Extending pre-measures to outer-measures
From Maths
- Caution:This page is currently being written and is not ready for being used as a reference, it's a notes quality page
Contents
[hide]Statement
Given a pre-measure, ˉμ, on a ring of sets, R, we can define a new function, μ∗ which is[1]:
- an extension of ˉμ and
- an outer-measure (on the hereditary σ-ring generated by R, written HσR(R))
Given by:
- μ∗:HσR(R)→ˉR≥0
- μ∗:A↦inf{∞∑n=1ˉμ(An)|(An)∞n=1⊆R∧A⊆∞⋃n=1An} - here inf denotes the infimum of a set.
The statement of the theorem is that this μ∗ is indeed an outer-measure
Proof
[Expand]
Proof notes
[Expand]
Recall the definition of an outer-measure, we must show μ∗ satisfies this.
For brevity we define the following shorthands:
- αA:={(An)∞n=1 | (An)∞n=1⊆R∧A⊆∞⋃n=1An}
- βA:={∞∑n=1ˉμ(An) | (An)∞n=1∈αA}
Now we may define μ∗ as:
- μ∗:A↦inf(βA)
Proof that μ∗ is an extension of ˉμ
- Let A∈R be given
- In order to prove ˉμ(A)=μ∗(A) we need only prove [ˉμ(A)≥μ∗(A)∧ˉμ(A)≤μ∗(A)][Note 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 A⊆⋃∞n=1An (as ⋃∞n=1An=A)
- As such this (An)∞n=1∈αA
- This means ∑∞n=1ˉμ(An)∈βA (as (An)∞n=1∈αA 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:
- ∀s∈S[inf(S)≤s] - this uses only the lower bound part of the infimum definition.
- for inf(S) (for a set, S) we see:
- By applying this to inf(βA)(=μ∗(A)) we see:
- We have shown μ∗(A)≤ˉμ(A) as required
- Consider the sequence (An)∞n=1 given by A1:=A and Ai:=∅ for i>1, so the sequence A,∅,∅,….
- Part 2: ˉμ(A)≤μ∗(A)
- SEE NOTEPAD. Define γA:={ˉμ(A)}, 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 ∀x∈βA∃y∈γA[y≤x] - we may now pass to the infimum.
- Part 1: ˉμ(A)≥μ∗(A)
- In order to prove ˉμ(A)=μ∗(A) we need only prove [ˉμ(A)≥μ∗(A)∧ˉμ(A)≤μ∗(A)][Note 1]
Proof that μ∗ is σ-subadditive
- Let (An)∞n=1⊆HσR(R) be given. We want to show that μ∗(⋃∞n=1An)≤∑∞n=1μ∗(An)
- Let ϵ>0 (with ϵ∈R) be given.
- We will now define a new family of sequences. For each An we will construct the sequence (Anm)∞m=1⊆R of sets such that:
- ∀n∈N[An⊆⋃∞m=1Anm] and
- ∀n∈N[∑∞m=1ˉμ(Anm)≤μ∗(An)+ϵ12n]
- Let n∈N be given (we will now define (Amn)∞m=1⊆R)
- 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
- Choose w:=inf(βAn)+ϵ2n
- By the definition of infimum:
- ∃s∈βAn[w>inf(βAn)⟹s<w]
- If s∈βAn then:
- ∃(Bn)∞n=1∈αAn 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:
- 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:
- \mu^*\left(\bigcup_{n=1}^\infty A_n\right)<\sum^\infty_{n=1}\mu^*(A_n)+\epsilon
- We will now define a new family of sequences. For each An we will construct the sequence (Anm)∞m=1⊆R of sets such that:
- 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)
- Let ϵ>0 (with ϵ∈R) be given.
- 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
- 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})
- 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:
- \mu^* being monotonic with respect to set inclusion and the usual ordering on the reals.
- \mu^*(\emptyset)=0 - this can come from the extension part as \bar{\mu} has this property already
Notes
- Jump up ↑ This is called the trichotomy rule or something, I should link to the relevant part of a partial order here
References
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