${\mathchoice{\raisebox{0.0pt}{$\displaystyle\chi$}}{\raisebox{0.0pt}{$\textstyle\chi$}}{\raisebox{0.0pt}{$\scriptstyle\chi$}}{\raisebox{0.0pt}{$\scriptscriptstyle\chi$}}}^{\phi}_{K,1},\ldots,{\mathchoice{\raisebox{0.0pt}{$\displaystyle\chi$}}{\raisebox{0.0pt}{$\textstyle\chi$}}{\raisebox{0.0pt}{$\scriptstyle\chi$}}{\raisebox{0.0pt}{$\scriptscriptstyle\chi$}}}^{\phi}_{K,n}$
are the anisotropic principal curvatures of $\partial E_{r}$ with respect to
the anisotropic normal $\bm{\nu}^{\phi}_{K}|\partial E_{r}$ (which points
towards $C$). It follows from (40) and (41) that

${\mathchoice{\raisebox{0.0pt}{$\displaystyle\chi$}}{\raisebox{0.0pt}{$\textstyle\chi$}}{\raisebox{0.0pt}{$\scriptstyle\chi$}}{\raisebox{0.0pt}{$\scriptscriptstyle\chi$}}}^{\phi}_{K,1}(x)=\ldots={\mathchoice{\raisebox{0.0pt}{$\displaystyle\chi$}}{\raisebox{0.0pt}{$\textstyle\chi$}}{\raisebox{0.0pt}{$\scriptstyle\chi$}}{\raisebox{0.0pt}{$\scriptscriptstyle\chi$}}}^{\phi}_{K,n}(x)=\frac{1}{r-\bm{r}^{\phi}_{K}(\bm{\xi}^{\phi}_{K}(x),\bm{\nu}^{\phi}_{K}(x))}\geq\frac{1}{r-\frac{n}{q}}=\Big{(}\frac{qr-n}{q}\Big{)}^{-1}$

for $\mathcal{H}^{n}$ a.e. $x\in\partial E_{r}$; in particular,
${\mathchoice{\raisebox{0.0pt}{$\displaystyle\chi$}}{\raisebox{0.0pt}{$\textstyle\chi$}}{\raisebox{0.0pt}{$\scriptstyle\chi$}}{\raisebox{0.0pt}{$\scriptscriptstyle\chi$}}}^{\phi}_{K,i}(x)<0$
for $i=1,\ldots,n$ and $\mathcal{H}^{n}$ a.e. $x\in\partial E_{r}$. Hence, an
application of [DRKS20, Lemma 3.2]111Notice that the last line of [DRKS20,
Lemma 3.2] contains a typo: one should replace the equality
$M=\partial\mathbf{B}^{F}(a,|\lambda|^{-1})$ with
$M=\partial\mathbf{B}^{F^{\ast}}(a,|\lambda|^{-1})$, which is what the proof
shows., to each of the at most countably many connected components of
$\partial E_{r}$ shows that there exist at most countably many points
$c_{1},c_{2},\ldots\in\mathbf{R}^{n+1}$ and numbers
$\lambda_{1},\lambda_{2}\ldots\geq\frac{n-qr}{q}$ so that

$E_{r}=\bigcup_{i=1}^{N}(c_{i}+\lambda_{i}\mathcal{W}^{\phi})\qquad\textrm{and}\qquad(c_{i}+\lambda_{i}\mathcal{W}^{\phi})\cap(c_{j}+\lambda_{j}\mathcal{W}^{\phi})=\varnothing\quad\textrm{for
$i\neq j$,}$

where $N\in\mathbf{N}\cup\\{\infty\\}$. Since
$\mathcal{L}^{n+1}(\operatorname{int}(C))<\infty$, it follows that $N<\infty$.
If $i\in\\{1,\ldots,N\\}$ and $z\in\partial\mathcal{W}^{\phi}$, we have

$\frac{\nabla\bm{\delta}^{\phi}_{K}(c_{i}+\lambda_{i}z)}{|\nabla\bm{\delta}^{\phi}_{K}(c_{i}+\lambda_{i}z)|}=-\bm{n}^{\phi}(z)$

and, noting (8) and (12), we conclude

$z=\nabla\phi(\bm{n}^{\phi}(z))=-\nabla\phi(\nabla\bm{\delta}^{\phi}_{K}(c_{i}+\lambda_{i}z))=-\bm{\nu}^{\phi}_{K}(c_{i}+\lambda_{i}z).$

For $0\leq s<r$ we define the bilipschitz homeomorphism $f_{s}:\partial
E_{r}\rightarrow\partial E_{r-s}$ by $f_{s}(x)=x-s\bm{\nu}^{\phi}_{K}(x)$ for
$x\in\partial E_{r}$. Then we get

$f_{s}(c_{i}+\lambda_{i}\partial\mathcal{W}^{\phi})=c_{i}+(\lambda_{i}+s)\partial\mathcal{W}^{\phi}\quad\textrm{for
$i\geq 1$,}\qquad\partial
E_{r-s}=\bigcup_{i=1}^{N}\big{(}c_{i}+(\lambda_{i}+s)\partial\mathcal{W}^{\phi}\big{)}$

and

$\big{(}c_{i}+(\lambda_{i}+s)\partial\mathcal{W}^{\phi})\cap\big{(}c_{j}+(\lambda_{j}+s)\partial\mathcal{W}^{\phi}\big{)}=\varnothing\quad\textrm{for
$i\neq j$}.$

Consequently, for $0\leq s<r$,

$E_{r-s}=\bigcup_{i=1}^{N}\big{(}c_{i}+(\lambda_{i}+s)\mathcal{W}^{\phi}\big{)},\qquad\big{(}c_{i}+(\lambda_{i}+s)\mathcal{W}^{\phi})\cap\big{(}c_{j}+(\lambda_{j}+s)\mathcal{W}^{\phi}\big{)}=\varnothing\quad\textrm{for
every $i\neq j$}.$

and the proof is complete. ∎

###  3.4 A general disintegration formula

For the next definition we need to recall that if $A$ is a closed set in
$\mathbf{R}^{n+1}$ and $a\in A$, then the set

$\operatorname{Dis}(A,a):=\\{u\in\mathbf{R}^{n+1}:\bm{\delta}_{A}(a+u)=|u|\\}$

is a closed convex set (see [Fed59, Theorem 4.8 (2)] or [MS19]). If $X$ is a
convex set, then $\dim X$ denotes the dimension of the affine hull of $X$.

######  Definition 3.21.

Let $\varnothing\neq A\subseteq\mathbf{R}^{n+1}$ be a closed set and
$i\in\\{0,\ldots,n+1\\}$. We define the $i$-th stratum of $A$ as

$A^{(i)}=\\{a\in A:\dim\operatorname{Dis}(A,a)=n+1-i\\}.$

######  Remark 3.22.

Evidently, we have $A=\bigcup_{i=0}^{n+1}A^{(i)}$ and
$A^{(n+1)}=A\setminus\bm{p}(N(A))$. Noting that
$N(A,a)=\\{u/|u|\in\mathbf{S}^{n}:u\in\operatorname{Dis}(A,a)\setminus\\{0\\}\\}$
for every $a\in\bm{p}(N(A))$, we can easily deduce from (13) that

$A^{(i)}=\\{a\in
A:0<\mathcal{H}^{n-i}(N^{\phi}(A,a))<\infty\\}\qquad\textrm{for
$i=0,\ldots,n$.}$

We recall that $A^{(i)}$ is a countably $i$-rectifiable Borel set which can be
covered outside a set of $\mathcal{H}^{i}$-measure zero by a countable union
of $\mathcal{C}^{2}$-submanifolds of dimension $i$, for $i=0,\ldots,n$; see
[MS19].

######  Lemma 3.23.

If $\varnothing\neq A\subseteq\mathbf{R}^{n+1}$ is a closed set,
$m\in\\{0,\ldots,n\\}$ and $S\subseteq\mathbf{R}^{n+1}$ is a countable union
of Borel subsets with finite $\mathcal{H}^{m}$ measure, then

$\mathcal{H}^{n}\big{(}(N^{\phi}(A)|A^{(j)})\setminus\textstyle\bigcup_{l=0}^{j}\widetilde{N}^{\phi}_{l}(A)\big{)}=0\qquad\textrm{for
$j=0,\ldots,n$}$ (42)

and

$\mathcal{H}^{n}\big{(}(N^{\phi}(A)|A^{(j)}\cap
S)\setminus\textstyle\bigcup_{l=0}^{m-1}\widetilde{N}^{\phi}_{l}(A)\big{)}=0\qquad\textrm{for
$j>m$.}$ (43)

###### Proof.

Recall that $N_{s}=\\{(a,\eta)\in N^{\phi}(A):\bm{r}^{\phi}_{A}(a,\eta)\geq
s\\}$, for every $s>0$, has finite $\mathcal{H}^{n}$ measure on each bounded
set (see Theorem 3.16) and therefore is $(\mathcal{H}^{n},n)$ rectifiable.

Fix $j\in\\{0,\ldots,n-1\\}$ and notice that $\mathcal{H}^{j+1}(A^{(j)})=0$.
Therefore we can apply Lemma 2.1 to conclude that

$\int_{N_{s}|A^{(j)}}\operatorname{ap}J_{j+1}^{N_{s}}\bm{p}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)=\int_{A^{(j)}}\mathcal{H}^{n-1-j}(N_{s}|\\{a\\})\,d\mathcal{H}^{j+1}(a)=0$

for every $s>0$. It follows that
$\operatorname{ap}J_{j+1}^{N_{s}}\bm{p}(a,\eta)=0$ for $\mathcal{H}^{n}$ a.e.
$(a,\eta)\in N_{s}|A^{(j)}$ and $s>0$. We infer from Lemma 3.9 that

$\kappa^{\phi}_{A,j+1}(a,\eta)=+\infty\qquad\textrm{for $\mathcal{H}^{n}$
a.e.\ $(a,\eta)\in N^{\phi}(A)|A^{(j)}$,}$

which is precisely the assertion in (42).

If $m=0$, then $S$ is a countable set and $\mathcal{H}^{n}(N^{\phi}(A,a))=0$
for every $a\in A^{(j)}$ and for every $j\geq 1$. This implies (43) for $m=0$.
Fix now $j>m\geq 1$. Noting that $\mathcal{H}^{n-m}(N^{\phi}(A,a))=0$ for
$a\in A^{(j)}$, we apply again the coarea formula in Lemma 2.1 to obtain

$\int_{N_{s}|S\cap
A^{(j)}}\operatorname{ap}J^{N_{s}}_{m}\bm{p}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)=\int_{S\cap
A^{(j)}}\mathcal{H}^{n-m}(N_{s}|\\{a\\})\,d\mathcal{H}^{m}(a)=0$

for $s>0$. As above this implies that $\kappa^{\phi}_{A,m}(a,\eta)=+\infty$
for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in N^{\phi}(A)|A^{(j)}\cap S$, which is
equivalent to the assertion in (43). ∎

######  Remark 3.24.

If $A$, $m$ and $S$ are as in Lemma 3.23 then

$\bm{H}^{\phi}_{n-m}(a,\eta)=\bm{1}_{\widetilde{N}^{\phi}_{m}(A)}(a,\eta)\qquad\textrm{for
$\mathcal{H}^{n}$ a.e.\ $(a,\eta)\in N^{\phi}(A)|A^{(m)}$},$
$\bm{H}^{\phi}_{n-m}(a,\eta)=0\qquad\textrm{for $\mathcal{H}^{n}$ a.e.\
$(a,\eta)\in N^{\phi}(A)|A^{(j)}$ and $j<m$}$

and

$\bm{H}^{\phi}_{n-m}(a,\eta)=0\qquad\textrm{for $\mathcal{H}^{n}$ a.e.\
$(a,\eta)\in N^{\phi}(A)|(A^{(j)}\cap S)$ and $j>m$.}$

######  Lemma 3.25.

For every closed set $A\subseteq\mathbf{R}^{n+1}$ the following statements
hold.

  1. (a)

$\mathcal{H}^{0}(N^{\phi}(A,a))\in\\{1,2\\}$ for
$a\in\bm{p}(\widetilde{N}^{\phi}_{n}(A))$.

  2. (b)

$\mathcal{H}^{m}\left(\big{\\{}a\in
A^{(m)}:\mathcal{H}^{n-m}\big{(}N^{\phi}(A,a)\setminus\widetilde{N}^{\phi}_{m}(A,a)\big{)}>0\big{\\}}\right)=0$
for $m\in\\{0,\ldots,n\\}$.

  3. (c)

$\mathcal{H}^{n}\big{(}\bm{p}\big{[}N^{\phi}(A)\setminus\widetilde{N}_{n}^{\phi}(A)\big{]}\big{)}=0$;
in particular, $\mathcal{H}^{0}(N^{\phi}(A,a))\in\\{1,2\\}$ for
$\mathcal{H}^{n}$ a.e. $a\in\bm{p}(N^{\phi}(A))$.

  4. (d)

$N^{\phi}(A,a)=-N^{\phi}(A,a)$ for $a\in A$ with
$\mathcal{H}^{0}(N^{\phi}(A,a))=2$.

  5. (e)

If
$X=\bm{p}(N^{\phi}(A)\setminus\widetilde{N}_{n}^{\phi}(A))\cap\bm{p}(\widetilde{N}^{\phi}_{n}(A))$,
then $\mathcal{H}^{n}(N^{\phi}(A)|X)=0$.

###### Proof.

(a) Let $(a,\eta)\in\widetilde{N}^{\phi}_{n}(A)$ and
$0<r<\underline{\bm{r}_{A}^{\phi}}(a,\eta)$. Then
$1-r\chi^{\phi}_{A,i}(a+r\eta)>0$ for $i=1,\ldots,n$ and, since these numbers
are the eigenvalues of
$\operatorname{D}\bm{\xi}^{\phi}_{A}(a+r\eta)|\operatorname{Tan}(S^{\phi}(A,r),a+r\eta)$,
we conclude (noting Remark 3.4 and [Fed69, 3.1.21])

$\operatorname{Tan}(\partial\mathcal{W}^{\phi},\eta)=\operatorname{D}\bm{\xi}^{\phi}_{A}(a+r\eta)[\operatorname{Tan}(S^{\phi}(A,r),a+r\eta)]\subseteq\operatorname{Tan}(A,a).$

Since
$N(A,a)\subseteq\operatorname{Nor}(A,a)\subseteq\operatorname{Nor}(\partial\mathcal{W}^{\phi},\eta)$
and $\dim\operatorname{Nor}(\partial\mathcal{W}^{\phi},\eta)=1$, it follows
that $\mathcal{H}^{0}(N^{\phi}(A,a))\in\\{1,2\\}$ and a is proved.

(b) First, let $m\in\\{1,\ldots,n\\}$, define
$P=\bigcup_{i=m}^{n}\widetilde{N}^{\phi}_{i}(A)$ and recall that $A^{(m)}$ is
a countably $m$-rectifiable set (see above). For $s>0$ define $N_{s}$ as in
Theorem 3.16 and set $W_{s}=[N_{s}\setminus P]|A^{(m)}$, which is an
$(\mathcal{H}^{n},n)$ rectifiable set. Noting from Lemma 3.9 that
$\operatorname{ap}J^{W_{s}}_{m}\bm{p}(a,\eta)=0$ for $\mathcal{H}^{n}$ a.e.
$(a,\eta)\in W_{s}$, we conclude from Lemma 2.1 that

$\int_{A^{(m)}}\mathcal{H}^{n-m}\big{(}(N_{s}|\\{x\\})\setminus(P|\\{x\\})\big{)}\,d\mathcal{H}^{m}(x)=0,$

and thus we infer that

$\mathcal{H}^{m}\left(\left\\{x\in
A^{(m)}:\mathcal{H}^{n-m}((N_{s}|\\{x\\})\setminus
P|\\{x\\})>0\right\\}\right)=0.$

Furthermore, we have

$\\{x\in
A^{(m)}:\mathcal{H}^{n-m}(N^{\phi}(A,x)\setminus(P|\\{x\\}))>0\\}=\bigcup_{i=1}^{\infty}\\{x\in
A^{(m)}:\mathcal{H}^{n-m}((N_{s_{i}}|\\{x\\})\setminus(P|\\{x\\}))>0\\}$

for any positive sequence $s_{i}\searrow 0$. Since also
$\mathcal{H}^{n}(N^{\phi}(A,a)\setminus\widetilde{N}^{\phi}(A,a))=0$ for $a\in
A^{(0)}$, we obtain that

$\mathcal{H}^{m}\left(\left\\{a\in
A^{(m)}:\mathcal{H}^{n-m}\bigg{(}N^{\phi}(A,a)\setminus\bigcup_{i=m}^{n}\widetilde{N}^{\phi}_{i}(A,a)\bigg{)}>0\right\\}\right)=0$
(44)

for $m\in\\{0,\ldots,n\\}$. Now let $m\in\\{0,\ldots,n-1\\}$. Since by Lemma
3.23 it holds that

$\mathcal{H}^{n}\bigg{(}\big{(}N^{\phi}(A)|A\big{)}\cap\bigcup_{i=m+1}^{n}\widetilde{N}^{\phi}_{i}(A)\bigg{)}=0,$

an application of the coarea formula in Lemma 2.1 yields that

$\int_{A^{(m)}}\mathcal{H}^{n-m}\left(N_{s}|\\{x\\}\cap\bigcup_{i=m+1}^{n}\widetilde{N}^{\phi}_{i}(A,x)\right)\,d\mathcal{H}^{m}(x)=0\qquad\textrm{for
every $s>0$,}$

whence, as above, we infer

$\mathcal{H}^{m}\left(\left\\{a\in
A^{(m)}:\mathcal{H}^{n-m}\bigg{(}N^{\phi}(A,a)\cap\bigcup_{i=m+1}^{n}\widetilde{N}^{\phi}_{i}(A,a)\bigg{)}>0\right\\}\right)=0.$
(45)

Now the assertion follows from (44) and (45).

c Since $\bm{p}(N^{\phi}(A))=\bigcup_{i=0}^{n}A^{(i)}$,
$\mathcal{H}^{n}(A^{(j)})=0$ for $j\in\\{0,\ldots,n-1\\}$ and
$\bm{p}\big{[}N^{\phi}(A)\setminus\widetilde{N}_{n}^{\phi}(A)\big{]}=\bm{p}(N^{\phi}(A))\cap\\{a:\mathcal{H}^{0}(N^{\phi}(A,a)\setminus\widetilde{N}^{\phi}_{n}(A,a))>0\\}$,
we conclude that c directly follows from b with $m=n$.

d The cone $\\{tu:t>0,\,u\in N(A,a)\\}$ is convex for
$a\in\bm{p}(N^{\phi}(A))$. Consequently, we have $N(A,a)=-N(A,a)$ for $a\in A$
with $\mathcal{H}^{0}(N(A,a))=2$. Since $\nabla\phi(N(A,a))=N^{\phi}(A,a)$ for
$a\in\bm{p}(N^{\phi}(A))$, we conclude that $N^{\phi}(A,a)=-N^{\phi}(A,a)$ for
$a\in A$ with $\mathcal{H}^{0}(N^{\phi}(A,a))=2$.

e Finally, let $X$ be the set defined in e. Employing again the sets $N_{s}$
as defined in Theorem 3.16, we notice that
$\operatorname{ap}J^{N_{s}}_{n}\bm{p}(a,u)>0$ for $\mathcal{H}^{n}$ a.e.
$(a,u)\in\widetilde{N}^{\phi}_{n}(A)$ and, by the coarea formula and c, we get

$\int_{(N_{s}\cap\widetilde{N}^{\phi}_{n}(A))|X}\operatorname{ap}J^{N_{s}}_{n}\bm{p}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)=0$

for every $s>0$. It follows that
$\mathcal{H}^{n}(\widetilde{N}^{\phi}_{n}(A)|X)=0$. Since by a and d we have
that

$N^{\phi}(A)|X=(\widetilde{N}^{\phi}_{n}(A)|X)\cup\\{(a,-\eta):(a,\eta)\in\widetilde{N}^{\phi}_{n}(A)|X\\},$

we obtain e. ∎

######  Remark 3.26.

If $A\subseteq\mathbf{R}^{n+1}$ is a convex body (with non-empty interior),
then $\mathcal{H}^{0}(N^{\phi}(A,a))=1$ for every
$a\in\bm{p}(\widetilde{N}_{n}^{\phi}(A))$.

We now prove a very general disintegration formula. This result, which is of
independent interest, plays a key role in the proof of Corollary 6.18 through
Lemma 6.6 and the subsequent Lemma 6.7.

######  Theorem 3.27.

Let $\varnothing\neq A\subseteq\mathbf{R}^{n+1}$ be a closed set and
$m\in\\{0,\ldots,n\\}$. Then there exists a positive real-valued
$\mathcal{H}^{n}\operatorname{\llcorner}N^{\phi}(A)$ measurable function
$\rho^{\phi}_{A,m}$ on $N^{\phi}(A)$ such that the following statements hold.

  1. (a)

$0<\rho^{\phi}_{A,m}(a,\eta)\leq c$ for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in
N^{\phi}(A)$, where $c$ depends only on $\phi$ and $n$.

  2. (b)

If $m\in\\{0,n\\}$ or if $\phi$ is the Euclidean norm, then
$\rho^{\phi}_{A,m}(a,\eta)=1$ for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in N(A)$.

  3. (c)

For every Borel set $B\subseteq\mathbf{R}^{n+1}$ with $\sigma$-finite
$\mathcal{H}^{m}$ measure it holds that

$\bm{H}^{\phi}_{A,n-m}(a,\eta)\,\bm{1}_{N^{\phi}(A)|B}(a,\eta)=\bm{1}_{\widetilde{N}^{\phi}_{m}(A)|(A^{(m)}\cap
B)}(a,\eta)$ (46)

for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in N^{\phi}(A)|B$ and

$\displaystyle\int_{N^{\phi}(A)|B}\mathbf{1}_{D}(a,\eta)\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{A}(a,\eta)\,\bm{H}^{\phi}_{A,n-m}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=\int_{B\cap
A^{(m)}}\int_{N^{\phi}(A,a)}\mathbf{1}_{D}(a,\eta)\phi(\bm{n}^{\phi}(\eta))\,\rho^{\phi}_{A,m}(a,\eta)\,d\mathcal{H}^{n-m}(\eta)\,d\mathcal{H}^{m}(a)$

for every Borel set $D\subseteq\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}$.

###### Proof.

Since $N^{\phi}(A)|B=\bigcup_{j=0}^{n}(N^{\phi}(A)|(B\cap A^{(j)}))$, the
equality in (46) follows from Remark 3.24.

Since the case $m=0$ is easy to check directly, we assume that $m\geq 1$ in
the following.

Let $\tau_{i}:\widetilde{N}^{\phi}(A)\rightarrow\mathbf{R}^{n+1}$ and
$\zeta_{i}:\widetilde{N}^{\phi}(A)\rightarrow\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}$
be the maps defined in Lemma 3.9 for $i=1,\ldots,n$. For
$(a,\eta)\in\widetilde{N}^{\phi}(A)$ we define $T(a,\eta)$ to be the linear
space generated by $\zeta_{1}(a,\eta),\ldots,\zeta_{n}(a,\eta)$ and we notice
that

$\displaystyle 1\geq\big{\|}\bm{p}|T(a,\eta)\|^{m}$
$\displaystyle\geq\big{\|}{\textstyle\bigwedge_{m}}\big{(}\bm{p}|T(a,\eta)\big{)}\big{\|}$
$\displaystyle\geq\bigg{|}{\textstyle\bigwedge_{m}}\bm{p}\bigg{(}\frac{\zeta_{1}(a,\eta)\wedge\ldots\wedge\zeta_{m}(a,\eta)}{|\zeta_{1}(a,\eta)\wedge\ldots\wedge\zeta_{m}(a,\eta)|}\bigg{)}\bigg{|}=\frac{|\tau_{1}(a,\eta)\wedge\ldots\wedge\tau_{m}(a,\eta)|}{|\zeta_{1}(a,\eta)\wedge\ldots\wedge\zeta_{m}(a,\eta)|}>0$
(47)

for $(a,\eta)\in\widetilde{N}^{\phi}_{m}(A)$. Therefore we define

$\varrho^{\phi}_{A,m}(a,\eta)=\begin{cases}\displaystyle{\frac{J^{\phi}_{A}(a,\eta)}{\big{\|}{\textstyle\bigwedge_{m}}(\bm{p}|T(a,\eta))\big{\|}}}&\textrm{for
$(a,\eta)\in\widetilde{N}^{\phi}_{m}(A)$},\\\\[8.61108pt] 1&\textrm{for
$(a,\eta)\in N^{\phi}(A)\setminus\widetilde{N}^{\phi}_{m}(A)$}.\end{cases}$

Notice that $\rho^{\phi}_{A,n}(a,\eta)=1$ for $\mathcal{H}^{n}$ a.e.
$(a,\eta)\in N^{\phi}(A)$ by Lemma 3.9.

If $m\in\\{1,\ldots,n-1\\}$, we combine

$|\tau_{1}(a,\eta)\wedge\ldots\wedge\tau_{n}(a,\eta)|\leq|\tau_{1}(a,\eta)\wedge\ldots\wedge\tau_{m}(a,\eta)|\cdot|\tau_{m+1}(a,\eta)\wedge\ldots\wedge\tau_{n}(a,\eta)|$

(see [Fed69, 1.7.5]) with the estimate in (3.10) from Remark 3.10 to get

$\displaystyle\rho^{\phi}_{A,m}(a,\eta)$
$\displaystyle\leq\frac{|\tau_{1}(a,\eta)\wedge\ldots\wedge\tau_{n}(a,\eta)|}{|\zeta_{1}(a,\eta)\wedge\ldots\wedge\zeta_{n}(a,\eta)|}\cdot\frac{|\zeta_{1}(a,\eta)\wedge\ldots\wedge\zeta_{m}(a,\eta)|}{|\tau_{1}(a,\eta)\wedge\ldots\wedge\tau_{m}(a,\eta)|}$
$\displaystyle\leq\frac{C}{c^{2}}\cdot\frac{|\tau_{m+1}(a,\eta)\wedge\ldots\wedge\tau_{n}(a,\eta)|}{|\zeta_{m+1}(a,\eta)\wedge\ldots\wedge\zeta_{n}(a,\eta)|}=\frac{C}{c^{2}}$

for $(a,\eta)\in\widetilde{N}^{\phi}_{m}(A)$, which provides the required
finite upper bound with constants $0<c\leq C<\infty$ depending only on
$n,\phi$.

Let $N_{s}$ be the set defined in Theorem 3.16 for $s>0$. By Lemma 3.9, we
have
$\operatorname{ap}J^{N_{s}}_{m}\bm{p}(a,\eta)=\big{\|}{\textstyle\bigwedge_{m}}\big{(}\bm{p}|T(a,\eta)\big{)}\big{\|}$
for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in N_{s}$, for $s>0$. An application of
the coarea formula shows that

$\displaystyle\int_{N_{s}|(B\cap
A^{(m)})}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{A}(a,\eta)\,\mathbf{1}_{D}(a,\eta)\,\mathbf{1}_{\widetilde{N}^{\phi}_{m}(A)}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=\int_{B\cap
A^{(m)}}\int_{N_{s}\cap\bm{p}^{-1}(\\{a\\})}\phi(\bm{n}^{\phi}(\eta))\varrho_{A,m}^{\phi}(a,\eta)\mathbf{1}_{\widetilde{N}^{\phi}_{m}(A)}(a,\eta)\,\mathbf{1}_{D}(a,\eta)\,d\mathcal{H}^{n-m}(a,\eta)\,d\mathcal{H}^{m}(a)$

for $s>0$. Applying the monotone convergence theorem in combination with Lemma
3.25 (b) and (46), we obtain

$\displaystyle\int_{N^{\phi}(A)|B}\mathbf{1}_{D}(a,\eta)\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{A}(a,\eta)\,\bm{H}^{\phi}_{A,n-m}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=\int_{\widetilde{N}^{\phi}_{m}(A)|(B\cap
A^{(m)})}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{A}(a,\eta)\,\mathbf{1}_{D}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=\int_{B\cap
A^{(m)}}\int_{\widetilde{N}^{\phi}_{m}(A,a)}\phi(\bm{n}^{\phi}(\eta))\varrho_{A,m}^{\phi}(a,\eta)\,\mathbf{1}_{D}(a,\eta)\,d\mathcal{H}^{n-m}(\eta)\,d\mathcal{H}^{m}(a)$
$\displaystyle\qquad=\int_{B\cap
A^{(m)}}\int_{N^{\phi}(A,a)}\phi(\bm{n}^{\phi}(\eta))\varrho_{A,m}^{\phi}(a,\eta)\,\mathbf{1}_{D}(a,\eta)\,d\mathcal{H}^{n-m}(\eta)\,d\mathcal{H}^{m}(a).$

Finally, suppose that $\phi$ is the Euclidean norm. Then
$\\{\zeta_{\lambda(1)}(a,\eta)\wedge\ldots\wedge\zeta_{\lambda(m)}(a,\eta):\lambda\in\Lambda(n,m)\\}$
is an orthogonal basis of $\bigwedge_{m}T(a,\eta)$ for every
$(a,\eta)\in\widetilde{N}(A)$. Fix now
$(a,\eta)\in\widetilde{N}^{\phi}_{m}(A)$ and $\xi\in\bigwedge_{m}T(a,\eta)$
such that

$|\xi|=1\qquad\textrm{and}\qquad\xi=\sum_{\lambda\in\Lambda(n,m)}c_{\lambda}\cdot\frac{\zeta_{\lambda(1)}(a,\eta)\wedge\ldots\wedge\zeta_{\lambda(m)}(a,\eta)}{|\zeta_{\lambda(1)}(a,\eta)\wedge\ldots\wedge\zeta_{\lambda(m)}(a,\eta)|}.$

It follows from the orthogonality that $|c_{\lambda}|\leq 1$ for every
$\lambda\in\Lambda(n,m)$. Therefore, denoting with
$\lambda_{0}\in\Lambda(n,m)$ the map such that $\lambda_{0}(i)=i$ for every
$i\in\\{1,\ldots,m\\}$, we notice that

${\textstyle\bigwedge_{m}}\bm{p}(\xi)=c_{\lambda_{0}}\frac{\tau_{1}(a,\eta)\wedge\ldots\wedge\tau_{m}(a,\eta)}{|\zeta_{1}(a,\eta)\wedge\ldots\wedge\zeta_{m}(a,\eta)|}$

and we infer in combination with (3.4) that

$\big{\|}{\textstyle\bigwedge_{m}}\big{(}\bm{p}|T(a,\eta)\big{)}\big{\|}=\frac{|\tau_{1}(a,\eta)\wedge\ldots\wedge\tau_{m}(a,\eta)|}{|\zeta_{1}(a,\eta)\wedge\ldots\wedge\zeta_{m}(a,\eta)|}.$

We can now easily conclude that $\varrho^{\phi}_{A,m}(a,\eta)=1$ for every
$(a,\eta)\in\widetilde{N}^{\phi}_{m}(A)$ is a suitable choice. ∎

###  3.5 Relation between Euclidean and anisotropic curvatures

We consider the map

$T:\mathbf{R}^{n+1}\times\partial\mathcal{W}^{\phi}\rightarrow\mathbf{R}^{n+1}\times\mathbf{S}^{n},\qquad(a,\eta)\mapsto
T(a,\eta)=(a,\bm{n}^{\phi}(\eta)).$

By equation (8) in the introduction, $T$ is a $\mathcal{C}^{1}$-diffeomorphism
whose inverse is $T^{-1}(a,u)=(a,\nabla\phi(u))$ for
$(a,u)\in\mathbf{R}^{n+1}\times\mathbf{S}^{n}$. In particular,
$\operatorname{D}T(a,\eta):\mathbf{R}^{n+1}\times\operatorname{Tan}(\partial\mathcal{W}^{\phi},\eta)\rightarrow\mathbf{R}^{n+1}\times\operatorname{Tan}(\mathbf{S}^{n},\bm{n}^{\phi}(\eta))$
is an isomorphism and

$\operatorname{D}T(a,\eta)(\tau,\upsilon)=(\tau,\operatorname{D}\bm{n}^{\phi}(\eta)(\upsilon))\quad\textrm{for
$(\tau,\upsilon)\in\mathbf{R}^{n+1}\times\operatorname{Tan}(\partial\mathcal{W}^{\phi},\eta)$}.$

As already recalled in (13), we have $T(N^{\phi}(A))=N(A)$.

######  Remark 3.28.

Suppose $X$ is a normed vector space, $\mu$ is a measure over $X$ and
$f,g:X\rightarrow Y$ are functions differentiable at $a$. If the $m$
dimensional density of $\mu\operatorname{\llcorner}\\{x:f(x)\neq g(x)\\}$ is
zero at $a$, then

$\operatorname{D}f(a)(v)=\operatorname{D}g(a)(v)\qquad\textrm{for every
$v\in\operatorname{Tan}^{m}(\mu,a)$,}$

where $\operatorname{Tan}^{m}(\mu,a)$ is the cone of the $(\mu,m)$ approximate
tangent vectors of $\mu$ at $a$ (see [Fed69, 3.2.16]). One can check this
remark from the definitions.

The following result expresses the mean curvature functions
$\bm{H}^{\phi}_{A,j}\cdot\bm{1}_{\widetilde{N}^{\phi}_{n}}=E^{\phi}_{A,j}\cdot\bm{1}_{\widetilde{N}^{\phi}_{n}(A)}$
in terms of the Euclidean generalized curvatures
$\kappa_{A,1},\ldots,\kappa_{A,n}$ of $A$.

######  Theorem 3.29.

Let $\varnothing\neq A\subseteq\mathbf{R}^{n+1}$ be a closed set, and let
$\phi$ be a uniformly convex $\mathcal{C}^{2}$ norm.

Then
$\mathcal{H}^{n}(T(\widetilde{N}^{\phi}_{d}(A))\triangle\widetilde{N}_{d}(A))=0$
for $d=0,\ldots,n$. Moreover, if
$e_{1},\ldots,e_{n}:\widetilde{N}(A)\rightarrow\mathbf{R}^{n+1}$ are maps such
that

$\operatorname{D}\bm{\nu}_{A}(a+ru)(e_{i}(a,u))={\mathchoice{\raisebox{0.0pt}{$\displaystyle\chi$}}{\raisebox{0.0pt}{$\textstyle\chi$}}{\raisebox{0.0pt}{$\scriptstyle\chi$}}{\raisebox{0.0pt}{$\scriptscriptstyle\chi$}}}_{A,i}(a+ru)e_{i}(a,u)\quad\textrm{for
$i=1,\ldots,n$, $0<r<\underline{\bm{r}_{A}}(a,u)$}\text{ and
}(a,u)\in\widetilde{N}(A),$

then

$\displaystyle E^{\phi}_{A,j}(a,\eta)$
$\displaystyle=\sum_{\lambda\in\Lambda(n,j)}\bigg{(}\prod_{i=1}^{j}\kappa_{A,\lambda(i)}(T(a,\eta))\bigg{)}$
$\displaystyle\qquad\times\bigg{(}\bigwedge_{i=1}^{j}\operatorname{D}(\nabla\phi)(\bm{n}^{\phi}(\eta))(e_{\lambda(i)}(T(a,\eta)))\bigg{)}\bullet\big{(}\bigwedge_{i=1}^{j}e_{\lambda(i)}(T(a,\eta))\big{)}\bigg{)}$

for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in\widetilde{N}^{\phi}_{n}(A)$ and
$j\in\\{1,\ldots,n\\}$.

###### Proof.

Given the maps $e_{1},\ldots,e_{n}$ as in the statement of the Theorem, we
define the maps
$z_{i}:\widetilde{N}(A)\rightarrow\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}$, for
$i=1,\ldots,n$, so that

$z_{i}(a,u)=\begin{cases}(e_{i}(a,u),\kappa_{A,i}(a,u)e_{i}(a,u)),&\textrm{if
$\kappa_{A,i}(a,u)<\infty$},\\\ (0,e_{i}(a,u)),&\textrm{if
$\kappa_{A,i}(a,u)=+\infty$.}\end{cases}$

Then we choose the maps $\tau_{1},\ldots,\tau_{n}$ and
$\zeta_{1},\ldots,\zeta_{n}$ as in Lemma 3.9. Notice that
$e_{1}(a,u).\ldots,e_{n}(a,u)$ form an orthonormal basis of
$\operatorname{Tan}(\mathbf{S}^{n},u)=\operatorname{Tan}(\partial\mathcal{W}^{\phi},\nabla\phi(u))$.

Suppose $W\subseteq N^{\phi}(A)$ is $\mathcal{H}^{n}$ measurable and
$\mathcal{H}^{n}(W)<\infty$ and define $W^{\ast}$ as the set of $(a,\eta)\in
W$ such that

$\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}W,(a,\eta))=\textrm{span}\\{\zeta_{1}(a,\eta),\ldots,\zeta_{n}(a,\eta)\\}$

and

$\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}T(W),T(a,\eta))=\textrm{span}\\{z_{1}(T(a,\eta)),\ldots,z_{n}(T(a,\eta))\\}.$

Since $T$ is a bi-lipschitz map it follows from Lemma 3.9 that
$\mathcal{H}^{n}(W\setminus W^{\ast})=0$. Fix now $(a,\eta)\in W^{\ast}$ and
define
$V=\ker[\bm{p}|\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}W,(a,\eta))]$,
$d=n-\dim V$,
$V^{\prime}=\ker[\bm{p}|\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}T(W),T(a,\eta))]$
and $d^{\prime}=n-\dim V^{\prime}$. By [San20][Lemma B.2] we infer that

$\operatorname{D}T(a,\eta)\big{(}\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}W,(a,\eta))\big{)}=\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}T(W),T(a,\eta)),$

whence we can easily deduce that $\operatorname{D}T(a,\eta)(V)=V^{\prime}$.
This implies in particular that $d=d^{\prime}$. Consequently it holds that

$T(W^{\ast}\cap\widetilde{N}^{\phi}_{d}(A))\subseteq\widetilde{N}_{d}(A)\quad\textrm{and}\quad\widetilde{N}_{d}(A)\cap
T(W^{\ast})\subseteq T(\widetilde{N}^{\phi}_{d}(A))$

for every $d=0,\ldots,n$. Since $N^{\phi}(A)$ is a countable union of
$\mathcal{H}^{n}$ measurable sets $W$ with finite $\mathcal{H}^{n}$ measure,
we conclude

$\mathcal{H}^{n}\big{(}T(\widetilde{N}^{\phi}_{d}(A))\triangle\widetilde{N}_{d}(A)\big{)}=0.$

For $r>0$ we define

$Q_{r}=\\{(a,\eta)\in N^{\phi}(A):\bm{r}^{\phi}_{A}(a,\eta)\geq
r,\;\bm{r}_{A}(T(a,\eta))\geq r\\}.$

From the upper semicontinuity of $\bm{r}^{\phi}_{A}$ and $\bm{r}_{A}$ (see
[KS21, Lemma 2.35]) it follows that $Q_{r}$ is relatively closed in
$N^{\phi}(A)$ (in particular it is a Borel set) and
$\mathcal{H}^{n}\operatorname{\llcorner}Q_{r}$ is finite on compact sets by
Theorem 3.16(a). Notice that $Q_{r}\subseteq Q_{s}$ is $s\leq r$ and
$\bigcup_{r>0}Q_{r}=N^{\phi}(A)$. In addition, for $r>0$ we define
$Q^{\ast}_{r}$ as the set of all $(a,\eta)\in Q_{r}$ such that
$\underline{\bm{r}_{A}}(a,\eta)=\bm{r}_{A}(a,\eta)$, the $n$-dimensional
density of
$\mathcal{H}^{n}\operatorname{\llcorner}(\mathbf{R}^{n+1}\times\mathbf{R}^{n+1})\setminus
Q_{r}$ is zero at $(a,\eta)$ and
$\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}Q_{r},(a,\eta))=\operatorname{lin}\\{\zeta_{1}(a,\eta),\ldots,\zeta_{n}(a,\eta)\\}$.
By Lemma 3.9, [Fed69, 2.10.19] and Lemma 2.7 it follows that
$\mathcal{H}^{n}(Q_{r}\setminus Q^{\ast}_{r})=0$ for every $r>0$; moreover,
one can check directly from the definitions (see [Fed69, 3.2.16]) that
$Q^{\ast}_{r}\subseteq Q^{\ast}_{s}$ if $s\leq r$.

Fix now $(\hat{a},\hat{\eta})\in\widetilde{N}^{\phi}_{d}(A)\cap
T^{-1}(\widetilde{N}_{d}(A))\cap Q^{\ast}_{r}$ for some $r>0$. For $s>0$
define
$L_{s}:\mathbf{R}^{n+1}\times\partial\mathcal{W}^{\phi}\rightarrow\mathbf{R}^{n+1}$
by $L_{s}(a,\eta)=a+s\bm{n}^{\phi}(\eta)$. Choose a function
$u:\mathbf{R}^{n+1}\rightarrow\mathbf{R}^{n+1}$ such that
$u(x)\in\bm{\nu}_{A}(x)$ for every $x\in\mathbf{R}^{n+1}\setminus A$. For
$0<s\leq r$,

$(\nabla\phi\circ u\circ L_{s})(a,\eta)=\bm{q}(a,\eta)=\eta\quad\textrm{for
$(a,\eta)\in Q_{s}$}$

and $\nabla\phi\circ u\circ L_{s}$ is differentiable at $(\hat{a},\hat{\eta})$
by Remark 3.7. Remark 3.28 yields $\operatorname{D}(\nabla\phi\circ u\circ
L_{s})(\hat{a},\hat{\eta})(\xi)=\bm{q}(\xi)$ for
$\xi\in\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}Q_{s},(\hat{a},\hat{\eta}))$.
Hence, for $1\leq i\leq d$ and $s<r$ we compute

$\big{[}\operatorname{D}(\nabla\phi)(\bm{n}^{\phi}(\hat{\eta}))\circ\operatorname{D}u(\hat{a}+s\bm{n}^{\phi}(\hat{\eta})\big{]}\big{(}\tau_{i}(\hat{a},\hat{\eta})+s\kappa^{\phi}_{A,i}(\hat{a},\hat{\eta})\operatorname{D}\bm{n}^{\phi}(\hat{\eta})(\tau_{i}(\hat{a},\hat{\eta}))\big{)}=\kappa^{\phi}_{A,i}(\hat{a},\hat{\eta})\tau_{i}(\hat{a},\hat{\eta}).$
(48)

Noting that

$\displaystyle
s\operatorname{D}u(\hat{a}+s\bm{n}^{\phi}(\hat{\eta}))\bigg{(}\sum_{j=1}^{d}\big{[}\operatorname{D}\bm{n}^{\phi}(\hat{\eta})(\tau_{i}(\hat{a},\hat{\eta}))\bullet
e_{j}(T(\hat{a},\hat{\eta}))\big{]}e_{j}(T(\hat{a},\hat{\eta}))\bigg{)}$
$\displaystyle\qquad\qquad=s\sum_{j=1}^{d}\big{[}\operatorname{D}\bm{n}^{\phi}(\hat{\eta})(\tau_{i}(\hat{a},\hat{\eta}))\bullet
e_{j}(T(\hat{a},\hat{\eta}))\big{]}\frac{\kappa_{A,j}(T(\hat{a},\hat{\eta}))}{1+s\kappa_{A,j}(T(\hat{a},\hat{\eta}))}e_{j}(T(\hat{a},\hat{\eta}))\to
0\quad\textrm{as $s\to 0$}$

and

$\displaystyle
s\operatorname{D}u(\hat{a}+s\bm{n}^{\phi}(\hat{\eta}))\bigg{(}\sum_{j=d+1}^{n}\big{[}\operatorname{D}\bm{n}^{\phi}(\hat{\eta})(\tau_{i}(\hat{a},\hat{\eta}))\bullet
e_{j}(T(\hat{a},\hat{\eta}))\big{]}e_{j}(T(\hat{a},\hat{\eta}))\bigg{)}$
$\displaystyle\qquad\qquad=\sum_{j=d+1}^{n}\big{[}\operatorname{D}\bm{n}^{\phi}(\hat{\eta})(\tau_{i}(\hat{a},\hat{\eta}))\bullet
e_{j}(T(\hat{a},\hat{\eta}))\big{]}e_{j}(T(\hat{a},\hat{\eta}))\quad\textrm{for
$s>0$,}$

we conclude from (48) that

$\displaystyle\lim_{s\to
0}\big{[}\operatorname{D}(\nabla\phi)(\bm{n}^{\phi}(\hat{\eta}))\circ\operatorname{D}u(\hat{a}+s\bm{n}^{\phi}(\hat{\eta})\big{]}\big{(}\tau_{i}(\hat{a},\hat{\eta}))$
(49)
$\displaystyle\qquad=\kappa^{\phi}_{A,i}(\hat{a},\hat{\eta})\tau_{i}(\hat{a},\hat{\eta})$
$\displaystyle\qquad\qquad-\kappa^{\phi}_{A,i}(\hat{a},\hat{\eta})\sum_{j=d+1}^{n}\big{[}\operatorname{D}\bm{n}^{\phi}(\hat{\eta})(\tau_{i}(\hat{a},\hat{\eta}))\bullet
e_{j}(T(\hat{a},\hat{\eta}))\big{]}\operatorname{D}(\nabla\phi)(\bm{n}^{\phi}(\hat{\eta}))(e_{j}(T(\hat{a},\hat{\eta})))$

for $i\leq d$.

Now choose $d=n$ in the previous paragraph and define the linear maps
$T_{s}:\operatorname{Tan}(\partial\mathcal{W}^{\phi},\hat{\eta})\rightarrow\operatorname{Tan}(\partial\mathcal{W}^{\phi},\hat{\eta})$
by

$T_{s}=\operatorname{D}(\nabla\phi)(\bm{n}^{\phi}(\hat{\eta}))\circ\operatorname{D}u(\hat{a}+s\bm{n}^{\phi}(\hat{\eta})$

and
$T_{0}:\operatorname{Tan}(\partial\mathcal{W}^{\phi},\hat{\eta})\rightarrow\operatorname{Tan}(\partial\mathcal{W}^{\phi},\hat{\eta})$
by
$T_{0}(\tau_{i}(\hat{a},\hat{\eta}))=\kappa^{\phi}_{A,i}(\hat{a},\hat{\eta})\tau_{i}(\hat{a},\hat{\eta})$
for $i=1,\ldots,n$. Denoting by $\|\cdot\|$ the operator norm, we notice that
$\sup_{s>0}\|T_{s}\|<\infty$ and $T_{s}(v)\rightarrow T_{0}(v)$ for each
$v\in\operatorname{Tan}(\partial\mathcal{W}^{\phi}.\hat{\eta})$ by (49).
Therefore, $\lim_{s\to\infty}\|T_{s}-T_{0}\|=0$ and by continuity

$\displaystyle H^{\phi}_{A,j}(\hat{a},\hat{\eta})$
$\displaystyle=\operatorname{trace}\big{(}{\textstyle\bigwedge_{j}}T_{0}\big{)}$
$\displaystyle=\operatorname{trace}\big{(}{\textstyle\bigwedge_{j}}\lim_{s\to
0}T_{s}\big{)}=\lim_{s\to
0}\operatorname{trace}\big{(}{\textstyle\bigwedge_{j}}T_{s}\big{)}.$

Computation of the
$\operatorname{trace}\big{(}{\textstyle\bigwedge_{j}}T_{s}\big{)}$ by means of
the orthonormal basis
$\\{\bigwedge_{i=1}^{j}e_{\lambda(i)}(T(\hat{a},\hat{\eta})):\lambda\in\Lambda(n,j)\\}$
of $\bigwedge_{j}\operatorname{Tan}(\partial\mathcal{W}^{\phi},\hat{\eta})$,
we get the conclusion. ∎

##  4 Differentiability of the volume function

In this section, employing the Steiner-type formula from the previous section,
we study the differentiability properties of the (localized) parallel volume
function $V$ of an anisotropic tubular neighbourhood around an arbitrary
compact (closed) set. In particular, we obtain an expression for the left and
right derivative of $V$ in terms of the anisotropic curvatures of the compact
set and we deduce a geometric characterization of the differentiability points
of $V$. The results of this section extend [HLW04, eq. (4.5) and (4.6),
Corollary 4.5] and complement some of the results in [CLV21], see also Remark
4.4 below and the preceding work [Sta76, RW10, Wi19].

######  Remark 4.1.

Notice that if $\rho>0$, $y\in S^{\phi}(A,\rho)$ and
$a\in\bm{\xi}^{\phi}_{A}(x)$, that is, $a\in A$ and
$\phi^{\ast}(y-a)=\bm{\delta}^{\phi}_{A}(y)$, then

$U^{\phi}(a,\rho)\cap S^{\phi}(A,\rho)=\varnothing\quad\textrm{and}\quad
B^{\phi}(a,\rho)\subseteq B^{\phi}(A,\rho).$

Hence we deduce that
$\bm{p}\big{(}N(B^{\phi}(A,\rho))\big{)}=\partial^{v}B^{\phi}(A,\rho)=\partial^{v}_{+}B^{\phi}(A,\rho)$
and the function $u$ defined in Theorem 4.3 satisfies
$u(x)=\bm{n}(B^{\phi}(A,\rho),x)$ for every
$x\in\partial^{v}_{+}B^{\phi}(A,\rho)$.

The following auxiliary result is used in the proof of Theorem 4.3. At the
same time it provides an interesting insight into the nature of the cut points
in $\operatorname{Cut}^{\phi}(A)\cap\operatorname{Unp}^{\phi}(A)$. In fact, in
this regard we recall that there are closed (compact) sets
$A\subseteq\mathbf{R}^{n+1}$ such that
$\mathcal{H}^{n}[S^{\phi}(A,\rho)\setminus(\partial
B^{\phi}(A,\rho)\cup\operatorname{Unp}^{\phi}(A))]>0$ for some $\rho>0$; for
instance, let $A\subseteq\mathbf{R}^{2}$ be the union of two parallel lines
(segments) at distance $2$ and $\rho=1$. In view of these simple examples, the
second equation in (53) (which holds for every $\rho>0$!) is quite surprising
since $\operatorname{Cut}^{\phi}(A)\cap\operatorname{Unp}^{\phi}(A)$ can be a
much larger set than
$\mathbf{R}^{n+1}\setminus(A\cup\operatorname{Unp}^{\phi}(A))$. In fact, since
$\mathbf{R}^{n+1}\setminus(A\cup\operatorname{Unp}^{\phi}(A))$ is always an
$n$-dimensional set (see section 2.4), it follows from the example in [BH08]
that the set $\operatorname{Cut}^{\phi}(A)\cap\operatorname{Unp}^{\phi}(A)$
can be an $(n+1)$-dimensional set!

######  Lemma 4.2.

Let $A\subseteq\mathbf{R}^{n+1}$ be a closed set and $\rho>0$. Let
$f_{\rho}:N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}\geq\rho\\}\rightarrow
S^{\phi}(A,\rho)$ be defined by $f_{\rho}(a,\eta)=a+\rho\eta$. Then

$\displaystyle f_{\rho}(N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}\geq\rho\\})$
$\displaystyle=S^{\phi}(A,\rho),$ (50) $\displaystyle
f_{\rho}(N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}>\rho\\})$
$\displaystyle=\partial^{v}_{+}B^{\phi}(A,\rho)\subseteq
S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A),$ (51)
$S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)\subseteq\partial
B^{\phi}(A,\rho),\quad\mathcal{H}^{n}(S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)\setminus\partial^{v}_{+}B^{\phi}(A,\rho))=0,$
(52)
$\partial^{v}_{+}B^{\phi}(A,\rho)\cap\operatorname{Cut}^{\phi}(A)=\varnothing\quad\textrm{and}\quad\mathcal{H}^{n}(S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)\cap\operatorname{Cut}^{\phi}(A))=0.$
(53)

###### Proof.

We start with (50). “$\subseteq$”: Let $(x,\eta)\in
N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}\geq\rho\\}$. Then
$\bm{\delta}^{\phi}_{A}(x+t\eta)=t$ for $0\leq t<\rho$, hence also for
$t=\rho$ since $\delta^{\phi}_{A}$ is continuous. Thus, $f_{\rho}(x,\eta)\in
S^{\phi}(A,\rho)$.

“$\supseteq$”: Let $z\in S^{\phi}(A,\rho)$, that is,
$\delta^{\phi}_{A}(z)=\rho$ and there is some $x\in A$ with
$\phi^{*}(z-x)=\rho$. Set $\eta=\rho^{-1}(z-x)$. Then
$\bm{\xi}^{\phi}_{A}(x+t\eta)=x$ for $0\leq t<\rho$ by [DRKS20, Lemma 2.38
(g)]. It follows that $(x,\eta)\in N^{\phi}(A)$,
$\bm{r}^{\phi}_{A}(x,\eta)\geq\rho$ and $f_{\rho}(x,\eta)=z$.

Next we deal with (51). Let $(x,\eta)\in
N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}>\rho\\}$. Then $f_{\rho}(x,\eta)\in
S^{\phi}(A,\rho)$ by (50). Since $\bm{r}^{\phi}_{A}(x,\eta)>\rho$ it follows
from (16) that $\bm{\rho}^{\phi}_{A}(x+\rho\eta)>1$, and therefore also
$f_{\rho}(x,\eta)=x+\rho\eta\in\operatorname{Unp}^{\phi}(A)$. This yields
$f_{\rho}(N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}>\rho\\})\subseteq
S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)$.

Now we show that
$f_{\rho}(N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}>\rho\\})=\partial^{v}_{+}B^{\phi}(A,\rho)$
by proving two inclusions. “$\subseteq$”: Let $(x,\eta)\in
N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}>\rho\\}$. In view of Remark 4.1 it
suffices to show that $x+\rho\eta\in\partial B^{\phi}(A,\rho)$. For this,
choose $\bar{\rho}\in(\rho,\bm{r}^{\phi}_{A}(x,\eta))$. Then we have already
shown that $x+\bar{\rho}\eta\in
S^{\phi}(A,\bar{\rho})\cap\operatorname{Unp}^{\phi}(A)$ and clearly
$\bm{\xi}^{\phi}_{A}(x+\bar{\rho}\eta)=x$. We now claim that
$\bm{\xi}^{\phi}_{B^{\phi}(A,\rho)}(x+\bar{\rho}\eta)=x+\rho\eta$, which will
imply the required inclusion. Since

$x+\rho\eta\in(x+\bar{\rho}\eta+(\bar{\rho}-\rho)B^{\circ})\cap
B^{\phi}(A,\rho)\qquad\text{and}\qquad(x+\bar{\rho}\eta+(\bar{\rho}-\rho)\operatorname{int}(B^{\circ}))\cap
B^{\phi}(A,\rho)=\varnothing,$

we get $x+\rho\eta\in\bm{\xi}^{\phi}_{B^{\phi}(A,\rho)}(x+\bar{\rho}\eta)$ and
$\delta^{\phi}_{B^{\phi}(A,\rho)}(x+\bar{\rho}\eta)=\bar{\rho}-\rho>0$. Let
$z\in\xi^{\phi}_{B^{\phi}(A,\rho)}(x+\bar{\rho}\eta)$ be arbitrarily chosen.
Then

$z\in(x+\bar{\rho}\eta+(\bar{\rho}-\rho)\partial B^{\circ})\cap(a+\rho\partial
B^{\circ})\qquad\text{for some }a\in A$

and

$(x+\bar{\rho}\eta+(\bar{\rho}-\rho)\operatorname{int}(B^{\circ}))\cap(a+\rho\operatorname{int}(B^{\circ}))=\varnothing.$

Hence $z=a+\rho\eta_{1}=x+\bar{\rho}\eta+(\bar{\rho}-\rho)\eta_{2}$ for some
$\eta_{1},\eta_{2}\in\partial B^{\circ}$. Since $B^{\circ}$ is smooth and
strictly convex, it follows that $\eta_{1}=-\eta_{2}$. This implies that
$a=x+\bar{\rho}\eta+\bar{\rho}\eta_{2}\in\bm{\xi}^{\phi}_{A}(x+\bar{\rho}\eta)$,
thus $a=x$ and $\eta=-\eta_{2}=\eta_{1}$, which yields $z=x+\rho\eta$.

“$\supseteq$”: Let $z\in\partial^{v}_{+}B(A,\rho)$. Then there is some $x\in
A$ with

$z\in x+\rho B^{\circ}\subseteq B^{\phi}(A,\rho)\qquad\text{and}\qquad z\in
S^{\phi}(A,\rho).$

Furthermore there are $y\notin B^{\phi}(A,\rho)$ and $\epsilon>0$ such that

$z\in y+\epsilon
B^{\circ}\qquad\text{and}\qquad(y+\epsilon\operatorname{int}(B^{\circ}))\cap
B^{\phi}(A,\rho)=\varnothing.$

This implies that $\bm{\delta}_{A}^{\phi}(y)=\rho+\epsilon$. Moreover,

$\rho+\epsilon\leq\phi^{*}(y-x)\leq\phi^{*}(y-z)+\phi^{*}(z-x)\leq\epsilon+\rho,$

and hence $\phi^{*}(y-x)=\rho+\epsilon$, $\phi^{*}(y-z)=\epsilon$ and
$\phi^{*}(z-x)=\rho$. Since $\phi^{*}$ is strictly convex, we also get
$y-z=s(z-x)$ with some $s>0$ and therefore $z=\frac{1}{1+s}y+\frac{s}{1+s}x$.
We set $\eta=(\rho+\epsilon)^{-1}(y-x)\in\partial B^{\circ}$ and thus get
$\bm{\delta}^{\phi}_{A}(y)=\bm{\delta}^{\phi}_{A}(x+(\rho+\epsilon)\eta)=\rho+\epsilon$.
From [DRKS20, Lemma 2.38 (g)] we now conclude that $(x,\eta)\in N^{\phi}(A)$
and $\bm{r}^{\phi}_{A}(x,\eta)\geq\rho+\epsilon>\rho$, which yields
$z=f_{\rho}(x,\eta)\in f_{\rho}(N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}>\rho\\})$.

Now we turn to (52). We fix an arbitrary $\rho>0$ and $x\in
S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)$. Recall that
$\bm{\delta}^{\phi}_{A}$ is semiconcave on $\mathbf{R}^{n+1}\setminus
B^{\phi}(A,s)$ for every $s>0$ (see section 2.4). Since $x\in
S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)$, the distance function
$\bm{\delta}^{\phi}_{A}$ is differentiable at $x$ with
$0\notin\\{\nabla\bm{\delta}^{\phi}_{A}(x)\\}=\partial\bm{\delta}^{\phi}_{A}(x)$,
where we use that the generalized subgradient coincides with the subgradient
from convex analysis for semiconcave functions (see [Fu85, Remark 1.4] and the
references to [Cl90] given there). Hence we infer from [Fu85, Theorem 3.3 and
Proposition 1.7] that there exist $\epsilon,\delta>0$, $u\in\mathbf{S}^{n}$
and a lipschitzian semiconvex function $f:x+u^{\perp}\rightarrow\mathbf{R}$
such that

$\operatorname{epi}(f)\cap U_{\epsilon,\delta}(x,u)=B^{\phi}(A,\rho)\cap
U_{\epsilon,\delta}(x,u)\quad\textrm{and}\quad\operatorname{graph}(f)\cap
U_{\epsilon,\delta}(x,u)=S^{\phi}(A,\rho)\cap U_{\epsilon,\delta}(x,u).$

(One can easily see that
$u=\nabla\bm{\delta}^{\phi}_{A}(x)/|\nabla\bm{\delta}^{\phi}_{A}(x)|$, but
this is not relevant here). Since semiconvex functions are pointwise twice-
differentiable almost everywhere by a classical theorem of Alexandrov, we
conclude that for $\mathcal{H}^{n}$ a.e. $y\in S^{\phi}(A,\rho)\cap
U_{\epsilon,\delta}(x,u)$ there exists an open ball $B\subseteq
U_{\epsilon,\delta}(x,u)$ such that $B\cap B^{\phi}(A,\rho)=\varnothing$ and
$y\in\operatorname{clos}(B)\cap B^{\phi}(A,\rho)$. Therefore it follows from
Remark 4.1 that

$\mathcal{H}^{n}(S^{\phi}(A,\rho)\cap
U_{\epsilon,\delta}(x,u)\setminus\partial^{v}_{+}B^{\phi}(A,\rho))=0.$

Since $x$ is arbitrarily chosen in
$S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)$, we obtain the assertions
in (52).

The assertions in (53) follow immediately from what we have already shown. ∎

For a closed set $A\subseteq\mathbf{R}^{n+1}$, a bounded Borel set
$D\subset\mathbf{R}^{n+1}\times B^{\circ}$ and $\rho>0$, we define

$P^{\phi}_{\rho}(A,D)=\left\\{x\in\operatorname{Unp}(A)\setminus
A:\bm{\psi}^{\phi}_{A}(x)\in D,\bm{\delta}_{A}^{\phi}(x)\leq\rho\right\\}$

and

$V(A,D;\rho):=\mathcal{L}^{n+1}(P^{\phi}_{\rho}(A,D)).$

Recall that $\mathcal{W}^{\phi}=B^{\circ}$. If $\xi:\mathbf{R}^{n+1}\setminus
A\rightarrow\partial A$ and $\nu:\mathbf{R}^{n+1}\setminus
A\rightarrow\partial\mathcal{W}^{\phi}$ are two Borel functions such that

$\xi(x)\in\bm{\xi}^{\phi}_{A}(x)\qquad\text{and}\qquad\nu(x)\in\bm{\nu}^{\phi}_{A}(x),$
(54)

then we also define

$\overline{P}^{\phi}_{\rho}(A,D)=\left\\{x\in\mathbf{R}^{n+1}\setminus
A:(\xi(x),\nu(x))\in D,\bm{\delta}_{A}^{\phi}(x)\leq\rho\right\\}$

and

$\overline{V}(A,D;\rho):=\mathcal{L}^{n+1}(\overline{P}^{\phi}_{\rho}(A,D)).$

Clearly, we have $\overline{V}(A,D;\rho)={V}(A,D;\rho)$, and if $D=A\times
B^{\circ}$, then
$\overline{V}(A,D;\rho)={V}(A,D;\rho)=\mathcal{L}^{n+1}(B^{\phi}(A,\rho)\setminus
A)$. Furthermore, note that if $x\in
S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)\setminus A$, then
$\xi(x)=x-\rho\cdot\nu(x)$.

######  Theorem 4.3.

Let $A\subseteq\mathbf{R}^{n+1}$ be a closed set, and let $\phi$ be a
uniformly convex $\mathcal{C}^{2}$-norm on $\mathbf{R}^{n+1}$. Let
$D\subset\mathbf{R}^{n+1}\times\mathcal{W}^{\phi}$ be a bounded Borel set, and
let $\xi:\mathbf{R}^{n+1}\setminus A\rightarrow\partial A$ and
$\nu:\mathbf{R}^{n+1}\setminus A\rightarrow\partial\mathcal{W}^{\phi}$ be
Borel functions as in (54). Define a Borel function
$u:\mathbf{R}^{n+1}\setminus A\rightarrow\mathbf{S}^{n}$ by
$u(x)=\bm{n}^{\phi}(\nu(x))$ for $x\in\mathbf{R}^{n+1}\setminus A$. Then, for
every $\rho>0$ the right derivative $V^{\prime}_{+}(A,D;\rho)$ and the left
derivative $V^{\prime}_{-}(A,D;\rho)$ of $V$ at $\rho$ exist and are given by

$\displaystyle V^{\prime}_{+}(A,D;\rho)$
$\displaystyle=\sum_{d=0}^{n}\int_{\widetilde{N}^{\phi}_{d}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}>\rho\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\,\rho^{n-d}\prod_{i=1}^{d}(1+\rho\kappa^{\phi}_{A,i}(a,\eta))\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=\sum_{i=0}^{n}\rho^{i}\int_{N^{\phi}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}>\rho\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\bm{H}^{\phi}_{A,i}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=\int_{\partial^{v}_{+}B^{\phi}(A,\rho)}\operatorname{\bm{1}}_{D}(\xi(x),\nu(x))\phi(u(x))\,d\mathcal{H}^{n}(x)$
(55)

and

$\displaystyle V^{\prime}_{-}(A,D;\rho)$
$\displaystyle=\sum_{d=0}^{n}\int_{\widetilde{N}^{\phi}_{d}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}\geq\rho\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\,\rho^{n-d}\prod_{i=1}^{d}(1+\rho\kappa^{\phi}_{A,i}(a,\eta))\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=\sum_{i=0}^{n}\rho^{i}\int_{N^{\phi}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}\geq\rho\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\bm{H}^{\phi}_{A,i}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=V^{\prime}_{+}(A,D;\rho)+\int_{S^{\phi}(A,\rho)\setminus\operatorname{Unp}^{\phi}(A)}\big{(}\bm{1}_{D}(x-\rho\nu(x),\nu(x))$
(56)
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\quad+\bm{1}_{D}(x+\rho\nu(x),-\nu(x)\big{)}\phi(u(x))\,d\mathcal{H}^{n}(x).$

Consequently, $V(A,D;\cdot)$ is differentiable at $\rho>0$ if

$\int_{N^{\phi}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}=\rho\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\bm{H}^{\phi}_{A,i}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)=0\qquad\textrm{for
$i=0,\ldots,n$},$

and this happens for all but countably many $\rho\in(0,\infty)$.

Furthermore, $V(A,D;\cdot)$ is differentiable at $\rho>0$ if and only if

$\int_{S^{\phi}(A,\rho)\setminus\operatorname{Unp}^{\phi}(A)}\big{(}\bm{1}_{D}(x-\rho\nu(x),\nu(x))+\bm{1}_{D}(x+\rho\nu(x),-\nu(x)\big{)}\phi(u(x))\,d\mathcal{H}^{n}(x).$

In particular, $V(A,(\mathbf{R}^{n+1})^{2};\cdot)$ is differentiable at
$\rho>0$ if and only if
$\mathcal{H}^{n}(S^{\phi}(A,\rho)\setminus\operatorname{Unp}^{\phi}(A))=0$.

###### Proof.

Note that if $(a,\eta)\in N^{\phi}(A)$ and $0<t<\bm{r}^{\phi}_{A}(a,\eta)$,
then $a+t\eta\in\operatorname{Unp}^{\phi}(A)$ and therefore
$\xi(a+t\eta)=\bm{\xi}^{\phi}_{A}(a+t\eta)=a$ and
$\nu(a+t\eta)=\bm{\nu}^{\phi}_{A}(a+t\eta)=\eta$. Therefore, choosing
$h(x)=\bm{1}_{D}(\xi(x),\nu(x))$ in Corollary 3.18, we get that

$\displaystyle\overline{V}(A,D;\rho)$
$\displaystyle\qquad=\sum_{d=0}^{n}\int_{\widetilde{N}^{\phi}_{d}(A)\cap
D}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\bigg{(}\int_{0}^{\rho}\bm{1}\\{t<\bm{r}^{\phi}_{A}(a,\eta)\\}\,t^{n-d}\prod_{i=1}^{d}(1+t\kappa^{\phi}_{A,i}(a,\eta))\,dt\bigg{)}\,d\mathcal{H}^{n}(a,\eta)$

for $\rho>0$. Since the integrand is non-negative, we can exchange the order
of integration by Fubini theorem [Fed69, Theorem2.6.2] and obtain

$\displaystyle\overline{V}(A,D;\rho)$
$\displaystyle\qquad=\int_{0}^{\rho}\left(\sum_{d=0}^{n}\int_{\widetilde{N}^{\phi}_{d}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}>t\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\,t^{n-d}\prod_{i=1}^{d}(1+t\kappa^{\phi}_{A,i}(a,\eta))\,d\mathcal{H}^{n}(a,\eta)\right)\,dt$

for $\rho>0$. For $\rho>0$, we define

$g(\rho)=\sum_{d=0}^{n}\int_{\widetilde{N}^{\phi}_{d}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}>\rho\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\,\rho^{n-d}\prod_{i=1}^{d}(1+\rho\kappa^{\phi}_{A,i}(a,\eta))\,d\mathcal{H}^{n}(a,\eta)$

and $f_{\rho}:N_{\rho}\rightarrow S^{\phi}(A,\rho)$ by
$f_{\rho}(a,\eta)=a+\rho\eta$ (where $N_{\rho}$ is defined as in Theorem
3.16). Since

$\bm{1}_{\\{\bm{r}^{\phi}_{A}>s\\}}(a,\eta)\nearrow\bm{1}_{\\{\bm{r}^{\phi}_{A}>t\\}}(a,\eta)\qquad\textrm{as
$s\searrow t$ for $(a,\eta)\in N^{\phi}(A)$,}$

an application of the dominated convergence theorem gives that
$\lim_{s\downarrow t}g(s)=g(t)$ for $t>0$. Using the formula (28) in the proof
of Theorem 3.16, we get

$g(t)=\int_{N^{\phi}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}>t\\}}\phi(\bm{n}^{\phi}(\eta))\,\operatorname{ap}J_{n}^{N_{t}}f_{t}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)<\infty$
(57)

for $t>0$. Now, noting that $\overline{V}(A,D;\rho)=\int_{0}^{\rho}g(s)\,ds$,
we readily obtain that

$\overline{V}^{\prime}_{+}(A,D;\rho)=g(\rho)\qquad\textrm{for $\rho>0$.}$ (58)

By Lemma 4.2, we have
$f_{\rho}(N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}>\rho\\})=\partial^{v}_{+}B^{\phi}(A,\rho)\subseteq\operatorname{Unp}^{\phi}(A)\cap
S^{\phi}(A,\rho)$; moreover,

$\bm{q}(f_{\rho}^{-1}(\\{x\\}))=\\{\nu(x)\\}=\\{\bm{\nu}^{\phi}_{A}(x)\\}\qquad\textrm{for
$x\in\partial^{v}_{+}B^{\phi}(A,\rho)$}.$

Noting (57), we can apply the coarea formula to conclude that

$g(\rho)=\int_{\partial^{v}_{+}B^{\phi}(A,\rho)}\bm{1}_{D}(\xi(x),\nu(x))\phi(u(x))\,d\mathcal{H}^{n}(x)\qquad\textrm{for
$\rho>0$}.$

Now we deal with the left derivative. Since

$\bm{1}_{\\{\bm{r}^{\phi}_{A}>s\\}}\searrow\bm{1}_{\\{\bm{r}^{\phi}_{A}\geq
t\\}}\quad\text{as }s\nearrow t,$

it follows again from the dominated convergence theorem that

$\lim_{s\uparrow t}g(s)=\sum_{d=0}^{n}\int_{\widetilde{N}^{\phi}_{d}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}\geq
t\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\,t^{n-d}\prod_{i=1}^{d}(1+t\kappa^{\phi}_{A,i}(a,\eta))\,d\mathcal{H}^{n}(a,\eta)$

for $t>0$. Hence we deduce again using the formula (28) in the proof of
Theorem 3.16 that

$\displaystyle\overline{V}^{\prime}_{-}(A,D;\rho)$
$\displaystyle=\sum_{d=0}^{n}\int_{\widetilde{N}^{\phi}_{d}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}\geq\rho\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\,\rho^{n-d}\prod_{i=1}^{d}(1+\rho\kappa^{\phi}_{A,i}(a,\eta))\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=\int_{N^{\phi}(A)\cap
D\cap\\{\bm{r}^{\phi}_{A}\geq\rho\\}}\phi(\bm{n}^{\phi}(\eta))\operatorname{ap}J^{N_{\rho}}_{n}f_{\rho}(a,u)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=\int_{S^{\phi}(A,\rho)}\int_{f_{\rho}^{-1}(\\{x\\})}\bm{1}_{D}(a,\eta)\phi(\bm{n}^{\phi}(\eta))\,d\mathcal{H}^{0}(a,\eta)\,d\mathcal{H}^{n}(x),$

where (50) and the coarea formula have been used in the last step.

If $x\in S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)$, then there is a
unique $(a,\eta)\in N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}\geq\rho\\}$ with
$f_{\rho}(a,\eta)=x$, and we have $\xi(x)=\bm{\xi}^{\phi}_{A}(x)=a$,
$\nu(x)=\bm{\nu}^{\phi}_{A}(x)=\eta$ and $u(x)=\bm{n}^{\phi}(\eta)$. On the
other hand, if $x\in S^{\phi}(A,\rho)\setminus\operatorname{Unp}^{\phi}(A)$,
then $\mathcal{H}^{0}\big{(}\bm{q}(f_{\rho}^{-1}(\\{x\\}))\big{)}>1$, and in
addition we have

$\bm{q}(f_{\rho}^{-1}(\\{x\\}))\subseteq
N^{\phi}(S^{\phi}(A,\rho),x)\qquad\textrm{for $x\in S^{\phi}(A,\rho)$.}$ (59)

Consequently, we deduce from Lemma 3.25 that

$\bm{q}(f_{\rho}^{-1}(\\{x\\}))=\\{\nu(x),-\nu(x)\\}\qquad\textrm{for
$\mathcal{H}^{n}$ a.e.\ $x\in
S^{\phi}(A,\rho)\setminus\operatorname{Unp}^{\phi}(A)$}.$ (60)

Thus we conclude that

$\displaystyle\overline{V}^{\prime}_{-}(A,D;\rho)$
$\displaystyle=\int_{S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)}\bm{1}_{D}(\xi(x),\nu(x))\phi(u(x))\,d\mathcal{H}^{n}(x)$
$\displaystyle\qquad+\int_{S^{\phi}(A,\rho)\setminus\operatorname{Unp}^{\phi}(A)}\big{(}\bm{1}_{D}(x-\rho\nu(x),\nu(x))\phi(\bm{n}^{\phi}(\nu(x))$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad+\bm{1}_{D}(x+\rho\nu(x),-\nu(x))\phi(\bm{n}^{\phi}(-\nu(x))\big{)}\,d\mathcal{H}^{n}(x).$

It follows from (51) and (52) that the first summand on the right side of the
preceding equation equals $\overline{V}^{\prime}_{+}(A,D;\rho)$. We now obtain
the asserted representation of $\overline{V}^{\prime}_{-}(A,D;\rho)$ by
observing that $\bm{n}^{\phi}(\pm\nu(x))=\pm u(x)$.

To check the second equality both in (4.3) and in (4.3) we notice that

$\sum_{d=0}^{n}\rho^{n-d}\prod_{i=1}^{d}(1+\rho\kappa^{\phi}_{A,i}(a,\eta))\,\bm{1}_{\widetilde{N}^{\phi}_{d}(A)}(a,\eta)=\sum_{i=0}^{n}\rho^{i}\bm{H}^{\phi}_{A,i}(a,\eta)\qquad\textrm{for
$(a,\eta)\in\widetilde{N}^{\phi}(A)$}$

and we use the integrability condition in (25) proved in Theorem 3.16 to
interchange summation and integral.

Finally, for $\epsilon>0$ and $s>0$ the set

$I_{\epsilon,s}=\bigg{\\{}t\in[s,\infty):\int_{N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}=t\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)|\bm{H}^{\phi}_{A,i}(a,\eta)|\,d\mathcal{H}^{n}(a,u)\geq\epsilon\bigg{\\}}$

is finite by the integrability property in (25). This readily implies that

$\int_{N^{\phi}(A)\cap\\{\bm{r}^{\phi}_{A}=\rho\\}}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{A}(a,\eta)\bm{H}^{\phi}_{A,i}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)\neq
0$

for at most countably many different values of $\rho$. ∎

######  Remark 4.4.

The derivative of the volume function has been the subject of several
investigations; see [Sta76], [HLW04], [HLW06], [RW10] and [CLV21].

In order to compare our characterization of the differentiability of the
parallel volume function of a compact set (in the non-localized setting) with
the one in [CLV21], we introduce the following notation. If
$F\subseteq\mathbf{R}^{n+1}$ is a measurable set, then we write $F^{1}$ for
the set of all $x\in\mathbf{R}^{n+1}$ for which the $(n+1)$-dimensional
density of $F$ at $x$ equals $1$. Let $A\subset\mathbf{R}^{n+1}$ be compact
and $\rho>0$. Then it follows from (59) and (60) that

$\mathcal{H}^{n}\left([S^{\phi}(A,\rho)\setminus\operatorname{Unp}^{\phi}(A)]\setminus
B^{\phi}(A,\rho)^{1}\right)=0.$

On the other hand, we also have

$\displaystyle\mathcal{H}^{n}\left(S^{\phi}(A,\rho)\cap
B^{\phi}(A,\rho)^{1}\cap\operatorname{Unp}^{\phi}(A)\right)=\mathcal{H}^{n}\left(S^{\phi}(A,\rho)\cap\operatorname{Unp}^{\phi}(A)\setminus\partial^{v}_{+}B^{\phi}(A,\rho)\right)=0,$

where we use (52) and the fact that
$B^{\phi}(A,\rho)^{1}\cap\partial^{v}_{+}B^{\phi}(A,\rho)=\varnothing$. Thus
we see that if $\Delta$ denotes the symmetric difference operator for subsets
of $\mathbf{R}^{n+1}$, then

$\mathcal{H}^{n}\left([S^{\phi}(A,\rho)\cap
B^{\phi}(A,\rho)^{1}]\Delta[S^{\phi}(A,\rho)\setminus\operatorname{Unp}^{\phi}(A)]\right)=0.$

More generally, our method allows us to study the differentiation of local
parallel volumes in an anisotropic setting. Such a localization was suggested
in an isotropic framework by results in [Wi19] (note however that [Wi19, Lemma
2.9] is not correct, which affects the proof of [Wi19, Proposition 2.10] for
instance).

######  Remark 4.5.

By the arguments in the proof of Theorem 4.3 it also follows that
$V_{+}^{\prime}(A,D;\cdot)$ is continuous from the right and
$V_{-}^{\prime}(A,D;\cdot)$ is continuous from the left.

##  5 Alexandrov points of sets of positive reach

Throughout this section, $\phi$ is a uniformly convex $\mathcal{C}^{2}$-norm
on $\mathbf{R}^{n+1}$ with corresponding gauge body $B$. For $\rho>0$ we set
$B^{\phi}(\rho)=\rho B^{\circ}=\rho\mathcal{W}^{\phi}$ and
$B^{\phi}=B^{\circ}=\mathcal{W}^{\phi}$. Recall that if $\phi$ is the
Euclidean norm, then the upper index $\phi$ is omitted.

#### Notation.

For $a\in\mathbf{R}^{n+1}$, $u\in\mathbf{S}^{n}$ and $\varepsilon,\delta>0$ we
define

$u^{\perp}=\\{x\in\mathbf{R}^{n+1}:x\bullet u=0\\},$
$U_{\varepsilon}(a,u)=\\{x\in a+u^{\perp}:|x-a|<\varepsilon\\},\quad
U_{\varepsilon,\delta}(a,u)=\\{x+\lambda u:x\in
U_{\epsilon}(a,u),\;\lambda\in(-\delta,\delta)\\}.$

If $f:a+u^{\perp}\to\mathbf{R}$ is a function (and $u$ is a given orienting
normal vector), we write

$\operatorname{graph}(f)=\\{x-f(x)u:x\in
a+u^{\perp}\\}\qquad\text{and}\qquad\operatorname{epi}(f)=\\{x-su:s\geq
f(x),\,x\in a+u^{\perp}\\}$ (61)

for the graph of $f$ and the epigraph of $f$ (with respect to $u$),
respectively.

###  5.1 Positive $\phi$-reach

######  Definition 5.1.

Let $A\subseteq\mathbf{R}^{n+1}$ be a closed set. The _$\phi$ -reach_ of $A$
is defined as the non-negative number

$\operatorname{reach}^{\phi}(A)=\sup\\{\rho\geq
0:B^{\phi}(A,\rho)\subseteq\operatorname{Unp}^{\phi}(A)\\}.$

The set $A$ is said to have _positive $\phi$-reach_ if
$\operatorname{reach}^{\phi}(A)>0$.

In the following, we say that a convex body $L\subset\mathbf{R}^{n+1}$ slides
freely inside a convex body $K\subset\mathbf{R}^{n+1}$ if for each
$x\in\partial K$ there is some $t\in\mathbf{R}^{n+1}$ such that $x\in
L+t\subseteq K$ (see [Sch14, Section 3.2]). In particular, $L$ slides freely
inside $K$ if and only if $L$ is a summand of $K$. It follows from [Sch14,
Theorem 3.2.12] that if $L,K\subset\mathbf{R}^{n+1}$ are convex bodies of
class $C^{2}_{+}$, then there is some $\rho>0$ (depending on $L,K$) such that
$\rho L$ slides freely inside $K$.

######  Lemma 5.2.

Let $A\subset\mathbf{R}^{n+1}$ be a closed set. Let $\phi,\bar{\phi}$ be
uniformly convex $C^{2}$ norms on $\mathbf{R}^{n+1}$ with corresponding gauge
bodies $B,\bar{B}$. Let $\rho>0$ be such that $B^{\phi}(\rho)=\rho B^{\circ}$
slides freely inside $B^{\bar{\phi}}=\bar{B}^{\circ}$. If
$\operatorname{reach}^{\bar{\phi}}(A)>r$, then
$\operatorname{reach}^{\phi}(A)>\rho r$.

###### Proof.

Assume that $\operatorname{reach}^{\bar{\phi}}(A)>r$ and $B^{\phi}(\rho)=\rho
B^{\circ}$ slides freely inside $B^{\bar{\phi}}=\bar{B}^{\circ}$. Then
$sB^{\circ}$ slides freely inside $r\bar{B}^{\circ}$ if $0<s\leq\rho r$.
Aiming at a contradiction, we assume that there is some
$z\in\mathbf{R}^{n+1}\setminus A$ and there are $x_{1},x_{2}\in\partial A$
with $x_{1}\neq x_{2}$ and such that
$\\{x_{1},x_{2}\\}\subseteq(z+s_{0}B^{\circ})\cap A$ and
$\operatorname{int}(z+s_{0}B^{\circ})\cap A=\emptyset$ for some
$s_{0}\in(0,\rho r]$. Then $x_{i}-z\in\partial(s_{0}B^{\circ})$ and
$N(s_{0}B^{\circ},x_{i}-z)=\\{v_{i}\\}$ for some unit vector $v_{i}$, for
$i=1,2$ (since $B^{\circ}$ is smooth). In particular, we have
$x_{i}-z=s_{0}\nabla h_{B^{\circ}}(v_{i})$ for $i=1,2$. Since $B^{\circ}$ is
of class $C^{2}_{+}$ (and hence a Euclidean ball slides freely inside
$B^{\circ}$), it follows that $-v_{i}\in N(A,x_{i})$ for $i=1,2$. Since
$\operatorname{reach}^{\bar{\phi}}(A)>r$, we conclude that

$(x_{i}-r\nabla h_{\bar{B}^{\circ}}(v_{i})+r\bar{B}^{\circ})\cap
A=\\{x_{i}\\}.$ (62)

Using first that $x_{i}-s_{0}\nabla h_{B^{\circ}}(v_{i})=z$ for $i=1,2$ and
then that $s_{0}B^{\circ}$ slides freely inside $r\bar{B}^{\circ}$, we get

$\\{x_{1},x_{2}\\}\subset z+s_{0}B^{\circ}\subseteq x_{i}-s_{0}\nabla
h_{B^{\circ}}(v_{i})+s_{0}B^{\circ}\subseteq x_{i}-r\nabla
h_{\bar{B}^{\circ}}(v_{i})+r\bar{B}^{\circ}.$

But then (62) yields

$x_{2}\in(x_{1}-r\nabla h_{\bar{B}^{\circ}}(v_{1})+r\bar{B}^{\circ})\cap
A=\\{x_{1}\\},$

a contradiction. ∎

######  Remark 5.3.

Assume that $\phi,\psi$ are any two uniformly convex $\mathcal{C}^{2}$-norms.
Then $\operatorname{reach}^{\phi}(A)>0$ if and only if
$\operatorname{reach}^{\psi}(A)>0$. We say that a closed set
$A\subseteq\mathbf{R}^{n+1}$ is a _set of positive reach_ if
$\operatorname{reach}(A)>0$, that is, if $A$ has positive reach with respect
to the Euclidean norm. Hence, a set has positive reach if and only if it has
positive $\phi$-reach for some (and then for any) uniformly convex
$\mathcal{C}^{2}$-norm $\phi$ on $\mathbf{R}^{n+1}$.

######  Remark 5.4.

The class of sets of positive reach as defined here is precisely the class of
sets of positive reach introduced in [Fed59, Definition 4.1]. We recall from
[Fed59, Theorem 4.8 (12)] that if $A\subseteq\mathbf{R}^{n+1}$ is a set with
positive reach, then

$N(A)=\\{(a,u):a\in A,u\in\operatorname{Nor}(A,a),|u|=1\\}.$

Moreover it follows from Lemma 2.6 that
$\bm{\xi}^{\phi}_{A}|\\{x\in\mathbf{R}^{n+1}:0<\bm{\delta}^{\phi}_{A}(x)<\operatorname{reach}^{\phi}(A)\\}$
is a locally Lipschitz map and $\bm{\psi}|S^{\phi}(A,r)$ is a locally
bilipschitz homeomorphism onto $N^{\phi}(A)$ for
$r\in(0,\operatorname{reach}^{\phi}(A))$. In particular, $S^{\phi}(A,r)$ is a
$\mathcal{C}^{1,1}$ closed hypersurface for
$r\in(0,\operatorname{reach}^{\phi}(A))$ and $N^{\phi}(A)$ is a closed
Lipschitz $n$-dimensional submanifold of
$\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}$.

In the next section we exploit the fundamental connection between sets of
positive reach and semiconvex functions. We will refer to the fundamental work
[Fu85] (see also the cited references therein). Here we recall the notion of a
semiconvex function.

######  Definition 5.5.

Suppose $U\subseteq\mathbf{R}^{k}$ is open and convex. Then a function
$f:U\rightarrow\mathbf{R}$ is called _semiconvex_ if there exists
$0\leq\kappa<\infty$ such that the function $U\ni y\mapsto\kappa|y|^{2}+f(y)$
is convex. We denote the _generalized Clarke gradient_ of $f$ at $a\in U$ by
$\partial f(a)$ (see [Fu85] or [Cl90]).

###  5.2 Alexandrov points and pointwise curvatures

The first main result of this section (see Theorem 5.6) states that for a set
$A$ of positive reach the set $\bm{p}(\widetilde{N}^{\phi}(A))$ (which is the
set of curvature points of $A$) can be partitioned as follows

$\bm{p}(\widetilde{N}^{\phi}(A))=\bm{p}(\widetilde{N}^{\phi}(A)\setminus\widetilde{N}^{\phi}_{n}(A))\cup\bm{p}(\widetilde{N}^{\phi}_{n}(A))$
(63)

For a convex body this partition is well known, as
$\bm{p}(\widetilde{N}^{\phi}_{n}(A))$ is the set of normal boundary points
(also known as Alexandrov points); see [Sch14, Notes for Sections 1.5 and
2.6], [Hu98, Lemma 3.1] and the literature cited there. The second goal of
this section is to extend the notion of an Alexandrov point to sets of
positive reach. This notion will play a central role in most of the subsequent
rigidity statements; see for instance Theorem 5.15 and Corollary 5.16 in the
next section.

######  Theorem 5.6.

Let $A\subseteq\mathbf{R}^{n+1}$ be a closed set with
$\operatorname{reach}(A)>0$.

If $(a,\eta)\in\widetilde{N}^{\phi}_{n}(A)$, $u\in N(A,a)$ with
$\nabla\phi(u)=\eta$ and $\nu\in\mathbf{S}^{n}\setminus u^{\perp}$, then the
following statements hold.

  1. (a)

There exist $\epsilon(\nu)>0$ and a function
$f_{\nu}:U_{\epsilon(\nu)}(a,\nu)\rightarrow\mathbf{R}$ which is
differentiable at $a$ and such that
$\operatorname{graph}(f_{\nu})\subseteq\partial A$.

  2. (b)

There exists a map
$\widetilde{\eta}:U_{\epsilon(u)}(a,u)\rightarrow\partial\mathcal{W}^{\phi}$
such that $\widetilde{\eta}(b)\in N^{\phi}(A,b+f_{u}(b)u)$ for every $b\in
U_{\epsilon(u)}(a,u)$, $\widetilde{\eta}(a)=\eta$, $\widetilde{\eta}$ is
differentiable at $a$ and the eigenvalues
$\lambda_{1}\leq\ldots\leq\lambda_{n}$ of
$\operatorname{D}\widetilde{\eta}(a)$ satisfy

$\lambda_{i}=\kappa^{\phi}_{A,i}(a,\eta)\qquad\textrm{for $i=1,\ldots,n$.}$

Moreover,
$\bm{p}(\widetilde{N}^{\phi}_{n}(A))\cap\bm{p}(\widetilde{N}^{\phi}(A)\setminus\widetilde{N}^{\phi}_{n}(A))=\varnothing$.

###### Proof.

a Suppose $a=0$ and $0<r<\operatorname{reach}^{\phi}(A)$. Fix
$\nu\in\mathbf{S}^{n}\setminus u^{\perp}$ and let $\pi_{\nu}$ be the
orthogonal projection onto $\nu^{\perp}$. We consider the lipschitz function
$\psi_{\nu}:S^{\phi}(A,r)\rightarrow\nu^{\perp}$,
$\pi_{\nu}\circ\bm{\xi}^{\phi}_{A}$. If $H$ is the halfspace orthogonal to $u$
which does not contain $u$, we notice the inclusions

$\operatorname{D}\bm{\xi}^{\phi}_{A}(r\eta)[\operatorname{Tan}(S^{\phi}(A,r),r\eta)]\subseteq\operatorname{Tan}(\bm{\xi}^{\phi}_{A}(S^{\phi}(A,r)),0)\subseteq\operatorname{Tan}(A,0)\subseteq
H.$

Since
$\operatorname{D}\bm{\xi}^{\phi}_{A}(r\eta)|\operatorname{Tan}(S^{\phi}(A,r),r\eta)$
is injective, we infer that

$\operatorname{Tan}(S^{\phi}(A,r),r\eta)=\operatorname{D}\bm{\xi}^{\phi}_{A}(r\eta)[\operatorname{Tan}(S^{\phi}(A,r),r\eta)]=u^{\perp}$

and $\operatorname{D}\psi_{\nu}(r\eta)$ is an isomorphism onto $\nu^{\perp}$.
Consequently, we can apply the (right-) inverse function theorem in [Sl00] to
infer the existence of a constant $\epsilon(\nu)>0$ and a map
$\varphi_{\nu}:U_{\epsilon(\nu)}(0,\nu)\rightarrow S^{\phi}(A,r)$ such that
$\varphi_{\nu}(0)=r\eta$, $\varphi_{\nu}$ is differentiable at $0$ with
$\operatorname{D}\varphi_{\nu}(0)=\operatorname{D}\psi_{\nu}(r\eta)^{-1}=\operatorname{D}\bm{\xi}^{\phi}_{A}(r\eta)^{-1}\circ(\pi_{\nu}|u^{\perp})^{-1}$
and $\psi_{\nu}(\varphi_{\nu}(b))=b$ for every $b\in
U_{\epsilon(\nu)}(a,\nu)$. We define
$f_{\nu}:U_{\epsilon(\nu)}(a,\nu)\rightarrow\mathbf{R}$ by

$f_{\nu}(b)=\bm{\xi}^{\phi}_{A}(\varphi_{\nu}(b))\bullet\nu\qquad\textrm{for
$b\in U_{\epsilon(\nu)}(a,\nu)$}$

and we readily check that $f_{\nu}(0)=0$,
$\operatorname{D}f_{\nu}(0)=(\pi_{\nu}|u^{\perp})^{-1}\bullet\nu$ and
$b+f_{\nu}(b)\nu=\bm{\xi}^{\phi}_{A}(\varphi_{\nu}(b))\in\partial A$ for every
$b\in U_{\epsilon(\nu)}(a,\nu)$.

b We define
$\widetilde{\eta}:U_{\epsilon(u)}(a,u)\rightarrow\partial\mathcal{W}^{\phi}$
by

$\widetilde{\eta}(b)=\frac{1}{r}(\varphi_{u}(b)-\bm{\xi}^{\phi}_{A}(\varphi_{u}(b)))\qquad\textrm{for
$b\in U_{\epsilon(u)}(a,u)$}$

and notice that $\widetilde{\eta}(0)=\eta$ and $\widetilde{\eta}(b)\in
N^{\phi}(A,b+f_{u}(b)u)$ for every $b\in U_{\epsilon(u)}(a,u)$ and

$\operatorname{D}\widetilde{\eta}(0)=\frac{1}{r}(\operatorname{D}\bm{\xi}^{\phi}_{A}(r\eta)^{-1}|u^{\perp}-\bm{1}_{u^{\perp}}).$

Noting that
$\\{(1-r{\mathchoice{\raisebox{0.0pt}{$\displaystyle\chi$}}{\raisebox{0.0pt}{$\textstyle\chi$}}{\raisebox{0.0pt}{$\scriptstyle\chi$}}{\raisebox{0.0pt}{$\scriptscriptstyle\chi$}}}^{\phi}_{A,i}(r\eta))^{-1}:i=1,\ldots,n\\}$
are the eigenvalues of
$\operatorname{D}\bm{\xi}^{\phi}_{A}(r\eta)^{-1}|u^{\perp}$, we conclude that
the eigenvalues of $\operatorname{D}\widetilde{\eta}(0)$ satisfy the equation
$\lambda_{i}=\kappa^{\phi}_{A,i}(a,\eta)$ for $i\in\\{1,\ldots,n\\}$.

Finally, since $\nabla\phi(u)\bullet u=\phi(u)\neq 0$ for $u\in\mathbf{S}^{n}$
(see 8), the remaining assertion follows from part a and Lemma 3.14. ∎

A more refined description of a set of positive reach around the points of the
viscosity boundary is given by the following result.

######  Theorem 5.7.

Let $A\subset\mathbf{R}^{n+1}$ be a closed set with
$\operatorname{reach}(A)>0$ and $a\in\partial^{v}A$. Assume that
$N(A,a)=\\{u\\}$ for some $u\in\mathbf{S}^{n}$. Then the following statements
hold.

  1. (a)

There are $\varepsilon,\delta>0$ and a semiconvex lipschitz function
$f:a+u^{\perp}\to\mathbf{R}$ such that $f(a)=0$, $f$ is differentiable at $a$
with $\operatorname{D}f(a)=0$ and (with respect to the orienting normal vector
$u$)

$\operatorname{graph}(f)\cap U_{\varepsilon,\delta}(a,u)=\partial A\cap
U_{\varepsilon,\delta}(a,u),\quad\operatorname{epi}(f)\cap
U_{\varepsilon,\delta}(a,u)=A\cap U_{\varepsilon,\delta}(a,u).$ (64)

  2. (b)

$(a,\nabla\phi(u))\in\widetilde{N}^{\phi}_{n}(A)$ if and only if $f$ is
pointwise twice differentiable at $a$. In this case, every map
$\nu:a+u^{\perp}\rightarrow\mathbf{S}^{n}$ such that $\nu(b)\in
N(\operatorname{epi}(f),b-f(b)u)$ for every $b\in a+u^{\perp}$ is
differentiable at $a$ and satisfies

$\operatorname{D}^{2}f(a)(\tau_{1},\operatorname{D}(\nabla\phi)(u)(\tau_{2}))=\operatorname{D}(\nabla\phi\circ\nu)(a)(\tau_{1})\bullet\tau_{2}\qquad\textrm{for
$\tau_{1},\tau_{2}\in u^{\perp}$.}$

###### Proof.

Let $0<r<\operatorname{reach}(A)$. We start with (a). By [Fed69, Theorem 4.8
(12)] we have $\operatorname{Tan}(A,a)=\\{v\in\mathbf{R}^{n+1}:v\bullet u\leq
0\\}$. Then $\mathcal{C}=\\{v\in\mathbf{R}^{n+1}:v\bullet
u\leq-\frac{1}{2}|v|\\}$ defines a closed convex cone such that
$\mathcal{C}\subseteq\operatorname{int}(\operatorname{Tan}(A,a))\cup\\{0\\}$.
Lemma 3.5 in [RZ17] shows that there is some $s\in(0,r/2)$ such that

$(a+\mathcal{C})\cap B(a,s)\subseteq A.$ (65)

We consider the set

$M_{u}=\\{z\in\partial A:|u-n_{z}|<1/4\text{ for some
}n_{z}\in\operatorname{Nor}(A,z)\cap\mathbf{S}^{n}\\}.$

It follows from [RZ17, Proposition 3.1 (iii)] that there is some
$s^{\prime}\in(0,s)$ such that if $z\in\partial A\cap B(a,s^{\prime})$, then
$|u-n_{z}|<1/4$ whenever $n_{z}\in\operatorname{Nor}(A,z)\cap\mathbf{S}^{n}$.
The proof of [RZ17, Theorem 5.9] then shows that $\partial A\cap
B(a,s^{\prime})\subset M_{u}\cap B(a,s^{\prime})$ is contained in the graph of
a lipschitz semiconvex function $f:a+u^{\perp}\to\mathbf{R}$ with $f(a)=0$,
and hence there are $\varepsilon,\delta>0$ such that (with respect to the
orienting normal vector $u$)

$\partial A\cap
U_{\varepsilon,\delta}(a,u)\subseteq\operatorname{graph}(f)\cap
U_{\varepsilon,\delta}(a,u).$ (66)

In order to verify the remaining assertions, we use [RZ17, Proposition 3.3]
(see [Fed59, Theorem 4.18]). Thus we get

$\displaystyle A\cap B(a,r)$
$\displaystyle\subseteq\left\\{a+x+tu:t\leq\tfrac{1}{2r}(|x|^{2}+t^{2}),x\in
u^{\perp},t\in\mathbf{R},|x+tu|\leq r\right\\}$
$\displaystyle\subseteq\left\\{a+x+tu:-r\leq t\leq\tfrac{1}{r}|x|^{2},x\in
u^{\perp},t\in\mathbf{R},|x|\leq r\right\\}.$ (67)

At this point the equalities in (64) follows from elementary topology by
combining (65)–(67). Moreover, since
$\operatorname{Nor}(\operatorname{epi}(f),a)=\operatorname{Nor}(A,a)=\\{tu:t\geq
0\\}$, it follows from [Fu85, Remarks 1.4, Lemma 2.9] that the generalized
Clarke gradient of $f$ at $a$ contains only $0$ and $f$ is differentiable at
$a$ with $\operatorname{D}f(a)=0$.

(b) Let $\mathcal{N}$ be the family of all maps
$\nu:a+u^{\perp}\rightarrow\mathbf{S}^{n}$ such that $\nu(b)\in
N(\operatorname{epi}(f),b-f(b)u)$ for every $b\in a+u^{\perp}$, where $f$ is
chosen according to (64) with respect to the orienting normal vector $u$. For
every $\nu\in\mathcal{N}$ we define the function
$g_{\nu}:U_{\epsilon}(a,u)\rightarrow u^{\perp}$ by

$g_{\nu}(b)=\frac{\nu(b)-(\nu(b)\bullet u)u}{\nu(b)\bullet u}\quad\textrm{for
$b\in U_{\epsilon}(a,u)$.}$

(Notice $\nu(b)\bullet u>0$ for every $b\in U_{\epsilon}(a,u)$ since $f$ is
lipschitz). Employing [Fu85, Lemma 2.9] we conclude that
$g_{\nu}(b)\in\partial f(b)$ for every $b\in U_{\epsilon}(a,u)$. In
particular, if we assume that
$(a,\nabla\phi(u))\in\widetilde{N}^{\phi}_{n}(A)$, then it follows from
Theorem 5.6 that there exists at least one map $\nu_{0}\in\mathcal{N}$ that is
differentiable at $a$. Therefore $g_{\nu_{0}}$ is differentiable at $a$ and
the classical theory of subgradients for (semi)convex functions (see [Ba79] or
[KS21, Lemma 2.39]) implies that $f$ is twice differentiable at $a$.

Suppose $f$ is twice differentiable at $a$, $a=0$ and $\nu\in\mathcal{N}$.
Then the aforementioned theory of subgradients for (semi)convex functions
implies that $\nu$ is differentiable at $0$ with
$\operatorname{D}\nu(0)\bullet\nu(0)=0$ and

$\operatorname{D}\nu(0)(\tau_{1})\bullet\tau_{2}=\operatorname{D}g_{\nu}(0)(\tau_{1})\bullet\tau_{2}=\operatorname{D}^{2}f(0)(\tau_{1},\tau_{2})$
(68)

for $\tau_{1},\tau_{2}\in u^{\perp}$ and for every $\nu\in\mathcal{N}$. It
follows that

$\operatorname{D}(\nabla\phi\circ\nu)(0)(\tau_{1})\bullet\tau_{2}=\operatorname{D}\nu(0)(\tau_{1})\bullet\operatorname{D}(\nabla\phi)(u)(\tau_{2})=\operatorname{D}^{2}f(0)(\tau_{1},\operatorname{D}(\nabla\phi)(u)(\tau_{2}))$

for $\tau_{1},\tau_{2}\in u^{\perp}$. Now we choose $r,\epsilon^{\prime}>0$ so
that $0<\epsilon^{\prime}<r<r+\epsilon^{\prime}<\operatorname{reach}(A)$ and
we define $J=(r-\epsilon^{\prime},r+\epsilon^{\prime})$,
$\eta=\nabla\phi\circ\nu$ and the function $F:U_{\epsilon}(0,u)\times
J\rightarrow\mathbf{R}^{n+1}$ by

$F(b,t)=b-f(b)u+t\eta(b)\qquad\textrm{for $b\in U_{\epsilon}(0,u)\times J$.}$

Then $F$ is differentiable at $(0,r)$ and, noting that
$U(\operatorname{reach}^{\phi}(A)\eta(0),\operatorname{reach}^{\phi}(A))\cap\operatorname{epi}(f)\cap
U_{\epsilon,\delta}(0,u)=\varnothing$, we conclude from (68), employing the
same comparison-of-curvatures argument as the one used in the proof of Lemma
3.14, that all eigenvalues of $\operatorname{D}\eta(0)$ are smaller or equal
than $\operatorname{reach}^{\phi}(A)^{-1}$. Consequently
$\operatorname{D}F(a,r)$ is invertible. Let $\pi:\mathbf{R}^{n+1}\rightarrow
u^{\perp}$ be the orthogonal projection onto $u^{\perp}$. Since the function
$G:(\bm{\delta}_{A}^{\phi})^{-1}(J)\rightarrow u^{\perp}\times\mathbf{R}$,
defined by $G(x)=(\pi(\bm{\xi}_{A}^{\phi}(x)),\bm{\delta}_{A}^{\phi}(x))$ for
$x\in(\bm{\delta}_{A}^{\phi})^{-1}(J)$, is Lipschitzian and satisfies

$F(U_{\epsilon}(0,u)\times
J)\subseteq(\bm{\delta}_{A}^{\phi})^{-1}(J)\quad\textrm{and}\quad G\circ
F=\bm{1}_{U_{\epsilon}(0,u)\times J},$

it follows from Lemma 2.5 that $G$ and consequently $\bm{\xi}_{A}$ is
differentiable at $ru$. Since
$\underline{\bm{r}_{A}^{\phi}}(0,\nabla\phi(u))\geq\operatorname{reach}^{\phi}(A)>r$
by [KS21, Lemma 4.16], we infer that $\bm{\xi}^{\phi}_{A}$ is differentiable
at $s\nabla\phi(u)$ for every
$0<s<\underline{\bm{r}_{A}^{\phi}}(0,\nabla\phi(u))$. Therefore
$(0,\nabla\phi(u))\in\widetilde{N}^{\phi}(A)$. Since $f$ is twice
differentiable at $0$, it follows that $0\in\partial^{v}_{+}A$ and
$(0,\nabla\phi(u))\in\widetilde{N}^{\phi}_{n}(A)$ by Lemma 3.14b. ∎

Subsequently, we prefer to write $C$ for a set of positive reach. Theorem 5.7
motivates the following definition.

######  Definition 5.8.

Suppose $C$ is a set of positive reach, $a\in\partial^{v}C$, $N(C,a)=\\{u\\}$
and $f:a+u^{\perp}\rightarrow\mathbf{R}$ is a semiconvex function locally
representing $C$ as in Theorem 5.7. Then $a$ is said to be an _Alexandrov
point of $C$_ if $f$ is twice differentiable at $a$. Moreover, if $\phi$ is a
uniformly convex $\mathcal{C}^{2}$-norm and $k\in\\{1,\ldots,n\\}$, then the
_pointwise $k$-th $\phi$-mean curvature_ of $C$ at $a$ is defined by

$\bm{h}^{\phi}_{C,k}(a)=S_{k}(\operatorname{D}(\nabla\phi\circ\nu)(a)),$

where $\nu:a+u^{\perp}\rightarrow\mathbf{S}^{n}$ is a map differentiable at
$a$ such that $\nu(b)\in N(\operatorname{epi}(f),b-f(b)u)$ for $b\in
a+u^{\perp}$ and $S_{k}(\operatorname{D}(\nabla\phi\circ\nu)(a))$ is the
$k$-th elementary symmetric function of the eigenvalues (counted with
multiplicities) of the endomorphism $\operatorname{D}(\nabla\phi\circ\nu)(a)$
on $u^{\perp}$.

We denote the set of Alexandrov points of $C$ by $\mathcal{A}(C)$.

######  Remark 5.9.

If $C$ is a convex body, then this notion of an Alexandrov point coincides
with the classical notion of a normal boundary point of $C$; see [Sch14, Notes
for Sections 1.5 and 2.6] for further background information.

######  Corollary 5.10.

Suppose $C\subseteq\mathbf{R}^{n+1}$ is a set of positive reach and $\phi$ is
a uniformly convex $\mathcal{C}^{2}$-norm. Then the following statements hold.

  1. (a)

$\mathcal{A}(C)=\bm{p}(\widetilde{N}^{\phi}_{n}(C))\cap\partial^{v}C=\bm{p}(\widetilde{N}^{\phi}(C))\cap\partial^{v}_{+}C$
and

$\bm{H}^{\phi}_{C,k}(a,\eta)=\bm{h}^{\phi}_{C,k}(a)\qquad\textrm{for
$a\in\mathcal{A}(C)$ and $N^{\phi}(C,a)=\\{\eta\\}$.}$

  2. (b)

$\mathcal{H}^{n}(\partial^{v}C\setminus\mathcal{A}(C))=0$ and

$\mathcal{P}^{\phi}(C)=\int_{\partial^{v}_{+}C}\phi(\bm{n}(C,a))\,d\mathcal{H}^{n}(a)=\int_{N^{\phi}(C)|\partial^{v}_{+}C}J^{\phi}_{C}(a,\eta)\phi(\bm{n}^{\phi}(\eta))\,d\mathcal{H}^{n}(a,\eta).$
(69)

  3. (c)

If $\mathcal{L}^{n+1}(C)<\infty$ then

$\operatorname{int}(C)\neq\varnothing\quad\iff\quad\mathcal{L}^{n+1}(C)>0\quad\iff\mathcal{H}^{n}(\partial^{v}_{+}C)>0.$

  4. (d)

$\mathcal{H}^{n}\big{[}\bm{p}(\widetilde{N}^{\phi}_{n}(C))\setminus\mathcal{A}(C)\big{]}=0$
if and only if $\mathcal{H}^{n}(\partial C\setminus\partial^{v}C)=0$.

###### Proof.

The assertions in a follow from Theorem 5.6, Theorem 5.7 and Lemma 3.14. Using
a we infer

$\partial^{v}C\setminus\mathcal{A}(C)=\partial^{v}C\setminus\bm{p}(\widetilde{N}^{\phi}_{n}(C))\subseteq\bm{p}(N^{\phi}(C))\setminus\bm{p}(\widetilde{N}^{\phi}_{n}(C))\subseteq\bm{p}\left(N^{\phi}(C)\setminus\widetilde{N}^{\phi}_{n}(C)\right),$

whence we obtain $\mathcal{H}^{n}(\partial^{v}C\setminus\mathcal{A}(C))=0$
from Lemma 3.25 c. Since $\partial C=\bm{p}(N^{\phi}(C))$ we obtain the first
equality in (69) from Remark 2.11 and the second equality by combining Remark
3.15, Lemma 3.9 and the coarea formula.

We prove c. Clearly, $\operatorname{int}(C)\neq\varnothing$ implies
$\mathcal{L}^{n+1}(C)>0$. Let us assume $\mathcal{L}^{n+1}(C)>0$. Then it
follows from Lemma 2.10 that $\mathcal{H}^{n}(\partial^{m}C)>0$. Since
$\partial^{v}_{+}C\subseteq\partial^{m}C\subseteq\partial^{v}C$ by Remark 2.11
and $\mathcal{H}^{n}(\partial^{v}C\setminus\partial^{v}_{+}C)=0$ by b, it
follows that $\mathcal{H}^{n}(\partial^{v}_{+}C)>0$. It is again clear that
$\mathcal{H}^{n}(\partial^{v}_{+}C)>0$ implies
$\operatorname{int}(C)\neq\varnothing$.

Finally, it follows by a that

$\partial C\setminus\partial^{v}C\subseteq\big{[}\partial
C\setminus\bm{p}(\widetilde{N}^{\phi}_{n}(C))\big{]}\cup\big{[}\bm{p}(\widetilde{N}^{\phi}_{n}(C))\setminus\partial^{v}C\big{]}=\big{[}\partial
C\setminus\bm{p}(\widetilde{N}^{\phi}_{n}(C))\big{]}\cup\big{[}\bm{p}(\widetilde{N}^{\phi}_{n}(C))\setminus\mathcal{A}(C)\big{]}$

and, since $\mathcal{H}^{n}\big{[}\partial
C\setminus\bm{p}(\widetilde{N}^{\phi}_{n}(C))\big{]}=0$ by Lemma 3.25 (c), we
obtain that
$\mathcal{H}^{n}\big{[}\bm{p}(\widetilde{N}^{\phi}_{n}(C))\setminus\mathcal{A}(C)\big{]}=0$
implies that $\mathcal{H}^{n}(\partial C\setminus\partial^{v}C)=0$. On the
other hand, if $\mathcal{H}^{n}(\partial C\setminus\partial^{v}C)=0$, then
Lemma 3.25 (c) implies again that
$\mathcal{H}^{n}\big{[}\bm{p}(\widetilde{N}^{\phi}_{n}(C))\setminus\partial^{v}C\big{]}=0$,
and hence
$\mathcal{H}^{n}\big{[}\bm{p}(\widetilde{N}^{\phi}_{n}(C))\setminus\mathcal{A}(C)\big{]}=0$
follows from part (a). ∎

######  Remark 5.11.

Suppose $C$ is a closed convex set with
$\operatorname{int}(C)\neq\varnothing$. Then
$C^{(n)}\setminus\partial^{v}C=\varnothing$ and, recalling that
$\bm{p}(\widetilde{N}^{\phi}_{n}(C))\subseteq C^{(n)}$ by Lemma 3.25a, we
infer from Corollary 5.10 that

$\mathcal{A}(C)=\bm{p}(\widetilde{N}_{n}^{\phi}(C)).$

###  5.3 Lower-bounded pointwise mean curvature and bubbling

After some preparations, we will deduce the Heintze-Karcher inequality for
sets of positive reach from the more general version for arbitrary closed
sets.

######  Lemma 5.12.

Suppose $C\subset\mathbf{R}^{n+1}$ is a set of positive reach with
$\operatorname{int}(C)\neq\varnothing$,
$K=\mathbf{R}^{n+1}\setminus\operatorname{int}(C)$ and

$\iota:\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}\rightarrow\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}$

is the linear map defined by $\iota(a,\eta)=(a,-\eta)$ for
$(a,\eta)\in\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}$.

Then the following statements hold.

  1. (a)

$\bm{p}(N^{\phi}(K))=\partial^{v}_{+}C=\partial^{v}_{+}K$ and
$N^{\phi}(K,a)=\\{-\nabla\phi(\bm{n}(C,a))\\}=-N^{\phi}(C,a)$ for
$a\in\bm{p}(N^{\phi}(K))$.

  2. (b)

$\widetilde{N}^{\phi}_{n}(K)=\widetilde{N}^{\phi}(K)$ and
$\mathcal{H}^{n}(N^{\phi}(K)|S)=0$ for every $S\subseteq\mathbf{R}^{n+1}$ with
$\mathcal{H}^{n}(S)=0$.

  3. (c)

$\kappa^{\phi}_{K,i}(a,\eta)=-\kappa^{\phi}_{C,n+1-i}(a,-\eta)$ for
$\mathcal{H}^{n}$ a.e. $(a,\eta)\in N^{\phi}(K)$ and $i=1,\ldots,n$.

  4. (d)

$\bm{H}^{\phi}_{K,1}(a,\eta)=-\bm{H}^{\phi}_{C,1}(a,-\eta)$ for
$\mathcal{H}^{n}$ a.e. $(a,\eta)\in N^{\phi}(K)$.

  5. (e)

$J^{\phi}_{K}(a,\eta)=\operatorname{ap}J_{n}^{N^{\phi}(K)}\iota(a,\eta)\,J^{\phi}_{C}(a,-\eta)$
for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in N^{\phi}(K)$.

###### Proof.

The statement in a readily follows from Remark 2.11.

For the statement in b we combine a and Lemma 3.14b to infer that
$\widetilde{N}^{\phi}_{n}(K)=\widetilde{N}^{\phi}(K)$. Moreover if
$\mathcal{H}^{n}(S)=0$, then we can combine Lemma 2.1 (first applied on
$N_{s}$ for some $s>0$) with Lemma 3.9 to see that

$\int_{\widetilde{N}^{\phi}_{n}(K)|S}\operatorname{ap}J_{n}^{N^{\phi}(K)}\bm{p}(a,u)\,d\mathcal{H}^{n}(a,u)=0$

and $\operatorname{ap}J_{n}^{N^{\phi}(K)}\bm{p}(a,u)>0$ for $\mathcal{H}^{n}$
a.e. $(a,u)\in\widetilde{N}^{\phi}_{n}(K)$. We conclude that
$\mathcal{H}^{n}(\widetilde{N}^{\phi}(K)|S)=0$ provided that
$\mathcal{H}^{n}(S)=0$. Since
$\mathcal{H}^{n}(N^{\phi}(K)\setminus\widetilde{N}^{\phi}(K))=0$, we obtain b.

Next we prove c. We define the open set
$U=\\{y\in\mathbf{R}^{n+1}:0<\bm{\delta}^{\phi}_{K}(y)<\operatorname{reach}^{\phi}(C)\\}$.
For $0<\lambda\leq 1$, $y\in U$ and $a\in\bm{\xi}^{\phi}_{K}(y)$ we notice
that

$a+\lambda\bm{\delta}^{\phi}_{K}(y)\frac{a-y}{\bm{\delta}^{\phi}_{K}(y)}=(1+\lambda)a-\lambda
y,$
$N^{\phi}(K,a)=\bigg{\\{}\frac{y-a}{\bm{\delta}^{\phi}_{K}(y)}\bigg{\\}},\qquad
N^{\phi}(C,a)=\bigg{\\{}\frac{a-y}{\bm{\delta}^{\phi}_{K}(y)}\bigg{\\}},$
$\bm{\nu}^{\phi}_{C}((1+\lambda)a-\lambda
y)=\frac{a-y}{\bm{\delta}^{\phi}_{K}(y)},\qquad-\bm{\nu}^{\phi}_{C}((1+\lambda)a-\lambda
y)\in\bm{\nu}^{\phi}_{K}(y).$

We infer that

$\bm{\nu}^{\phi}_{K}(y)=\\{-\bm{\nu}^{\phi}_{C}((1+\lambda)a-\lambda
y):a\in\bm{\xi}^{\phi}_{K}(y)\\}\qquad\textrm{for $y\in U$ and $0<\lambda\leq
1$.}$ (70)

Define
$S=\bm{p}\big{(}\iota(N^{\phi}(K))\setminus\widetilde{N}^{\phi}(C)\big{)}$ and
notice that $\mathcal{H}^{n}(S)=0$ by Remark 3.7. It follows from b that
$\mathcal{H}^{n}(N^{\phi}(K)|S)=0$. Fix now
$(a,\eta)\in\widetilde{N}^{\phi}_{n}(K)$ with $a\notin S$,
$0<r<\inf\\{\bm{r}^{\phi}_{K}(a,\eta),\operatorname{reach}^{\phi}(C)\\}$ and,
noting that $1-r\chi^{\phi}_{K,i}(a+r\eta)>0$ for $i=1,\ldots,n$, we select
$0<\lambda\leq 1$ so that

$\chi^{\phi}_{K,i}(a+r\eta)<\frac{1}{(1+\lambda)r}\qquad\textrm{for
$i=1,\ldots,n$.}$

Since $a\notin S$, then $(a,-\eta)\in\widetilde{N}^{\phi}(C)$ and
$\bm{\nu}^{\phi}_{C}$ is differentiable at $a-t\eta$ for every
$0<t<\operatorname{reach}^{\phi}(C)$. Since $a+r\eta\in U$ and
$\bm{\nu}^{\phi}_{C}$ is differentiable at
$(1+\lambda)a-\lambda(a+r\eta)=a-\lambda r\eta$, we differentiate at $a+r\eta$
the equality in (70), and thus we get

$\operatorname{D}\bm{\nu}^{\phi}_{K}(a+r\eta)=-\operatorname{D}\bm{\nu}^{\phi}_{C}(a-\lambda
r\eta)\circ((1+\lambda)\operatorname{D}\bm{\xi}^{\phi}_{K}(a+r\eta)-\textrm{Id}_{\mathbf{R}^{n+1}}).$

If $\tau_{1},\ldots,\tau_{n}$ form a basis of
$\operatorname{Tan}(\partial\mathcal{W}^{\phi},\eta)$ such that
$\operatorname{D}\bm{\nu}^{\phi}_{K}(a+r\eta)(\tau_{i})=\chi^{\phi}_{K,i}(a+r\eta)\tau_{i}$
for $i=1,\ldots,n$, then we infer

$\operatorname{D}\bm{\nu}^{\phi}_{C}(a-\lambda
r\eta)(\tau_{i})=\frac{\chi^{\phi}_{K,i}(a+r\eta)}{(1+\lambda)r\chi^{\phi}_{K,i}(a+r\eta)-1}\tau_{i}\qquad\textrm{for
$i=1,\ldots,n$}.$

Note that
$\operatorname{Tan}(\partial\mathcal{W}^{\phi}_{1},-\eta)=\operatorname{Tan}(\partial\mathcal{W}^{\phi}_{1},\eta)$
and recall that $(1+\lambda)r\chi^{\phi}_{K,i}(a+r\eta)-1<0$ for $i=1,\ldots
n$. Hence, we conclude that

$\chi^{\phi}_{C,n+1-i}(a-\lambda
r\eta)=\frac{\chi^{\phi}_{K,i}(a+r\eta)}{(1+\lambda)r\chi^{\phi}_{K,i}(a+r\eta)-1}$

for $i=1,\ldots,n$. Therefore,

$\kappa^{\phi}_{C,n+1-i}(a,-\eta)=\frac{\chi^{\phi}_{C,n+1-i}(a-\lambda
r\eta)}{1-\lambda r\chi^{\phi}_{C,n+1-i}(a-\lambda
r\eta)}=\frac{\chi^{\phi}_{K,i}(a+r\eta)}{r\chi^{\phi}_{K,i}(a+r\eta)-1}=-\kappa^{\phi}_{K,i}(a,\eta)$

for $i=1,\ldots,n$.

To prove d we use b and c, which yields that

$\bm{H}^{\phi}_{K,1}(a,\eta)=\sum_{i=1}^{n}\kappa^{\phi}_{K,i}(a,\eta)=-\sum_{i=1}^{n}\kappa^{\phi}_{C,i}(a,-\eta)=-\bm{H}^{\phi}_{C,1}(a,-\eta)$

for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in\widetilde{N}^{\phi}_{n}(K)$.

Finally, we prove e. Let $\tau_{1},\ldots,\tau_{n},\zeta_{1},\ldots,\zeta_{n}$
be $\mathcal{H}^{n}\operatorname{\llcorner}N^{\phi}(K)$-measurable functions
satisfying the hypothesis of Lemma 3.9. Noting b and Lemma 3.5, we observe
that the proof of c shows that

$\operatorname{D}\bm{\nu}^{\phi}_{C}(a-t\eta)(\tau_{i}(a,\eta))={\mathchoice{\raisebox{0.0pt}{$\displaystyle\chi$}}{\raisebox{0.0pt}{$\textstyle\chi$}}{\raisebox{0.0pt}{$\scriptstyle\chi$}}{\raisebox{0.0pt}{$\scriptscriptstyle\chi$}}}^{\phi}_{C,n+1-i}(a-t\eta)\tau_{i}(a,\eta)$

for $\mathcal{H}^{n}$ a.e. $(a,u)\in N^{\phi}(K)$ and
$0<t<\operatorname{reach}^{\phi}(C)$. Since
$\kappa^{\phi}_{C,i}(a,-\eta)<\infty$ for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in
N^{\phi}(K)$ by b and c, we infer that

$J^{\phi}_{C}(a,-\eta)=\frac{|\tau_{1}(a,\eta)\wedge\ldots\wedge\tau_{n}(a,\eta)|}{|\iota(\zeta_{1}(a,\eta))\wedge\ldots\wedge\iota(\zeta_{n}(a,\eta))|}$

for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in N^{\phi}(K)$. Since

$\operatorname{ap}J_{n}^{N^{\phi}(K)}\iota(a,\eta)=\frac{|\iota(\zeta_{1}(a,\eta))\wedge\ldots\wedge\iota(\zeta_{n}(a,\eta))|}{|\zeta_{1}(a,\eta)\wedge\ldots\wedge\zeta_{n}(a,\eta)|},$

the equation in e follows. ∎

######  Remark 5.13.

The second statement in Lemma 5.12 b can also be obtained as follows. Let
$0<s<\operatorname{reach}(C)$. From a we get

$N^{\phi}(K)|S=\bigcup_{\ell\in\mathbb{N}}\left\\{(x,-\nabla\phi(\bm{n}(C,x))):x\in
S\cap X_{\frac{s}{4\ell},s}(C)\right\\}.$

The assertion now follows from Remark 2.14.

######  Remark 5.14.

We also outline an alternative argument for Lemma 5.12 c. First, we obtain
that for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in\widetilde{N}^{\phi}_{n}(K)$ also
$(a,-\eta)\in\widetilde{N}^{\phi}(C)$ and

$T=\operatorname{Tan}^{n}(\mathcal{H}^{n}\operatorname{\llcorner}\widetilde{N}^{\phi}_{n}(K),(a,\eta))=\text{lin}\\{\zeta_{1}^{K}(a,\eta),\ldots,\zeta_{n}^{K}(a,\eta)\\}=\text{lin}\\{\zeta_{1}^{C}(a,\eta),\ldots,\zeta_{n}^{C}(a,\eta)\\},$

where

$\zeta_{i}^{K}(a,\eta)=\left(\tau_{i}^{K}(a,\eta),\kappa_{K,i}^{\phi}(a,\eta)\tau_{i}^{K}(a,\eta)\right)\qquad\text{for
$i=1,\ldots,n$},$

the linearly independent vectors
$\tau_{1}^{K}(a,\eta),\ldots,\tau_{n}^{K}(a,\eta)$ span an $n$-dimensional
linear subspace $V$ of $\mathbf{R}^{n+1}$ and
$\kappa_{K,i}^{\phi}(a,\eta)\in\mathbf{R}$, and where

$\zeta_{i}^{C}(a,\eta)=\begin{cases}\left(\tau_{i}^{C}(a,-\eta),-\kappa_{C,i}^{\phi}(a,-\eta)\tau_{i}^{C}(a,-\eta)\right),&\text{
if }\kappa_{C,i}^{\phi}(a,-\eta)<\infty,\\\\[6.45831pt]
\left(0,-\kappa_{C,i}^{\phi}(a,-\eta)\tau_{i}^{C}(a,-\eta)\right),&\text{ if
}\kappa_{C,i}^{\phi}(a,-\eta)=\infty,\end{cases}$

with linearly independent vectors
$\tau_{1}^{C}(a,-\eta),\ldots,\tau_{n}^{C}(a,-\eta)$ which span an
$n$-dimensional linear subspace $V^{\prime}$ of $\mathbf{R}^{n+1}$.

Since the number of curvatures which are infinite equals the dimension of the
kernel of the image of the linear map $\bm{p}|T$, a comparison of the two
representations of $T$ shows that $\kappa_{C,i}^{\phi}(a,-\eta)<\infty$ for
$i=1,\ldots,n$. Hence $V=V^{\prime}$ and $\bm{p}|T$ is an injective linear
map. Therefore the linear map $L:V\to V$ with $L=\bm{q}\circ\bm{p}^{-1}$ is
well defined and its eigenvalues are $\kappa_{K,i}^{\phi}(a,\eta)$ with
corresponding eigenvectors $\tau_{i}^{K}(a,\eta)$, but also
$-\kappa_{C,i}^{\phi}(a,-\eta)$ with corresponding eigenvectors
$\tau_{i}^{C}(a,-\eta)$ for $i=1,\ldots,n$. This implies the assertion.

We can now state the Heintze–Karcher inequality for sets of positive reach in
the following form. Recall that
$\mathcal{H}^{n}(\partial^{v}C\setminus\mathcal{A}(C))=0$ by Corollary 5.10.

######  Theorem 5.15.

Suppose $\varnothing\neq C\subseteq\mathbf{R}^{n+1}$ is a set of positive
reach with finite volume and assume that

$\bm{h}^{\phi}_{C,1}(a)\geq 0\quad\textrm{ for $\mathcal{H}^{n}\ \text{a.e.}\
a\in\mathcal{A}(C)$}.$

Then

$(n+1)\mathcal{L}^{n+1}(C)\leq
n\int_{\partial^{v}C}\frac{\phi(\bm{n}(C,a))}{\bm{h}^{\phi}_{C,1}(a)}\,d\mathcal{H}^{n}(a).$
(71)

If $\operatorname{int}(C)\neq\varnothing$, equality holds in (71) and there
exists $q<\infty$ so that $\bm{h}^{\phi}_{C,1}(a)\leq q$ for $\mathcal{H}^{n}$
a.e. $a\in\mathcal{A}(C)$, then there are $N\in\mathbb{N}$,
$c_{1},\ldots,c_{N}\in\mathbf{R}^{n+1}$ and
$\rho_{1},\ldots,\rho_{N}\geq\frac{n}{q}$ such that

$\operatorname{int}(C)=\bigcup_{i=1}^{N}\operatorname{int}(c_{i}+\rho_{i}\mathcal{W}^{\phi}),\qquad\operatorname{dist}^{\phi}\big{(}c_{i}+\rho_{i}\mathcal{W}^{\phi},c_{j}+\rho_{j}\mathcal{W}^{\phi}\big{)}\geq
2\operatorname{reach}^{\phi}(C)\quad\textrm{for $i\neq j$.}$

###### Proof.

We assume $\operatorname{int}(C)\neq\varnothing$ (otherwise there is nothing
to prove) and we define $K=\mathbf{R}^{n+1}\setminus\operatorname{int}(C)$.
Note that $\mathcal{L}^{n+1}(\partial C)=0$ and
$\iota(N^{\phi}(K))=N^{\phi}(C)|\partial^{v}_{+}C$. By Lemma 5.12 and the
assumption, we infer that

$\bm{H}^{\phi}_{K,1}(a,\eta)=-\bm{H}^{\phi}_{C,1}(a,-\eta)\leq
0\quad\textrm{for $\mathcal{H}^{n}$ a.e.\ $(a,\eta)\in N^{\phi}(K)$.}$

Therefore, applying Theorem 3.20, Lemma 5.12 and the coarea formula, we obtain

$\displaystyle(n+1)\mathcal{L}^{n+1}(C)$ $\displaystyle\leq
n\int_{N^{\phi}(K)}J^{\phi}_{K}(a,\eta)\,\frac{\phi(\bm{n}^{\phi}(\eta))}{|\bm{H}^{\phi}_{K,1}(a,\eta)|}\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=n\int_{N^{\phi}(K)}\operatorname{ap}J^{N^{\phi}(K)}_{n}\iota(a,\eta)\,J^{\phi}_{C}(a,-\eta)\,\frac{\phi(\bm{n}^{\phi}(\eta))}{\bm{H}^{\phi}_{C,1}(a,-\eta)}\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=n\int_{N^{\phi}(C)|\partial^{v}_{+}C}J^{\phi}_{C}(a,\eta)\,\frac{\phi(\bm{n}^{\phi}(\eta))}{\bm{H}^{\phi}_{C,1}(a,\eta)}\,d\mathcal{H}^{n}(a,\eta).$

Since
$\phi(\bm{n}^{\phi}(\eta))=\phi(\bm{n}^{\phi}(\nabla\phi(\bm{n}(C,a))))=\phi(\bm{n}(C,a))$
for $(a,\eta)\in N^{\phi}(C)|\partial^{v}_{+}C$, recalling Remark 3.15 and
Corollary 5.10, we apply coarea formula in combination with Lemma 3.9 to
obtain

$\int_{N^{\phi}(C)|\partial^{v}_{+}C}J^{\phi}_{C}(a,\eta)\,\frac{\phi(\bm{n}^{\phi}(\eta))}{\bm{H}^{\phi}_{C,1}(a,\eta)}\,d\mathcal{H}^{n}(a,\eta)=\int_{\partial^{v}C}\frac{\phi(\bm{n}(C,a))}{\bm{h}^{\phi}_{C,1}(a)}\,d\mathcal{H}^{n}(a),$

which yields the first part of the assertion of the theorem.

Assume now that $\bm{h}^{\phi}_{C,1}(a)\leq q$ for $\mathcal{H}^{n}$ a.e.
$a\in\mathcal{A}(C)$ and $\operatorname{int}(C)\neq\varnothing$. Combining
Corollary 5.10 with Lemma 5.12 we get that

$-\bm{H}^{\phi}_{K,1}(a,\eta)=\bm{H}^{\phi}_{C,1}(a,-\eta)=\bm{h}^{\phi}_{C,1}(a)\leq
q\qquad\textrm{for $\mathcal{H}^{n}$ a.e.\ $(a,\eta)\in N^{\phi}(K)$.}$

Therefore if the equality holds in (71) then the conclusion follows from the
characterization provided by Theorem 3.20. ∎

From Theorem 5.15 we obtain a geometric rigidity result for a set $C$ of
positive reach with positive and finite volume under the assumption of a sharp
lower bound on the pointwise $\phi$ mean-curvature at almost all points in
$\partial^{v}C$. For the set $C$ in the next theorem we notice that
$\mathcal{P}^{\phi}(C)>0$ and $\operatorname{int}(C)\neq\varnothing$ by
Corollary 5.10.

######  Corollary 5.16.

Suppose $C\subset\mathbf{R}^{n+1}$ is a set of positive reach with finite and
positive volume and define
$\rho=\frac{(n+1)\mathcal{L}^{n+1}(C)}{\mathcal{P}^{\phi}(C)}$. Assume that

$\bm{h}^{\phi}_{C,1}(a)\geq\frac{n}{\rho}\qquad\textrm{for $\mathcal{H}^{n}$
a.e.\ $a\in\mathcal{A}(C)$.}$ (72)

Then there exist $N\in\mathbb{N}$ and $c_{1},\ldots,c_{N}\in\mathbf{R}^{n+1}$
such that

$\operatorname{int}(C)=\bigcup_{i=1}^{N}\operatorname{int}(c_{i}+\rho\mathcal{W}^{\phi}),\quad\operatorname{dist}^{\phi}\big{(}c_{i}+\rho\mathcal{W}^{\phi},c_{j}+\rho\mathcal{W}^{\phi}\big{)}\geq
2\operatorname{reach}^{\phi}(C)\quad\textrm{for $i\neq j$.}$

###### Proof.

Notice that there is at least one point $a\in\partial^{v}_{+}C$ with
$\bm{h}^{\phi}_{C,1}(a)<\infty$, hence we obtain $\rho>0$. Therefore
$0<\mathcal{P}^{\phi}(C)<\infty$.

For $\epsilon>0$ we set

$Z_{\epsilon}=\left\\{a\in\mathcal{A}(C):\bm{h}^{\phi}_{C,1}(a)\geq(1+\epsilon)\frac{n}{\rho}\right\\}.$

We claim that $\mathcal{H}^{n}(Z_{\epsilon})=0$ for $\epsilon>0$. Suppose that
$\mathcal{H}^{n}(Z_{\epsilon})>0$ for some $\epsilon>0$. Then we deduce

$\displaystyle
n\int_{\partial^{v}C}\frac{\phi(\bm{n}(C,a))}{\bm{h}^{\phi}_{C,1}(a)}\,d\mathcal{H}^{n}(a)=n\int_{\partial^{v}C\setminus
Z_{\epsilon}}\frac{\phi(\bm{n}(C,a))}{\bm{h}^{\phi}_{C,1}(a)}\,d\mathcal{H}^{n}(a)+n\int_{Z_{\epsilon}}\frac{\phi(\bm{n}(C,a))}{\bm{h}^{\phi}_{C,1}(a)}\,d\mathcal{H}^{n}(a)$
$\displaystyle\qquad\leq\rho\int_{\partial^{v}C\setminus
Z_{\epsilon}}\phi(\bm{n}(C,a))\,d\mathcal{H}^{n}(a)+(1+\epsilon)^{-1}\rho\int_{Z_{\epsilon}}\phi(\bm{n}(C,a))\,d\mathcal{H}^{n}(a)$
$\displaystyle\qquad<\rho\mathcal{P}^{\phi}(C)=(n+1)\mathcal{L}^{n+1}(C),$

where we used (72) on $\partial^{v}C\setminus Z_{\epsilon}$ and the lower
bound for $\bm{h}^{\phi}_{C,1}(a)$ on $Z_{\epsilon}$. This contradicts the
inequality in Theorem 5.15 and thus proves the claim.

Since $\mathcal{H}^{n}(Z_{\epsilon})=0$ for $\epsilon>0$, we infer that

$\bm{h}^{\phi}_{C,1}(a)=\frac{n\mathcal{P}^{\phi}(C)}{(n+1)\mathcal{L}^{n+1}(C)}\qquad\textrm{for
$\mathcal{H}^{n}$ a.e.\ $a\in\partial^{v}C$},$

whence we infer that

$n\int_{\partial^{v}C}\frac{\phi(\bm{n}(C,a))}{\bm{h}^{\phi}_{C,1}(a)}\,d\mathcal{H}^{n}(a)=(n+1)\mathcal{L}^{n+1}(C),$

thus (71) holds with equality. We obtain now the conclusion of the theorem by
employing the second part of Theorem 5.15. ∎

######  Remark 5.17.

Corollary 5.16 is sharp already in the special isotropic case and for convex
bodies. In fact, if we consider the union of two congruent proper antipodal
spherical caps of the unit sphere, we obtain a convex body $K$ whose $k$-th
mean curvature on the smooth part of its boundary is constant and smaller than
$\frac{\mathcal{H}^{n}(\partial K)}{(n+1)\mathcal{L}^{n+1}(K)}\binom{n}{k}$.
We provide the details for completeness. Let $P\in\mathbf{G}(n+1,n)$ and
$\eta\in P^{\perp}$ with $|\eta|=1$. For every $0<\epsilon<1$ we define

$\Sigma^{+}_{\epsilon}=\mathbf{S}^{n}\cap\\{x:x\bullet\eta\geq\epsilon\\},\quad\Sigma^{-}_{\epsilon}=\mathbf{S}^{n}\cap\\{x:x\bullet\eta\leq-\epsilon\\},$
$P^{+}_{\epsilon}=\mathbf{B}(0,1)\cap\\{x:x\bullet\eta=\epsilon\\},\quad
P^{-}_{\epsilon}=\mathbf{B}(0,1)\cap\\{x:x\bullet\eta=-\epsilon\\},$

and we denote by $K^{+}_{\epsilon}$ and $K^{-}_{\epsilon}$ the convex bodies
enclosed by $\Sigma^{+}_{\epsilon}\cup P^{+}_{\epsilon}$ and
$\Sigma^{-}_{\epsilon}\cup P^{-}_{\epsilon}$ respectively. Then we define

$K_{\epsilon}=\\{x-\epsilon\eta:x\in
K^{+}_{\epsilon}\\}\cup\\{x+\epsilon\eta:x\in K^{-}_{\epsilon}\\}.$

Let $X(x)=x$ for every $x\in\mathbf{R}^{n+1}$. Since $K^{+}_{\epsilon}$ is a
set of finite perimeter, we denote by $\eta_{\epsilon}$ the exterior unit
normal and we compute by means of the divergence theorem [Fed69, Gauss–Green
Theorem 4.5.6] (alternatively by noting that
$K_{\epsilon}=\operatorname{conv}(\\{o\\}\cup\Sigma_{\epsilon}^{+})\setminus\operatorname{conv}(\\{o\\}\cup
P_{\epsilon})$, where $\operatorname{conv}$ denotes the convex hull operator)

$\displaystyle(n+1)\mathcal{L}^{n+1}(K^{+}_{\epsilon})$
$\displaystyle=\int_{K^{+}_{\epsilon}}\mathrm{div}X\,d\mathcal{L}^{n+1}$
$\displaystyle=\int_{\Sigma^{+}_{\epsilon}\cup
P^{+}_{\epsilon}}\eta_{\epsilon}(x)\bullet
X(x)\,d\mathcal{H}^{n}(x)=\mathcal{H}^{n}(\Sigma^{+}_{\epsilon})-\epsilon\mathcal{H}^{n}(P^{+}_{\epsilon}).$

We conclude that

$\frac{\mathcal{H}^{n}(\partial
K_{\epsilon})}{(n+1)\mathcal{L}^{n+1}(K_{\epsilon})}=\frac{\mathcal{H}^{n}(\Sigma^{+}_{\epsilon})}{(n+1)\mathcal{L}^{n+1}(K^{+}_{\epsilon})}=1+\epsilon\frac{\mathcal{H}^{n}(P_{\epsilon}^{+})}{(n+1)\mathcal{L}^{n+1}(K^{+}_{\epsilon})}>1$

for $0<\epsilon<1$. Finally, we notice that the $k$-th mean curvature of
$K_{\epsilon}$ equals $\binom{n}{k}$ on the smooth part of $\partial
K_{\epsilon}$.

##  6 Curvature measures and soap bubbles

###  6.1 Curvature measures and Minkowski formulae

In the following, we write $C$ for a non-empty set with positive reach in
$\mathbf{R}^{n+1}$. For sets with positive reach, the Steiner formula
simplifies in the following way (also in the anisotropic setting).

######  Corollary 6.1 (Anisotropic Steiner formula for sets of positive
reach).

Let $\varnothing\neq C\subset\mathbf{R}^{n+1}$ be a set of positive reach. Let
$\varphi:N^{\phi}(C)\rightarrow\mathbf{R}$ be a bounded Borel function with
compact support. Then

$\int_{\\{x\in\mathbf{R}^{n+1}:0<\bm{\delta}^{\phi}_{C}(x)\leq\rho\\}}(\varphi\circ\bm{\psi}^{\phi}_{C})\,d\mathcal{L}^{n+1}\\\
=\sum_{i=0}^{n}\frac{\rho^{i+1}}{i+1}\int_{N^{\phi}(C)}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,i}(a,\eta)\;\varphi(a,u)\,d\mathcal{H}^{n}(a,\eta)$

for $0<\rho<\operatorname{reach}^{\phi}(C)$.

###### Proof.

The assertion is a straightforward consequence of Theorem 3.16. ∎

We can now introduce the generalized curvature measures of a set of positive
reach with respect to $\phi$. These are real-valued Radon measures (see [Za86,
RZ19, Hu99, HL00] and the references cited there).

######  Definition 6.2.

Let $\varnothing\neq C\subset\mathbf{R}^{n+1}$ be a set of positive reach and
$m\in\\{0,\ldots,n\\}$. The _$m$ -th generalized curvature measure of $C$ with
respect to $\phi$_ is the real-valued Radon measure
$\Theta^{\phi}_{m}(C,\cdot)$ on $\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}$ such
that

$\Theta^{\phi}_{m}(C,B)=\frac{1}{n-m+1}\int_{N^{\phi}(C)\cap
B}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,n-m}(a,\eta)\;d\mathcal{H}^{n}(a,\eta)$

for any bounded Borel subset $B\subset\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}$.
Moreover, we set

$\mathcal{V}^{\phi}_{m}(C)=\Theta^{\phi}_{m}(C,N^{\phi}(C))\qquad\textrm{for
$m\in\\{0,\ldots,n\\}$.}$

######  Remark 6.3.

Let $C\subseteq\mathbf{R}^{n+1}$ be a set of positive reach and
$m\in\\{0,\ldots,n\\}$. The _$m$ -th curvature measure of $C$ with respect to
$\phi$_ is the real-valued Radon measure $\mathcal{C}^{\phi}_{m}(C,\cdot)$ on
$\mathbf{R}^{n+1}$ such that

$\mathcal{C}^{\phi}_{m}(C,B)=\Theta^{\phi}_{m}(C,B\times\mathbf{R}^{n+1})$

for any bounded Borel subset $B\subset\mathbf{R}^{n+1}$.

######  Lemma 6.4.

Suppose $C$ is a set of positive reach, $m\in\\{0,\ldots,n\\}$ and let
$\nu:C^{(n)}\rightarrow\mathbf{S}^{n}$ be a Borel map such that $\nu(a)\in
N(C,a)$ for every $a\in C^{(n)}$. Let $\eta(a)=\nabla\phi(\nu(a))$. Then

$\displaystyle(n-m+1)\Theta^{\phi}_{m}(C,B\cap\widetilde{N}^{\phi}_{n}(C))$
$\displaystyle\qquad=\int_{\mathcal{A}(C)}\mathbf{1}_{B}(a,\eta(a))\phi(\nu(a))\,\bm{h}^{\phi}_{C,n-m}(a)\,d\mathcal{H}^{n}(a)$
$\displaystyle\qquad\quad+\int_{C^{(n)}\setminus\partial^{v}C}\phi(\nu(a))\big{[}\mathbf{1}_{B}(a,\eta(a))\bm{H}^{\phi}_{C,n-m}(a,\eta(a))+\mathbf{1}_{B}(a,-\eta(a))\bm{H}^{\phi}_{C,n-m}(a,-\eta(a))\big{]}\,d\mathcal{H}^{n}(a)$

for every Borel set $B\subseteq N^{\phi}(C)$.

###### Proof.

Let $B\subseteq N^{\phi}(C)$ be a Borel set. Combining Lemma 3.9 and the
coarea formula, we get

$(n-m+1)\Theta^{\phi}_{m}(C,B\cap\widetilde{N}^{\phi}_{n}(C))=\int_{\bm{p}(\widetilde{N}^{\phi}_{n}(C))}\int_{N^{\phi}(C,a)}\mathbf{1}_{B}(a,\eta)\phi(\bm{n}^{\phi}(\eta))\,\bm{H}^{\phi}_{C,n-m}(a,\eta)\,d\mathcal{H}^{0}(\eta)\,d\mathcal{H}^{n}(a).$

Since $\bm{n}^{\phi}(\nabla\phi(\nu(a)))=\nu(a)$ for $a\in C^{(n)}$, the
argument is completed by applying Lemma 3.25 c and Corollary 5.10, since
$N^{\phi}(C,a)=\\{\pm\nabla\phi(\nu(a))\\}$ for $a\in
C^{(n)}\setminus\partial^{v}C$. ∎

######  Remark 6.5.

The Lebesgue decomposition of the curvature measure
$\mathcal{C}^{\phi}_{m}(C,\cdot)$ with respect to
$\mathcal{H}^{n}\llcorner\partial C$ is given by
$\Theta^{\phi}_{m}(C,(\cdot\times\mathbf{R}^{n+1})\cap\widetilde{N}^{\phi}_{n}(C))$
(the absolutely continuous part) and
$\Theta^{\phi}_{m}(C,(\cdot\times\mathbf{R}^{n+1})\setminus\widetilde{N}^{\phi}_{n}(C))$
(the singular part). It follows from Lemma 3.25 e that these parts are indeed
singular with respect to each other. For the absolutely continuous part, Lemma
6.4 yields an explicit description. In the case of convex bodies where
$C^{(n)}\setminus\partial^{v}C=\emptyset$, a corresponding analysis can be
found in [Hu98] in the isotropic framework.

The following lemma extends [Fa96, Lemma 2.1] from Euclidean curvature
measures of convex bodies to generalized curvature measures with respect to a
$C^{2}$-norm $\phi$ and sets with positive reach (compare also [CH00, Section
3]). The non-negative Radon measure
$|\Theta^{\phi}_{m}(C,\cdot)\llcorner(A\times\partial\mathcal{W}^{\phi})|$ in
the next lemma is the total variation of the real-valued Radon measure
$\Theta^{\phi}_{m}(C,\cdot)\llcorner(A\times\partial\mathcal{W}^{\phi})$; see
[AFP00, Definition 1.4].

######  Lemma 6.6.

Let $\varnothing\neq C\subset\mathbf{R}^{n+1}$ be a set of positive reach. Let
$A\subset\mathbf{R}^{n+1}$ be a $\mathcal{H}^{m}$ measurable set and
$m\in\\{0,\ldots,n\\}$. Then there is a non-negative constant $c$, depending
only on $n,\phi$, such that

$|\Theta^{\phi}_{m}(C,\cdot)\llcorner(A\times\partial\mathcal{W}^{\phi})|\leq
c\cdot\mathcal{H}^{m}(A).$

###### Proof.

For the proof, one can assume that $\mathcal{H}^{m}(A)<\infty$. An application
of Theorem 3.27 to the positive and the negative part of
$\Theta^{\phi}_{m}(C,\cdot)\llcorner(A\times\partial\mathcal{W}^{\phi})$ then
yields the assertion. ∎

The following lemma is now an immediate consequence of Lemma 6.6. We do not
include the case $m=n$ in the statement of the lemma, since in this case the
hypothesis is always satisfied by Lemma 3.25 c and the conclusion holds
essentially by definition; see Remark 3.12.

######  Lemma 6.7.

Let $\varnothing\neq C\subset\mathbf{R}^{n+1}$ be a set of positive reach. Let
$m\in\\{0,\ldots,n-1\\}$ and assume that

$\mathcal{H}^{m}[\bm{p}(\widetilde{N}^{\phi}(C)\setminus\widetilde{N}_{n}^{\phi}(C))]=0.$

Then

$(n-m+1)\cdot\Theta^{\phi}_{m}(C,B)=\int_{\widetilde{N}_{n}^{\phi}(C)\cap
B}\phi(\bm{n}^{\phi}(\eta))J^{\phi}_{C}(a,\eta)\bm{H}^{\phi}_{C,n-m}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$

for every Borel set $B\subseteq N^{\phi}(C)$.

###### Proof.

An application of Lemma 6.6 with
$A=\bm{p}(\widetilde{N}^{\phi}(C)\setminus\widetilde{N}_{n}^{\phi}(C))$ yields
that

$\Theta^{\phi}_{m}(C,\cdot)\llcorner(\bm{p}(\widetilde{N}^{\phi}(C)\setminus\widetilde{N}_{n}^{\phi}(C))\times\partial\mathcal{W}^{\phi})=0,$

and hence
$\Theta^{\phi}_{m}(C,B\cap\widetilde{N}^{\phi}(C)\setminus\widetilde{N}_{n}^{\phi}(C))=0$,
which is the desired conclusion. ∎

We now prove the anisotropic Minkowski formulae for sets of positive reach.
The case of convex bodies has been treated in a different way in [Hu99].

######  Theorem 6.8.

If $\varnothing\neq C\subset\mathbf{R}^{n+1}$ is a set of positive reach with
finite volume and $r\in\\{1,\ldots,n\\}$, then

$\displaystyle(n-r+1)\int_{N^{\phi}(C)}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,r-1}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=r\int_{N^{\phi}(C)}[a\bullet\bm{n}^{\phi}(\eta)]J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,r}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$

and

$\int_{N^{\phi}(C)}a\bullet\bm{n}^{\phi}(\eta)\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,0}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)=(n+1)\mathcal{L}^{n+1}(C).$

###### Proof.

We set
$u(x)=\frac{\nabla\bm{\delta}^{\phi}_{C}(x)}{|\nabla\bm{\delta}^{\phi}_{C}(x)|}$
for $x\in\operatorname{Unp}^{\phi}(C)$ and notice that for
$0<\rho<\operatorname{reach}^{\phi}(C)$ the set $B^{\phi}(C,\rho)$ is a domain
with $\mathcal{C}^{1,1}$-boundary $\partial B^{\phi}(C,\rho)=S^{\phi}(C,\rho)$
whose exterior unit normal is given by $u|\partial B^{\phi}(C,\rho)$.
Moreover, for $0<\rho<\operatorname{reach}^{\phi}(C)$ the map
$f_{\rho}:N^{\phi}(C)\rightarrow S^{\phi}(C,\rho)$ defined by

$f_{\rho}(a,\eta)=a+\rho\eta\qquad\textrm{for $(a,\eta)\in N^{\phi}(C)$}$

is a bi-lipschitz homeomorphism by Remark 5.4. We observe (see proof of
Theorem 3.16) that

$J_{n}^{N^{\phi}(C)}f_{\rho}(a,\eta)=J^{\phi}_{C}(a,\eta)\sum_{m=0}^{n}\rho^{n-m}\,\bm{H}^{\phi}_{C,n-m}(a,\eta)$

for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in N^{\phi}(C)$. We set

$I_{m}(C)=\int_{N^{\phi}(C)}a\bullet\bm{n}^{\phi}(\eta)\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,n-m}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)\qquad\textrm{for
$m=0,\ldots,n$.}$

The divergence theorem and Remark 3.4 yield

$\displaystyle(n+1)\mathcal{L}^{n+1}(B^{\phi}(C,\rho))$
$\displaystyle=\int_{S^{\phi}(C,\rho)}x\bullet u(x)\,d\mathcal{H}^{n}(x)$
$\displaystyle=\int_{N^{\phi}(C)}[(a+\rho\eta)\bullet
u(a+\rho\eta)]\,J_{n}^{N^{\phi}(C)}f_{\rho}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle=\sum_{m=0}^{n}\rho^{n-m}I_{m}(C)+\sum_{m=0}^{n}(n-m+1)\rho^{n-m+1}\mathcal{V}^{\phi}_{m}(C)$

for $0<\rho<\operatorname{reach}^{\phi}(C)$. Employing the Steiner formula
6.1, we get

$(n+1)\mathcal{L}^{n+1}(B^{\phi}(C,\rho))=(n+1)\mathcal{L}^{n+1}(C)+(n+1)\sum_{m=0}^{n}\rho^{n-m+1}\mathcal{V}^{\phi}_{m}(C)$

for $0<\rho<\operatorname{reach}^{\phi}(C)$. Hence, we infer

$\sum_{m=0}^{n-1}[I_{m}(C)-(m+1)\mathcal{V}^{\phi}_{m+1}(C)]\rho^{n-m}+I_{n}(C)-(n+1)\mathcal{L}^{n+1}(C)=0$

for $0<\rho<\operatorname{reach}^{\phi}(C)$. It follows that
$I_{m}(C)=(m+1)\mathcal{V}^{\phi}_{m+1}(C)$ for $m=0,\ldots,n-1$ and in
addition we have $I_{n}(C)=(n+1)\mathcal{L}^{n+1}(C)$. ∎

###  6.2 The soap bubble theorem for sets of positive reach

The following notion of $k$-convexity generalizes the classical analogous
notion used in the Euclidean setting to study isoperimetric-type inequalities
for Querrmassintegrals (see [Tru94] or the more recent [CW13]). Analogous
concepts also arise in the context of elliptic differential operators (see
[TW99] and [Sal99] and the references given there to earlier work for instance
by Caffarelli, Nirenberg, Spruck (’85), Garding (’59), Ivochkina (’83, ’85),
Li (’90)).

######  Definition 6.9.

Let $\varnothing\neq C\subset\mathbf{R}^{n+1}$ be a set of positive reach with
$\mathcal{P}^{\phi}(C)>0$, and let $r\in\\{0,\ldots,n\\}$. We say that $C$ is
_ $(r,\phi)$-mean convex _ if

$\bm{h}^{\phi}_{C,i}(a)\geq 0\qquad\textrm{for $\mathcal{H}^{n}$ a.e.\
$a\in\mathcal{A}(C)$ and $i=1,\ldots,r-1$}$ (73)

and

$\bm{H}^{\phi}_{C,r}(a,u)\geq 0\qquad\textrm{for $\mathcal{H}^{n}$ a.e.\
$(a,u)\in{N}^{\phi}(C)$.}$ (74)

######  Remark 6.10.

Suppose that $\varnothing\neq C\subset\mathbf{R}^{n+1}$ is a set of positive
reach with finite volume. Then $\mathcal{P}^{\phi}(C)>0$ if and only if
$\mathcal{H}^{n}(\partial^{v}_{+}C)>0$ by Corollary 5.10. This in turn is
equivalent to $\operatorname{int}(C)\neq\varnothing$.

######  Remark 6.11.

The $(0,\phi)$-mean convex sets are all sets of positive reach with non-empty
interior, since by definition we have
$\bm{H}^{\phi}_{C,0}=\bm{1}_{\widetilde{N}^{\phi}_{n}(C)}\geq 0$. Moreover, if
$C$ is a set of positive reach and positive perimeter such that
$\Theta^{\phi}_{n-k}(C.\cdot)$ are non-negative measures for all
$k\in\\{1,\ldots,r\\}$, then $C$ is $(r,\phi)$-mean convex.

Before we can add another remark, we need some preparations. Let
$d_{H}^{\phi}$ denote the Hausdorff distance on the space of closed subsets of
the metric space $(\mathbf{R}^{n+1},\phi^{*})$. We say that a sequence of
closed sets $C_{i}\subseteq\mathbf{R}^{n+1}$, $i\in\mathbb{N}$, converges to a
closed set $C\subseteq\mathbf{R}^{n+1}$ as $i\to\infty$ if
$d_{H}^{\phi}(C_{i},C)\to 0$ as $i\to\infty$, which is equivalent to the
uniform convergence of $\delta^{\phi}_{C_{i}}$ to $\delta^{\phi}_{C}$ as
$i\to\infty$ on $\mathbf{R}^{n+1}$ (see [Be85]). The following lemma is well
known in the Euclidean setting (see Theorem 4.13, Remark 4.14 and Theorem 5.9
in [Fed59] and [RZ01, Section 3.1, pp. 7–9]).

######  Lemma 6.12.

Let $\varnothing\neq C_{i}\subset\mathbf{R}^{n+1}$ for $i\in\mathbb{N}$ be a
sequence of closed sets converging to a closed set $C\subset\mathbf{R}^{n+1}$.
Suppose there is a constant $\rho>0$ such that
$\operatorname{reach}^{\phi}(C_{i})\geq\rho$ for all $i\in\mathbb{N}$. Then
$\operatorname{reach}^{\phi}(C)\geq\rho$ and for each $k\in\\{0,\ldots,n\\}$
the Radon measures $\Theta^{\phi}_{k}(C_{i},\cdot)$ converge vaguely to the
Radon measure $\Theta^{\phi}_{k}(C,\cdot)$ as $i\to\infty$.

###### Proof.

Let $\rho_{1}\in(0,\rho)$. Under the assumptions of the lemma, we show that
$\operatorname{reach}(C)\geq\rho_{1}$. Let $x\in\mathbf{R}^{n+1}\setminus C$
with $\delta^{\phi}_{C}(x)\leq\rho_{1}$. There is some $i_{1}\in\mathbb{N}$
such that if $i\geq i_{1}$, then
$0<\delta^{\phi}_{C_{i}}(x)\leq\rho_{2}<\rho$, where
$\rho_{2}:=\frac{1}{2}(\rho+\rho_{1})$. Define
$x_{i}:=\xi_{C_{i}}^{\phi}(x)+\rho_{2}\cdot\nu^{\phi}_{C_{i}}(x)$, hence
$\delta^{\phi}_{C_{i}}(x_{i})=\rho_{2}$,
$\delta^{\phi}_{C_{i}}(x_{i})=\delta^{\phi}_{C_{i}}(x)$ and
$U^{\phi}(x_{i},\rho_{2})\cap C_{i}=\varnothing$, since
$\bm{r}^{\phi}_{C_{i}}(\xi_{C_{i}}^{\phi}(x),\nu_{C_{i}}^{\phi}(x))\geq\operatorname{reach}(C_{i})$.
By compactness, we can find an infinite subset $I\subseteq\mathbb{N}$ such
$\xi^{\phi}_{C_{i}}(x_{i})\to\xi\in\partial C$, $x_{i}\to z$,
$\delta^{\phi}_{C}(z)=\rho_{2}$, $U^{\phi}(z,\rho_{2})\cap C=\varnothing$ and
$x\in(\xi,z)$, where $I\ni i\to\infty$. But then clearly
$U^{\phi}(x,\rho_{2})\cap C=\varnothing$, $x\in\operatorname{Unp}^{\phi}(C)$
and $\xi=\xi^{\phi}_{C}(x)$. This proves the first assertion.

In view of Corollary 6.1, it is sufficient to show that if
$\varphi:\mathbf{R}^{n+1}\times\mathbf{R}^{n+1}\to\mathbf{R}$ is a continuous
function with compact support and $t\in(0,\rho)$, then

$\int\mathbf{1}\\{0<\delta^{\phi}_{C_{i}}(x)\leq
t\\}\varphi(\xi^{\phi}_{C_{i}}(x),\nu^{\phi}_{C_{i}}(x))\,d\mathcal{L}^{n+1}(x)\to\int\mathbf{1}\\{0<\delta^{\phi}_{C}(x)\leq
t\\}\varphi(\xi^{\phi}_{C}(x),\nu^{\phi}_{C}(x))\,d\mathcal{L}^{n+1}(x)$

as $i\to\infty$.

We consider an arbitrary point $x\in\mathbf{R}^{n+1}$. If
$0<\delta_{C}^{\phi}(x)<t$, then also $0<\delta_{C_{i}}^{\phi}(x)<t$ if
$i\in\mathbb{N}$ is large enough. Moreover,
$\xi_{C_{i}}^{\phi}(x)\to\xi_{C}^{\phi}(x)$ and
$\nu_{C_{i}}^{\phi}(x)\to\nu_{C}^{\phi}(x)$ as $i\to\infty$, which can be
obtained by minor adjustments of the proof for [HL00, Lemma 2.2]. Therefore,

$\mathbf{1}\\{0<\delta^{\phi}_{C_{i}}(x)\leq
t\\}\varphi(\xi^{\phi}_{C_{i}}(x),\nu^{\phi}_{C_{i}}(x))\to\mathbf{1}\\{0<\delta^{\phi}_{C}(x)\leq
t\\}\varphi(\xi^{\phi}_{C}(x),\nu^{\phi}_{C}(x))$ (75)

as $i\to\infty$.

If $\delta_{C}^{\phi}(x)>t$, then also $\delta_{C_{i}}^{\phi}(x)>t$ if
$i\in\mathbb{N}$ is large enough, hence (75) holds trivially. The same is true
if $x\in\operatorname{int}(C)$ which implies that $x\in C_{i}$ for all
sufficiently large $i\in\mathbb{N}$.

Since
$\mathcal{L}^{n+1}(\\{x\in\mathbf{R}^{n+1}:\delta_{C}^{\phi}(x)=t\\}\cup\partial
C)=0$, the assertion follows from the dominated convergence theorem. ∎

The preceding lemma implies that a non-negativity condition closely related to
Definition 6.9 is stable with respect to converging sequences of sets of
positive reach, as described in the following corollary.

######  Corollary 6.13.

Let $\rho>0$ be a fixed constant. Let $\varnothing\neq
C_{i}\subset\mathbf{R}^{n+1}$ for $i\in\mathbb{N}$ be a sequence of closed
sets with $\operatorname{reach}(C_{i})\geq\rho>0$ converging to a closed set
$C\subset\mathbf{R}^{n+1}$. Then the following statements hold.

  1. (a)

If $\Theta^{\phi}_{k}(C_{j},\cdot)\geq 0$ for all $j\in\mathbb{N}$, then also
$\Theta^{\phi}_{k}(C,\cdot)\geq 0$.

  2. (b)

If $C_{j}$ is $(r,\phi)$-mean convex and smooth for all $j\in\mathbb{N}$ (so
that $\Theta^{\phi}_{n-k}(C_{j},\cdot)\geq 0$ holds for $k=1,\ldots,r$), then
$\Theta^{\phi}_{n-k}(C,\cdot)\geq 0$ for $k=1,\ldots,r$; hence, $C$ is
$(r,\phi)$-mean convex.

A major ingredient for the proof of the Alexandrov theorem for sets with
positive reach is the next lemma.

######  Lemma 6.14.

Let $\varnothing\neq C\subset\mathbf{R}^{n+1}$ be a set of positive reach with
finite and positive volume. Let $\lambda\in\mathbf{R}$ and
$r\in\\{1,\ldots,n\\}$ be such that

$\Theta^{\phi}_{n-r}(C,\cdot)=\lambda\,\Theta^{\phi}_{n}(C,\cdot).$ (76)

Then $\bm{H}^{\phi}_{C,r}(a,\eta)=0$ for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in
N^{\phi}(C)\setminus\widetilde{N}^{\phi}_{n}(C)$ and

$\bm{H}^{\phi}_{C,r}(a,\eta)=(r+1)\lambda=\frac{n-r+1}{(n+1)\mathcal{L}^{n+1}(C)}\mathcal{V}^{\phi}_{n-r+1}(C)\qquad\textrm{for
$\mathcal{H}^{n}$ a.e.\ $(a,\eta)\in\widetilde{N}^{\phi}_{n}(C)$}.$

Furthermore, the following two statements hold.

  1. (a)

If $\lambda\neq 0$, $r\geq 2$ and $C$ is $(r-1,\phi)$-mean convex, then

$\infty>\frac{\bm{h}^{\phi}_{C,1}(a)}{\binom{n}{1}}\geq\Bigg{(}\frac{\bm{h}^{\phi}_{C,2}(a)}{\binom{n}{2}}\Bigg{)}^{\frac{1}{2}}\geq\ldots\geq\Bigg{(}\frac{\bm{h}^{\phi}_{C,r}(a)}{{\binom{n}{r}}}\Bigg{)}^{\frac{1}{r}}\geq\frac{\mathcal{P}^{\phi}(C)}{(n+1)\mathcal{L}^{n+1}(C)}$

for $\mathcal{H}^{n}$ a.e. $a\in\partial^{v}C$.

  2. (b)

If $r=1$, then
$\bm{H}^{\phi}_{C,1}(a,\eta)\geq\frac{n\mathcal{P}^{\phi}(C)}{(n+1)\mathcal{L}^{n+1}(C)}$
for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in\widetilde{N}^{\phi}_{n}(C)$.

###### Proof.

From the equality
$\Theta^{\phi}_{n-r}(C,\cdot)=\lambda\,\Theta^{\phi}_{n}(C,\cdot)$ we get

$\displaystyle\int_{B\cap\widetilde{N}^{\phi}_{n}(C)}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,\Big{(}\frac{1}{r+1}\bm{H}^{\phi}_{C,r}(a,\eta)-\lambda\Big{)}\;d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad+\frac{1}{r+1}\int_{(N^{\phi}(C)\setminus\widetilde{N}_{n}^{\phi}(C))\cap
B}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,r}(a,\eta)\;d\mathcal{H}^{n}(a,\eta)=0$

for every bounded Borel set $B\subset N^{\phi}(C)$. Since
$J^{\phi}_{C}(a,\eta)\phi(\bm{n}^{\phi}(\eta))>0$ for $\mathcal{H}^{n}$ a.e.
$(a,\eta)\in N^{\phi}(C)$, we conclude that $\bm{H}^{\phi}_{C,r}(a,\eta)=0$
for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in
N^{\phi}(C)\setminus\widetilde{N}^{\phi}_{n}(C)$ and
$\bm{H}^{\phi}_{C,r}(a,\eta)=(r+1)\lambda$ for $\mathcal{H}^{n}$ a.e.
$(a,\eta)\in\widetilde{N}^{\phi}_{n}(C)$. Noting that
$\bm{H}^{\phi}_{C,0}=\bm{1}_{\widetilde{N}^{\phi}_{n}(C)}$ and employing
Theorem 6.8, we derive

$\displaystyle(n-r+1)\int_{N^{\phi}(C)}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,r-1}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=r\int_{N^{\phi}(C)}[a\bullet\bm{n}^{\phi}(\eta)]J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,r}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=r(r+1)\lambda\int_{\widetilde{N}_{n}^{\phi}(C)}[a\bullet\bm{n}^{\phi}(\eta)]J^{\phi}_{C}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=r(r+1)\lambda(n+1)\mathcal{L}^{n+1}(C),$ (77)

from which we infer that

$\bm{H}^{\phi}_{C,r}(a,\eta)=\frac{n-r+1}{(n+1)\mathcal{L}^{n+1}(C)}\mathcal{V}^{\phi}_{n-r+1}(C)\qquad\textrm{for
$\mathcal{H}^{n}$ a.e.\ $(a,\eta)\in\widetilde{N}^{\phi}_{n}(C)$,}$ (78)

since $\mathcal{L}^{n+1}(C)\in(0,\infty)$. In particular, using Corollary
5.10, we obtain

$\bm{h}^{\phi}_{C,r}(a)=\frac{n-r+1}{(n+1)\mathcal{L}^{n+1}(C)}\mathcal{V}^{\phi}_{n-r+1}(C)\qquad\textrm{for
$\mathcal{H}^{n}$ a.e.\ $a\in\partial^{v}C$.}$

Notice that the asserted conclusion for $r=1$ already follows from (78), since
$\mathcal{V}^{\phi}_{n}(C)\geq\mathcal{P}^{\phi}(C)$.

We assume now $r\geq 2$, $\lambda\neq 0$ and that $C$ is $(r-1,\phi)$-mean
convex. The non-negativity property of $\bm{H}^{\phi}_{C,r-1}$ in combination
with (6.2) implies that $\lambda\geq 0$, that means $\lambda>0$. Therefore
$\bm{h}^{\phi}_{C,r}(a)>0$ is satisfied for $\mathcal{H}^{n}$ a.e.
$a\in\partial^{v}C$. Hence we can apply Lemma 2.2 to conclude that

$\frac{\bm{h}^{\phi}_{C,1}(a)}{\binom{n}{1}}\geq\ldots\geq\left(\frac{\bm{h}^{\phi}_{C,r-1}(a)}{\binom{n}{r-1}}\right)^{\frac{1}{r-1}}\geq\left(\frac{\bm{h}^{\phi}_{C,r}(a)}{\binom{n}{r}}\right)^{\frac{1}{r}}=\left(\frac{(r+1)\lambda}{\binom{n}{r}}\right)^{\frac{1}{r}}$
(79)

for $\mathcal{H}^{n}$ a.e. $a\in\partial^{v}C$. Using again that
$\bm{H}^{\phi}_{C,r-1}(a,\eta)\geq 0$ for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in
N^{\phi}(C)$ and the lower bound for $\bm{H}^{\phi}_{C,r-1}(a,\eta)$ from
(79), we get

$\displaystyle(n-r+1)\int_{N^{\phi}(C)}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,r-1}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad\geq(n-r+1)\int_{N^{\phi}(C)|\partial^{v}_{+}C}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,\bm{H}^{\phi}_{C,r-1}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad\geq(n-r+1)[(r+1)\lambda]^{\frac{r-1}{r}}{\binom{n}{r}}^{\frac{1-r}{r}}{\binom{n}{r-1}}\int_{N^{\phi}(C)|\partial^{v}_{+}C}\phi(\bm{n}^{\phi}(\eta))\,J^{\phi}_{C}(a,\eta)\,d\mathcal{H}^{n}(a,\eta)$
$\displaystyle\qquad=(n-r+1)[(r+1)\lambda]^{\frac{r-1}{r}}{\binom{n}{r}}^{\frac{1-r}{r}}{\binom{n}{r-1}}\mathcal{P}^{\phi}(C).$
(80)

Combining (6.2) and (6.2), noting that
$(n-r+1){\binom{n}{r}}^{-1}{\binom{n}{r-1}}\frac{1}{r}=1$ and recalling that
$\lambda>0$, we conclude

$\infty>[\lambda(r+1)]^{\frac{1}{r}}\geq\frac{\mathcal{P}^{\phi}(C)}{(n+1)\mathcal{L}^{n+1}(C)}{\binom{n}{r}}^{\frac{1}{r}}.$

Now the remaining assertion follows from (79). ∎

The Alexandrov theorem for sets of positive reach is now a corollary of our
previous results.

######  Theorem 6.15.

Let $r\in\\{1,\ldots,n\\}$, and let $C\subset\mathbf{R}^{n+1}$ be an
$(r-1,\phi)$-mean convex set with positive and finite volume such that

$\Theta^{\phi}_{n-r}(C,\cdot)=\lambda\,\Theta^{\phi}_{n}(C,\cdot)\quad\textrm{for
some $\lambda\in\mathbf{R}\setminus\\{0\\}$}.$ (81)

Then $\lambda>0$ and there exist a finite natural number $N\geq 1$ and
$c_{1},\ldots,c_{N}\in\mathbf{R}^{n+1}$ such that

$\operatorname{int}(C)=\bigcup_{i=1}^{N}\operatorname{int}(c_{i}+\rho\mathcal{W}^{\phi}),\quad\operatorname{dist}^{\phi}\big{(}c_{i}+\rho\mathcal{W}^{\phi},c_{j}+\rho\mathcal{W}^{\phi}\big{)}\geq
2\operatorname{reach}^{\phi}(C)\quad\textrm{for $i\neq j$,}$

where $\rho$ satisfies the relations
$\binom{n}{r}^{\frac{1}{r}}\left(\lambda(r+1)\right)^{-\frac{1}{r}}=\rho=\frac{(n+1)\mathcal{L}^{n+1}(C)}{\mathcal{P}^{\phi}(C)}$
and $\lambda>0$.

If $r=1$, then the same conclusion is obtained for any $\lambda\in\mathbf{R}$
and any set $C$ of positive reach with positive and finite volume.

###### Proof.

The assertion is implied by a combination of Lemma 6.14 and Corollary 5.16, in
particular it follows that $\lambda>0$.

To check the equation
$\rho=\binom{n}{r}^{\frac{1}{r}}\left(\lambda(r+1)\right)^{-\frac{1}{r}}$, we
first observes that
$\kappa^{\phi}_{\rho\mathcal{W}^{\phi}}(a,\eta)=\frac{1}{\rho}$ for
$(a,\eta)\in N^{\phi}(\rho\mathcal{W}^{\phi})$ (see for instance [DRKS20,
Corollary 2.33]); since $C$ is the disjoint union of $N$ translated copies of
$\rho\mathcal{W}^{\phi}$, one infers that
$\bm{H}^{\phi}_{C,r}(a,\eta)=\binom{n}{r}\rho^{-r}$ for $(a,\eta)\in
N^{\phi}(C)$. As $\bm{H}^{\phi}_{C,r}(a,\eta)=\lambda(r+1)$ for $(a,\eta)\in
N^{\phi}(C)$ by Lemma 6.14, we obtain the aforementioned equation. ∎

In Theorem 6.15 we deal with sets which are non-convex, hence it is natural to
state the proportionality assumption (81) on the $r$-th $\phi$-mean curvatures
in terms of generalized curvature measures. However, under a slightly more
restrictive mean convexity assumption we also get the following variant of
Theorem 6.15 in which a corresponding assumption on the proportionality of a
curvature measure is imposed in (82). The proof is based on a modified version
of Lemma 6.14, whose proof only requires minor adjustments (and uses (63) in
first part of the argument).

######  Theorem 6.16.

Let $r\in\\{1,\ldots,n\\}$, and let $C\subset\mathbf{R}^{n+1}$ be an
$(r-1,\phi)$-mean convex set with positive and finite volume and such that
$\bm{H}^{\phi}_{C,r}(a,\eta)\geq 0$ for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in
N^{\phi}(C)$. Assume that

$\mathcal{C}^{\phi}_{n-r}(C,\cdot)=\lambda\,\mathcal{C}^{\phi}_{n}(C,\cdot)\quad\textrm{for
some $\lambda>0$.}$ (82)

Then there exist a finite natural number $N\geq 1$ and
$c_{1},\ldots,c_{N}\in\mathbf{R}^{n+1}$ such that

$\operatorname{int}(C)=\bigcup_{i=1}^{N}\operatorname{int}(c_{i}+\rho\mathcal{W}^{\phi}),\quad\operatorname{dist}^{\phi}\big{(}c_{i}+\rho\mathcal{W}^{\phi},c_{j}+\rho\mathcal{W}^{\phi}\big{)}\geq
2\operatorname{reach}^{\phi}(C)\quad\textrm{for $i\neq j$,}$

where $\rho$ is given as in Theorem 6.15.

We conclude this section discussing the validity of the hypothesis of Theorem
6.15 in terms of Alexandrov points and pointwise curvatures for a large
subclass of sets of positive reach, namely those sets $C$ for which
$\mathcal{H}^{n}(\partial C\setminus\partial^{v}C)=0$. This class includes all
convex bodies, and more generally all closed sets that can be locally
represented as the epigraph of a semiconvex function; see Lemma 6.20. But it
includes much more; indeed, it is easy to construct sets of positive reach $C$
for which $\mathcal{H}^{n}(\partial C\setminus\partial^{v}C)=0$, but the
boundary is not a topological manifold (see [RZ17, Example 7.12] or [ACV08,
Example 1]).

The hypotheses in the next statement should be seen in connection with the
disjoint union displayed in (63).

######  Lemma 6.17.

If $k\in\\{1,\ldots,n\\}$, $\lambda\in\mathbf{R}$ and $\varnothing\neq
C\subset\mathbf{R}^{n+1}$ is a set of positive reach such that
$\mathcal{H}^{n}(\partial C\setminus\partial^{v}C)=0$ and
$\mathcal{P}^{\phi}(C)>0$, then the following two statements hold.

  1. (a)

If $\bm{h}^{\phi}_{C,k}(a)=\lambda$ for $\mathcal{H}^{n}$ a.e.
$a\in\mathcal{A}(C)$ and
$\mathcal{H}^{n-k}\big{[}\bm{p}(\widetilde{N}^{\phi}(C)\setminus\widetilde{N}^{\phi}_{n}(C))\big{]}=0$,
then

$\Theta^{\phi}_{n-k}(C,\cdot)=\lambda\,\Theta^{\phi}_{n}(C,\cdot).$

  2. (b)

If $\bm{h}^{\phi}_{C,1}(a)\geq 0,\ldots,\bm{h}^{\phi}_{C,k-1}(a)\geq 0$ for
$\mathcal{H}^{n}$ a.e. $a\in\mathcal{A}(C)$ and
$\mathcal{H}^{n-k+1}\big{[}\bm{p}(\widetilde{N}^{\phi}(C)\setminus\widetilde{N}^{\phi}_{n}(C))\big{]}=0$,
then $C$ is $(k-1,\phi)$-mean convex.

###### Proof.

(a) Let $B\subseteq N^{\phi}(C)$ be a Borel set. Noting that
$\Theta^{\phi}_{n}(C,B)=\Theta^{\phi}_{n}\big{(}C,B\cap\widetilde{N}^{\phi}_{n}(C)\big{)}$
by definition of $\bm{H}^{\phi}_{C,0}$, we can use Lemma 6.7, Lemma 6.4 and
Corollary 5.10 (a), (b) together with $\eta(a)=\nabla\phi(\bm{n}(C,a))$ for
$a\in\partial^{v}C$ to compute

$\displaystyle\Theta^{\phi}_{n-k}(C,B)$
$\displaystyle=\Theta^{\phi}_{n-k}\big{(}C,B\cap\widetilde{N}^{\phi}_{n}(C)\big{)}$
$\displaystyle=\int_{\partial^{v}C}\mathbf{1}_{B}(a,\eta(a))\phi(\bm{n}(C,a))\bm{h}^{\phi}_{C,k}(a)\,d\mathcal{H}^{n}(a)$
$\displaystyle=\lambda\int_{\partial^{v}C}\mathbf{1}_{B}(a,\eta(a))\phi(\bm{n}(C,a))\,d\mathcal{H}^{n}(a)$
$\displaystyle=\lambda\,\Theta^{\phi}_{n}(C,B).$

(b) Arguing as in (a) we can compute

$\displaystyle\Theta^{\phi}_{n-k+1}(C,B)$
$\displaystyle=\Theta^{\phi}_{n-k+1}\big{(}C,B\cap\widetilde{N}^{\phi}_{n}(C)\big{)}$
$\displaystyle=\int_{\partial^{v}C}\mathbf{1}_{B}(a,\eta(a))\phi(\bm{n}(C,a))\bm{h}^{\phi}_{C,k-1}(a)\,d\mathcal{H}^{n}(a)\geq
0$

for every Borel set $B\subseteq N^{\phi}(C)$. This means that
$\bm{H}^{\phi}_{C,k-1}(a,\eta)\geq 0$ for $\mathcal{H}^{n}$ a.e. $(a,\eta)\in
N^{\phi}(C)$ and consequently $C$ is $(\phi,k-1)$-mean convex. ∎

Now with the help of Lemma 6.17, the following result can be easily deduced as
a special case of Theorem 6.15.

######  Corollary 6.18.

Suppose $k\in\\{1,\ldots,n\\}$, $\lambda\in\mathbf{R}\setminus\\{0\\}$ and
$\varnothing\neq C\subset\mathbf{R}^{n+1}$ is a set of positive reach with
finite and positive volume such that

  1. (1)

$\mathcal{H}^{n}(\partial C\setminus\partial^{v}C)=0$ and
$\mathcal{H}^{n-k}\big{[}\bm{p}(\widetilde{N}^{\phi}(C)\setminus\widetilde{N}^{\phi}_{n}(C))\big{]}=0$,

  2. (2)

$\bm{h}^{\phi}_{C,k}(a)=\lambda$ for $\mathcal{H}^{n}$ a.e.
$a\in\mathcal{A}(C)$,

  3. (3)

$\bm{h}^{\phi}_{C,1}(a)\geq 0,\ldots,\bm{h}^{\phi}_{C,k-1}(a)\geq 0$ for
$\mathcal{H}^{n}$ a.e. $a\in\mathcal{A}(C)$.

Then the conclusion of Theorem 6.15 holds. If $k=1$, then the same conclusion
is true for every $\lambda\in\mathbf{R}$.

######  Remark 6.19.

Corollary 6.18 includes as very special cases the soap bubble theorems of
Alexandrov ([Ale58]), Korevaar–Ros ([Ros87] and [Ros88]) and He–Li–Ma–Ge
([HLMG09]). In fact for connected and compact domains with
$\mathcal{C}^{2}$-boundary the hypothesis (c) of Corollary 6.18 can be easily
deduced from the existence of an elliptic point, the continuity of the
principal curvatures and the Garding theory on hyperbolic polynomials (see
[Ros88, page 450] for further details). The continuity of the principal
curvatures and the assumption of connectedness play a key role in this
argument.

Recall the definition of epigraph from (61).

######  Lemma 6.20.

Suppose $C\subseteq\mathbf{R}^{n+1}$ is a compact set such that for every
$a\in\partial C$ there exists $u\in\mathbf{S}^{n}$, $\epsilon,\delta>0$ and a
semiconvex function $f:a+u^{\perp}\rightarrow\mathbf{R}$ such that

$\operatorname{epi}(f)\cap U_{\varepsilon,\delta}(a,u)=C\cap
U_{\varepsilon,\delta}(a,u).$ (83)

Then $\operatorname{reach}(C)>0$ and $\mathcal{H}^{n}(\partial
C\setminus\partial^{v}C)=0$.

###### Proof.

For $a\in\partial C$, let $\varepsilon(a),\delta(a)>0$ and the local
representation in terms of a semiconvex function $f_{a}$ be as in (83). By
[Fu85, Theorem 2.3] we know that
$\operatorname{reach}(\operatorname{epi}(f_{a}))\geq r(a)>0$. Define
$\rho(a)=\frac{1}{4}\min\\{\varepsilon(a),\delta(a),r(a)\\}$. Then we have
$U(a,\rho(a))\subseteq\operatorname{Unp}(C)$ for every $a\in\partial C$. If
$a\in\operatorname{int}(C)$, then there is also a positive number $\rho(a)$
such that
$U(a,\rho(a))\subset\operatorname{int}(C)\subseteq\operatorname{Unp}(C)$.
Since the sets $U(a,\rho(a))$, for $a\in C$, are an open cover of the compact
set $C$, we get a finite number of points $a_{1},\ldots,a_{N}\in C$ such that
$C\subseteq\bigcup_{i=1}^{N}U(a_{i},\frac{\rho(a_{i})}{2})$. Then for
$0<\tau<\inf\\{\frac{\rho(a_{1})}{2},\ldots,\frac{\rho(a_{N})}{2}\\}$, it
holds that for every $c\in C$ there is some $a_{i}$ such that
$U(c,\tau)\subseteq U(a_{i},\rho(a_{i}))\subseteq\operatorname{Unp}(C)$. This
shows that $\operatorname{reach}(C)\geq\tau>0$.

Note that $\partial C=\bm{p}(N(C))$ (see e.g. [RZ19, Corollary 4.12(a)]).
Since $\mathcal{H}^{0}(N(C,a))\neq 2$ for $a\in\partial C$ due to (83), it
follows from Lemma 3.25 (c) that
$\mathcal{H}^{n}(\bm{p}(N(C))\setminus\partial^{v}C)=0$, which gives the
remaining assertion. ∎

Acknowledgements. D. Hug was supported by DFG research grant HU 1874/5-1 (SPP
2265) and gratefully acknowledges support by ICERM (Brown University).

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