View source for Notes:Outer-measures to measures

==Notation on this page==
I made a bad choice using {{M|\bar{\mu}^*}} as the measure induced by an outer measure on the class of all outer-measurable sets. So on this page:
* {{M|\bar{\mu} }} - pre-measure
* {{M|\mu^*}} - outer measure
* {{M|\bar{\mu}^* }} - measure induced on {{M|\bar{S} }}
* {{M|\bar{S} }} - the set of all outer-measurable sets, WRT {{M|\mu^*}} (for the definition of outer-measurability see top of Halmos' section 11 (in this document or in book))
==Problem==
Once we have [[extension (function)|extended]] a [[pre-measure]], {{M|\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }} to an [[outer-measure]], {{M|\mu^*:\mathcal{H}_{\sigma R}(\mathcal{R})}} we must then "contract" {{M|\mu^*}} to a normal [[measure]] (which is defined on a [[sigma-ring]].

Halmos introduces a chain of theorems and does this rather indirectly, I hope to "distil" it on this page.

Halmos also does a lot of "bad" things, like:
# [[Sequence|Sequences]]: {{M|1=\{E_n\}_{n=1}^\infty}} for sequences! It's not a set, it is an ordered set! There is no "nth term" operator of a set, and if he defines {{M|E_n}} as being a term "{{M|E_n}}" that happens to be in a set.... it's iffy at best (then {{M|E_j}} wouldn't make sense)
# Operators: He writes (page 44, outer-measurability of a set) {{M|A\cap E'}} - meaning complementation, but of course on a [[ring of sets]] even a [[hereditary system of sets]] - there is no [[complementation|complementation operator]] - he means {{M|A-E}} - which is "the same thing".
==Halmos==
===Section 11===
* '''Outer-measurability'''<ref group="Note">He calls it "''{{M|\mu^*}}-measurability''"</ref> - for outer-measure {{M|\mu^*}} on hereditary sigma-ring, {{M|H}} a set {{M|A\in H}} is ''outer-measurable WRT {{M|\mu^*}}''<ref group="Note">Again my term.</ref> if:
** {{M|1=\forall B\in H[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B\cap A')]}} - see what I mean by abuse? {{M|A'}} isn't defined on a ring.
*** He means: {{M|A\in H}} is ''outer measurable'' if: {{M|1=\forall B\in H[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)]}}
* '''Theorem A:''' if {{M|\bar{S} }} is the class of all outer-measurable sets then {{M|\bar{S} }} is a [[ring of sets]]
* '''Theorem B:''' {{M|\bar{S} }} is a [[sigma-ring|{{sigma|ring}}]]
* '''Theorem C:''' (two parter)
*# '''Theorem C (i): ''' Every set of outer-measure {{M|0}} belongs to {{M|\bar{S} }}
*# '''Theorem C (ii): ''' The set function {{M|\bar{\mu}^*:\bar{S}\rightarrow\bar{\mathbb{R} }_{\ge 0} }}<ref group="Note">{{M|\bar{\mu}^*}} is my notation as I have taken MIAM's notation of {{M|\bar{\mu} }} for [[pre-measure|pre-measures]]. Halmos uses {{M|\bar{\mu} }}</ref> defined by {{M|\bar{\mu}^*:A\mapsto\mu^*(A)}} is a [[complete measure]] on {{M|\bar{S} }}
{{Begin Blue Notebox}}
Why are we not done here? Short answer: '''we still know nothing about {{M|\bar{S} }} and the ring {{M|\mathcal{R} }} we started with'''
{{Begin Blue Notebox Content}}
We started with a [[pre-measure]], {{M|\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }} and we have now got some [[measure]]:
* {{M|\bar{\mu}^*:\bar{S}\rightarrow\bar{\mathbb{R} }_{\ge 0} }}
However we know nothing about the relationship between {{M|\bar{S} }} and {{M|\mathcal{R} }}.

Our goal is to construct a measure on {{M|\sigma(\mathcal{R})}} - the sigma-ring generated by the ring {{M|\mathcal{R} }}.

(Halmos ends chapter 2 and section 11 here)
{{End Notebox Content}}{{End Notebox}}
===Section 12===
* '''Theorem A: ''' Every set in {{M|\sigma(\mathcal{R})}} is ''outer-measurable WRT {{M|\mu^*}}''. This tells us {{M|\sigma(\mathcal{R})\subseteq\bar{S} }}.
** We still do not know that {{M|\bar{\mu}^*}} restricted to {{M|\sigma(\mathcal{R})}} would even be a measure.
* '''Theorem B: ''' If {{M|A\in\mathcal{H}_{\sigma R}(\mathcal{R})}} then: {{M|1=\begin{array}{rcl}&&\\ \mu^*(A) & = & \text{inf}\{\bar{\mu}^*(B)\ \vert\ A\subseteq B\wedge B\in\bar{S}\} \\ & = & \text{inf}\{\bar{\mu}^*(F)\ \vert\ E\subseteq F\wedge F\in \sigma(\mathcal{R})\}\end{array} }} {{Caution|3 line array looks pretty bad - TODO - this better}}
** This is equivalent to: 
*** The outer measure induced by {{M|\bar{\mu}^*}} on {{M|\sigma(\mathcal{R})}} and the outer measure induced by {{M|\bar{\mu}^*}} on {{M|\bar{S} }} agree.
**** {{Warning|Not quite sure I agree! This doesn't look exactly like an outer measure now does it?}}
* '''Measurable cover: ''' if {{M|E\in\mathcal{H}_{\sigma R}(\mathcal{R})}} and {{M|F\in\sigma(\mathcal{R})}} - we shall say that {{M|F}} is a ''measurable cover'' of {{M|E}} if:
** {{M|E\subseteq F}} and if {{M|1=\forall G\in\sigma(\mathcal{R})[G\subseteq F-E\implies \bar{\mu}^*(G)=0]}} {{Caution|What does this ''actually'' mean?}}
* '''Theorem C: ''' If a set {{M|E\in\mathcal{H}_{\sigma R}(\mathcal{R})}} has {{sigma|finite}} ''outer'' measure then:
** {{M|1=\exists F\in\sigma(\mathcal{R})[\mu^*(E)=\bar{\mu}^*(F)\wedge F\text{ is a measurable cover of }E]}} {{Note|Could be a good time to try some formal logic substitution!}}
* '''Theorem D: ''' (two parter)
*#* '''Theorem D (i): ''' If {{M|E\in\mathcal{H}_{\sigma R}(\mathcal{R})}} and {{M|F}} is a measurable cover of {{M|E}} then:
*#** {{M|1=\mu^*(E)=\bar{\mu}^*(F)}}.
*#* '''Theorem D (ii): ''' additionally, if both {{M|F_1}} and {{M|F_2}} are measurable covers of {{M|E}} then:
*#** {{M|1=\bar{\mu}^*(F_1\triangle F_2)=0}}
* '''Theorem E: ''' If {{M|\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }} is {{sigma|finite}} then so are the measures {{M|\bar{\mu}^*}} on {{M|\sigma(\mathcal{R})}} and {{M|\bar{S} }}.
===Section 13===
* '''Theorem A: ''' If {{M|\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }} is a {{sigma|finite}} [[pre-measure]] then there is a unique measure on {{M|\bar{\mu}^* }} (defined as above) on {{M|\sigma(\mathcal{R})}} such that for {{M|E\in\mathcal{R} }} we have {{M|1=\bar{\mu}^*(E)=\mu(E)}} - the measure {{M|\bar{\mu}^*}} is {{sigma|finite}}
I've gone as far as page 55 (from 44) and ''still'' not hit "and here is a measure on {{M|\sigma(\mathcal{R})}} - stopping for now.
==Measures, Integrals and Martingales==
Not much better, Schilling doesn't do the "pre-measure {{M|\rightarrow}} outer-measure {{M|\rightarrow}} measure" that'd I'd hoped for.
==Notes==
<references group="Note"/>
{{Notes|Measure Theory}}