The answer to the particular question is about to be a kinda straigtforward one, but i have a feeling that i'm missing something import here.
Let $\psi: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be the support function of a polytope, given by $$\psi(x) = \text{max}_{i = 1, \ldots, L}{[a_{i} \cdot x + b_{i}]}$$
Here $x \in \mathbb{R}^{n}$, $a_{i} \in \mathbb{R}^{n}$ and $b_{i} \in \mathbb{R}^{n}$ as well, though, considered as scalars. (In fact those quantites define the supporting hyperplanes to the polytope and that is why it's called so)
With each convex function $\psi$ let's assosiate the measure, given by the density: $$d \mu(x) = e^{-\psi(x)} dx$$
Let $\nu$ be the pushforward of $\mu$ under the gradient map $T$: $$\psi \mapsto \nabla \psi(x)$$
(here the gradient is treated in a weak sence, by Alexandrov theorem the gradient of a convex function exists almost everywhere w.r.t to the Hausdorff measure on $\mathbb{R}^{n}$.)
The term "push-forward" means that for any test function $f \in C_{0}^{\infty}(\mathbb{R}^{n})$
$$\int_{\mathbb{R}^{n}}{f d \nu} = \int_{\mathbb{R}^{n}}{f(\nabla \psi(x))e^{-\psi(x)} dx}$$
How to figure out the density (if exists) of $\nu$?
The paper i'm skimming through indicates that $\nu$ is a discrete measure, supported at the points $${a_{1}, a_{2}, \ldots, a_{n}}$$
How to convince oneself that the mentioned statement indeed holds?
It looks quite reasonable to work with the equation above on "elementary" sets, like, since every test function can be approximated with step functions, let us suppose for a moment that
$$\int_{U \subset {\mathbb{R}^{n}}}{d\nu} = \int_{T(U) \subset \mathbb{R}^{n}}{\nabla \psi(x) e^{-\psi(x)} dx}$$ then
$$d \nu = c \cdot e^{-\psi} dx$$ where $c$ is the gradient of $\psi$ which is a locally constant function.
What am i missing here? How to derive the proposed statement?
The function $\psi$ is not only convex, but piecewise linear. In fact, you have $$ \nabla \psi(x) \in \{a_1, a_2, \ldots, a_L\} $$ for almost all $x \in \mathbb R^n$. The set where this does not hold is a null-set and we can ignore it. Let us call $S_i = \{x \mid \nabla \psi(x) = a_i$. Then, $$ \int_{\mathbb R^n} f \, \mathrm d x = \int_{\mathbb R^n} f(\nabla \psi(x)) \, e^{-\psi(x)} \, \mathrm d x = \sum_{i=1}^L \int_{S_i} f(a_i) \, e^{-\psi(x)} \, \mathrm{d}x = \sum_{i=1}^L f(a_i) \, \int_{S_i} e^{-\psi(x)} \, \mathrm{d}x. $$ This shows that $\nu$ is a linear combination of Diracs supported at $a_i$.