Additivity of outer measure if one of the sets is open.Suppose A and G are...

Additivity of outer measure if one of the sets is open.
Suppose $A$ and $G$ are disjoint subsets of $\mathbb{R}$ and $G$ is open. Then $|A\cup G|=|A|+|G|$.
*Note that: |$A$| and l($I$) denote the outer measure of $A$ and the length of interval $I$ respectively.
I wonder that, for the part of proof to the case $G=(a,b)$ and the deduced inequality:
$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}l(I_n)& =\sum _{n=1}^{\mathrm{\infty }}(l(J_n)+l(L_n))+\sum _{n=1}^{\mathrm{\infty }}l(K_n)\\ & \ge |A|+|G|\end{array}$
How can the above inequality show that $|A\cup G|\ge |A|+|G|$?
I can only deduce from $(A\cup G)\subset \bigcup _{n=1}^{\mathrm{\infty }}{I}_n$, that:
$|A\cup G|\le \left|\bigcup _{n=1}^{\mathrm{\infty }}{I}_n\right|\le \sum _{n=1}^{\mathrm{\infty }}l(I_n)$
However, along with:
$\sum _{n=1}^{\mathrm{\infty }}l(I_n)\ge |A|+|G|$
then both the inequalities are on the same side, so I cannot link them to prove what I desire. How did author reach his conclusion?