Given a finite measure $\mu\colon \mathcal{A}\rightarrow [0,\infty]$ on $(X,\mathcal{A})$ and a signed measure $\nu$ which admits a density $$\frac{\mathrm{d}\nu}{\mathrm{d}\mu}=f$$ with $f\colon X\rightarrow \mathbb{R^+_0}$, I want to find a condition such that $\mu\ll \nu$, meaning that $\mu$ is absolutely continuous with respect to $\nu$.
My solution/attempt:
Using the change of variables theorem which states that for every $g\colon X\rightarrow \mathbb{R}$ we have that $$\int_A g\,\mathrm{d}\nu=\int_A g f\,\mathrm{d}\mu$$
in our situation above, I make the choice $g=f$ and obtain for a $\nu$-nullset $B$ that $$\nu(B)=\int_B f^2\,\mathrm{d}\mu=0$$ which is equivalent to $f=0$ $\mu$-a.e. Now I want this nullset to be exactly $B$ so I require that $f$ satisfies $f>0$ that $\mu \ll \nu$ holds.
Is my reasoning correct and are there any weaker conditions? Note the condition does not necessarily be imposed on $f$.