[proofplan] We prove the identity by reducing it to a coordinate calculation on the two local models for a [manifold with boundary](/page/Manifold%20with%20Boundary): open subsets of $\mathbb{R}^n$ and open subsets of the closed half-space $\mathbb{H}^n = \{x \in \mathbb{R}^n : x_n \geq 0\}$. A [partition of unity](/page/Partition%20of%20Unity) subordinate to oriented coordinate charts decomposes the compactly supported form into finitely many pieces, each supported in a single chart. For each piece, naturality of the [exterior derivative](/page/Exterior%20Derivative) and the definition of [integration of differential forms](/page/Integration%20of%20Differential%20Forms) transfer the desired identity to the Euclidean local model. The half-space computation is where the [boundary orientation](/page/Boundary%20Orientation) convention fixes the sign. [/proofplan]

[step:Choose an oriented finite chart decomposition of the compact support]Let $K := \operatorname{supp}\omega \subset M$ denote the compact support of $\omega$. Since $M$ is oriented, there exists an oriented smooth atlas by charts \begin{align*} \varphi_i: U_i &\to V_i \end{align*} where each $V_i$ is either an open subset of $\mathbb{R}^n$ or an open subset of the closed half-space \begin{align*} \mathbb{H}^n := \{x = (x_1,\dots,x_n) \in \mathbb{R}^n : x_n \geq 0\}. \end{align*} For a boundary chart, $\varphi_i(U_i \cap \partial M) = V_i \cap \partial \mathbb{H}^n$, where \begin{align*} \partial \mathbb{H}^n := \{x \in \mathbb{R}^n : x_n = 0\}. \end{align*} By the [Existence of Smooth Partitions of Unity](/page/Partition%20of%20Unity) theorem, applied after passing to a locally finite refinement of the coordinate cover on a neighbourhood of $K$, choose a smooth partition of unity $(\rho_i)_{i \in I}$ on a neighbourhood of $K$ subordinate to the chart domains, with $\operatorname{supp}\rho_i \subset U_i$ for each $i$. Since $K$ is compact and the partition is locally finite, the set \begin{align*} I_K := \{i \in I : \operatorname{supp}\rho_i \cap K \neq \varnothing\} \end{align*} is finite. For each $i \in I_K$, define the compactly supported smooth $(n-1)$-form \begin{align*} \omega_i := \rho_i \omega \in \Omega_c^{n-1}(U_i). \end{align*} Then, on $M$, \begin{align*} \omega = \sum_{i \in I_K} \omega_i, \end{align*} because $\sum_{i \in I_K}\rho_i = 1$ on $K$ and $\omega$ vanishes outside $K$.[/step]

[guided]The purpose of this step is to convert a global statement into finitely many local statements. Let \begin{align*} K := \operatorname{supp}\omega \end{align*} be the support of $\omega$. The hypothesis $\omega \in \Omega_c^{n-1}(M)$ means precisely that $K$ is compact. An orientation on $M$ lets us choose coordinate charts that preserve the chosen orientation. Thus we take an oriented smooth atlas with chart maps \begin{align*} \varphi_i: U_i &\to V_i, \end{align*} where each chart image $V_i$ is one of the two local models for manifolds with boundary: either an open subset of $\mathbb{R}^n$, for interior points, or an open subset of \begin{align*} \mathbb{H}^n := \{x = (x_1,\dots,x_n) \in \mathbb{R}^n : x_n \geq 0\}, \end{align*} for points near the boundary. In the boundary case, the chart respects the boundary in the sense that \begin{align*} \varphi_i(U_i \cap \partial M) = V_i \cap \partial \mathbb{H}^n, \qquad \partial \mathbb{H}^n := \{x \in \mathbb{R}^n : x_n = 0\}. \end{align*} We now use the [Existence of Smooth Partitions of Unity](/page/Partition%20of%20Unity) theorem. Its hypotheses are satisfied because the coordinate domains $(U_i)_{i \in I}$ form an open cover of a neighbourhood of $K$ in the smooth paracompact manifold-with-boundary $M$; after passing to a locally finite refinement near $K$, the theorem gives smooth functions $\rho_i: M \to [0,1]$ with $\operatorname{supp}\rho_i \subset U_i$ and with locally finite sum equal to $1$ on a neighbourhood of $K$. Compactness of $K$ is important here: local finiteness implies that only finitely many supports meet $K$. Define \begin{align*} I_K := \{i \in I : \operatorname{supp}\rho_i \cap K \neq \varnothing\}. \end{align*} Then $I_K$ is finite. For each $i \in I_K$, define \begin{align*} \omega_i := \rho_i \omega \in \Omega_c^{n-1}(U_i). \end{align*} The support of $\omega_i$ is contained in $\operatorname{supp}\rho_i \cap K$, hence is compact and lies in $U_i$. Since the partition sums to $1$ on $K$ and $\omega$ is zero outside $K$, we obtain the finite decomposition \begin{align*} \omega = \sum_{i \in I_K} \omega_i. \end{align*} This is the point where compact support turns the partition-of-unity argument into a finite sum, so no convergence issue arises when we integrate.[/guided]

[step:Reduce the global identity to compactly supported forms in one chart] Let $\mathcal{I}_M$ denote the orientation-defined [integration of differential forms](/page/Integration%20of%20Differential%20Forms) functional on compactly supported $n$-forms on $M$, and let $\mathcal{I}_{\partial M}$ denote the corresponding functional on compactly supported $(n-1)$-forms on $\partial M$ with the induced [boundary orientation](/page/Boundary%20Orientation). For every oriented open submanifold $W \subset M$, let $\mathcal{I}_W$ denote the same integration functional restricted to compactly supported top-degree forms on $W$; for every oriented open subset $A$ of $\mathbb{R}^n$ or of $\mathbb{H}^n$, let $\mathcal{I}_A$ denote integration of the coefficient of the oriented Euclidean top form with respect to the relevant Lebesgue measure. By linearity of the [exterior derivative](/page/Exterior%20Derivative) and of these integration functionals, \begin{align*} \mathcal{I}_M(d\omega) &= \sum_{i \in I_K} \mathcal{I}_M(d\omega_i) = \sum_{i \in I_K} \mathcal{I}_{U_i}(d\omega_i), \\ \mathcal{I}_{\partial M}(\omega) &= \sum_{i \in I_K} \mathcal{I}_{\partial M}(\omega_i) = \sum_{i \in I_K} \mathcal{I}_{U_i \cap \partial M}(\omega_i). \end{align*} Thus it is enough to prove, for each $i \in I_K$, \begin{align*} \mathcal{I}_{U_i}(d\omega_i) = \mathcal{I}_{U_i \cap \partial M}(\omega_i). \end{align*} Fix such an index $i$. Let \begin{align*} \eta_i := (\varphi_i^{-1})^*\omega_i \in \Omega_c^{n-1}(V_i) \end{align*} be the coordinate expression of $\omega_i$ on $V_i$. Naturality of the [exterior derivative](/theorems/1525) gives \begin{align*} d\eta_i = d((\varphi_i^{-1})^*\omega_i) = (\varphi_i^{-1})^*(d\omega_i). \end{align*} Because $\varphi_i$ is orientation preserving, the definition of [integration of differential forms](/theorems/1529) in oriented charts gives \begin{align*} \mathcal{I}_{U_i}(d\omega_i) = \mathcal{I}_{V_i}(d\eta_i), \end{align*} where $\mathcal{I}_{V_i}(d\eta_i)$ means integration of the coefficient of the Euclidean top form with respect to $\mathcal{L}^n$. If $V_i \subset \mathbb{R}^n$, then $U_i \cap \partial M = \varnothing$ and the desired identity becomes $\mathcal{I}_{V_i}(d\eta_i) = 0$. If $V_i \subset \mathbb{H}^n$, then the induced boundary orientation is transported by $\varphi_i$ to the standard boundary orientation on $\partial\mathbb{H}^n$, and \begin{align*} \mathcal{I}_{U_i \cap \partial M}(\omega_i) = \mathcal{I}_{V_i \cap \partial\mathbb{H}^n}(\eta_i), \end{align*} where $\mathcal{I}_{V_i \cap \partial\mathbb{H}^n}(\eta_i)$ means integration of the boundary coefficient with respect to $\mathcal{L}^{n-1}$ in the oriented boundary coordinates. It remains to prove the corresponding local formula in these two Euclidean models. [/step]

[step:Prove the local formula in an open subset of $\mathbb{R}^n$]Let $V \subset \mathbb{R}^n$ be open, and let $\eta \in \Omega_c^{n-1}(V)$. Write \begin{align*} \eta = \sum_{j=1}^{n} (-1)^{j-1} a_j\, dx_1 \wedge \cdots \wedge \widehat{dx_j} \wedge \cdots \wedge dx_n, \end{align*} where each coefficient $a_j: V \to \mathbb{R}$ is smooth and compactly supported in $V$. Because $\operatorname{supp} a_j$ is compact and contained in the [open set](/page/Open%20Set) $V$, it has positive distance from $\mathbb{R}^n \setminus V$ on every sufficiently small neighbourhood of its support; hence extension by zero across $\mathbb{R}^n \setminus V$ gives a smooth compactly supported function on $\mathbb{R}^n$, still denoted $a_j$. Then \begin{align*} d\eta = \sum_{j=1}^{n} \partial_{x_j}a_j\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Choose $R > 0$ such that $\operatorname{supp}a_j \subset (-R,R)^n$ for every $j$. By [Fubini's theorem](/theorems/2961) and the one-dimensional [fundamental theorem of calculus](/theorems/632) applied along the $x_j$ variable, \begin{align*} \mathcal{I}_V(d\eta) &= \sum_{j=1}^{n} \int_{\mathbb{R}^n} \partial_{x_j}a_j(x)\,d\mathcal{L}^n(x) \\ &= \sum_{j=1}^{n} \int_{(-R,R)^{n-1}} \left( \int_{-R}^{R} \partial_{x_j}a_j(x_1,\dots,x_n)\,d\mathcal{L}^1(x_j) \right) d\mathcal{L}^{n-1}(x_1,\dots,\widehat{x_j},\dots,x_n) \\ &= \sum_{j=1}^{n} \int_{(-R,R)^{n-1}} \left( a_j(x_1,\dots,R,\dots,x_n) - a_j(x_1,\dots,-R,\dots,x_n) \right) d\mathcal{L}^{n-1} \\ &= 0. \end{align*} Thus $\mathcal{I}_V(d\eta) = 0$ in the interior local model, where the preceding computation is the defining $\mathcal{L}^n$-coordinate integral of the top-form coefficient.[/step]

[guided]We prove the local statement directly. Let $V \subset \mathbb{R}^n$ be open, and let $\eta \in \Omega_c^{n-1}(V)$. Every smooth $(n-1)$-form on $V$ can be written uniquely as \begin{align*} \eta = \sum_{j=1}^{n} (-1)^{j-1} a_j\, dx_1 \wedge \cdots \wedge \widehat{dx_j} \wedge \cdots \wedge dx_n, \end{align*} where $a_j: V \to \mathbb{R}$ is a smooth function. Since $\eta$ is compactly supported in $V$, each $a_j$ has compact support in $V$. Extending $a_j$ by zero outside $V$ gives a smooth compactly supported function on $\mathbb{R}^n$, because the support is contained in the interior of $V$. We compute the exterior derivative. In the $j$-th summand, all terms in $da_j$ vanish after wedging with \begin{align*} dx_1 \wedge \cdots \wedge \widehat{dx_j} \wedge \cdots \wedge dx_n \end{align*} except the $\partial_{x_j}a_j\,dx_j$ term. The sign was chosen so that \begin{align*} (-1)^{j-1} dx_j \wedge dx_1 \wedge \cdots \wedge \widehat{dx_j} \wedge \cdots \wedge dx_n = dx_1 \wedge \cdots \wedge dx_n. \end{align*} Therefore \begin{align*} d\eta = \sum_{j=1}^{n} \partial_{x_j}a_j\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Choose $R > 0$ such that every support $\operatorname{supp}a_j$ is contained in $(-R,R)^n$. The functions $\partial_{x_j}a_j$ are continuous and compactly supported, so they are integrable with respect to $\mathcal{L}^n$, and Fubini's theorem applies. For each fixed $j$, \begin{align*} \int_{\mathbb{R}^n} \partial_{x_j}a_j(x)\,d\mathcal{L}^n(x) &= \int_{(-R,R)^{n-1}} \left( \int_{-R}^{R} \partial_{x_j}a_j(x_1,\dots,x_n)\,d\mathcal{L}^1(x_j) \right) d\mathcal{L}^{n-1}(x_1,\dots,\widehat{x_j},\dots,x_n) \\ &= \int_{(-R,R)^{n-1}} \left( a_j(x_1,\dots,R,\dots,x_n) - a_j(x_1,\dots,-R,\dots,x_n) \right) d\mathcal{L}^{n-1} \\ &= 0. \end{align*} The last equality holds because $a_j$ vanishes near the boundary of the cube $(-R,R)^n$. Summing over $j$ gives \begin{align*} \mathcal{I}_V(d\eta) = 0. \end{align*} This is the Stokes formula for a compactly supported form in an interior chart: there is no boundary term because the support does not meet any artificial boundary of the coordinate patch.[/guided]

[step:Prove the local formula in the closed half-space]Let $V \subset \mathbb{H}^n$ be open in the relative topology, and let $\eta \in \Omega_c^{n-1}(V)$. Write \begin{align*} \eta = \sum_{j=1}^{n} (-1)^{j-1} a_j\, dx_1 \wedge \cdots \wedge \widehat{dx_j} \wedge \cdots \wedge dx_n, \end{align*} where each $a_j: V \to \mathbb{R}$ is smooth and compactly supported in $V$. Because $\operatorname{supp} a_j$ is compact in the relative open set $V \subset \mathbb{H}^n$, extension by zero across $\mathbb{H}^n \setminus V$ gives a smooth compactly supported function on $\mathbb{H}^n$, still denoted $a_j$. Then \begin{align*} d\eta = \sum_{j=1}^{n} \partial_{x_j}a_j\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Choose $R > 0$ such that $\operatorname{supp}a_j \subset (-R,R)^{n-1}\times[0,R)$ for every $j$. For $1 \leq j \leq n-1$, Fubini's theorem and the one-dimensional fundamental theorem of calculus in the $x_j$ variable give \begin{align*} \int_{\mathbb{H}^n} \partial_{x_j}a_j(x)\,d\mathcal{L}^n(x) = 0. \end{align*} For $j=n$, the same argument in the normal variable gives \begin{align*} \int_{\mathbb{H}^n} \partial_{x_n}a_n(x)\,d\mathcal{L}^n(x) &= \int_{(-R,R)^{n-1}} \left( \int_0^R \partial_{x_n}a_n(x_1,\dots,x_{n-1},x_n)\,d\mathcal{L}^1(x_n) \right) d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}) \\ &= -\int_{(-R,R)^{n-1}} a_n(x_1,\dots,x_{n-1},0)\, d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} Hence \begin{align*} \mathcal{I}_V(d\eta) = -\int_{V \cap \partial\mathbb{H}^n} a_n(x_1,\dots,x_{n-1},0)\, d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} It remains to identify this with the boundary integral of $\eta$. The standard orientation on $\mathbb{H}^n$ is represented by \begin{align*} dx_1 \wedge \cdots \wedge dx_n. \end{align*} The outward normal direction along $\partial\mathbb{H}^n$ is $-\partial_{x_n}$. With the outward-normal-first convention, the induced boundary orientation is represented by \begin{align*} \beta := i_{-\partial_{x_n}}(dx_1 \wedge \cdots \wedge dx_n) = (-1)^n dx_1 \wedge \cdots \wedge dx_{n-1}. \end{align*} Here $i_v$ denotes the interior product, or contraction, of a differential form by the vector $v$. Restricting $\eta$ to $\partial\mathbb{H}^n$, all summands containing $dx_n$ vanish, so \begin{align*} \eta|_{\partial\mathbb{H}^n} = (-1)^{n-1}a_n(x_1,\dots,x_{n-1},0)\, dx_1 \wedge \cdots \wedge dx_{n-1} = -a_n(x_1,\dots,x_{n-1},0)\,\beta. \end{align*} Therefore \begin{align*} \mathcal{I}_{V \cap \partial\mathbb{H}^n}(\eta) &= -\int_{V \cap \partial\mathbb{H}^n} a_n(x_1,\dots,x_{n-1},0)\, d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}) \\ &= \mathcal{I}_V(d\eta). \end{align*}[/step]

[guided]The boundary chart has one new feature: integration in the normal variable starts at $x_n=0$, and this produces exactly the boundary term. Let $V \subset \mathbb{H}^n$ be open in the relative topology, and let $\eta \in \Omega_c^{n-1}(V)$. Write \begin{align*} \eta = \sum_{j=1}^{n} (-1)^{j-1} a_j\, dx_1 \wedge \cdots \wedge \widehat{dx_j} \wedge \cdots \wedge dx_n, \end{align*} where each $a_j: V \to \mathbb{R}$ is smooth and compactly supported. Since the support is compactly contained in $V$, each $a_j$ extends by zero to a smooth compactly supported function on $\mathbb{H}^n$. As in the interior computation, the sign convention in the expansion gives \begin{align*} d\eta = \sum_{j=1}^{n} \partial_{x_j}a_j\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Choose $R > 0$ such that \begin{align*} \operatorname{supp}a_j \subset (-R,R)^{n-1}\times[0,R) \end{align*} for every $j$. For tangential directions $1 \leq j \leq n-1$, integration in the $x_j$ variable still has two vanishing endpoint terms, so Fubini's theorem and the one-dimensional fundamental theorem of calculus give \begin{align*} \int_{\mathbb{H}^n} \partial_{x_j}a_j(x)\,d\mathcal{L}^n(x) = 0. \end{align*} The normal direction is different because the half-space has the lower endpoint $x_n=0$. For $j=n$, \begin{align*} \int_{\mathbb{H}^n} \partial_{x_n}a_n(x)\,d\mathcal{L}^n(x) &= \int_{(-R,R)^{n-1}} \left( \int_0^R \partial_{x_n}a_n(x_1,\dots,x_{n-1},x_n)\,d\mathcal{L}^1(x_n) \right) d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}) \\ &= \int_{(-R,R)^{n-1}} \left( a_n(x_1,\dots,x_{n-1},R) - a_n(x_1,\dots,x_{n-1},0) \right) d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}) \\ &= -\int_{(-R,R)^{n-1}} a_n(x_1,\dots,x_{n-1},0)\, d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} The endpoint at $R$ vanishes because $a_n$ is supported in $(-R,R)^{n-1}\times[0,R)$. Therefore \begin{align*} \mathcal{I}_V(d\eta) = -\int_{V \cap \partial\mathbb{H}^n} a_n(x_1,\dots,x_{n-1},0)\, d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} Now we check the sign coming from the boundary orientation. The standard orientation form on $\mathbb{H}^n$ is \begin{align*} dx_1 \wedge \cdots \wedge dx_n. \end{align*} Along $\partial\mathbb{H}^n$, the outward normal direction is $-\partial_{x_n}$. The outward-normal-first convention says that the boundary orientation is represented by \begin{align*} \beta := i_{-\partial_{x_n}}(dx_1 \wedge \cdots \wedge dx_n) = (-1)^n dx_1 \wedge \cdots \wedge dx_{n-1}. \end{align*} Here $i_v$ denotes contraction of a differential form by the vector $v$. When we restrict $\eta$ to the boundary, every summand containing $dx_n$ pulls back to zero. Only the $j=n$ summand remains: \begin{align*} \eta|_{\partial\mathbb{H}^n} &= (-1)^{n-1}a_n(x_1,\dots,x_{n-1},0)\, dx_1 \wedge \cdots \wedge dx_{n-1} \\ &= -a_n(x_1,\dots,x_{n-1},0)\,\beta. \end{align*} Thus, with the induced boundary orientation, \begin{align*} \mathcal{I}_{V \cap \partial\mathbb{H}^n}(\eta) = -\int_{V \cap \partial\mathbb{H}^n} a_n(x_1,\dots,x_{n-1},0)\, d\mathcal{L}^{n-1}(x_1,\dots,x_{n-1}). \end{align*} This equals the value computed for $\mathcal{I}_V(d\eta)$, so \begin{align*} \mathcal{I}_V(d\eta) = \mathcal{I}_{V \cap \partial\mathbb{H}^n}(\eta). \end{align*}[/guided]

[step:Sum the local identities to obtain the global Stokes formula] Applying the local formula to each coordinate form $\eta_i = (\varphi_i^{-1})^*\omega_i$ gives \begin{align*} \mathcal{I}_{U_i}(d\omega_i) = \mathcal{I}_{U_i \cap \partial M}(\omega_i) \end{align*} for every $i \in I_K$. Summing over the finite set $I_K$, \begin{align*} \mathcal{I}_M(d\omega) &= \sum_{i \in I_K} \mathcal{I}_{U_i}(d\omega_i) \\ &= \sum_{i \in I_K} \mathcal{I}_{U_i \cap \partial M}(\omega_i) \\ &= \mathcal{I}_{\partial M}\left(\sum_{i \in I_K}\omega_i\right) \\ &= \mathcal{I}_{\partial M}(\omega). \end{align*} This proves \begin{align*} \mathcal{I}_M(d\omega) = \mathcal{I}_{\partial M}(\omega), \end{align*} which is the desired identity. [/step]