The following is part of a theorem and proof in Folland's Real Analysis: Modern Techniques and Their Applications:
Let $\mu$ be a complete Lebesgue-Stieltjes measure on $\mathbb{R}$ associated to the increasing right continuous function $F$, and denote by $\mathcal{M}_{\mu}$ the domain of $\mu$.
If $E \in \mathcal{M}_{\mu}$, and if $E$ is bounded, then \begin{equation} \mu(E) = \sup \left\{ \mu(K): K \subset E \text{ and $K$ is compact} \right\} \end{equation}
Proof:
If $E$ is closed then $E$ is compact and the equality is obvious. Otherwise, given $\varepsilon>0$ we can choose an open $U$ such that $\bar{E}\setminus E \subset U$ such that $\mu(U) \leq \mu(\bar{E}\setminus E) + \epsilon$. Let $K=\bar{E} \setminus U$. Then $K$ is compact and $K \subset E$, and
\begin{align} \mu(K) &= \mu(E) - \mu (E \cap U)\\ &=\mu(E) - \left[ \mu(U) - \mu(U\setminus E) \right]\\ &\geq \mu(E) - \mu(U) + \mu(\bar{E} \setminus E)\\ &\geq \mu(E) - \epsilon\\ \end{align}
This concludes the proof.
However, I cannot understand either the first or second equality above (although I feel the answer must be obvious).
How do we get that $\mu(K) = \mu(E) - \mu (E \cap U)$?
How do we get that $\mu(K)= \mu(E) - \left[ \mu(U) - \mu(U\setminus E) \right]$?
For the first, notice that $E=K\cup(E\cap U)$ is a disjoint union.
For the second, notice that $U=(E\cap U)\cup (U\setminus E)$ is a disjoint union.
All sets are measurable, so the measure is additive.