Let $M$ be an oriented smooth $n$-manifold with boundary $\partial M$. Then its interior $\mathrm{Int}(M)$ is an open submanifold of $M$. For any $n$-form $\omega$ on $M$, does the following equality hold because $\partial M$ is measure-zero? $$ \int_{\mathrm{Int}(M)} \omega = \int_M \omega. $$ For example, if $\alpha$ is an $(n-1)$-form on $M$, and if I have to compute $\int_M \mathrm{d} \alpha$, then by Stokes' theorem $$ \int_M \mathrm{d} \alpha = \int_{\partial M} \alpha. $$ But if I have to compute $$ \int_{\mathrm{Int}(M)} \mathrm{d} \alpha, $$ which way is correct?
- Since $\mathrm{Int}(M)$ does not have boundary, $$ \int_{\mathrm{Int}(M)} \mathrm{d} \alpha = \int_{\partial \mathrm{Int}(M)} \alpha = \int_{\varnothing} \alpha = 0. $$
- Since $\partial M$ is measure-zero in $M$, $$ \int_{\mathrm{Int}(M)} \mathrm{d} \alpha = \int_M \mathrm{d} \alpha = \int_{\partial M} \alpha. $$
Thank you.
Integration on an oriented manifold is defined only for compactly supported differential forms $\alpha$, and Stokes theorem is also stated for compactly supported one. If $\alpha$ is a compactly supported differential form on $M$, $\alpha|_{\operatorname{Int}M}$ might not be of compact support on $\operatorname{Int} M$. So in general you cannot apply Stokes theorem to $\alpha|_{\operatorname{Int}M}$ (There are cases where integration are defined and Stokes theorem are applicable for more general differential forms, but definitely not in full generality).