Let $(\Omega, F, \mu)$ be a complete measure space, $\mu(\Omega)=1$ and $\mu$ takes values 0 or 1.Let $\nu$ positive measura, $\sigma$-finite and absolutely continuous with respect to $\mu$. Show that then $f=\frac{d\nu}{d\mu}$ is constant a.e. equal to $\nu(\Omega)$ that is a finite number.
Hint:prove that there exists a value $\alpha\in \mathbb{R}$ such that $\mu(f^{-1}(\alpha))=1$.
Hint: define $F(\alpha):=\mu(f^{-1}(-\infty,\alpha]))$ (the Radon-Nikodym derivative exists since $\nu$ is assumed $\sigma$-finite). It's an increasing function which takes its values in $\{0,1\}$. Define $\alpha_0:=\inf\{t\mid F(t)=1\}$.