Question: I understand that Lebesgue outer measure on $\mathbb R$ is not countably additive. But if there are two disjoint sets, does the outer measure of their union equal the sum of their outer measure? Can someone give me a counterexample?

Answer: 
Use transfinite induction to construct $2$ (or $2^{\aleph\_0}$) disjoint subsets $A\_i$ of the unit interval $[0,1]$, such that each $A\_i$ has nonempty intersection with every uncountable closed subset of $[0,1]$. (These are called [Bernstein sets](http://en.wikipedia.org/wiki/Bernstein_set).) Then each $A\_i$ has outer measure $1$, as does the union of all the $A\_i$.
For a non-measurable set, the outer measure will necessarily be positive. Therefore it suffices to consider a non-Lebesgue measurable subset $A$ of the circle, and let $B$ be its complement in the circle. Then additivity must fail for this pair of sets for the following example. 

Consider a set $A$ of representatives for equivalence classes under the action of rational rotations on the circle (the usual example of a non-measurable set). Recall that the circle is a countable union of the sets $A\_\alpha$ obtained by rotating $A$ by a rational angles $\alpha$.

Let $\epsilon=\mu(A)>0$ where $\mu$ is the outer measure. Let $N=1+\lfloor\frac{1}{\epsilon}\rfloor$. Now take $N$ rational translates of $A$, namely $A\_1,A\_2,\ldots,A\_N$. These are all disjoint by construction. If $\mu$ satisifed finite additivity, then the total outer measure would be greater than $1$:
$$
\mu(A\_1\cup\cdots\cup A\_N)=N\cdot\epsilon>1,
$$
yielding the desired contradiction. This shows that the outer measure is not even finitely additive.