A problem from Real analysis-Royden regarding the finite additivity of bounded disjoint sets for the Lebesgue outer measure Let $A$ and $B$ be bounded sets for which there exists an $\alpha > 0$ s.t $|a-b|\geq \alpha $ $\forall a \in A, b\in B$. Prove that 
$$m^{*}(A\cup B)=m^{*}(A)+m^{*}(B)$$.
Where, $m^{*}$ is the  Lebesgue outer measure.
Now we already have $$m^{*}(A\cup B)\leq m^{*}(A)+m^{*}(B)$$.
But, I couldn't proceed further! Now the author says that the lebesgue outer measure fails to be Countably additive, infact it fails to finitely additive, that is,
there are disjoint set $A$ and $B$ such that
 $$m^{*}(A\cup B)< m^{*}(A)+m^{*}(B)$$.
But the condition in the question precisely says that $A$ and $B$ are disjoint. So, if the question is true, then the boundedness $A$ and $B$ is essential.
Am I true in this regard? How do I solve the problem?
Thanks in advance for any help! 

Hints: The point is that $A$ and $B$ are not only disjoint but they are well separated; so they can be covered by balls (or cubes or whatever you use to generate the Lebesgue outer measure in your favorite definition) independently. Draw a picture. That should help a lot.
