Estimate volume of Tubes around a smooth hypersurface Suppose $\Sigma \subset \mathbb{R}^n$ is a smooth compact hypersurface and the boundary of some set $\Omega$.  Now set $\Sigma_r:=\{x \in \Omega^c: \inf_{y\in \Sigma} |x-y|\leq r\}$ to be a tube of radius $r$ around the hypersurface. I was wondering if it is possible to somehow estimate the volume of these tubes in terms of the area of $\Sigma$, like for example
$$
vol(\Sigma_r) \leq C_{\Sigma} \cdot r \cdot area(\Sigma). 
$$
It should be possible for small $r$ due to this formula. There is of course a trivial bound like if I take some ball $B_R(y)$ such that $\Sigma \in B_R(y)$ then
$
vol(\Sigma_r) \leq vol(B_{R+r}(y)).
$
Is there something better?
 A: First of all, suppose that $\Omega$ is orientable and that $\nu$ is a unit normal vector field on $\Omega$ (in fact, a compact hypersurface in the euclidean spane is forced to be orientable). The map $(p,t)\in \Omega \times \mathbb{R} \to p+t\nu(p)\in\mathbb{R}^n$ is smooth and its differential at a point $(p,0)$ is invertible. By compactness assumption and by the inverse function theorem, there exists $r_0>0$ such that it is a diffeomorphism from $\Omega\times (-r_0,r_0)$ onto its image that is the open tube around $\Omega$ with radius $r_0$.
Thus, we have a chart $f : \Omega \times (-r_0,r_0) \to \mathbb{R}^n$. If $\Sigma_r$ is the tube of radius $r$ around $\Omega$, then
$$
vol(\Sigma_r) = \int_{\Sigma_r} 1 ~\mathrm{d}vol_{eucl} = \int _{f^{-1}(\Sigma_r)}1 \left(f^*\mathrm{d}vol_{eucl}\right) = \int_{\Omega\times (-r,r)} 1 
\left(f^*\mathrm{d}vol_{eucl}\right)$$
Thus, it suffices to compute the voume form $\left(f^*\mathrm{d}vol_{eucl}\right)$ to have an estimate on the volume of $\Sigma_r$.
First of all, the volume form $\left(f^*\mathrm{d}vol_{eucl}\right)$ on $\Omega\times (-r_0,r_0)$ is proportionnal to the volume form $\mathrm{d}vol_{\Omega}\wedge \mathrm{d}t$ because the top bundle of a manifold is of rank $1$. Thus, there exists a positive function $\lambda(p,t)$ such that
$$
\left(f^*\mathrm{d}vol_{eucl}\right) = \lambda\cdot \mathrm{d}vol_{\Omega}\wedge \mathrm{d}t
$$
In every good Riemannian geometry book, you can find a proof of the following:
$$
\partial_t \lambda = \lambda \times \mathrm{trace}S
$$
where $S$ is the shape operator of the level hypersurface $\Omega_t = f\left( \Omega\times \{t\} \right)$. We are now reduced to evaluating the trace of $S$. But in every good Riemannian geometry book, there is this well known fact:
$$
\partial_t S = -S \circ S
$$
in the euclidean space (in general, one has to add a curvature term on the RHS): this is the Riccati equation for the shape operator. Taking the trace and using the fact that $\mathrm{trace}(S^2) \geqslant \frac{(\mathrm{trace}S)^2}{n-1}$, it follows that
$$
\partial_t\left(\frac{\mathrm{trace} S}{n-1}\right) + \left(\frac{\mathrm{trace}S}{n-1}\right)^2 \leqslant 0
$$
From this, one can recover some information on $\mathrm{trace}S$, then on $\lambda$ and finally, one can show an estimate on the volume of $\Sigma_r$.
Note: here are listed some good Riemannian geometry books :

*

*Riemannian Geometry, Gallot, Hulin, Lafontaine

*Riemannian Geometry, Petersen

*Introduction to Riemannian manifolds, Lee

One can find in them references for the above claims in the sections dedicated to extrinsic geometry and/or comparison estimates.
In some editions of Petersen's book, the function $\mathrm{trace}S$ does not appear explicitly; what is used is $\Delta r$, the Laplacian of $r$, where $r = d(\cdot,\Omega)$. But in fact this function is equal to $\mathrm{trace}S$.
