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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.

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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 $\nu(A) = \int_A f d\mu < \epsilon$.

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I don't think one needs the Radon-Nikodym derivative here. One can prove this directly as follows:

Assume the contrary, then there exists a $\varepsilon > 0$ such that for all $\delta > 0$, there exists a measurable set $A$ such that
$\mu(A) < \delta$, but $\nu(A) > \varepsilon$. Thus for each $k \geq 1$, there exists a measurable set $A_{k}$ such that $\mu(A_{k}) < 2^{-k}$ and $\nu(A_{k}) > \varepsilon$. Consider $B = \cap_{n = 1}^{\infty}\cup_{k = n}^{\infty}A_{k}$, and observe that $\mu(B) = 0$, but $\nu(B) > \varepsilon$- a contradiction.

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