Zero-integral compactly supported top forms on Euclidean space have compactly supported primitives

By [F9], the increasing open cubes have a finite subfamily covering $\mathrm{supp}f$; take $R>1$ larger than the largest radius in such a finite family. Thus $f$ vanishes outside a compact box strictly inside $[-R,R{]}^n$. By [F11] and [F12], the given integral is the ordinary integral of $f$ on that rectangle, and all sections used below are integrable. If $n=1$, put $a(x)={\int }_{-R}^{x}f(t)dt$ using oriented interval integrals. The fundamental theorem gives $a^{\mathrm{\prime }}=f$; repeated differentiation makes $a$ smooth. It vanishes for $x\le -R$, because the integrand is zero there, and for $x\ge R$, because the total integral is zero. Hence $\eta =a$ has compact support by [F8] and $d\eta =fdx$.

Fix one bump $\rho$ from [F2]. By [F3], $1\le c:={\int }_{-1}^1\rho (t)dt\le 2,$ since $\rho =1$ on the middle interval of length one and is between zero and one elsewhere. Integrability follows from [F11]. Put $b=\rho \mathrm{/}c$. It is smooth, has compact support in $(-1,1)$ and has integral one. This normalization uses one bump and an explicit positive integral bound, not a global positivity theorem or an infinite choice.

Let $n\ge 2$, write $x^{\mathrm{\prime }}=(x_1,\dots ,x_{n-1})$, $t=x_n$, and define $f_0(x^{\mathrm{\prime }})={\int }_{-R}^{R}f(x^{\mathrm{\prime }},s)ds,g(x^{\mathrm{\prime }},t)=f(x^{\mathrm{\prime }},t)-f_0(x^{\mathrm{\prime }})b(t),h(x^{\mathrm{\prime }},t)={\int }_{-R}^{t}g(x^{\mathrm{\prime }},s)ds\mathrm{.}$ The function $f_0$ is smooth: repeatedly apply [F4] in each one parameter coordinate on smaller closed parameter rectangles. Each resulting derivative is the integral of the corresponding derivative of $f$. Joint continuity follows from [F10] and [F3], since the change of that integral is bounded by $2R$ times the uniform change of the integrand. Hence $g$ is smooth. For joint smoothness of $h$, use $h(x^{\mathrm{\prime }},t)=(t+R){\int }_0^{1}g(x^{\mathrm{\prime }},-R+u(t+R))du\mathrm{.}$ This formula is valid also at $t=-R$ and for $t<-R$, by [F13] for $t\mathrm{\ne }-R$ (reverse the interval when $t<-R$); at $t=-R$ both sides are zero. On any compact parameter neighbourhood the integrand and all parameter derivatives are continuous on its product with $[0,1]$; repeated [F4], [F10] and the same bound prove all joint derivatives continuous. Finally [F5] gives ${\mathrm{\partial }}_{t}h=g$.

The function $f_0$ vanishes outside $[-R,R{]}^{n-1}$ and has compact support by [F8]. For every $x^{\mathrm{\prime }}$, ${\int }_{-R}^{R}g(x^{\mathrm{\prime }},s)ds=f_0(x^{\mathrm{\prime }})-f_0(x^{\mathrm{\prime }}){\int }_{-R}^{R}b(s)ds=0,$ since $R>1$ and $b$ is supported in $(-1,1)$. Therefore $h$ vanishes when $t\le -R$ or $t\ge R$. It also vanishes outside the stated $x^{\mathrm{\prime }}$ box, because both $f$ and $f_0$ vanish there. Thus $h$ has closed support inside $[-R,R{]}^n$ and is compactly supported by [F8]. By [F6] with all smooth sections and by [F12], ${\int }_{{\mathbb{R}}^{n-1}}{f}_{0}d{x}^{\mathrm{\prime }}={\int }_{{\mathbb{R}}^n}fdx=0.$

Apply the induction hypothesis in dimension $n-1$ to the compactly supported top form $f_{0}d{x}_1\wedge \cdots \wedge d{x}_{n-1}$. It gives a compactly supported $(n-2)$-form $\gamma$ with $d\gamma =f_{0}d{x}_1\wedge \cdots \wedge d{x}_{n-1}$. Let $\pi :{\mathbb{R}}^n\to {\mathbb{R}}^{n-1}$ be projection and put $\eta =(-1{)}^{n-1}hd{x}_1\wedge \cdots \wedge d{x}_{n-1}+{\pi }^{\ast }\gamma \wedge b(t)dt\mathrm{.}$ For the first summand, every $x^{\mathrm{\prime }}$ derivative repeats a differential and vanishes; moving $dt$ past $n-1$ factors cancels $(-1{)}^{n-1}$, so its derivative is $gd{x}_1\wedge \cdots \wedge d{x}_n$. For the second, [F7] gives ${\pi }^{\ast }d\gamma \wedge bdt$, since $d(bdt)=b^{\mathrm{\prime }}dt\wedge dt=0$. Its coefficient is $f_{0}b$. Thus $d\eta =(g+f_{0}b)d{x}_1\wedge \cdots \wedge d{x}_n=\omega$.

The first summand of $\eta$ is supported in $[-R,R{]}^n$ by step 3.1. The second is supported in $\mathrm{supp}\gamma \times [-1,1]$. By [F9] the first factor is bounded; the product support is closed and bounded, so [F8] makes it compact. Their finite union is bounded, and the closed support of the sum lies in it, again compact by [F8]. This completes the induction with an actual compact primitive; projection pullback by itself was not claimed to preserve compact support.

At $n=0$, [F1] and [F12] identify the integral on the standard oriented point with its scalar value; zero integral forces zero, the derivative of the zero negative-degree element. For zero input the construction gives $f_0=g=h=0$ and one may take $\gamma =0$, hence $\eta =0$. The interval endpoints, zero fibres and vanishing support were checked in steps 1.1 and 3.1. The induction chooses one normalized bump and, at each of finitely many dimensions for a given input, one previously established primitive. There is no choice of primitives over an infinite family and no AC.