Radon-Nikodym Derivative and Push Forward of Measures I am wondering how the Radon-Nikodym derivative is affected by push-forwarding with a random variable. Formally, let $(\Omega, \mathcal{F})$ be a measurable space, and $\mathbb{P}$ and $\mathbb{Q}$ be two probability measures on this space such that $\mathbb{Q} \ll \mathbb{P}$. Radon-Nikodym theorem tells us that there exists a $\mathcal{F}$-measurable function $f: \Omega \rightarrow [0, \infty)$ denoted by $\frac{d\mathbb{Q}}{d\mathbb{P}}$, such that
$$ \mathbb{Q}(E) = \int_E f(\omega) d\mathbb{P}(\omega) $$
for all $E \in \mathcal{F}$.
If somehow we know the expression for $f(\omega)$ already, and $X: \Omega \rightarrow \mathbb{R}^d$ is a random vector, is there a way to derive the Radon-Nikodym derivative $\frac{d\mathbb{Q}^{X}}{d\mathbb{P}^X}$for the induced distribution measures $\mathbb{Q}^X$ and $\mathbb{P}^X$ on $(\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d))$? My rough guess was $\frac{d\mathbb{Q}^{X}}{d\mathbb{P}^X}(X(\omega)) = f(\omega)$ but am not sure how to prove/disprove it.
 A: Your guess that
$$
\frac{d\mathbb Q^X}{d\mathbb P^X}\circ X=f
$$
holds is correct but to define properly that Radon-Nikodym derivative we need one
more restriction on $X$, namely, hat $X$ is invertible.
By definition:
\begin{align}\tag{1}
{\mathbb Q}^X(B)&=\mathbb Q(X\in B)=\mathbb Q(X^{-1}(B))=\int_{X^{-1}(B)}f(\omega)\,d\mathbb P(\omega)\,.
\end{align}
When $X$ is invertible then $R=f\circ X^{-1}$ is a map from $\mathbb R^d$ to $[0,\infty)\,.$
This $R$ is the Radon-Nikodym derivative you are looking for
$$\tag{2}\boxed{
\quad\frac{d\mathbb Q^X}{d\mathbb P^X}=R=f\circ X^{-1}\,.\quad}
$$
To see this observe first that for every $B\in{\cal B}(\mathbb R^d)$,
$$\tag{3}
\mathbb P^X(B)=\int_{\mathbb R^d} 1_B\,d\mathbb P^X=\int_\Omega 1_{X^{-1}(B)}\,d\mathbb P=\int_\Omega (1_B\circ X)\,d\mathbb P\,. 
$$
The function $R$ is the supremum of simple functions of the form
$$
\sum_{i=1}^nR_i1_{C_i}\,,\quad\quad R_i\in\mathbb R\,,\quad C_i\in{\cal B}(\mathbb R^d)\,.
$$
By monotone convergence, and using (3),
\begin{align}
&\int_B R\,d\mathbb P^X=\sup_n\sum_{i=1}^nR_i\int_B1_{C_i}\,d\mathbb  P^X=\sup_n\sum_{i=1}^nR_i\int_\Omega (1_{B\cap C_i}\circ X)\,d\mathbb  P\\
&=\sup_n\sum_{i=1}^nR_i\int_{X^{-1}(B)} (1_{C_i}\circ X)\,d\mathbb  P
\end{align}
Note however that
$$
\sup_n\sum_{i=1}^nR_i(1_{C_i}\circ X)=R\circ X=f.
$$
Therefore, using (1)
$$
\int_{B} R\,d\mathbb P^X=\int_{X^{-1}(B)} f\,d\mathbb  P=\mathbb Q^X(B)\,.
$$
This shows (2).
