Show that $\mu(A) = \sup \{ \mu(K) : K \subset A, K \text{ compact}\}$ is a measure

First, $\mu$ is $\sigma$-additive iff it is continuous at $\emptyset$. Suppose that $\mu$ is not $\sigma$-additive, i.e. there exists a decreasing sequence of sets $\{A_n\}$ s.t. $\cap_{n\ge 1}A_n=\emptyset$ but $\inf_n\mu(A_n)=\epsilon>0$. Let $K_n\subset A_n$ compact s.t. $\mu(K_n)\le \mu(A_n)+\epsilon2^{-(n+1)}$. Then, letting $B_n=\cap_{k=1}^nK_n$, we get $$ \mu(A_n\setminus B_n)\le \sum_{k=1}^n\mu(A_k\setminus K_k)<\frac{\epsilon}{2}. $$ Thus, $\mu(B_n)>0$ so that $\{B_n\}$ is a decreasing sequence of nonempty compact sets. Hence, $\cap_{n\ge 1}K_n\ne \emptyset$, a contradiction.