In the proof of Theorem 1.23 in Real Analysis by Bruckner it is used the following without a proof
Let $G$ be an open subset of an interval $[a, b]$ and write $K =[a, b]\backslash G$. Then $m^{*}{(K)}=b − a- m^{*}{(G)}$.
If it were measure, yes $G$ and $K$ are disjoint thus $m(K)=b − a- m(G)$ holds but how to prove the equality for outer measure that is defined by $m^{*}{(E)}=\inf{\{ \sum_{i=1}^{\infty} |I_i| \ :\ E\subset\cup_{i=1}^{\infty}I_i }\}$. I even know $G$ can be written as a countable union of open intervals and $K$ is a compact set but those don't help too. A detailed explanation would be much appreciated.
Added: Actually $m^{*}([a, b]\backslash G)=b − a- m_{*}{(G)}$ holds for every set $G$ and the claim to be proved is $m^{*}{(G)}=m_{*}{(G)}$. The book hasn't introduced much things yet!
It suffices to prove that, if $G$ is an open subset of $[a, b]$, then we have $m_*(G) = m^*(G)$, where $m_*(G)$ is defined as $\sup\{m^*(K): K \subseteq G \text{ compact}\}$.
We write $G$ as a countable disjoint union of open intervals $G = \bigcup_{i = 1}^\infty I_i$. I assume that you know that $m^*(G) = \sum_{i = 1}^\infty |I_i|$.
Since $m_*(G) \leq m^*(G)$ always holds, it suffices to show the other direction.
In the following, we assume that $m^*(G) < \infty$. The case $m^*(G) = \infty$ can be treated similarly.
For a given $\epsilon > 0$, we choose a finite subset $S \subseteq \{1, 2, \dots\}$ such that $m^*(G) - \sum_{i \in S} |I_i| < \frac \epsilon 2$.
Without loss of generality, we may assume that $S$ is the set $\{1, \dots, n\}$. We write $I_i = (a_i, b_i)$.
We then construct a compact set $K = \bigcup_{i = 1}^n [a_i + \frac \epsilon{4n}, b_i - \frac \epsilon{4n}]$. Since $K$ contains the disjoint union $\bigcup_{i = 1}^n (a_i + \frac \epsilon{4n}, b_i - \frac \epsilon{4n})$, we have $$m^*(K) \geq m^*(\bigcup_{i = 1}^n (a_i + \frac \epsilon{4n}, b_i - \frac \epsilon{4n})) = \sum_{i = 1}^n (|I_i| - \frac \epsilon {2n}) = (\sum_{i = 1}^n |I_i|) - \frac \epsilon 2 > m^*(G) - \epsilon. $$
Therefore we have shown that $m_*(G) > m^*(G) - \epsilon$ for any $\epsilon > 0$.