View source for Extending pre-measures to outer-measures

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==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\} }} - here {{M|\text{inf} }} denotes the [[infimum]] of a set.
The statement of the theorem is that this {{M|\mu^*}} is indeed an [[outer-measure]]
==Proof==
{{Begin Inline Theorem}}
Proof notes
{{Begin Inline 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)}}
#*#* By [[passing to the infimum]] we see that {{M|\bar{\mu}(A)\le\mu^*(A)}} as required.

'''Problems with proof'''
* How do we know the [[infimum]] even exists!
** Was being silly, any set of real numbers bounded below has an infimum, as {{M|\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} }} we see that {{M|-1}} is a lower bound for example. Having a lot of silly moments lately.
* For the application of ''[[passing to the infimum]]'' how do we know that the [[infimum]] involving {{M|\bar{\mu} }} even exists (this probably uses [[monotonic|monotonicity]] of {{M|\bar{\mu} }} and should be easy to show)
{{End Proof}}{{End Theorem}}
{{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}}
For brevity we define the following shorthands:
# {{MM|1=\alpha_A:=\left\{(A_n)_{n=1}^\infty\ \Big\vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}^\infty A_n\right\} }}
# {{MM|1=\beta_A:=\left\{\sum^\infty_{n=1}\bar{\mu}(A_n)\ \Big\vert\ (A_n)_{n=1}^\infty\in\alpha_A \right\} }}
Now we may define {{M|\mu^*}} as:
* {{M|1=\mu^*:A\mapsto\text{inf}(\beta_A)}}
===Proof that {{M|\mu^*}} is an extension of {{M|\bar{\mu} }}===
* Let {{M|A\in\mathcal{R} }} be given
** In order to prove {{M|1=\bar{\mu}(A)=\mu^*(A)}} we need only prove {{M|[\bar{\mu}(A)\ge\mu^*(A)\wedge\bar{\mu}(A)\le\mu^*(A)]}}<ref group="Note">This is called the trichotomy rule or something, I should link to the relevant part of a [[partial order]] here</ref>
**# '''Part 1: ''' {{M|\bar{\mu}(A)\ge\mu^*(A)}}
**#* Consider the sequence {{MSeq|A_n}} given by {{M|1=A_1:=A}} and {{M|1=A_i:=\emptyset}} for {{M|i>1}}, so the sequence {{M|A,\emptyset,\emptyset,\ldots}}.
**#** Clearly {{M|1=A\subseteq\bigcup^\infty_{n=1}A_n}} (as {{M|1=\bigcup^\infty_{n=1}A_n=A}})
**#** As such this {{MSeq|A_n|post=\in\alpha_A}}
**#** This means {{M|1=\sum^\infty_{n=1}\bar{\mu}(A_n)\in\beta_A}} (as {{MSeq|A_n|post=\in\alpha_A}} and {{M|\beta_A}} is the sum of all the pre-measures {{WRT}} {{M|\bar{\mu} }} of the sequences of sets in {{M|\alpha_A}})
**#** Recall that the [[infimum]] of a set is, among other things, a [[lower bound]] of the set. So:
**#*** for {{M|\text{inf}(S)}} (for a [[set]], {{M|S}}) we see:
**#**** {{M|\forall s\in S[\text{inf}(S)\le s]}} - this uses only the [[lower bound]] part of the [[infimum]] definition.
**#** By applying this to {{M|1=\text{inf}(\beta_A)\big(=\mu^*(A)\big)}} we see:
**#*** {{M|1=\mu^*(A):=\text{inf}(\beta_A)\le\sum^\infty_{n=1}\bar{\mu}(A_n)=\bar{\mu}(A)}} 
**#**** as {{M|1=\sum^\infty_{n=1}\bar{\mu}(A_n)\in\beta_A}} and {{M|\text{inf}(S)}} remember and
**#**** By definition of a (''[[pre-measure|pre]]''-)[[measure]], {{M|1=\mu(\emptyset)=0}}, so: {{M|1=\sum^\infty_{n=1}\bar{\mu}(A_n)=\bar{\mu}(A)+\bar{\mu}(\emptyset)+\bar{\mu}(\emptyset)+\cdots=\bar{\mu}(A)}}
**#* We have shown {{M|1=\mu^*(A)\le\bar{\mu}(A)}} as required
**# '''Part 2:''' {{M|\bar{\mu}(A)\le\mu^*(A)}}
**#* SEE NOTEPAD. Define {{M|1=\gamma_A:=\left\{\bar{\mu}(A)\right\} }}, then 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]] we see {{M|\forall x\in\beta_A\exists y\in\gamma_A[y\le x]}} - we may now [[passing to the infimum|pass to the infimum]].
===Proof that {{M|\mu^*}} is {{sigma|subadditive}}===
*Let {{MSeq|A_n|in=\mathcal{H}_{\sigma R}(\mathcal{R})}} be given. We want to show that {{M|1=\mu^*(\bigcup_{n=1}^\infty A_n)\le\sum^\infty_{n=1}\mu^*(A_n)}}
** Let {{M|\epsilon>0}} (with {{M|\epsilon\in\mathbb{R} }}) be given.
*** We will now define a new family of {{plural|sequence|s}}. For each {{M|A_n}} we will construct the sequence {{MSeq|A_{nm}|m|in=\mathcal{R} }} of sets such that:
***# {{M|1=\forall n\in\mathbb{N}[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}]}} and
***# {{M|1=\forall n\in\mathbb{N}[\sum^\infty_{m=1}\bar{\mu}(A_{nm})\le\mu^*(A_n)+\epsilon\frac{1}{2^n}]}}
*** Let {{M|n\in\mathbb{N} }} be given (we will now define {{MSeq|A_{mn}|m|in=\mathcal{R} }})
**** Recall that {{M|1=\mu^*(A_n):=\text{inf}(\beta_{A_n})}}
**** Any value greater than the [[infimum|{{M|\text{inf}(\beta_{A_n})}}]], say {{M|w}}, is not a [[lower bound]] so there must exist an element in {{M|\beta_{A_n} }} less that {{M|w}} (so {{M|w}} cannot be a lower bound)
***** Choose {{M|1=w:=\text{inf}(\beta_{A_n})+\frac{\epsilon}{2^n} }}
****** As {{M|\epsilon>0}} and {{M|\frac{1}{2^n}>0}} we see {{M|\frac{\epsilon}{2^n}>0}}, thus {{M|\mu^*(A_n)<\mu^*(A_n)+\frac{\epsilon}{2^n} }}
**** By the definition of [[infimum]]:
***** {{M|1=\exists s\in\beta_{A_n}[w>\text{inf}(\beta_{A_n})\implies s< w]}}
**** If {{M|s\in\beta_{A_n} }} then:
***** {{M|1=\exists(B_n)_{n=1}^\infty\in\alpha_{A_n} }} such that {{M|1=s=\sum^\infty_{n=1}\bar{\mu}(B_n)}}.
**** As {{M|1=s<w=\text{inf}(\beta_{A_n})+\frac{\epsilon}{2^n}=\mu^*(A_n)+\frac{\epsilon}{2^n} }} and {{M|1=s=\sum^\infty_{n=1}\bar{\mu}(B_n)}} we see:
***** {{M|1=\sum^\infty_{n=1}\bar{\mu}(B_n)<\mu^*(A_n)+\frac{\epsilon}{2^n} }}
**** {{Caution|1=This doesn't show that {{MM|1=A_n\subseteq\bigcup_{m=1}^\infty A_{nm} }} - don't forget!}}
**** Define a new sequence, {{MSeq|A_{nm}|m|in=\mathcal{R} }} to be the sequence {{MSeq|B_n|post=\in\alpha_{A_n} }} we just showed to exist
*** Since {{M|n\in\mathbb{N} }} was arbitrary for each {{M|1=A_n\in(A_k)_{k=1}^\infty\subseteq\mathcal{H}_{\sigma R}(\mathcal{R})}} we now have a new sequence: {{MSeq|A_{nm}|m|in=\mathcal{R} }} such that:
**** {{MM|1=\forall n\in\mathbb{N}\left[\sum^\infty_{m=1}\bar{\mu}(A_{nm})<\mu^*(A_n)+\frac{\epsilon}{2^n}\right]}} and {{MM|1=\forall n\in\mathbb{N}\left[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}\right]}}
*** Recall now that a [[union of subsets is a subset of the union]], thus:
**** {{MM|1=\bigcup_{n=1}^\infty A_n\subseteq \bigcup_{n=1}^\infty\left(\bigcup_{m=1}^\infty A_{nm}\right)}}
*** So {{MM|1=\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\left(\sum_{m=1}^\infty \bar{\mu}(A_{nm})\right)<\sum_{n=1}^\infty\left(\mu^*(A_n)+\frac{\epsilon}{2^n}\right)}}{{MM|1==\sum^\infty_{n=1}\mu^*(A_n)+\sum^\infty_{n=1}\frac{\epsilon}{2^n} }}
**** Note that {{M|1=\sum^\infty_{n=1}\frac{\epsilon}{2^n}=\epsilon\sum^\infty_{n=1}\frac{1}{2^n} }} and that {{M|1=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots}} is a classic example of a [[geometric series]], we see easily that:
***** {{M|1=\epsilon\sum^\infty_{n=1}\frac{1}{2^n}=1\epsilon=\epsilon}} thus:
*** {{MM|1=\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon}}
** Since {{M|\epsilon>0}} (with {{M|\epsilon\in\mathbb{R} }} was arbitrary we see:
*** {{M|1=\forall\epsilon>0\left[\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon\right]}}
** Recall that {{M|1=\left(\forall\epsilon>0[a<b+\epsilon]\right)\iff\left(a\le b\right)}} (from the [[epsilon form of inequalities]])
** Thus: {{MM|1=\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\mu^*(A_n)}}
* Since {{MSeq|A_n|in=\mathcal{H}_{\sigma R}(\mathcal{R})}} was arbitrary we have shown that:
** {{MM|1=\forall(A_n)_{n=1}^\infty\subseteq\mathcal{H}_{\sigma R}(\mathcal{R})\left[\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\mu^*(A_n)\right]}}
This completes the proof that {{M|\mu^*}} is {{sigma|subadditive}}
====Caveats====
# Halmos starts with a set {{M|A\in\mathcal{H}_{\sigma R}(\mathcal{R})}} and a [[sequence]] {{MSeq|A_n|in=\mathcal{H}_{\sigma R}(\mathcal{R})}} such that:
#* {{M|1=A\subseteq\bigcup_{n=1}^\infty A_n}}
#: where as I just start with a sequence, as {{M|\mathcal{H}_{\sigma R}(\mathcal{R})}} is a [[sigma-algebra|{{sigma|algebra}}]], their union is also in {{M|\mathcal{H}_{\sigma R}(\mathcal{R})}}
# {{Warning|I never consider the case where a measure measures a set to be infinite. Where this happens things like {{M|\infty<\infty}} make no sense}}
===The rest===
Still to do:
# {{M|\mu^*}} being monotonic with respect to set inclusion and the usual ordering on the reals.
# {{M|1=\mu^*(\emptyset)=0}} - this can come from the extension part as {{M|\bar{\mu} }} has this property already

==Notes==
<references group="Note"/>
==References==
<references/>
{{Measure theory navbox|plain}}
{{Theorem Of|Measure Theory}}