Extending pre-measures to outer-measures

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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:

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

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

Proof

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Proof notes

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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 $\bar{\mu}(A)\le\sum_i\bar{\mu}(A_i)$
        • By passing to the infimum we see that $\bar{\mu}(A)\le\mu^*(A)$ as required.

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

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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

For brevity we define the following shorthands:

1. $$\alpha_A:=\left\{(A_n)_{n=1}^\infty\ \Big\vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R}\wedge A\subseteq\bigcup_{n=1}^\infty A_n\right\}$$
2. $$\beta_A:=\left\{\sum^\infty_{n=1}\bar{\mu}(A_n)\ \Big\vert\ (A_n)_{n=1}^\infty\in\alpha_A \right\}$$

Now we may define $\mu^*$ as:

• $\mu^*:A\mapsto\text{inf}(\beta_A)$

Proof that $\mu^*$ is an extension of $\bar{\mu}$

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