Why must there be disjoint sets $A$ and $B$ such that $m^*(A \cup B) < m^*(A) + m^*(B)$? (Theorem 18, Royden) I am trying to follow Theorem 18 in Royden's Real Analysis book (fourth edition). It says the following: 
Proof
The "preceding theorem" is Vitali's theorem which says that any set with positive outer measure contains a non-measurable subset. 
I am trying to follow the proof but am not understanding why the definition of measurability (combined with the assumption that $m^*(A \cup B) = m^*(A) + m^*(B)$) implies that every set must then be measurable. I feel like I'm missing something obvious here.. because he doesn't elaborate on that fact. 
Second, why does the contradiction for assuming equality prove the theorem? Couldn't it still be the case that 
$m^*(A \cup B) > m^*(A) + m^*(B)$?
for two disjoint sets $A$ and $B$? 

If you always have $m^*(A\cup B)=m^*(A)+m^*(B)$, then, for any two disjoint subsets $A$ and $B$ of $\mathbb R$,$$m^*(A)=m^*\bigl((A\cap B)\cup(A\cap B^\complement)\bigr)=m^*(A\cap B)+m^*(A\cap B^\complement).$$So, indeed, every set $A$ is measurable.
And for the outer measure the inequality $m^*(A\cup B)\leqslant m^*(A)+m^*(B)$ always holds and it is easy to prove from the definition of $m^*$.
