Spatial BV, finite perimeter, and TV conventions

If $\mathrm{\Omega }=\mathrm{\varnothing }$, equality is immediate. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, its singleton indicator is a smooth compactly supported test and determines every measure. Assume henceforth $n\ge 1$. Set $\lambda :=\mu -\nu$. If $K\subset \mathrm{\Omega }$ is compact and nonempty, then for every sufficiently large $m$, rational inner balls with doubled closures in $\mathrm{\Omega }\cap \{x:\mathrm{dist}(x,K)<1\mathrm{/}m\}$ cover $K$; [F2] gives a finite subcover, and ordering finite rational lists by natural-number codes gives a least such cover. Put the fixed bump [F3] on each selected ball and combine them as ${\eta }_m=1-{\prod }_j(1-{\psi }_{m,j})$. Then $0\le {\eta }_m\le 1$, it equals one on $K$, is supported in the $1\mathrm{/}m$-neighborhood of $K$, and ${\eta }_m(x)\to {\mathbf{1}}_K(x)$. Since each ${\eta }_m$ is a smooth compactly supported test function, $\int {\eta }_{m}d\lambda =0$; dominated convergence [F4] with respect to the finite regular total-variation measure [F1] gives $\lambda (K)=0$. The empty compact set has measure zero directly.

For any Borel $A\subseteq \mathrm{\Omega }$ and $\epsilon >0$, inner regularity of $\mathrm{\mid }\lambda \mathrm{\mid }$ [F1] gives a compact $K\subseteq A$ with $\mathrm{\mid }\lambda \mathrm{\mid }(A\setminus K)<\epsilon$. Since step 1.1 gives $\lambda (K)=0$, we have $\mathrm{\mid }\lambda (A)\mathrm{\mid }\le \mathrm{\mid }\lambda \mathrm{\mid }(A\setminus K)<\epsilon$. Thus $\lambda (A)=0$ for every Borel $A$, so $\mu =\nu$.

If $m=0$, the vector measure and $V$ vanish and the zero field gives the asserted supremum. If $\mathrm{\Omega }=\mathrm{\varnothing }$, the conclusion is immediate. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, then $\mathrm{\Omega }$ is a singleton, $V(\mathrm{\Omega })=\mathrm{\mid }\mu (\mathrm{\Omega })\mathrm{\mid }$, and the constant field $\mu (\mathrm{\Omega })\mathrm{/}\mathrm{\mid }\mu (\mathrm{\Omega })\mathrm{\mid }$ attains the supremum when the vector is nonzero (use the zero field otherwise). Assume henceforth $n,m\ge 1$. For every Borel $A$ and every countable Borel partition $(A_j)$ of $A$, the Euclidean norm satisfies $\mathrm{\mid }\mu (A_j)\mathrm{\mid }\le {\sum }_{i=1}^m\mathrm{\mid }{\mu }_i(A_j)\mathrm{\mid }$. Summing over $j$ gives $V(A)\le \lambda (A)$ for $\lambda :={\sum }_{i=1}^m\mathrm{\mid }{\mu }_i\mathrm{\mid }$, so $V$ is finite and $V(\mathrm{\Omega })<\mathrm{\infty }$.

If $A={⨆}_{k\ge 1}{E}_k$ is a disjoint Borel union, then intersecting any partition of $A$ with each $E_k$ and using countable additivity of each coordinate shows $V(A)\le {\sum }_{k}V(E_k)$. Conversely, for fixed $N$ and $\epsilon >0$, choose for each of the finitely many $k\le N$ a partition of $E_k$ whose sum exceeds $V(E_k)-\epsilon \mathrm{/}N$. Their union, together with $A\setminus {\bigcup }_{k\le N}{E}_k$, is a partition of $A$, so $V(A)\ge {\sum }_{k\le N}V(E_k)-\epsilon$. Letting $\epsilon \downarrow 0$ and then $N\to \mathrm{\infty }$ proves countable additivity. Thus $V$ is a finite positive Borel measure.

For every $f\in C_c(\mathrm{\Omega };{\mathbb{R}}^m)$ with $\mathrm{\mid }f\mathrm{\mid }\le 1$, approximate $f$ uniformly by finite-valued Borel simple functions whose values have norm at most one. The partition definition bounds each simple integral by $V(\mathrm{\Omega })$, and finiteness of the coordinate measures lets the uniform errors tend to zero. Hence the displayed supremum is at most $V(\mathrm{\Omega })$.

Since $V\le \lambda$ and $\lambda ={\sum }_i\mathrm{\mid }{\mu }_i\mathrm{\mid }$ is a finite regular Borel measure by [F1], for every Borel $A$ and $\epsilon >0$ there is compact $K\subseteq A$ with $\lambda (A\setminus K)<\epsilon$, whence $V(A\setminus K)<\epsilon$. There is also an open $G\supseteq A$ with $\lambda (G\setminus A)<\epsilon$, whence $V(G\setminus A)<\epsilon$. Thus $V$ is a regular Borel measure in the sense of Regular Borel measure on an LCH space, and in particular a Radon measure.

If $V(\mathrm{\Omega })=0$, the zero field proves the reverse inequality. Otherwise fix $0<\epsilon <V(\mathrm{\Omega })\mathrm{/}4$ and choose a countable Borel partition $(A_j)$ with sum greater than $V(\mathrm{\Omega })-\epsilon$; truncate after $N\ge 1$ so the tail is below $\epsilon$. Omit indices with $\mu (A_j)=0$ or $V(A_j)\le \epsilon \mathrm{/}(4N)$; the sum of the omitted partition values is at most $\epsilon \mathrm{/}4$. For every retained index, compact inner regularity [F2] gives a nonempty compact $K_j\subset A_j$ with $V(A_j\setminus K_j)<\epsilon \mathrm{/}(8N)$. Since there are finitely many pairwise disjoint compact sets $K_j$ in the metric space $\mathrm{\Omega }$, choose pairwise disjoint open neighborhoods $N_j$ with compact closures in $\mathrm{\Omega }$. Outer regularity [F2] gives open $G_j\supset K_j$ with $V(G_j\setminus K_j)<\epsilon \mathrm{/}(8N)$. Put $U_j=G_j\cap N_j$. Then $K_j\subset U_j$, $V(U_j\setminus K_j)<\epsilon \mathrm{/}(8N)$, and the closures of the $U_j$ remain pairwise disjoint compact subsets of $\mathrm{\Omega }$. Define ${\psi }_j(x):=\frac{\mathrm{dist}(x,\mathrm{\Omega }\setminus U_j)}{\mathrm{dist}(x,\mathrm{\Omega }\setminus U_j)+\mathrm{dist}(x,K_j)}\mathrm{.}$ This is continuous and compactly supported, equals one on $K_j$, and vanishes outside $U_j$. Set $e_j=\mu (A_j)\mathrm{/}\mathrm{\mid }\mu (A_j)\mathrm{\mid }$ and $f={\sum }_j{e}_j{\psi }_j$. The disjoint closures of the $U_j$ imply disjoint supports and $\mathrm{\mid }f\mathrm{\mid }\le 1$. For bounded Borel vector fields, the partition definition and uniform simple-function approximation give $\mathrm{\mid }\int g\cdot d\mu \mathrm{\mid }\le \int \mathrm{\mid }g\mathrm{\mid }dV$. Thus replacing $e_j{\mathbf{1}}_{A_j}$ by $e_j{\psi }_j$ loses at most $V(A_j\setminus K_j)+V(U_j\setminus K_j)$ for each retained $j$. Summing these errors, the omitted terms, and the tail gives $\int f\cdot d\mu >V(\mathrm{\Omega })-3\epsilon$. Letting $\epsilon \downarrow 0$ proves the reverse bound.

If $\mathrm{\Omega }=\mathrm{\varnothing }$, the empty family proves the claim. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, then $\mathrm{\Omega }={\mathbb{R}}^0$ is a singleton; choose one cover member containing it and the single function $\eta =1$, which is smooth and compactly supported. Thus assume $n\ge 1$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$. For $m\ge 1$ set $K_m=\{x:\mathrm{\mid }x\mathrm{\mid }\le m,\mathrm{dist}(x,{\mathrm{\Omega }}^c)\ge 1\mathrm{/}m\}$ when $\mathrm{\Omega }\mathrm{\ne }{\mathbb{R}}^n$, and set $K_m=\bar{B}(0,m)$ when $\mathrm{\Omega }={\mathbb{R}}^n$; put $K_m=\mathrm{\varnothing }$ for $m\le 0$. Each $K_m$ is compact by [F1], the sets are nested, $K_m\subset \mathrm{int}{K}_{m+1}$, and their interiors cover $\mathrm{\Omega }$. The compact shells $S_m:=K_m\setminus \mathrm{int}{K}_{m-1}$ cover $\mathrm{\Omega }$ and lie in the open sets $U_m:=\mathrm{int}{K}_{m+1}\setminus K_{m-2}$. The family $(U_m)$ is locally finite: any compact subset of $\mathrm{\Omega }$ lies in some $\mathrm{int}{K}_N$ and hence in $K_{m-2}$ for all sufficiently large $m$.

Fix once a smooth nonnegative cutoff $\theta$ equal to one on $\bar{B}(0,1)$ and supported in $B(0,2)$, whose existence is [F3]. In each $U_m$, consider the rational inner balls $B(q,r)$ for which $\bar{B}(q,2r)\subset U_m\cap V_{\mathrm{\ell }}$ for some cover index $\mathrm{\ell }$. Indexed by $(\mathrm{\ell },q,r)$, these balls form a countable cover of $S_m$: for each point choose a cover member containing it, then density [F2] gives a rational center and radius with the required doubled closure. Compactness of $S_m$ gives a finite subcover. Order all finite lists of eligible indexed balls by natural-number codes and take the first list covering $S_m$; thus no choice principle is needed. For each selected ball use the fixed rescaled cutoff ${\psi }_{m,j}(x):=\theta ((x-q)\mathrm{/}r)$. The family is locally finite because each shell contributes finitely many cutoffs supported in $U_m$, and each support also lies in its assigned cover member. Its sum $s:={\sum }_{m,j}{\psi }_{m,j}$ is smooth and positive on $\mathrm{\Omega }$ because the shells cover $\mathrm{\Omega }$.

Define ${\eta }_{m,j}:={\psi }_{m,j}\mathrm{/}s$. Then each function is smooth, nonnegative and compactly supported in $\mathrm{\Omega }$, the family is locally finite, and ${\sum }_{m,j}{\eta }_{m,j}=1$. Each support remains in its assigned cover member, and enumeration gives the required countable family.

If $\mathrm{\Omega }=\mathrm{\varnothing }$, the zero function suffices. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, then its singleton supports every function and every function is smooth, so take the approximant $f$ itself. Assume $n\ge 1$. Let $f\in C_0(\mathrm{\Omega })$ and $\epsilon >0$. The set $K:=\{x\in \mathrm{\Omega }:\mathrm{\mid }f(x)\mathrm{\mid }\ge \epsilon \mathrm{/}3\}$ is compact by the definition of vanishing at infinity. If $K=\mathrm{\varnothing }$, the zero function approximates $f$ within $\epsilon \mathrm{/}3$. Otherwise cover $K$ by finitely many inner balls whose doubled closed balls lie in $\mathrm{\Omega }$, using compactness [F1]. Choose the corresponding bumps from [F2] and combine them as $\chi :=1-{\prod }_j(1-{\chi }_j)$. Then $\chi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and $\chi =1$ on $K$, so $\mathrm{\parallel }f-\chi f{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon \mathrm{/}3$.

Put $g=\chi f$ and $L=\mathrm{supp}g⋐\mathrm{\Omega }$. If $L=\mathrm{\varnothing }$, take the zero approximant. Otherwise, for each $x\in L$, continuity gives $r_x>0$ such that $\bar{B}(x,2r_x)\subset \mathrm{\Omega }$ and $\mathrm{\mid }g(y)-g(x)\mathrm{\mid }<\epsilon \mathrm{/}3$ on $B(x,2r_x)$. Finitely many inner balls $B(x_j,r_j)$ cover $L$. By [F2], let ${\psi }_j$ be nonnegative smooth bumps equal to one on those inner balls and supported in their doubled balls. The sum $s={\sum }_j{\psi }_j$ is positive on a neighborhood of $L$. Repeating the finite cutoff construction of step 1.1 inside the open set $\{s>0\}$ gives $\zeta \in C_c^{\mathrm{\infty }}(\{s>0\})$ with $0\le \zeta \le 1$ and $\zeta =1$ on $L$. Define $a=\zeta s^{-1}{\sum }_{j}g(x_j){\psi }_j$ on $\{s>0\}$ and extend it by zero. Its compact support lies in $\mathrm{\Omega }$, and the extension is smooth because $\mathrm{supp}\zeta ⋐\{s>0\}$. Where $s>0$, the weighted average differs from $g(y)$ by at most $\epsilon \mathrm{/}3$, since each nonzero weight has $\mathrm{\mid }g(x_j)-g(y)\mathrm{\mid }<\epsilon \mathrm{/}3$. Also $\zeta g=g$ everywhere, because $\zeta =1$ on $\mathrm{supp}g$. Therefore $\mathrm{\parallel }a-g{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon \mathrm{/}3$, and step 1.1 gives $\mathrm{\parallel }a-f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2\epsilon \mathrm{/}3<\epsilon$. This uses only finitely many balls, bumps and sampled values, with no integration or infinite selection.

First replace the $g_i$ by their Borel representatives from [F3], setting each representative to zero on its Borel set of nonfinite or negative values. The exceptional sets lie in Borel null sets $N_i$. Their coordinate lifts are null in ${\mathbb{R}}^n$: intersect $N_i\times \mathbb{R}$ with slabs $N_i\times [-k,k]$, $k\ge 1$, and apply the product rectangle formula in [F1], then take their countable union and permute coordinates. Completeness in [F1] therefore makes the original product Lebesgue-measurable with the same integral as the Borel product; the right-hand integrals are unchanged as well. We may henceforth assume all inputs Borel. For $n=2$, Tonelli [F1] factors the left side as $({\int }_{\mathbb{R}}{g}_1)({\int }_{\mathbb{R}}{g}_2)$, which is the right side.

For $n\ge 3$, assume [A1]. For $i<n$ define $h_i(x_1,\dots ,\widehat{x_i},\dots ,x_{n-1}):={\int }_{\mathbb{R}}{g}_i(x_1,\dots ,\widehat{x_i},\dots ,x_{n-1},x_n)d{x}_n$. Apply Tonelli [F1] using the Borel sigma-algebras: each $h_i$ is Borel and integrable, possibly infinite on a Borel null set. Set it to zero there. The lifted exceptional sets in ${\mathbb{R}}^{n-1}$ are null by the slab argument of step 1.1. Consequently all the following sectional inequalities hold almost everywhere, and the modified $h_i$ are admissible finite inputs for [A1]. Applying Hölder [F2] in $x_n$ to the first $n-1$ factors and then Tonelli gives an upper bound for the left side of ${\int }_{{\mathbb{R}}^{n-1}}{g}_n(x^{\mathrm{\prime }}{)}^{1\mathrm{/}(n-1)}{\prod }_{i<n}{h}_i({\widehat{x_i}}^{\mathrm{\prime }}{)}^{1\mathrm{/}(n-1)}d{x}^{\mathrm{\prime }}$. Set $H(x^{\mathrm{\prime }}):={\prod }_{i<n}{h}_i({\widehat{x_i}}^{\mathrm{\prime }}{)}^{1\mathrm{/}(n-2)}$. Hölder in $x^{\mathrm{\prime }}$ with exponents $(n-1)\mathrm{/}(n-2)$ and $n-1$ bounds this by $(\int H{)}^{(n-2)\mathrm{/}(n-1)}(\int g_n{)}^{1\mathrm{/}(n-1)}$. The induction hypothesis [A1] bounds $\int H\le {\prod }_{i<n}(\int h_i{)}^{1\mathrm{/}(n-2)}$, while Tonelli gives $\int h_i=\int g_i$. Substitution yields the claimed product bound in dimension $n$.

Induction from the base case [step 1.1] through step 2.1 proves the inequality for every integer $n\ge 2$.

If $\mathrm{\Omega }=\mathrm{\varnothing }$, choose $c=0$. If $n=0$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, then $\mathrm{\Omega }={\mathbb{R}}^0$ is a singleton and its function $u$ is constant. Henceforth assume $n\ge 1$ and $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$. Let $B⋐\mathrm{\Omega }$ be any open ball and choose concentric balls $B⋐B^{\mathrm{\prime }}⋐B^{\mathrm{\prime }\mathrm{\prime }}⋐\mathrm{\Omega }$. By [F1] take $\chi \in C_c^{\mathrm{\infty }}(B^{\mathrm{\prime }\mathrm{\prime }})$ equal to one on $B^{\mathrm{\prime }}$, and extend $w:=\chi u$ by zero to ${\mathbb{R}}^n$. For sufficiently small $\epsilon$, every mollifier centered at a point of $B$ is supported in $B^{\mathrm{\prime }}$. For such a point $x$, differentiating the convolution gives ${\mathrm{\partial }}_i(w\ast {\rho }_{\epsilon })(x)=-{\int }_{B^{\mathrm{\prime }}}u(y){\mathrm{\partial }}_{y_i}{\rho }_{\epsilon }(x-y)dy=0$ by the distributional identity [F2]. Thus $w\ast {\rho }_{\epsilon }$ is constant on the convex ball $B$ by the one-dimensional fundamental theorem along each segment.

Since $w=u$ on $B^{\mathrm{\prime }}$, the approximate-identity property [F4] gives $w\ast {\rho }_{{\epsilon }_j}\to u$ in $L^1(B)$ for a sequence ${\epsilon }_j\downarrow 0$. Write each constant from step 1.1 as $c_j$. The $L^1(B)$ convergence makes $(c_j)$ Cauchy because $\mathrm{\mid }B\mathrm{\mid }\mathrm{\mid }{c}_j-c_k\mathrm{\mid }\le \mathrm{\parallel }{c}_j-u{\mathrm{\parallel }}_{L^1(B)}+\mathrm{\parallel }{c}_k-u{\mathrm{\parallel }}_{L^1(B)}$. Its limit $c_B$ satisfies $u=c_B$ almost everywhere on $B$.

If $\mathrm{\Omega }=\mathrm{\varnothing }$, any constant satisfies the conclusion. Otherwise fix one ball and let $c$ be its almost-everywhere value. Let $S$ be the set of points of $\mathrm{\Omega }$ having a neighborhood on which $u=c$ almost everywhere. It is nonempty and open. If $x\in \bar{S}\cap \mathrm{\Omega }$, choose a ball $B⋐\mathrm{\Omega }$ containing $x$ and a point of $S$; step 2.1 makes $u$ almost everywhere constant on $B$, and its positive-measure intersection with a neighborhood from $S$ forces that constant to be $c$. Thus $x\in S$, so $S$ is relatively closed. Connectedness [F5] gives $S=\mathrm{\Omega }$. By the countable rational base [F6], choose a countable cover of $\mathrm{\Omega }$ by rational balls contained in these neighborhoods; the union of their exceptional null sets is null, hence $u=c$ almost everywhere on all of $\mathrm{\Omega }$. The constants $c_j$ and $c_B$ may be complex; completeness of $\mathbb{C}$ justifies their limits. When $u$ is real almost everywhere on a nonempty domain, its a.e. constant $c$ is real because each ball has positive measure.

For $\delta >0$, tile ${\mathbb{R}}^n$ by the half-open cubes $Q=\delta (z+[0,1{)}^n)$, $z\in {\mathbb{Z}}^n$, and let $A_{\delta }v$ equal ${\delta }^{-n}{\int }_{Q}v$ on $Q$. The triangle inequality, followed by [F1] and the substitution $y=x+h$ using [F2], gives $\mathrm{\parallel }v-A_{\delta }v{\mathrm{\parallel }}_1\le {\delta }^{-n}{\sum }_Q{\int }_Q{\int }_Q\mathrm{\mid }v(x)-v(y)\mathrm{\mid }dydx\le {\delta }^{-n}{\int }_{[-\delta ,\delta {]}^n}\mathrm{\parallel }v(\cdot +h)-v{\mathrm{\parallel }}_{1}dh$. Thus, writing $\omega (r)={\sup }_{k,\mathrm{\mid }h\mathrm{\mid }\le r}\mathrm{\parallel }{v}_k(\cdot +h)-v_k{\mathrm{\parallel }}_1$, one has ${\sup }_k\mathrm{\parallel }{v}_k-A_{\delta }{v}_k{\mathrm{\parallel }}_1\le 2^n\omega (\sqrt{n}\delta )\to 0$ as $\delta \downarrow 0$.

For each $m\ge 1$, only finitely many cubes of side $2^{-m}$ meet a fixed ball containing the common supports; all other cell averages vanish. Enumerate all these cells at all levels by first increasing $m$ and then lexicographic order of their integer coordinates. Each cell-average coefficient $a_j(k)$ is a bounded real sequence, with bound $M$ divided by that cell's volume. Set $S_0=\mathbb{N}$ and initialize $k_0=0$. Recursively, at stage $m\ge 1$, partition the bounded ranges of the first $m$ coefficients into finitely many half-open dyadic bins of width $2^{-m}$, adding an endpoint bin if necessary. Among the finitely many vector bins containing infinitely many indices in $S_{m-1}$, take the lexicographically first; let $S_m$ be exactly that infinite set of indices. The finite pigeonhole principle guarantees existence and the specified order makes the rule unique. Then let $k_m$ be the least member of $S_m$ larger than $k_{m-1}$; such a member exists because $S_m$ is infinite. This deterministic successor rule and initial pair $(S_0,k_0)$ define the sequence recursively by [F3], with no dependent-choice instance. For fixed $j$ and $m\ge j$, all selected indices $k_q$ with $q\ge m$ lie in $S_m$, so their $j$th coefficients differ by at most $2^{-m}$. Consequently every coefficient converges and, at each fixed cell level $\mathrm{\ell }$, $(A_{2^{-\mathrm{\ell }}}{v}_{k_m}{)}_m$ is Cauchy in $L^1$, being a finite weighted sum of these coefficients.

Given $\epsilon >0$, choose a cell level $\mathrm{\ell }$ for which $2^n\omega (\sqrt{n}{2}^{-\mathrm{\ell }})<\epsilon \mathrm{/}3$, using step 1.1. For sufficiently large $a,b$, step 1.2 gives $\mathrm{\parallel }{A}_{2^{-\mathrm{\ell }}}{v}_{k_a}-A_{2^{-\mathrm{\ell }}}{v}_{k_b}{\mathrm{\parallel }}_1<\epsilon \mathrm{/}3$. The two approximation errors from step 1.1 then show $\mathrm{\parallel }{v}_{k_a}-v_{k_b}{\mathrm{\parallel }}_1<\epsilon$. Thus the subsequence is Cauchy.

For completeness at exactly the stated axiom strength, choose representatives of this countable sequence by [A1]. Select a further subsequence $g_j$ with $\mathrm{\parallel }{g}_{j+1}-g_j{\mathrm{\parallel }}_1<2^{-j}$, always choosing the least sufficiently late index, which exists by the Cauchy property. By [F1], ${\sum }_j\mathrm{\mid }{g}_{j+1}-g_j\mathrm{\mid }$ is integrable and finite almost everywhere. Hence $g_j$ has a measurable almost-everywhere limit $g$; moreover $\mathrm{\mid }g\mathrm{\mid }\le \mathrm{\mid }{g}_1\mathrm{\mid }+{\sum }_j\mathrm{\mid }{g}_{j+1}-g_j\mathrm{\mid }$, so $g\in L^1$, and $\mathrm{\parallel }g-g_j{\mathrm{\parallel }}_1\le {\sum }_{i\ge j}{2}^{-i}\to 0$. The original Cauchy subsequence also converges to this $L^1$ class by the triangle inequality. This proves the assertion using only [A1].

We first need the integral identity for Lipschitz interval functions without invoking the stronger published general absolutely continuous FTC. If $b$ is $L$-Lipschitz on $[a,d]$, extend it constantly past the endpoints. Its difference quotients are bounded by $L$ and converge to $b^{\mathrm{\prime }}$ almost everywhere by [F1]. For any $a\le s<t\le d$, dominated convergence gives ${\int }_s^t{b}^{\mathrm{\prime }}(r)dr={\lim }_{ϵ\downarrow 0}\frac{1}{ϵ}({\int }_t^{t+ϵ}b(r)dr-{\int }_s^{s+ϵ}b(r)dr)=b(t)-b(s)\mathrm{.}$ Coordinate difference quotients of $h$ are measurable; [F1] and [F2] show their limits $g_i$ exist almost everywhere, with $\mathrm{\mid }{g}_i\mathrm{\mid }\le L$. Define them to be zero on the exceptional sets. Applying the preceding identity and one-dimensional integration by parts on each line, then [F2], gives $\int h{\mathrm{\partial }}_i\eta =-\int g_i\eta$ for smooth compactly supported $\eta$. The same line identities hold for every coordinate line of $h$, outside the appropriate line-null exceptional set of derivatives.

For a cube $C$ of side $2r$ and a Lipschitz $w$, join $x$ to $y$ by the $m$ coordinate segments replacing their coordinates in order. Step 1.1 bounds $\mathrm{\mid }w(x)-w(y)\mathrm{\mid }$ by the sum of the line integrals of $\mathrm{\mid }{\mathrm{\partial }}_{i}w\mathrm{\mid }$. Integrate in $x,y\in C$. For coordinate $i$, the unused coordinates contribute $(2r{)}^{m-1}$, while the measure of endpoint pairs whose interval contains a fixed coordinate value is at most $(2r{)}^2$. Thus, after division by $\mathrm{\mid }C\mathrm{\mid }$, the triangle inequality yields ${\int }_C\mathrm{\mid }w-w_C\mathrm{\mid }\le 2r{\sum }_{i=1}^m{\int }_C\mathrm{\mid }{\mathrm{\partial }}_{i}w\mathrm{\mid },$ where $w_C$ is its mean. Outside a null set, $g=(g_i)$ has strong Lebesgue points: apply [F3] to its coordinates and to $\mathrm{\mid }g-q\mathrm{\mid }$ for every rational vector $q$, then approximate $g(x)$ by such $q$. The countable union of the exceptional null sets is null under Countable Choice. Ball differentiation also gives cube differentiation by the fixed volume comparison. At such a point $x$, set $p=g(x)$ and $w(y)=h(y)-h(x)-p\cdot (y-x)$. The displayed inequality gives ${\int }_{C(x,r)}\mathrm{\mid }w-w_C\mathrm{\mid }=o(r^{m+1})$. As $w(x)=0$ and $w$ has a fixed Lipschitz bound, a mean with $\mathrm{\mid }{w}_C\mathrm{\mid }\ge ϵr$ would give $\mathrm{\mid }w-w_C\mathrm{\mid }\ge ϵr\mathrm{/}2$ on a cube of side proportional to $r$ centered at $x$, contradicting this estimate. Thus $w_C=o(r)$. If $\mathrm{\mid }w(y)\mathrm{\mid }\ge ϵr$ at some $y$ in the half cube, the Lipschitz bound gives the same contradiction on a cube of side proportional to $r$ centered at $y$. Hence ${\sup }_{C(x,r\mathrm{/}2)}\mathrm{\mid }w\mathrm{\mid }=o(r)$, proving differentiability with derivative $p$. Difference quotients then give $\mathrm{\mid }p\mathrm{\mid }\le L$. Countably many rational boxes cover $Q$, so the assertion is global.

We prove the area identity first for bounded Borel base sets within one $L$-Lipschitz box. On the differentiability set, for any $0<ϵ<1\mathrm{/}4$, form countably many Borel pieces on each of which $Dh$ lies within $ϵ$ of a fixed rational vector $p$, the differentiability remainder is at most $ϵ\mathrm{\mid }z-y\mathrm{\mid }$ whenever $\mathrm{\mid }z-y\mathrm{\mid }<1\mathrm{/}k$, and the points lie in one cube of diameter less than $1\mathrm{/}k$. Such pieces cover the differentiability set: the remainder condition follows from differentiability, can be tested on countably many rational $z$ by continuity, and is measurable; use increasing integers $k$ and a rational-vector enumeration. Subtract earlier pieces to make them disjoint. For two points $y,z$ of a single piece $E$, set $A_{p}v=(v,p\cdot v)$. Then $\mathrm{\mid }F(z)-F(y)-A_p(z-y)\mathrm{\mid }\le 2ϵ\mathrm{\mid }z-y\mathrm{\mid }\le 2ϵ\mathrm{\mid }{A}_p(z-y)\mathrm{\mid }$. Consequently $F\circ A_p^{-1}:A_p(E)\to F(E)$ is at most $(1+2ϵ)$-Lipschitz, and its inverse is at most $(1-2ϵ{)}^{-1}$-Lipschitz. The sets $F(E)$ are Borel because $F$ is a homeomorphism from $Q$ onto its graph.

Choose an orthonormal coordinate system on the range plane of $A_p$. Its square coordinate matrix has determinant of absolute value $J_p=\sqrt{\det (A_p^T{A}_p)}=\sqrt{1+\mathrm{\mid }p{\mathrm{\mid }}^2}$. Thus [F4] gives ${\sigma }_m(A_p(E))=J_p\mathrm{\mid }E\mathrm{\mid }$. Apply [F4] to the two maps in step 3.1 to obtain $(1-2ϵ{)}^m{J}_p\mathrm{\mid }E\mathrm{\mid }\le {\sigma }_m(F(E))\le (1+2ϵ{)}^m{J}_p\mathrm{\mid }E\mathrm{\mid }\mathrm{.}$ The function $p\mapsto \sqrt{1+\mathrm{\mid }p{\mathrm{\mid }}^2}$ is $1$-Lipschitz, so $\mathrm{\mid }{J}_p-\sqrt{1+\mathrm{\mid }Dh(y){\mathrm{\mid }}^2}\mathrm{\mid }\le ϵ$ on $E$. The excluded base-null set has graph image of zero ${\sigma }_m$ by [F4], since $F$ is Lipschitz. Sum the displayed inequalities over the disjoint pieces, using $\mathrm{\mid }p\mathrm{\mid }\le L+ϵ$. The difference between ${\sigma }_m(F(B))$ and ${\int }_B\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}$ is bounded by $C_{m,L}ϵ\mathrm{\mid }B\mathrm{\mid }$. Letting $ϵ\downarrow 0$ proves equality. Exhaustion and countable localization extend it to arbitrary Borel $B\subseteq Q$; indicator functions, simple functions and monotone convergence prove the asserted integral formula.

For a compactly supported smooth scalar $b$ in the chart, vertical integration gives ${\int }_G{\mathrm{\partial }}_{z}b=-{\int }_{Q}b(y,h(y))dy$. For a horizontal coordinate put $B_i(y)={\int }_{h(y)}^{\mathrm{\infty }}b(y,z)dz$. This is Lipschitz and compactly supported in $Q$; step 2.1 and ordinary differentiation of the smooth integral give ${\mathrm{\partial }}_i{B}_i={\int }_{h(y)}^{\mathrm{\infty }}{\mathrm{\partial }}_{i}bdz-b(y,h(y))g_i$ almost everywhere. Its weak derivative integrates to zero by step 1.1 and [F2]. Apply these identities to the components of $\varphi$ and sum. Step 4.1 converts the resulting flux into the unit-normal formula. The graph measure is a graph pushforward of the locally bounded measurable density $\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}dy$; compact and open approximation in the base show it is locally Radon on the graph with its relative topology. The density is between $1$ and $\sqrt{1+L^2}$ on every local box, which also proves the null-set assertion.