Notes:Outer-measures to measures

Notation on this page

I made a bad choice using [ilmath]\bar{\mu}^*[/ilmath] as the measure induced by an outer measure on the class of all outer-measurable sets. So on this page:

• [ilmath]\bar{\mu} [/ilmath] - pre-measure
• [ilmath]\mu^*[/ilmath] - outer measure
• [ilmath]\bar{\mu}^* [/ilmath] - measure induced on [ilmath]\bar{S} [/ilmath]
• [ilmath]\bar{S} [/ilmath] - the set of all outer-measurable sets, WRT [ilmath]\mu^*[/ilmath] (for the definition of outer-measurability see top of Halmos' section 11 (in this document or in book))

Problem

Once we have extended a pre-measure, [ilmath]\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] to an outer-measure, [ilmath]\mu^*:\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] we must then "contract" [ilmath]\mu^*[/ilmath] to a normal measure (which is defined on a sigma-ring.

Halmos introduces a chain of theorems and does this rather indirectly, I hope to "distil" it on this page.

Halmos also does a lot of "bad" things, like:

1. Sequences: [ilmath]\{E_n\}_{n=1}^\infty[/ilmath] for sequences! It's not a set, it is an ordered set! There is no "nth term" operator of a set, and if he defines [ilmath]E_n[/ilmath] as being a term "[ilmath]E_n[/ilmath]" that happens to be in a set.... it's iffy at best (then [ilmath]E_j[/ilmath] wouldn't make sense)
2. Operators: He writes (page 44, outer-measurability of a set) [ilmath]A\cap E'[/ilmath] - meaning complementation, but of course on a ring of sets even a hereditary system of sets - there is no complementation operator - he means [ilmath]A-E[/ilmath] - which is "the same thing".

Halmos

Section 11

• Outer-measurability - for outer-measure [ilmath]\mu^*[/ilmath] on hereditary sigma-ring, [ilmath]H[/ilmath] a set [ilmath]A\in H[/ilmath] is outer-measurable WRT [ilmath]\mu^*[/ilmath] if:
  • [ilmath]\forall B\in H[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B\cap A')][/ilmath] - see what I mean by abuse? [ilmath]A'[/ilmath] isn't defined on a ring.
    • He means: [ilmath]A\in H[/ilmath] is outer measurable if: [ilmath]\forall B\in H[\mu^*(B)=\mu^*(B\cap A)+\mu^*(B-A)][/ilmath]
• Theorem A: if [ilmath]\bar{S} [/ilmath] is the class of all outer-measurable sets then [ilmath]\bar{S} [/ilmath] is a ring of sets
• Theorem B: [ilmath]\bar{S} [/ilmath] is a [ilmath]\sigma[/ilmath]-ring
• Theorem C: (two parter)
  1. Theorem C (i): Every set of outer-measure [ilmath]0[/ilmath] belongs to [ilmath]\bar{S} [/ilmath]
  2. Theorem C (ii): The set function [ilmath]\bar{\mu}^*:\bar{S}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] defined by [ilmath]\bar{\mu}^*:A\mapsto\mu^*(A)[/ilmath] is a complete measure on [ilmath]\bar{S} [/ilmath]

Why are we not done here? Short answer: we still know nothing about [ilmath]\bar{S} [/ilmath] and the ring [ilmath]\mathcal{R} [/ilmath] we started with

We started with a pre-measure, [ilmath]\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] and we have now got some measure:

• [ilmath]\bar{\mu}^*:\bar{S}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath]

However we know nothing about the relationship between [ilmath]\bar{S} [/ilmath] and [ilmath]\mathcal{R} [/ilmath].

Our goal is to construct a measure on [ilmath]\sigma(\mathcal{R})[/ilmath] - the sigma-ring generated by the ring [ilmath]\mathcal{R} [/ilmath].

(Halmos ends chapter 2 and section 11 here)

Section 12

• Theorem A: Every set in [ilmath]\sigma(\mathcal{R})[/ilmath] is outer-measurable WRT [ilmath]\mu^*[/ilmath]. This tells us [ilmath]\sigma(\mathcal{R})\subseteq\bar{S} [/ilmath].
  • We still do not know that [ilmath]\bar{\mu}^*[/ilmath] restricted to [ilmath]\sigma(\mathcal{R})[/ilmath] would even be a measure.
• Theorem B: If [ilmath]A\in\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] then: [ilmath]\begin{array}{rcl}&&\\ \mu^*(A) & = & \text{inf}\{\bar{\mu}^*(B)\ \vert\ A\subseteq B\wedge B\in\bar{S}\} \\ & = & \text{inf}\{\bar{\mu}^*(F)\ \vert\ E\subseteq F\wedge F\in \sigma(\mathcal{R})\}\end{array}[/ilmath] Caution:3 line array looks pretty bad - TODO - this better
  • This is equivalent to:
    • The outer measure induced by [ilmath]\bar{\mu}^*[/ilmath] on [ilmath]\sigma(\mathcal{R})[/ilmath] and the outer measure induced by [ilmath]\bar{\mu}^*[/ilmath] on [ilmath]\bar{S} [/ilmath] agree.
      • Warning:Not quite sure I agree! This doesn't look exactly like an outer measure now does it?
• Measurable cover: if [ilmath]E\in\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] and [ilmath]F\in\sigma(\mathcal{R})[/ilmath] - we shall say that [ilmath]F[/ilmath] is a measurable cover of [ilmath]E[/ilmath] if:
  • [ilmath]E\subseteq F[/ilmath] and if [ilmath]\forall G\in\sigma(\mathcal{R})[G\subseteq F-E\implies \bar{\mu}^*(G)=0][/ilmath] Caution:What does this actually mean?
• Theorem C: If a set [ilmath]E\in\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] has [ilmath]\sigma[/ilmath]-finite outer measure then:
  • [ilmath]\exists F\in\sigma(\mathcal{R})[\mu^*(E)=\bar{\mu}^*(F)\wedge F\text{ is a measurable cover of }E][/ilmath] Could be a good time to try some formal logic substitution!
• Theorem D: (two parter)
  1. • Theorem D (i): If [ilmath]E\in\mathcal{H}_{\sigma R}(\mathcal{R})[/ilmath] and [ilmath]F[/ilmath] is a measurable cover of [ilmath]E[/ilmath] then:
       • [ilmath]\mu^*(E)=\bar{\mu}^*(F)[/ilmath].
     • Theorem D (ii): additionally, if both [ilmath]F_1[/ilmath] and [ilmath]F_2[/ilmath] are measurable covers of [ilmath]E[/ilmath] then:
       • [ilmath]\bar{\mu}^*(F_1\triangle F_2)=0[/ilmath]
• Theorem E: If [ilmath]\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] is [ilmath]\sigma[/ilmath]-finite then so are the measures [ilmath]\bar{\mu}^*[/ilmath] on [ilmath]\sigma(\mathcal{R})[/ilmath] and [ilmath]\bar{S} [/ilmath].

Section 13

• Theorem A: If [ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0} [/ilmath] is a [ilmath]\sigma[/ilmath]-finite pre-measure then there is a unique measure on [ilmath]\bar{\mu}^* [/ilmath] (defined as above) on [ilmath]\sigma(\mathcal{R})[/ilmath] such that for [ilmath]E\in\mathcal{R} [/ilmath] we have [ilmath]\bar{\mu}^*(E)=\mu(E)[/ilmath] - the measure [ilmath]\bar{\mu}^*[/ilmath] is [ilmath]\sigma[/ilmath]-finite

I've gone as far as page 55 (from 44) and still not hit "and here is a measure on [ilmath]\sigma(\mathcal{R})[/ilmath] - stopping for now.

Measures, Integrals and Martingales

Not much better, Schilling doesn't do the "pre-measure [ilmath]\rightarrow[/ilmath] outer-measure [ilmath]\rightarrow[/ilmath] measure" that'd I'd hoped for.