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	Emanuele Paolini: 
	Regularity for minimal boundaries in $R^n$
	with mean curvature in $L^n$}
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	Manuscripta Mathematica
	MM-No.: 64
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\markboth{Regularity for minimal boundaries
with mean curvature in $L^n$
}{Emanuele Paolini}
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\begin{LARGE}
{Regularity for minimal boundaries in $\R^n$ with mean curvature in $L^n$}
\end{LARGE}
\mbox{}\vspace{0.5 cm}\\
\begin{Large}Emanuele Paolini\end{Large}
\mbox{}\vspace{0.1 cm}\\
{Scuola Normale Superiore, Piazza dei Cavalieri 7, 56126 PISA, Italy}
\\
\\
\\}
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	{}
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\item
\noindent
\begin{small}
\\
Let $E_0\subset\R^n$ be a minimal set with mean curvature in $L^n$
that is a minimum of the functional
$E\mapsto P(E,\Omega)+\int_{E\cap\Omega} H$, where $\Omega\subset\R^n$
is open and $H\in L^n(\Omega)$. We prove that if $2\le n\le 7$ then
$\partial E_0$ can be parametrized over the $(n-1)$-dimensional disk with a
$\C^{0,\alpha}$ mapping with $C^{0,\alpha}$ inverse.
\end{small}
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\section*{Introduction}
We are concerned with the study of the functional
\[ %\label{eq:fun}
	F_H(E) = P(E,\Omega)+\int_{E\cap\Omega} H
\] 
where $\Omega$ is an open set of $\R^n$, $H\in L^1(\Omega)$, $E$ is a
\emph{Caccioppoli set} (i.e. the distributional derivative $D\chi_E$ of the
characteristic function of $E$ is a Radon vector measure)
and $P(E,\Omega)=|D\chi_E|$ (the total variation of $D\chi_E$) is the 
\emph{perimeter} of $E$ in $\Omega$.
If $E$ is a local minimizer of $F_H$ (i.e. if $F_H(E)\le F_H(X)$ for
all $X$ such that $E\diffsim X$\footnote{Here $E\diffsim X=(E\setminus X) \cup (X\setminus E)$ denotes the symmetric difference between $E$ and $X$.} has compact closure in $\Omega$), then we will say that
\emph{$E$ is a minimal set with mean curvature $H$ in $\Omega$}.
As a matter of fact, if $H$ and $\partial E$ are
sufficiently regular near a point $x\in \partial E\cap \Omega$, then
$-H(x)/(n-1)$ is the usual mean curvature of $\partial E$ in $x$. 

In 1974 U.~Massari \cite{Mas74,Mas75} proved that if $E$ is a
minimal set with mean curvature $H\in L^p$
and $p>n$ then, up to a set of dimension not greater than $n-8$,
$\partial E$ is a hypersurface of class $C^{1,\alpha}$ with $\alpha=(p-n)/4p$.
Then, in 1987 E.~Barozzi, E.~H.~A.~Gonzales and I.~Tamanini
\cite{BarGonTam87,BarTam88} proved that given any Caccioppoli set
$E$ it is possible to find a suitable $H\in L^1$ such that $E$
is a minimal set with mean curvature $H$. Moreover they noticed that
if $E$ is a minimal set with curvature
$H\in L^p$ with $1\le p <n$ then $\partial E$ may contain many
singular points.
Nothing was said about the case $H\in L^n$ until 1992 when 
E.~De~Giorgi conjectured the existence of a singular
curve in $\R^2$ which is the boundary of a Caccioppoli set and which
is a minimal set with mean curvature $H\in L^2$. This conjecture was proved by
E.~H.~A.~Gonzales, U.~Massari and I.~Tamanini in \cite{GonMasTam93}.

A still open conjecture by De~Giorgi is that if $E$ is a minimal set
with mean curvature
$H\in L^n(\Omega)$ then (up to a set of dimension not greater than $n-8$) $\partial E$ can be parametrized by a
bi-lipschitz map. The aim of this paper is to give a partial answer to
this conjecture proving the following

\begin{theorem}\label{teo}
Let $\Omega$ be an open set of $\R^n$.
If $H\in L^n(\Omega)$, $E_0$ is a minimal set with mean curvature
$H$ in $\Omega$ and $2\le n\le 7$, then
given any $x_0\in \partial E_0$ and given $0<\alpha<1$ there exists a
neighbourhood of $x_0$ in $\partial E$ in which $\partial E_0$ may be
parametrized by an $\alpha$-H{\"o}lder map
with $\alpha$-H{\"o}lder inverse.
\end{theorem}

\noindent
{\bf Acknowledgments.} I would like to thank Mariano Giaquinta for his
invaluable advice; I also thank Luigi Ambrosio for many useful suggestions. 

\section{Fundamental tools}
To prove this theorem we follow the techniques used by
Reifenberg~\cite{Rei60} in the study of minimal surfaces. In his work
Reifenberg represented surfaces simply as closed
sets whose area is the Hausdorff measure and whose boundaries are
given by homological properties.
A fundamental tool used by Reifenberg is the topological-disk theorem
(Theorem~\ref{teo:Reif}), which gives a sufficient condition for
parametric 
$C^{0,\alpha}$-regularity for a closed set.

To apply Reifenberg's techniques, we take advantage of the theory of
Caccioppoli sets. It's well known that a minimum of $F_H$ does exist but, as we have seen, regularity
results hold only if $H\in L^p$ with $p>n$. Anyway in~\cite{GonMasTam93}
many properties for minima are given even in the case $H\in L^n$. In
Theorem~\ref{teo:spirale} we summarize the properties we need in
this paper. 

Given $X$ a subset of $\R^n$ and $\rho>0$ we denote by
\[
	(X)_\rho=\{y\in\R^n\: \exists x\in X\ |x-y|\le \rho\}
\]
the closed $\rho$-neighbourhood of $X$.

\begin{definition}\label{def:reif}
Let $S$ be a closed set of $\R^n$, $x_0\in S$, $R>0$, $\epsilon>0$,
and $m\le n$ an integer.
We say that $S$ satisfies \emph{$(\epsilon,R,m)$-Reifenberg property} in $x_0$ if given any ball $B(x,r)\subset B(x_0,R)$
with $x\in S$ there exists an affine $m$-plane $\Sigma$ through $x$ such that
\bea
	S\cap B(x,r)&\subset& \left(\Sigma\cap B(x,r)\right)_{\epsilon
	r},\\
	\Sigma\cap B(x,r) &\subset& \left(S\cap
	B(x,r)\right)_{\epsilon r}.
\eea
\end{definition}

The following theorem can be found in~\cite[Chapter~4, page~64]{Rei60}.
\begin{theorem}[topological disk]\label{teo:Reif}
Given $0<\alpha<1$ there exists $\epsilon_0>0$ such that if $S$ is a
closed set in $\R^n$ satisfying the 
$(\epsilon,R,m)$-Reifenberg property in $x_0\in S$ for some $R>0$,
some integer $0\le m \le n$ and some
$0<\epsilon<\epsilon_0$ then there exists a neighbourhood of
$x_0$ in $S$ which is homeomorphic
to the $m$-dimensional disk through an homeomorphism $\tau$ such
that both $\tau$ and $\tau^{-1}$ are $\alpha$-H{\"o}lder maps.
\end{theorem}

\begin{theorem}\label{teo:spirale}
Let $E\subset\R^n$ be a minimal set with mean curvature
$H$ in $\Omega$, with $H\in L^n(\Omega)$. 
Then we can find a closed set $E_0$  which differs from $E$
by a set of Lebesgue measure zero, which is a minimal set with mean
curvature $H$ in $\R^n$ and
satisfies the following properties:
\begin{enumerate}
\item 
$\partial E_0$ is $(n-1)$-rectifiable 
and 
$P(E_0,A)=\H^{n-1}(A\cap\partial E_0)$ for any open set $A\subset\R^n$;
\item
given a sequence of points $x_k\to x_0$ with $x_0,x_k\in\partial
 E_0\cap \Omega$, and a sequence $\lambda_k\to
 \infty$ ($\lambda_k>0$) there exist a subsequence 
 $k_j$ and a minimal set $E_\infty$ with  mean curvature $0$ in $\R^n$  such that
\[
	\lim_{j\to\infty}\lambda_{k_j}(E_0-x_{k_j})= E_\infty
\]
in $L^1_\loc(\R^n)$, and
\[
	\lim_{j\to\infty} P(\lambda_{k_j}(E_0-x_{k_j}),B(0,1))
	= P(E_\infty,B(0,1)).
\]	
Moreover if $n\le 7$ then $E_\infty$ is an half-space.
\end{enumerate}
\end{theorem}
\begin{proof}
See Theorem~1.1 and Theorem~1.2 of~\cite{GonMasTam93}.
\end{proof}

The following lemma gives a link between minimal surfaces and surfaces
with prescribed mean curvature, specifing in which sense the latter
may be called \emph{quasi-minimal surfaces.}
%
\begin{lemma}\label{lemma:omegamin}
Let $H\in L^n(\Omega)$ and $E_0$ be a minimal set with mean curvature
$H$ in $\Omega$.
Then, there exists an increasing function $\omega\:\R^+\to\R^+$ with 
\[
	\lim_{r\to 0^+}\omega(r)=0
\]
such that if $E$ is a Caccioppoli set and
$\overline{E\diffsim E_0} \subset B(x,r)\subset \Omega$ then
\[
	P(E_0,B(x,r))\le P(E,B(x,r))+ r^{n-1} \omega(r).
\]
If only
$E\diffsim E_0\subset B(x,r)\subset\Omega$ then
\[
	P(E_0,\overline{B(x,r)})\le P(E,\overline{B(x,r)}) +r^{n-1}\omega(r).
\]	
\end{lemma}
\begin{proof}
If $\overline{E\diffsim E_0}\subset B(x,r)$, in view of the locality
of perimeter, one can find an open set which contains
$\Omega\setminus B(x,r)$ such that the measures $P(E_0,\cdot)$
and $P(E,\cdot)$ coincide on it. So $P(E_0,\Omega\setminus
B(x,r))=P(E,\Omega\setminus B(x,r))$.
From the minimality of $E_0$ we obtain
\[
	P(E_0,B(x,r))+\int_{B(x,r)\cap E_0} H
	\le P(E,B(x,r))+\int_{B(x,r)\cap E} H
\]
hence
\[
	P(E_0,B(x,r))
	\le P(E,B(x,r))+\int_{B(x,r)}|H|
\]
and from H{\"o}lder inequality we get
\[
	\int_{B(x,r)}|H|
	\le ||H||_{L^n(B(x,r))} |B(x,r)|^{n-1\over n}
	= \gamma_n^{n-1\over n} r^{n-1} ||H||_{L^n(B(x,r))}
\]
(in the whole paper we denote by $\gamma_k$ the measure of the
 $k$-dimensional unit ball and use the notation $|X|=\H^n(X)$).
So if we set
\begin{equation}\label{eq:omega}
	\omega(r)=\sup\{\gamma_n^{n-1\over n} ||H||_{L^n(B(x,\rho))}\:
	B(x,\rho)\subset\Omega,\ \rho\le r\}
\end{equation}
using $H\in L^n$ we obtain, as needed, that $\omega$ is increasing and
with limit $0$ as $r\to 0$.

If $E\diffsim E_0\subset B(x,r)$ we know that $P(E_0,\cdot)$ and
$P(E,\cdot)$ coincide on the open set
$\Omega\setminus \overline{B(x,r)}$. With similar considerations as
before, we obtain the second statement.
\end{proof}

\section{Technical Lemmas}

Let us fix some notations.
Let $\Omega\subset\R^n$, $H\in L^n(\Omega)$, $E_0$ be
a minimum of $F_H$ satisfying the properties given in 
Theorem~\ref{teo:spirale} and $\omega$ the function
described in Lemma~\ref{lemma:omegamin}. 
Let $2\le n\le 7$ and $m=n-1$.
We adopt the following notations 
\bea
	S_0 &=& \partial E_0;\\
	\phi(x,r) &=& \H^m(S_0\cap B(x,r));\\
	\psi(x,r) &=& \int_0^r \H^{m-1}(S_0\cap\partial B(x,t))\d t.
\eea
Here by $\H^k$ we mean the $k$-dimensional spherical Hausdorff
measure that is
\[
	\H^k(X) = \lim_{\delta\to 0} \H_\delta^k(X)
\]
where
\[
	\H^k_\delta(X) = \inf\left\{\sum_i \gamma_m r_i^m\:
	\{B(x_i,r_i)\}_i\ \mbox{is a covering of}\ X,\ 
	r_i<\delta \right\}.
\]
Notice that Theorem~\ref{teo:spirale} was proved for Hausdorff
measure, but it also holds for spherical Hausdorff measure 
as these two measures coincide on rectifiable sets (see \cite[3.2.26]{Fed69}).
On the other hand, for a general Caccioppoli set
$E$ we will only use the following property
	\begin{equation}\label{eq:diseq}
		P(E,A)\le \H^m(\partial E\cap A)
	\end{equation}
which holds for Hausdorff measure and then also for spherical
Hausdorff measure, the second being bigger than the first (see
\cite[2.10.2]{Fed69}).

\begin{lemma}\label{lemma:coarea} 
Let $S$ be a closed set of $\R^n$ and denote
by $S_t$ the set of the points of $S$ which have distance $t$ from a
given $k$-dimensional plane $\Sigma$ where $0\le k \le n$. 
Then, for every integer $m\le n$,
\begin{equation}\label{eq:coarea}
\int_0^\infty \H^{m-1}(S_t)\d t \le \H^m(S)
\end{equation}
\end{lemma}
\begin{proof}
See~\cite[Theorem~10.2.3]{Mor66}.
\end{proof}

\begin{lemma}\label{lemma:cono}
Given $x\in\Omega$, for almost every $r$ with
$B(x,r)\subset\Omega$ and every $z\in B(x,r)$ there exists a 
set $E$ such that 
\[
	E\diffsim E_0 \subset  B(x,r)
\]
and such that for every open set $A\subset\R^n$ the following estimate holds
\bea
	P(E,\overline{B(x,r)}) 
	&\le& {r_1\over m} \H^{m-1}(S_0\cap\partial B(x,r)\cap A)\\
	\mbox{}&&+ {r_2\over m} \H^{m-1}(S_0\cap\partial B(x,r)\setminus A)
\eea
where
\bea
	r_1&=&\sup\{|y-z|\: y\in S_0\cap\partial B(x,r)\cap A\};\\
	r_2&=&\sup\{|y-z|\: y\in (S_0\cap\partial B(x,r))\setminus A\}.
\eea
In particular, 
if we choose $A=\R^n$ and $z=x$ we obtain
\[
	P(E,\overline{B(x,r)})
	\le {r\over m} \H^{m-1}(S_0\cap\partial B(x,r)).
\]

\end{lemma}
\begin{proof}
As $S_0$ is an $m$-rectifible set, for almost every $r$
the set $S_0\cap\partial B(x,r)$ is $(m-1)$-rectifiable.
Let $C=C(z,E_0\cap\partial B(x,r))$ be the cone over $E_0\cap\partial
B(x,r)$ with vertex $z$ and let $E=(E_0\setminus B(x,r))\cup C$.
We can easily find that
	\[
	\partial E = (S_0 \setminus B(x,r)) \cup C(z,S_0\cap\partial B(x,r))
	\]
so $\partial E$ is an $m$-rectifiable set with finite $m$-dimensional
measure, that is $E$ is a Caccioppoli set.

Moreover, from~(\ref{eq:diseq}),
\bea
\lefteqn{P(E,\overline{B(x,r)})}\\
	&\le& \H^m(\partial E\cap\overline{B(x,r)}) \\
	&=& \H^m\left(C(z,S_0\cap\partial B(x,r))\right)\\
	&=& \H^m\left(C(z,S_0\cap\partial B(x,r)\cap A)\right)\\
	&&\mbox{}+ \H^m\left(C(z,(S_0\cap\partial B(x,r))\setminus
	A)\right)\\
	&\le& {r_1\over m} \H^{m-1}(S_0\cap\partial B(x,r)\cap A) 
		+{r_2\over m} \H^{m-1}((S_0\cap\partial B(x,r))\setminus
	A)
\eea
where in the last inequality we use the fact that if
$X\subset B(z,r)$ is  an  $(m-1)$-rectifiable set then
\[
	\H^{m}\left(C(z,X)\right) \le {r\over m}\H^{m-1}(X).
\]
\end{proof}

\begin{lemma}\label{lemma:1*} 
If $x\in \Omega$ and $B(x,r)\subset\Omega$ then
\[
	\psi(x,r)\le\phi(x,r)
\]
and for almost every $r$ such that $B(x,r)\subset\Omega$ 
\[	
	\phi(x,r)\le {r\H^{m-1}(S_0\cap\partial
	B(x,r))\over m}+ r^m \omega(r).
\]
\end{lemma}
\begin{proof}
First inequality is given by Lemma~\ref{lemma:coarea}. 
To prove the second inequality
we consider
the set $E$ given by the second part of Lemma~\ref{lemma:cono}. Then
\[
	P(E,\overline{B(x,r)}) \le {\frac r m}\H^{m-1}(S_0\cap\partial B(x,r))
\]
and, on the other hand, 
from Lemma~\ref{lemma:omegamin} we get
\bea
	\phi(x,r)&=&\H^m(S_0\cap B(x,r))=P(E_0,B(x,r))\\
	&\le& P(E,\overline{B(x,r)})+r^m\omega(r).
\eea
\end{proof}


\begin{lemma}\label{lemma:omegamul}
If $B(x,r)\subset \Omega$ and $E=E_0\setminus B(x,r)$ then, for almost
every $r$,
\[
P(E_0,\overline{B(x,r)}) 
\le {1+c_n\omega(r)\over 1-c_n\omega(r)} P(E,\overline{B(x,r)}).
\]
where $c_n$ is a constant depending only on $n$.
\end{lemma}
\begin{proof}
As in Lemma~\ref{lemma:omegamin} we have (being $E=E_0\setminus B(x,r)$)
	\[
	P(E_0,\overline{B(x,r)})
	\le P(E,\overline{B(x,r)}) + ||H||_{L^n(B(x,r))}|E_0\cap B(x,r)|^{n-1\over n}.
	\]	
From the isoperimetric inequality we obtain
	\bea
	|E_0\cap B(x,r)| 	
	&\le& c'_n P(E_0\cap B(x,r),\R^n)^{n\over n-1}\\
	&\le&
	c'_n\left(P(E_0,\overline{B(x,r)})+P(E,\overline{B(x,r)})\right)^{n\over n-1}.
	\eea
So
	\bea
	\lefteqn{
	||H||_{L^n(B(x,r))}|E_0\cap B(x,r)|^{n-1\over n}}\\
	&\le& ||H||_{L^n(B(x,r))}c''_n 
	\left( P(E_0,\overline{B(x,r)})+P(E,\overline{B(x,r)})\right)
	\eea
and by the definition of $\omega$ we have
	\bea
	P(E_0,\overline{B(x,r)})
	&\le& P(E,\overline{B(x,r)}) \\
	\mbox{}&&+ c_n\omega(r) \left(P(E_0,\overline{B(x,r)})+P(E,\overline{B(x,r)})\right)
	\eea
which is the claim of the lemma.
\end{proof}

\begin{lemma}\label{lemma:spirale}
Let $n\le 7$ and $x_0\in S_0\cap \Omega$. 
Given $\alpha>0$ there exists
$r_0=r_0(\alpha)>0$ such that if $x\in S_0$, $B(x,r)\subset B(x_0,r_0)$ then 
\[
1-\alpha 
\le {\phi(x,r)\over \gamma_m r^m}
\le 1+\alpha.
\]
In particular every point of $S_0\cap\Omega$ has $m$-density $1$ in
$S_0$.
\end{lemma}
\begin{proof}
Suppose
there exist $\alpha>0$,
a sequence of points $x_k\to
x_0$, $x_k\in S_0$ and a sequence of radii
 $r_k\to 0$ such that
\begin{equation}\label{eq:spirale}
	\left|\frac{\H^m(\partial E_0\cap B(x_k,r_k))}{\gamma_m
	r_k^m}-1 \right|
	\ge \alpha.
\end{equation}
By Theorem~\ref{teo:spirale}, up to a subsequence, the sequence 
$(1/ r_k)(E_0-x_k)$ converges to an halfspace $E_\infty$.
So we obtain
	\bea
	\gamma_m
	&=&P(E_\infty,B(0,1))
	= \lim_{k\to\infty} P((1/r_k)(E_0-x_k),B(0,1)) \\
	&=&\lim_{k\to\infty} {1\over r_k^m}P(E_0,B(x_k,r_k))
	=\lim_{k\to\infty} {\H^m(\partial E_0\cap B(x_k,r_k))\over r_k^m}
	\eea
which is a contradiction with~(\ref{eq:spirale}).
\end{proof}

\begin{lemma}\label{lemma:schiacciata}
Let $E$ be a closed Caccioppoli set in $\Omega$, 
$B(x,r)\subset\Omega$,
$\Sigma$ an affine hyperplane through $x$, $\pi_\Sigma$ the orthogonal
projection onto $\Sigma$ and
$0<\epsilon<{1\over 2}$. If $\partial E\cap
\overline{B(x,r)}$ is contained in the interior of $(\Sigma)_{\epsilon r}$, 
then either
	\begin{equation}\label{eq:schiacciata1}
	\pi_\Sigma(\partial E\cap B(x,r))\supset\Sigma\cap
	B(x,(1-\epsilon)r)
\end{equation}
or, if $E'=E\setminus B(x,r)$, 
	\begin{equation}\label{eq:schiacciata2}
	P(E',\overline{B(x,r)})
	\le 4\epsilon r\H^{m-1}(\partial E\cap\partial B(x,r)).
	\end{equation}
\end{lemma}
\begin{proof}
Let $x_1,\ldots,x_n$ be a system of coordinates such that
$\Sigma=\{x_n=0\}$ and define $E^+=\{x_n\ge\epsilon
r\}\cap\overline{B(x,r)}$ and $E^-=\{x_n\le -\epsilon
r\}\cap\overline{B(x,r)}$. 
From the assumptions on $E$ 
we know that $\partial E$ itersects
 neither $E^+$ nor $E^-$. So, being $\partial E$ closed and
$E^+$, $E^-$ compact sets, we conclude that $\partial E$ has positive
distance from $E^+$ and $E^-$.
So one of the following statements must hold:
\begin{enumerate}
\item\label{poss1}
$E\supset E^-$, $E\cap E^+=\emptyset$;
\item\label{poss2}
$E\supset E^+$, $E\cap E^-=\emptyset$;
\item\label{poss3}
$ E\cap E^+=\emptyset$, $E\cap E^-=\emptyset$;
\item\label{poss4}
$ E\supset E^+\cup E^-$.
\end{enumerate}

\emph{Suppose that~\ref{poss1} is true,} than we claim
that~(\ref{eq:schiacciata1}) holds.
Let $\Sigma^+=\{x_n=\epsilon r\}\cap\overline{B(x,r)}$ and
$\Sigma^-=\{x_n=-\epsilon r\}\cap\overline{B(x,r)}$.
Given $y\in\Sigma\cap B(x,(1-\epsilon)r)$ denote by $y^+$ and $y^-$
respectively the intersections of $\pi^{-1}_\Sigma(x)$ with $\Sigma^+$
and $\Sigma^-$. 
Since $y^-\in E$ while $y^+\not\in E$ 
there must certainly be a point in the segment
 $[x^-,x^+]$ which belongs to
$\partial E$ (and that is contained in $B(x,r)$) so
$\pi_\Sigma(\partial E\cap B(x,r))\ni y$.

\emph{If statement~\ref{poss2} holds} with a symmetry with respect to
$\Sigma$ we obtain~(\ref{eq:schiacciata1}) again.

\emph{Suppose that~\ref{poss3} is true}.
This time we claim that~(\ref{eq:schiacciata2}) holds.
Let $E'=E\setminus B(x,r)$. Obviously $E'$ is a closed Caccioppoli set,
$E\diffsim E'\subset B(x,r)$ and we get
\begin{equation}\label{eq:sch1}
	\partial E'\cap\overline{B(x,r)}
	=\partial E'\cap\partial B(x,r)
	= E \cap \partial B(x,r).
\end{equation}
Consider now the projection 
\[
\pi\:\partial
B(x,r)\cap(\Sigma)_{\epsilon r}\to \partial B(x,r)\cap \Sigma
\]
which maps a point $z$ to the point of $\partial B(x,r)\cap\Sigma$
with minimal distance from $z$. 
Then we have
\[
	\pi(E\cap \partial B(x,r))\subset \pi(\partial E\cap\partial
	B(x,r))
\]
so that
\begin{eqnarray}\label{eq:sch2}
	\H^m(E\cap\partial B(x,r))
	&\le& \H^m(\pi^{-1}(\pi(\partial E\cap\partial B(x,r)))) \\
	&\le& 2\epsilon r\H^{m-1}(\pi(\partial E\cap \partial B(x,r)))\nonumber
	\\
	&\le& 4\epsilon r\H^{m-1}(\partial E\cap\partial B(x,r)).\nonumber
\end{eqnarray}
We conclude, using~(\ref{eq:sch1}) and~(\ref{eq:sch2}),
\bea
	P(E',\overline{B(x,r)})
	&\le& \H^m(\partial E'\cap\overline{B(x,r)})\\
	&=& \H^m(E\cap \partial B(x,r))
	\le 4\epsilon r \H^{m-1}(\partial E\cap\partial B(x,r)).
\eea

\emph{If statement~\ref{poss4} holds} 
the proof is that of case~\ref{poss3} where instead of $E$ we consider
its complementary set.
\end{proof}

\begin{lemma}\label{lemma:7*}
Let $S_0$ be a closed set, $\theta>0$, $e>0$, $\alpha>0$ and $0<\epsilon<
1-\cos\theta$.
Let $B(x,r)$ be a ball such that given any arbitrarily small
$\delta>0$ there exist a family of disjoint balls $\{B(x_i,r_i)\}$
and a family of $m$-dimensional affine planes $\{\Sigma_i\}$ with $x_i\in\Sigma_i$
such that every plane $\Sigma_i$ makes an angle greater than $\theta$
with the segment $[x_i;x]$
and, moreover, the following properties hold
\begin{eqnarray*}
B(x_i,r_i)&\subset& B(x,r) \label{eq:7*15} \\
r_i  &<& \delta \label{eq:7*16}\\
\phi(x_i,r_i)&\ge&(1-\alpha)\gamma_m r_i^m \label{eq:7*17}\\
\sum_i \gamma_m r_i^m &>& e \label{eq:7*18}\\
S_0\cap B(x_i,r_i)&\subset& (\Sigma_i)_{\epsilon r_i}. \label{eq:7*19}
\end{eqnarray*}
Then
\[
\psi(x,r)\le \frac{\phi(x,r)}{1-\alpha}-e\left({1-\cos\theta-\epsilon \over
2}\right)^m.
\]
\end{lemma}

\begin{proof} 
(See~{\cite[Lemma $7^*$]{Rei60}}).
Suppose $\Sigma$ is an $m$-dimensional plane and that
$y_1$ and $y_2$ are two points of $\Sigma$ at a distance $l$ apart:
define $d(l,\rho,t)$ and $\sigma(l,\rho,t)$ to be respectively the
diameter and the $\H^{m-1}$ measure of the set $\Sigma\cap B(y_1,\rho)\cap
\partial B(y_2,t)$. It follows immediately from the definition of the
symbols concerned that
\begin{equation}\label{eq:7*21}
	\int_0^r \sigma(l,\rho,t)\d t
	= \H^m(\Sigma\cap B(y_1,\rho)\cap B(y_2,r)),
\end{equation}
and that for $\delta > \rho$
\begin{equation}\label{eq:7*22}
	H_\delta^{m-1} B(y_1,\rho)\cap\partial B(y_2,t)
	\le \gamma_{m-1}\big(\frac 1 2 d(l,\rho,t)\big)^{m-1}
	\le \sigma(l,\rho,t).
\end{equation}

From the definition of spherical Hausdorff measure we can choose a set
of balls $B(x_j',r_j')$ such that for all $j$
\bea
	r_j'&<&\delta \\
	S_0\cap B(x,r)\setminus \bigcup_i (S_0\cap B(x_i,r_i))
		&\subset& \bigcup_i B(x_j',r_j') \\
	\H^m\Big((S_0\cap B(x,r))\setminus \bigcup_i(S_0\cap B(x_i,r_i))\Big)
		+ \delta &\ge& \sum_j \gamma_m {r_j'}^m.
\eea
By~(\ref{eq:7*22}) and~(\ref{eq:7*21}) for any $m$-plane $\Sigma\supset\{x,x_j'\}$
\begin{eqnarray}\label{eq:7*25}
\lefteqn{\int_0^r \H^{m-1}_\delta(S_0\cap\partial B(x,t)\cap
	B(x_j',r_j')) \d t}\\
	&\le& \int_0^r \H^{m-1}_\delta(\partial B(x,t)\cap
	B(x_j,r_j')) \d t \nonumber\\
	&\le& \int_0^r \sigma(|x-x_j'|,r_j',t)\d t \nonumber\\
	&=& \H^m (\Sigma\cap B(x_j',r_j')\cap B(x,r)) \nonumber\\
	&\le& \H^m(\Sigma\cap B(x_j',r_j')) 
	= \gamma_m {r_j'}^m.\nonumber
\end{eqnarray}

Consider now any $i$ such that $|x_i-x|\ge r_i$.
If we let $R=|x-x_i|-r_i\cos\theta -r_i \epsilon$ we obtain that
$S_0\cap B(x_i,r_i)\cap B(x,R)=\emptyset$ so, for any $m$-plane
$\Sigma\supset\{x,x_i\}$ and
by~(\ref{eq:7*22}) and~(\ref{eq:7*21}), we obtain
\bea
	\lefteqn{\int_0^r \H^{m-1}_\delta(S_0\cap\partial B(x,t)\cap
		B(x_i,r_i)) \d t} \\
	&\le& \int_R^r \H^{m-1}_\delta(\partial B(x,t)\cap B(x_i,r_i))\d t\\
	&\le& \int_R^r \H^{m-1}(\partial B(x,t)\cap B(x_i,r_i)) \d t\\
	&=& \int_R^r \sigma(|x-x_i|,r_i,t)\d t.
\eea
As the set $B(x,R)\cap B(x_i,r_i)$ contains a ball of radius $\frac 1
2 (1-\cos\theta-\epsilon) r_i$ we get
\bea
	\int_0^R \sigma(|x-x_i|,r_i,t)\d t 
	&=& \H^m(\Sigma\cap B(x_i,r_i)\cap B(x,R))\\
	&\ge& \gamma_m\left(\frac{1-\cos\theta-\epsilon}{2}\right)^m
	r_i^m
\eea
and, on the other hand, 
\[
	\int_0^r \sigma(|x-x_i,r_i,t)\d t
	= \H^m(\Sigma\cap B(x_i,r_i)\cap B(x,t))
	\le \gamma_m r_i^m
\]
so that
\bea
	\lefteqn{\int_0^r \H^{m-1}_\delta (S_0\cap\partial B(x,t)\cap
	B(x_i,r_i))\d t}\\
	&\le& \gamma_m r_i^m-\gamma_m\left(\frac{1-\cos\theta
	-\epsilon}{2} \right)^m r_i^m.
\eea

Since the balls $B(x_i,r_i)$ are disjoint, there is at most one
$i=i_0$ such that $|x-x_i|<r_i\le \delta$  
and we obtain, as in~(\ref{eq:7*25}) 
\[
	\int_0^r \H^{m-1}_\delta(S_0\cap\partial B(x,t)\cap
	B(x_{i_0},r_{i_0})) \d t
	\le \gamma_m r_{i_0}^m \le \gamma_m\delta^m.
\]
Thus 
\bea
	\lefteqn{\int_0^r\H^{m-1}_\delta(S_0\cap\partial B(x,t))\d t}\\
	&=& \int_0^r \H^{m-1}_\delta
	\Big(\big(\bigcup_j B(x_j',r_j')\cup\bigcup_i B(x_i,r_i)\big)
\cap S_0 \cap\partial B(x,t)\Big) \d t \\
	&\le& \sum_j\int_0^r\H^{m-1}_\delta(S_0\cap\partial B(x,t)\cap
B(x_j',r_j'))\d t \\
	&&\mbox{} +\sum_i\int_0^r \H^{m-1}_\delta(S_0\cap\partial
B(x,t) \cap B(x_i,r_i))\d t \\
	&\le& \sum_j \gamma_m {r_j'}^m + \sum_i \gamma_m r_i^m\left(
1- \left(\frac{1-\cos \theta - \epsilon}{2}\right)^m\right) +
\gamma_m\delta^m\\
	&\le& \H^m\big(S_0\cap B(x,r)\setminus\bigcup_i(S_0\cap
B(x_i,r_i))\big) + \delta \\
	&&\mbox{}+ \frac{1}{1-\alpha}\H^m\big(\bigcup_i(S_0\cap
B(x_i,r_i))\big) - e \left(\frac{1-\cos\theta-\epsilon}{2}\right)^m 
	+ \gamma_m\delta^m\\
	&\le& \frac{\H^m(S_0\cap B(x,r))}{1-\alpha} -
e\left(\frac{1-\cos\theta -\epsilon}{2}\right)^m
+\delta+\gamma_m\delta^m
\eea
and letting $\delta\to 0$ we conclude the proof.

\end{proof}

\section{A fundamental Lemma}

\begin{lemma}\label{lemma:6*}
Given $\epsilon>0$, there exist positive numbers $R_0=R_0(\omega,\epsilon,n)$,
$\lambda=\lambda(n,\epsilon)$, $\alpha=\alpha(n,\epsilon)$ 
such that for all $0<r_0<R_0$, $x_0\in S_0$ and
$B(x_0,r_0)\subset \Omega$ the following assertion holds: if the inequality 
	\begin{equation}\label{eq:beta}
	1-\alpha
	\le \frac{\psi(x,r)}{\gamma_m r^m}
	\le \frac{\phi(x,r)}{\gamma_m r^m}
	\le 1+\alpha,
	\end{equation}
is true for each ball $B(x,r)\subset B(x_0,r_0)$ with center $x\in
S_0$, then there exist a point $x^*\in S_0\cap
B(x_0,r_0)$ and a hyperplane $\Sigma$ through $x^*$ such that
$B(x^*,\lambda r_0)\subset B(x_0,r_0)$ and
	\[
	S_0\cap B(x^*,\lambda r_0)\subset (\Sigma)_{\epsilon\lambda r_0}.
	\]
\end{lemma}
\begin{proof}
%[lemma~\ref{lemma:5*}]
Suppose $\epsilon <1$ and choose
\begin{equation}\label{eq:defalpha}
	\alpha=\left(\frac{\epsilon^m}{64\cdot 8^m}\right)^6
\end{equation}
and $R_0$ such that $\omega(R_0)\le \gamma_m\alpha$.

\emph{Step one.}
Let $r_1=\lambda_1 r_0$ with $\lambda_1=\frac{4^m}{2(16\sqrt n)^n}$.
We will estimate how many
disjoint balls of radius $2 r_1$
can fit into $B(x_0,r_0/4)$.
First of all note that $B(x_0,r_0/4)$ contains an
$n$-cube $K$ with side 
${r_0\over 2\sqrt n}$ while a ball of radius $2 r_1$ is contained
in an $n$-cube of side $4 r_1$.
So, splitting the cube $K$ into small cubes of side
$4 r_1$, and putting one 
ball into every small cube we find at least
\[
\left\lfloor{{r_0\over \sqrt n}\over 8 r_1}\right\rfloor^n
\ge \left({{r_0\over\sqrt n}\over 16 r_1}\right)^n
= \big(16 \sqrt{n}\frac{r_1}{r_0}\big)^ {-n}
\]
disjoint balls of radius $2 r_1$ that fit into $B(x_0,r_0/4)$. 
Let $\{x_i\}$ be the set of the centers of these balls.
We have determined the radius $r_1$ such that at least one of the
balls $B(x_i,r_1)$ does not intersect $S_0$.
In fact, suppose that for every $i$ there exists $y_i\in B(x_i,r_1)\cap
S_0$. $B(y_i,r_1)$ are disjoint balls all contained
in $B(x_0,r_0/4)$, so we get (using~(\ref{eq:beta}))
\bea
(1+\alpha)\gamma_m \left({r_0\over 4}\right)^m 
&\ge& \H^m(S_0\cap B(x_0,{1\over 4}r_0)) \\
&\ge& \sum_i \H^m(S_0\cap B(y_i,r_1))\\
&\ge& \sum_i (1-\alpha) \gamma_m r_1^m \\
&\ge& \left(16\sqrt{n}{r_1\over
r_0}\right)^{-n} (1-\alpha) \gamma_m r_1^m.
\eea
which is a contradiction being $\alpha<1/3$.

So, we have found a ball $B(y,r_1)\subset
B(x_0,r_0/2)$
which does not intersect $S_0$.
Let
\[
r_2=\max \{r\in\R\quad\: \quad B(y,r)\cap S_0 =\emptyset\}
\]
so that $B(y,r_2)\cap S_0=\emptyset$ while there exists a point 
$x^*\in S_0\cap\partial B(y,r_2)$. 
Since $x_0\in S_0$ we have $r_2\le |x_0-y| \le r_0/4$ so
$B(y,r_2)\subset B(x_0,r_0/2)$. 

\emph{Step two.}
Define
\begin{eqnarray*}
	r_3&=&\alpha^{\frac 1 3} r_2 \\
	r_4&=&\alpha ^{\frac 1 6} r_3\\
	  d&=& \alpha^{\frac 2 3} r_3\\ 
	\theta &=& \arccos (\alpha^{\frac 1 6}).
\end{eqnarray*}
\begin{figure*}[t]
\center{%
%\input{fig5}
\psset{xunit=0.5, yunit=0.5, runit=0.5}
\pspicture*(-9,-6)(9,9)


\SpecialCoor

\pscircle(0,0){5}
\psdots(0,0)
\uput[ur](0,0){$x^*$}
\psline[linewidth=0.5pt]{->}(0,0)(5;120)
\uput[dl](2.5;120){$r$}

\pscircle(0,8){8}
\psdots(0,8)
\uput[r](0,8){$y$}
\psline[linewidth=0.5pt]{->}(0,8)(-8,8)
\uput[d](-4,8){$r_2$}

\psline[linewidth=0.5pt](0,8)(0,-7)

\psdots(5;18)
\uput[dr](5;18){$p$}

\psline[linewidth=2pt](-10,0)(10,0)
\uput[d](-7,0){$\Sigma$}
\rput(7,1.2){$X$}
\rput(7,-1.2){$Y$}

\psdots(0,-1.5)
\uput[dr](0,-1.5){$z$}

\psline(0,0)(10;200)
\psline(0,0)(10;-20)
\psarc{<-}(0,0){1}{200}{270}
\uput[dl](0.2;235){$_\theta$}
\rput(6;235){$C_\theta$}

\psdots(5;-20)
\uput[ur](5;-20){$q$}

\pscurve[linewidth=2pt](-10,1)(-7,1)(5;175)(-0.2,0)(0,0)(0.3,-0.1)(2,-3)(5;-25)(7,-2)(10,-1)
\uput[u](-7,1){$S_0$}

\psline(0,-1.5)(5;-25)
\uput[d](5.5;-25){$l_\theta(r)$}

\psline(0,-1.5)(5;175)
\uput[ul](5;175){$l(r)$}

\endpspicture
%
}
\caption{The surface $S_0$, the plane $\Sigma$ and the Cone $C_\theta$.}
\label{fig:Ctheta}
\end{figure*}

Let $\Sigma$ be the $m$-dimensional plane tangent to $B(y,r_2)$ in
$x^*$ (see Figure~\ref{fig:Ctheta}), consider the cone $C_\theta$
with vertex $x^*$ and amplitude $\theta$ defined as follows
\[
	C_\theta=\{v\in\R^n\: (v-x^*)\cdot(x^*-y)\ge |v-x^*||x^*-y| \cos\theta\}
\]
and take a point $z\in C_\theta$ lying on the line $x^*y$ and at a
distance $d$ from $x^*$.
For a radius $r\in[r_4,r_3]$ define
\bea
	l(r)&=&S_0\cap \partial B(x^*,r);\\
	l_\theta(r)&=& S_0\cap\partial B(x^*,r)\cap C_\theta; \\
	d_1(r)&=& \sup\{|z-x|\: x\in l(r)\setminus l_\theta(r)\};\\
	d_2(r)&=& \sup\{|z-x|\: x\in l_\theta(r)\}.
\eea
For almost every $r\in[r_4,r_3]$, in view of Lemma~\ref{lemma:cono}
(with $A=\R^n\setminus C_\theta$),
we can find a Caccioppoli set $E$  such that
\[
	P(E,\overline{B(x^*,r)})
	\le \frac{d_1(r)}{m} \H^{m-1}(l(r)\setminus l_\theta(r))
	+ \frac{d_2(r)}{m} \H^{m-1}(l_\theta(r))
\]
and from Lemma~\ref{lemma:omegamin} we obtain
\bea
	\lefteqn {m\H^m(S_0\cap B(x^*,r))=mP(E_0,B(x^*,r))}\\
	&\le& m P(E,B(x^*,r)) + m r^m \omega(r) \\
	&\le& d_1(r)\H^{m-1}(l(r)\setminus l_\theta(r))
		+d_2(r)\H^{m-1}(l_\theta(r)) + m r^m \omega(r).
\eea
Given any point $p\in\partial B(y,r_2)\cap \partial B(x^*,r)$ and
$q\in \partial C_\theta \cap \partial B(x^*,r)$, with simple geometric
considerations we deduce
\bea
	d_1(r) &\le& |p-z| = \sqrt{r^2+d^2+d\frac{r^2}{r_2}}
	\le r\sqrt{1+\frac{d^2}{r_4^2}+\frac{d}{r_2}} \\
	&=& r\sqrt{1+2\alpha}
	\le r(1+\alpha),
		\\
	d_2(r) &\le& |q-z| = \sqrt{r^2-2dr\cos\theta+d^2} 
	\le r \sqrt{1-2\frac{d}{r_3}\cos\theta+\frac{d^2}{r_4^2}} \\
	&=& r\sqrt{1-2\alpha^{\frac 5 6}+\alpha}
	\le r\sqrt{1-\alpha^{\frac 5 6}}
	\le r(1-\frac 1 2 \alpha^{\frac 5 6}),
\eea
so that we obtain
\bea
	\lefteqn{m\H^m(S_0\cap B(x^*,r))- m r^m \omega(r)} \\
	&\le& r (1+\alpha)\H^{m-1}(l(r)\setminus l_\theta(r))
	+ r (1-\frac 1 2\alpha^{\frac 5 6})\H^{m-1}(l_\theta(r)) \\
	&=& r(1+\alpha)\H^{m-1}(l(r)) 
    	-r(1+\alpha-1+\frac{1}{2}\alpha^{\frac 5 6})\H^{m-1}(l_\theta(r))\\
	&\le& r(1+\alpha)\H^{m-1}(l(r))
	-\frac{1}{2}r\alpha^{\frac 5 6}\H^{m-1}(l_\theta(r)).
\eea
Dividing by $r$, integrating and using hypothesis~(\ref{eq:beta}) we obtain
\bea
	\lefteqn{\frac 1 2 \alpha^{\frac 5
	6}\int_{r_4}^{r_3}\H^{m-1}(l_\theta(r))\d r} \\
	&\le& (1+\alpha)\int_{r_4}^{r_3}\H^{m-1}(l(r))\d r\\
	&&\mbox{}- \int_{r_4}^{r_3} \frac m r \Big(\H^m(S_0\cap
	B(x^*,r))-r^m\omega(r)\Big) \d r \\
	&\le& (1+\alpha)\Big((1+\alpha)\gamma_m r_3^m -
	(1-\alpha)\gamma_m r_4^m\Big)\\
	&&\mbox{} - (1-\alpha - \frac
	{\omega(R_0)}{\gamma_m}) \gamma_m(r_3^m - r_4^m) \\
	&=& (1+\alpha)(1+\alpha-\alpha^{\frac m 6} + \alpha
	^{1+\frac m 6})\gamma_m r_3^m \\
	&&\mbox{}-(1-\alpha-\frac{\omega(R_0)}{\gamma_m}-\alpha^{\frac m 6}
	+ \alpha^{1+\frac m 6} + \alpha ^{\frac m
	6}\frac{\omega(R_0)}{\gamma_m}) \gamma_m r_3^m \\
	&=& (3\alpha + \alpha^2+\alpha^{2+\frac m 6}
	+\frac{\omega(R_0)}{\gamma_m}(1-\alpha^{\frac m 6})
	-\alpha^{1+\frac m 6})\gamma_m r_3^m 
	\le 6 \alpha \gamma_m r_3^m
\eea
that is
\[
	\int_{r_4}^{r_3}\H^{m-1}(l_\theta(r))\d r
	\le 12 \alpha^{\frac{1}{6}}\gamma_m r_3^m.
\]
So, by Lemma~\ref{lemma:coarea} and~(\ref{eq:beta})
\begin{eqnarray} \label{eq:ltheta}
	\lefteqn{\H^m(S_0\cap B(x^*,r_3)\cap C_\theta)}\\
	&=& \H^m(S_0\cap B(x^*,r_3)) - \H^m((S_0\cap
	B(x^*,r_3))\setminus C_\theta) \nonumber \\
	&\le& (1+\alpha)\gamma_m r_3^m
	-\int_0^{r_3}\H^{m-1}(l(r)\setminus l_\theta(r))\d r\nonumber \\
	&=& (1+\alpha)\gamma_m r_3^m 
	 -\int_0^{r_3} \H^{m-1}(l(r)) \d r \nonumber \\
	&&\mbox{}+\int_{r_4}^{r_3} \H^{m-1}(l_\theta(r))\d r
	+ \int_0^{r_4} \H^{m-1}(l_\theta(r))\d r \nonumber\\
	&\le& (1+\alpha)\gamma_m r_3^m- (1-\alpha)\gamma_m r_3^m
	  + 12\alpha^{\frac 1 6}\gamma_m r_3^m + \alpha^{\frac m
	6}(1+\alpha)\gamma_m r_3^m \nonumber\\ 
	&\le& 16 \alpha^{\frac 1 6}\gamma_m r_3^m. \nonumber
\end{eqnarray}
\emph{Step three.}
Now we claim that
every point $w\in S_0\cap B(x^*,r_3/2)$ is not further than
${1\over 4}\epsilon r_3$ from $\Sigma$.
First of all note that $\Sigma$ divides the space $\R^n$
into two halfspaces $X$ and $Y$, one of which (suppose $X$) contains
the ball $B(y,r_2)$ while the other ($Y$)
contains the cone $C_\theta$.

Suppose $w\in X$.
Since the ball
$B(y,r_2)$ doesn't intersect $S_0$, if we consider a point
$p\in\partial B(x^*,r_3/2)\cap\partial B(y,r_2)$, with simple
geometrical considerations we obtain
\[
	d(w,\Sigma)\le d(p,\Sigma)= {\frac 1 2}\frac{{(\frac {r_3} {2}})^2}{r_2}
	= {1\over 8}\alpha^{\frac 1 3} r_3 \le {1\over 4}\epsilon r_3.
\]

Consider now $w\in Y$ and suppose 
$d(w,\Sigma)>{1\over 4}\epsilon r_3$.
Choose $q\in \partial B(x^*,r_3/2)\cap\partial C_\theta$
and denote by $\Sigma'$ the hyperplane parallel to $\Sigma$ 
through $q$. Again with a geometrical argument we obtain
\bea
	d(w,\partial C_\theta)&\ge& d(w,\Sigma') \ge {1\over 4}\epsilon
	r_3-{1\over 2} r_3 \cos\theta \\
	&=& ({1\over 4}\epsilon - {1\over 2}\cos\theta)
	r_3
\eea
so that
\[
B\big(w,\big({1\over 4}\epsilon-{1\over
2}\cos\theta\big)r_3\big)\subset B(x^*,r_3)\cap C_\theta.
\]
From inequality~(\ref{eq:ltheta}) we obtain
(using also~(\ref{eq:beta}))
\bea
\lefteqn{\gamma_m(1-\alpha) r_3 ^m\big({1\over 4}\epsilon-{1\over 2}\cos\theta\big)^m}\\
&\le&
\H^m\left(S_0\cap B\big(w,({1\over
4}\epsilon-{1\over 2} \cos\theta)r_3\big)\right) \\
&\le&  \H^m(S_0\cap B(x^*,r_3)\cap C_\theta) \\
&\le& 16\alpha^{\frac 1 6}\gamma_m r_3^m
\eea
so that (recall~(\ref{eq:defalpha}))
\[
32\alpha^{\frac 1 6} = \frac{\epsilon^m}{2\cdot 8^m}\le 
(1-\alpha)\big({\frac 1 4}\epsilon-{\frac 1 2}\alpha^{\frac 1 6}\big)^m
 \le 16 \alpha^{\frac 1 6} 
\]
which is a contradiction.

Thus, we've proved that $S_0\cap B(x^*,r_3/2)$ is within a distance of
${1\over 4}\epsilon r_3$ from $\Sigma$ that is the lemma for
$\lambda={1\over 2}\lambda_1\alpha^{\frac 1 3}$.
\end{proof}

\section{Main theorem}

In view of Theorem~\ref{teo:Reif} to
prove Theorem~\ref{teo} we only need the following

\begin{theorem}\label{teo:valeReif}
If $x_0\in S_0=\partial E_0$ and $\epsilon_0>0$ then there exist $R>0$,
$\epsilon'<\epsilon_0$ such that $(\epsilon',R,n-1)$-Reifenberg property
holds in $x_0$ for the set $S_0$.
\end{theorem}

\begin{proof}
Recall that $m=n-1$ and choose positive numbers $\epsilon'$, $\epsilon$,
$\lambda$, $\alpha$ and $r_0$ such that the following hypotheses hold:
\[
\begin{array}{ccc}
\begin{array}{rcl}
	\epsilon' &<& \frac{1}{32\cdot 2^m} \\
	\epsilon' &<& \epsilon_0 \\
	\epsilon  &=& \left(\frac{\epsilon'}{320(m+1)(m+2)}\right)^2\\
	\lambda   &=& \lambda(n,\epsilon) \\
	\alpha    &<& \alpha(n,\epsilon) \\
	\alpha    &<& \frac{1}{12(m+3)}
			\left(\frac{\lambda \epsilon}{4}\right)^m 
\end{array}&\hspace{1cm}&
\begin{array}{rcl}
	\alpha    &<& \frac{1}{5} \\
	r_0	  &<& r_0(\alpha/2) \\
	\omega(r_0)&<&\gamma_m\frac{\alpha}{2} \\
	\omega(r_0)&<&\frac{1}{3 c_n} \\
	r_0	  &<& R_0(\omega,\epsilon,n)
\end{array}
\end{array}
\]
where $\lambda(n,\epsilon)$, $\alpha(n,\epsilon)$ and
$R_0(\omega,\epsilon,n)$ are the constants given by
Lemma~\ref{lemma:6*}, $c_n$ is the constant given by
Lemma~\ref{lemma:omegamul} and $r_0(\alpha/2)$ is the constant given in 
Lemma~\ref{lemma:spirale}.

By Lemma~\ref{lemma:spirale} 
for any ball $B(x,r)$ $\subset B(x_0,r_0)$ with center $x\in S_0$ we have
\[
1-{\alpha\over 2}
\le {\phi(x,r)\over \gamma_m r^m}
\le 1+{\alpha\over 2}.
\]
On the other hand, by Lemma~\ref{lemma:1*} 
\bea
\psi(x,r) 
	&=& \int_0^r\H^{m-1}(S_0\cap\partial B(x,t))\d t\\
	&\ge& \int_0^r {m\over t}\H^m(S_0\cap B(x,t)) - m \omega(t) t^{m-1}\d
	t \\
	&\ge& (1-{\alpha\over 2})\gamma_m r^m - \omega(r_0)r^m \\
	&=& \left(1-\frac{\alpha}{2} -{\omega(r_0)\over \gamma_m}\right) \gamma_m r^m
\eea
and hence the following holds:
\begin{equation}\label{eq:*53}
1-\alpha
	\le {\psi(x,r)\over \gamma_m r^m}
	\le {\phi(x,r)\over \gamma_m r^m}
	\le 1+\alpha.
\end{equation}
By Lemma~\ref{lemma:6*}, given any ball
$B(x,r)$ $\subset B(x_0,r_0)$ with center $x\in S_0$ there exist
a point $x^*\in S_0$ and an $m$-dimensional plane
$\Sigma$ through $x^*$ such that 
\bea
S_0\cap B(x^*,\lambda r) &\subset& (\Sigma)_{\epsilon\lambda r}\\
B(x^*,\lambda r)&\subset& B(x,r).
\eea

Now take any ball $B(x',r')\subset B(x_0,r_0)$ with $x'\in S_0$ and consider
$m+2$ points $y_1,\ldots,$ $y_{m+2}$ contained in $S_0\cap
B(x',r'/4)$.
By well known convering theorems, given $\delta>0$ we can find a
finite family $\{B(x_{i\delta},\rho_{i\delta})\}_{i\in I_\delta}$ of disjoint balls 
with center on $S_0$, contained in $B(x',r'/4)$ and with
radius $\rho_{i\delta}< \delta$, such that 
\[
\sum_{i\in I_\delta} \H^m(S_0\cap B(x_{i\delta},\rho_{i\delta}))\ge {1\over
2}\H^m(S_0\cap B(x',r'/8)).
\]
Thus, by~(\ref{eq:*53})
\begin{equation}\label{eq:*57}
(1+\alpha)\sum_i \gamma_m\rho_{i\delta}^m 
\ge {1\over 2}\H^m(S_0\cap B(x',r'/8))
\ge {(1-\alpha)\gamma_m\over 2}\left({r'\over 8}\right)^m.
\end{equation}
Obviously for each $j\le m+2$ and for each $i\in I_\delta$ 
\[
B(x_{i\delta},\rho_{i\delta})\subset B(x',r'/4) \subset
B(y_j,r'/2)\subset B(x',r') 
\]
so by Lemma~\ref{lemma:6*} we can find points
$x^*_{i\delta}\in S_0$ and $m$-dimensional planes $\Sigma_{i\delta}$
such that
\bea
S_0\cap B(x^*_{i\delta},\lambda\rho_{i\delta})
&\subset&\left(\Sigma_{i\delta}\right)_{\epsilon\lambda\rho_{i\delta}}\\
B(x^*_{i\delta},\lambda\rho_{i\delta})&\subset& B(x_{i\delta},\rho_{i\delta}).
\eea
Choose $\theta$ so that $1-\cos\theta=3\epsilon$, and let
$I_{j\delta}$ be the set of indices $i$ such that the segment
$[x^*_{i\delta},y_j]$ makes an angle greater than $\theta$ with
$\Sigma_{i\delta}$.
Let
\[
e={\gamma_m\lambda^m\over 4(m+3)(1-\alpha)}\left({r'\over 8}\right)^m.
\]
If for fixed $j$ there were arbitrarily small $\delta$ such that
\[
\sum_{i\in I_{j\delta}}\gamma_m \rho_{i\delta}^m \lambda^m 
> e
\]
by Lemma~\ref{lemma:7*} we had
\[
\psi(y_j,r'/2)\le \frac {\phi(y_j,r'/2)}{1-\alpha}- e\left({1-\cos\theta
-\epsilon\over 2}\right)^m
\]
hence by~(\ref{eq:*53})
\bea
(1-\alpha)\gamma_m \left({r'\over2}\right)^m
&\le& \psi(y_j,r'/2)
\le \frac{\phi(y_j,r'/2)}{1-\alpha}-e\epsilon^m \\
&\le& \frac{1+\alpha}{1-\alpha}\left({r'\over 2}\right)^m\gamma_m - e\epsilon^m
\eea
that is
\[
12\alpha (m+3)
\ge \left({\lambda\epsilon\over 4}\right)^m.
\]
But by hypotheses on $\alpha$ the previous inequality is false
so we deduce that for a sufficiently small $\delta$ (using also~(\ref{eq:*57}))
\bea
\sum_{i\in I_{j\delta}}\gamma_m\rho_{i\delta}^m\lambda^m 
&\le& {\gamma_m\lambda^m\over 4(m+3)(1-\alpha)}\left({r'\over 8}\right)^m\\
&\le& \frac{\lambda^m (1+\alpha)}{2(1-\alpha)^2(m+3)}\sum_{i\in
I_\delta}\gamma_m \rho_{i\delta}^m
\eea
so that (since $\alpha<\frac{1}{5}$)
\[
	\sum_{i\in I_{j\delta}}\rho_{i\delta}^m 
	\le \frac{1}{m+3}\sum_{i\in I_\delta}\rho_{i\delta}^m
\]
and
\[
	\sum_{j=1}^{m+2}\sum_{i\in I_{j\delta}}\rho_{i\delta}^m
	< \sum_{i\in I_\delta}\rho_{i\delta}^m.
\]

Therefore there exists an index $i_0\in I_\delta$ contained in none 
of the sets $I_{j\delta}$.
So we have proved that given any $m+2$ points $y_j$ of $S_0\cap
B(x',r'/4)$ it is possible to find an $m$-dimensional plane
$\Sigma=\Sigma_{i_0\delta}$ such that every $y_j$
lies within a distance of ${1\over 2}r'\sin\theta$ from $\Sigma$.
Being $1-\cos\theta=3\epsilon$ we have
\bea
{1\over 2}r'\sin\theta
&=& {1\over 2}r'\sqrt{1-\cos^2\theta}
= {1\over 2}r'\sqrt{3\epsilon (1+\cos\theta)}\\
&<& {5\over 4}r'\sqrt{\epsilon}.
\eea

Let's consider all $m$-simplexes\footnote
{Here an $m$-simplex is meant to be the convex envelope of $m+1$ points,
which are the vertices of the simplex. The $m$-area of the simplex
is its $m$-dimensional measure.}
with each vertex in $S_0\cap B(x',r'/4)$.
Let then $y_1,\ldots,y_{m+1}$ be the vertices of the $m$-simplex of greatest
$m$-area among all the simplexes considered, and call $a$ its area.
Let $y_{m+2}$ be any other point of $S_0\cap B(x',r'/4)$ and let $\Sigma$
be the $m$-dimensional plane, corresponding to the points $y_1,\ldots,y_{m+2}$
as we have seen before.
If we denote by $\Delta$ the $(m+1)$-simplex with  vertices
$y_1,\ldots,y_{m+2}$, we find that
the $m$-area of the faces of $\Delta$ could be at most $a$,
so the area of the projection of $\Delta$ onto $\Sigma$ cannot be
greater than $(m+2)a$ and the $(m+1)$-area of $\Delta$ would be less
than ${5\over 2}(m+2) a r' \sqrt{\epsilon}$.
Thus $y_{m+2}$ is within a distance of 
${5\over 2}(m+1)(m+2)r'\sqrt{\epsilon}= \epsilon' r'/128$ 
from the $m$-dimensional plane
through $y_1,\ldots,y_{m+1}$. 
If $\Sigma'$ is the $m$-dimensional plane through $x'$ parallel to
the plane through $y_1,\ldots,y_{m+1}$, we know that
every point of  $S_0\cap B(x',r'/4)$ is whithin a distance of $\epsilon'
r'/64$ from $\Sigma'$.

So we have proved that given any  $B(x',r')\subset B(x_0,r_0)$ with
$x'\in S_0$ it is
possible to find an $m$-dimensional plane $\Sigma'$ through $x'$ such that
\begin{equation}\label{eq:*60}
	S_0\cap B\left(x',{r'\over 4}\right)
\subset \left(\Sigma'\right)_{\epsilon' {r'\over 64}}.
\end{equation}

Since
\[
\int_{{1\over 8}r'}^{{1\over 4}r'}\H^{m-1}(S_0\cap\partial B(x',t))\d t
\le \psi(x',r'/4)
\le (1+\alpha)\left({r'\over 4}\right)^m\gamma_m
\]
we are able to find a set of positive measure of radii
$r$ with ${1\over 8}r'<r<{1\over 4}r'$ such that
\[
\H^{m-1}(S_0\cap \partial B(x',r)) 
\le (1+\alpha)\gamma_m{\left({r'\over 4}\right)^m \over {1\over 8} r'}
\]
and by~(\ref{eq:*53})
\bea
\H^m(S_0\cap B(x',r))
&\ge&(1-\alpha) \gamma_m r^m
= \gamma_m \left({r'\over 4}\right)^m 4^m (1-\alpha)
	\left({r\over r'}\right)^m\\
&\ge& \H^{m-1}(S_0\cap\partial B(x',r)) 
	{1\over 4\cdot 2^m} r. 
\eea
So, by the hypotheses on $\epsilon'$ and $r_0$ we have
\[
\H^m(S_0\cap B(x',r))
> 4\epsilon' r {1+c_n\omega(r)\over 1-c_n\omega(r)}\H^{m-1}(S_0\cap\partial B(x',r)).
\]
If now we let $E=E_0\setminus B(x',r)$ by Lemma~\ref{lemma:omegamul}
there exist $r$ such that
\bea
P(E,\overline{B(x',r)}) 
&\ge& {1-c_n \omega(r)\over 1+c_n \omega(r)} P(E_0,B(x',r)) \\
&=& {1-c_n \omega(r)\over 1+c_n \omega(r)}\H^m(S_0\cap B(x',r)) \\
&>& 4\epsilon' r \H^{m-1}(S_0\cap\partial B(x',r))
\eea
and then, by Lemma~\ref{lemma:schiacciata},
we get that the projection of  $S_0\cap B(x',r)$ on $\Sigma'$ must contain
$\Sigma'\cap B(x',(1-{1\over 4}\epsilon')r)$.

We claim that
\begin{equation}\label{eq:*64}
\Sigma'\cap B\left(x',{1\over 16}r'\right)
\subset \left(S_0\cap B\left(x',{1\over 16}r'\right)\right)_{\epsilon' r'\over 32}.
\end{equation}
Given any point $p\in \Sigma'\cap B(x',{1\over 16}r')$ let
$p'\in\Sigma'\cap B(x',{1\over 16}r' - {\epsilon' r'\over 64})$ be a
point with minimal distance from $p$. 
Obviously $|p-p'|\le {\epsilon' r'\over 64}$.
As we have seen previously, being
$(1-{1\over 4}\epsilon') r> {r'\over 16}$, we know that
there exists a point $p''\in S_0\cap B(x',r)$ whose projection is $p'$. 
Equation~(\ref{eq:*60}) gives $|p'-p''|\le {\epsilon' r'\over
64}$ so that $p''\in B(x',{r'\over 16})$.
Thus we've found a point $p''\in S_0\cap B(x',{r'\over 16})$ whose
distance from $p$ is less than ${\epsilon' r'\over 32}$ 
as desired.

By~(\ref{eq:*60}) and~(\ref{eq:*64}) we have proved that the 
$(\epsilon',R,m)$-Reifen\-berg property holds in $x_0$ if we let $R={r_0\over 16}$.
\end{proof}

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\end{document}
