View source for Extending pre-measures to outer-measures

==Statement==
Given a [[pre-measure]], {{M|\bar{\mu} }}, on a [[ring of sets]], {{M|\mathcal{R} }}, we can define a new [[function]], {{M|\mu^*}} which is{{rMTH}}:
* an [[extension]] of {{M|\bar{\mu} }} and
* an [[outer-measure]] (on the ''[[hereditary sigma-ring|hereditary {{sigma|ring}}]]'' [[hereditary sigma-ring generated by|generated by]] {{M|\mathcal{R} }}, written {{M|\mathcal{H}_{\sigma_R}(\mathcal{R})}})
Given by:
* {{M|\mu^*:\mathcal{H}_{\sigma_R}(\mathcal{R})\rightarrow\bar{\mathbb{R} }_{\ge0} }}
** {{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\} }}
The statement of the theorem is that this {{M|\mu^*}} is indeed an [[outer-measure]]
==Proof==
{{Begin Notebox}}
Recall the definition of an [[outer-measure]], we must show {{M|\mu^*}} satisfies this.
{{Begin Notebox Content}}
{{:Outer-measure/Definition}}
{{End Notebox Content}}{{End Notebox}}
# We claimed that {{M|\mu^*}} is an [[extension]] of {{M|\bar{\mu} }}, this means that: {{M|1=\forall A\in\mathcal{R}[\mu^*=\bar{\mu}]}}. Let us check this.
#* Let {{M|A\in\mathcal{R} }} be given.
#*# First we must bound {{M|\mu^*}} above. This is because {{M|1=[\mu^*(A)=\bar{\mu}(A)]\iff[\mu^*(A)\ge\bar{\mu}(A)\wedge\bar{\mu}(A)\ge\mu^*(A)]}}
#*#* Remember that {{M|\emptyset\in\mathcal{R} }} as {{mathcal|R}} is a [[ring of sets]]
#*#** We can now define a [[sequence]], {{MSeq|A_n|in=\mathcal{R} }} as follows:
#*#*** {{M|1=A_1=A}}
#*#*** {{M|1=A_n=\emptyset}} for {{M|n\ge 2}}
#*#** So {{MSeq|A_n|in=\mathcal{R} }} is {{M|(A,\emptyset,\emptyset,\ldots)}}
#*#** Now {{M|1=\sum_{n=1}^\infty \bar{\mu}(A_n)=\bar{\mu}(A)+\bar{\mu}(\emptyset)+\bar{\mu}(\emptyset)+\ldots=\bar{\mu}(A)+0+0+\ldots=\bar{\mu}(\emptyset)}}
#*#* So {{M|1=\mu^*(A)\le\bar{\mu}(A)}} (as {{M|\mu^*}} is the defined as the [[infimum]] of such expressions, all we have done is find an [[upper-bound]] for it)
#*# Now we must bound {{M|\mu^*}} below (by {{M|\bar{\mu}(A)}}) to show they're equal.
#*#* Using [[the (pre-)measure of a set is no more than the sum of the (pre-)measures of the elements of a covering for that set]], which states, symbolically:
#*#** Given a set {{M|A}} and a [[countably infinite]] or [[finite]] ''[[sequence]]'' of sets, {{M|(A_i)}} such that {{M|A\subseteq\bigcup_i A_i}} then {{M|\bar{\mu}(A)\le\sum_i\bar{\mu}(A_i)}}
#*#* '''Halmos handwaves here''', we want to show that {{M|\bar{\mu}(A)\le\mu^*(A)}}, however all we know is {{M|\mu^*(A)\le\sum_i\bar{\mu}(A_i)}} and {{M|\bar{\mu}(A)\le\sum_i\bar{\mu}(A_i)}} - this isn't very helpful, the solution will be in the nature of the [[infimum]] I am sure
{{Todo|Finish proof}}
==References==
<references/>
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{{Theorem Of|Measure Theory}}