Notes:Outer-measures to measures

Notation on this page

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

• $\bar{\mu}$ - pre-measure
• $\mu^*$ - outer measure
• $\bar{\mu}^*$ - measure induced on $\bar{S}$
• $\bar{S}$ - the set of all outer-measurable sets, WRT $\mu^*$ (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, $\bar{\mu}:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge 0}$ to an outer-measure, $\mu^*:\mathcal{H}_{\sigma R}(\mathcal{R})$ we must then "contract" $\mu^*$ 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: $\{E_n\}_{n=1}^\infty$ 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 $E_n$ as being a term "$E_n$" that happens to be in a set.... it's iffy at best (then $E_j$ wouldn't make sense)
2. Operators: He writes (page 44, outer-measurability of a set) $A\cap E'$ - meaning complementation, but of course on a ring of sets even a hereditary system of sets - there is no complementation operator - he means $A-E$ - which is "the same thing".

Halmos

Section 11

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

[Expand]

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

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

• $\bar{\mu}^*:\bar{S}\rightarrow\bar{\mathbb{R} }_{\ge 0}$

However we know nothing about the relationship between $\bar{S}$ and $\mathcal{R}$.

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

(Halmos ends chapter 2 and section 11 here)

Section 12

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

Section 13

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

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

Measures, Integrals and Martingales

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