Let $(X, \rho)$ be a metric space and $\mu^*$ and outer measure on $X$. Then, if every open subset of $X$ is $\mu^*$-measurable, $\mu^*$ is a metric outer measure.
I'm stuck with this problem. I'm supposed to prove that for every $A$, $B$ subsets of $X$ with $d(A, B) = \inf\{d(a,b) \mid a \in A, b \in B \} > 0$, $$\mu^*(A \cup B) = \mu^*(A) + \mu^*(B)$$ and the hypothesis says that if $U \subset X$ is open, then $$\mu^*(E) = \mu^*(E\cap U) + \mu^*(E\cap U^c)$$ for every set $E$ but I don't know how to do it. I thought of letting $E = A \cup B$ and then try to find some choice of $U$. With that choice of $E$, we have that $$ \begin{align*} \mu^*(A\cup B) &= \mu^*((A\cup B) \cap U)+\mu^*((A\cup B)\cap A^c)\\ & =\mu^*((A\cap U) \cup (B \cap U))+ \mu^*((a \cap U^c)\cap (B\cap U^c)) \end{align*}$$ So at first sight, it looks like the open set $U$ we seek for has to be of one of the following two forms, $U \subset A$ and $B \subset U^c$ or the inverse, $U \subset B$ and $A \subset U^c$. The problem is that, even though the existence of an open set $U \subset A$ is always guaranteed, I don't know why it should also satisfy the property that $B \subset U^c$ as well. My guess is that there is some property of metric spaces that I'm not aware off (or maybe I am really off the track here). Any help would be appreciated. Thank you!
For $x\in A$, let $U_x=\{y\in X:d(y,x)<\frac12 d(A,B)\}$. Clearly this set is open, so $U:=\bigcup_{x\in A}U_x$ is open. Obviously $A\subseteq U$. If $y\in U$, then there exists $x\in A$ such that $$d(x,y)<\frac12 d(A,B)\le\frac12\inf_{z\in B}d(x,z),$$ so in particular $y\notin B$. Hence $$\mu^*(A\cup B)=\mu^*\big((A\cup B)\cap U\big)+\mu^*\big((A\cup B)\cap U^c\big)=\mu^*(A)+\mu^*(B).$$