Suppose that $\nu$ is a $\sigma$-finite signed measure and $\mu, \lambda$ are $\sigma$-finite positive measures on $(X, \mathcal{M})$ such that $\nu \ll \mu$ and $\mu \ll \lambda$. Then, the following hold:
(a) If $g \in L^{1}(\nu),$ then $g(d \nu / d \mu) \in L^{1}(\mu)$ (the function $g$ multiplied by $d \nu / d \mu$) and $$ \int g d \nu=\int g \frac{d \nu}{d \mu} d \mu $$ (b) We have $\nu \ll \lambda$ (this is obvious from $\nu \ll \mu$ and $\mu \ll \lambda$), and $$ \frac{d \nu}{d \lambda}=\frac{d \nu}{d \mu} \frac{d \mu}{d \lambda} \text{ holds for } \lambda \text {-a.e. } $$
I have no qualms with the proof of (a). Here is the proof for (b), and I am having difficulty understanding the part in bold:
For any measurable set $E$, note that by definition we have $$ \nu(E) = \int_E \left(\frac{d\nu}{d\lambda}\right)d\lambda. $$ Use (a) by replacing measures $\nu,\mu$ by the measures $\mu, \lambda$, and we obtain that for $g \in L^1(\mu)$, $$\int g d\mu = \int g \left(\frac{d\mu}{d\lambda}\right)d\lambda.$$ $\textbf{Letting $g = \chi_E \left(\frac{d\nu}{d\mu}\right)$, we obtain that (how do we show $g \in L^1(\mu)$? This is needed for us to use (a).)}$ $$\int \chi_E \left(\frac{d\nu}{d\mu}\right) d\mu = \int \chi_E \left(\frac{d\nu}{d\mu}\right) \left(\frac{d\mu}{d\lambda}\right)d\lambda.$$ We also have by definition: $$ \nu(E) = \int_E \left(\frac{d\nu}{d\mu}\right)d\mu := \int \chi_E \left(\frac{d\nu}{d\mu}\right)d\mu. $$ Thus, we conclude: $$\nu(E) = \int \chi_E \left(\frac{d\nu}{d\mu}\right) d\mu = \int \chi_E \left(\frac{d\nu}{d\mu}\right) \left(\frac{d\mu}{d\lambda}\right)d\lambda.$$ Note that we have shown $$ \nu(E) = \int_E \left(\frac{d\nu}{d\lambda}\right)d\lambda = \int \chi_E \left(\frac{d\nu}{d\mu}\right) \left(\frac{d\mu}{d\lambda}\right)d\lambda. $$ We can now apply proposition 1 to give the desired conclusion.
(This is irrelevant to the question, but) Proposition 1 states:
(a) If $f \in L^1$, then $\{ x:f(x) \neq 0\}$ is $\sigma$-finite.
(b) If $f,g \in L^1$, then $\int_E f = \int_E g$ for all $E \in \mathcal{M}$ $\iff$ $\int ||f-g|| = 0$ $\iff$ $f = g$ a.e..