Euclidean balls, annuli and spheres as bounded smooth domains and hypersurfaces

Because $N$ is continuous and $(0,R)$ and $(\epsilon ,R)$ are open, $B_R=N^{-1}((-\mathrm{\infty },R))$ and $A_{\epsilon ,R}=N^{-1}((\epsilon ,R))$ are open, nonempty (they contain the origin, respectively a point of norm $(\epsilon +R)\mathrm{/}2$) and bounded, being contained in the closed ball of radius $R$ ([F1], [F2]). If $\mathrm{\mid }x\mathrm{\mid }>R$, continuity of $N$ gives a ball about $x$ on which $\mathrm{\mid }y\mathrm{\mid }>R$, so no point outside ${\bar{B}}_R$ is a limit of points of $B_R$; if $\mathrm{\mid }x\mathrm{\mid }=R$, the points $(1-1\mathrm{/}n)x$ lie in $B_R$ and converge to $x$; hence the closure of $B_R$ is ${\bar{B}}_R$ and, since $B_R$ is open, its boundary is $S_R$ ([F1], [F2]). The same argument with $N^{-1}((\epsilon ,R))$, together with the identification of its closure with $\{x:\epsilon \le \mathrm{\mid }x\mathrm{\mid }\le R\}$, gives that the boundary of $A_{\epsilon ,R}$ is $S_{\epsilon }\cup S_R$, and both sets have nonempty interior equal to themselves because they are open.

Fix $k\in \{1,2,3\}$ and $\sigma \in \{\pm 1\}$, let $z=\sigma x_k$ and let $y$ collect the two remaining coordinates in a fixed order, labelled $y=(y_1,y_2)$; the map $x\mapsto (y,z)$ is a rigid change of coordinates (a permutation with one sign change). For $r>0$ put $U_{r,k,\sigma }:=\{x:\sigma x_k>r\mathrm{/}(2\sqrt{3})\}$ and $h(y):=\sqrt{R^2-\mathrm{\mid }y{\mathrm{\mid }}^2}$, which is $C^{\mathrm{\infty }}$ on the open set $\mathrm{\mid }y\mathrm{\mid }<R$ with $Dh(y)=-y\mathrm{/}h(y)$ ([F3]). On $U_{R,k,\sigma }\cap \{\mathrm{\mid }y\mathrm{\mid }<R\}$ one has $z>0$ and $\mathrm{\mid }x{\mathrm{\mid }}^2<R^2⟺z^2<R^2-\mathrm{\mid }y{\mathrm{\mid }}^2⟺z<h(y),$ so $S_R\cap U_{R,k,\sigma }$ is the graph $z=h(y)$ over the open parameter set $V_R:=\{y:\mathrm{\mid }y\mathrm{\mid }<R,h(y)>R\mathrm{/}(2\sqrt{3})\}$ and $B_R\cap U_{R,k,\sigma }\cap \{\mathrm{\mid }y\mathrm{\mid }<R\}$ is its subgraph; this is the local presentation required of a bounded $C^1$ domain ([F4]). The graph normal of [F4] is $\frac{(-Dh(y),1)}{\sqrt{1+\mathrm{\mid }Dh(y){\mathrm{\mid }}^2}}=\frac{(y,h(y))}{R},$ because $-Dh=y\mathrm{/}h$ and $1+\mathrm{\mid }Dh{\mathrm{\mid }}^2=R^2\mathrm{/}{h}^2$; transporting back through the rigid coordinate map gives $\nu (x)=x\mathrm{/}R=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ on $S_R$. The six coordinate neighbourhoods $U_{R,k,\sigma }$ cover the sphere $S_R$: for $x\mathrm{\ne }0$ some coordinate satisfies $\mathrm{\mid }{x}_k\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }\mathrm{/}\sqrt{3}$, so with $\sigma :=\mathrm{sign}(x_k)$ one has $\sigma x_k=\mathrm{\mid }{x}_k\mathrm{\mid }\ge R\mathrm{/}\sqrt{3}>R\mathrm{/}(2\sqrt{3})$ whenever $\mathrm{\mid }x\mathrm{\mid }\ge R$, and in particular the sphere $S_R$ is covered by the six graph patches.

Fix again $k,\sigma$ and put $h_{\epsilon }(y)=\sqrt{{\epsilon }^2-\mathrm{\mid }y{\mathrm{\mid }}^2}$ on $\mathrm{\mid }y\mathrm{\mid }<\epsilon$. In the coordinates $(y,z)$ of step 2.1 the complement of the closed ball satisfies $z>h_{\epsilon }(y)$ inside the patch $U_{\epsilon ,k,\sigma }$ with $\mathrm{\mid }y\mathrm{\mid }<\epsilon$, and the second rigid change $w=-z$ turns this into the subgraph $w<-h_{\epsilon }(y)$ of the smooth function $g(y):=-h_{\epsilon }(y)$ with $Dg(y)=y\mathrm{/}{h}_{\epsilon }(y)$ ([F3]). The graph normal of [F4] in the $(y,w)$ coordinates is $\frac{(-Dg(y),1)}{\sqrt{1+\mathrm{\mid }Dg(y){\mathrm{\mid }}^2}}=\frac{(-y,h_{\epsilon }(y))}{\epsilon },$ and transporting back through $w=-\sigma x_k$ gives $\nu (x)=-x\mathrm{/}\epsilon =-x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ on $S_{\epsilon }$. Near a point of $S_{\epsilon }$ the ball $B_R$ occupies a full neighborhood because $\mathrm{\mid }x\mathrm{\mid }=\epsilon <R$, so $A_{\epsilon ,R}=B_R\setminus {\bar{B}}_{\epsilon }$ is locally exactly the subgraph just described, while near a point of $S_R$ it is locally the subgraph of $h$ from step 2.1 with normal $x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$. Hence $A_{\epsilon ,R}$ is a bounded $C^1$ domain with boundary $S_{\epsilon }\cup S_R$ and the stated outward normals ([F4]).

Each $S_r$ is closed, being the preimage of the closed set $\{r\}$ under the continuous norm, and bounded, hence compact ([F1], [F5]). The restrictions of the six maps $y\mapsto (y,h_r(y))$ from steps 2.1 and 3.1, composed with the inverse rigid coordinate maps, are regular injective $C^{\mathrm{\infty }}$ parametrizations of the open patches $S_r\cap U_{r,k,\sigma }$ with $C^{\mathrm{\infty }}$ transitions (each transition is the coordinate projection from one graph chart composed with the other smooth graph parametrization) ([F3]). Therefore each $S_r$ is a compact embedded $C^1$ hypersurface in the sense of [F6], and the two graph normal computations of steps 2.1 and 3.1 exhibit both unit normals $\pm x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$: the positive sign from the ball chart and the negative sign from the complement chart, continuously across overlaps.

For $r=\epsilon ,R$ define the twelve compact faces $F_{r,k,\sigma }=\{x\in S_r:\sigma x_k\ge \mathrm{\mid }{x}_j\mathrm{\mid }\text{ for all }j\}$, $k=1,2,3$, $\sigma =\pm 1$. They cover $S_r$: choose $k$ with $\mathrm{\mid }{x}_k\mathrm{\mid }={\max }_j\mathrm{\mid }{x}_j\mathrm{\mid }$; then $x_k\mathrm{\ne }0$ and $\sigma =\mathrm{sign}(x_k)$ gives $\sigma x_k\ge \mathrm{\mid }{x}_j\mathrm{\mid }$ for all $j$. Each $F_{r,k,\sigma }$ is closed in the compact $S_r$, hence compact, and on it $\sigma x_k\ge r\mathrm{/}\sqrt{3}>r\mathrm{/}(2\sqrt{3})$, so it is contained in the single regular patch $U_{r,k,\sigma }$ of steps 2.1-3.1 with parameter set contained in the compact set $\{\mathrm{\mid }y\mathrm{\mid }\le r\sqrt{2\mathrm{/}3}\}\subset \{\mathrm{\mid }y\mathrm{\mid }<r\}$ ([F5]).

Let $E$ be the union over $r=\epsilon ,R$ of the seams $\{x\in S_r:\sigma x_k=\mathrm{\mid }{x}_j\mathrm{\mid }\text{ for some }j\mathrm{\ne }k,\sigma =\pm 1\}$, a finite union of closed subsets of the compact spheres, hence compact. Every relative boundary point of a face $F_{r,k,\sigma }$ has $\sigma x_k=\mathrm{\mid }{x}_j\mathrm{\mid }$ for some $j$, and every overlap $F_{r,k,\sigma }\cap F_{r,k^{\mathrm{\prime }},{\sigma }^{\mathrm{\prime }}}$ with $(k,\sigma )\mathrm{\ne }(k^{\mathrm{\prime }},{\sigma }^{\mathrm{\prime }})$ lies in such a seam, so $E$ contains all relative face boundaries and all overlaps ([F7]). To see that $F_{r,k,\sigma }\cap E$ has zero surface measure, work in the chart $z=\sigma x_k=h_r(y)$ of $F_{r,k,\sigma }$: a seam involving the maximal coordinate reads either $\mathrm{\mid }{y}_1\mathrm{\mid }=h_r(y)$ or $\mathrm{\mid }{y}_2\mathrm{\mid }=h_r(y)$. The first is $y_1^2=(r^2-y_2^2)\mathrm{/}2$, the second $y_2^2=(r^2-y_1^2)\mathrm{/}2$. For either case take the other coordinate as the fixed section parameter: the seam meets the vertical line in at most two points, a finite union of singletons and hence a ${\lambda }_1$-null set ([F8]); Fubini applied to the bounded Borel indicator on a containing rectangle gives the planar measure of these parameter sets as zero ([F8]). The other seams in $E$ equate the absolute values of the two nonmaximal coordinates; in this face chart they lie on $y_1=\pm y_2$, whose sections are again finite. The same argument gives planar measure zero for them. Thus the full finite union of seams meeting this face is planar null. The graph density $\sqrt{1+\mathrm{\mid }D{h}_r{\mathrm{\mid }}^2}=r\mathrm{/}{h}_r$ is continuous and bounded on the compact parameter set $\{\mathrm{\mid }y\mathrm{\mid }\le r\sqrt{2\mathrm{/}3}\}$, so integrating it against the indicator of a planar null set gives zero ([F9]); by the chart-independence of the surface measure this says exactly that $S_j\cap E$ has zero surface measure in each face, as required by [F7].

Statement (a) is steps 2.1 and 3.1; statement (b) is steps 4.1, 5.1 and 6.1, the twelve compact faces with their null compact seam set satisfying exactly the requirements of [F7] for the annulus $A_{\epsilon ,R}$, whose exposed faces carry the outward normals computed in steps 2.1 and 3.1 off $E$. Finally, (c): the divergence theorem and Green's first identity of [F10] apply on $B_R$ and on $A_{\epsilon ,R}$ because both are bounded $C^1$ domains with the outward normals of (a), and $S_r$ carries a finite Borel surface measure with the standard graph charts by [F6].