The following is a quick outline of the proof of Stokes' Theorem as found in a Comprehensive Introduction to Differential Geometry Vol. 1 by Spivak.
Theorem (Local Stokes' Theorem). Let $M$ be a smooth manifold, $c$ a singular $k$-chain and $\omega$ a $k – 1$-form on $M$. Then $$\int_c d\omega = \int_{\partial c} \omega.$$
Theorem (Stokes' Theorem). Let $M^n$ be an oriented smooth manifold with boundary and $\omega \in \Omega^{n – 1}_c(M)$. Then $$\int_M d\omega = \int_{\partial M}\omega$$ where $\partial M$ is given the induced orientation.
Proof. Suppose that the support of $\omega$ is contained in the interior of some positively oriented singular cube $c$ with $\operatorname{im} c \cap \partial M = \varnothing$. Then we can apply the local Stokes' theorem to conclude. Indeed, we have that $$\int_M d\omega = \int_c d\omega = \int_{\partial c}\omega = 0.$$
Shouldn't it really be $\operatorname{int} M$ in the first integral instead of just $M$? Because the local Stokes' theorem only applies for manifolds with boundaries. However, the next step is to consider a singular cube $c$ such that $\partial M \cap \operatorname{im}c = \operatorname{im}F_1c$, where $F_1c$ is the first front face. Spivak then proceeds again by using the local version of Stokes' theorem: $$\int_M d\omega = \int_c d\omega = \int_{\partial c}\omega = …$$ Why can we use the local version here? Again, I mean, the local version applies for manifolds without boundary.
Best Answer
The local version of Stokes Theorem is one of those results which can be generalised to hold for the case of a manifold with boundary with pretty low effort.