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 $A\_i$.
Countably additive holds only If the sets $\{ E\_n \} $ are measurable and pairwise disjoint: That is. If $\{ E\_n \}\_{n\geq 1} $ are measurable, then 

$$ \mu^{\*} ( \bigcup E\_n ) = \sum \mu^\*(E\_n) $$