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?
 A: 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 $A_i$.
A: 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,
$$
  \mu^*(\{a,b\}) = 1
  \neq
  2 = \mu^*(\{a\}) + \mu^*(\{b\}).
$$
A: 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 
$$ \mu^{*} ( \bigcup E_n ) = \sum \mu^*(E_n)  $$
A: 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.
