I have come across an uncertainty in a proof I have just read:
Let $(X,\mathcal{A})$ be measure space and $\mu, \nu, \eta$ be $\sigma-$finite measures such that $\eta \ll \nu$ and $\nu \ll \mu$
Show that $\frac{d\eta}{d\mu}=\frac{d\eta}{d\nu}\frac{d\nu}{d\mu}, \mu-$a.e.
My professor defines $(A_{n}^s)_{s}\subseteq X$ s.t. $S(A_{n}^s)<\infty, \forall n \in \mathbb N$, whereby $A_{n}^{s}\subseteq A_{n+1}^{s}$and $\bigcup_{n \in \mathbb n}A_{n}^s=X, \forall s \in \{\mu, \nu, \eta\}$ and $A_{n}:=A_{n}^{\mu}\cap A_{n}^{\nu}\cap A_{n}^{\eta}$
Then following continuity from below, for all $A \in \mathcal{A}$
$\eta(A)=\lim_{n\to \infty}\eta(A \cap A_{n})=\lim_{n\to \infty}\int_{A \cap A_{n}}\frac{d\eta}{d\nu}d\nu=\lim_{n\to \infty}\int_{A}\chi_{A_{n}}\frac{d\eta}{d\nu}{d\nu}$
Problem: Then it states $\chi_{A_{n}}\frac{d\eta}{d\nu}\in \mathcal{L}(X,\mu)$ because $\eta(A_{n})<\infty$. Surely this is not correct. If anything it should be $\chi_{A_{n}}\frac{d\eta}{d\nu}\in \mathcal{L}(X,\nu)$ since by definition of the Radon-Nikodym derivative $\frac{d\eta}{d\nu}$ is measurable and $\geq 0$. Furthermore, is the statement of this fact even necessary as $A\cap A_{n}\in \mathcal{A}$ and is therefore by definition measurable w.r.t. $\eta$.
I am only interested in the problem I laid out, not the RTP.