# Extending pre-measures to outer-measures

Caution:This page is currently being written and is not ready for being used as a reference, it's a notes quality page

## 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]:

Given by:

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

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.

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:

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 [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]
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.
• 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]
• We have shown [ilmath]\mu^*(A)\le\bar{\mu}(A)[/ilmath] as required
2. Part 2: [ilmath]\bar{\mu}(A)\le\mu^*(A)[/ilmath]

### Proof that [ilmath]\mu^*[/ilmath] is [ilmath]\sigma[/ilmath]-subadditive

• Let [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R}) [/ilmath] be given. We want to show that [ilmath]\mu^*(\bigcup_{n=1}^\infty A_n)\le\sum^\infty_{n=1}\mu^*(A_n)[/ilmath]
• Let [ilmath]\epsilon>0[/ilmath] (with [ilmath]\epsilon\in\mathbb{R} [/ilmath]) be given.
• We will now define a new family of sequences. For each [ilmath]A_n[/ilmath] we will construct the sequence [ilmath] ({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] of sets such that:
1. [ilmath]\forall n\in\mathbb{N}[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}][/ilmath] and
2. [ilmath]\forall n\in\mathbb{N}[\sum^\infty_{m=1}\bar{\mu}(A_{nm})\le\mu^*(A_n)+\epsilon\frac{1}{2^n}][/ilmath]
• Let [ilmath]n\in\mathbb{N} [/ilmath] be given (we will now define [ilmath] ({ A_{mn} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath])
• Recall that [ilmath]\mu^*(A_n):=\text{inf}(\beta_{A_n})[/ilmath]
• Any value greater than the [ilmath]\text{inf}(\beta_{A_n})[/ilmath], say [ilmath]w[/ilmath], is not a lower bound so there must exist an element in [ilmath]\beta_{A_n} [/ilmath] less that [ilmath]w[/ilmath] (so [ilmath]w[/ilmath] cannot be a lower bound)
• Choose [ilmath]w:=\text{inf}(\beta_{A_n})+\frac{\epsilon}{2^n}[/ilmath]
• As [ilmath]\epsilon>0[/ilmath] and [ilmath]\frac{1}{2^n}>0[/ilmath] we see [ilmath]\frac{\epsilon}{2^n}>0[/ilmath], thus [ilmath]\mu^*(A_n)<\mu^*(A_n)+\frac{\epsilon}{2^n} [/ilmath]
• By the definition of infimum:
• [ilmath]\exists s\in\beta_{A_n}[w>\text{inf}(\beta_{A_n})\implies s< w][/ilmath]
• If [ilmath]s\in\beta_{A_n} [/ilmath] then:
• [ilmath]\exists(B_n)_{n=1}^\infty\in\alpha_{A_n}[/ilmath] such that [ilmath]s=\sum^\infty_{n=1}\bar{\mu}(B_n)[/ilmath].
• As [ilmath]s<w=\text{inf}(\beta_{A_n})+\frac{\epsilon}{2^n}=\mu^*(A_n)+\frac{\epsilon}{2^n}[/ilmath] and [ilmath]s=\sum^\infty_{n=1}\bar{\mu}(B_n)[/ilmath] we see:
• [ilmath]\sum^\infty_{n=1}\bar{\mu}(B_n)<\mu^*(A_n)+\frac{\epsilon}{2^n}[/ilmath]
• Caution:This doesn't show that $A_n\subseteq\bigcup_{m=1}^\infty A_{nm}$ - don't forget!
• Define a new sequence, [ilmath] ({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] to be the sequence [ilmath] ({ B_n })_{ n = 1 }^{ \infty } \in\alpha_{A_n} [/ilmath] we just showed to exist
• Since [ilmath]n\in\mathbb{N} [/ilmath] was arbitrary for each [ilmath]A_n\in(A_k)_{k=1}^\infty\subseteq\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] we now have a new sequence: [ilmath] ({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} [/ilmath] such that:
• $\forall n\in\mathbb{N}\left[\sum^\infty_{m=1}\bar{\mu}(A_{nm})<\mu^*(A_n)+\frac{\epsilon}{2^n}\right]$ and $\forall n\in\mathbb{N}\left[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}\right]$
• Recall now that a union of subsets is a subset of the union, thus:
• $\bigcup_{n=1}^\infty A_n\subseteq \bigcup_{n=1}^\infty\left(\bigcup_{m=1}^\infty A_{nm}\right)$
• So $\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\left(\sum_{m=1}^\infty \bar{\mu}(A_{nm})\right)<\sum_{n=1}^\infty\left(\mu^*(A_n)+\frac{\epsilon}{2^n}\right)$$=\sum^\infty_{n=1}\mu^*(A_n)+\sum^\infty_{n=1}\frac{\epsilon}{2^n}$
• Note that [ilmath]\sum^\infty_{n=1}\frac{\epsilon}{2^n}=\epsilon\sum^\infty_{n=1}\frac{1}{2^n}[/ilmath] and that [ilmath]\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots[/ilmath] is a classic example of a geometric series, we see easily that:
• [ilmath]\epsilon\sum^\infty_{n=1}\frac{1}{2^n}=1\epsilon=\epsilon[/ilmath] thus:
• $\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon$
• Since [ilmath]\epsilon>0[/ilmath] (with [ilmath]\epsilon\in\mathbb{R} [/ilmath] was arbitrary we see:
• [ilmath]\forall\epsilon>0\left[\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon\right][/ilmath]
• Recall that [ilmath]\left(\forall\epsilon>0[a<b+\epsilon]\right)\iff\left(a\le b\right)[/ilmath] (from the epsilon form of inequalities)
• Thus: $\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\mu^*(A_n)$
• Since [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R}) [/ilmath] was arbitrary we have shown that:
• $\forall(A_n)_{n=1}^\infty\subseteq\mathcal{H}_{\sigma R}(\mathcal{R})\left[\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\mu^*(A_n)\right]$

This completes the proof that [ilmath]\mu^*[/ilmath] is [ilmath]\sigma[/ilmath]-subadditive

#### Caveats

1. Halmos starts with a set [ilmath]A\in\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] and a sequence [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R}) [/ilmath] such that:
• [ilmath]A\subseteq\bigcup_{n=1}^\infty A_n[/ilmath]
where as I just start with a sequence, as [ilmath]\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] is a [ilmath]\sigma[/ilmath]-algebra, their union is also in [ilmath]\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath]
2. Warning:I never consider the case where a measure measures a set to be infinite. Where this happens things like [ilmath]\infty<\infty[/ilmath] make no sense

### The rest

Still to do:

1. [ilmath]\mu^*[/ilmath] being monotonic with respect to set inclusion and the usual ordering on the reals.
2. [ilmath]\mu^*(\emptyset)=0[/ilmath] - this can come from the extension part as [ilmath]\bar{\mu} [/ilmath] has this property already

## Notes

1. This is called the trichotomy rule or something, I should link to the relevant part of a partial order here