Given an outer measure $m$ on the power set of $\mathbb R$.
$m(A) = 0$ if $A$ is countable.
$m(A) = 1/2$ if $A$ and $A^c$ are uncountable.
$m(A) = 1$ if $A^c$ is countable.
I showed that if $A$ or $A^c$ is countable, then $A$ is $m$-measurable. But why are the rest (i.e. $A$ and $A^c$ are both uncountable) not $m$-measurable?
Assume you find an $A$, such that $A$, $A^c$ are uncountable and $A$ is measurable. Now we need to find a set $S$, which contradicts Caratheodory's criterion, i.e. $$m(s)\neq m(S\setminus A)+m(A\cap S).$$ Since $A$ is uncountable we find an $R>0$, such that $A\cap B_R(0)$ is uncountable as well. Furthermore $B_R(0)\setminus A\subset A^c$ and $$\bigcup_{r>0}B_r(0)\setminus A = A^c.$$ Hence $A\cap B_R(0)$, $(A\cap B_R(0))^c$, $B_R(0)\setminus A$, $(B_R(0)\setminus A)^c$ are all uncountable, if $R>0$ is big enough. This implies $$m(B_R(0))=\frac{1}{2},\ m(B_R(0)\setminus A)=\frac{1}{2},\ m(B_R(0)\cap A)=\frac{1}{2},$$ which contradicts Caratheodory.