Suppose $E$ measurable and let $\varepsilon>0$. Then, there is closed cube $\{Q_j\}_{j=1}^N$ such that $$m^*(E\Delta F)\leq \varepsilon$$ where $$F=\bigcup_{j=1}^NQ_j.$$ I denote $A\Delta B:= (A\setminus B)\cup (B\setminus A)$.
I tried to make a draw, but I don't see what it mean concretely. Could someone show me on a draw ? Thank you very much.
Think of inscribing a square S in a circle C. now make the square a tiny bit bigger so it is now S'. now the square will not cover the whole circle, there will be a piece p1 left over. also the circle will not cover the whole square, there will be a piece p2 left over. $S' \Delta C = p1 \cup p2$
I tried to do my best to make the draw on the computer, but it's not that easy. Your formula says that if $E$ is measurable (and of finite measure), $E$ differ by a finite union of cube by a set of measure $\varepsilon$ (in the draw, the difference between $E$ and $F$ is the set in red and the measure of this set is $\varepsilon$.)
I don't know if you can imagine how strong (and important) is this result, but to vulgarize it, it mean that $E$ is a very nice set with a margin of $\varepsilon$.
In fact, $$E\subset E\Delta F\cup F\implies m(E)\leq m(F)+m(E\Delta F)$$ and $$F\subset E\Delta F\cup E\implies m(F)\leq m(E)+m(E\Delta F),$$ therefore, if $m(E\Delta F)<\varepsilon$, $$m(E)<m(F)<m(E)+2\varepsilon$$
and thus, you can approximate every measurable set by a finite union of cube.
