Extending pre-measures to outer-measures

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Statement

Given a pre-measure, ˉμ, on a ring of sets, R, we can define a new function, μ which is[1]:

Given by:

  • μ:HσR(R)ˉR0
    • μ:Ainf{n=1ˉμ(An)|(An)n=1RAn=1An} - here inf denotes the infimum of a set.

The statement of the theorem is that this μ 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 μ satisfies this.

For brevity we define the following shorthands:

  1. αA:={(An)n=1 | (An)n=1RAn=1An}
  2. βA:={n=1ˉμ(An) | (An)n=1αA}

Now we may define μ as:

  • μ:Ainf(βA)

Proof that μ is an extension of ˉμ

  • Let AR be given
    • In order to prove ˉμ(A)=μ(A) we need only prove [ˉμ(A)μ(A)ˉμ(A)μ(A)][Note 1]
      1. Part 1: ˉμ(A)μ(A)
        • Consider the sequence (An)n=1 given by A1:=A and Ai:= for i>1, so the sequence A,,,.
          • Clearly An=1An (as n=1An=A)
          • As such this (An)n=1αA
          • This means n=1ˉμ(An)βA (as (An)n=1αA and βA is the sum of all the pre-measures Template:WRT ˉμ of the sequences of sets in αA)
          • Recall that the infimum of a set is, among other things, a lower bound of the set. So:
            • for inf(S) (for a set, S) we see:
              • sS[inf(S)s] - this uses only the lower bound part of the infimum definition.
          • By applying this to inf(βA)(=μ(A)) we see:
            • μ(A):=inf(βA)n=1ˉμ(An)=ˉμ(A)
              • as n=1ˉμ(An)βA and inf(S) remember and
              • By definition of a (pre-)measure, μ()=0, so: n=1ˉμ(An)=ˉμ(A)+ˉμ()+ˉμ()+=ˉμ(A)
        • We have shown μ(A)ˉμ(A) as required
      2. Part 2: ˉμ(A)μ(A)

Proof that μ is σ-subadditive

  • Let (An)n=1HσR(R) be given. We want to show that μ(n=1An)n=1μ(An)
    • Let ϵ>0 (with ϵR) be given.
      • We will now define a new family of sequences. For each An we will construct the sequence (Anm)m=1R of sets such that:
        1. nN[Anm=1Anm] and
        2. nN[m=1ˉμ(Anm)μ(An)+ϵ12n]
      • Let nN be given (we will now define (Amn)m=1R)
        • Recall that μ(An):=inf(βAn)
        • Any value greater than the inf(βAn), say w, is not a lower bound so there must exist an element in βAn less that w (so w cannot be a lower bound)
          • Choose w:=inf(βAn)+ϵ2n
            • As ϵ>0 and 12n>0 we see ϵ2n>0, thus μ(An)<μ(An)+ϵ2n
        • By the definition of infimum:
          • sβAn[w>inf(βAn)s<w]
        • If sβAn then:
          • (Bn)n=1αAn such that s=n=1ˉμ(Bn).
        • As s<w=inf(βAn)+ϵ2n=μ(An)+ϵ2n and s=n=1ˉμ(Bn) we see:
          • n=1ˉμ(Bn)<μ(An)+ϵ2n
        • Caution:This doesn't show that A_n\subseteq\bigcup_{m=1}^\infty A_{nm} - don't forget!
        • Define a new sequence, ({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} to be the sequence ({ B_n })_{ n = 1 }^{ \infty } \in\alpha_{A_n} we just showed to exist
      • Since n\in\mathbb{N} was arbitrary for each A_n\in(A_k)_{k=1}^\infty\subseteq\mathcal{H}_{\sigma R}(\mathcal{R}) we now have a new sequence: ({ A_{nm} })_{ m = 1 }^{ \infty }\subseteq \mathcal{R} 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 \sum^\infty_{n=1}\frac{\epsilon}{2^n}=\epsilon\sum^\infty_{n=1}\frac{1}{2^n} and that \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots is a classic example of a geometric series, we see easily that:
          • \epsilon\sum^\infty_{n=1}\frac{1}{2^n}=1\epsilon=\epsilon thus:
      • \mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon
    • Since \epsilon>0 (with \epsilon\in\mathbb{R} was arbitrary we see:
      • \forall\epsilon>0\left[\mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon\right]
    • Recall that \left(\forall\epsilon>0[a<b+\epsilon]\right)\iff\left(a\le b\right) (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 ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}_{\sigma R}(\mathcal{R}) 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 \mu^* is \sigma-subadditive

Caveats

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

The rest

Still to do:

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

Notes

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

References

  1. Jump up to: 1.0 1.1 1.2 Measure Theory - Paul R. Halmos