Notes:The set of all mu*-measurable sets is a ring

Overview

In Measure Theory Halmos does something weird for section 11, theorem B. I have yet to "crack" what he's doing, and this is the point of this page.

Halmos' theorem:

Section 11 - Theorem B - page 46:

• If [ilmath]\mu^*:\mathcal{H}\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath] is an outer-measure on a hereditary sigma-ring, [ilmath]\mathcal{H} [/ilmath] and if [ilmath]\mathcal{S} [/ilmath] is the set of all [ilmath]\mu^*[/ilmath]-measurable sets then [ilmath]\mathcal{S} [/ilmath] is a sigma-ring.
• Furthermore, if [ilmath]A\in\mathcal{H} [/ilmath] and if [ilmath] ({ E_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{S} [/ilmath] is a sequence of pair-wise disjoint sets, and [ilmath]E:=\bigcup_{n=1}^\infty E_n[/ilmath] then:
  • [math]\mu^*(A\cap E)=\sum^\infty_{n=1}\mu^*(A\cap E_n)[/math]

Context:

• The previous theorem proved was that [ilmath]\mathcal{S} [/ilmath] is a ring of sets.
• I have proved that [ilmath]\mu^*\Big\vert_\mathcal{S} [/ilmath] - the restriction (function) of [ilmath]\mu^*[/ilmath] to [ilmath]\mathcal{S} [/ilmath] - is a pre-measure.

His proof

Step	Comment
Notice we have: [ilmath]\mu^*(A\cap(E_1\cup E_2))=\mu^*(A\cap E_1)+\mu^*(A\cap E_2)[/ilmath]	As [ilmath]A\cap(E_1\cup E_2)\subseteq A\cap E_1[/ilmath] and [ilmath]A\cap(E_2\cup E_2)\subseteq A\cap E_2[/ilmath] we see: [ilmath]\mu^*(A\cap(E_1\cup E_2))=[/ilmath] [ilmath]\mu^*(\underbrace{(A\cap(E_1\cup E_2))\cap E_1}_{=A\cap E_1})[/ilmath] [ilmath]+\mu^*(\underbrace{(A\cap(E_1\cup E_2))-E_1}_{=A\cap E_2})[/ilmath] (as [ilmath]E_1[/ilmath] is [ilmath]\mu^*[/ilmath]-measurable) (This requires that they are disjoint for the subtraction side to be true) We could just as well have used [ilmath]E_2[/ilmath] to split.
It follows by induction that: [ilmath]\mu^*\left(A\cap\bigcup_{n=1}^\infty E_n\right)=\sum^\infty_{n=1}\mu^*(A\cap E_n)[/ilmath] For every [ilmath]n\in\mathbb{N}_{\ge 0} [/ilmath]	Agreed
Define: [ilmath]F_n:=\bigcup_{i=1}^n E_i[/ilmath]
"Then it follows from theorem A" that: [ilmath]\begin{array}{rcl}\mu^*(A) & = & \mu^*(A\cap F_n)+\mu^*(A-F_n) \\ &\ge&\sum^n_{i=1}\mu^*(A\cap E_i)+\mu^*(A - E)\end{array}[/ilmath]	First note that: [ilmath]\mu^*(A\cap F_n)=\sum^n_{i=1}\mu^*(A\cap E_i)[/ilmath], this is equality. So: [ilmath]\mu^*(A)=\sum^n_{i=1}\mu^*(A\cap E_i)+\mu^*(A-F_n)[/ilmath] Next note that [ilmath]A-F_n\supseteq A-F_m[/ilmath] for [ilmath]m\ge n[/ilmath], so: [ilmath]A-E\subseteq A-F_n[/ilmath], by the subadditive property of [ilmath]\mu*[/ilmath] (of all outer-measures we see: [ilmath]\mu^*(A-E)\le\mu^*(A-F_n)[/ilmath] Thus: [ilmath]\mu^*(A)\ge\sum_{i=1}^n\mu^*(A\cap E_i)+\mu^*(A-E)[/ilmath]
"Since it is true for every [ilmath]n[/ilmath] we obtain": [ilmath]\begin{array}{rcl}\mu^*(A)&\ge&\sum_{n=1}^\infty\mu^*(A\cap E_i)+\mu^*(A-E)\\&\ge&\mu^*(A\cap E)+\mu^*(A-E)\end{array}[/ilmath]	Witchcraft