Is the outer measure of $A\cup B$ equal to the sum of their outer measures if $A\cap B=\varnothing$? 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?
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.) 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 

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