Difference between revisions of "Extending pre-measures to outer-measures"
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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 29: | Line 32: | ||
#*#* By [[passing to the infimum]] we see that {{M|\bar{\mu}(A)\le\mu^*(A)}} as required. | #*#* 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! | * 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. | ** 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) | * 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]]. | ||
| + | ==Notes== | ||
| + | <references group="Note"/> | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Measure theory navbox|plain}} | {{Measure theory navbox|plain}} | ||
{{Theorem Of|Measure Theory}} | {{Theorem Of|Measure Theory}} | ||
Revision as of 22:41, 31 July 2016
- Caution:This page is currently being written and is not ready for being used as a reference, it's a notes quality page
Contents
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]:
- an extension of [ilmath]\bar{\mu} [/ilmath] and
- an outer-measure (on the hereditary [ilmath]\sigma[/ilmath]-ring generated by [ilmath]\mathcal{R} [/ilmath], written [ilmath]\mathcal{H}_{\sigma_R}(\mathcal{R})[/ilmath])
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:
- an extended real valued countably subadditive set function that is monotonic and non-negative with the property: [ilmath]\mu^*(\emptyset)=0[/ilmath] defined on a hereditary [ilmath]\sigma[/ilmath]-ring
- 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.
- 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]
- We can now define a sequence, [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] as follows:
- 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)
- Remember that [ilmath]\emptyset\in\mathcal{R} [/ilmath] as [ilmath]\mathcal{ R } [/ilmath] is a ring of sets
- Now we must bound [ilmath]\mu^*[/ilmath] below (by [ilmath]\bar{\mu}(A)[/ilmath]) 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 [ilmath]A[/ilmath] and a countably infinite or finite sequence of sets, [ilmath](A_i)[/ilmath] such that [ilmath]A\subseteq\bigcup_i A_i[/ilmath] then [ilmath]\bar{\mu}(A)\le\sum_i\bar{\mu}(A_i)[/ilmath]
- By passing to the infimum we see that [ilmath]\bar{\mu}(A)\le\mu^*(A)[/ilmath] as required.
- 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:
- 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]
- Let [ilmath]A\in\mathcal{R} [/ilmath] be given.
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:
- an extended real valued countably subadditive set function that is monotonic and non-negative with the property: [ilmath]\mu^*(\emptyset)=0[/ilmath] defined on a hereditary [ilmath]\sigma[/ilmath]-ring
For brevity we define the following shorthands:
- [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]
- [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]
- 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.
- for [ilmath]\text{inf}(S)[/ilmath] (for a set, [ilmath]S[/ilmath]) we see:
- 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]
- [ilmath]\mu^*(A):=\text{inf}(\beta_A)\le\sum^\infty_{n=1}\bar{\mu}(A_n)=\bar{\mu}(A)[/ilmath]
- We have shown [ilmath]\mu^*(A)\le\bar{\mu}(A)[/ilmath] as required
- 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].
- Part 2: [ilmath]\bar{\mu}(A)\le\mu^*(A)[/ilmath]
- SEE NOTEPAD. Define [ilmath]\gamma_A:=\left\{\bar{\mu}(A)\right\}[/ilmath], 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 [ilmath]\forall x\in\beta_A\exists y\in\gamma_A[y\le x][/ilmath] - we may now pass to the infimum.
- Part 1: [ilmath]\bar{\mu}(A)\ge\mu^*(A)[/ilmath]
- 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]
Notes
- ↑ This is called the trichotomy rule or something, I should link to the relevant part of a partial order here
References
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