Let $A$ and $B$ be bounded subsets of $\mathbb{R}$ for which there is an $\alpha>0$ such that $|a-b|\geq \alpha$ for all $a\in A,b\in B$. Prove that $m^{*}(A\cup B)=m^*(A)+m^*(B)$.
My attempt :
By the hypothesis, we conclude that $A\cap B=\phi$ otherwise if $c\in A\cap B$, then $|c-c|=0\geq \alpha>0$ which is absurd. If $m^*(A)=0$, then $m^*(A\cup B)=m^*(B)$ and we are done. Now we assume that $m^*(A)>0$ and $m^*(B)>0$. Therefore both $A$ and $B$ are uncountable sets.
Now I don't know how to go further. I have an intuition that the following result can be used here.
Result : Let $E$ be any bounded subset of $\mathbb{R}$. Then there is a $G_{\delta}$ set $G$ for which $E\subseteq G$ and $m^*(E)=m^*(G)$.
It will be really helpful if someone give me an outline of the proof of the above result, too.
I hope it can help you.
We define $$U=\bigcup_{x\in A}\left(x-\frac{\alpha}2,x+\frac{\alpha}2\right)$$ we have $ A \subset U $ and, by the condition given in the problem $ B \cap U = \emptyset$. Since U is open, hence measurable, we have $m^*(A \cup B) = m^*((A \cup B) \cap U) + m^*((A \cup B) \cap U^c)$.
Since $(A\cup B)\cap U=A$ and $ (A\cup B)\cap U^c =B$.
=> $m^{*}(A\cup B)=m^*(A)+m^*(B)$.
proof of the result: (Problem 6)
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