Prove: If $m^*(E)=0$ Then $E$ Is Measurable
If $A\subset E$ then $A\cap E\subseteq E$ and by the monotonicity of the outer measure $$m^*(A\cap E)\leq m^*(E)=0$$
On the other hand $A\cap E^C\subseteq A$ then by the monotonicity of the outer measure $$m^*(A\cap E^C)\leq m^*(A)$$
Using the sufficient condition of Cartheodory: $E$ is measurable if for all $A$, $m^*(A)\geq m^*(A\cap E)+m^*(A\cap E^C)$
We have:
$m^*(A)\geq m^*(A\cap E^C)+0=m^*(A\cap E^C)+m^*(A\cap E)$
Why in the first place can we assume that $A\subset E$? shouldn't it hold for any $A$?