The question is as follows :
Show that for any set $A$ and $\epsilon > 0$, there is an open set $O$ containing $A$ and such that $m^*(O) \le m^*(A)+ \epsilon$.
Solution given in the book is as follows :
Choose a sequence of intervals $I_n$ such that $A \subseteq \bigcup_{n=1}^\infty I_n$ and
$\sum_{n=0}^\infty l(I_n)-\epsilon/2 \le m^*(A)$.
If $I_n$=$[a_n,b_n)$, let $I_n' = (a-\epsilon/2^{n+1},b_n)$, so that ,
$A \subseteq \bigcup_{n=1}^\infty I_n'$.
Hence if $O =\bigcup_{n=1}^\infty I_n'$, $O$ is an open set and
$m^*(O) \le \sum_{n=0}^\infty l(I_n') = \sum_{n=0}^\infty l(I_n)+\epsilon/2 \le m^*(A)+\epsilon$
There are a few things that i didn't quite follow in the solution given :
First, why has $I_n$ been chosen as $[a_n,b_n)$ particularly ? Can't i just let $I_n$ to be a sequence of open intervals and choose $O$ as the union of all $I_n$'s ?
Second,how is $I_n'$ being formulated ? why particularly $a-\epsilon/2^{n+1}$ ? How would one get to that ?