let $v$ be a signed measure on $(X,\Sigma)$. let $A\in \Sigma$ such that $v(A)>0$. show that there is a $v$-positive set $B\subset A$ such that $v(A)\le v(B)$.
I believe this statement can be derived from Hahn's decomposition theorem. I couldn't find it on the site so I don't know if my proof is correct.
Proof: Let $P$ be $v$-positive and $N$ be $v$-negative so that $X=P\cup N$. Then $B=P\cap A \subset P$ so it is $v$-positive (as a subset of $v$-positive set). $A\cap N \subset N$ so $v(A\cap N) <0$. now:
$v(A)=v(A\cap P)+v(A\cap N)=v(B) + v(A\cap N)\le v(B)$.
my only question is, why do I need the fact that $v(A)>0$?
You don't need $\nu (A) >0$ However, the case $\nu (A) \leq 0$ is uninteresting: we can take $B=\emptyset$ in this case.