measures 10

On an arbitrary space $\Omega$ , if there is an outer measure $m^o:a\subset\Omega \rightarrow \mathbb{R}$ , the collection of sets
$\mathcal{F}=\left\{a\subset\Omega|\forall e\subset\Omega,m^o(e)=m^o(e\cap a)+m^o(e\cap a^c) \right\}$
could be defined.

It has been proved that for the elements in this collection $\mathcal{F}$ ,
if $a_1,a_2\in\mathcal{F},a_1\cap a_2=\phi$ , then $(a_1\cup a_2)\in\mathcal{F}$ .

Although the proof is not easy, the number of set in the collection can be extended from 2 to $\infty$ . Namely, for $a_1,a_2,\cdots$  and $a_i\cap a_j=\phi$ , then $(\cup a_i)\in \mathcal{F}$ .

The collection $\mathcal{F}$ is called $\sigma$ algebra, and
on $\sigma$ algebra, an outer measure $m^o$  becomes a measure $m$ .

measures 9

[the preceding definition] You have to remember;
(1)$m^o$  is an outer measure,
(2)$\mathcal{F}=\left\{a\subset\Omega | \forall e\subset\Omega, m^o(e)=m^o(e\cap a)+m^o(e\cap a^c)   \right\}$

For $a_1,a_2\in\mathcal{F}$ , $a_1\cap a_2=\phi$ ,and  $\forall e\subset\Omega$ , we shall prove that
$m^o(e)=m^o(e\cap (a_1\cup a_2))+m^o(e\cap (a_1\cup a_2)^c)$ .

If $A=a_1\cup a_2$ , then $A^c=(a_1\cup a_2)^c=a_1^c\cap a_2^c$ .
As $e=(e\cap A)\cup (e\cap A^c)=(e\cap(a_1\cup a_2))\cup (e\cap(a_1\cup a_2)^c)$ ,and
$m^o$  is an outer measure,
$m^o(e)\leq m^o(e\cap(a_1\cup a_2))+m^o(e\cap(a_1\cup a_2)^c)$ .

You know that as $a_1,a_2\in\mathcal{F}$ ,
$m^o(e)= m^o(e\cap a_1)+m^o(e\cap a_1^c)$  and $m^o(e)= m^o(e\cap a_2)+m^o(e\cap a_2^c)$ .

For any $e\subset\Omega$ , as this assertion is always true, please replace $e$  to $e\cap a_1^c$ in last equation . Hereby,
$$\begin{eqnarray*} m^o(e\cap a_1^c)&=&m^o((e\cap a_1^c)\cap a_2)+m^o((e\cap a_1^c)\cap a_2^c)\\ &=&m^o(e\cap(a_1^c\cap a_2))+m^o(e\cap(a_1^c\cap a_2^c)) \\ &=& m^o(e\cap a_2)+m^o(e\cap(a_1\cup a_2)^c ) \end{eqnarray*}$$
where as $a_1\cap a_2=\phi$ , $e\cap(a_1^c\cap a_2)=e\cap a_2$ .
Therefore,
$$m^o(e)=m^o(e\cap a_1)+m^o(e\cap a_2) +m^o(e\cap(a_1\cup a_2)^c) .$$

As $m^o$  is an outer measure,
$$\begin{eqnarray*} m^o(e\cap a_1)+m(e\cap a_2)&\geq&m^o((e\cap a_1)\cup(e\cap a_2)) \\ &=&m^o(e\cap(a_1\cup a_2)) . \end{eqnarray*}$$

We get
$$m^o(e)\geq m^o(e\cap (a_1\cup a_2))+m^o(e\cap(a_1\cup a_2)^c) .$$

This assertion reverses an inequality sign on an outer masure.

Since, if $'\geq'$  and $'\leq'$  happen at same time in a equation, then just only $'='$ is available,
$$m^o(e)= m^o(e\cap(a_1\cup a_2))+m^o(e\cap(a_1\cup a_2)^c) .$$
Hence, $a_1,a_2\in\mathcal{F}\quad (a_1\cap a_2=\phi)$ , then
$$(a_1\cup a_2)\in \mathcal{F}$$