Completing A proof of a question related to Radon-Nikodym 
Let $\nu$ be a absolutely continous with respect to the measure $\mu$, where both of them are $\sigma$ finite and are on $( \omega , X)$. Prove that then. $\forall$ $\epsilon>0$ $\exists$ $\delta>0$ such that $\mu(A)<\delta$ $\implies$ $\nu(A)<\epsilon$, $\forall A \in X$

I want to prove this directly (its rather easy if we go from converse). Given $\epsilon >0$, for arbitrary $A\in X$, $\nu(A)=\int_{A} fd\mu<\epsilon$ where $f$ is the Radon-Nikodym derivative. Then by using simple functions we can write this as $\sum_{k=1}^{n}a_{k}\mu(A_{k})$ where $A_{k}$'s are partition of $A$.Hence for any $k<n+1$, $a_{k}\mu(A_{k})<\epsilon$. How we can find the $\delta$ from ther? Or how we can prove this direclty, if this proof doesn't work.

You certainly don't have to use Radon-Nikodym to prove this, as the other answer pointed out. But if you want to use it, then the result follows immediately from the fact that $f$ is $\mu$-integrable. See here (and note that the proof is essentially the same as the one given by harmonicuser).
Indeed, integrability implies that for all $\epsilon > 0$, there exists $\delta > 0$ such that $\mu(A) < \delta$ implies $\int_A f d\mu < \epsilon$. Hence, $\mu(A) < \delta$ implies $\nu(A) = \int_A f d\mu < \epsilon$.
