I attached the book link here, you may check the proof on Page 47.
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{align*} \sum\limits^{\infty}_{n=1}l(I_{n}) &= \sum\limits^{\infty}_{n=1}(l(J_{n}) + l(L_{n})) + \sum\limits^{\infty}_{n=1}l(K_{n})\\ & \geq |A| + |G| \end{align*}
How can the above inequality show that $|A \cup G| \geq |A| + |G|$?
I can only deduce from $(A \cup G) \subset \bigcup^{\infty}_{n=1}I_{n}$, that:
\begin{equation} |A \cup G| \leq \left| \bigcup\limits^{\infty}_{n=1}I_{n}\right| \leq \sum\limits^{\infty}_{n=1}l(I_{n})\\ \end{equation}
However, along with: \begin{equation} \sum\limits^{\infty}_{n=1}l(I_{n}) \geq |A|+ |G| \end{equation}
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?
Recall the definition of outer measure: $$|A\cup G|=\inf\{\sum l(I_n):A\cup G\subset\cup I_n\}.$$
Then you already know for all $\{I_n\}$ covering $A\cup G$, $$\sum l(I_n)\geq|A|+|G|,$$
after taking infimum over these intervals you get the result.