Regularity of intersection of Lipschitz boundaries

Question

I am interested in the regularity of the "$n-1$ dimensional boundary" of the intersection of two Lipschitz boundaries, in particular I would like to know if this boundary always has zero $n-1$ dimensional measure.

For $i = 1, 2$ let $\Omega_i \subset \mathbb{R}^n$ open and disjoint with Lipschitz boundaries, i.e. for $x \in \partial \Omega_i$ there is a bi-Lipschitz map $$ \Phi_{i, x}: U(x) \to B $$ where $U(x)$ is an open neighbourhood of $x$, $B$ is the unit ball with center $0$ and $\Phi_{i, x}$ maps $U(x) \cap \Omega_i$ to the lower half space and $U(x) \cap \partial \Omega_i$ to the hyperplane $\{x^n = 0\}$. The case of interest is $\partial \Omega_1 \cap \partial \Omega_2$ non empty. Consider the sets $$ F_i = \{ x \in \partial \Omega_i \mid \exists \epsilon > 0: B_\epsilon(x) \cap cl(\Omega_i)^c \subset \Omega_j \} \\ G_i = \{ x \in \partial \Omega_i \mid \exists \epsilon > 0: B_\epsilon(x) \cap cl(\Omega_i)^c \subset cl(\Omega_j)^c \} \\ N_i = \partial \Omega_i \setminus (F_i \cup G_i) $$ where $j \neq i$. Now, intuitively $N_i$ is an $n-1$ dimensional boundary. Indeed, we see that $$ F_i = \{ x \in \partial \Omega_i \cap \partial \Omega_j \mid \exists \epsilon > 0: B_\epsilon(x) \cap (\partial \Omega_i \cup \partial \Omega_j) \subset \partial \Omega_i \cap \Omega_j\} $$ and $N_i = (\partial \Omega_i \cap \partial \Omega_j) \setminus F_i$. Thus $F_i$ is the interior of $\partial \Omega_i \cap \Omega_j$ in the subspace topology of $\partial \Omega_i \cup \partial \Omega_j$ and $N_i$ is the associated boundary. Does $N_i$ have surface measure zero?

Edit: fedja gave a counterexample in the comments. What about the situation in which the additional assumption that $\overline{\Omega_1} \cup \overline{\Omega_2}$ has Lipschitz boundary itself is made? I can't come up with a modification of the example which works under this additional assumption. Seems like this assumption always gives a coordinate system in which a neighbourhood of $N$ can be written as a Lipschitz graph and which is "linearly independent" of the {t = 0} plane implying that $U_+$ has Lipschitz boundary, hence $N$ has zero 2d measure. I can't make this argument precise in the general case though.

In general boundaries do not have measure zero, however I wonder if the Lipschitz regularity of $\partial \Omega_i$ somehow induces enough regularity on the intersection $\partial \Omega_1 \cap \partial \Omega_2$ to imply this. I was trying to find a cone in the $\{x^n = 0\}$ plane such that the preimage under $\Phi_{i, x}$ is always contained either in $F_i$ or $G_i$, which would imply the desired result. If the result is not true in general, what other useful conditions are there besides the existence of such a cone?

Background: I consider broken Sobolev spaces, i.e. functions which are Sobolev functions in each of the $\Omega_i$ but not necessarily in the whole space. To examine the jump of their traces at the interface I would like to work locally as if the interface was the interface of two polygons (up to a set of zero surface measure). Usually this is considered for polygonal cells, for which you can find the desired cone (anyway, $N_i$ will be the union of $n-2$ dimensional affine spaces, hence it has $n-1$ dimensional measure zero) but I would like to know if this works in a more general setting.

Answer 1

Ah, now it makes sense. Your approach is perfectly sound and you would figure out the technicalities youself in finite time, but to spare you some effort, I'll still post a sketch of the solution. Take any point on the intersection of the boundaries. Then in some neighborhood of that point, at every point of $N$, you can draw 3 pairwise disjoint open cones: one to each of $\Omega$'s and one to the complement to the union. After some general mumbo-jumbo about Rademacher's theorem, density of rationals and countable unions, you can just assume that the shape and direction of each of three cones is fixed and their axial vectors are linearly independent plus all points of $N$ lie on the graph $x_d=f(x'), x'\in \mathbb R^{d-1}$ of a Lipschitz function whose Lipschitz constant is much smaller than the aperture of each cone. Two of 3 cones (placed at the origin) have their axial vectors looking to the same side of the plane. Take any two points $a,b$ in them with the same $x_d$. Then for any $x,y\in N$ with $x'-y'$ parallel (or even almost parallel) to $a-b$, the distance $|x'-y'|$ is bounded from below, so the projection of $N$ to $\mathbb R^{d-1}$ is a countable union of (Lipschitz, if you want, but it is useless because you lose more on Rademacher) graphs. The rest should be clear.