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...") |
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#*#* '''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 | #*#* '''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}} | {{Todo|Finish proof}} | ||
| + | * Suppose it is not true, that is that we have {{M|\bar{\mu}(A)>\mu^*(A)}}, then {{M|\sum_i\bar{\mu}(A_i)>\mu^*(A)}}... then what.... | ||
| + | ** This confirms it has something to do with the infimum. Investigate in the morning | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Measure theory navbox|plain}} | {{Measure theory navbox|plain}} | ||
{{Theorem Of|Measure Theory}} | {{Theorem Of|Measure Theory}} | ||
Revision as of 23:26, 8 April 2016
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]
The statement of the theorem is that this [ilmath]\mu^*[/ilmath] is indeed an outer-measure
Proof
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]
- Halmos handwaves here, we want to show that [ilmath]\bar{\mu}(A)\le\mu^*(A)[/ilmath], however all we know is [ilmath]\mu^*(A)\le\sum_i\bar{\mu}(A_i)[/ilmath] and [ilmath]\bar{\mu}(A)\le\sum_i\bar{\mu}(A_i)[/ilmath] - this isn't very helpful, the solution will be in the nature of the infimum I am sure
- 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.
TODO: Finish proof
- Suppose it is not true, that is that we have [ilmath]\bar{\mu}(A)>\mu^*(A)[/ilmath], then [ilmath]\sum_i\bar{\mu}(A_i)>\mu^*(A)[/ilmath]... then what....
- This confirms it has something to do with the infimum. Investigate in the morning
References
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