Let $f, g \colon (0, 1) \to \mathbb R$ be two functions (both spaces are equipped with their respective Borel $\sigma$ algebras). What is the Radon-Nikodym derivative of $f_{\#} \lambda$ with respect to $g_{\#} \lambda$, where $\lambda$ is the restriction of the Lebesgue measure on $\mathbb R$ to $(0, 1)$ and $f_{\#} \lambda := \lambda \circ f^{-1}$ is the push-forward of $\lambda$ by $f$?
To be more precise, my set-up is the following: let $\mu, \nu \in \mathcal P(\mathbb R)$ be probability measures on $\mathbb R$ and $Q_{\mu}, Q_{\nu}$ their respective quantile functions. If $\mu \ll \nu$ (that is, $\mu$ is absolutely continuous with respect to $\nu$), is there an expression for the Radon-Nikodym derivative $\frac{d \mu}{d \nu} = \frac{d (Q_{\mu})_{\#} \lambda}{d (Q_{\nu})_{\#} \lambda}$?
Writing out the definition of the Radon-Nikodym derivative I arrive at $$ \frac{d (Q_{\mu})_{\#} \lambda}{d (Q_{\nu})_{\#} \lambda}(t) \mathbb 1_{E}(t) = \mathbb 1_{Q_{\nu} \circ Q_{\mu}^{-1}(E)}(t) \qquad \forall t \in \mathbb R, \quad \forall E \in \mathcal B(\mathbb R), $$ which I can't resolve.
If I additionally assume that $\mu$ and $\nu$ both have densities $f_{\mu}$ and $f_{\nu}$ with respect to the Lebesgue measure on $\mathbb R$, respectively, is the question answerable?
