r(y-x) + x \in E_2 \} = E_2^r.$$

We keep the same notation as in Theorem \[tw:blowup\] and use it to obtain

$$\chi_{H_1^-(x)} \leftarrow \chi_{E_1^r} \leq \chi_{E_1^r} \rightarrow \chi_{H_2^-(x)},$$

where the convergence holds in $L^1_{loc}$ topology. Thus $H_1^-(x) = H_2^-(x)$, so $\nu_{E_2}(x) = \nu_{E_2}(x)$.

\[stw:zbiorskokow\] For $u \in BV(\Omega)$ the structure of its jump set is as follows:

$$J_u = \bigcup_{s,t \in \mathbb{Q}; s \neq t} (\partial^{*} \{ u > s \} \cap \partial^{*} \{ u > t \}).$$

Let $x \in J_u$. By definition of $J_u$ the normal vector at $x$ is well defined. The same applies to the trace values from both sides: let us denote them by $u^{-} (x) < u^{+} (x)$. But then there exist $s,t \in \mathbb{Q}$ such that $u^{-} (x) < s < t < u^{+} (x)$, so $x \in \partial^{*} \{ u > s \} \cap \partial^{*} \{ u > t \}$.

On the other hand, let $x \in \partial^{*} \{ u > s \} \cap \partial^{*} \{ u > t \}$. From the previous proposition the normal vectors coincide, so the normal at $x$ does not depend on $t$ and we may define traces from both sides as

$$u^{+}(x) = \sup \{ 