Problem:
Let $ \nu $ be a signed measure and $ \mu,\lambda $ be measures on $ (\Omega,\mathcal{F}) $ such that $ \lambda,\mu,\nu $ are $ \sigma $-finite, $ \nu\ll\mu $ and $ \mu \ll \lambda $. Prove that then, $$ \dfrac{d\nu}{d\lambda}=\dfrac{d\nu}{d\mu}\cdot \dfrac{d\mu}{d\lambda},\text{ } \lambda\text{-a.e.} $$
My proof:
Since $ \nu \ll \mu $ $$ \nu (E)=\int_{E}fd\mu \text{ for } E\in\mathcal{F} $$ and since $ \mu \ll \lambda $ $$ \mu (E)=\int_{E}gd\lambda \text{ for } E\in\mathcal{F}. $$ Combining these gives $$ \nu (E)=\int_{E}fgd\lambda=\int_{E}hd\lambda \text{ for } E\in\mathcal{F} $$ where $ h=fg $ $ \lambda $-a.e. Since $$ f=\dfrac{d\nu}{d\mu}, $$ $$ g=\dfrac{d\mu}{d\lambda} $$ and $$ h=\dfrac{d\nu}{d\lambda} $$ we get that $$ \dfrac{d\nu}{d\lambda}=\dfrac{d\nu}{d\mu}\cdot \dfrac{d\mu}{d\lambda},\text{ } \lambda\text{-a.e.} $$
Question:
Is my proof correct? I have a strong feeling of that it is not correct. For example I don't really argue why the sentence starting with "Combining these gives" is true, and I don't even know if it is. Can someone help me out?
Let $$ f=\dfrac{d\nu}{d\mu} \text{ and } g=\dfrac{d\mu}{d\lambda} $$ be measurable functions. Then $ f_+ $ and $ f_- $ are measurable (and non-negative), with $$ f=f_+-f_-. $$ We can therefore assume that $ f $ is non-negative. Let $ f_n $ be an increasing sequence of non-negative simple functions such that $$ f_n\to f. $$ Then $$ \int_{E}f_n d\mu \to \int_{E}fd\mu$$ and $$ \int_{E}f_n gd\lambda \to \int_{E}fgd\lambda $$ for every measurable set $ E $. Let $$ 1_F= \begin{cases} 1 & \text{for } x\in F\\ 0 & \text{for } x\in \Omega\backslash F. \\ \end{cases} $$ Since \begin{align*} \int_{E}1_Fd\mu &= \mu(E\cap F) \\ &= \int_{E\cap F}gd\lambda\\ &=\int_{E}1_Fgd\lambda \end{align*} for every measurable set $ F $, it follows that $$ \int_{E}f_nd\mu=\int_{E}f_ngd\lambda $$ for each $ n $, and therefore that (using the Randon-Nikodym theorem) $$ \nu(E)=\int_{E}fd\mu=\int_{E}fgd\lambda $$ Hence $$ \dfrac{d\nu}{d\lambda}=\dfrac{d\nu}{d\mu}\cdot \dfrac{d\mu}{d\lambda},\text{\space\space} \lambda\text{-a.e.} $$