measures 8

Given a map of measurement $m^o$ in a $\mathcal{C}$ which is a collection of sets in $\Omega$ .
$m^o: a\in\mathcal{C}\rightarrow x\in\mathbb{R}$

Suppose $m^o$  satisfies; 
  
(1)for any $a\in\mathcal{C}$ ,  $m^o(a)\geq 0$ ,

(2) $m^o(\phi)=0$ ,

(3) if $a_1\subset a_2 (\in \mathcal{C})$ , then $m^o(a_1)\leq m^o(a_2)$ , 
(4) if $a_1,a_2,\cdots\in\mathcal{C}$  ,  $a_i\cap a_j=\phi (i\ne j)$ , then $m^o(\cup a_i)\leq \sum m^o(a_i)$  ($i,j=1,2,\cdots$)  . 

We calls $m^o$  an outer measure. By using $m^o$ , a following collection $\mathcal{F}$  of sets can be defined. 

$\mathcal{F}=\left\{ a\subset\Omega | \mbox{any } e\subset\Omega, m^o(e)=m^o(e\cap a)+m^o(e\cap a^c)  \right\}\subset\mathcal{C}$ , 

where $e$  is an arbitrary set in $\Omega$  and $a^c$  is the complementary set of $a$  in $\Omega$. 

What properties $\mathcal{F}$  has?

(i)$\Omega, \phi\in \mathcal{F}$
Any collection of sets and $\Omega$  have the empty set $\phi$ .If $a$  is $\phi$ , then $a^c$  is $\Omega$ . 
Therefore, when $a$  is $\phi$ , for any $e$ , 
$m^o(e)=m^o(e\cap \phi)+m^o(e\cap \Omega)$  is always true. 

(ii)if $a\in \mathcal{F}$ , then $a^c\in\mathcal{F}$ . 
As for any set $b\in\mathcal{F}$  $(b^c)^c=b$ , $m^o(b)=m((b^c)^c)$ . 
Therefore, when $a$  is $b^c$ ,  for any $e$ , 
$m^o(e)=m^o(e\cap b^c)+m^o(e\cap (b^c)^c)$  is always true. 

(iii)if $a_1,a_2\in\mathcal{F}$ , and $a_1\cap a_2=\phi$ , then  $a_1\cup a_2\in \mathcal{F}$ . 
 (　→　We will prove this proposition in next post. )