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I am reading Analysis on Manifolds by Munkres, this is from Section 37
Stokes' Theorem: Let M be a compact oriented $k$-manifold in $\mathbb{R}^n$, $A \subseteq \mathbb{R}^n$ open s.t. $M \subseteq A$, $\omega$ a $(k-1)$-form defined on $A$

  • If $\partial M \neq \emptyset$, give it the induced orientation, then $\int_M d \omega =\int_{ \partial M} \omega$
  • If $\partial M = \emptyset$, then $\int_M d \omega$=0.

I just have a bookkeeping question: in order to show this, do we need to show $M$ compact implies $\partial M$ is compact? Munkres only defines integrals of $k$-forms over compact, so it seems we need that to make sense of $\int_{ \partial M} \omega$. But it's not mentioned in the proof, and looking back through earlier sections I don't think Munkres mentions it either. Since I am new to this subject, I am genuinely confused. Any help is greatly appreciated.

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    The boundary is a closed set, and closed subsets of compact sets are still compact. – 2017-02-07
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    The boundary of a manifold is always a relatively closed subset. So, if the manifold is compact, the boundary is compact as well. – 2017-02-07

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If $M$ is a manifold with boundary, then the interior of $M$ is an open set (since every point has a neighborhood diffeomorphic to $\mathbb{R}^n$, and then every point in that neighborhood is also in the interior). So the boundary $\partial M$ is a closed subset of $M$. If $M$ is compact, this means $\partial M$ is automatically compact as well.