A discontinuous inclusion in the planar BV class

The disk $B(x_0,1\mathrm{/}16)$ is contained in $D^{\mathrm{\prime }}$ because $1\mathrm{/}16<1\mathrm{/}8$, and $\gamma =1$ on $D\setminus D^{\mathrm{\prime }}$. Almost everywhere its only values are $1$ and $3\mathrm{/}2$, so $1\mathrm{/}2\le \gamma \le 3\mathrm{/}2$. Thus the support and ellipticity conditions in [F1] hold.

Since $\mathrm{\mid }D\mathrm{\mid }=\pi \mathrm{/}16$ and $\mathrm{\mid }B(x_0,1\mathrm{/}16)\mathrm{\mid }=\pi \mathrm{/}256$, the integral satisfies $\mathrm{\parallel }\gamma {\mathrm{\parallel }}_{L^1(D)}=\pi \mathrm{/}16+\pi \mathrm{/}512=33\pi \mathrm{/}512<1$. By [F2] and homogeneity of distributional variation, $TV(\gamma ;D)\le \frac{1}{2}(2\pi \mathrm{/}16)=\pi \mathrm{/}16<1$; restriction to $D$ cannot increase the full-space variation. Therefore the BV norm is at most $1$, so [F1] gives $\gamma \in X_2(1)$.

At every point of the circle $\mathrm{\partial }B(x_0,1\mathrm{/}16)$ lying inside $D$, every neighborhood meets the interior disk and its exterior in sets of positive measure. The function equals $3\mathrm{/}2$ almost everywhere on the former and $1$ almost everywhere on the latter. A continuous representative would be constant $3\mathrm{/}2$ on the interior and constant $1$ on the exterior by continuity and almost-everywhere equality, contradicting continuity across the circle. Hence the conductivity class is discontinuous.