If $A$ and $B$ are separated, is $m^*(A \cup B)=m^*(A)+m^*(B)$? Suppose that $A,B$ are separated sets of real numbers, that is
$$\inf \{|a-b|:a \in A,b \in B\}>0.$$
Is it then true that 
$$m^*(A \cup B)=m^*(A)+m^*(B),$$
where $m^*$ is the Lebesgue outer measure? Does this relation hold for all outer measures?
 A: Yes, it is true for $m^{*}$. Let $A$ and $B$ be such that $d(A,B)>0$. This in particular implies that $\mathrm{cl}(A)\cap B=\emptyset$, where $\mathrm{cl}(A)$ denotes the closure of $A$. As a closed set $\mathrm{cl}(A)$ is Lebesgue-measurable, so we may apply measurability of $\mathrm{cl}(A)$ and take $A\cup B$ as a test set:
\begin{align*}
m^{*}(A\cup B)&=m^{*}((A\cup B)\cap\mathrm{cl}(A))+m^{*}((A\cup B)\setminus\mathrm{cl}(A)) \\
              &=m^{*}((A\cap\mathrm{cl}(A))\cup (B\cap\mathrm{cl}(A))+m^{*}((A\setminus\mathrm{cl}(A))\cup (B\setminus\mathrm{cl}(A)) \\
              &=m^{*}(A\cup \emptyset)+m^{*}(\emptyset\cup B)\\
              &=m^{*}(A)+m^{*}(B).
\end{align*}
In fact, as you can see we only used the fact that $m^{*}$ was an outer-measure and $\mathrm{cl}(A)$ was $m^{*}$-measurable. So the same proof works for any Borel outer-measure on a metric space $(X,d)$.
As for the latter question. An outer-measure on a metric space $(X,d)$ that satisfies this condition is called a metric outer-measure. One may show that an outer measure $\mu$ is metric if and only if all Borel sets are $\mu$-measurable. One direction was proven above, and the other direction is a bit more involved. So I guess you could produce a counter-example by using any non-Borel outer-measure.
A: This is true for Lebesgue outer measure. See, for example, Lemma 5 in this blogpost by T. Tao.
If you are curious about other, particular outer measures, please ask about them specifically.
