measures 11

If you carefully watch the proof (in 'measure 9') of
"if $a_1,a_2\in\mathcal{F},a_1\cap a_2=\phi$ , then $(a_1\cup a_2)\in\mathcal{F}$ ",
you will know it contains that an outer measure becomes a measure.

Namely,
" $m^o(e)\leq m^o(e\cap a)+m^o(e\cap a^c)$ "(means $m^o$  is an outer measure),
and
" $m^o(e)\geq m^o(e\cap a)+m^o(e\cap a^c)$ ".

Therefore, we understand that
" $m^o(e) =m^o(e\cap a)+m^o(e\cap a^c)$ "
is satisfied.

However, this proposition is true on just only the already defined collection of sets.

The collection of sets

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

is a $\sigma$ algebra and on the collection of sets, an  outer measure becomes a measure.