% December 4, 2001; version accepted to RAEx

\documentclass{rae}

\usepackage{amsmath}\usepackage{amssymb}
\MathReviews{26A21,28A05}

\keywords{%Key words and phrases:
sets, algebraic sum, measurability.}

\date{}

\markboth{K. Ciesielski, H. Fejzi\'c, C. Freiling} {Measurable
$A$ with non-measurable $A+A$}


\newcommand{\real}{{\mathbb R}}
\newcommand{\rational}{{\mathbb Q}}
\newcommand{\mathZ}{{\mathbb Z}}
\newcommand{\mathN}{{\mathbb N}}
\newcommand{\U}{{\mathcal U}}
\newcommand{\E}{{\mathcal E}}
\newcommand{\K}{{\mathcal K}}
\renewcommand{\P}{{\mathcal P}}

\newcommand{\cl}{\operatorname{cl}}
\newcommand{\bd}{\operatorname{bd}}
\newcommand{\lin}{{{\rm LIN}_{\rational}}}

\newcommand{\la}{{\langle}}
\newcommand{\ra}{{\rangle}}
\newcommand{\e}{{\varepsilon}}

\newcommand{\chKC}[1]{\marginpar{{\tiny KC: #1}}}



\def\continuum{{\mathfrak c}}

\def\proof{\noindent {\sc Proof. }}
\def\qed{\hfill\vrule height6pt width6pt depth1pt\medskip}



\newtheorem{theorem}{Theorem}%[section]
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{definition}[theorem]{Definition}


\newcommand{\thm}[2]{\begin{theorem}\label{#1}{\sl #2}\end{theorem}}
\newcommand{\conj}[2]{\begin{conjecture}\label{#1}{\sl
#2}\end{conjecture}}
\newcommand{\prob}[2]{\begin{problem}\label{#1}{\sl #2}\end{problem}}

\newcommand{\half}{\frac{1}{2}}
\newcommand{\third}{\frac{1}{3}}
\newcommand{\twothirds}{\frac{2}{3}}
\newcommand{\Chalf}{C_\frac{1}{2}}
\newcommand{\Cthird}{C_\frac{1}{3}}

\author{%
Krzysztof Ciesielski%
\thanks{The work of the first author was partially supported by
NATO Grant PST.CLG.977652.\endgraf
\ \ Papers authored or co-authored by a Contributing Editor are
managed by a Managing Editor or one of the other Contributing Editors.},
\ Department of Mathematics, West Virginia
University, Morgantown, WV 26506-6310, USA (e-mail:
K\_Cies@math.wvu.edu\\
web page: {\tt
http://www.math.wvu.edu/\~{}kcies})
\and Hajrudin Fejzi\'c,
\ Department of Mathematics, California State University, San
Bernardino, CA 92407, USA (e-mail: hfejzic@csusb.edu)
\and Chris Freiling, \
Department of Mathematics, California State University, San
Bernardino, CA 92407, USA (e-mail: cfreilin@csusb.edu)
}

\title{Measure zero sets with non-measurable sum}

\begin{document}\maketitle


\begin{abstract}
For any $C\subseteq \real$ there is a subset $A\subseteq C$ such that
$A+A$ has inner measure zero and outer measure the same as $C+C$. Also,
there is a subset $A$ of the Cantor middle third set such that $A+A$ is
Bernstein in
$[0,2]$. On the other hand there is a perfect set $C$ such
that $C+C$ is an interval $I$ and there is no subset $A\subseteq C$  with $A+A$
Bernstein in $I$.
\end{abstract}

%_______________________________________________________________________

\section{Introduction.}
It is not at all surprising that there should be measure zero
sets, $A$, whose sum $A+A=\{x+y\colon x \in A, y\in A\}$ is
non-measurable.  Ask a typical mathematician why this should be so
and you are likely to get the following response:
\begin{quote}
The Cantor middle-third set, when added to itself gives an entire
interval, $[0,2]$. So certainly there exists a measure zero set
that when added to itself gives a non-measurable set.
\end{quote}
The intuition being that an interval has much more content than is
needed for a non-measurable set.

Indeed such sets do exist (in ZFC).  Sierpi\'nski (1920) seems to be
the first to address this issue. Actually, he shows the existence
of measure zero sets $X,Y$ such that $X+Y$ is non-measurable (see
\cite{Sierpinski}). The paper by Rubel (see~\cite{Rubel}) in 1963
contains the first proof that we could find for the case $X=Y$
(see also ~\cite{Muk}). Ciesielski~\cite{Ciesielski} extends these
results to much greater generality, showing that $A$ can be a
measure zero Hamel basis, or it can be a (non-measurable)
Bernstein set and that $A+A$ can also be Bernstein.  He also
establishes similar results for multiple sums, $A+A+A$ etc.

This paper is mainly about the statement above and the intuition
behind it.  Below we list four conjectures, each of which seems
justified by extending this line of reasoning.

\begin{enumerate}
\item Not only does such a set exist, but it can be taken to be a
subset of the Cantor middle-third
set, $\Cthird$. (This does not seem to
immediately follow from any of the above proofs.
Thomson~\cite[p. 136]{Thomson}
claims this to be true, but without proof.)

\item The intuition really has nothing to do with the precise
structure of the Cantor set, which might lead one to conjecture
the following.  Suppose $C$ is any set with the property that
$C+C$ contains a set of positive measure. Then there must exist a
subset $A\subseteq C$ such that $A+A$ is non-measurable.

\item The intuition relies on the fact that non-measurable
sets can have far less content
than an entire interval. Therefore,
the claim should also hold when {\it non-measurable} is replaced
by other similar qualities.  Recall that if $I$ is a set then a
set $S$ is called {\it Bernstein in I}\/ if and only if both $S$ and
its complement intersect every non-empty perfect subset of $I$.
Constructing a set that is Bernstein in an interval is one of the
standard ways of establishing non-measurability. Certainly, any
set that is Bernstein in an interval has far less content than the
interval itself. Therefore, we might conjecture that there is a
subset $A\subseteq \Cthird$ with $A+A$ Bernstein in [0,2].

\item Combining the reasoning behind the Conjectures 2 and 3,
let $C$ be any set with the property that $C+C$ contains an
interval, $I$. We might conjecture that there must exist a subset
$A\subseteq C$ such that $A+A$ is Bernstein in $I$.
\end{enumerate}

We will settle these four conjectures in the next four sections.
In particular, in Section~\ref{Conj1} we will give a proof of the
first conjecture using transfinite induction. This provides what
is possibly the simplest proof of the original assertion.
However, the proof depends on a particular property of symmetric
perfect sets. In Section~\ref{Conj2} we give a slightly more
complicated argument to prove the second conjecture. We show that
any set $C$ contains a subset, $A$, such that the inner measure of
$A+A$ is zero and the outer measure is the same as $C+C$. In
Section~\ref{Conj3} we will settle the third conjecture, finding a
subset $A$ of the Cantor set such that $A+A$ is Bernstein in
[0,2].  Finally, in Section~\ref{Conj4} we will give a
counterexample to the fourth conjecture, showing that the above
lines of reasoning are indeed limited. In all of these proofs, we
will be assuming the axioms of ZFC.
No additional set theoretical assumptions will be used.

The existence of these sets is interesting historically. Suppose
$E$ is a measurable set and $f$ is a measurable function. Several
researchers in
the area of generalized derivatives have taken for granted the
measurability of sets such as:
$$
E_j=\left\{x\in E\colon \left|\frac{f(x+h)-2f(x)+f(x-h)}{h^2}\right|<j
\mbox{ for all }0<|h|<\frac{1}{j} \right\}.
$$
It wasn't until 1960 that Stein and Zygmund~\cite{Stein-Zygmund}
pointed out that the measurability of these sets is not automatic,
and not until 1993 that Fejzi\'c and Weil re-proved these results
without this assumption.  In their paper~\cite{Fejzic-Weil} they
also show that this measurability assumption can be reduced to the
(false) claim that $A+A$ is measurable whenever $A$ is measurable
(see also~\cite{Muk}).

%_______________________________________________________________________

\section{Conjecture 1 is true.}
\label{Conj1}

It is well known that every real $x$ in $[0,2]$ can be expressed
as $x=a+b$ where $a,b$ are in the Cantor set $\Cthird$. It is also
well known (a proof occurs below in Lemma~\ref{cmany}) that for
almost every such $x$ there are continuum many such
representations.  The following theorem therefore gives a positive
answer to Conjecture 1.


\begin{theorem}
\label{thm0} Let $C$ be any subset of $\real$
and let
$$
R=\{x\in\real\colon x \mbox{ has }\continuum \mbox{ many representations }
x=a+b \mbox{ with }a,b \in C\} \, .
$$
Then there is a subset $A \subseteq C$ such that $A+A$ is
Bernstein in $R$.
\end{theorem}

\proof If $R$ has no non-empty perfect subsets then we are done.
Otherwise it has continuum
many such sets. Let $\{P_\xi \colon  \xi
< \continuum \}$ be the family of all non-empty perfect subsets of
$R$.  We will find an $A\subseteq C$ such that each $P_\xi$
intersects both $A+A$ and its complement. Construct a sequence,
$$
\la \la a_\xi, b_\xi, c_\xi, d_\xi \ra \in C\times C\times P_\xi
\times P_\xi \colon  \xi < \continuum \ra
$$
such that for each $\xi<\continuum$,

\begin{itemize}
\item[($*$)]
$c_\xi = a_\xi + b_\xi $ $\mbox{and } \{d_\eta \colon  \eta \leq \xi \}
\cap (A_{\xi +1}+A_{\xi +1}) = \emptyset,$
\end{itemize}
where $A_{\xi}=\bigcup_{\eta < \xi}\{ a_\eta, b_\eta \}$.
  This will ensure that
$A=A_\continuum$ has the desired properties, since then $\{ c_\xi
\colon  \xi < \continuum \}\subseteq A+A \subseteq \real \setminus \{
d_\xi \colon  \xi < \continuum \}$.

To make an inductive step, assume that for some $\alpha <
\continuum$ we have already constructed a partial sequence
satisfying ($*$) for all $\xi < \alpha$.  First choose a $c_\alpha
\in P_\alpha \setminus \{  d_\xi \colon  \xi < \alpha \}$. Then choose
$a_\alpha, b_\alpha$ in $C$ such that $a_\alpha + b_\alpha =
c_\alpha$ and neither $a_\alpha$ nor $b_\alpha$ is in $\{ d_\xi \colon
\xi < \alpha \} - A_\alpha$.  Construction is finished by choosing
$d_\alpha \in P_\alpha - (A_{\alpha +1}+A_{\alpha +1})$.
 \qed

%_______________________________________________________________________
\section{Conjecture 2 is true.}
\label{Conj2}


Conjecture 2 is settled by the following theorem.
Note that it proves more than what is needed for Conjecture 2.
However, as is the case with Theorem~\ref{thm0}, it falls short of
establishing that $A+A$ is Bernstein in $C+C$.

\thm{th1}{For every $C\subset\real$ there exists an $A\subset C$
such that $A+A$ has inner measure zero and outer measure the same
as $C+C$.}

\proof  We can assume that the inner measure of $C+C$ is positive,
since otherwise $A=C$ is as desired.

Let $G$ be a $G_{\delta }$-set containing $C+C$ such that
$G\setminus (C+C)$ has inner measure zero. We will construct a set
$A\subset C$ such that every perfect set $P\subset G$ of positive
measure intersects both $G\setminus (A+A)$ and $A+A$. This implies
that $A+A$ has inner measure zero and the same outer measure as
$C+C$.

Let $\bar{\E}$ be the family of all perfect subsets $P$ of $G$ of
positive measure such that $|P\cap (C+C)|<\continuum$.\footnote{
Since $P\cap (C+C)$ must have positive outer measure, this set
must be uncountable. Also $\bar{\E}$ is empty if either $C+C$ is
measurable or Martin's axiom holds. However, there are models of
ZFC containing sets $C$ of cardinality less than $\continuum$ with
full outer measure.  In that case, $C+C$ also has these properties
and so $\bar{\E}$ contains all perfect subsets of 
$G$ of positive measure.}
Let $\E$ be
a maximal subfamily of $\bar{\E}$ of pairwise disjoint sets. Then
$\E$ is at most countable, so $E=\bigcup \E$ is an $F_{\sigma}$-set 
and $E\cap (C+C)$ has cardinality less than $\continuum$.
For every $e\in E\cap (C+C)$ fix $c_e,d_e\in C$ such that
$e=c_e+d_e$. Let $Z=\bigcup\{\{c_e,d_e\}\colon e\in E\cap
(C+C)\}$. Then $Z\subset C$
also has cardinality less than $\continuum$
and $E\cap (C+C)\subset Z+Z$. Notice that by the maximality
of~$\E$,

\begin{itemize}
\item[(a)]  $|P\cap (C+C)|=\continuum$ for every perfect set $P\subset
G\setminus E$ of positive measure.
\end{itemize}

Next, let $\bar{\K}$ be the family of all perfect subsets $F$ of
$G\setminus E$ of positive measure for which there exists an
$X_{F}\subset C$ such that

\begin{center}
$|X_{F}|<\continuum$ and $F\cap (C+C)\subset X_{F}+C$.\footnote{
The family $\bar{\K}$ may be 
non-empty even if $C$ has measure zero. 
Although this cannot happen under Martin's axiom,
it happens in any model of ZFC in which there is 
a non-meager measure zero set of cardinality
less than $\continuum$. (See e.g. \cite[thms. 2.7.3 and 2.1.7]{BJ}.)}
\end{center}
Let $\K$ be a maximal subfamily of $\bar{\K}$ of pairwise disjoint
sets. Then $\K$ is at most countable, so $K=\bigcup \K$ is an
$F_{\sigma }$-set and $X=\bigcup_{F\in \K}X_{F}\subset C$ has
cardinality less than
$\continuum$.
Clearly $K\cap (C+C)\subset X+C$. Moreover, by the maximality of
$\K$,

\begin{itemize}
\item[(b)]
for every $P\subset G\setminus (E\cup K)$ of positive measure
and any set $X\subseteq C$ of cardinality less than $\continuum$,
$(P\cap (C+C)) \nsubseteq (X+C)$.
\end{itemize}

Let $\P$ be the family of all non-empty perfect subsets $P$ of
$G\setminus E$ of positive measure such that either $P\subset K$
or $P\cap K=\emptyset$.
Let $\{P_{\xi }\colon \xi <\continuum\}$
be an enumeration of $\P $. We will find an $A\subset C$
containing $Z$ such that each $P_{\xi }$ intersects
both
$A+A$ and $G\setminus (A+A)$. First notice that such a set
will be as desired.

Indeed, take a perfect set $P\subset G$ of positive measure. We
have to show that $P$ intersects both $G\setminus (A+A)$ and
$A+A$. If $P\cap E$ has positive measure then 
$\emptyset \neq P\cap E\cap (C+C)\subset Z+Z\subset A+A$, while
$P\cap E\not\subset A+A$
since $E\cap (A+A)\subset E\cap (C+C)$ has cardinality less than
$\continuum$.
On the other hand, if
$P\cap E$ has measure zero then $P\setminus E$ contains one of
sets $P_{\xi }$, since in this case either $(P\setminus E)\cap K$
or $(P\setminus E)\setminus K$ has to have positive measure.

To define $A$ construct $\la \la a_\xi,b_\xi,c_\xi,d_\xi\ra\in
C\times C\times P_\xi\times P_\xi\colon \xi<\continuum\ra$ such
that for every $\xi<\continuum$

\begin{itemize}
\item  $c_{\xi }=a_{\xi }+b_{\xi }$ and $\{d_{\eta }\colon \eta \leq \xi
\}\cap (A_{\xi +1}+A_{\xi +1})=\emptyset $,
\end{itemize}
where $A_\beta=Z\cup X\cup\bigcup_{\eta<\beta}\{a_\eta,b_\eta\}$.
This will ensure that $A=A_\continuum$ has the desired properties
since then $\{c_\xi\colon\xi<\continuum\}\subset A+A$ and
$\{d_\xi\colon\xi<\continuum\}\cap(A+A)=\emptyset$.

To make an inductive step assume that for some $\xi<\continuum$ we
have already constructed a partial sequence for which
$\{d_\eta\colon\eta<\xi\}\cap(A_\xi+A_\xi)=\emptyset$.
Let $Y=\lin\left(A_\xi\cup\{d_\eta\colon\eta<\xi\}\right)$, 
where $\lin(S)$ denotes the set of finite linear
combinations of elements in $S$ with rational coefficients. Notice
that $|Y|<\continuum$ since $|S|<\continuum$ implies
$|\lin(S)|<\continuum$. Consider two cases.

\bigskip

\noindent {\bf Case 1:} $P_\xi\subset K$.
Choose $c_\xi\in P_\xi\cap(C+C)\setminus Y$. 
The choice can be made by (a), since
$|P_\xi\cap(C+C)|=\continuum$ as $P_\xi\subset G\setminus E$ has
positive measure.

As $c_{\xi }\in P_{\xi }\cap (C+C)\subset K\cap (C+C)\subset X+C$
there are $a_{\xi }\in X\subset A_{\xi }$ and $b_{\xi }\in C$ such
that $c_{\xi}=a_{\xi}+b_{\xi}$.
We now show that $d_{\eta} \notin A_{\xi +1}+A_{\xi +1}$
for each $\eta<\xi$.  
Assume otherwise.  Since $A_{\xi+1}=A_\xi\cup\{b_\xi\}$ and 
$d_{\eta}\notin A_{\xi}+A_{\xi}$, it
must be that $d_{\eta}\in\{b_{\xi}\}+(A_\xi\cup\{b_\xi\})$.  But then
$b_{\xi}\in Y=\lin\left(A_\xi\cup\{d_\eta\colon\eta<\xi\}\right)$ 
and so also is $c_{\xi}=a_{\xi}+b_{\xi}$,
contradicting the choice of $c_{\xi}$.

\bigskip

\noindent {\bf Case 2:} $P_\xi\subset G\setminus (E\cup K)$. We
will first show that

\begin{itemize}
\item[($*$)]  there exist $a,b\in C\setminus Y$
such that $a+b\in P_{\xi}\setminus Y$.
\end{itemize}

To see this, for each $p\in P_\xi\cap(C+C)$ choose $a_p,b_p\in C$
with $p=a_p+b_p$. If ($*$) is false then for every $p\in
P_\xi\cap(C+C)\setminus Y$ we would have $\{a_p,b_p\}\cap
Y\neq\emptyset$. But then 
$P_\xi\cap(C+C)\setminus Y\subset (Y\cap C)+C$. So, for
%KC I see no reason for making your complicated correction, 
% since it is obvious that the original form is fine,
% as $P_{\xi}\cap (C+C) \cap Y\subset\{a_p\colon p\in Y\}+C$
%
$X_0=(Y\cap C)\cup\{a_p\colon p\in Y\}$
%%cf
%$X_0=(Y\cap C)\cup\{a_p\colon p\in P_{\xi}\cap (C+C) \cap Y\}$
%%old way:
%%$X_0=(Y\cap C)\cup\{a_p\colon p\in Y\}$
%%end
we have $P_\xi\cap(C+C)\subset X_0+C$ contradicting~(b).

Now, take $a$ and $b$ as in ($*$) and put $a_\xi=a$, $b_\xi=b$,
and $c_\xi=a+b$. Then $a_\xi,b_\xi,c_\xi\notin Y$. In particular,
$d_\eta\notin\{a_\xi,b_\xi,c_\xi\}+\lin\left(A_\xi\right)$ for
every $\eta<\xi$. So,
$\{d_\eta\colon\eta<\xi\}\cap(A_{\xi+1}+A_{\xi+1})=\emptyset$.

\smallskip

In both cases we finish
the inductive step
by choosing a $d_\xi\in P_\xi\setminus\lin(A_{\xi+1})$. \qed

%_______________________________________________________________________

\section{Conjecture 3 is true.}
\label{Conj3}

In this section we embellish the argument in Section 1 to settle
the third conjecture.
\begin{definition}
Two real numbers $x,y$ will be called {\it equivalent} and we
write $x\sim y$ if and only if there is a ternary expansion of $x$
and a ternary expansion of $y$ such that the two expansions
disagree on only finitely many digits.
\end{definition}

Note that if $x$ is a ternary rational then it will have two
possible ternary expansions. According to the definition, all such
ternary rationals are equivalent.  Every other real $x$ has a
unique ternary expansion.  The following theorem fulfills the
promise made in Section~\ref{Conj1}, showing that almost every
$x\in [0,2]$ has $\continuum$ many representations as the sum of
elements in the Cantor set.

\begin{lemma}
\label{cmany}
Let $x\in[0,2]$.  If $x/2$ has infinitely many ones
in its ternary expansion, then there are $\continuum$ many
representations of $x$ as the sum of two Cantor-set elements.
Otherwise, $x$  has only finitely many such representations and
all of the elements of $\Cthird$ used to represent $x$ are
equivalent to $x/2$.
\end{lemma}
\proof Every element of $\Cthird$ has a ternary expansion
consisting of only even digits.
Fix $x\in[0,2]$ and
let $c=x/2$.  If $x=a+b$ with $a,b\in \Cthird$, then $c$ is the
average of $a$ and $b$. If $c_i,a_i,b_i$ are the $i^{th}$ digits
of $c,a$, and $b$, respectively, then either
\begin{itemize}
\item $c_i=0$ and $a_i=b_i=0$,
\item $c_i=1$ and $a_i=0,b_i=2$,
\item $c_i=1$ and $a_i=2,b_i=0$, or
\item $c_i=2$ and $a_i=b_i=2$.
\end{itemize}
Suppose first that $c$ is a ternary rational. Then the digits of
$c$ must end in either a sequence of zeros or a sequence of two's.
In either case, the digits of $a$ and $b$ must do likewise and so
they are also ternary rationals. Therefore, $a\sim b\sim c$. Now
consider the case when $c$ is not a ternary rational, so there is
a unique ternary expansion of $c$. Let us construct the numbers
$a$ and $b$ using only even digits.  For each $c_i$ that is zero
or two, we must have $a_i=b_i=c_i$. But for each $c_i$ that is
one, we have a choice, either $a_i=0,b_i=2$ or $a_i=2,b_i=0$. Thus
if $k\in\{0,1,2,\ldots,\omega\}$ is the number of digits in $c$
that have the value 1, then there are $2^k$ possible choices for
the pair $a,b$.  In particular, if $c$ has infinitely many ones in
its expansion, then there are $2^{\omega}=\continuum$ many
representations for $x$.  If there are only finitely many ones,
then the digits of $a,b,c$ will all agree on a tail end.\qed

\begin{theorem}
There is a set $A\subseteq \Cthird$ such that $A+A$ is Bernstein
in $[0,2]$.
\end{theorem}
\proof Let $R_0$ be the set of elements of $[0,2]$ that can be
expressed in $\continuum$ many ways as the sum of elements of
$\Cthird$, and let $R_1$ be the elements that can be expressed in
only finitely many ways.  Let $\{P_\xi \colon  \xi < \continuum \}$ be
the family of all non-empty perfect subsets of $[0,2]$.  We will
find an $A\subseteq \Cthird$ such that each $P_\xi$ intersects
both $A+A$ and its complement. Construct
$$
\la \la a_\xi, b_\xi, c_\xi, d_\xi \ra \in C\times C\times P_\xi
\times P_\xi \colon  \xi < \continuum \ra
$$
such that for each $\xi<\continuum$,
\begin{itemize}
\item[($*$)]
$c_\xi = a_\xi + b_\xi $ $\mbox{and } D_{\xi} \cap (A_{\xi
+1}+A_{\xi +1}) = \emptyset,$
\end{itemize}
where $A_{\xi}=\bigcup_{\eta < \xi}\{ a_\eta, b_\eta \}$ and $
D_{\xi}=\{d_\eta \colon  \eta \leq \xi \}$.
  This will ensure that
$A=A_\continuum$ has the desired properties, since then $\{ c_\xi
\colon  \xi < \continuum \}\subseteq A+A \subseteq
\real \setminus D_{\xi}$.
To make the inductive step, assume that for some $\alpha
< \continuum$ we have already constructed a partial sequence
satisfying ($*$) for all $\xi < \alpha$.  We first choose
$a_{\alpha}, b_{\alpha}, c_{\alpha}$ such that $a_{\alpha}+
b_{\alpha}=c_{\alpha}$ and such that neither $a_\alpha$ nor
$b_\alpha$ is in $D_{\alpha} - A_\alpha$.  We distinguish two
cases.

{\bf Case 1:}  $P_\alpha$ intersects $R_0$ in a set of cardinality
$\continuum$.  First choose
$c_\alpha \in P_\alpha \cap R_0
\setminus D_{\alpha}$. Then choose $a_\alpha, b_\alpha$ in
$\Cthird$ such that $a_\alpha + b_\alpha = c_\alpha$ and neither
$a_\alpha$ nor $b_\alpha$ is in $D_{\alpha} - A_\alpha$.

{\bf Case 2:}  $P_\alpha$ intersects $R_1$ in a set of cardinality
$\continuum$.  First choose
$c_{\alpha}\in P_\alpha \cap R_1\setminus D_{\alpha}$ such
that $c_{\alpha}/2$ is not equivalent to any element of $D_\alpha -
A_\alpha$. Then choose $a_\alpha, b_\alpha \in \Cthird$ such that
$a_\alpha + b_\alpha = c_\alpha$.
Since, by Lemma~\ref{cmany}, both $a_{\alpha}$ and
$b_{\alpha}$ are equivalent to $c_{\alpha}/2$, neither of them is
in $D_{\alpha} - A_\alpha$.

Construction is finished by choosing $d_\alpha \in P_\alpha
\setminus (A_{\alpha +1}+A_{\alpha +1})$.\qed

%_______________________________________________________________________

\section{Conjecture 4 is false.}
\label{Conj4}

In this section we will construct a set $A$ such that $A+A$
contains an interval, $I$, yet there is no subset $B\subseteq A$
with $B+B$ Bernstein in $I$. Let $\Chalf$ be the Cantor
middle-half set, which will be the basis of our construction. That
is, $\Chalf$ is the set of points, $x$, in the unit interval such
that there is a base four expansion of $x$ that uses only zeros and
threes.  Note that if the expansion of $x$ ends in a sequence of
zeros, then there will be an equivalent expansion ending in a
sequence of threes.  In this case we will say that $x$ is a {\em
quaternary rational}.  In all other cases, the base 4 decimal
expansion is unique.

When we dealt with sums from $\Cthird +\Cthird$, it was easier to
think in terms of averages, so that the work could be carried out
digit-wise. Similarly, when we work with $\Chalf$ it will be
easier to think in terms of the following auxiliary sets.  Let $U$
be the set of elements of $[0,1]$ that use only zeros and twos in
one of its base four expansions, and let $V$ be the set of
elements that use only zeros and ones.  Since threes are not
allowed in either set, each element of $U$ and $V$ has only one
valid expansion.  Furthermore, the sums in $U+V$ and the sums in
$V+V$ can be carried out digit-wise since there will be no
carries.

Our construction will be based on the following three lemmas.
The proofs of the first two of them can be
seen geometrically by examining Figure 1.



\begin{lemma} \label{lem1}
$\Chalf + \Chalf$ has measure zero.
\end{lemma}
\proof Let $x=a+b$ with $a,b\in \Chalf$.  Let $c=\third x = \third
a + \third b$. Using a base 4 expansion of $a$ and $b$ where all
digits are divisible by three, the computation of $\third a +
\third b$ can be carried out digit-wise and results in an
expansion where the digit 3 is not used. Therefore, unless $c$ is
a quaternary rational, its expansion will never use the digit
three.
Thus
$\third(\Chalf + \Chalf)$ has measure zero and so does
$\Chalf + \Chalf$.
\qed

Now consider a sum $c=u+v$ where $u\in U$ and $v\in V$.  There are
four possibilities for the $i^{th}$ digits of this calculation.
\begin{itemize}
\item $c_i=0$, $u_i=v_i=0$
\item $c_i=1$, $u_i=0$, $v_i=1$
\item $c_i=2$, $u_i=2$, $v_i=0$
\item $c_i=3$, $u_i=2$, $v_i=1$
\end{itemize}
Unlike the sums of $\Cthird + \Cthird$ this time there is no
ambiguity. Every choice of the digits of $c$ gives a unique choice
for the digits of $u$ and $v$.  The next lemma relates this to the
sum $\Chalf + \half\Chalf$.

\begin{lemma}\label{lem2}
$\Chalf + \half\Chalf = [0,1.5]$.  Furthermore, each element in
$[0,1.5]$ can be expressed as such a sum in at most two ways, and
except for a countable set, each element in $[0,1.5]$ can be
expressed in a unique way.
\end{lemma}
\proof
Fix an
$x\in [0,1.5]$, and suppose $x=a+b$ with $a\in \Chalf$
and $b\in \half\Chalf$. Let $c=\twothirds x = \twothirds a +
\twothirds b$. Using the fact that all of the digits of $a$ are
divisible by three, the computation of $\twothirds a$ can be
carried out digit-wise and results in an element of $U$.
Similarly, $\twothirds b \in V$. But each such $c$ has at most two
such representations and except when $c$ is a quaternary rational,
each such $c$ has a unique representation.

The equality is justified by
$\twothirds(\Chalf + \half\Chalf) = U+V=[0,1]=\twothirds[0,1.5]$.
\qed


%_______________________________________________________________________

\begin{figure}
\begin{picture}(150,140)(0,-10)
\put(0,0){\vector(1,0){120}}
\put(0,0){\vector(0,1){120}}
\put(0,0){\framebox(25,25)}
\put(75,0){\framebox(25,25)}
\put(0,75){\framebox(25,25)}
\put(75,75){\framebox(25,25)}
\put(0,-10){0} \put(100,-10){1}
\put(25,25){\vector(1,-1){25}}
\put(0,75){\vector(1,-1){75}}
\put(48,5){hole}
\end{picture}
\begin{picture}(200,140)(0,-10)
\put(0,0){\vector(1,0){170}}
\put(0,0){\vector(0,1){120}}
\put(0,0){\framebox(25,25)}
\put(75,0){\framebox(25,25)}
\put(0,75){\framebox(25,25)}
\put(75,75){\framebox(25,25)}
\put(0,-10){0} \put(100,-10){1} \put(150,-10){1.5}
\put(0,75){\vector(1,-2){37.5}}
\put(25,100){\vector(1,-2){50}}
\put(75,75){\vector(1,-2){37.5}}
\put(100,100){\vector(1,-2){50}}
\end{picture}
\caption{
Consider the pairs of reals from the unit interval with the middle
half removed.  The
projection along the direction
with a slope $-1$
has holes, while the
projection along the direction
with a slope $-2$
fills the interval
$[0,1.5]$ without any overlaps.}
\end{figure}


%_______________________________________________________________________



We are now ready to define the set $A$ that will serve as our
counterexample. Let $A=\Chalf \cup \half \Chalf$.

\begin{lemma}
There are two non-empty perfect subsets $P$ and $Q$ of $A$ such that
every element of $P+Q$ can be expressed uniquely as the sum of two
elements in $A$.
\end{lemma}
\proof $A+A$ is the union of three closed sets: $\Chalf+\Chalf$,
$\half\Chalf+\half\Chalf$, and $\Chalf+\half \Chalf$.  By
Lemma~\ref{lem1}, the first and second
sets are both measure
zero. Choose an open interval $I\subseteq [0,1.5]$ that is
disjoint from the first two sets. By Lemma~\ref{lem2}, the third
set is the interval $[0,1.5]$. Furthermore,
this set can be
partitioned into two sets $X$ and $Y$ such that $X$ is countable
and every element in $Y$ has a unique representation as a sum of
two elements, one in $\Chalf$ and the other in $\half\Chalf$.
Choose an $x$ in $Y\cap I$ and let $x=a+b$ with $a\in \Chalf$ and
$b\in \half\Chalf$. Now choose a neighborhood $J$ of $a$ and a
neighborhood $K$ of $b$ small enough that
the closure
$cl(J+K)$ of $J+K$ is a subset of $I$.
Let $R$ be the intersection of $A$ with $cl(J)$ and let $S$ be the
intersection of $A$ with $cl(K)$. Then both $R$ and $S$ are
non-empty perfect subsets of $A$.  Let $D$ be the countable set
consisting of all numbers used in the representations of elements
of $X$, and let $P$ and $Q$ be non-empty perfect subsets of
$R\setminus D$ and $S\setminus D$ respectively.

Fix $x\in P+Q$.  We must show that $x$ has a unique representation
as a sum of elements of $A$. Since $x\in (R\setminus D) +
(S\setminus D)$ then there
exist $a\in R\setminus D$ and $b\in
S\setminus D$ with $x=a+b$. Since $R \subseteq cl(J)$ and $S
\subseteq cl(K)$, this gives us that $x\in I$.  But then $x$ is
not in the first two pieces of $A+A$ so it must be that one of the
elements $a,b$ is in $\Chalf$ and the other is in $\half\Chalf$.
Since $a$ and $b$ are not in $D$, $x$ cannot be in $X$. Therefore,
$x \in Y$, and we are done. \qed

\begin{theorem}
There is no subset $B\subseteq A$ such that $B+B$ is Bernstein in
$[0,1.5]$.
\end{theorem}
\proof Suppose that such a set $B$ exists.  Let $P$ and $Q$ be as
in the previous lemma. $B$ can't contain a non-empty perfect
subset, since that would imply $B+B$ also contains a non-empty
perfect subset
of $[0,1.5]$.  Therefore, there is some element $x$ in $P
\setminus B$. Then $x+Q$ is a perfect subset of $P+Q$ and so each
element of $x+Q$ has a unique representation as a sum of elements
in $A$. But then since $x \notin B$ no element of $x+Q$ is in
$B+B$.
So, $B+B$ is not Bernstein in $[0,1.5]$, which is a
contradiction.
 \qed

%_______________________________________________________________________

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\end{document}
