Let $\mu, \nu$ be nonnegative Borel measures and $f, g$ their corresponding Radon-Nikodym derivatives w.r.t. Lebesgue's measure. $$ \frac{d \mu}{d \lambda} = f, \frac{d \nu}{d \lambda} = g $$
Suppose $$ \nu << \mu $$
What relationship is there then between $f$ and $g$?
I think there must be $$f(x) = 0 \implies g(x) = 0 \text{ $\lambda$ a.e.}$$ or equivalently if we set $A = \{x: g(x)>0\}, B = \{x: f(x)=0\}$ then $$ \lambda(A \cap B) = 0 $$
I know that $$ \int_{A \cap B}fd \lambda = \int f \mathbb{1}_{A \cap B}d \lambda \leq \int f \mathbb{1}_{B}d \lambda = 0 $$ because $f = 0$ for a.e. x $\in B$.
I think it is not enough to conclude that $\lambda(A \cap B\}) = 0$ though.
Could anyone finish this in the right direction?
If $\ \nu\ll\mu\ $, then $\ \nu\ $ has a Radon-Nikodym derivative, $\ \frac{d\nu}{d\mu}\ $, with respect to $\ \mu\ $, which satisfies $$ \nu(A)=\int_A \frac{d\nu}{d\mu}d\mu=\int_A \frac{d\nu}{d\mu} \frac{d\mu}{d\lambda}d\lambda= \int_A \frac{d\nu}{d\mu} fd\lambda $$ for all Borel sets $\ A\ $. Therefore, by the uniqueness of the Radon-Nikodym derivative, $\ g = \frac{d\nu}{d\mu}f\ \ \lambda$-a.e. Thus, $$ \lambda\left(\left\{x: g(x)-\frac{d\nu}{d\mu}f>0\right\}\right)=0, $$ and since $$ \ A\cap B\subseteq \left\{x: g(x)-\frac{d\nu}{d\mu}f>0\right\}\ , $$ it follows that $ \lambda(A\cap B)=0\ $.