What is the difference between proving countable subaddivity and Proving that $m^*(A \cup B) < m^*(A) + m^*(B)$?

Question

The proof of countable subadditivity of outer measure is given in pg.33 in Royden fourth edition as follows:

And the proof of theorem 18,pg.49, in the same edition is given by:

And this is the preceding theorem:

My question is:

I do not understand the proof of theorem 18, I want to prove it in details, specifically:

1- I do not understand how "by the very definition of a measurable set, every set must be measurable"?

here the definition of the measurable set we use:

So, could anyone provide me with a detailed proof of this theorem, please?

2- Will the proof similar to the above given for countable subadditivity? what is the difference in the proof and why?

Answer 1

You know that there are non-measurable set (for instance, Vitali set $\mathcal V$). If $m^*(A\cup B)=m^*(A)+m^*(B)$ for all disjoints set $A$ and $B$, then, in particular, $$m^*(A)=m^*(A\cap (\mathcal V\cup\mathcal V^c))$$ $$=m^*(A\cap \mathcal V\cup A\cap \mathcal V^c)\underset{(*)}{=}m^*(A\cap \mathcal V)+m^*(A\cap \mathcal V^c),$$ where $(*)$ come from the fact that $A\cap \mathcal V$ and $A\cap \mathcal V^c$ are disjoints. Thus $\mathcal V$ is measurable. Contradiction.

Answer 2

Suppose $m^{*}(A\cup B)=m^{*}(A)+m^{*}(B)$ $\cdots$ (1) whenever $A$ and $B$ are disjoint. Take any set $C$. The equation $m^{*}(C)=m^{*}(C \cap E)+m^{*}(C \cap E^{c})$ is satisfied for any $E$: to see this take $A=C \cap E$ and $B=C \cap E^{c}$ in (1). This proves that $C$ is measurable.