Prove that in $\mathbb{R}$, if $|a-b|>\alpha$ for all $a\in A$ and $b\in B$, then outer measure $m^*(A\cup B)=m^*(A)+ m^*(B)$

Prove that for sets $A,B$ bounded in $\mathbb{R}$:

If there exists $\alpha > 0$ such that $|a-b|>\alpha$ for all $a\in A$ and $b\in B$, then outer measure $m^*(A\cup B)=m^*(A)+m^*(B)$.

This comes out of section 2.2 of Royden's Real Analysis. I'm really having trouble with this one for some reason. The only theorem that I can see that might be of some help is that outer measure is preserved under set translation. But I would have to translate each point of one of these sets a different amount, so that seems hopeless.

Because I have so few theorems to work with my hunch is that I need to go back to the very definition of outer measure and do something clever with it, but so far I haven't had any luck. Can anyone help me? Thanks.

Solution 1:

HINT: Let $$U=\bigcup_{a\in A}\left(a-\frac{\alpha}2,a+\frac{\alpha}2\right)$$ and $$V=\bigcup_{b\in B}\left(b-\frac{\alpha}2,b+\frac{\alpha}2\right)\;.$$ Suppose that $x\in U\cap V$; then there are $a\in A$ and $b\in B$ such that $$|x-a|,|x-b|<\frac{\alpha}2\;.$$ Is this actually possible?