Reparametrization trick Note: all bold math symbols denote vectors.
I am reading this paper. In it, section 2.4 describes so called reparametrization trick, which essentially is as follows:
Let $\boldsymbol{Z}$ be a continous random vector, conditionally distributed as $\boldsymbol{Z} \sim q_{_{\boldsymbol{\phi}}}(\boldsymbol{Z} \,|\, \boldsymbol{X})$, with $\boldsymbol{\phi}$ denoting parameter of the parametric distribution $q_{_{\boldsymbol{\phi}}}$. Suppose that
\begin{alignat*}{2}
 \boldsymbol{Z} = g_{_{\boldsymbol{\phi}}}(\boldsymbol{\epsilon}, \boldsymbol{X})
\end{alignat*}
where


*

*$\boldsymbol{\epsilon}$ is an auxiliary random variable that is independent of $\boldsymbol{X},\; \boldsymbol{\epsilon} \perp \!\!\!\perp \boldsymbol{X}$, and is distributed with $\boldsymbol{\epsilon} \sim p(\boldsymbol{\epsilon})$

*$g_{_{\boldsymbol{\phi}}}$ is a vector-valued function parameterized by $\boldsymbol{\phi}$


Then they show that
\begin{alignat*}{2}
 \int q_{_{\boldsymbol{\phi}}}(\boldsymbol{z} \,|\, \boldsymbol{X}) f(\boldsymbol{z}) d\boldsymbol{z} = \int p(\boldsymbol{\epsilon}) f\left( g_{_{\boldsymbol{\phi}}}(\boldsymbol{\epsilon}, \boldsymbol{X}) \right) d\boldsymbol{\epsilon}
\end{alignat*}
stating that
\begin{alignat*}{2}
 q_{_{\boldsymbol{\phi}}}(\boldsymbol{z} \,|\, \boldsymbol{X}) d\boldsymbol{z} = p(\boldsymbol{\epsilon}) d\boldsymbol{\epsilon}
\end{alignat*}
and I don't see why this statment is true.
So my question is how to show this rigorously?
 A: As far as I know, the paper does not state it is always possible to find such a mapping $g$. 
But let us assume we have one and that it is a bijective (i.e. invertible) deterministic mapping.
Consider a differential area $d\textbf{z}$ in the domain of the r.v $\textbf{Z}$ and a differential area $d\boldsymbol{\epsilon}$ in the domain of the r.v $\boldsymbol{\epsilon}$, so that $d\textbf{z}$ is the image of $d\boldsymbol{\epsilon}$ through $g$. 
Both $\textbf{Z}$ and $\boldsymbol{\epsilon}$ being related through a deterministic mapping $g$, the probability attached to the event "$\boldsymbol\epsilon$ is in $d\boldsymbol{\epsilon}$" must be the same as that of its image through $g$ being in $d\textbf{z}$.
In other words, $p(\boldsymbol\epsilon)d\boldsymbol{\epsilon}=p(\textbf{z})d\textbf{z}$. From here, the proposition of the paper follows.
On wikipedia, the same statement is made, but absolute value braces are set on each side of the above equality.
https://en.wikipedia.org/wiki/Probability_density_function
My guess is that the wiki page refers to $dx$ and $dy$ as small variations which may be negative or positive.
My notation indicates differential areas (i.e. areas of very small regions), which are positive quantities, hence no need for absolute value braces.
