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 $A\_i$.
Take $\Omega = \{a, b\}$. Define
$$
 \mu^\*(A)
 =
 \left\{
 \begin{array}{ll}
 0, & A = \emptyset
 \\
 1, & A \neq \emptyset
 \end{array}
 \right.
$$
This is an outer measure.

In this case,
$$
 \mu^\*(\{a,b\}) = 1
 \neq
 2 = \mu^\*(\{a\}) + \mu^\*(\{b\}).
$$