Assume you have a (Borel) probability measure $\mu$ on $[0,1]$ with full support, $\mathrm{supp}(\mu)=[0,1]$, which is absolutely continuous with respect to Lebesgue, i.e. there exists a measurable map $f:[0,1]\to\mathbb R$ such that $d\mu=f\, dx$.
Is is true that the set $\{f=0\}$ has zero Lebesgue measure?
It seems to be true but I don't manage to prove it. Somehow the question's spirit is "what can you say on the support $\{f\neq 0\}$ of $f$ from the support of $\mu$?"
It is well known that there exists a measurable set $E$ in $(0,1)$ such that $0<m(E\cap I)<m(I)$ for every open interval $I$. If $\mu(A)=m(A\cap E)$ then $\mu (I) =m(I\cap E)>0$ for every open interval $I$ so $\mu$ has full support. But $\frac {d\mu} {dm} =I_E$ which vanishes on $E^{c}$ and $m(E^{c}) >0$ because $m(E\cap (0,1))<m((0,1))=1$.