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 $N$.

Take the famous Vitali Set $\mathcal V$. Then, if you set $F=[0,1]\backslash \mathcal N$, you'll get $$m^*(F\cup\mathcal V)\neq m^*(\mathcal V)+m^*(F).$$
To prove it, I give you a hint : If $m^*(F)<1$, there is an open set $O$ s.t. $m^*(O)<1$.
