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

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==Skeleton==
 
==Skeleton==
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* [[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".
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* [[Sigma-ring]]
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The "core" of Halmos' measure theory book is the following:
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* [[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)
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* ''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} }}
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* ''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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# [[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} }}
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** ''hereditary ring generated by''
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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} }}
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* ''subadditivity''
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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)]}}<!--
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* ''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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QUESTION 1 (regarding goal 1)
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*** {{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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--><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><!--
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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:
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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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END OF QUESTION 1, back after pre-measure
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*** '''PROBLEM: How can we do [[complementation]] in a ring?'''
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 +
-->
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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]}}
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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:
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* {{M|1=\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))}}?
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The book makes it clear that it intends to use:
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* {{M|\mathcal{H}(\sigma_R(S))}}
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</ref>
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# [[Outer-measure]], {{M|\mu^*}} - extended real valued countably subadditive monotonic set function with {{M|1=\mu^*(\emptyset)=0}}
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#* '''Theorem: ''' for a [[pre-measure]], {{M|\bar{\mu} }} on a ring {{M|\mathcal{R} }} the function:
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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\} }}
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#*: is an ''outer measure''
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# [[outer-measurable sets|{{M|\mu^*}}-measurable sets]]
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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
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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
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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>
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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!
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==Questions==
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<references group="Question"/>
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==Old notes==
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{{Begin Inline Theorem}}
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These notes were "too long" I need to compress it into steps.
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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, [ilmath]\mathcal{R} [/ilmath] - DONE - Alec (talk) 20:31, 3 April 2016 (UTC)
  2. [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath]
  3. Measure, [ilmath]\mu[/ilmath], countably additive extended real valued set function on a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath], [ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath]
  4. Pre-measure, [ilmath]\bar{\mu} [/ilmath], countably additive extended real valued set function defined on a ring of sets, [ilmath]\mathcal{R} [/ilmath], [ilmath]\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath]
    • Goal 1: "extend" a pre-measure, [ilmath]\bar{\mu} [/ilmath] to a measure, [ilmath]\mu[/ilmath] such that (for a ring of sets [ilmath]\mathcal{R} [/ilmath]): [ilmath]\forall A\in\mathcal{R}[\bar{\mu}(A)=\mu(A)][/ilmath][Question 1]
  5. Hereditary set system - a system of sets, say [ilmath]H[/ilmath], such that [ilmath]\forall A\in H\forall B\in\mathcal{P}(A)[B\in H][/ilmath]
  6. Outer-measure, [ilmath]\mu^*[/ilmath] - extended real valued countably subadditive monotonic set function with [ilmath]\mu^*(\emptyset)=0[/ilmath]
    • Theorem: for a pre-measure, [ilmath]\bar{\mu} [/ilmath] on a ring [ilmath]\mathcal{R} [/ilmath] the function:
      • [math]\mu^*:\mathcal{H}\rightarrow\bar{\mathbb{R} }_{\ge0}[/math] given by [math]\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\}[/math]
      is an outer measure
  7. [ilmath]\mu^*[/ilmath]-measurable sets
  8. Theorem: - the set of all [ilmath]\mu^*[/ilmath]-measurable sets is a [ilmath]\sigma[/ilmath]-ring and [ilmath]\mu^*[/ilmath] is a measure on this sigma-ring
  9. Theorem: - every set in [ilmath]\sigma_R(\mathcal{R})[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable and [ilmath]\mu^*[/ilmath] is a measure on this sigma-ring
  10. Theorem: - the measure induced on the sigma-ring of [ilmath]\mu^*[/ilmath]-measurable sets is the same as the outer measure induced by the outer-measure when restricted to the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]\mathcal{R} [/ilmath][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. ↑ Why specifically a measure? An outer-measure extends a measure to be able to measure every subset of every set in [ilmath]\mathcal{R} [/ilmath] - 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 [ilmath]\implies[/ilmath] subadditivity
  2. ↑ Suppose [ilmath]\mathcal{H}(S)[/ilmath] is the hereditary system generated by a collection of subsets, [ilmath]S[/ilmath], and [ilmath]\sigma_R(S)[/ilmath] the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]S[/ilmath], is it true that:
    • [ilmath]\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))[/ilmath]?
    The book makes it clear that it intends to use:
    • [ilmath]\mathcal{H}(\sigma_R(S))[/ilmath]
  3. ↑ I could phrase this better

Old notes

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


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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