Difference between revisions of "Extending pre-measures to outer-measures"

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(Created page with "==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 exte...")
 
(→‎Proof: Adding sigma-subadditive part.)
 
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: {{Caution|This page is currently being written and is not ready for being used as a reference, it's a ''notes quality'' page}}
 
==Statement==
 
==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}}:
 
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}}:
Line 5: Line 6:
 
Given by:
 
Given by:
 
* {{M|\mu^*:\mathcal{H}_{\sigma_R}(\mathcal{R})\rightarrow\bar{\mathbb{R} }_{\ge0} }}
 
* {{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\} }}
+
** {{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]]
 
The statement of the theorem is that this {{M|\mu^*}} is indeed an [[outer-measure]]
 
==Proof==
 
==Proof==
 +
{{Begin Inline Theorem}}
 +
Proof notes
 +
{{Begin Inline Proof}}
 
{{Begin Notebox}}
 
{{Begin Notebox}}
 
Recall the definition of an [[outer-measure]], we must show {{M|\mu^*}} satisfies this.
 
Recall the definition of an [[outer-measure]], we must show {{M|\mu^*}} satisfies this.
Line 26: Line 30:
 
#*#* 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:
 
#*#* 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)}}
 
#*#** 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
+
#*#* By [[passing to the infimum]] we see that {{M|\bar{\mu}(A)\le\mu^*(A)}} as required.
−
{{Todo|Finish proof}}
+
 
 +
'''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==
 
<references/>
 
<references/>
 
{{Measure theory navbox|plain}}
 
{{Measure theory navbox|plain}}
 
{{Theorem Of|Measure Theory}}
 
{{Theorem Of|Measure Theory}}

Latest revision as of 16:59, 17 August 2016

Caution:This page is currently being written and is not ready for being used as a reference, it's a notes quality page

Statement

Given a pre-measure, [ilmath]\bar{\mu} [/ilmath], on a ring of sets, [ilmath]\mathcal{R} [/ilmath], we can define a new function, [ilmath]\mu^*[/ilmath] which is[1]:

Given by:

  • [ilmath]\mu^*:\mathcal{H}_{\sigma_R}(\mathcal{R})\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath]
    • [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] - here [ilmath]\text{inf} [/ilmath] denotes the infimum of a set.

The statement of the theorem is that this [ilmath]\mu^*[/ilmath] is indeed an outer-measure

Proof

Proof notes

Recall the definition of an outer-measure, we must show [ilmath]\mu^*[/ilmath] satisfies this.

An outer-measure, [ilmath]\mu^*[/ilmath] is a set function from a hereditary [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{H} [/ilmath], to the (positive) extended real values, [ilmath]\bar{\mathbb{R} }_{\ge0} [/ilmath], that is[1]:

  • [ilmath]\forall A\in\mathcal{H}[\mu^*(A)\ge 0][/ilmath] - non-negative
  • [ilmath]\forall A,B\in\mathcal{H}[A\subseteq B\implies \mu^*(A)\le\mu^*(B)][/ilmath] - monotonic
  • [ilmath] \forall ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H} [\mu^*(\bigcup_{n=1}^\infty A_n)\le\sum^\infty_{n=1}\mu^*(A_n)] [/ilmath] - countably subadditive

In words, [ilmath]\mu^*[/ilmath] is:

  1. We claimed that [ilmath]\mu^*[/ilmath] is an extension of [ilmath]\bar{\mu} [/ilmath], this means that: [ilmath]\forall A\in\mathcal{R}[\mu^*=\bar{\mu}][/ilmath]. Let us check this.
    • Let [ilmath]A\in\mathcal{R} [/ilmath] be given.
      1. First we must bound [ilmath]\mu^*[/ilmath] above. This is because [ilmath][\mu^*(A)=\bar{\mu}(A)]\iff[\mu^*(A)\ge\bar{\mu}(A)\wedge\bar{\mu}(A)\ge\mu^*(A)][/ilmath]
        • Remember that [ilmath]\emptyset\in\mathcal{R} [/ilmath] as [ilmath]\mathcal{ R } [/ilmath] is a ring of sets
          • We can now define a sequence, [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] as follows:
            • [ilmath]A_1=A[/ilmath]
            • [ilmath]A_n=\emptyset[/ilmath] for [ilmath]n\ge 2[/ilmath]
          • So [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] is [ilmath](A,\emptyset,\emptyset,\ldots)[/ilmath]
          • Now [ilmath]\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)[/ilmath]
        • So [ilmath]\mu^*(A)\le\bar{\mu}(A)[/ilmath] (as [ilmath]\mu^*[/ilmath] is the defined as the infimum of such expressions, all we have done is find an upper-bound for it)
      2. Now we must bound [ilmath]\mu^*[/ilmath] below (by [ilmath]\bar{\mu}(A)[/ilmath]) to show they're equal.

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 [ilmath]\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] we see that [ilmath]-1[/ilmath] 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 [ilmath]\bar{\mu} [/ilmath] even exists (this probably uses monotonicity of [ilmath]\bar{\mu} [/ilmath] and should be easy to show)

Recall the definition of an outer-measure, we must show [ilmath]\mu^*[/ilmath] satisfies this.

An outer-measure, [ilmath]\mu^*[/ilmath] is a set function from a hereditary [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{H} [/ilmath], to the (positive) extended real values, [ilmath]\bar{\mathbb{R} }_{\ge0} [/ilmath], that is[1]:

  • [ilmath]\forall A\in\mathcal{H}[\mu^*(A)\ge 0][/ilmath] - non-negative
  • [ilmath]\forall A,B\in\mathcal{H}[A\subseteq B\implies \mu^*(A)\le\mu^*(B)][/ilmath] - monotonic
  • [ilmath] \forall ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H} [\mu^*(\bigcup_{n=1}^\infty A_n)\le\sum^\infty_{n=1}\mu^*(A_n)] [/ilmath] - countably subadditive

In words, [ilmath]\mu^*[/ilmath] is:

For brevity we define the following shorthands:

  1. [math]\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\}[/math]
  2. [math]\beta_A:=\left\{\sum^\infty_{n=1}\bar{\mu}(A_n)\ \Big\vert\ (A_n)_{n=1}^\infty\in\alpha_A \right\}[/math]

Now we may define [ilmath]\mu^*[/ilmath] as:

  • [ilmath]\mu^*:A\mapsto\text{inf}(\beta_A)[/ilmath]

Proof that [ilmath]\mu^*[/ilmath] is an extension of [ilmath]\bar{\mu} [/ilmath]

  • Let [ilmath]A\in\mathcal{R} [/ilmath] be given
    • In order to prove [ilmath]\bar{\mu}(A)=\mu^*(A)[/ilmath] we need only prove [ilmath][\bar{\mu}(A)\ge\mu^*(A)\wedge\bar{\mu}(A)\le\mu^*(A)][/ilmath][Note 1]
      1. Part 1: [ilmath]\bar{\mu}(A)\ge\mu^*(A)[/ilmath]
        • Consider the sequence [ilmath] ({ A_n })_{ n = 1 }^{ \infty } [/ilmath] given by [ilmath]A_1:=A[/ilmath] and [ilmath]A_i:=\emptyset[/ilmath] for [ilmath]i>1[/ilmath], so the sequence [ilmath]A,\emptyset,\emptyset,\ldots[/ilmath].
          • Clearly [ilmath]A\subseteq\bigcup^\infty_{n=1}A_n[/ilmath] (as [ilmath]\bigcup^\infty_{n=1}A_n=A[/ilmath])
          • As such this [ilmath] ({ A_n })_{ n = 1 }^{ \infty } \in\alpha_A [/ilmath]
          • This means [ilmath]\sum^\infty_{n=1}\bar{\mu}(A_n)\in\beta_A[/ilmath] (as [ilmath] ({ A_n })_{ n = 1 }^{ \infty } \in\alpha_A [/ilmath] and [ilmath]\beta_A[/ilmath] is the sum of all the pre-measures Template:WRT [ilmath]\bar{\mu} [/ilmath] of the sequences of sets in [ilmath]\alpha_A[/ilmath])
          • Recall that the infimum of a set is, among other things, a lower bound of the set. So:
            • for [ilmath]\text{inf}(S)[/ilmath] (for a set, [ilmath]S[/ilmath]) we see:
              • [ilmath]\forall s\in S[\text{inf}(S)\le s][/ilmath] - this uses only the lower bound part of the infimum definition.
          • By applying this to [ilmath]\text{inf}(\beta_A)\big(=\mu^*(A)\big)[/ilmath] we see:
            • [ilmath]\mu^*(A):=\text{inf}(\beta_A)\le\sum^\infty_{n=1}\bar{\mu}(A_n)=\bar{\mu}(A)[/ilmath]
              • as [ilmath]\sum^\infty_{n=1}\bar{\mu}(A_n)\in\beta_A[/ilmath] and [ilmath]\text{inf}(S)[/ilmath] remember and
              • By definition of a (pre-)measure, [ilmath]\mu(\emptyset)=0[/ilmath], so: [ilmath]\sum^\infty_{n=1}\bar{\mu}(A_n)=\bar{\mu}(A)+\bar{\mu}(\emptyset)+\bar{\mu}(\emptyset)+\cdots=\bar{\mu}(A)[/ilmath]
        • We have shown [ilmath]\mu^*(A)\le\bar{\mu}(A)[/ilmath] as required
      2. Part 2: [ilmath]\bar{\mu}(A)\le\mu^*(A)[/ilmath]

Proof that [ilmath]\mu^*[/ilmath] is [ilmath]\sigma[/ilmath]-subadditive

  • Let [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R}) [/ilmath] be given. We want to show that [ilmath]\mu^*(\bigcup_{n=1}^\infty A_n)\le\sum^\infty_{n=1}\mu^*(A_n)[/ilmath]
    • Let [ilmath]\epsilon>0[/ilmath] (with [ilmath]\epsilon\in\mathbb{R} [/ilmath]) be given.
      • We will now define a new family of sequences. For each [ilmath]A_n[/ilmath] we will construct the sequence [ilmath] ({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] of sets such that:
        1. [ilmath]\forall n\in\mathbb{N}[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}][/ilmath] and
        2. [ilmath]\forall n\in\mathbb{N}[\sum^\infty_{m=1}\bar{\mu}(A_{nm})\le\mu^*(A_n)+\epsilon\frac{1}{2^n}][/ilmath]
      • Let [ilmath]n\in\mathbb{N} [/ilmath] be given (we will now define [ilmath] ({ A_{mn} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath])
        • Recall that [ilmath]\mu^*(A_n):=\text{inf}(\beta_{A_n})[/ilmath]
        • Any value greater than the [ilmath]\text{inf}(\beta_{A_n})[/ilmath], say [ilmath]w[/ilmath], is not a lower bound so there must exist an element in [ilmath]\beta_{A_n} [/ilmath] less that [ilmath]w[/ilmath] (so [ilmath]w[/ilmath] cannot be a lower bound)
          • Choose [ilmath]w:=\text{inf}(\beta_{A_n})+\frac{\epsilon}{2^n}[/ilmath]
            • As [ilmath]\epsilon>0[/ilmath] and [ilmath]\frac{1}{2^n}>0[/ilmath] we see [ilmath]\frac{\epsilon}{2^n}>0[/ilmath], thus [ilmath]\mu^*(A_n)<\mu^*(A_n)+\frac{\epsilon}{2^n} [/ilmath]
        • By the definition of infimum:
          • [ilmath]\exists s\in\beta_{A_n}[w>\text{inf}(\beta_{A_n})\implies s< w][/ilmath]
        • If [ilmath]s\in\beta_{A_n} [/ilmath] then:
          • [ilmath]\exists(B_n)_{n=1}^\infty\in\alpha_{A_n}[/ilmath] such that [ilmath]s=\sum^\infty_{n=1}\bar{\mu}(B_n)[/ilmath].
        • As [ilmath]s<w=\text{inf}(\beta_{A_n})+\frac{\epsilon}{2^n}=\mu^*(A_n)+\frac{\epsilon}{2^n}[/ilmath] and [ilmath]s=\sum^\infty_{n=1}\bar{\mu}(B_n)[/ilmath] we see:
          • [ilmath]\sum^\infty_{n=1}\bar{\mu}(B_n)<\mu^*(A_n)+\frac{\epsilon}{2^n}[/ilmath]
        • Caution:This doesn't show that [math]A_n\subseteq\bigcup_{m=1}^\infty A_{nm}[/math] - don't forget!
        • Define a new sequence, [ilmath] ({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] to be the sequence [ilmath] ({ B_n })_{ n = 1 }^{ \infty } \in\alpha_{A_n} [/ilmath] we just showed to exist
      • Since [ilmath]n\in\mathbb{N} [/ilmath] was arbitrary for each [ilmath]A_n\in(A_k)_{k=1}^\infty\subseteq\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] we now have a new sequence: [ilmath] ({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] such that:
        • [math]\forall n\in\mathbb{N}\left[\sum^\infty_{m=1}\bar{\mu}(A_{nm})<\mu^*(A_n)+\frac{\epsilon}{2^n}\right][/math] and [math]\forall n\in\mathbb{N}\left[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}\right][/math]
      • Recall now that a union of subsets is a subset of the union, thus:
        • [math]\bigcup_{n=1}^\infty A_n\subseteq \bigcup_{n=1}^\infty\left(\bigcup_{m=1}^\infty A_{nm}\right)[/math]
      • So [math]\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)[/math][math]=\sum^\infty_{n=1}\mu^*(A_n)+\sum^\infty_{n=1}\frac{\epsilon}{2^n}[/math]
        • Note that [ilmath]\sum^\infty_{n=1}\frac{\epsilon}{2^n}=\epsilon\sum^\infty_{n=1}\frac{1}{2^n}[/ilmath] and that [ilmath]\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots[/ilmath] is a classic example of a geometric series, we see easily that:
          • [ilmath]\epsilon\sum^\infty_{n=1}\frac{1}{2^n}=1\epsilon=\epsilon[/ilmath] thus:
      • [math]\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon[/math]
    • Since [ilmath]\epsilon>0[/ilmath] (with [ilmath]\epsilon\in\mathbb{R} [/ilmath] was arbitrary we see:
      • [ilmath]\forall\epsilon>0\left[\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon\right][/ilmath]
    • Recall that [ilmath]\left(\forall\epsilon>0[a<b+\epsilon]\right)\iff\left(a\le b\right)[/ilmath] (from the epsilon form of inequalities)
    • Thus: [math]\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\mu^*(A_n)[/math]
  • Since [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R}) [/ilmath] was arbitrary we have shown that:
    • [math]\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][/math]

This completes the proof that [ilmath]\mu^*[/ilmath] is [ilmath]\sigma[/ilmath]-subadditive

Caveats

  1. Halmos starts with a set [ilmath]A\in\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] and a sequence [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R}) [/ilmath] such that:
    • [ilmath]A\subseteq\bigcup_{n=1}^\infty A_n[/ilmath]
    where as I just start with a sequence, as [ilmath]\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] is a [ilmath]\sigma[/ilmath]-algebra, their union is also in [ilmath]\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath]
  2. Warning:I never consider the case where a measure measures a set to be infinite. Where this happens things like [ilmath]\infty<\infty[/ilmath] make no sense

The rest

Still to do:

  1. [ilmath]\mu^*[/ilmath] being monotonic with respect to set inclusion and the usual ordering on the reals.
  2. [ilmath]\mu^*(\emptyset)=0[/ilmath] - this can come from the extension part as [ilmath]\bar{\mu} [/ilmath] has this property already

Notes

  1. ↑ This is called the trichotomy rule or something, I should link to the relevant part of a partial order here

References

  1. ↑ 1.0 1.1 1.2 Measure Theory - Paul R. Halmos

v • d • e
 
Measure Theory
Overview of the basic and important objects studied in measure theory
Set-structures
Measures
Pre-measure [ilmath]\leadsto[/ilmath] outer-measure [ilmath]\leadsto[/ilmath] measure (see Extending pre-measures to measures (via: Extending pre-measures to outer-measures [ilmath]\rightarrow[/ilmath] "Restricting outer-measures to measures" maybe?). Alternatively: inner-measure can be used.
measure space
Functions
Spaces
Integrals



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