of sets, {{M|\mathcal{R} }}, {{M|\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }}
* ''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}]}}
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#* '''Goal 1: ''' "extend" a pre-measure, {{M|\bar{\mu} }} to a measure, {{M|\mu}} such that (for a ring of sets {{M|\mathcal{R} }}): {{M|1=\forall A\in\mathcal{R}[\bar{\mu}(A)=\mu(A)]}}<!--
** ''hereditary ring generated by''
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* ''subadditivity''
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QUESTION 1 (regarding goal 1)
* ''outer measure'', {{M|\mu^*}} (p42) - [[extended real valued]], non-negative, monotone and countably subadditive set function on an ''hereditary'' {{sigma|ring}} with {{M|1=\mu^*(\emptyset)=0}}
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** Theorem: If {{M|\mu}} is a measure on a ring {{M|\mathcal{R} }} and if:
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--><ref group="Question">Why specifically a measure? An [[outer-measure]] extends a measure to be able to measure every subset of every set in {{M|\mathcal{R} }} - at the cost of it no longer being [[additive set function|additive]] but instead [[subadditive set function|subadditive]] - why do we want additivity so much? Why is subadditivity not good enough? Obviously it's a weaker property as additivity {{M|\implies}} subadditivity</ref><!--
*** {{M|1=\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\}]}} then {{M|\mu^*}} is an ''extension'' of {{M|\mu}} to an outer measure on {{M|\mathbf{H}(\mathcal{R})}}
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** {{M|\mu^*}} is the ''outer measure induced by the measure {{M|\mu}}''
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END OF QUESTION 1, back after pre-measure
* {{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:
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** {{M|1=\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B\cap A')]}}
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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:
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# [[Hereditary set system]] - a system of sets, say {{M|H}}, such that {{M|1=\forall A\in H\forall B\in\mathcal{P}(A)[B\in H]}}
* {{M|1=A\in\mathcal{H} }} is {{M|\mu^*}}-measurable if:
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#* [[Hereditary sigma-ring|Hereditary {{sigma|ring}}]], {{M|\mathcal{H} }}<ref group="Question">Suppose {{M|\mathcal{H}(S)}} is the hereditary system generated by a collection of subsets, {{M|S}}, and {{M|\sigma_R(S)}} the {{sigma|ring}} generated by {{M|S}}, is it true that:
** {{M|1=\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)]}}</ref>
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* {{M|1=\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))}}?
*** '''Solution: ''' Note that {{M|1=S\cap T'=S-T}}, so Halmos is really saying:
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The book makes it clear that it intends to use:
**** {{M|1=A\in\mathcal{H} }} is {{M|\mu^*}}''-measurable'' if:
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* {{M|\mathcal{H}(\sigma_R(S))}}
***** {{M|1=\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B - A)]}}
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</ref>
* ''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]]''
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# [[Outer-measure]], {{M|\mu^*}} - extended real valued countably subadditive monotonic set function with {{M|1=\mu^*(\emptyset)=0}}
* ''Theorem: '' (p46) - {{M|\mathcal{S} }} is a [[sigma-ring]]
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#* '''Theorem: ''' for a [[pre-measure]], {{M|\bar{\mu} }} on a ring {{M|\mathcal{R} }} the function:
* ''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^*}}).
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#** {{MM|1=\mu^*:\mathcal{H}\rightarrow\bar{\mathbb{R} }_{\ge0} }} given by {{MM|1=\mu^*:A\mapsto\text{inf}\left.\left\{\sum^\infty_{n=1}\bar{\mu}(A_n)\ \right\vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}^\infty A_n\right\} }}
* ''Theorem'' - Every set in {{M|\sigma_R(\mathcal{R})}} is {{M|\mu^*}}''-measurable'' (the {{sigma|ring}} generated by {{M|\mathcal{R} }})
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#*: is an ''outer measure''
** Alternatively: {{M|\sigma_R(\mathcal{R})\subseteq\mathcal{S} }}
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# [[outer-measurable sets|{{M|\mu^*}}-measurable sets]]
* ''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} }]}}
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# '''Theorem: ''' - the set of all {{M|\mu^*}}-measurable sets is a {{sigma|ring}} and {{M|\mu^*}} is a [[measure]] on this sigma-ring
** That is to say that:
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# '''Theorem: ''' - every set in {{M|\sigma_R(\mathcal{R})}} is {{M|\mu^*}}-measurable and {{M|\mu^*}} is a [[measure]] on this sigma-ring
*** Outer measure induced by {{M|\bar{\mu} }} on {{M|\sigma_R(\mathcal{R})}} AND
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# '''Theorem: ''' - the measure induced on the sigma-ring of {{M|\mu^*}}-measurable sets is the same as the outer measure induced by the outer-measure when restricted to the {{sigma|ring}} generated by {{M|\mathcal{R} }}<ref group="Question">I could phrase this better</ref>
*** the outer measure induced by {{M|\bar{\mu} }} on {{M|\mathcal{S} }}
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#* We haven't shown anything to do with equality of these two sigma-rings! We only show that the outer measure each induce is the same!
*** agree with {{M|\mu^*}}
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==Questions==
* ''completion of a measure'' - p55
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<references group="Question"/>
* ''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:
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==Old notes==
** 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
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{{Begin Inline Theorem}}
==Solutions==
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These notes were "too long" I need to compress it into steps.
<references group="Solution"/>
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{{Begin Inline Proof}}
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{{:Notes:Halmos measure theory skeleton/Old page}}
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{{End Proof}}{{End Theorem}}

Latest revision as of 20:31, 3 April 2016

Skeleton

If we were to cut out all the "extra" (and useful) theorems into just the core that let us get from pre-measures to measures we would be left with what I call the "skeleton". The "core" of Halmos' measure theory book is the following:

  1. Ring of sets, R - DONE - Alec (talk) 20:31, 3 April 2016 (UTC)
  2. σ-ring, R
  3. Measure, μ, countably additive extended real valued set function on a σ-ring, R, μ:RˉR0
  4. Pre-measure, ˉμ, countably additive extended real valued set function defined on a ring of sets, R, ˉμ:RˉR0
    • Goal 1: "extend" a pre-measure, ˉμ to a measure, μ such that (for a ring of sets R): AR[ˉμ(A)=μ(A)][Question 1]
  5. Hereditary set system - a system of sets, say H, such that AHBP(A)[BH]
  6. Outer-measure, μ - extended real valued countably subadditive monotonic set function with μ()=0
    • Theorem: for a pre-measure, ˉμ on a ring R the function:
      • μ:HˉR0 given by μ:Ainf{n=1ˉμ(An) | (An)n=1RAn=1An}
      is an outer measure
  7. μ-measurable sets
  8. Theorem: - the set of all μ-measurable sets is a σ-ring and μ is a measure on this sigma-ring
  9. Theorem: - every set in σR(R) is μ-measurable and μ is a measure on this sigma-ring
  10. Theorem: - the measure induced on the sigma-ring of μ-measurable sets is the same as the outer measure induced by the outer-measure when restricted to the σ-ring generated by R[Question 3]
    • We haven't shown anything to do with equality of these two sigma-rings! We only show that the outer measure each induce is the same!

Questions

  1. Jump up Why specifically a measure? An outer-measure extends a measure to be able to measure every subset of every set in R - at the cost of it no longer being additive but instead subadditive - why do we want additivity so much? Why is subadditivity not good enough? Obviously it's a weaker property as additivity subadditivity
  2. Jump up Suppose H(S) is the hereditary system generated by a collection of subsets, S, and σR(S) the σ-ring generated by S, is it true that:
    • H(σR(S))=σR(H(S))?
    The book makes it clear that it intends to use:
    • H(σR(S))
  3. Jump up I could phrase this better

Old notes

[Expand]

These notes were "too long" I need to compress it into steps.