Suppose that $M$ is a (compact, oriented, smooth) manifold with corners. Let $M_{\leq 1}$ the manifold with boundary of all points of index $\leq 1$ (following the notations here). This manifold is an open subset of $M$.
Intuitively, it seems to me that the "measure" of $M\backslash M_{\leq 1}$ is zero (even though this is not really well-defined). More precisely, is it true that for any volume form $\omega$ on $M$, $$ \int_M \omega = \int_{M_{\leq 1}} \omega \qquad ? $$ If this is correct, (unless I'm mistaken) this would directly imply the Stokes theorem for manifolds with corner by reducing immediately to the "manifold with boundary" situation.
Also, if this is true, does it really depend on the fact that $M$ is a manifold with corners ? I could imagine it holds even if the singularities are more complex than "corners". For example, suppose that :
- $N$ is $k$-submanifold of $\mathbb{R^n}$ that is bounded and without boundary (with $k\geq 2$)
- there is a $k-2$-submanifold $C$ of $\mathbb{R}^n$ such that $\overline N\backslash C$ is a manifold with boundary containing $N$ as an open subset.
Do we have then $\int_{\overline{N}\backslash C} \omega = \int_N \omega$, for any $k$-form $\omega$ defined on an open subset containing $\overline N$ ?
