Exterior measure does not satisfy additivity. Exterior measure $m^*$ satisfy subadditivity.
$$m^*(A\cup B) \leq m^*(A) + m^*(B).$$
But for disjoint sets $A$ and $B$ it may be that $m^*(A\cup B) < m^*(A)+m^*(B)$.
So I learned that It is one reason measurable concepts come up.
I want such disjoint sets $A$, $B$.
I guess I may find this set with respect to nonmeasurable set $N$.

Consider a Bernstein set  $B \subset \mathbb R$.  This has the property that both $B \cap A$ and $B^c \cap A$ are nonempty for any uncountable closed subset of $\mathbb R$.  In particular, both $B$ and its complement have full outer measure: $m^*(B \cap [a,b]) + m^*(B^c \cap [a,b]) = 2 (b-a)$ for any $a < b$.
