Euclidean Surface Measure, Divergence, and Green Identities

For $0<r<R$ set $s=(r+R)\mathrm{/}2$ and $b_{\ast }(x)=\sigma ((s^2-\mathrm{\mid }x-a{\mathrm{\mid }}^2)\mathrm{/}(s^2-r^2))$. F1 makes this smooth, between zero and one, equal to one for $\mathrm{\mid }x-a\mathrm{\mid }\le r$, and zero for $\mathrm{\mid }x-a\mathrm{\mid }\ge s$. Its support lies in the closed s-ball, a compact subset of the open R-ball. Taking a=0,r=1,s=3/2 defines the fixed b with support inside B_2.

F2 gives $D{b}_{a,r}(x)=r^{-1}Db((x-a)\mathrm{/}r)$. F3 and F4 imply $\int f((x-a)\mathrm{/}r)dx=r^n\int f$ for every nonnegative Borel f: this is first the set-measure identity for indicators, then a finite sum for nonnegative simple f, and finally F5 applied to $2^{-k}\lfloor 2^k\min (f,k)\rfloor \uparrow f$. Apply it to the continuous compactly supported $f=\mathrm{\mid }Db\mathrm{\mid }$. Its integral is finite because Db is bounded and vanishes outside a bounded ball. Multiplication by $r^{-1}$ yields $C_n{r}^{n-1}$ as asserted.

If K is empty take all chi_j zero. Otherwise consider all pairs of concentric balls with positive rational radii r<R whose closed outer ball lies in some U_j and whose inner ball meets K. Their inner balls cover K, because every point has a positive neighborhood inside a member of the given open cover. Compactness (F2) retains finitely many such inner balls covering K. F1 gives corresponding bumps b_l, equal to one on those inner balls and compactly supported in their assigned U_j.

Put $s={\sum }_l{b}_l$. Then s is smooth with compact support, and s at least one on K. Define $\theta =\sigma (4s-1)$. By F3 theta=1 wherever s at least one half, a neighborhood of K, and its support lies in the compact set where s at least one quarter. On s>0 put $f_l=\theta b_l\mathrm{/}s$, and extend by zero on s=0. Because theta vanishes on s at most one quarter, this extension is smooth. Each f_l has compact support in the assigned U_j, is nonnegative, and their sum is theta.

For each j sum f_l over the finitely many bumps assigned to U_j, using zero if no bump is assigned. These sums are the required chi_j: their supports are finite unions of compact subsets of U_j, and their total is theta=1 near K. Since all constructions took place in the ambient Euclidean space, no differentiability of K was required.

First let k be a real continuous compactly supported function on an open Euclidean set W. Its zero extension is Borel by F8 once continuity below is established. Its zero extension is continuous: its compact support has a positive distance from the closed complement of W, and k vanishes outside that support. On a closed nondegenerate bounding box Q the extension is Riemann integrable by F1 and Darboux integrable by F2. For a grid partition, assign each point to one adjacent cell to disjointify the cells; all removed faces are null by F4. The step functions formed with the infimum and supremum of k on each closed cell bound k, and their integrals are precisely the lower and upper Darboux sums by F3. Add a constant making k and both step functions nonnegative. F5 squeezes its Lebesgue integral between the Darboux sums, whose gap tends to zero by F2. All are bounded on a finite-measure box, so F6 subtracts the added constant. Thus the Riemann and Lebesgue integrals of k agree.

For Borel A in U set $\mu (A)={\lambda }_m(T(A))$ and $\nu (A)={\int }_{A}Jd{\lambda }_m$. Since $T(A)=(T^{-1}{)}^{-1}(A)$ is Borel by F8 and T is injective, images preserve disjoint unions; hence mu is a Borel measure. F9 makes nu a Borel measure. For compact K, T(K) is compact and J is bounded on K, so F10 gives $\mu (K)<\mathrm{\infty }$ and $\nu (K)\le {\sup }_{K}J{\lambda }_m(K)<\mathrm{\infty }$ (empty K gives zero). U is second-countable and locally compact Hausdorff as an open Euclidean set. F11 therefore makes both measures Radon.

For $f\in C_c(V)$, $k(x)=f(T(x))J(x)$ with $J=\mathrm{\mid }\det DT\mathrm{\mid }$ has compact support contained in $T^{-1}(\mathrm{supp}f)$ and is continuous, since J is continuous and T is a homeomorphism. Extend f and k by zero. T is injective C1 with invertible derivative, so F7 applies to these Riemann-integrable extensions. Step 1.1 identifies both resulting integrals as Lebesgue integrals and gives ${\int }_{V}f={\int }_U(f\circ T)J$.

For nonnegative Borel psi on U, the definitions give $\int \psi d\mu ={\int }_V\psi (T^{-1}(y))dy$ and $\int \psi d\nu ={\int }_U\psi Jdx$ first when psi is an indicator, then by finite additivity for nonnegative simple psi. For general psi use $s_k=2^{-k}\lfloor 2^k\min (\psi ,k)\rfloor$, taking min(infinity,k)=k. These are Borel simple, increase to psi, and their compositions and products with positive J increase to the required integrands. F12 proves both identities. F8 and F14 verify the measurability of every composition and product. Subtracting positive and negative parts extends the identities to real compact-support continuous psi, whose absolute integrals are finite by step 1.2.

For $\phi \in C_c(U)$ take $f=\phi \circ T^{-1}\in C_c(V)$. Step 2.1 and the two identities in step 2.2 give $\int \phi d\mu ={\int }_{V}f={\int }_U(f\circ T)J=\int \phi d\nu$. The Radon hypotheses were proved in step 1.2, so F13 yields mu=nu on all Borel subsets of U.

For the stated nonnegative Borel h put $\psi =h\circ T$, Borel by F8. The first identity in step 2.2 gives ${\int }_{V}h={\int }_U\psi d\mu$; step 3.1 replaces mu by nu, and the second identity gives ${\int }_U\psi d\nu ={\int }_U(h\circ T)J$. These are identities of nonnegative extended integrals and involve no subtraction of infinities. If U is empty then V is empty and both integrals are zero.

On an overlap write $X=Y\circ T$, where T is the C1 coordinate transition. F2 gives $DX=(DY\circ T)DT$, so F3–F5 yield $J_X(y)=J_Y(T(y))\mathrm{\mid }\det DT(y)\mathrm{\mid }$. Both Gram determinants are positive because the tangent columns have full rank. F6 in dimension n-1, applied to any nonnegative Borel f supported in this overlap, now gives $\int f(X(y))J_X(y)dy=\int f(Y(z))J_Y(z)dz$. Restriction to Borel subsets is obtained by multiplying f by their indicators. Continuity of DX and the polynomial determinant, followed by the positive square root, also proves continuity of each J. For a two-dimensional surface with Gram matrix $\left(\begin{array}{cc}E& F\\ F& G\end{array}\right)$, the same definition reads $J=\sqrt{EG-F^2}$.

For two partitions chi_j and eta_k, insert ${\sum }_k{\eta }_k=1$ into each chi_j integral in F1. This produces the finite sum of integrals of ${\chi }_j{\eta }_{k}f$ on overlaps. Step 1.1 transfers each term to the eta_k chart. Summing first in j gives ${\sum }_j{\chi }_j=1$, recovering exactly the second proposed integral. Every term is nonnegative, so this works also for infinite integrals. Applying the conclusion to positive and negative parts gives agreement for integrable signed f.

Each chart term defines a measure: for disjoint Borel sets the indicators of the finite partial unions increase to the indicator of the union, and monotone convergence of the weighted chart integrals gives countable additivity. For finiteness, the parameter preimage of the support of chi_j on S is compact inside V_j because X_j is a homeomorphism onto the chart image. J_X is continuous and bounded there and the compact set is bounded, so its weighted integral is finite. There are finitely many charts. Empty S gives the zero measure.

For a graph the tangent columns are $(e_i,{\mathrm{\partial }}_{i}h)$, so $D{X}^{T}DX=I+v{v}^T$ with $v=Dh$. Expanding its determinant by columns leaves the identity term one and the terms with exactly one replaced column, namely $v_i^2$; terms with two replaced columns vanish because those columns are proportional to v. Thus its determinant is $1+\mathrm{\mid }Dh{\mathrm{\mid }}^2$, including v=0. The vector $(-Dh,1)$ is orthogonal to every tangent column, and points out of the subgraph because its derivative on $z-h(y)$ is $1+\mathrm{\mid }Dh{\mathrm{\mid }}^2>0$. The two unit vectors normal to the common tangent space have opposite sides, so the exterior-side condition selects the same one in every chart. The displayed formula is continuous, proving normal continuity. In dimension three the cross product ${\mathrm{\partial }}_{1}X\times {\mathrm{\partial }}_{2}X$ is orthogonal to both tangent vectors and has squared length equal to their Gram determinant. Its normalized value gives the outward normal exactly when it points to the exterior side; otherwise its negative does. Thus a parametrization alone does not fix the outward sign.

All integrands are bounded on the compact localized support. Extend by zero across the cylinder sides, where F already vanishes, and multiply interior derivatives by the subgraph indicator. These are integrable functions on bounded boxes. F1–F3 and the zero lower trace give ${\int }_Q{\int }_a^{h(y)}{\mathrm{\partial }}_z{F}_n(y,z)dzdy={\int }_Q{F}_n(y,h(y))dy$. The endpoint derivative is not needed: F2 uses only the interior derivative and its continuous trace.

For i<n first replace the upper endpoint by $h(y)-\epsilon$ for small positive epsilon on a compact base box containing the support projection. Put $A_i^{\epsilon }(y)={\int }_a^{h(y)-\epsilon }{F}_i(y,z)dz$. In a neighborhood of each y the varying interval lies strictly inside Omega. Split its increment into a fixed-interval integral and the short endpoint interval. F4 applies to the former because the spatial derivative is uniformly bounded on a compact subcylinder; continuity gives the latter derivative $F_i(y,h(y)-\epsilon ){\mathrm{\partial }}_{i}h(y)$. Hence ${\mathrm{\partial }}_i{A}_i^{\epsilon }={\int }_a^{h(y)-\epsilon }{\mathrm{\partial }}_i{F}_{i}dz+F_i(y,h(y)-\epsilon ){\mathrm{\partial }}_{i}h$.

A_i^epsilon vanishes near the sides of the base box. F1–F3 along its ith coordinate show ${\int }_Q{\mathrm{\partial }}_i{A}_i^{\epsilon }=0$. In step 1.2 the integral over the omitted strip is bounded by epsilon times the derivative bound, and the endpoint values converge uniformly by continuity of F on the compact closure; Dh is bounded on the base. Letting epsilon decrease to zero gives ${\int }_Q{\int }_a^{h(y)}{\mathrm{\partial }}_i{F}_{i}dzdy=-{\int }_Q{F}_i(y,h(y)){\mathrm{\partial }}_{i}hdy$. Add this for i<n to step 1.1 to obtain the asserted identity.

For an interior compactly supported field extend it by zero to a containing box. This extension is C1, since its support has positive distance from the domain complement. Fubini and FTC integrate each coordinate derivative to the difference of its two zero endpoint values. Summing these zero integrals gives zero total divergence.

Compactness of the boundary gives finitely many smaller graph cylinders covering it. Together with the open set Omega these cover the compact closure of Omega. F1 gives an ambient partition chi_j subordinate to this cover. For each boundary term chi_j F, F2 and F3 identify its divergence integral with its outward surface flux, since $\nu dS=(-Dh,1)dy$. The interior term has divergence integral zero and boundary trace zero by F2. Rigid coordinate changes preserve this calculation: the transformed field is $Q^{T}F(a+Qy)$, its derivative is $Q^{T}DFQ$ and has the same trace, its dot products are unchanged, and its volume Jacobian has modulus one.

The continuous F and DF are bounded on the compact closure; Omega is bounded of finite volume and F3 gives finite boundary area. Thus all terms are integrable. The product rule gives ${\sum }_j\mathrm{div}({\chi }_{j}F)=({\sum }_j{\chi }_j)\mathrm{div}F+({\sum }_{j}D{\chi }_j)\cdot F=\mathrm{div}F$, because the partition sum is one on a neighborhood of the closure. The boundary flux sum likewise equals $F\cdot \nu$. F4 sums the local identities from step 1.1 to give the stated theorem.

If E is empty use eta=0. Otherwise subdivide the compact chart preimages into finitely many smaller closed boxes contained in chart domains, with interiors covering them. On each box the derivative has a bound L, enlarged to at least one, and its density J has a positive lower bound c by regularity and compactness (F1). The parameter subset mapping into E is compact and null: $c{\lambda }_{n-1}(A)\le {\int }_{A}J=0$. The segment integral of DX in the convex box gives $\mathrm{\mid }X(y)-X(z)\mathrm{\mid }\le L\mathrm{\mid }y-z\mathrm{\mid }$.

Fix delta,A_0>0. F2 covers each null preimage by closed cubes with sum of side lengths to the power n-1 as small as desired. Make them open by enlarging the kth side by a positive amount with added volume below a prescribed geometric error $2^{-k}$ times the budget. Subdivide beforehand if needed so every resulting side is smaller than a prescribed positive bound. Intersect with the chart box. Compactness of each null preimage retains finitely many covering cubes; discard empty intersections with that preimage and choose one such point in each retained cube. The Lipschitz bound from step 1.1 places its image in a ball centered at the chosen image point of E with radius $r_j=2L\sqrt{n-1}{\mathrm{\ell }}_j$. Choosing the finitely many chart budgets and side bounds sufficiently small gives a finite open ball cover of E with $r_j<\delta$ and ${\sum }_j{r}_j^{n-1}<A_0$.

Use the fixed bump of F3 and put $b_j(x)=b((x-c_j)\mathrm{/}{r}_j)$ and $\eta =1-{\prod }_j(1-b_j)$. It lies between zero and one and is one on the union of the covering balls, hence near E. Its support lies in the union of the doubled balls, within distance $2\delta$ of E. The finite product rule and $0\le 1-b_j\le 1$ give $\mathrm{\mid }D\eta \mathrm{\mid }\le {\sum }_j\mathrm{\mid }D{b}_j\mathrm{\mid }$, so F3 yields $\int \mathrm{\mid }D\eta \mathrm{\mid }\le C_n{\sum }_j{r}_j^{n-1}<C_n{A}_0$.

The support volume is at most $2^n\mathrm{\mid }{B}_1\mathrm{\mid }{\sum }_j{r}_j^n\le 2^n\mathrm{\mid }{B}_1\mathrm{\mid }\delta A_0$ by volume dilation F4 and translation invariance. Take $\delta =\epsilon \mathrm{/}4$ and $A_0=\epsilon \mathrm{/}(2\max (1,C_n))$. Then the support is within epsilon of E, its gradient integral is less than epsilon, and its volume is bounded by a fixed dimensional constant times ${\epsilon }^2$, tending to zero.

For epsilon decreasing to zero use F2 and set $F_{\epsilon }=(1-{\eta }_{\epsilon })F$. Its support in the compact closure avoids a neighborhood of E. The remaining boundary support has a finite cover by the single-graph neighborhoods of F1; add an interior open set. F3 partitions this support and F4 applies to each localized term. Summing the product rules gives ${\int }_{\mathrm{\Omega }}\mathrm{div}{F}_{\epsilon }={\sum }_j{\int }_{S_j}(1-{\eta }_{\epsilon })F\cdot {\nu }_{j}dS$. Null overlaps in F1 prevent double counting.

The product rule gives $\mathrm{div}{F}_{\epsilon }=(1-{\eta }_{\epsilon })\mathrm{div}F-D{\eta }_{\epsilon }\cdot F$. The omitted first term has integral bounded by $\mathrm{\parallel }\mathrm{div}F{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mathrm{supp}{\eta }_{\epsilon }\mathrm{\mid }\to 0$ and the second by $\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}\int \mathrm{\mid }D{\eta }_{\epsilon }\mathrm{\mid }\to 0$ by F2. Thus the bulk integrals tend to ${\int }_{\mathrm{\Omega }}\mathrm{div}F$.

For every point of a face outside E, its positive distance from compact E exceeds epsilon eventually, so eta_epsilon vanishes there. Each face has finite area, and the flux is bounded by $\mathrm{\parallel }F{\mathrm{\parallel }}_{\mathrm{\infty }}$. Since E is facewise null, F5 gives convergence of each face integral in step 1.1 to its full flux. There are finitely many faces, so steps 1.1–2.1 yield the claimed identity.

In a specified finite gluing, apply this formula to every piece. Shared faces have the same surface measure and opposite normals by F1, so their two integrands sum pointwise to zero off the null edge sets. Each regular face has n-dimensional volume zero: in graph coordinates Fubini gives zero from its singleton vertical sections, and a rigid coordinate change preserves volume. Thus adding the piece volumes counts the final domain integral exactly, and only exposed face fluxes remain. This proves the cancellation statement.

Write $m$ for chart measure and take a finite sphere atlas $X_j:U_j\to S^{n-1}$ with a subordinate partition ${\chi }_j$. Since $\mathrm{\mid }{X}_j{\mathrm{\mid }}^2=1$, differentiation gives $X_j\cdot {\mathrm{\partial }}_i{X}_j=0$. For $T_j(r,y)=r{X}_j(y)$, $1<r<2$, the derivative has columns $X_j,r{\mathrm{\partial }}_1{X}_j,\dots ,r{\mathrm{\partial }}_{n-1}{X}_j$. Its Gram determinant is $r^{2n-2}\det (D{X}_j^{T}D{X}_j)$, hence $\mathrm{\mid }\det D{T}_j\mathrm{\mid }=r^{n-1}{J}_j(y)$ with $J_j=\sqrt{\det (D{X}_j^{T}D{X}_j)}>0$. The inverse is $z\mapsto (\mathrm{\mid }z\mathrm{\mid },X_j^{-1}(z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }))$, so $T_j$ is a $C^1$ diffeomorphism onto its open annular sector.

For a Borel set $A\subset S^{n-1}$, apply F3 on each sector to $h_j(z)=1_A(z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }){\chi }_j(z\mathrm{/}\mathrm{\mid }z\mathrm{\mid })$. Its pullback times the Jacobian is the separated nonnegative function $r^{n-1}{1}_A(X_j(y)){\chi }_j(X_j(y))J_j(y)$. Integrating this product gives $C{\int }_{U_j}{1}_A(X_j){\chi }_j(X_j)J_{j}dy$, where $C={\int }_1^2{r}^{n-1}dr=(2^n-1)\mathrm{/}n>0$. The product integration identity follows first for simple functions of y by the defining product measure of rectangles, then by increasing simple approximation. Summing j, the partition identity shows that the volume of $\{z:1<\mathrm{\mid }z\mathrm{\mid }<2,z\mathrm{/}\mathrm{\mid }z\mathrm{\mid }\in A\}$ equals $Cm(A)$.

Applying F2 to the indicator of that same annular cone gives its volume as $C\sigma (A)$. Thus $Cm(A)=C\sigma (A)$ for every Borel A, and division by the strictly positive C gives $m=\sigma$. For an orthogonal O, $D(OX{)}^{T}D(OX)=D{X}^{T}X$, while $D(a+RX{)}^{T}D(a+RX)=R^{2}D{X}^{T}DX$. Applying F1 chartwise proves orthogonal invariance and the factor $R^{n-1}$, also for integrals by simple approximation.

Use F2 with $1_{B_1}$: $\mathrm{\mid }{B}_1\mathrm{\mid }={\int }_0^1{r}^{n-1}\sigma (S^{n-1})dr=\sigma (S^{n-1})\mathrm{/}n$. A possible contribution at the origin is absent because the polar formula already applies to this indicator. Step 3.1 now yields the asserted chart area $\mathrm{\mid }{S}^{n-1}\mathrm{\mid }=n\mathrm{\mid }{B}_1\mathrm{\mid }$.

Set $F_i=v{\mathrm{\partial }}_{i}u$. On each coordinate line inside Omega the product rule F4 gives ${\mathrm{\partial }}_i{F}_i=({\mathrm{\partial }}_{i}v)({\mathrm{\partial }}_{i}u)+v{\mathrm{\partial }}_i^{2}u$; the same rule for every ${\mathrm{\partial }}_j{F}_i$ shows that F and all its first derivatives extend continuously to the closure. Thus $F\in C^1(\bar{\mathrm{\Omega }};{\mathbb{R}}^n)$ and summing over i gives $\mathrm{div}F=Du\cdot Dv+v\mathrm{\Delta }u$.

On every regular boundary point F3 gives $F\cdot \nu =v(Du\cdot \nu )=v{\mathrm{\partial }}_{\nu }u$. Apply F1 in the C1 case and F2 in the specified piecewise case to the field from step 1.1. Substituting its divergence and flux gives exactly the displayed identity. Continuous derivatives and functions on the compact closure are bounded; finite volume and the finite surface measures in F1/F2 prove absolute integrability. The edge set contributes zero to each face integral.

Both functions are C1 as well as C2, so apply F1 to (u,v) and (v,u). This gives ${\int }_{\mathrm{\Omega }}(v\mathrm{\Delta }u+Du\cdot Dv)={\int }_{\mathrm{\partial }\mathrm{\Omega }}v{\mathrm{\partial }}_{\nu }u$ and ${\int }_{\mathrm{\Omega }}(u\mathrm{\Delta }v+Dv\cdot Du)={\int }_{\mathrm{\partial }\mathrm{\Omega }}u{\mathrm{\partial }}_{\nu }v$, using the same specified outward normal and faces in both applications.

All four integrals in step 1.1 are finite by F1, so subtraction is legitimate. Since $Du\cdot Dv={\sum }_i({\mathrm{\partial }}_{i}u)({\mathrm{\partial }}_{i}v)=Dv\cdot Du$ pointwise, those terms cancel. Linearity leaves exactly the asserted volume and boundary differences, with the unchanged outward normals on every component.