Agreement of Borel overlap integrals

For $n\ge 1$ put $W=x(U\cap V\cap \mathrm{Int}M)$ and $Z=y(U\cap V\cap \mathrm{Int}M)$. These are open in ${\mathbb{R}}^n$. Boundary invariance makes $T=y\circ x^{-1}:W\to Z$ a smooth diffeomorphism. The images of $E\cap \mathrm{Int}M$ are Borel: charts are homeomorphisms and the trace sigma-algebra agrees with relative Borel sets.

On $Z$ take $h={\mathbf{1}}_{y(E\cap \mathrm{Int}M)}{r}_y$, with the zero-times-infinity convention. It is nonnegative Borel. For $u\in W$, $h(Tu)\mathrm{\mid }\det DT(u)\mathrm{\mid }={\mathbf{1}}_{x(E\cap \mathrm{Int}M)}(u)r_x(u)$. Applying the Borel substitution formula to these exact domains and this integrand equates the two interior integrals.

The remaining coordinate pieces of $E$ lie in the face $u_n=0$, or are empty for an interior chart. Their integrals vanish by nullity, without boundedness of $r_x$ or $r_y$. Adding them back proves the formula for $n\ge 1$.

For $n=0$ each nonempty chart has one point. Thus $E$ is empty or the common singleton. Both integrals are respectively zero or the same weight, because the transition determinant is one. This proves the formula in every dimension, including zero or infinite weight.