The proof is here. The questions asks us to prove
If $ E=\cup I_n$ is a countable union of pairwise disjoint intervals ,prove that $m^*(E)=\sum _{n=1}^\infty l(I_n)$.
and the interested direction is $m^{*}(E)\geq \sum_n l(I_n)$.
The proof is the following:
\begin{align*} m^*(E)&=m^*\left( \bigcup_{n} I_n \right)\\ & = m^{*} \left( \bigcup_n (a_n,b_n) \right)\\ &\geq m^{*}\left( \bigcup_{n=1}^{\infty}\left( a_n+\frac{1}{2^{n}}\varepsilon, b_n-\frac{1}{2^{n}}\varepsilon \right) \right)\\ & = \sum_{n=1}^{\infty} m^{*}\left( \left( a_{n}+\frac{1}{2^{n}}\varepsilon, b_{n}-\frac{1}{2^n}\varepsilon\right) \right)\\ & = \sum_{n=1}^{\infty}\left[l(I_n)-\frac{1}{2^{n-1}}\varepsilon\right]\\ & = \sum_{n}l(I_n)-2\varepsilon. \end{align*}
I don't understand the fourth line, $m^{*}\left( \bigcup_{n=1}^{\infty}\left( a_n+\frac{1}{2^{n}}\varepsilon, b_n-\frac{1}{2^{n}}\varepsilon \right) \right) = \sum_{n=1}^{\infty} m^{*}\left( \left( a_{n}+\frac{1}{2^{n}}\varepsilon, b_{n}-\frac{1}{2^n}\varepsilon\right) \right)$. What's the justification for this line? Isn't this something we're supposed to prove?