Assume that $\mu, \nu, \rho$ are $\sigma$-finite positive measures. If $\nu << \mu, \rho << \nu$, then $\rho << \mu$ and $$\frac{d\rho}{d\mu} = \frac{d\rho}{d\nu}\frac{d\nu}{d\mu}.$$ ($\nu << \mu$ := $\nu$ is absolutely continuous with respect to $\mu.$)
Proof It is clear that $\rho << \nu.$ By applying Radon-Nikodym decomposition to $\nu << \mu, \rho << \nu, \rho << \mu,$ we have $$d\nu = \frac{d\nu}{d\mu}d\mu, d\rho = \frac{d\rho}{d\nu}d\nu, d\rho = \frac{d\rho}{d\mu}d\mu.$$ Then I conclude that $$d\rho = \frac{d\rho}{d\nu}\frac{d\nu}{d\mu}d\mu$$ which says that $$\rho(A) = \int_A \frac{d\rho}{d\nu}\frac{d\nu}{d\mu} d\mu := \int_A fg d\mu.$$ By $\rho << \mu$, $$d\rho = h d\mu$$ for some non-negative measurable function $h$. So $$\rho(A) = \int_A h d\mu.$$ This yields that $$0 = \rho(A) - \rho(A) = \int_A fg -h d\mu \ (\mbox{Here I am not sure if} \ \rho(A) - \rho(A) \neq \infty - \infty) $$ for any measurable set $A$ which yields that $$\frac{d\rho}{d\mu} = h = fg = \frac{d\rho}{d\nu}\frac{d\nu}{d\mu}\ \mbox{a.e.}.$$ I do not know how to move from here to conclude $$\frac{d\rho}{d\mu} = h = fg = \frac{d\rho}{d\nu}\frac{d\nu}{d\mu}$$ and I am not sure if my proof works (e.g. at $\rho(A) - \rho(A) = \infty - \infty).$ Any suggestion please ?