In Folland's Real Analysis, there is this theorem:
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 $$ \frac{d \nu}{d \lambda}=\frac{d \nu}{d \mu} \frac{d \mu}{d \lambda} \text{ holds for } \lambda \text {-a.e. } $$
The proof is given as follows: Let $E \in \mathcal{M}$. By considering $\nu^{+}$ and $\nu^{-}$ separately, we may assume that $\nu \geq 0$. The equation $\int g d \nu=\int g(d \nu / d \mu) d \mu$ is true when $g=\chi_{E}$ by definition of $d \nu / d \mu$. That is, noting that $d \nu = f d \mu$, we have $\int \chi_E d \nu = \int \chi_E f d \mu$. This is exactly $\int g d \nu=\int g(d \nu / d \mu) d \mu$ for $g=\chi_{E}$. It is therefore true for simple functions by linearity of integrals, then for nonnegative measurable functions by the monotone convergence theorem, and finally for functions in $L^{1}(\nu)$ by linearity (adding/subtracting the positive/negative parts of the real/imaginary parts of $g, g \frac{d \nu}{d \mu}$, etc.) again. Replacing $\nu, \mu$ by $\mu, \lambda$ and setting $g=\chi_{E}(d \nu / d \mu),$ we obtain from (a): $$ \nu(E)=\int_{E} \frac{d \nu}{d \mu} d \mu=\int_{E} \frac{d \nu}{d \mu} \frac{d \mu}{d \lambda} d \lambda $$ for all $E \in \mathcal{M},$ $\textbf{therefore $(d \nu / d \lambda)=(d \nu / d \mu)(d \mu / d \lambda)$ holds for $\lambda$-a.e. by proposition 1}$.
I don't understand the last part, the part in boldface. For reference, 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..
Surely, in order to use this proposition, one must have $\int_{E} \frac{d \nu}{d \mu} d \lambda=\int_{E} \frac{d \nu}{d \mu} \frac{d \mu}{d \lambda} d \lambda $? Is there some connection between $\mu$ and $\lambda$ that I am missing?