Suppose that $\nu$ is a $\sigma$-finite positive measure, and that $\rho$ is a measurable function that's nonzero $\nu$-a.e. Define $\mu(A) = \int_{A} \rho\, d\nu$ for all $\nu$-measurable $A$, so $\mu$ is a measure on the same measurable sets with $\rho$ its Radon-Nikodym derivative with respect to $\nu$.

By construction, $\mu$ is absolutely continuous with respect to $\nu$.

Given that $\rho$ is nonzero $\nu$-a.e., is $\nu$ absolutely continuous with respect to $\mu$?

If we assume that it is, then the chain rule says its Radon-Nikodym derivative is equal to $1/\rho$ $\nu$-a.e. (or, equivalently in that case, $\mu$-a.e.). But if we remove that assumption, then the chain rule doesn't apply, so I don't know whether or how one can show that $\nu(A) = \int_{A} \frac{1}{\rho}\, d\mu$, which would suffice to show that $\nu$ is absolutely continuous with respect to $\mu$. I presume there is a counterexample, though, since the chain rule takes as an assumption that $\mu$ and $\nu$ are equivalent.

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    $\begingroup$ Assume $\mu(A)=0$. By construction, $\int_A \rho \mathrm{d}\nu=\int_{A\cap(\rho\neq 0)}\rho\mathrm{d}\nu=0$. This implies $\nu(A\cap (\rho\neq 0))=0$. Since $\nu(\rho=0)=0$, we have $\nu(A)=0$. $\endgroup$ – user111463 Apr 20 '15 at 23:22

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