Extending pre-measures to outer-measures

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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] - 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

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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]
        • By passing to the infimum we see that [ilmath]\bar{\mu}(A)\le\mu^*(A)[/ilmath] 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 [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:

• [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

For brevity we define the following shorthands:

1. [math]\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\}[/math]
2. [math]\beta_A:=\left\{\sum^\infty_{n=1}\bar{\mu}(A_n)\ \Big\vert\ (A_n)_{n=1}^\infty\in\alpha_A \right\}[/math]

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]
    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]
       • SEE NOTEPAD. Define [ilmath]\gamma_A:=\left\{\bar{\mu}(A)\right\}[/ilmath], 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 [ilmath]\forall x\in\beta_A\exists y\in\gamma_A[y\le x][/ilmath] - we may now pass to the infimum.

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 [math]A_n\subseteq\bigcup_{m=1}^\infty A_{nm}[/math] - 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:
      • [math]\forall n\in\mathbb{N}\left[\sum^\infty_{m=1}\bar{\mu}(A_{nm})<\mu^*(A_n)+\frac{\epsilon}{2^n}\right][/math] and [math]\forall n\in\mathbb{N}\left[A_n\subseteq\bigcup_{m=1}^\infty A_{nm}\right][/math]
    • Recall now that a union of subsets is a subset of the union, thus:
      • [math]\bigcup_{n=1}^\infty A_n\subseteq \bigcup_{n=1}^\infty\left(\bigcup_{m=1}^\infty A_{nm}\right)[/math]
    • So [math]\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)[/math][math]=\sum^\infty_{n=1}\mu^*(A_n)+\sum^\infty_{n=1}\frac{\epsilon}{2^n}[/math]
      • 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:
    • [math]\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon[/math]
  • 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: [math]\mu^*\left(\bigcup_{n=1}^\infty A_n\right)\le\sum^\infty_{n=1}\mu^*(A_n)[/math]
• Since [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R}) [/ilmath] was arbitrary we have shown that:
  • [math]\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][/math]

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