Let $\mu: \mathcal P(\mathbb R^n) \to [0, \infty]$ be an outer Radon-measure. This means that every Borel-set $B \subset \mathbb R^n$ is $\mu$-measurable, $\mu$ is Borel-regular, i.e. for every set $A \subset \mathbb R^n$ there exists a Borel-set $B \supset A$ with $\mu(A) = \mu(B)$, and $\mu(K) < \infty$ for every compact set $K \subset \mathbb R^n$.
A proof in my book states now that for each $\mu$-measurable set $A \subset \mathbb R^n$ there exists a Borel-set $B \supset A$, such that $\mu(B \backslash A) = 0$. I managed to understand this for the case where $\mu(A) < \infty$. By the Borel-regularity of $\mu$ we can find a Borel-set $B \supset A$ with $\mu(A) = \mu(B)$ and the $\mu$-measurability of $A$ gives us $$\mu(B) = \mu(B \cap A) + \mu(B \backslash A) = \mu(A) + \mu(B \backslash A) \; ,$$ i.e. $$\mu(B \backslash A) = 0 \; .$$ But why is this statement true in the case where $\mu(A) = \infty$? I tried to use that $$\mu(A) = \sup \{ \mu(K) \; | \; K \subset \mathbb R^n \text{compact}, \, K \subset A \} \; ,$$ but I could not find a solution...
Thanks, I think I've found a solution. Let $K_i := \overline {\mathbb B}(0, i) \subset \mathbb R^n$ for $i \in \mathbb N^\times$. Then each $K_i$ is compact. Let $A_i := A \cap K_i$ for $i \in \mathbb N^\times$. We have $\mu(A_i) \leq \mu(K_i) < \infty$ for $i \in \mathbb N^\times$, each $A_i$ is $\mu$-measurable and $A = \bigcup_{i=1}^\infty A_i$. Now for each $A_i$ there exists a Borel set $B_i \supset A_i$, such that $\mu(B_i \backslash A_i) = 0$. Let $B := \bigcup_{i=1}^\infty B_i$. $B$ is a Borel set and $B \supset A$. It follows, that $$\mu(B \backslash A) = \mu \left( \left[ \bigcup_{i=1}^\infty B_i \right] \big\backslash \left[ \bigcup_{i=1}^\infty A_i \right] \right) \leq \mu\left( \bigcup_{i=1}^\infty \left[ B_i \backslash A_i \right] \right) \leq \sum_{i=1}^\infty \mu(B_i \backslash A_i) = 0 \; .$$
Is this ok?