Let $\mu,\nu,\omega$ $\sigma$-finite measures over the $\sigma$-algebra $\mathfrak{A}$ such that $\mu\ll\nu\ll\omega$.
I am asked to show that $\mu\ll\omega$ and that \begin{align*} \frac{d\mu}{d\omega} = \frac{d\nu}{d\omega} \frac{d\mu}{d\nu}. \end{align*}
So I know that I'm supposed to show that these two functions are equal but I don't really know how to do it.
$\int_E f \frac {d\nu} {d\omega} \, d\omega =\int_E f \, d\nu$ for any non-negative measurable function $f$. To see this note that this holds when $f=I_A$ by definition; now extend to simple functions and use Monotone Convergence Theorem to show that it holds whenever $f$ is a non-negative measurable function. Now put $f= \frac {d\mu} {d\nu}$. You get $\int_E \frac {d\mu} {d\nu} \frac {d\nu} {d\omega} \, d\omega =\int_E \frac {d\mu} {d\nu} \, d\nu=\mu (E)$. By definition of Radon Nikodym derivative this implies $\frac {d\mu} {d\nu} \frac {d\nu} {d\omega}=\frac {d\mu} {d\omega}$