I have a sum over a finite set $X$ composed of $q$s, which are continuous variables :
$$ I = \sum_{q \in X} f(q) \, , $$
that I know I can rewrite in terms of a discrete measure
$$I = \int_X d\mu(q) f(q) \, . $$
Nothing really changed up to now. However, my objective in writing this sum as an integral is in exploiting its parameterization as a function of a continuous variable $p$, emerging from considering $q$ (which is continuous) as a function of $p$:
$$q = q(p) \Rightarrow dq = \left\vert \frac{dq(p)}{dp} \right\vert dp \, , $$
so that $I$ is brought to
$$I = \sum_{q \in X} f(q) \longmapsto \int_X d\mu(q) f(q) \longmapsto \int_\mathbb{R} dp \left\vert \frac{dq(p)}{dp} \right\vert f(q(p)) \, .$$
Is this allowed? I know it works numerically (I am integrating over a continuous variable and yes, the result is equivalent to performing the summation), but I would like to know if it's a mathematically sound procedure.
No, this is not allowed. To do this you need to use $\frac{\text{d}\mu}{\text{d}p}$ instead of $\frac{\text{d}q}{\text{d}p}$: $$\int_\mathbb{R} f \text{d}\mu=\int_\mathbb{R} f\cdot\frac{\text{d}\mu}{\text{d}p} \text{d} p$$ But $\mu$ is atomic it is not absolutely continuous wrt $\text{d}p$ (the Lebesgue measure) therefore $\frac{\text{d}\mu}{\text{d}p}$ does not exist.