Recalling (A.4) and (A.5), we have that $p \in \tilde{\Pi}_1(x_0, r_1, r_2)$. Moreover, by construction, we further have $\delta(p, q) = |p - q| \in [r_1, r_2]$. Therefore, for all $e \in S(x_0)$ and all $q \in [x_0 - \varkappa e, x_0 + \varkappa e] \cap \text{star}(x_0)$, there exists $p \in \tilde{\Pi}_1(x_0, r_1, r_2)$ such that $r_1 \le \delta(p, q) \le r_2$. Consequently,

$$ \bigcup_{e \in S(x_0)} [x_0 - \varkappa e, x_0 + \varkappa e] \cap \text{star}(x_0) \subset \tilde{\Pi}_2(x_0, r_1, r_2). $$

But since $S(x_0)$ contains a half-sphere, the left-hand side in the above equation is nothing but $\overline{B_\varkappa(x_0)} \cap \text{star}(x_0)$. Hence, we have that

$$ (A.6) \qquad \overline{B_\varkappa(x_0)} \cap \text{star}(x_0) \subset \tilde{\Pi}_2(x_0, r_1, r_2). $$

Let us now prove that $\overline{B_\varkappa(x_0)} \cap \Omega \setminus \text{star}(x_0) \subset \tilde{\Pi}_2(x_0, r_1, r_2)$. We may suppose, without loss of generality, that $\overline{B_\varkappa(x_0)} \cap \Omega \setminus \text{star}(x_0) \ne \emptyset$, since otherwise there is nothing to prove. So, we have, in particular, that $\overline{B_\varkappa(x_0)} \cap K \ne \emptyset$. Let $m \in \partial K$ be the orthogonal projection of $x_0$ to $\partial K$. Then, by construction, we have $x_0 = m + |x_0 - m| \nu(m)$, where $\nu(m)$ denotes the outward unit normal to $\partial K$ at $m$. Set $x_0^\perp := x_0 + r_1 \nu(m)$. Notice that $x_0^\perp \in \overline{\mathcal{A}(x_0, r_1, r_2)} \setminus \mathcal{C}(x_0)$ (by construction of $x_0^\perp$), so that $x_0^\perp \in \tilde{\Pi}_1(x_0, r_1, r_2)$ (remember (A.4) and (A.5)). Moreover, we have $\overline{B_\varkappa(x_0)} \cap \Omega \setminus \text{star}(x_0) \subset \mathbb{R}^N \setminus B_{r_1}(x_0^\perp)$ and $\overline{B_\varkappa(x_0)} \subset \overline{B_{r_2}(x_0^\perp)}$. Therefore, we have

$$ (A.7) \qquad \overline{B_\varkappa(x_0)} \cap \Omega \setminus \text{star}(x_0) \subset \overline{\mathcal{A}(x_0^\perp, r_1, r_2)}. $$

Since $x_0 = p + \ell \nu(p)$ and $x_0^\perp = p + (\ell + r_1) \nu(p)$ for some $\ell \in [0, \varkappa]$ and some $p \in \partial K$, we may apply (A.3), which then yields $\overline{B_\varkappa(x_0)} \cap \Omega \subset \text{star}(x_0^\perp)$. Hence, using (A.7), it follows that

$$ \overline{B_\varkappa(x_0)} \cap \Omega \setminus \text{star}(x_0) \subset \text{star}(x_0^\perp) \cap \overline{\mathcal{A}(x_0^\perp, r_1, r_2)}. $$

Since $\delta(x_0^\perp, y) = |x_0^\perp - y|$ for all $y \in \text{star}(x_0^\perp)$, this then implies that

$$ \overline{B_\varkappa(x_0)} \cap \Omega \setminus \text{star}(x_0) \subset \tilde{\Pi}_1(x_0^\perp, r_1, r_2) \subset \tilde{\Pi}_2(x_0, r_1, r_2), $$

where, in the last inclusion, we have used that $x_0^\perp \in \tilde{\Pi}_1(x_0, r_1, r_2)$. Together with (A.6), this yields that $\overline{B_\varkappa(x_0)} \setminus \Omega \subset \tilde{\Pi}_2(x_0, r_1, r_2)$. At this stage, we may conclude exactly as in the proof of Proposition A.1 (remember that $\tilde{\Pi}_j(x, r_1, r_2) \subset \Pi_j(J, x)$ for all $x \in \overline{\Omega}$) and we therefore obtain that $(\Omega, \delta)$ has the J-covering property, as desired. $\square$

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ÉCOLE DES HAUTES ÉTUDES EN SCIENCES SOCIALES, PSL RESEARCH UNIVERSITY, CNRS, CENTRE D'ANALYSE ET DE MATHÉMATIQUE SOCIALES, PARIS, FRANCE

E-mail address: julien.brasseur@ehess.fr, julienbrasseur@wanadoo.fr

BIOSP, INRA, 84914, AVIGNON, FRANCE

E-mail address: jerome.coville@inra.fr