Stokes formula for finite ordinary surface corners

If $D=\mathrm{\varnothing }$, both integrals are zero. For nonempty $D$, every boundary arc is locally a $C^2$ graph, so the finite boundary has content zero in each chart meeting it. Thus a smooth top-form coefficient multiplied by $1_D$ is Riemann integrable in every relatively compact chart.

Cover $D$ by positive coordinate rectangles lying inside $\mathrm{Int}D$, rectangles where a smooth boundary arc is a graph and $D$ lies on one side, and corner rectangles where the two incident arcs form a continuous piecewise-$C^2$ graph and $D$ lies on one side. At a corner, the two oriented one-sided tangent vectors are not negative multiples by [F1]; after normalizing them in any linear coordinates, their sum defines a linear coordinate whose differential is positive on both. Thus the boundary is a continuous graph across the corner. Compactness gives a finite refinement by smaller chart neighbourhoods $V_i⋐U_i$ of these types still covering $D$. Use [F2] to choose $b_i=1$ near $\overline{V_i}$ with compact support in $U_i$; [F3] justifies these finitely many choices. The sum $B={\sum }_i{b}_i$ is positive on a neighbourhood of $D$. Use [F2] once more to choose $\psi =1$ near $D$ with compact support in $\{B>0\}$, and set ${\rho }_i=\psi b_i\mathrm{/}B$ on $\{B>0\}$, extended by zero. Then ${\sum }_i{\rho }_i=1$ near $D$, each ${\rho }_i$ is smooth and compactly supported in $U_i$, and $\eta ={\sum }_i{\rho }_i\eta$ near $D$. This finite partition defines ${\int }_D$ as the sum of the Riemann chart integrals of the localized top forms over $D$. If a second such partition is used, the products ${\rho }_i{\tilde{\rho }}_j$ form a finite partition near $D$ subordinate to chart overlaps. On each overlap the transition is a diffeomorphism with positive Jacobian, so [F4] and compactly supported change of variables identify the corresponding terms and the resulting sums agree. The coordinate boundary integral is $\int \eta (\dot{\gamma })$ on each supplied arc, so its finite chart localization is the same intrinsic integral.

Let $\alpha ={\rho }_i\eta$ be one localized term whose chart lies in $\mathrm{Int}D$. Its coordinate representative has compact support there; it vanishes near every point outside that support, so its coordinate derivative does too. Thus $\mathrm{supp}d\alpha \subseteq \mathrm{supp}\alpha \subset \mathrm{Int}D$, and [F7] gives ${\int }_{D}d\alpha ={\int }_{{\mathbb{R}}^2}d\alpha =0$. Its boundary integral is zero because its support misses $\mathrm{\partial }D$.

For a localized term in a smooth-boundary or corner chart, choose a positive coordinate rectangle $[a,b]\times [c,d]$ containing its support, with the form zero near the rectangle's artificial sides. Write $\alpha =Pdx+Qdy$ and the local boundary as $y=f(x)$. At a corner $f$ is continuous and piecewise $C^2$. Suppose first that $D$ is the side $y\le f(x)$. By [F5], ${\int }_{D}d\alpha ={\int }_a^b{\int }_c^{f(x)}(Q_x-P_y)dydx$. On each smooth piece of $f$, the difference quotient for $H(x)={\int }_c^{f(x)}Q(x,y)dy$ gives $H^{\mathrm{\prime }}(x)={\int }_c^{f(x)}{Q}_x(x,y)dy+Q(x,f(x))f^{\mathrm{\prime }}(x)$; this follows by splitting the increment at $f(x)$ and using continuity of $Q_x$ and $f$. Split $[a,b]$ at the finitely many corner abscissas and apply [F6] on each smooth subinterval. The intermediate endpoint values of $H$ cancel by continuity, and $H$ vanishes near $a,b$, so ${\int }_a^b{\int }_c^{f(x)}{Q}_{x}dydx=-{\int }_a^{b}Q(x,f(x))f^{\mathrm{\prime }}(x)dx$. Applying [F6] in the $y$ variable to $P$ gives $-{\int }_a^b{\int }_c^{f(x)}{P}_{y}dydx=-{\int }_a^{b}P(x,f(x))dx$. Hence ${\int }_{D}d\alpha =-{\int }_a^b(P(x,f(x))+Q(x,f(x))f^{\mathrm{\prime }}(x))dx$. By [F8] and [F9] the positive boundary direction on this side runs from right to left, so this is ${\int }_{\mathrm{\partial }D}\alpha$.

If instead $D$ is the side $y\ge f(x)$, use [F5] on ${\int }_a^b{\int }_{f(x)}^d(Q_x-P_y)dydx$. For $H(x)={\int }_{f(x)}^{d}Q(x,y)dy$, the same difference-quotient argument gives $H^{\mathrm{\prime }}(x)={\int }_{f(x)}^d{Q}_x(x,y)dy-Q(x,f(x))f^{\mathrm{\prime }}(x)$. Split at the finitely many corner abscissas and apply [F6] on each smooth subinterval; the intermediate $H$ values cancel by continuity and $H$ vanishes near $a,b$, giving ${\int }_a^b{\int }_{f(x)}^d{Q}_{x}dydx={\int }_a^{b}Q(x,f(x))f^{\mathrm{\prime }}(x)dx$. Integrating $-P_y$ in $y$ gives ${\int }_a^{b}P(x,f(x))dx$. Thus ${\int }_{D}d\alpha ={\int }_a^b(P(x,f(x))+Q(x,f(x))f^{\mathrm{\prime }}(x))dx$. By [F8] and [F9] the positive boundary direction is left to right, so this equals ${\int }_{\mathrm{\partial }D}\alpha$. These two graph-side calculations cover both convex and reflex corners.

Sum the local equalities of steps 3.1, 3.2, and 4.1 over the finite partition. Linearity and ${\sum }_i{\rho }_i=1$ near $D$ give ${\int }_{D}d\eta ={\int }_{\mathrm{\partial }D}\eta$. The empty case is in step 1.1; a single boundary component, an empty boundary, a smooth boundary with no corners, the zero form, and endpoints of the supplied arcs are covered by the same finite calculation (endpoints have measure zero). The regular-region hypothesis excludes degenerate arcs and lower-dimensional nonempty regions. Only finite choices were made in step 2.1 by [F3]; no countable or arbitrary choice is used. [F1, F3, F8, step 2.1, step 3.1, step 3.2, step 4.1, algebra, discharge-construct: Stokes formula] $\mathrm{\square }$