The definition for $\mathbb P$ to be a (probability) measure on a set $(\Omega ,\mathcal F)$ one axiom is the $\sigma -$additivity, i.e. $$\mathbb P\left(\bigcup_{i=1}^\infty A_i\right)=\sum_{i=1}^\infty \mathbb P(A_i),$$ for $A_i$ disjoints.
But if we give as axiom the additivity of the measure instead of the $\sigma -$additivity, i.e. $$\mathbb P\left(A_1\cup A_2\right)=\mathbb P(A_1)+\mathbb P(A_2)$$ for $A_1,A_2$ disjoint, can't we deduce the $\sigma -$additivity ? I guess it is not true, but I can't find a counter example.
I know that for example, on $([0,1], 2^{[0,1]},m)$ and $m$ be the Lebesgue measure, then there are disjoints sets $A,B$ in $2^{[0,1]}$ s.t. $$m(A\cup B)<m(B)-m(A),$$ so since additivity doesn't hold, $m$ can't be a measure, but I can't find an application $\nu:2^{[0,1]}\to \mathbb R$ s.t. $\nu$ is additive but not $\sigma -$additive. Any example ?