Let $E = \cup_{n=1}^\infty E_n$, where the $E_n$ can be taken, without loss of generality, to be pairwise disjoint and $\mu E_n < \infty $. Then by Theorem 2.14, $\mu E_n = \sup \{ \mu K | K \subset E_n, K \text{compact} \}$.
Let $\epsilon> 0$ and choose $K_n \subset E_n$ compact so that $\mu E_n \leq \mu K_n + \frac{\epsilon}{2\cdot 2^n}$. Then we have $\mu E = \sum_{n=1}^\infty \mu E_n \leq \sum_{n=1}^\infty \mu K_n+ \frac{\epsilon}{2}$. (Note that since $K_n \subset E_n$, we have $ \sum_{n=1}^\infty \mu K_n \leq \mu E$.)
Suppose $\sum_{n=1}^\infty \mu K_n = \infty$. Then let $C_N = \cup_{n=1}^N K_n \subset E$. $C_N$ is compact, $\mu C_n = \sum_{n=1}^N K_n$, and $\mu C_N \to \infty$ which gives $\mu E = \sup \{ \mu K | K \subset E, K \text{compact} \}$.
If $\sum_{n=1}^\infty \mu K_n < \infty$, we have $\sum_{n=1}^N \mu K_n \to \sum_{n=1}^\infty \mu K_n$, we can choose $N$ sufficiently large so that $\mu E \leq \sum_{n=1}^N \mu K_n+ \epsilon$. As above, let $C_N = \cup_{n=1}^N K_n \subset E$. Then $\mu E \leq \mu C_N + \epsilon$. Again, this gives $\mu E = \sup \{ \mu K | K \subset E, K \text{compact} \}$.