Does this property of Radon-Nykodym derivatives exist Suppose I have $\pi \ll \lambda$ so that
$$
f = \frac{d \pi}{d\lambda}
$$
and $\lambda \ll \eta$ so that
$$
g = \frac{d \lambda}{d \eta}
$$
Can we say that
$$
\frac{d \pi}{d\lambda} \frac{d \lambda}{d\eta} = \frac{d \pi}{d \eta} \qquad ?
$$
and if so, what can we say about this density? And does it have a name?
 A: I do not think this has a specific name, but it is a commonly known result.
Suppose that $\nu$ is a $\sigma$ -finite signed measure and $\mu, \lambda$ are $\sigma$ -finite measures on $(X, \mathcal{M})$ such that $\nu \ll \mu$ and $\mu \ll \lambda$
If $g \in L^{1}(\nu),$ then $g(d \nu / d \mu) \in L^{1}(\mu)$ and
$$
\int g d \nu=\int g \frac{d \nu}{d \mu} d \mu .
$$
We have $\nu \ll \lambda,$ and
$$
\frac{d \nu}{d \lambda}=\frac{d \nu}{d \mu} \frac{d \mu}{d \lambda} \quad \lambda-a . e
$$
I give you here a proof you can find in Folland's book:
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 $d \nu / d \mu .$ It is therefore true for simple functions by linearity, then for nonnegative measurable functions by the monotone convergence theorem, and finally for functions in $L^{1}(\nu)$ by linearity again. Replacing $\nu, \mu$ by $\mu, \lambda$ and setting $g=\chi_{E}(d \nu / d \mu),$ we obtain
$$
\nu(E)=\int_{E} \frac{d \nu}{d \mu} d \mu=\int_{E} \frac{d \nu}{d \mu} \frac{d \mu}{d \lambda} d \lambda
$$
for all $E \in \mathcal{M},$ whence $(d \nu / d \lambda)=(d \nu / d \mu)(d \mu / d \lambda) \lambda$ -a.e.
