# Show that the measure is absolutely continuous w.r.t Lebesgue measure [duplicate]

Let $\mu$ be a measure on Borel-sigma algebra on $[0,1]$ which has the property that for every $f$ continuously differentiable, we have

$$\bigg| \int f' \ d\mu \bigg| \leq \left (\int_{0}^{1} f(x)^2 \ dx \right )^{1/2}$$

Then prove that :

(i) $\mu$ is absolutely continuous w.r.t Lebesgue measure on $[0,1]$.

(ii) If $g$ is the Radon-Nikodym derivative of $\mu$ w.r.t Lebesgue measure, then there exists a constant $c$ such that $$|g(x)-g(y)| \leq c |x-y|^{1/2}$$

Can anyone help me with this ? I am unable to proceed in any fruitful direction.

If I have $m(A)=0$, who do I use the above condition to show $\mu(A)=0$ ? All I have is an integral over $[0,1]$.