Notes:Halmos measure theory skeleton
From Maths
Skeleton
If we were to cut out all the "extra" (and useful) theorems into just the core that let us get from pre-measures to measures we would be left with what I call the "skeleton". The "core" of Halmos' measure theory book is the following:
- Ring of sets, [ilmath]\mathcal{R} [/ilmath] - DONE - Alec (talk) 20:31, 3 April 2016 (UTC)
- [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath]
- Measure, [ilmath]\mu[/ilmath], countably additive extended real valued set function on a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath], [ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath]
- Pre-measure, [ilmath]\bar{\mu} [/ilmath], countably additive extended real valued set function defined on a ring of sets, [ilmath]\mathcal{R} [/ilmath], [ilmath]\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath]
- Goal 1: "extend" a pre-measure, [ilmath]\bar{\mu} [/ilmath] to a measure, [ilmath]\mu[/ilmath] such that (for a ring of sets [ilmath]\mathcal{R} [/ilmath]): [ilmath]\forall A\in\mathcal{R}[\bar{\mu}(A)=\mu(A)][/ilmath]^{[Question 1]}
- Hereditary set system - a system of sets, say [ilmath]H[/ilmath], such that [ilmath]\forall A\in H\forall B\in\mathcal{P}(A)[B\in H][/ilmath]
- Hereditary [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{H} [/ilmath]^{[Question 2]}
- Outer-measure, [ilmath]\mu^*[/ilmath] - extended real valued countably subadditive monotonic set function with [ilmath]\mu^*(\emptyset)=0[/ilmath]
- Theorem: for a pre-measure, [ilmath]\bar{\mu} [/ilmath] on a ring [ilmath]\mathcal{R} [/ilmath] the function:
- [math]\mu^*:\mathcal{H}\rightarrow\bar{\mathbb{R} }_{\ge0}[/math] given by [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]
- is an outer measure
- Theorem: for a pre-measure, [ilmath]\bar{\mu} [/ilmath] on a ring [ilmath]\mathcal{R} [/ilmath] the function:
- [ilmath]\mu^*[/ilmath]-measurable sets
- Theorem: - the set of all [ilmath]\mu^*[/ilmath]-measurable sets is a [ilmath]\sigma[/ilmath]-ring and [ilmath]\mu^*[/ilmath] is a measure on this sigma-ring
- Theorem: - every set in [ilmath]\sigma_R(\mathcal{R})[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable and [ilmath]\mu^*[/ilmath] is a measure on this sigma-ring
- Theorem: - the measure induced on the sigma-ring of [ilmath]\mu^*[/ilmath]-measurable sets is the same as the outer measure induced by the outer-measure when restricted to the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]\mathcal{R} [/ilmath]^{[Question 3]}
- We haven't shown anything to do with equality of these two sigma-rings! We only show that the outer measure each induce is the same!
Questions
- ↑ Why specifically a measure? An outer-measure extends a measure to be able to measure every subset of every set in [ilmath]\mathcal{R} [/ilmath] - at the cost of it no longer being additive but instead subadditive - why do we want additivity so much? Why is subadditivity not good enough? Obviously it's a weaker property as additivity [ilmath]\implies[/ilmath] subadditivity
- ↑ Suppose [ilmath]\mathcal{H}(S)[/ilmath] is the hereditary system generated by a collection of subsets, [ilmath]S[/ilmath], and [ilmath]\sigma_R(S)[/ilmath] the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]S[/ilmath], is it true that:
- [ilmath]\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))[/ilmath]?
- [ilmath]\mathcal{H}(\sigma_R(S))[/ilmath]
- ↑ I could phrase this better
Old notes
These notes were "too long" I need to compress it into steps.
Skeleton
- Ring of sets
- Sigma-ring
- additive set function
- measure, [ilmath]\mu[/ilmath] - extended real valued, non negative, countably additive set function defined on a ring of sets
- complete measure - [ilmath]\mu[/ilmath] is complete if:
- [ilmath]\forall A\in\mathcal{R}\forall B[B\subseteq A\wedge\mu(A)=0\implies B\in\mathcal{R} ][/ilmath]
- complete measure - [ilmath]\mu[/ilmath] is complete if:
- hereditary system - a system of sets, [ilmath]\mathcal{E} [/ilmath] such that if [ilmath]E\in\mathcal{E} [/ilmath] then [ilmath]\forall F\in\mathcal{P}(E)[F\in\mathcal{E}][/ilmath]
- hereditary ring generated by
- subadditivity
- outer measure, [ilmath]\mu^*[/ilmath] (p42) - extended real valued, non-negative, monotone and countably subadditive set function on an hereditary [ilmath]\sigma[/ilmath]-ring with [ilmath]\mu^*(\emptyset)=0[/ilmath]
- Theorem: If [ilmath]\mu[/ilmath] is a measure on a ring [ilmath]\mathcal{R} [/ilmath] and if:
- [ilmath]\forall A\in\mathbf{H}(\mathcal{R})[\mu^*(A)=\text{inf}\{\sum^\infty_{n=1}\mu(A_n)\ \vert\ (A_n)_{n=1}^\infty\subseteq\mathcal{R} \wedge A\subseteq \bigcup^\infty_{n=1}A_n\}][/ilmath] then [ilmath]\mu^*[/ilmath] is an extension of [ilmath]\mu[/ilmath] to an outer measure on [ilmath]\mathbf{H}(\mathcal{R})[/ilmath]
- [ilmath]\mu^*[/ilmath] is the outer measure induced by the measure [ilmath]\mu[/ilmath]
- Theorem: If [ilmath]\mu[/ilmath] is a measure on a ring [ilmath]\mathcal{R} [/ilmath] and if:
- [ilmath]\mu^*[/ilmath]-measurable - given an outer measure [ilmath]\mu^*[/ilmath] on a hereditary [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{H} [/ilmath] a set [ilmath]A\in\mathcal{H} [/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
- [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B\cap A')][/ilmath]
- PROBLEM: How can we do complementation in a ring?^{[Solution 1]}
- Solution: Note that [ilmath]S\cap T'=S-T[/ilmath], so Halmos is really saying:
- [ilmath]A\in\mathcal{H}[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
- [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B - A)][/ilmath]
- [ilmath]A\in\mathcal{H}[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
- [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(A\cap B)+\mu^*(B\cap A')][/ilmath]
- Theorem: if [ilmath]\mu^*[/ilmath] is an outer measure on a hereditary [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{H} [/ilmath] and if [ilmath]\mathcal{S} [/ilmath] is the class of all [ilmath]\mu^*[/ilmath]-measurable sets, then [ilmath]\mathcal{S} [/ilmath] is a ring of sets
- Theorem: (p46) - [ilmath]\mathcal{S} [/ilmath] is a sigma-ring
- Theorem: - Every set of [ilmath]\mu^*=0[/ilmath] belongs to [ilmath]\mathcal{S} [/ilmath] and the set function [ilmath]\bar{\mu}:\mathcal{S}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] is a complete measure (AKA: [ilmath]\bar{\mu} [/ilmath] is the measure induced by [ilmath]\mu^*[/ilmath]).
- Theorem - Every set in [ilmath]\sigma_R(\mathcal{R})[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable (the [ilmath]\sigma[/ilmath]-ring generated by [ilmath]\mathcal{R} [/ilmath])
- Alternatively: [ilmath]\sigma_R(\mathcal{R})\subseteq\mathcal{S} [/ilmath]
- Theorem - [ilmath]\forall A\in\mathbf{H}(\mathcal{R})[\underbrace{\mu^*(A)}_{\text{Outer} }=\underbrace{\text{inf}\{\bar{\mu}(B)\ \vert\ A\subseteq B\in\mathcal{S}\} }_{\text{Induced by } \bar{\mu} }=\underbrace{\{\bar{\mu}(B)\ \vert\ A\subseteq B\in\sigma_R(\mathcal{R})\} }_{\text{Induced by } \bar{\mu} }][/ilmath]
- That is to say that:
- Outer measure induced by [ilmath]\bar{\mu} [/ilmath] on [ilmath]\sigma_R(\mathcal{R})[/ilmath] AND
- the outer measure induced by [ilmath]\bar{\mu} [/ilmath] on [ilmath]\mathcal{S} [/ilmath]
- agree with [ilmath]\mu^*[/ilmath]
- That is to say that:
- completion of a measure - p55
- Theorem: if [ilmath]\mu[/ilmath] is a [ilmath]\sigma[/ilmath]-finite measure on a [ilmath]\sigma[/ilmath]-ring [ilmath]\mathbb{R} [/ilmath] and if [ilmath]\mu^*[/ilmath] is the outer measure it induces, then:
- The completion of the extension of [ilmath]\mu[/ilmath] to [ilmath]\sigma_R(\mathcal{R})[/ilmath] is identical with [ilmath]\mu^*[/ilmath] on the class of all [ilmath]\mu^*[/ilmath]-measurable sets
Solutions
- ↑ We only have to deal with [ilmath]A\cap B'[/ilmath] which could just be Halmos abusing notation. If we take it literally ("[ilmath]B':=\{x(\in?)\vert x\notin B\}[/ilmath]") we may be able to work through this. Note that the intersection of sets is a subset of each set so [ilmath]A\cap B'\subseteq A[/ilmath] and is in fact [ilmath]=A-B[/ilmath], so what Halmos is really saying is:
- [ilmath]A\in\mathcal{H}[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if:
- [ilmath]\forall B\in\mathcal{H}[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)][/ilmath]
- [ilmath]A\in\mathcal{H}[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable if: