measures 12

You may think it is strange that an outer measure becomes a measure.

An outer measure is a measure of a certain set to cover its area fully.

This means to allow measurement of a given set by approximating it from the outside of its perimeter.

Namely, an outer measure contains all elements of the set and some elements outside of the set.

However, an inner measure contains just only the elements in the set although it is not able to cover fully.

$\cup I^k\subset C\subset \cup I^l$ .

The theorem of Darboux needs that an outer measure will be consistent with an inner measure.

$m(\cup I^k)\leq m(C)\leq m(\cup I^l)$  .

How do we understand an inner measure?

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.