Why must there be disjoint sets $A$ and $B$ such that $m^*(A \cup B) < m^*(A) + m^*(B)$? (Theorem 18, Royden)

Question

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 $B$?

Answer 1

The first part sounds like he's just using Caratheodory's criterion for measurability.

For the second part, $m^*$ needs to be subadditive, so we always have $$m^*(A\cup B) \leq m^*(A) + m^*(B).$$ For instance, with the standard outer measure, you can take the union of covers of A and B to give a cover of the union (the standard proof of its subadditivity)

Answer 2

Any outer measure (like $m^*$) must satisfy countable subadditivity for all sets, in particular for $A$ and $B$, which is why $m^*(A\cup B)>m^*(A)+m^*(B)$ cannot occur. Regarding your first question, the equality for all sets means that the outer measure $m^*$ actually satisfies all the conditions of a measure, and moreover it inherits from the outer measure the property that it is defined for all sets. So combining the "best" of both worlds, it is a measure defined for all sets, which is a contradiction.

Answer 3

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 $m^*$.