The integral of the divergence of an integrable C1 field vanishes

Insert the cutoff: by [F2] with $r=1$, $R=2$ choose $\chi :{\mathbb{R}}^n\to [0,1]$ smooth with $\chi =1$ on ${\bar{B}}_1(0)$ and $\chi =0$ outside $B_2(0)$; for $R>0$ set ${\chi }_R(x):=\chi (x\mathrm{/}R)$, so that ${\chi }_R$ is $C^1$ with $0\le {\chi }_R\le 1$, ${\chi }_R=1$ on ${\bar{B}}_R(0)$ and ${\chi }_R=0$ outside $B_{2R}(0)$; the chain rule [F4] applied to the map $x\mapsto x\mathrm{/}R$ gives $D{\chi }_R(x)=R^{-1}(D\chi )(x\mathrm{/}R)$, so with $C_{\chi }:={\sup }_{{\mathbb{R}}^n}\mathrm{\mid }D\chi \mathrm{\mid }<\mathrm{\infty }$ by [F10] on ${\bar{B}}_2(0)$, since $D\chi =0$ off that ball one has $\mathrm{\mid }D{\chi }_R(x)\mathrm{\mid }\le C_{\chi }\mathrm{/}R$ for every $x$ and every $R>0$.

The cutoff divergence integrates to zero: fix $R>0$; the field ${\chi }_{R}F$ is $C^1$ on ${\mathbb{R}}^n$ and vanishes outside the compact set ${\bar{B}}_{2R}(0)$; if $n\ge 2$, apply [F1] on the open ball $\mathrm{\Omega }=B_{3R}(0)$, a bounded $C^1$ domain whose closure contains $\mathrm{supp}({\chi }_{R}F)$ in its interior, where the boundary term vanishes because ${\chi }_{R}F=0$ on $\mathrm{\partial }\mathrm{\Omega }$, so that ${\int }_{\mathrm{\Omega }}\mathrm{div}({\chi }_{R}F)d{\lambda }_n=0$ and hence, the integrand vanishing off $\mathrm{\Omega }$, also ${\int }_{{\mathbb{R}}^n}\mathrm{div}({\chi }_{R}F)d{\lambda }_n=0$; if $n=1$, put $g:={\chi }_{R}F\in C^1(\mathbb{R})$, so $g$ vanishes outside $(-2R,2R)$, and with $a:=-3R$, $b:=3R$ the derivative $g^{\mathrm{\prime }}$ is continuous on $[a,b]$, hence bounded and Darboux integrable, [F6] gives ${\int }_a^b{g}^{\mathrm{\prime }}=g(b)-g(a)=0$, [F7] makes this Darboux integral equal to the Riemann integral of $g^{\mathrm{\prime }}$ over $[a,b]$, [F8] applied on the box $Q=[a,b]$ makes that Riemann integral equal to the Lebesgue integral ${\int }_{[a,b]}{g}^{\mathrm{\prime }}d{\lambda }_1$, and $g^{\mathrm{\prime }}=0$ on $\mathbb{R}\setminus [a,b]$, so [F9] applied to the positive and negative parts gives ${\int }_{\mathbb{R}}{g}^{\mathrm{\prime }}d{\lambda }_1={\int }_{[a,b]}{g}^{\mathrm{\prime }}d{\lambda }_1=0$; in both cases ${\int }_{{\mathbb{R}}^n}\mathrm{div}({\chi }_{R}F)d{\lambda }_n=0$.

Expand and let $R\to \mathrm{\infty }$: by [F3] one has pointwise $\mathrm{div}({\chi }_{R}F)=⟨D{\chi }_R,F⟩+{\chi }_R\mathrm{div}F$, and each of the three functions lies in $L^1({\lambda }_n)$, the left side because it is continuous with compact support, ${\chi }_R\mathrm{div}F$ because $\mathrm{\mid }{\chi }_R\mathrm{div}F\mathrm{\mid }\le \mathrm{\mid }\mathrm{div}F\mathrm{\mid }$, and $⟨D{\chi }_R,F⟩$ because step 1.1 gives $\mathrm{\mid }⟨D{\chi }_R,F⟩\mathrm{\mid }\le (C_{\chi }\mathrm{/}R)\mathrm{\mid }F\mathrm{\mid }$; by linearity of the Lebesgue integral on $L^1({\lambda }_n)$, ${\int }_{{\mathbb{R}}^n}\mathrm{div}({\chi }_{R}F)d{\lambda }_n={\int }_{{\mathbb{R}}^n}{\chi }_R\mathrm{div}Fd{\lambda }_n+{\int }_{{\mathbb{R}}^n}⟨D{\chi }_R,F⟩d{\lambda }_n$ for every $R>0$, with left-hand side $0$ by step 2.1; as $R\to \mathrm{\infty }$ through positive integers (so $R\ge 1$), the functions ${\chi }_R\mathrm{div}F$ converge pointwise to $\mathrm{div}F$ and are dominated by $\mathrm{\mid }\mathrm{div}F\mathrm{\mid }\in L^1({\lambda }_n)$, while $⟨D{\chi }_R,F⟩$ converge pointwise to $0$ and are dominated by $C_{\chi }\mathrm{\mid }F\mathrm{\mid }\in L^1({\lambda }_n)$, so [F5] gives ${\int }_{{\mathbb{R}}^n}{\chi }_R\mathrm{div}Fd{\lambda }_n\to {\int }_{{\mathbb{R}}^n}\mathrm{div}Fd{\lambda }_n$ and ${\int }_{{\mathbb{R}}^n}⟨D{\chi }_R,F⟩d{\lambda }_n\to 0$; passing to the limit gives $0={\int }_{{\mathbb{R}}^n}\mathrm{div}Fd{\lambda }_n+0$, which is the claim.