In their interesting paper ["Integration des fonctions sous-analytiques et volumes des sous-ensembles sous-analytiques"](https://eudml.org/doc/75301) 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.



why use Crofton's formula in order to prove log-analyticity of volume?
This is an embarrasing confusion, really, the answer to this question seems to be less surprising than I expected. 


Of course, the integral of the indicator function mentioned in the question has nothing to do with the volume of $Y\_x$ which is a $k$-dimensional submanifold of $\mathbb{R}^n$, so one has to integrate a $k$-dimensional volume form inhereted from $\mathbb{R}^n$ over it. One can of course proceed formally by cutting $Y\_x$ according to some atlas, and taking care that one ends up with an integral over an analytic domain of an analytic volume form, but Crofton's formula just does this in one sweep.


