Extending pre-measures to outer-measures

Statement

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

• an extension of $\bar{\mu}$ and
• an outer-measure (on the hereditary $\sigma$-ring generated by $\mathcal{R}$, written $\mathcal{H}_{\sigma_R}(\mathcal{R})$)

Given by:

• $\mu^*:\mathcal{H}_{\sigma_R}(\mathcal{R})\rightarrow\bar{\mathbb{R} }_{\ge0}$
  • $$\mu^*:A\mapsto\text{inf}\left\{\left.\sum^\infty_{n=1}\bar{\mu}(A_n)\right\vert(A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}^\infty A_n\right\}$$

The statement of the theorem is that this $\mu^*$ is indeed an outer-measure

Proof

[Expand]

Recall the definition of an outer-measure, we must show $\mu^*$ satisfies this.

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

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

In words, $\mu^*$ is:

• an extended real valued countably subadditive set function that is monotonic and non-negative with the property: $\mu^*(\emptyset)=0$ defined on a hereditary $\sigma$-ring

1. We claimed that $\mu^*$ is an extension of $\bar{\mu}$, this means that: $\forall A\in\mathcal{R}[\mu^*=\bar{\mu}]$. Let us check this.
   • Let $A\in\mathcal{R}$ be given.
     1. First we must bound $\mu^*$ above. This is because $[\mu^*(A)=\bar{\mu}(A)]\iff[\mu^*(A)\ge\bar{\mu}(A)\wedge\bar{\mu}(A)\ge\mu^*(A)]$
        • Remember that $\emptyset\in\mathcal{R}$ as $\mathcal{ R }$ is a ring of sets
          • We can now define a sequence, $({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R}$ as follows:
            • $A_1=A$
            • $A_n=\emptyset$ for $n\ge 2$
          • So $({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{R}$ is $(A,\emptyset,\emptyset,\ldots)$
          • Now $\sum_{n=1}^\infty \bar{\mu}(A_n)=\bar{\mu}(A)+\bar{\mu}(\emptyset)+\bar{\mu}(\emptyset)+\ldots=\bar{\mu}(A)+0+0+\ldots=\bar{\mu}(\emptyset)$
        • So $\mu^*(A)\le\bar{\mu}(A)$ (as $\mu^*$ 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 $\mu^*$ below (by $\bar{\mu}(A)$) 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 $A$ and a countably infinite or finite sequence of sets, $(A_i)$ such that $A\subseteq\bigcup_i A_i$ then $\bar{\mu}(A)\le\sum_i\bar{\mu}(A_i)$
        • Halmos handwaves here, we want to show that $\bar{\mu}(A)\le\mu^*(A)$, however all we know is $\mu^*(A)\le\sum_i\bar{\mu}(A_i)$ and $\bar{\mu}(A)\le\sum_i\bar{\mu}(A_i)$ - this isn't very helpful, the solution will be in the nature of the infimum I am sure

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TODO: Finish proof

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• Suppose it is not true, that is that we have $\bar{\mu}(A)>\mu^*(A)$, then $\sum_i\bar{\mu}(A_i)>\mu^*(A)$... then what....
  • This confirms it has something to do with the infimum. Investigate in the morning