Let $A$ and $B$ be bounded sets for which there is an $c > 0$ such that $|a – b| ≥ c$ for all $a \in A$, $b \in B$.
Prove that $m^*(A \cup B) = m^*(A) + m^*(B)$.
I saw this question in Royden 4th edition, from $c$ condition's we know that $A$ and $B$ are disjoint.
$A$ and $B$ are subsets of the real number, and $m^*$ is the Lebesgue outer measure.
Best Answer
Not only are $A$ and $B$ disjoint but we can find open sets $U \supset A$ and $V \supset B$ such that $U$ and $V$ are disjoint (every point in $A$ has a open neighbourhood which does not intersect $B$ and vice-versa).
Now take a tight enough open cover of $A \cup B$ and split it by $U$ and $V$.