In their interesting paper "Integration des fonctions sous-analytiques et volumes des sous-ensembles sous-analytiques" Lion and Rollin show that the volume of a fibre $Y_x$ of a globally subanalytic set $Y \subset \mathbb{R}^{m+n}$ is a function of $x \in \mathbb{R}^n$ which is of the form $$ P(t_1, \ldots, t_d, \log(t_1), \ldots, \log(t_d)) $$ where $P$ is a polynomial, and $t_1, \ldots, t_d$ are globally subanalytic functions of $x$.
I will only speak about the situation where all $Y_x$ are uniformly bounded (the general case can be reduced to it).
The first step in the proof is to apply Crofton's formula in order to write the volume of $Y_x$ as the integral $$ \int_{e \in \mathrm{Gr(k,m)}} \delta(x,e) d\mu $$ where $\mu$ is some measure (with analytic density) on the Grassmanian of affine $k$-subspaces of $\mathbb{R}^m$, and $\delta(x,e)$ is the number of intersections of $Y_x$ with $k$-dimenional affine subspace $e$ (if it is finite; otherwise $\delta=0$).
The authors then say that, by application of Gabrielov's complement theorem, and uniform finiteness of subanalytic sets, the above integral is just an integral of a subanalytic function on the Grassmanian, which in turn can be reduced to an integral over a unit box of appropriate dimension by an analytic base-change. The proof then proceeds to use integration and elemination theorems to conclude.
My question is: why does one have to resort to Cauchy-Crofton's formula in order to reduce calculation of the volume of $Y_x$ (as a function of $x$) to an integral of this type? It seems that if one takes just the integral of the indicator function of $Y_x$, then such integral can be reduced to the integral as in the previous paragraph after an appropriate base change.
