View source for Notes:Halmos measure theory skeleton

==Skeleton==
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".
The "core" of Halmos' measure theory book is the following:
# [[Ring of sets]], {{M|\mathcal{R} }} - '''DONE''' - [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 20:31, 3 April 2016 (UTC)
# [[Sigma-ring|{{sigma|ring}}]], {{M|\mathcal{R} }}
# [[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} }}
# [[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} }}
#* '''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)]}}<!--

QUESTION 1 (regarding goal 1)

--><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><!--

END OF QUESTION 1, back after pre-measure

-->
# [[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]}}
#* [[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=\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))}}? 
The book makes it clear that it intends to use:
* {{M|\mathcal{H}(\sigma_R(S))}}
</ref>
# [[Outer-measure]], {{M|\mu^*}} - extended real valued countably subadditive monotonic set function with {{M|1=\mu^*(\emptyset)=0}}
#* '''Theorem: ''' for a [[pre-measure]], {{M|\bar{\mu} }} on a ring {{M|\mathcal{R} }} the function:
#** {{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\} }}
#*: is an ''outer measure''
# [[outer-measurable sets|{{M|\mu^*}}-measurable sets]]
# '''Theorem: ''' - the set of all {{M|\mu^*}}-measurable sets is a {{sigma|ring}} and {{M|\mu^*}} is a [[measure]] on this sigma-ring
# '''Theorem: ''' - every set in {{M|\sigma_R(\mathcal{R})}} is {{M|\mu^*}}-measurable and {{M|\mu^*}} is a [[measure]] on this sigma-ring
# '''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>
#* 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==
<references group="Question"/>
==Old notes==
{{Begin Inline Theorem}}
These notes were "too long" I need to compress it into steps.
{{Begin Inline Proof}}
{{:Notes:Halmos measure theory skeleton/Old page}}
{{End Proof}}{{End Theorem}}