$m^*(A \cup B) = m^*(A) + m^*(B)$ Let $A$ and $B$ be bounded sets for which there is an $\alpha >0$ such that for all $a \in A$ and all $b \in B$, $|a-b| \geq \alpha$. Prove that  $m^*(A \cup B) = m^*(A) + m^*(B)$.
I can't use Caratheodory because it is in the next chapter. Just properties of outer measure.
 A: By general properties of outer measures, $m^\ast (A \cup B) \leq m^\ast (A) + m^\ast (B)$, so that we only have to prove the reverse estimate.
In the case $m^\ast(A\cup B) =\infty$, this is trivial, so assume $m^\ast(A \cup B) < \infty$. Let $\varepsilon > 0$ be arbitrary and choose a covering $(I_k)_k$ of $A \cup B$ such that
$$
\sum_k \ell(I_k) \leq m^\ast(A \cup B) + \varepsilon. \qquad (\dagger)
$$
Note that we can assume $I_k \cap (A \cup B) \neq \emptyset$ for each $k$ (why?).
Also, we can replace each interval $I_k = [a_k, b_k]$ by a suitable finite union
$$
I_k = \bigcup_{m=1}^{m_k} [a_{k,m}, b_{k,m}]
$$
with $b_{k,m} - a_{k,m} \leq \frac{\alpha}{2}$ and $\sum\limits_{m=1}^{m_k} (b_{k,m} - a_{k,m}) = \ell(I_k)$, without changing $(\dagger)$, i.e. we can assume $\ell(I_k) \leq \frac{\alpha}{2}$ for all $k$.
Now, let
$$
I_A := \{k \in \Bbb{N} \mid I_k \cap A \neq \emptyset\}, \\
I_B := \{k \in \Bbb{N} \mid I_k \cap B \neq \emptyset\}.
$$
Because of $I_k \cap (A\cup B) \neq \emptyset$ for each $k$, we see $\Bbb{N} = I_A \cup I_B$. Furthermore, $I_A \cap I_B = \emptyset$ follows from your assumption that $|x-y| \geq \alpha$ for all $x \in A$, $y \in B$ together with $\ell(I_k) \leq \alpha/2$ (why?).
Hence,
$$
m^\ast(A) + m^\ast(B) \leq \sum_{k \in I_A} \ell(I_k) + \sum_{k \in I_B} \ell(I_k) = \sum_k \ell(I_k) \leq m^\ast(A \cup B) + \varepsilon.
$$
As $\varepsilon > 0$ was arbitrary, this completes the proof.
