Finite star-shaped cover of a bounded Lipschitz domain

Fix one graph chart, with Lipschitz constant $L$, and a smaller horizontal ball $C$ whose closure lies in its base. Shrink its radius $r$ until the chart's upper height $H_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{t}}$ satisfies $H_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{t}}-{\sup }_C\phi >4(L+1)r$. Choose $c$ with ${\sup }_C\phi +3(L+1)r<c<H_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{t}}-(L+1)r$ and $0<\rho <r$. Then the closed ball $\bar{B}((x_0^{\mathrm{\prime }},c),\rho )$ stays in the cylinder. Set $U=\{(x^{\mathrm{\prime }},x_d):x^{\mathrm{\prime }}\in C,\phi (x^{\mathrm{\prime }})<x_d<H\}$ with $c+\rho <H<H_{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{t}}$.

If $x\in U$, $z\in \bar{B}((x_0^{\mathrm{\prime }},c),\rho )$, and $w=(1-t)x+tz$ with $0<t<1$, then $z^{\mathrm{\prime }}\in C$ and convexity keeps $w^{\mathrm{\prime }}\in C$. Since $\mathrm{\mid }{z}^{\mathrm{\prime }}-x^{\mathrm{\prime }}\mathrm{\mid }\le 2r$, Lipschitz continuity gives $\phi (w^{\mathrm{\prime }})\le \phi (x^{\mathrm{\prime }})+2Lrt$. Also $c-\rho >{\sup }_C\phi +2Lr$, so $w_d>(1-t)\phi (x^{\mathrm{\prime }})+t(c-\rho )>\phi (x^{\mathrm{\prime }})+2Lrt\ge \phi (w^{\mathrm{\prime }})$, and $w_d<H$ by convexity of the vertical interval. Thus $w\in U$; the endpoints belong to $U$ as well, proving star-shapedness with respect to the closed ball.

The boundary $\mathrm{\partial }D$ is closed and bounded, hence compact by [F1]. At any $p\in \mathrm{\partial }D$, a chart containing $p$ writes the boundary locally as $x_d=\phi (x^{\mathrm{\prime }})$, with $p_d=\phi (p^{\mathrm{\prime }})$ strictly below the chart's upper height. Shrinking a horizontal ball $C$ about $p^{\mathrm{\prime }}$ makes its closure lie in the chart base and makes the height-gap inequality of step 1.1 hold; the choices there then produce a patch $U=\{(x^{\mathrm{\prime }},x_d):x^{\mathrm{\prime }}\in C,\phi (x^{\mathrm{\prime }})<x_d<H\}$ star-shaped with respect to its selected closed ball by step 2.1. Since $p$ lies inside the chart and inside the open set with $x^{\mathrm{\prime }}\in C$ and $x_d<H$, there is an ambient open ball $N$ about $p$ whose closure lies in that chart and open set. For every $x\in N\cap D$, the chart description gives $x_d>\phi (x^{\mathrm{\prime }})$, so $x\in U$ and $N\cap D\subset U$. Let $\mathcal{N}$ be the family of all pairs $(N,U)$ obtained this way from the finite chart atlas and admissible local choices; the preceding existence argument shows that its first components cover $\mathrm{\partial }D$, without choosing a patch simultaneously for every boundary point. Compactness supplies a finite subfamily $(N_i,U_i)$, $1\le i\le m$, whose neighborhoods cover $\mathrm{\partial }D$. Their union is an open neighborhood of the compact boundary, so some relative collar $\{x\in D:\mathrm{dist}(x,\mathrm{\partial }D)<\delta \}$ lies in that union and hence in ${\bigcup }_{i=1}^m{U}_i$. The remainder $K_{\delta }=\{x\in \bar{D}:\mathrm{dist}(x,\mathrm{\partial }D)\ge \delta \}$ is closed and bounded, hence compact by [F1], and is contained in $D$ because every point of $\bar{D}\setminus D$ lies on $\mathrm{\partial }D$. Finitely many balls with closures in $D$ cover $K_{\delta }$; each ball is convex and star-shaped with respect to a smaller concentric ball. Together these sets form a finite star-shaped cover of $D$.

The intersection graph is connected: if it split into two nonempty graph components, the unions of the corresponding open sets would be disjoint nonempty relatively open subsets whose union is $D$, contradicting connectedness. Order the cover and set $E_i=U_i\setminus {\bigcup }_{j<i}{U}_j$. Then the $E_i$ are disjoint measurable sets with $E_i\subset U_i$ and union $D$. A finite connected graph has a spanning tree.

Each tree-edge intersection is nonempty and open, so it contains a ball. Choose a nonnegative smooth bump supported in that ball. Its integral is strictly positive; dividing by that integral gives ${\psi }_{ij}\in C_c^{\mathrm{\infty }}(U_i\cap U_j)$ with integral one.