Extending pre-measures to outer-measures

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:

• 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:

• [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

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.
        • 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

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TODO: Finish proof

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• 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