Scaling the perimeter of a disk inclusion

Put $B=B(0,1)$ and first prove its flux formula directly from [F2]. Cover $\mathrm{\partial }B$ by finitely many graph neighborhoods, each expressing $B$ locally as an epigraph or subgraph of a smooth function, and add an interior open set compactly contained in $B$ to cover $\bar{B}$. Rigid changes of coordinates preserve volume, divergence and normal flux by [F3]–[F5], so the graph identities apply in these coordinates. Choose smooth nonnegative functions ${\eta }_i$ compactly supported in those sets and with ${\sum }_i{\eta }_i>0$ on $\bar{B}$: finitely many smaller balls cover this compact set, and ordinary Euclidean smooth bumps positive on those balls supply the functions. Normalize their sum on a neighborhood of $\bar{B}$, using one additional smooth cutoff equal to one near $\bar{B}$ and supported where that sum is positive. This gives functions ${\chi }_i$ subordinate to the same sets with ${\sum }_i{\chi }_i=1$ near $\bar{B}$. For any smooth compactly supported vector field $\psi$, apply [F2] to ${\chi }_i\psi$ in each boundary graph patch. Its divergence integral over $B$ is its outward flux over $\mathrm{\partial }B$ by [F4], because the graph domain and $B$ agree on the support. The interior piece has divergence integral zero: its support is compact inside $B$, so choose a horizontal line below that support and apply [F2] on the half-plane above that line. The field vanishes on the line and its half-plane integral equals its whole-space integral, which equals its integral over $B$. Summing the finitely many identities, the derivatives of the partition cancel since its sum is one near $\bar{B}$. Consequently ${\int }_B\mathrm{div}\psi ={\int }_{\mathrm{\partial }B}\psi \cdot {\nu }_{B}d{\mathcal{H}}^1$. This proves the required unit-disk identity without using a B-example or a stronger-choice divergence theorem.

Let $T(y)=x_0+ry$ and, for arbitrary $\varphi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^2;{\mathbb{R}}^2)$, put $\psi =\varphi \circ T$. By [F5], ${\mathrm{div}}_y\psi (y)=r{\mathrm{div}}_x\varphi (T(y))$. Applying [F3] to the positive and negative parts of $\mathrm{div}\varphi$, whose integrals are finite because $\varphi$ is smooth and compactly supported, gives ${\int }_E\mathrm{div}\varphi (x)dx=r^2{\int }_B\mathrm{div}\varphi (T(y))dy=r{\int }_B\mathrm{div}\psi (y)dy\mathrm{.}$ By step 1.1 this is $r{\int }_{\mathrm{\partial }B}\psi \cdot {\nu }_{B}d{\mathcal{H}}^1$. The circle parametrization in [F4] has outward normal $(\cos \theta ,\sin \theta )$ and arclength element $rd\theta$, so this last expression equals ${\int }_{\mathrm{\partial }E}\varphi \cdot {\nu }_{E}d{\mathcal{H}}^1$. Thus ${\int }_E\mathrm{div}\varphi ={\int }_{\mathrm{\partial }E}\varphi \cdot {\nu }_{E}d{\mathcal{H}}^1$ for every test field; [F1] yields $D{\chi }_E=-{\nu }_E{\mathcal{H}}^1{\mathrm{\mid }}_{\mathrm{\partial }E}$.

By [F4], the circle is parametrized with speed $r$, so ${\mathcal{H}}^1(\mathrm{\partial }E)=2\pi r$. For any Borel partition $(A_j)$ of $\mathrm{\partial }E$, [F1] and $\mathrm{\mid }{\nu }_E\mathrm{\mid }=1$ give ${\sum }_j\mathrm{\mid }D{\chi }_E(A_j)\mathrm{\mid }={\sum }_j\mathrm{\mid }{\int }_{A_j}{\nu }_{E}d{\mathcal{H}}^1\mathrm{\mid }\le 2\pi r$. For the reverse bound, partition the circle into $N$ equal arcs and their finitely many endpoints. The endpoints have zero arclength; on each arc, direct integration of $(\cos \theta ,\sin \theta )r$ gives a vector of norm $2r\sin (\pi \mathrm{/}N)$. Thus the partition sum is $2rN\sin (\pi \mathrm{/}N)$, which tends to $2\pi r$ as $N\to \mathrm{\infty }$. By the partition definition in [F1], $TV({\chi }_E;{\mathbb{R}}^2)=2\pi r$. This finite variation proves ${\chi }_E\in BV({\mathbb{R}}^2)$.

Because $\bar{E}\subset D$, the full derivative measure in step 1.2 is supported inside $D$. Hence $TV({\chi }_E;D)=\mathrm{\mid }D{\chi }_E\mathrm{\mid }(D)=\mathrm{\mid }D{\chi }_E\mathrm{\mid }({\mathbb{R}}^2)=2\pi r$.