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\title{On Besicovitch's $\frac12$-problem [after Preiss and Ti\v{s}er]}
\author{Pierre Pansu, Universit\'e Paris-Saclay}
\date{November 5th, 2020}


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\titlepage
\end{frame}

\section{Density and rectifiability}

\subsection{The problem}

\begin{frame}

$M$ metric space, $E\subset M$. The lower density at $x$ is
$$
\underline{D}(E,x)=\liminf_{r\to 0} \frac{\mathrm{H}^1(E\cap B(x,r))}{2r}.
$$

\begin{thm}[Besicovitch 1938]
If $M=\R^2$, any subset $E$ of finite $\mathcal{H}^1$-measure such that $\underline{D}(E,x)>\frac{3}{4}$ at $\mathcal{H}^1$-almost every point of $E$ is rectifiable.
\end{thm}

Besicovitch gives examples showing that $\frac{3}{4}$ cannot be improved beyond $\frac{1}{2}$ and conjectures that the optimal bound is $\frac{1}{2}$.\pause

\begin{thm}[Preiss, Ti\v{s}er 1992]
For arbitrary metric spaces $M$, any subset $E\subset M$ of finite $\mathcal{H}^1$-measure such that $\underline{D}(E,x)>\frac{2+\sqrt{46}}{12}$ at $\mathcal{H}^1$-almost every point of $E$ is rectifiable.
\end{thm}

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\subsection{The method}

\begin{frame}

\textbf{Goal}: construct a \emph{continuum} (compact connected subset) in $E$. \pause Indeed (Eilenberg-Harrold 1943), continua of finite $\mathcal{H}^1$-measure are rectifiable.\pause

\medskip

By contradiction, if construction fails, one encounters a pair of disjoint compact sets $E_1$ and $E_2$ of rather high density. Then one uses the \emph{Besicovitch pair condition} to get a contradiction.\pause

\begin{dfi}
Say that $M$ satisfies the \emph{Besicovitch pair condition} with parameter $\sigma\in(0,1)$ if whenever $\mu$ is a measure on $M$ satisfying $\mu(S)\leq\mathrm{diam}(S)$ for every subset $S$, then there exists $\tau>0$ such that $\forall\lambda>0$, $\exists \delta>0$ such that for every pair of Borel subsets $E_1$ and $E_2$ for which \begin{itemize}
  \item $0<dist(E_1,E_2)<\delta$, and 
  \item $\mu(B(x,s))>2\sigma s$ for every $x\in E_1\cup E_2$ and every $0<s<\lambda$,
\end{itemize}
there exists a subset $U\subset M$ intersecting both $E_1$ and $E_2$ and such that 
$$
\mu(U\setminus(E_1\cup E_2))>\tau\,\mathrm{diam}(U).
$$
\end{dfi}\pause



The proof of the theorem splits into two steps :
\begin{enumerate}
\item Every metric space satisfies BPC$(\sigma)$ for $\sigma=\frac{2+\sqrt{46}}{12}$.
  \item BPC$(\sigma)\implies$ the theorem with bound $\sigma$.
\end{enumerate}


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\section{Proof of BPC}

\begin{frame}

\textbf{Proof of BPC}

\medskip

Let $\mu$ be a measure on $M$ satisfying $\mu(S)\leq\mathrm{diam}(S)$ for every subset $S$. Fix $\lambda>0$. Let $E_1$ and $E_2$ be subsets for which 
\begin{itemize}
  \item $0<dist(E_1,E_2)<\delta:=\lambda$, and 
  \item $\mu(B(x,s))>2\sigma s$ for every $x\in E_1\cup E_2$ and every $0<s<\lambda$,
\end{itemize}\pause

\smallskip

Pick $x\in E_1$, $y\in E_2$ with $d(x,y)=d(E_1,E_2):=r$. Then (up two switching $x$ and $y$),
\begin{align}\label{1}
\forall t\geq 0, \, \mu(B(x,t)\cap E_1)\leq t+\frac{1}{2}r.
\end{align}
\pause

For BPC$(\sigma)$, it suffices to set $U=B(x,2r)$ and prove that $\mu(B(x,2r)\setminus(E_1\cup E_2)) > 4\tau\,r$ for a $\tau>0$ that depends only on $\sigma$.\pause

\smallskip

Assume instead that $\mu(B(x,2r)\setminus(E_1\cup E_2)) \leq 4\tau\,r$. 
\begin{align*}
d:=\mathrm{diam}(B(x,r)\cap E_1)
&=\mu(B(x,r)\cap E_1)\\
&\geq \mu(B(x,r))-\mu(B(x,2r)\setminus(E_1\cup E_2))\\
&>2\sigma r-4\tau r.
\end{align*}\pause
Thus one can pack inside $B(x,2r)$ two disjoint balls of not too small radii centered at points of $E_1$. Their measures are not too small, this contradicts (\ref{1}) if $\sigma\geq\frac{2+\sqrt{46}}{12}$.

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\section{Proof of Theorem}

\subsection{Besicovitch's upper bound on density}

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When the BPC will be used, the measure $\mu$ will be a constant multiple of $\mathcal{H}^1\lefthalfcup A$ for a suitable $A$. The fact that $\mu(S)\leq \mathrm{diam}(S)$ for all $S\subset M$ comes from the

\begin{pro}[Besicovitch]
Let $M$ be a metric space with $\mathcal{H}^1(M)<\infty$. At $\mathcal{H}^1$-almost all points $x\in M$,
$$
\limsup_{\mathrm{diam}(S)\to 0}\frac{\mathcal{H}^1(S)}{\mathrm{diam}(S)}\leq 1.
$$
\end{pro}\pause

For some $\tilde\sigma>\sigma$, this allows to pick compact sets $B\subset A\subset M$ of positive $\mathcal{H}^1$ measure such that $\mathcal{H}^1(A\setminus B)<\frac{\tau}{15}\,\mathcal{H}^1(B)$ (where $\tau$ is the constant in BPC$(\sigma)$) and
\begin{align*}
S\cap A\not=\emptyset,\,\mathrm{diam}(S)\text{ small} &\implies \mathcal{H}^1(S)\leq \frac{\tilde\sigma}{\sigma}\,\mathrm{diam}(S),\\
x\in B,\,s\text{ small} &\implies \mathcal{H}^1(B(x,s)\cap A)\geq 2\tilde\sigma s,
\end{align*}
and set $\mu(S)=\frac{\sigma}{\tilde\sigma}\mathcal{H}^1(S\cap A)$.

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\subsection{Filling holes}

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Embed $A$ isometrically in a separable Banach space $X$ ($X=C^0(A)$).\pause

\medskip

\textbf{Strategy}. Add to $B$ the convex hulls of its intersections with a suitable collection of convex sets. Show that the result is connected (or at least contains a large connected subset). Replace convex hulls with line segments. The result is a continuum of finite $\mathcal{H}^1$ measure, hence rectifiable. Then remove (slightly larger) convex sets and show that the remainder, which is contained in $M$, has positive $\mathcal{H}^1$ measure.\pause

\bigskip

\textbf{Notation}. $W^p=\{x\in X\,;\, d(x,W)<p\,\mathrm{diam}(W)\}$.\pause

\medskip

Among the family $\mathcal{V}$ of convex subsets $W$ of $X$ intersecting $B$ and for which $\mu(W\setminus B)>\tau\,\mathrm{diam}(W)$, pick a disjointed sequence $W_i$ such that
$$
\forall W\in\mathcal{V},~\exists i,~W\cap W_i\not=\emptyset \text{ and }\mathrm{diam}(W)<2\,\mathrm{diam}(W_i).
$$\pause

Then $\sum_i \mathrm{diam}(W_i)<\frac{1}{\tau}\sum_{i}\mu(W_i\setminus B)\leq \frac{1}{\tau}\mu(A\setminus B)<\frac{1}{15}\mu(B)$. \pause This implies that
$$
\mu(\bigcup_{i}W^7_i)\leq \sum_i \mathrm{diam}(W^7_i)<\mu(B).$$\pause 
Hence there exists a point $z$ in $B\setminus\bigcup_i W^7_i$ where the density of $A\setminus B$ vanishes. In particular, for some small $s$, $\mu(B(z,2s)\setminus B)<\frac{\sigma\tau}{14}s$.

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\subsection{Applying BPC}

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Let $Q=B\cup\bigcup_i conv(B\cap W^2_i)$. This is a compact set. One uses BPC$(\sigma)$ to show that the connected component $C$ of $Q$ containing $z$ has diameter $>\frac{\sigma s}{2}$.\pause

\bigskip

Otherwise, there is a clopen $H$ of $Q$ containing $z$ and contained in $B(z,\frac{\sigma s}{2})$. Then BPC$(\sigma)$ applies to $E_1=B\cap H$ and $E_2=B\setminus H$, yielding a set $U$ intersecting both $E_1$ and $E_2$ and such that
$$
\mu(U\setminus B)=\mu(U\setminus(E_1\cup E_2))>\tau\,\mathrm{diam}(U).\pause
$$
Then $W=conv(U)\in\mathcal{V}$. Thus $\exists i$, $W\cap W_i\not=\emptyset$ and $\mathrm{diam}(W)<2\,\mathrm{diam}(W_i)$, so $W\subset W^2_i$.\pause

\medskip

But inside $W^2_i$, $Q$ is convex, so its disconnectedness cannot happen there, contradiction.

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\subsection{Removing additions}

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In the sequel, one sticks to those $W_i$'s such that $W^3_i\cap C\not=\emptyset$. Then 
$$
\sum_i \mathrm{diam}(W^3_i)\leq \frac{7}{\tau}\sum_i \mu(W_i\setminus B)\leq \frac{7}{\tau}\mu(B(z,2s)\setminus B)<\frac{7}{\tau}\frac{\sigma\tau s}{14}\leq\mathrm{diam}(C).\pause
$$
Replace each $W^3_i\cap C$ with a line segment. This produces a continuum $D$ satisfying 
\begin{itemize}
  \item $\mathrm{diam}(D)=\mathrm{diam}(C)$,
  \item $\mathcal{H}^1(D\cap \bigcup_i W^3_i)<\mathrm{diam}(C)$.
\item $K:=D\setminus\bigcup_i W^3_i \subset C$.
\end{itemize}\pause

\medskip
Then
$$
\mathcal{H}^1(K)\geq \mathrm{diam}(C)-\mathcal{H}^1(D\cap \bigcup_i W^3_i)>0.
$$ \pause

\medskip

By Eilenberg-Harrold's theorem, $D$ is rectifiable, thus so is $K$. 
\smallskip

Thus $M$ contains a rectifiable subset of positive $\mathcal{H}^1$ measure. \pause

\bigskip

Remove from $M$ the union of a $\mathcal{H}^1$ measure maximizing sequence of rectifiable subsets. Apply previous result to remainder, conclude that it has measure $0$, so $M$ is rectifiable.

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\end{document}




