A smooth-curve jump has perimeter total variation

Extend ${\mathrm{\partial }}_1{\varphi }_1$ by zero outside $B$. It is integrable because it is bounded and supported in a finite-measure disk [F5]. Fubini [F4] expresses its integral as the integral over $y\in (-1,1)$ of the slice integrals over $[-a(y),a(y)]$, where $a(y)=\sqrt{1-y^2}$. Each slice integrand is continuous, so [F6] identifies its Lebesgue integral with its Riemann integral, and [F7] evaluates that integral as ${\varphi }_1(a(y),y)-{\varphi }_1(-a(y),y)$. In the right graph $x=a(y)$, [F8] gives $d{\mathcal{H}}^1=dy\mathrm{/}a(y)$ and ${\nu }_1=a(y)$; in the left graph $x=-a(y)$, it gives $d{\mathcal{H}}^1=dy\mathrm{/}a(y)$ and ${\nu }_1=-a(y)$. Thus the slice difference equals ${\int }_{\mathrm{\partial }B}{\varphi }_1{\nu }_{1}d{\mathcal{H}}^1$. The same argument with the coordinates exchanged gives the second component, hence ${\int }_B\mathrm{div}\varphi dx={\int }_{\mathrm{\partial }B}\varphi \cdot \nu d{\mathcal{H}}^1$.

By the distributional derivative convention [F1], step 1.1 yields $D{\chi }_B=-\nu {\mathcal{H}}^1{\mathrm{\mid }}_{\mathrm{\partial }B}$. The circle has finite arclength, so this is a finite vector Radon measure and ${\chi }_B\in BV({\mathbb{R}}^2)$.

The parametrization $\theta \mapsto (\cos \theta ,\sin \theta )$ has speed one on $[0,2\pi ]$, so ${\mathcal{H}}^1(\mathrm{\partial }B)=2\pi$. For every admissible test field, step 1.1 gives ${\int }_B\mathrm{div}\varphi dx={\int }_{\mathrm{\partial }B}\varphi \cdot \nu d{\mathcal{H}}^1\le 2\pi \mathrm{.}$ Choose a smooth radial cutoff $q$ supported in $(1\mathrm{/}2,3\mathrm{/}2)$, with $0\le q\le 1$ and $q(1)=1$, and set $\varphi (x)=q(\mathrm{\mid }x\mathrm{\mid })x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ on that annulus and zero outside. By [F3] this is smooth, compactly supported and has norm at most one; on $\mathrm{\partial }B$ it equals $\nu$. Its flux is $2\pi$. The supremum in [F2] is therefore exactly $2\pi$, proving the asserted total variation and perimeter.