Integration of Forms and the General Stokes Theorem

If $K=\mathrm{\varnothing }$, take the empty neighborhood and empty index set. Otherwise cover $K$ by open sets $V$ each meeting only finitely many $C_i$. Ambient compactness gives a finite subcover $V_1,\dots ,V_m$. Their union $V$ meets only a finite set $J$ of indices.

Let $J_0=\{i:C_i\cap K\mathrm{\ne }\mathrm{\varnothing }\}\subseteq J$. Since each $C_i$ is closed, $V\setminus {\bigcup }_{i\in J\setminus J_0}{C}_i$ is open, contains $K$, and misses every $C_i$ for $i\mathrm{\notin }{J}_0$.

Apply this to $C_i=\mathrm{supp}{\rho }_i$ and $K=\mathrm{supp}\omega$. Outside $K$, $\omega =0$; if $i\mathrm{\notin }{J}_0$, the two supports are disjoint, so ${\rho }_i\omega =0$. The argument includes a singleton support and the zero form.

By compact support, $f$ vanishes on $U\setminus K$ and is bounded on $K$. A point at an artificial edge of $U$ lies outside the closed Euclidean compact set $K$; a neighborhood missing $K$ has zero extended coefficient. Inside $U$ the coefficient is smooth up to the genuine face. Thus $\tilde{f}$ is smooth off that face and supported in $K$. This includes $f=0$.

Auxiliary extensions can be constructed near $K$: choose finitely many extension neighborhoods, smooth Euclidean bump functions supported there and positive on smaller neighborhoods covering $K$, and divide by their sum near $K$. The weighted extensions agree with $f$ on the half-space near $K$. Cut off on a smaller neighborhood of $K$ to obtain a compactly supported smooth Euclidean function there. The cutoff is one near $K$; its restriction to the half-space, extended by zero at artificial edges, is $f$. Such cutoffs follow by applying the closed-neighborhood extension lemma to the constant function one and, if needed, composing with a smooth nonnegative function.

Choose $R>0$ so $K\subset (-R,R{)}^n$. Partition the first $n-1$ coordinates of $[-R,R{]}^{n-1}$ into at most $(2R\mathrm{/}\delta +2{)}^{n-1}$ cells of side at most $\delta$. Center a closed cube of side $2\delta$ on each face cell. They cover the face in the bounding cube and have total volume at most $2^n(2R+2\delta {)}^{n-1}\delta$. This tends to zero; reciprocal integers give arbitrarily small $\delta$. For $n=1$ this is a single interval of length $2\delta$.

The discontinuities of $\tilde{f}$ in $[-R,R{]}^n$ lie in that content-zero, hence null, face. Boundedness and the null-discontinuity criterion imply Riemann integrability. The criterion is used in its sufficient direction only.

The compact-support integral lemma makes the value independent of any larger bounding rectangle. Every auxiliary extension after restriction to $H^n$ gives the same zero-extended function, hence the same integral. With an interior chart there is no genuine face, so the artificial-edge argument proves smoothness everywhere.

The face maps into the face and the interior into the interior. The half-space chain rule applied to $G^{-1}G$ shows that $D{G}_p$ is invertible. Derivatives do not depend on the smooth extensions chosen near $p$.

Write the last component of a local extension as $h(x^{\mathrm{\prime }},t)$. On a small face disk $h(x^{\mathrm{\prime }},0)=0$, so its tangential derivatives vanish. Since $h(x^{\mathrm{\prime }},t)>0$ for small $t>0$, ${\mathrm{\partial }}_{t}h(p)\ge 0$. Invertibility and the zero tangential entries in the last row exclude zero, so ${\mathrm{\partial }}_{t}h(p)>0$.

After shrinking to a product neighborhood, continuity makes ${\mathrm{\partial }}_{t}h$ positive there. The one-variable fundamental theorem gives $h(x^{\mathrm{\prime }},t)={\int }_0^t{\mathrm{\partial }}_{t}h(x^{\mathrm{\prime }},s)ds=ta(x^{\mathrm{\prime }},t)$ with $a>0$, also for negative $t$. Hence $h$ has exactly the sign of $t$.

Apply the Euclidean inverse theorem to the extension and shrink its inverse neighborhoods inside that product neighborhood. The inverse is smooth: its derivative is the inverse derivative matrix composed with the inverse map, and repeated differentiation bootstraps the stated $C^1$ inverse to every finite order. The sign identity gives both inclusions of each side equality. For $n=1$ the tangential row is empty and the same positive derivative argument applies.

For $n\ge 1$, let $G=\psi {\varphi }^{-1}$ on the overlap, and write the coefficients as $f_x=(f_y\circ G)\det DG$. Pullback and wedge functoriality give this determinant formula. The chart signs obey ${\sigma }_{\varphi }\det DG={\sigma }_{\psi }\mathrm{\mid }\det DG\mathrm{\mid }$.

Cover the compact support by overlap neighborhoods on which the transition is a Euclidean diffeomorphism, using the side-preserving extension lemma at face points and the transition itself at interior points. A subordinate smooth partition yields finitely many nonzero localized forms with compact support in those neighborhoods. The partition existence uses ${\mathrm{A}\mathrm{C}}_{\omega }$.

For each piece choose the extension neighborhoods large enough to contain its compact coordinate support. Its zero-extended target coefficient is compactly supported Riemann integrable by the chart-integral definition. The side-preserving extension carries its zero extension to the corresponding source zero extension, including zero values on the negative side. Apply compact-support Euclidean change of variables on the open Euclidean extension domain; its injectivity, invertible derivative, and target-support containment all hold. Multiply the equality by ${\sigma }_{\psi }$ and use the sign identity to identify the signed source integral.

Add the finitely many piece equalities using linearity of the underlying Riemann integral. If the support is empty every coefficient is zero. For $n=0$ a nonempty connected chart is the same single point in either description, and both values are $\epsilon (p)\omega (p)$. Thus all cases agree.

Let $({\rho }_i)$ and $({\tau }_j)$ be two subordinate partitions. Near $K=\mathrm{supp}\omega$ only finitely many indices from either family occur. Hence ${\rho }_i\omega ={\sum }_j{\rho }_i{\tau }_j\omega$ and ${\tau }_j\omega ={\sum }_i{\rho }_i{\tau }_j\omega$ are finite identities, including when $K$ is empty.

The support of ${\rho }_i{\tau }_j\omega$ is compact and contained in the intersection of its two chart domains. Its integral can therefore be computed in either chart with the same value. By linearity of chart integrals, the two original sums both equal ${\sum }_{i,j}I({\rho }_i{\tau }_j\omega )$. This also proves invariance under refinement.

For locality take the charts near $K$ inside $U$ and complete their cover by $M\setminus K$; terms supported in the latter vanish. Equivalently the same product-partition argument compares a partition on $U$ to one on $M$ near $K$. In dimension zero both sides are the same finite signed sum over $K$, including individual points and empty sums.

Choose a common chart partition for the compact union of the two supports. In positive dimension, each chart coefficient for $a\omega +b\eta$ is the corresponding linear combination. Riemann linearity, followed by summation of finitely many terms, gives the first formula. Empty supports and zero scalars cause no exception.

Manifolds have connected small ball or half-ball neighborhoods. Every connected component is therefore open. Its components form an open cover, so a compact support meets only finitely many of them. Choose the chart cover inside the components, group the finite sum accordingly, and use locality.

In dimension zero use the finite sums ${\sum }_p\epsilon (p)\omega (p)$; both distributivity and grouping are identities of finite sums. A singleton contributes its one signed value.

Compute both integrals with the same chart cover and partition, as independence permits. In positive dimension each chart sign changes from $\sigma$ to $-\sigma$, with its coefficient and Riemann integral unchanged.

In dimension zero each point sign changes from $\epsilon$ to $-\epsilon$. Factoring $-1$ out of either finite sum proves the formula; for empty support or the zero form it reads $0=-0$.

In a signed chart, nonnegativity means ${\sigma }_{\varphi }f\ge 0$. Multiplying by nonnegative partition weights and using Riemann monotonicity shows every chart contribution is nonnegative, so their finite sum is nonnegative.

If $n\ge 1$ and ${\omega }_p\mathrm{\ne }0$, some partition weight is positive at $p$. Its signed coefficient is continuous and positive there, hence at least $c>0$ on a sufficiently small rectangle, or on a half-rectangle at a face. Inside this neighborhood choose a nondegenerate rectangle of positive volume; monotonicity and rectangle additivity bound that chart integral below by $c$ times its positive volume. The other terms are nonnegative.

For $n=0$ every summand $\epsilon (p)\omega (p)$ is nonnegative and a nonzero form has a strictly positive summand. The zero form and the empty manifold give zero. These observations prove all assertions.

The support of $F^{\ast }\omega$ is $F^{-1}(\mathrm{supp}\omega )$ because the tangent maps are isomorphisms. It is compact, being the continuous image of the compact support under the inverse homeomorphism. Pull back the target partition and use the charts ${\varphi }_i\circ F$; these remain subordinate and locally finite.

If orientation is preserved, the corresponding source and target charts have the same sign and the same localized coefficient: $({\varphi }_{i}F{)}^{-1}=F^{-1}{\varphi }_i^{-1}$ and functoriality cancels the pullbacks. Thus their finite chart sums agree. Choice independence makes this the asserted intrinsic equality.

For global reversal, reverse the source orientation to apply the preceding equality and then negate its integral. More generally the sign is locally constant because a continuous nonzero determinant has constant sign on each connected component; apply that argument componentwise. In dimension zero the diffeomorphism is a bijection of finite supports with the corresponding point signs; empty support and zero forms give zero.

The specified embedding is smooth, so the pointwise formula $j^{\ast }\omega (v_1,\dots ,v_k)=\omega (dj{v}_1,\dots ,dj{v}_k)$ defines a smooth top form on the oriented manifold $S$. Its assumed compact support makes its intrinsic integral available. This works for empty $S$, the zero form, and $k=0$.

The pointwise pullback formula gives $(jF{)}^{\ast }\omega =F^{\ast }(j^{\ast }\omega )$. Its support is compact because $F$ is a diffeomorphism. Change of variables gives ${\int }_T{F}^{\ast }(j^{\ast }\omega )={\int }_S{j}^{\ast }\omega$. A nonproper inclusion does not supply compact support by itself; that condition was explicitly assumed.

First record boundary control. Compactness and continuity give ${\bar{W}}_i=F_i({\bar{D}}_i)$ and ${\mathrm{\partial }}_M{W}_i\subset F_i(\mathrm{\partial }{D}_i)$: a limit of interior image points has a convergent parameter subsequence, and an interior parameter limit has image in $W_i$. Each $\mathrm{\partial }{D}_i$ is compact of content zero. Cover it by finitely many parameter neighborhoods with smooth coordinate extensions. Intersect smaller closed neighborhoods with $\mathrm{\partial }{D}_i$ and apply the $C^1$ null-image lemma on each extension domain. Thus $F_i(\mathrm{\partial }{D}_i)$ is content zero in every fixed relatively compact target chart, after finite localization. No derivative rank condition is used here.

By a finite chart partition of the compact support and linearity, it suffices to consider a form supported compactly inside a small chart $U$ whose coordinate domain $Y$ is a bounded rectangle or half-rectangle and whose chart extends past its artificial edges. Such charts come from restricting a larger chart; the Euclidean boundary of $Y$ has content zero. Put $C_i=\varphi (U\cap W_i)$. The boundaries of the bounded $C_i$ lie in $\mathrm{\partial }Y$ together with the chart images of ${\mathrm{\partial }}_M{W}_i$, so $C_i$ are Jordan measurable. The localized coefficient $f$ is bounded, zero near artificial edges, and Riemann integrable, including the genuine face.

For that localized coefficient, $f=0$ outside ${\bigcup }_i{\bar{C}}_i$. The disjoint $C_i$ overlap only on null boundaries after closure. Apply finite almost-partition additivity to the pieces $C_i$ and $Y\setminus {\bigcup }_i{C}_i$ (whose integral is zero since $f$ vanishes there except on those boundaries). Consequently the signed chart integral is ${\sigma }_{\varphi }{\sum }_i{\int }_{C_i}f$.

Fix $i$ and write $g=\varphi F_i$ on $A_i=F_i^{-1}(U\cap W_i)\subset D_i$. This is a diffeomorphism onto $C_i$. To justify substitution despite possible singularities at parameter boundary, let $h$ be the coefficient of $F_i^{\ast }\omega$ on $D_i$; it extends continuously to the compact ${\bar{D}}_i$ and is bounded, say by $B$. Choose a finite union $K\subset D_i$ of grid cubes, with disjoint interiors, covering all but a collar of $\mathrm{\partial }{D}_i$ of arbitrarily small volume. Images of that collar have arbitrarily small chart volume as well: finitely many smooth coordinate extensions have bounded derivatives and are Lipschitz on smaller convex neighborhoods; a cube of side $\delta$ maps into a cube of side at most $c\delta$, so total covering volume increases by at most a fixed factor. Such collars exist because $\mathrm{\partial }{D}_i$ has content zero.

On the compact part $K$, the nonzero support of $h$ lies in a compact subset of $A_i$, since the localized form is supported inside $U$. Subdivide or cover this compact part by finitely many cubes compactly contained in $A_i$, splitting overlaps along their faces. On each such compact Jordan piece, the published substitution theorem applies to $g$: it is injective, $C^1$, and has invertible derivative on the surrounding open subset of $A_i$. Pieces where the form vanishes contribute zero. Add these equalities. The omitted integrals on the parameter side are bounded by $B$ times collar volume; on the image side they are bounded by $\sup \mathrm{\mid }f\mathrm{\mid }$ times the image-collar covering volume. Let those bounds tend to zero. This proves ${\int }_{D_i}h={\sigma }_{\varphi }{\int }_{C_i}f$, with the sign supplied by orientation preservation.

Sum over $i$ and then over the finite chart localization. Empty support gives only zero coefficients; for $n=1$ the same collar estimate uses intervals and point boundaries. Degenerate Jacobians at boundary points are harmless because substitution was used only on compact subsets of the diffeomorphism domains.

Fix a basis $e$ (the empty basis when $n=0$). Every basis is uniquely $eA$ for $A\in GL(n,\mathbb{R})$. A density is determined by $c=\delta (e)$ since $\delta (eA)=c\mathrm{\mid }\det A\mathrm{\mid }$. Conversely this formula, with zero on dependent tuples for $n\ge 1$, satisfies the required transformation law by multiplicativity of determinants.

The formula is linear in $c$, so evaluation and its displayed inverse are linear bijections. Since $\mathrm{\mid }\det A\mathrm{\mid }>0$, positivity is equivalent to $c>0$ and nonnegativity to $c\ge 0$, independently of the chosen basis. Thus the nonzero nonnegative densities are exactly one ray. For $n=0$ this says precisely that scalars form $\mathbb{R}$.