Let $\mu,\lambda$ and $\nu$ be $\sigma$-finite measures on $(X,M)$ such the $\nu\ll \mu$. Let $\mu= \nu + \lambda$. Then if $f$ is the Radon-Nikodym derivate of $\nu$ wrt $\mu$, we have $0\leq f\lt 1~\mu$-a.e. where $f = \frac{d\nu}{d\mu}$.
Approach.
Suppose to the contrary that $f\geq 1$. Then since I can infer $\nu \ll \mu$, $\nu(E) = \int_E f~d\mu \geq \int_E 1 d\mu = \mu(E).$ But then $\nu(E) \leq \nu(E) + \lambda(E) = \mu(E)$. So a contradiction. Thus $f\lt 1~\mu$-a.e.
Please, does this look okay?