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

Context:

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

His proof

Step	Comment
Notice we have: $\mu^*(A\cap(E_1\cup E_2))=\mu^*(A\cap E_1)+\mu^*(A\cap E_2)$	As $A\cap(E_1\cup E_2)\subseteq A\cap E_1$ and $A\cap(E_2\cup E_2)\subseteq A\cap E_2$ we see: $\mu^*(A\cap(E_1\cup E_2))=$ $\mu^*(\underbrace{(A\cap(E_1\cup E_2))\cap E_1}_{=A\cap E_1})$ $+\mu^*(\underbrace{(A\cap(E_1\cup E_2))-E_1}_{=A\cap E_2})$ (as $E_1$ is $\mu^*$-measurable) (This requires that they are disjoint for the subtraction side to be true) We could just as well have used $E_2$ to split.
It follows by induction that: $\mu^*\left(A\cap\bigcup_{n=1}^\infty E_n\right)=\sum^\infty_{n=1}\mu^*(A\cap E_n)$ For every $n\in\mathbb{N}_{\ge 0}$	Agreed
Define: $F_n:=\bigcup_{i=1}^n E_i$
"Then it follows from theorem A" that: $\begin{array}{rcl}\mu^*(A) & = & \mu^*(A\cap F_n)+\mu^*(A-F_n) \\ &\ge&\sum^n_{i=1}\mu^*(A\cap E_i)+\mu^*(A - E)\end{array}$	First note that: $\mu^*(A\cap F_n)=\sum^n_{i=1}\mu^*(A\cap E_i)$, this is equality. So: $\mu^*(A)=\sum^n_{i=1}\mu^*(A\cap E_i)+\mu^*(A-F_n)$ Next note that $A-F_n\supseteq A-F_m$ for $m\ge n$, so: $A-E\subseteq A-F_n$, by the subadditive property of $\mu*$ (of all outer-measures we see: $\mu^*(A-E)\le\mu^*(A-F_n)$ Thus: $\mu^*(A)\ge\sum_{i=1}^n\mu^*(A\cap E_i)+\mu^*(A-E)$
"Since it is true for every $n$ we obtain": $\begin{array}{rcl}\mu^*(A)&\ge&\sum_{n=1}^\infty\mu^*(A\cap E_i)+\mu^*(A-E)\\&\ge&\mu^*(A\cap E)+\mu^*(A-E)\end{array}$	Unsure: This suggests $\sum_{n=1}^\infty\mu^*(A\cap E_i)\ge\mu^*(A\cap E)$, I don't see where this has come from I'm uncomfortable jumping from a finite sum to an infinite (See details below)

Details

I am completely happy with:

• $\mu^*(A\cap\bigcup_{i=1}^nA_i)=\sum^n_{i=1}\mu^*(A\cap E_i)$
• $\mu^*(A-E)\le\mu^*(A-\bigcup^n_{i=1}A_i)$
• $\mu^*(A)=\mu^*(A\cap F_n)+\mu^*(A-F_n)$ (this is okay because $\mathcal{S}$ is a ring, thus the finite union, $F_n\in\mathcal{S}$.)
  • This is: $\mu^*(A)=\mu^*(A\cap\bigcup_{i=1}^nE_i)+\mu^*(A-\bigcup_{i=1}^nE_i)$ by definition of $F_n$.

Combining these we see:

• $\begin{array}{rcl}\mu^*(A)&=&\mu^*(A\cap F_n)+\mu^*(A-F_n)\\&\ge&\mu^*(A\cap F_n)+\mu^*(A-E)\\&=&\sum^n_{i=1}\mu^*(A\cap E_i)+\mu^*(A-E)\end{array}$

I've also consulted Measures Integrals and Martingales, it has the same thing in it. Both books say (and require):

• $$\sum^\infty_{n=1}\mu^*(A\cap E_n)\ge\mu^*(A\cap E)$$

I could believe $\forall m\in\mathbb{N}\left[\sum^m_{i=1}\mu^*(A\cap E_n)\le\mu^*(A\cap E)\right]$ (by monotonicity of the outer-measure) but I cannot justify this.

I am also nervous about: $\forall m\in\mathbb{N}\left[\mu^*(A)\ge\sum^m_{i=1}\mu^*(A\cap E_i)+\mu^*(A-E)\right]\implies\mu^*(A)\ge\sum^\infty_{n=1}\mu^*(A\cap E_n)+\mu^*(A-E)$ however:

• The sequence $({ \text{(no content) } })_{ n = 1 }^{ \infty }$ is monotonically increasing, and it is bounded above (if $\mu^*(A)$ is finite), thus is finite, and there's a similar case to be made in the infinite case. I don't know how to jump from the finite to the infinite case.