minimal surfaces.

From [@Giu Chapter 10] it follows that for $t \in \mathbb{R}$ the set $\partial \{ u > t \}$ is a sum of minimal surfaces $S_{t,i}$ and $\partial \{ u > t \} = \partial^* \{ u > t \}$. Obviously $\partial \{ u > t \}$ $($boundary in topology of $\Omega)$ is closed in $\Omega$, so $\partial S_{t,i} \cap \Omega = \emptyset$ $($boundary in topology of $\partial \{ u > t \})$; suppose otherwise. Let $x \in \partial S_{t,i} \cap \Omega$. Then in $x$ the blow-up theorem does not hold, so $x \notin \partial \{ u > t \}$, contradiction.

Now suppose that $S_{t,i}$ and $S_{t,j}$ are not disjoint in $\Omega$. Then from the Proposition \[stw:sternbergpowmin\] applied to $E_1 = E_2 = \{ u > t \}$ we get $S_{t,i} = S_{t,j}$.

Let $E_1 \subset E_2$ and suppose that $E_1$ and $E_2$ are sets of locally bounded perimeter and let $x \in \partial^{*} E_1 \cap \partial^{*} E_2$. Then $\nu_{E_1}(x) = \nu_{E_2}(x)$.

We are going to use the blow-up theorem (Theorem \[tw:blowup\]). First notice that the inclusion $E_1 \subset E_2$ implies

$$E_1^r = \{ y \in \mathbb{R}^N: r(y-x) + x \in E_1 \} \subset \{ y \in \mathbb{R}^N: 