measures 4

For measuring various kinds of figures $C$  in $\Omega=([0,1]\times [0,1])\subset \mathbb{R}^2$ , infinite operations are required;
\[  \sup\cup I^l \subset C\subset \inf\cup I^k  \]
,where $I$s  are squares whose area is precisely known and measurable.

You might think this way is a matter of course.  However, it is not always promised that the infinite operations for subsets in $\Omega$  will bring the desired outcomes.
Then, the following things are put in axioms.

If all $I^j\subset \Omega$ , then $\cup I^j\subset \Omega$  $(i=1,2,\cdots)$ .

In other words,

If $I^j\in \mathcal{F}$ , then $\cup I^j\in\mathcal{F}$   $(i=1,2,\cdots)$ ,
where $\mathcal{F}$  is a family of subsets in $\Omega$ .

It indicates $\cup I^j$  is measurable and $m(\cup I^j)$  exists, including $\pm\infty$ .
 

measures 3

There is an arbitrary figure $C$ in $\Omega=( \left[0,1\right]\times\left[0,1\right] )\subset \mathbb{R}^2$ . Then, we want to measure the area of $C$ . If the set function $m$  is a measure of the area in $\Omega$ ,
\[  m : C\in\Omega\rightarrow x\in\mathbb{R}, \]
it is clear that $m(C)\leq 1$ .

However, as we want to measure more precisely any figures $C$, we keep making many various kinds of  small squares $I^k$ whose horizontal size is $c_h$  and vertical size is $c_v$ . Then,  $m(I^k)=c_h^k\times c_h^k,\quad (0\lt c_h,c_v\lt 1,k=1,2,\cdots)$ .

Covering $C$  by some $I^k$ , we can put $C\subset \cup I^k$ and covering some $I^l$  by $C$, we can put $\cup I^l\subset C$ . Hence,
\[  \cup I^l\subset C\subset \cup I^k  \]

Reducing the size of squares and increasing the number of squares,
\[  \sup\cup I^l\subset C\subset \inf\cup I^k  \]
Please note that $\sup\cup I^l$  and $\inf\cup I^k$  must be measurable.

If $l,k\rightarrow \infty$  and
\[  m(\sup\cup I^l)=m(\inf\cup I^k)=m^* , \]
then we will define $m(C)=m^*$ .

You will remember the theorem of Darboux in integral.