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\begin{document}
\title{Topics in the theory of uniqueness of trigonometrical series}
\author{Paul J. Cohen\thanks{Current address: Mathematics Department,
Stanford University.}}
\date{Ph.D. Thesis, University of Chicago, 1958\\
$\;$\\ $\;$\\
$\;$\\
with notes by Ilan Vardi\thanks{Address: IHES,
BuressurYvette France,~ilan@ihes.fr}}
\maketitle
%%\pagenumbering{roman}
\begin{center}
{\large ACKNOWLEDGMENT}
\end{center}
\bigskip
\medskip
The author wishes to express his deepest gratitude to
Professor A.~Zygmund for his constant aid and encouragement
during the preparation of this dissertation.
$\;$
\newpage
\chapter*{Introduction}
%%\pagenumbering{arabic}
%%\pagestyle{myheadings}
%%\markboth{INTRODUCTION}{INTRODUCTION}
The theory of uniqueness of trigonometrical series can
be regarded as arising from the question of deciding in
what sense the Fourier series of a function may be considered
as the legitimate expansion of the function in an infinite
trigonometrical series. We know, of course, that if the
series converges boundedly to the function, then indeed
the coefficients of the series must be given by the
EulerFourier Formulas. However, in the absence of such
a condition, we may ask ourselves whether two trigonometrical
series may converge to the same function everywhere.
The answer to this question is in the negative and was essentially
proved so by Riemann, the proof being completed by Cantor.
It is with the replacement of the condition of convergence
everywhere with that of convergence almost everywhere, that
the theory of sets of uniqueness is concerned. The situation
here is not quite as simple. It was shown by Mensov, that
there exist trigonometrical series which converge to zero
almost everywhere, but which are not identically zero. By
the theorem of RiemannCantor mentioned above, this series
does not converge to zero almost everywhere. The set at
which it fails to converge to zero is an example of what
is called a set of multiplicity. To be more precise, we
say that a set $E$ of measure zero, is a set of multiplicity
or an Mset, if there exists a trigonometrical series
converging to zero everywhere outside $E$, but not identically
zero. A set of measure zero is called a set of uniqueness
or a Uset, if it is not an Mset.
Young [13] showed that every denumerable set is a Uset.
Rajchman and Bary [13] independently discovered
the existence of nondenumerable perfect Usets. The
main problem in the subject might be considered to be that of finding
necessary and sufficient conditions for a set to be a Uset.
In practice, attention has been largely restricted to closed sets.
It seems that metrical properties such as capacity have little
to do with the question [7]. In general one
can say that the problem is related to the arithmetical structure
of the set.
Our knowledge of Usets may be summed up in the following
way. On the one hand a theorem of Bary says that a countable
union of closed Usets is a Uset. On the other, certain
sets designated as $H^r$ sets are known to be closed Usets.
Here $r$ may be any positive integer. $H^1$ sets were discovered
by Rajchman and Bary while $H^r$ sets, which are generalizations
of the $H^1$ sets, were found by PyatetskiiShapiro. In his
paper [5], PyatetskiiShapiro proved that there were
$H^2$ sets which were not countable unions of $H^1$ sets. He also
stated that for each $r$, there exists a set of type $H^r$,
which is not a countable union of $H^{r1}$ sets. One of our
theorems will be a complete proof of this fact. It depends upon
a certain arithmetical property of $H^r$ sets, which is fairly
obvious for $H^1$ sets, but rather involved for $r>1$.
PyatetskiiShapiro also introduced Banach space methods to give
necessary and sufficient conditions for a closed set to be
a Uset. This condition cannot be applied to any known sets,
however, because it is stated in terms of certain Banach space
notions and not in terms of the set itself. Because of the
unavailability of the paper in English, I have included
a rather brief outline of the proof of his results. Also
I give sufficient criteria for Usets which are more
arithmetical in nature and are in terms of the set itself.
PyatetskiiShapiro's criterion may be used to reprove
a result of Zygmund and Marcinkiewicz [4] in the case
of closed sets. In the first section is also included a
theorem which is a necessary condition for Usets to be
of a certain special type. As no necessary conditions, to my knowledge,
have been given for Usets up to now, I think this may be
of some interest.
In the theory of several variables, no analogue of Riemann's
uniqueness theorem has ever been proved, without certain strong
additional hypotheses. These hypotheses concern the rate at
which the coefficients are allowed to grow. In particular
the results of [9], imply that if the coefficients
of a double series tend to zero, and the series is circularly
summable to zero everywhere, then it is identically zero.
It seems that no results concerning the rate of growth of the
coefficients of a convergent trigonometrical series have been
published. In the second chapter I prove a result which implies
that the rate of growth is smaller than exponential. This
result applies to a general type of method of summation
which includes both circular and square summation.
There is a notion of Msets of restricted type which will be
discussed in the third chapter. This is a set on which there
exists a measure, whose FourierStieltjes series converges
to zero outside the set. The classical Msets were of this
type and the trigonometrical series given were the series
associated with the measures. In this section I will give a
constructive example of a series which converges to zero almost
everywhere, but which is not the Fourier series of a measure.
Also I have included a generalization of a theorem of Wiener
concerning the Fourier series of measures, to several dimensions.
In the last chapter, I consider Green's theorem for the plane
in what might be considered as a best possible form. Bochner
in [3], and Shapiro in [10], considered
this question, but the theorem that I prove is stronger
than Bochner's, and has fewer hypotheses than the one in
[10], but it does not allow the same type of exceptional
sets as in this latter work.
%%%\pagestyle{headings}
\chapter{}
\noindent
{\bf 1.} We first make several definitions which will occur
in the course of this chapter. We denote by $W$ the ring of
absolutely convergent Fourier series on the interval
$[0,1]$. We may think of $W$ as the set of all sequence
$c_n$, $\infty < n < \infty$, such that
$\sum_n c_n < \infty$. Then $W$ forms a Banach space
under the norm $\sum_n c_n$. Let ${\cal S}$
denote the Banach space of all sequences $a_n$,
$\infty < n < \infty$, such that
$\displaystyle \mathop{{\rm Lim}}_{n\to \infty} a_n = 0$.
The norm of the sequence $\{a_n\}$ is defined to be
${\rm Max} _n a_n$. Then $W$ is the dual space of ${\cal S}$
where the sequence $\{c_n\}$ represents a linear functional
on ${\cal S}$ which, when applied to an element $\{a_n\}$
of ${\cal S}$, yields $\sum_n a_n c_{n}$. In $W$ we have
the weak star topology induced by ${\cal S}$. In this topology
a sequence of elements of $W$ $\{c_n^{(m)}\}$,
$1\le m < \infty$, converges to the element $\{c_n'\}$
if and only if (a) $\sum_n c_n^{(m)}$ is uniformly bounded
in $m$, and (b) $c_n^{(m)}$ approaches $c_n'$ for each $n$
as $m$ tends to infinity. In [5], PyatetskiiShapiro proves
the following theorem:
\medskip\noindent
{\bf Theorem 1.} Let $E$ be a closed set in the interval
$[0, 1]$. Let $L$ be the space of all absolutely converging
Fourier series vanishing on open neighborhoods of $E$.
Then $E$ is a Uset if and only if the weak closure
of $L$ is all of $W$.
\medskip\noindent
{\bf Proof.} We assume familiarity with the concept of formal
multiplication as expounded for example in [13].
We further need a little known lemma of Zygmund which is essentially
reproved in [5] by means of idealtheoretic considerations.
\medskip\noindent
{\bf Lemma 1.} Let
$P = \sum_n c_n e^{2\pi i n x}$ and
$Q = \sum_n d_n e^{2\pi i n x}$ be two trigonometrical series
where $P$ belongs to $W$ and $Q$ belongs to ${\cal S}$. Assume
that either $P$ or $Q$ vanishes on some open set $U$. Then the
formal product $R = \sum_n \gamma_n e^{2\pi i n x}$
where $\gamma_n = \sum_p c_p d_{np}$ vanishes on $U$.
\medskip
Postponing the proof of the lemma, we proceed to prove the theorem.
Let $f = \sum_n c_n e^{2\pi i n x}$ belong to ${\cal S}$.
A necessary and sufficient condition for $f$ to converge
to zero outside $E$ is that $f$ annihilate the subspace $L$.
For, if $f$ does converge to zero outside $E$, and
$g = \sum_{n} d_n e^{2\pi i n x}$ belongs to $L$, then by the
lemma the formal product of $f$ and $g$ converges to zero outside
$E$ since $f$ converges to zero there, and converges to zero on
some neighborhood of $E$, since $g$ does so. Hence this formal
product is identically zero, and in particular the constant
term $\sum_n c_n d_{n}$ equals zero. Thus $f$ annihilates
the subspace $L$. Conversely, assume that $f$ does annihilate
$L$. This means then that the constant term of the formal
product of $f$ and $g$ is zero. Now every term of this formal
product is the constant term of the product of $f$ and
$e^{2\pi i k x}\, g$, for some $k$. Since $e^{2\pi i kx}\, g$
lies in $L$ whenever $g$ does, it follows that the formal
product of $f$ and $g$ is zero. Let $y$ not be in $E$.
We can then choose $g$ to be rapidly convergent in the sense
of [13], and $g(y)\ne 0$. The theorem on formal
multiplication tells us that since this product is zero, $f$
converges to zero at $y$. Hence $E$ is a Uset if and only
if no $f$ annihilates $L$. By the HahnBanach theorem applied
to $W$, this is equivalent to saying that the weak closure of
$L$ is all of $W$.
It remains to prove the lemma. Assume for example that $Q$
vanishes on $U$. For any $x_0$ in $U$, let
$T(x) = \sum_n t_n e^{2\pi i n x}$ be a rapidly converging
Fourier series such that $T(x_0)\ne 0$ and vanishing outside
$U$. Then the triple product $PQT$, by the associative law,
may be evaluated in two different ways. On the one hand
$QT$ is identically zero since the formal product vanishes
everywhere. On the other hand, since $T(x_0) \ne 0$, it follows
that $PQ$ vanishes at $x_0$. A similar proof holds if $P$
vanishes on $U$.
\medskip
The subspace $L$ is of course an ideal in $W$. Its weak closure
$\bar L$ is also an ideal. This may be seen as follows.
If $f$ belongs to $L$, then so does $e^{2\pi i k x}\, f$,
for any $k$. Hence $\bar L$ is closed under multiplication
by finite trigonometric polynomials and hence by approximating
an arbitrary element of $W$ by trigonometric polynomials we see
that $\bar L$ is an ideal. Thus we may rephrase the condition
of Theorem~1, namely that $\bar L$ is all of $W$, and say
that $\bar L$ contains 1.
A theorem of Banach [1] states that if $W$ is the dual
space of a separable Banach space, and if $L$ is a convex set,
closed under sequential limits in the weak star topology, then
it is closed in that topology. If $E$ is a set of uniqueness
we know that $\bar L = W$. For each ordinal number
$\alpha$ set $L_\alpha = \cup_{\beta < \alpha} L_\beta$
if%%\footnote{The typo ``is'' appears in the original text.}
$\alpha$ is
a limit ordinal and set $L_{\alpha+1}$ equal to the closure
with respect to sequential limits of $L_\alpha$. For some
$\alpha$ then, we have $L_\alpha = W$. The least such $\alpha$
is an invariant of the set. The $H^r$ sets mentioned in the
introduction have the ordinal $\alpha = 1$ associated to them,
and there are no sets known which have any other value of
$\alpha$ associated to them. It would seem reasonable that a
countable union of $H^r$ sets for increasing values of $r$ would
be an example of such a set.
\bigskip\noindent
{\bf 2.} Let $E$ be a Uset situated in the interval
$[0, 1]$. Let $\lambda$ be a positive real number. Consider
the set $\lambda E$ consisting of all points of the form
$\lambda x$ where $x$ belongs to $E$. A theorem of Zygmund
and Marcinkiewicz says that $\lambda E$ considered modulo~1
is also a Uset. In this section we shall apply the results
of the previous section to obtain a new proof of this fact
in the case of closed Usets. First we need the following lemma.
\medskip\noindent
{\bf Lemma 2.} Assume that the intervals
$[\alpha,\beta]$ and $[\lambda \alpha, \lambda \beta]$ are
both contained in $[0, 1]$. There exists an absolute constant
$A$, such that if $f(x) = \sum_n c_n e^{2\pi i n x}$ is a member
of $W$ vanishing outside of $[\alpha, \beta]$, then
$f\left(\frac{x}{\lambda}\right) = \sum_n c_n' e^{2\pi i n x}$
also belongs to
$W$ and $\sum_n c_n' \le A \sum_n c_n$.
\medskip\noindent
{\bf Proof.} Let $D(x) = \sum_n d_n e^{2\pi i n x}$ belong to
$W$, have two continuous derivatives, and be equal to $1$
on $[\alpha, \beta]$. Further, let $D(x)$ vanish outside
a small enough neighborhood of $[\alpha, \beta]$ so that
$D\left(\frac{x}{\lambda}\right) $ is unambiguously defined. Now we have
$$
c_n' = \lambda\, \int_0^1 f(x) e^{2\pi i n x \lambda} dx\, ,
\leqno(1)
$$
so
$$
c_n' = \lambda\, \int_0^1 f(x)\, D(x) e^{2\pi i n x \lambda} dx\, .
\leqno(2)
$$
Also
$$
D(x) \, e^{2\pi i n \lambda x}
=
\sum_m d_m ^{(n)} e^{2\pi i m x}
\leqno(3)
$$
where
$$
d_m^{(n)} = \int_0^1 D(x) e^{2\pi i (n\lambda + m)x} dx\,.
\leqno(4)
$$
Remembering that $D(x)$ has two continuous derivatives, we
get after integrating by parts twice that
$$
d_m^{(n)} \le \frac{C_1}{C_2 + (n\lambda + m)^2}
\leqno(5)
$$
where $C_1$ and $C_2$ are suitable positive constants. Applying
Parseval's formula to (2), we obtain
$$
c_n' \le \lambda\, \sum_m c_m\,
\frac{C_1}{C_2 + (n\lambda + m)^2}\,.
\leqno(6)
$$
Hence,
$$
\sum_n c_n'\le \sum_m c_m \sum_n
\frac{\lambda C_1}{C_2 + (n\lambda + m)^2}
\le A\, \sum_m c_m
\leqno(7)
$$
where $A$ is some absolute constant.
If $E$ is a Uset, then we know by \S 1, that the function $1$
belongs to $\bar L$. Let $D(x)$ denote the same function as in the
proof of the lemma. For any $f(x)$ in $W$, we have that $f(x)$
belongs to the weak closure of $f(x)\, L$. This we see as follows.
If $f(x) = e^{2\pi i k x}$, then it is clear. It then follows
easily for a finite linear combination of exponential functions,
and finally by approximation in the norm, for arbitrary elements
of $W$. In particular $D(x)$ belongs to the closure of
$D(x) \, L$. The elements of $D(x)\, L$ all satisfy the hypothesis
of Lemma~2. Since the elements of $D(x)\, L$ vanish outside a
neighborhood of $[\alpha, \beta]$ it follows that any function
in the closure of $D(x)\, L$ also vanishes there. By the theorem
of Banach quoted above, $D(x)$ is the iterated sequential limit
of suitable elements of $D(x)\, L$. For each $f$ vanishing outside
a neighborhood of $[\alpha, \beta]$ let $\tilde f$ denote
$f\left(\frac{x}{\lambda}\right) $.
If we can verify that whenever $f_n$ tends
to $g$, then $\tilde f_n$ tends to $\tilde g$, we will have
proved that $D\left(\frac{x}{\lambda}\right) $
is in the weak closure of the space
consisting of all $\tilde f$, where $\tilde f$ lies in
$D(x)\, L$. In turn this implies that $D\left(\frac{x}{\lambda}\right) $
is in
the closure of the ideal of all functions vanishing on some
neighborhood of the set $\lambda E$. Now, with the aid of
Lemma~2, we can easily prove this assertion. For, if $f_n$
tends to $g$, this means precisely two things. First, the norms
of the $f_n$ are bounded, and second, each coefficient of $f_n$
approaches the corresponding coefficient of $g$. Lemma~2
tells us that if the first condition is satisfied for $f_n$,
it is still satisfied for $\tilde f_n$. Since the functions
$f_n$ are bounded in $L^1$ norm, it follows that the second
condition implies that each Fourier coefficient of $\tilde f_n$
approaches the corresponding coefficient of $\tilde g$.
Now, if the Uset $E$ is contained in the interval
$[\alpha,\beta]$ it follows that $\lambda E$ is a Uset.
For, $D\left(\frac{x}{\lambda}\right) $ is a function which is equal
to one on a neighborhood of $\lambda E$, and is in the
closure of $L'$, where $L'$ denotes the ideal of all functions
of $W$ vanishing on some neighborhood of $\lambda E$. If
$h(x)$ is an element of $L'$ equal to one on an open set containing
all the points where $D\left(\frac{x}{\lambda}\right) $
is not one, we have then
that $D\left(\frac{x}{\lambda}\right)
+ h(x)  D\left(\frac{x}{\lambda}\right) h(x) = 1$ belongs to
the closure of $L'$. Hence $\lambda E$ is a Uset. Since
we can always subdivide the set $E$ into sufficiently many
portions $E_i$, such that each is contained in an interval
$[\alpha, \beta]$ as above, and since a finite union of
closed Usets is a Uset, the theorem of Zygmund
and Marcinkiewicz follows.
\bigskip\noindent
{\bf 3.} As the simplest examples of Usets, we shall
consider those sets which have the property that, as explained
in \S 1, the ordinal 1 is invariantly associated with them. Recalling
the definition, this means that there exists a sequence of functions
in $W$, each vanishing on some
neighborhood%%\footnote{The typo ``meighborhood'' appears in the original text.}
of the set, and such that $f_n$ weakly approaches 1. Such a set
we shall call a U$_1$ set.
\medskip\noindent
{\bf Theorem 2.} A necessary and sufficient condition for
a closed set $E$ to be a U$_1$ set, is the following. %%??
There exists $\epsilon > 0$, such that for no integer
$N$ and real number $\delta>0$ is it true that for every
open set $O$ containing $E$, and every sequence $c_n > 0$,
$n \ne 0$, with $c_j < \delta$ for $j=1,\ldots, N$ and
$\sum c_n = 1$ there exists a finite number of points
in $O$, $x_1,x_2,\ldots, x_k$ and constants $\lambda_1,\ldots,\lambda_k$
such that $\sum \lambda_j = 1$ and
$$
\sum_{n\ne 0} c_n\, \left\sum_{j=1}^k \lambda_j e^{2\pi i n x_j}\right^2
< \epsilon\,.
\leqno(8)
$$
\smallskip\noindent
{\bf Remark:} Since the statement of the theorem is rather involved,
something should be said concerning the point of the theorem.
Equation (8) expresses the fact that the points $x_j$ are such
that $nx_j$ are evenly distributed for many values of $n$.
More specifically, the left side of $(8)$ is an average of quantities
which will be small if $nx_j$ are well distributed. Thus our
theorem might be paraphrased by saying that a set $E$ is a
U$_1$ set if one cannot find points arbitrarily close to $E$
having more or less random distributions modulo $\frac{1}{n}$ for many
values of $n$. The significance of the number $\delta$ is that one
may ignore what happens for small values of $n$.
\medskip\noindent
{\bf Proof.} The condition is necessary. If $E$ is a U$_1$ set
this means that there exists a sequence $f_m$ belonging to $W$,
each function vanishing on an open neighborhood $O_m$ of $E$,
and tending weakly to one. We may assume that
$$
f_m = 1 + \sum_{n\ne 0} c_n^{(m)} e^{2\pi i n x}
\leqno(9)
$$
while
$$
\sum_n c_n^{(m)} < A
$$
where $A$ is an absolute constant. Also, we have that
$c_n^{(m)}$ tends to zero for fixed $n$ as $m$ tends to
infinity. In the following we will drop the superscript $m$
when it is convenient. Let $\mu = \sum_n c_n$ and
set $\bar c_n = \frac{c_n}{\mu}$, so that $\sum \bar c_n = 1$.
Set
$$
g_1 = 1 + \sum_n \bar c_n e^{2\pi i n x}\, ,
\quad
g_2 = 1 + \sum_n \frac{c_n}{\sqrt{\bar c_n}}\, e^{2\pi i n x}\, ,
\leqno(10)
$$
and $\epsilon = \frac{1}{2 A^2}$. If the theorem were false
there would exist $N$ and $\delta$ as stated in the theorem.
Furthermore for some value of $m$ sufficiently large, the
sequence $c_n$ would satisfy the hypothesis $\bar c_n< \delta$
for $n \le N$. Hence there would exist $x_1,\ldots,x_k$,
lying in $O_m$, and $\lambda_1,\ldots,\lambda_k$ such that
$\sum \lambda_j = 1$ and
$$
\sum_{n\ne 0} \bar c_n \,
\left\sum_{j=1}^k \lambda_j e^{2\pi i n x_j}\right^2
< \epsilon\,.
\leqno(11)
$$
Now we put
$$
h(x) = \sum \lambda_j \, g_1(x + x _j)\,.
$$
Then we will have
$$
\int_0^1 g_2(x)\, h(x) dx = 0
$$
by Parseval's formula, remembering that $f_m$ vanishes at $x_j$
since they lie on $O_m$. On the other hand
$$
\int_0^1 g_2(x)\, h(x) dx
=
1 + \sum_{n\ne 0} \left(\sum_{j=1}^k \sqrt{\bar c_n}\, \lambda_j
e^{2\pi i n x_j}\right)\, \frac{c_n}{\sqrt{\bar c_n}}\,.
$$
By Schwarz's inequality this is greater in absolute value than
$$
1  \left[\epsilon \, \mu\, \sum c_n\right]^{1/2}
=
1  \mu \, \sqrt{\epsilon} > 0
$$
and hence we have a contradiction. Thus the necessity is proved.
\smallskip
The condition is sufficient. The condition means that there
exists an $\epsilon > 0$, and a sequence $c_n^{(m)}$, such
that $\sum_n c_n^{(m)} = 1$, and $c_n^{(m)}$ tends to zero
for a fixed $m$ as $n$ tends to infinity, and open sets $O_m$
containing $E$, such that
$$
\sum_n c_n^{(m)}\, \left\sum_{j=1}^k \lambda_j e^{2\pi i n x_j}\right^2
> \epsilon
\leqno(12)
$$
for all $x_j$ in $O_m$ where $\lambda_j$ are numbers such that
$\sum \lambda_j = 1$. Let
$$
f_m = 1 + \sum_n \sqrt{c_n^{(m)}}\, e^{2\pi i n x}\,.
\leqno(13)
$$
Clearly $f_m$ belongs to $L^2$ of the interval $[0, 1]$. Furthermore
$\int_0^1 f_m^2(x) dx = 2$. Let $M$ be the subspace of
$L^2$ consisting of all functions with constant term 1 in their
Fourier expansions. Again we drop the index $m$ where convenient.
Let $M_0$ be the subspace of $M$ generated by the functions
$f_m(x + t)$ where $t$ belongs to $O_m$. Let $g$ be that
element of $M_0$ such that $\int_0^1 g^2 dx$ is minimum.
Set $h= \alpha + \beta \, g(x)$ where $\alpha, \beta$ are so
chosen that
$$
\int_0^1 h\, \bar g dx = 0\,, \quad
\int_0^1 hdx = 1\,.
\leqno(14)
$$
This will be possible provided $\int_0^1 g^2 dx > 1$, for in
that case we take
$$
\alpha = \frac{\displaystyle \int_0^1 g^2 dx}
{\displaystyle 1  \int_0^1 g^2 dx}\,,
\quad
\beta = \frac{1}{\displaystyle 1  \int_0^1 g^2 dx}\,.
\leqno(15)
$$
Now the hypotheses of the theorem tell us that
$$
\int_0^1 g^2 dx > 1 + \varepsilon\,.
$$
Hence we see that
$$
\int_0^1 h^2 dx < A
$$
where $A$ depends only on $\epsilon$. On the other hand,
$$
\int_0^1 h(x)\, \bar j(x)dx= 0
$$
for any $j(x)$ belonging to $M_0$. For we have
$$
\int_0^1 (\alpha +\beta\, g)\, \bar j dx
=
\int_0^1 (\alpha +\beta\, g)\, \bar g dx
+
\int_0^1 (\alpha +\beta\, g)\, (\overline{jg}) dx \,.
\leqno(16)
$$
The first integral on the right is zero because of $(14)$,
whereas the second integral equals $\beta\, \int_0^1 g\,
(\overline{jg})dx$, which is zero by the minimal property of $g$.
Hence if
$$
h_m(x) = 1 + \sum_n d_n^{(m)}\, e^{2\pi i n x}\,,
$$
the function
$$
k_m(x) = 1 + \sum_n d_n^{(m)}\, \sqrt{c_n^{(m)}}\, e^{2\pi i n x}
\leqno(17)
$$
belongs to $W$, and has its norm in that ring, by Schwarz's inequality,
less than $1 + \sqrt{A}$. Furthermore since
$\int_0^1 h(x)\, \bar j(x) dx = 0$ for all $j$ in $M_0$ and in
particular for $f_m(x+ t)$ where $t$ lies in $O_m$, it follows
from Parseval's formula that $k_m(x) = 0$ for $x$ in $O_m$. Thus it
is clear that the sequences%%\footnote{Typo ``sequence'' in original text.}
$k_m$ tend weakly to one and vanish on neighborhoods of
$E$, so that $E$ is a U$_1$ set. The fact that the coefficients
of $k_m$ tend to zero as $m$ tends to infinity follows from the
fact that the $c_n^{(m)}$ do so and $d_n^{(m)}$ are all bounded
by $\sqrt{A}$.
\bigskip
Now we shall turn our attention to the $H^r$ sets which were
discovered by PyatetskiiShapiro. We need a preliminary
definition.
\medskip\noindent
{\bf Definition.} A sequence of $r$tuples of integers,
$n_i^j$, $1\le j \le r$, $1\le i< \infty$, is said to be
normal if whenever $a_j$, $1\le j \le r$, are integers not all
zero,
$$
\left\sum_j a_j n_i^j\right
\leqno(18)
$$
tends to infinity with $i$.
\smallskip
Then we have:
\medskip\noindent
{\bf Definition.} A set $S$ is an $H^r$ set if and only if there
exist $r$ intervals $L_1,\ldots,L_r$ on the interval $[0, 1]$
and a normal sequence of $r$tuples of integers $n_i^j$, such
that for all $x$ in $S$, and all $i$, there exists $j$ such
that $n_i^j x$ does not lie in $L_j$ modulo~1.
\smallskip
We now prove that the $H^r$ sets are sets of uniqueness. For
each $L_j$, let $\lambda_j(x)$ be a function in $W$ vanishing
everywhere except in an interval interior to $L_j$, and such
that the constant term of $\lambda_j(x)$ is 1. Let
$$
f_i(x) = \prod_{j=1}^r \lambda_j(n_i^jx)\,.
$$
These functions then vanish on a neighborhood of $E$, and their norms
as elements of $W$ are uniformly bounded by the product of the norms
of $\lambda_j$. Let
$$
\lambda_j(x) = \sum c_{\alpha_j}^{(j)} \, e^{2\pi i \alpha_j x}
\leqno(19)
$$
where $\alpha_1,\ldots,\alpha_r$ range from $\infty$ to $\infty$.
Then the constant term of $f_i(x)$
is\footnote{Original contained
$\sum_{n_i^{(1)}\alpha_1\ldots n_i^{(r)} \alpha_r = 0}
c_1^{(1)} \cdot \ldots \cdot c_r^{(r)}$.}
$$
\sum_{n_i^{(1)}\alpha_1+\ldots +n_i^{(r)} \alpha_r = 0}
c^{(1)}_{\alpha_1} \cdot \ldots \cdot c^{(r)}_{\alpha_r}\,.
\leqno{(20)}
$$
This last sum contains the term
$c_0^{(1)} \cdot \ldots \cdot c_0^{(r)} = 1$. For any fixed
choice of $\alpha_1,\ldots,\alpha_r$ not all zero there will
be an $i$ such that no term corresponding to these values of
$\alpha_j$ occurs in the sum $(20)$ because of the normality
condition $(18)$, for that value of $i$ or thereafter. Hence it
is clear that this constant term $(20)$ approaches one, and a
similar argument shows that the nonconstant terms approach zero.
Thus $f_i(x)$ tend weakly to one, and since they all vanish on
open sets containing $E$ by definition of an $H^r$ set, it
follows that $E$ is a set of uniqueness. Moreover $E$ is obviously
a U$_1$ set. The next theorem will have as its object to show that
in some sense $H^r$ sets are the simplest U$_1$ sets.
\medskip\noindent
{\bf Theorem 3.} Let
$f_m= \sum_n c_n^{(m)}\, e^{2\pi i n x}$, $c_0^{(m)} = 1$,
be functions in $W$, vanishing on a set $E$, such that
$f_m$ tends weakly to one. Assume that there exists
$\epsilon < 1$, and an integer $N$ such that for all $m$
sufficiently large we can find a set $J_m$ of $N$ indices
$n_1^{(m)},\ldots , n_N^{(m)}$, such that
$$
\sum_{n\ne 0, \ n \not \in J_m} c_n^{(m)} < \epsilon\,.
\leqno(21)
$$
Then $E$ is a finite union of $H^r$ sets.
\smallskip\noindent
{\bf Remark:} In the proof that $H^r$ sets are sets of uniqueness,
we constructed exactly such a sequence of elements of $W$. On the
other hand it is possible by taking appropriate linear combinations
of the functions in this sequence, to construct a sequence vanishing
on neighborhoods of a $H^r$ set, tending weakly to one, but
not satisfying the hypotheses of Theorem~3.
\medskip\noindent
{\bf Proof.} The elements of the sets $J_m$ are not uniformly bounded,
because, since $f_m$ tend weakly to one, this would imply that the
coefficients corresponding to the indices in $J_m$ tend to zero, and
hence by $(21)$ that $\sum_{n\ne 0} c_n^{(m)} < 1$ for $m$
sufficiently large. This in turn implies that the functions $f_m$
are never zero. Thus we see that after possible rearranging
$n_j^{(m)}$, we have that $n_1^{(m)}$ forms a normal sequence of
$1$tuples in the sense of our definition. Assume that after
further rearrangement $n_1^{(m)},\ldots, n_r^{(m)}$ form a
normal sequence of $r$tuples, while
$n_1^{(m)},\ldots, n_r^{(m)}, n_{r+1}^{(m)}$ do not form
a normal sequence of $r+1$tuples. By restricting ourselves to
a suitable subsequence, this implies that there exist integers
$a_{11},a_{12},\ldots,a_{1r}$, $b_1\ne 0$, $c_1$, such
that\footnote{Typo ``$n_r^{(m)}{}_1$'' in manuscript.}
$$
\sum_{j=1}^r a_{1j} n_j^{(m)} + b_1n_{r+1}^{(m)}
=
c_1\,.
$$
Now with this new sequence of $n_j^{(m)}$, we consider the sequence
of $r+1$tuples
$n_1^{(m)},\ldots, n_r^{(m)}, n_{r+2}^{(m)}$. If this
sequence is not a normal sequence of $r+1$tuples we proceed
exactly as before. If it is a normal sequence we adjoin
$n_{r+2}^{(m)}$ to $n_j^{(m)}$, $1 \le j \le r$. Thus eventually
we will find that for a suitable subsequence of our original
sequence and suitable rearrangements, there will be a number
$r$ and integers $a_{kj}$ where $1\le j\le r$, $1\le k \le Nr$,
$b_k\ne 0$, $c_k$, $1\le k \le Nr$, such that
$$
\sum_{j=1}^r a_{kj} n_j^{(m)} + b_k n_{r+k}^{(m)} = c_k\,.
\leqno(23)
$$
Now set
$$
B = b_1\cdot \ldots \cdot b_{Nr}\,,
\quad
\bar n_j^{(m)} = \left[\frac{n_j^{(m)}}{B}\right]\,,
\quad \bar a_{kj} = \frac{a_{kj}\, B}{b_k}\,.
\leqno(24)
$$
We have then
$$
\sum_{j=1}^r \bar a_{kj} \bar n_j^{(m)}
+ \bar n_{r+k}^{(m)} = \bar c_k^{(m)}
\leqno(25)
$$
where $\bar c_k^{(m)}$ depends on $m$ but remains bounded,
$\bar c_k^{(m)} \le K$. The sequence $\bar n_j^{(m)}$,
$1\le j \le r$, is clearly still a normal sequence. We shall
show now that the set $E$ is a finite union of $H^r$ sets,
each defined relative to some subsequence of this sequence. For
each integer $d$, let $I_1,\ldots, I_d$ be the $d$ consecutive
intervals of length $\frac{1}{d}$ which cover the unit interval.
If $E\cap I_1$ were not an $H^r$ set, then for each choice
of $\alpha_1,\ldots, \alpha_r$ from among the set of positive
integers less than or equal to $d$, the relations
$$
\bar n_1^{(m)}x\in I_{\alpha_1},\ldots,\bar n_r^{(m)}\in I_{\alpha_r}
\leqno(26)
$$
hold for some $x$ contained in $E\cap I_1$, for some $m$
sufficiently large, since otherwise $E\cap I_1$ would be an
$H^r$ set. There are $d^r$ possible choices of the numbers
$\alpha_1,\ldots, \alpha_r$, and so for each $m$ sufficiently
large we have a set of $d^r$ points which we denote by $S_m$,
belonging to $I_1\cap E$ and satisfying $(26)$. Our next object
will be to show that the sums
$$
\frac{1}{d^r}\, \sum_{x\in S_m} e^{2\pi i n_j^{(m)} x}\,,
\quad 1\le j \le N\,,
$$
are quite small.
First we have
$$
\left
\frac{1}{d^r}\, \sum_{x\in S_m} e^{2\pi i n_j^{(m)} x}

\frac{1}{d^r}\, \sum_{x\in S_m} e^{2\pi i \bar n_j^{(m)} Bx}
\right
< \left1  e^{\frac{2\pi i B}{d}}\right\,,
\leqno(27)
$$
remembering that $I_1$ consists of the interval
$\left[0, \frac{1}{d}\right]$ and $x$ belongs to $I_1$. For
each $\bar n_j^{(m)} x$ where $x$ belongs to $S_m$, has
$d^{r1}$ points in each of the intervals $I_1,\ldots,I_d$.
Therefore we have that the average value of the function
$e^{2\pi i Bx}$ taken over these points differs from the integral
extended over the interval $[0, 1]$ by at most
$$
\mathop{{\rm Max}}_{xy\le \frac{1}{d}}\,
\lefte^{2\pi i Bx}  e^{2\pi i By}\right\,.
$$
Hence, since the integral of $e^{2\pi i Bx}$ is zero, we obtain that
$$
\left\frac{1}{d^r}\, \sum_{x\in S_m} e^{2\pi i B\bar n_j^{(m)} x}
\right
\le \left1e^{\frac{2\pi i B}{d}}\right\,,
\leqno(28)
$$
or, from $(27)$
$$
\left\frac{1}{d^r}\, \sum_{x\in S_m} e^{2\pi i n_j^{(m)} x}
\right
\le
2\, \left1e^{\frac{2\pi i B}{d}}\right
\le 2\, \frac{2\pi B}{d}\,.
\leqno(29)
$$
Now, if $j = r+ k $, we have\footnote{Typo ``$\bar n_r^{(m)}{}_k$''
in manuscript.}
$\bar n_{r+k}^{(m)} = \bar c_k^{(m)}
(\bar a_{k1}\bar n_{1}^{(m)} +\cdots + \bar a_{kr} \bar n_r^{(m)})$.
Assume that $a_{k1}\ne 0$. Consider any set of indices
$\alpha_2,\ldots, \alpha_r$. Let $T$ denote the set of $d$ points
in $S_m$ for which
$$
\bar n_2^{(m)}x \in I_{\alpha_2},\ldots,\bar n_r^{(m)} x \in I_{\alpha_r}\,.
\leqno(30)
$$
Then the points
$$
\bar c_k^{(m)} x  (\bar a_{k2}\bar n _2^{(m)} + \ldots +
\bar a_{kr} \bar n_r^{(m)})\, x
$$
all lie within distance
$$
\frac{\bar c_k^{(m)} + \bar a_{k2}+\cdots + \bar a_{kr}}
{d} \le \frac{A}{d}
$$
of each other and hence of a single point $x_0$. Here $A$ denotes
an absolute constant. Hence
$$
\left\frac{1}{d}\, \sum_{x\in T}e^{2\pi i \bar n_{r+k}^{(m)} x}
\frac{1}{d}\, \sum_{x\in T}
e^{2\pi i (\bar a_{k1}\bar n_1^{(m)} x  x_0)}
\right
\le \frac{2\pi A}{d}\,.
\leqno(31)
$$
But, exactly as above, since the points $\bar n_1^{(m)}x$ lie
successively in each of the intervals $I_1,\ldots,I_d$, we have
$$
\left\frac{1}{d}\, \sum_{x\in T}
e^{2\pi i \bar a_{k1} \bar n_1^{(m)}x}\right
\le \frac{2\pi \bar a_{k1}}{d}\,.
\leqno(32)
$$
Since $S_m$ consists of $d^{r1}$ sets each exactly like our
set $T$, it follows that by combining $(31)$ and $(32)$ we obtain
$$
\left\frac{1}{d^r}\, \sum_{x\in S_m} e^{2\pi i n_j^{(m)} x}
\right \le \frac{K}{d}
\leqno(33)
$$
where $K$ is some constant, independent of $m$ and $d$.
Now the theorem follows easily. Since for all points of the set
$S_m$, $f_m(x)$ equals zero, we have
$$
\leqalignno{
0&= \frac{1}{d^r}\, \sum_{x\in S_m}\sum_n c_n^{(m)} e^{2\pi i nx}
= 1 + \sum_{n\not \in J_m}\frac{1}{d^r}
\sum_{x\in S_m} c_n^{(m)} e^{2\pi i n x}
& (34)
\cr\noalign{\vskip 5pt}
&\qquad + \sum_{n\in J_m}\sum_{x\in S_m} \frac{1}{d^r}
\, c_n^{(m)} e^{2\pi i n x}\,.
\cr
}
$$
The first sum on the right is bounded by $1\epsilon$ in
absolute value, while the second sum does not exceed
$\sum_n c_n^{(m)}\, \frac{K}{d}$. Hence we have
$$
1\le (1\epsilon) + \frac{K}{d}\, \sum_n c_n^{(m)}
\leqno(35)
$$
which is clearly impossible if $d$ is chosen large enough.
\medskip
In the next theorem the symbol $E$ will be used to denote the
measure of $E$.
\medskip\noindent
{\bf Theorem 4.} Let $E_n$ be a sequence of closed sets in
$[0, 1]$, such that $E_n > \delta > 0$. Assume that for each fixed
interval $I$,
$$
\frac{I\cap E_n}{I}  E_n
\leqno{(36)}
$$
tends to zero as $n$ tends to infinity. Let $F_n$ denote the set
of all $x$ such that $E_n+x$, which means $E_n$ translated by $x$,
does not not intersect $E_n$. Then any closed set contained in the
intersection of all the $F_n$ is a U$_1$ set.
\medskip\noindent
{\bf Proof.} Let $\chi_n(x)$ be the characteristic function of the
set $E_n$. Set
$$
f_n(x) = \frac{\chi_n(x)}{E_n}\,.
$$
$f_n$ then has constant term one in its Fourier expansion.
Since each $f_n(x)$ belongs to $L^2$, and $\ f_n\_2 =
\frac{1}{\sqrt{\delta}}$, it follows that
$g_n(x) = f_n(x) \star f_n(x)$, where $\star$ denotes convolution,
lies in $W$, and its norm in that ring does not exceed
$\frac{1}{\delta}$. For any $k\ne 0$, and $\epsilon > 0$, let
$d$ be so large that
$e^{2\pi i k x}  e^{2\pi i k y}  < \epsilon$ if
$xy < \frac{1}{d}$. Let $I_1,\ldots,I_d$ be as in the proof
of Theorem~3. Pick $n$ so large that
$$
\left\frac{I_\alpha \cap E_n}{I_\alpha}
E_n\right < \epsilon,
\quad 1\le \alpha \le d\,.
\leqno(37)
$$
Then we have
$$
\left\int_{I_\alpha\cap E_n} e^{2\pi i k x dx} dx

\frac{I_\alpha \cap E_n}{I_\alpha}\,
\int_{I_\alpha} e^{2\pi i k x} dx\right < \epsilon\, I_\alpha\,,
\leqno(38)
$$
or by $(37)$,
$$
\left\int_{I_\alpha\cap E_n} e^{2\pi i k x dx} dx

E_n \int_{I_\alpha} e^{2\pi i k x} dx\right
< 2\epsilon\, I_\alpha\,.
$$
Summing over all $\alpha$, we obtain
$$
\left\int_{E_n} e^{2\pi i kx}\right < 2\epsilon\,.
$$
It follows that the functions $g_n(x)$ tend weakly to one.
On the other hand we have
$$
g_n(x) = \int f_n(t) \, f_n(xt) dt\,,
$$
so it is clear that $g_n(x)$ vanishes on $F_n$. Hence the
theorem is proved.
We now remark that those $H^r$ sets which are defined with
respect to normal sequences where the ratio of successive terms
in each $r$tuple tends to infinity, obviously satisfy the
hypotheses of our theorem~4. The choice of the set $E_n$ is rather
obvious.
\bigskip\noindent
{\bf 4.} Now we shall consider the question of whether there
exist $H^r$ sets which are not countable unions of $H^{r1}$ sets.
We will first prove that the complement of an $H^{r1}$ set has
a certain metric property, and then we will find an $H^r$ set
whose complement does not have this property. We need
some preliminary definitions and lemmas.
\medskip\noindent
{\bf Definition:} Let $S$ be any set in $[0, 1]$. For
$c,\epsilon> 0$, we say that an interval $I$ is of type
$(1, c, \epsilon)$ relative to $S$, if to every point $x$
belonging to $I$, there exists an interval contained in $S$
of length $\delta$, with its midpoint at $m$, such that
$$
xm < \epsilon , \quad \frac{\delta}{xm} > c
\leqno(39)
$$
both hold. Similarly, we say that an interval $I$ is of type
$(r, c, \epsilon)$ relative to $S$, if to every point
$x$ belonging to $I$, there exists an interval of type
$(r1, c, \epsilon)$ relative to $S$, of length $\delta$, with
its midpoint at $m$, such that
$$
xm < \epsilon , \quad \frac{\delta}{xm} > c\,.
$$
Finally, if the interval $[0, 1]$ itself is of type $(r, c, \epsilon)$
relative to $S$, we say that $S$ is of type $(r, c, \epsilon)$.
\smallskip
Roughly speaking, this definition says that a set is of type
$1$, if every point is close to a relatively long interval of the
set. It is of type~2, if every point is close to a relatively long
interval of points, each one of which is close to a long interval
of the set, etc. Such sets as those in the definition occur in the
definition of $H^r$ sets, and the purpose of Theorem~5, is precisely
to prove that a certain set if of type $(r, c, \epsilon)$.
We first have a lemma.
\medskip\noindent
{\bf Lemma.} Let $L_i$ be the intervals $(h_i, h_i + d)$
where $0\le h_i < 1$, $0 < d < 1$, $1\le i \le r$.
There exist constants $B(d)$ and $C(d)$ depending only on
$d$, as follows: For any $\epsilon > 0$, there exists an
integer $N$ such that if
$$
s_1 = \frac{p_1}{n}\,, \; s_2 = \frac{p_2}{n}\, \ldots,
\; s_r = \frac{p_r}{n}
$$
are $r$ rational numbers, with $s_i\le 1$, and having
the property that for all integers $a_j$, $0\le j \le r$, not
all zero, satisfying $a_j\le N$, we have
$$
\lefta_0 n + \sum_{j=1}^r a_j p_j \right > N\,,
\leqno(40)
$$
then there exists an integer $q \le B(d)$ such that the set
of all intervals of the form $\left[\frac{k}{n}, \frac{k+1}{n}\right]$
where $k$ runs through all the integers satisfying
$kqs_i \in L_i$~(modulo~1), $1\le i \le r$, forms a set of
type $(r, C(d), \epsilon)$.
\medskip\noindent
{\bf Proof.} In the statement of the lemma, the number $N$ in
reality depends upon both $d$ and $r$. Therefore, at times we
shall denote it by $N(r, \epsilon, d)$. The proof proceeds
by induction on $r$. We assume the lemma true for $r1$.
Set $C' = C\left(\frac{d}{4}\right)$ and $B'= B\left(\frac{d}{4}\right)$,
where we consider these quantities defined for the case $r1$.
Let $\epsilon$ be an arbitrary positive quantity. Then if we set
$N' = N\left(\epsilon, \frac{d}{4}\right)$, again for the case
$r1$, we shall prove the lemma for the case $r$ with the following
determination of constants,
$$
\leqalignno{
&C = {\rm Min}\, \left(\frac{d}{32 B'}, C'\right)\,
\quad
B = \left(\frac{8B'}{d} + 1\right)\, B'\,,
&(41)
\cr\noalign{\vskip 5pt}
&\qquad N = {\rm Max}\, \left(\frac{5}{\epsilon}, r\, (B+1)N'\right)\,.
&\cr
}
$$
Therefore, assume that $s_i = \frac{p_i}{n}$ are $r$ rational
numbers which satisfy the condition $(40)$ of the lemma where
$N$ is given by $(41)$. Thus in particular for no $k$ satisfying
$$
k\le \left(\frac{8B'}{d} + 1\right)^r
\leqno(42)
$$
do we have
$$
k\, (s_1,s_2, \ldots, s_r) = 0\; ({\rm modulo}\; 1)\,,
\leqno(43)
$$
by which we mean that the corresponding vector does not have
all its components integral. This is so since otherwise
$k\, s_1 = t$, where $t$ is an integer and both $k$ and $t$
are smaller than $N$, so that we have $kp_1  tn = 0$ which
violates $(40)$. Now we apply the well known box principle of
Dirichlet. We divide the unit interval up into
$\frac{8 B'}{d} + 1$
equal intervals. If we do this for
each axis in $r$dimensional space, we will have subdivided
the $r$dimensional cube into at most
$\left(\frac{8 B'}{d} + 1\right)^r$ cubes.
Now, consider all the vectors $k\, (s_1,\ldots,s_r)$, where
$k$ satisfies $(42)$. There must exist two values of $k$
such that the corresponding vectors lie in the same cube.
Their difference then will be an integer $k_1$ such that
$$
k_1 \le \left(\frac{8 B'}{d} + 1\right)^r\, ,
\quad
k_1 s_i = \alpha_i + E_i\,,
\quad
\alpha_i  \le \frac{d}{8B'}\,,
\leqno(44)
$$
and $E_i$ are integers. Now let $\alpha_1$ be the largest of
the $\alpha_i$ in absolute value, and assume that $\alpha_1$
is positive.
Let $C$ denote the cube in $r1$ dimensional space defined by
the following inequalities,
$$
\eqalign{
&{\rm if}\ \; \alpha_i > 0\,,
\quad h_i  \frac{h_1}{\alpha_1}\, \alpha_i + \frac{d}{4}
\le y_i \le y_i  \frac{h_1}{\alpha_1}\, \alpha _i + \frac{d}{2}\,,
\cr\noalign{\vskip 5pt}
&{\rm if}\ \; \alpha_i < 0\,,
\quad h_i  \frac{h_1}{\alpha_1}\, \alpha_i + \frac{d}{2}
\le y_i \le y_i  \frac{h_1}{\alpha_1}\, \alpha _i + \frac{3d}{4}\,,
\cr
}
\leqno(45)
$$
where $y_2,\ldots, y_r$ are the variables. If $(y_2,\ldots, y_r)$
lies in the cube $C$, and $\lambda$ is an integer satisfying
$$
h_1 + \frac{d}{8}\le \alpha_1 \lambda \le h_1 +\frac{3d}{8}\,,
\leqno(46)
$$
then it will follow that
$$
\lambda\, (\alpha_1,\ldots, \alpha_r)
+(0, y_2,\ldots,y_r)
\leqno(47)
$$
lies in the cube $L'$ which is defined as the direct product of
the intervals $L_i' = \left(h_i + \frac{d}{8}, h_i + \frac{7d}{8}\right)$.
This is so since the first component of the vector $(47)$
lies in the interval $L_i'$ by $(46)$, while if $\alpha_i$ is
greater than zero we have
$$
h_1\, \frac{\alpha_i}{\alpha_1}\le \lambda\, \alpha_i
\le h_1\, \frac{\alpha_i}{\alpha_1} + \frac{3d}{8}
\leqno(48)
$$
since $\alpha_i  \le \alpha _1$ and for $\alpha_i < 0$ we
have
$$
h_1\, \frac{\alpha_i}{\alpha_1}  \frac{3d}{8}\le \lambda\, \alpha_i
\le h_1\, \frac{\alpha_i}{\alpha_1}\, .
\leqno(49)
$$
Thus comparing $(48)$, $(49)$, and $(46)$ we see that $(47)$
lies in the cube $C$. Now set
$$
\beta_i = \frac{\alpha_i}{\alpha_1} =
\frac{k_1p_i  E_i n}{k_1p_1  E_1 n} = \frac{p_i'}{n'}\,,
\quad
2\le i \le r\,,
\leqno(50)
$$
where
$$
n' = k_1 p_1  E_1 n \, , \quad
p_i' = k_1 p_i  E_i n\,.
$$
Now if $a_i$, $1\le i \le r$, are $r$ integers not all zero and
such that $a_i\le N'$, where we recall that $N'$ was defined
as $N'\left(r1, \epsilon, \frac{d}{4}\right)$, we have
$$
\left\sum_{i=2}^r a_i p_i' + a_1 n'\right
=
\left\sum_{i=1}^r a_i k_1 p_i  n \, \left(\sum_{i=1}^r
E_i a_i\right)\, \right\,.
\leqno(51)
$$
Using the inequality $E_i\le B+1$, we see that we have a linear
combination of $p_i$ and $n$ with coefficients less than or equal
to $r(B+1)\, N'$ which does not exceed $N$. Hence the quantity
in $(51)$ exceeds $N$ and therefore $N'$. We also note that
$\beta_i\le 1$ and that the length of each side of the cube
$C$ is $\frac{d}{4}$. Thus applying our lemma to these numbers and
the cube $C$ for the case $r1$, we deduce that there exists a number
$q'\le B'$ such that the set of intervals of the form
$\left[\frac{k'}{n'}, \frac{k'+1}{n'}\right]$
where $k'$ ranges over all integers such that
$q'k'\left(\frac{\alpha_i}{\alpha_1}\right)$ lies in the cube
$C$ modulo~1 for $2\le i \le r$, forms a set of type
$(r1, C',\epsilon)$. Remembering that $C\le C'$, it follows
that to prove the lemma we must merely show that each one
of the intervals
$\left[\frac{k'}{n'}, \frac{k'+1}{n'}\right]$
is of type $(1, C, \epsilon)$ relative to the original set mentioned
in the lemma. Let us then consider a particular $k'$ such that
$q'k'\left(\frac{\alpha_i}{\alpha_1}\right)$ lies in $C$
modulo~1. Set $m = \left[\frac{k'}{\alpha_1}\right]$.
By what was said above concerning the vector in $(47)$, it follows
that if an integer $\lambda$ satisfies
$$
h_1 + \frac{d}{8} \le \alpha_1 q'\, \lambda \le h_1 +\frac{3d}{8}\,,
\leqno(52)
$$
then the vector
$$
q'\, \left(\lambda + \frac{k'}{\alpha_1}\right)\,
(\alpha_1,\ldots, \alpha_r)
\leqno(53)
$$
lies in the cube $L'$ modulo~1. The vector $(53)$, however, differs
in every component from the vector
$$
q'\, (\lambda + m)\, (\alpha_1,\ldots,\alpha_r)
\leqno(54)
$$
by at most $q'\, \alpha_i\le \frac{d}{8}$. Hence the vector
$(54)$ lies in the cube $L$~modulo~1, where $L$ is defined as the
direct product of the intervals $L_i$. Now set $q = q'\, k$, so
that we have $q\le B$, and all intervals of the form
$\left[\frac{\lambda+m}{n}, \frac{\lambda+m+1}{n}\right]$ belong
to the original set described in the lemma. The condition
$(52)$ on $\lambda$ can be rewritten as follows:
$$
\frac{1}{\alpha_1\, q'}\, \left(h_1+\frac{d}{8}\right)
\le
\lambda \le
\frac{1}{\alpha_1\, q'}\, \left(h_1+\frac{3d}{8}\right)\,.
$$
Obviously then, there is a string of consecutive such
$\lambda$ numbering at least\footnote{Original: ``$\alpha$'' missing
on right hand side.}
$$
\frac{d}{4\alpha_1\, q'}  1\ge \frac{d}{8\alpha_1\, q'}
$$
since
$$
\frac{d}{4\alpha_1\, q'}\ge 2\,.
$$
The corresponding intervals
$\left[\frac{\lambda+m}{n}, \frac{\lambda+m+1}{n}\right]$
which we know belong to the original set of the lemma, form
on large interval of length at least
$\frac{d}{8\alpha_1\, q'n} = \frac{d}{8\, q'n'}$.
This large interval is contained in
$\left[\frac{m}{n}, \frac{m+1}{n} + \frac{2}{n'q'}\right]$
since $h_1 + \frac{3d}{8} < 2$.
In turn, this interval is of length at most $\frac{3}{n'}$.
We also have
$\frac{k'}{n'} \ge \frac{m}{n} \ge \frac{k'}{n'}  \frac{1}{n}$.
Thus we conclude that the distance from any point in
$\left[\frac{k'}{n'}, \frac{k'+1}{n'}\right]$
to the midpoint of our block of intervals is at most
$\frac{4}{n'}$. Now $n' \ge N$ so that this distance is smaller
than $\epsilon$, and the ratio of the length of the block
of intervals to this distance is at least
$\frac{d}{32q'} \ge \frac{d}{32 B'}$. Thus the lemma is
proved for the case $r$ under the assumption that it holds
for the case $r1$. Notice that at this last step if we took each
point in the interval
$\left[\frac{k'}{n'}, \frac{k'+1}{n'}\right]$
and examined its distance to the block of intervals of the original
set which we associated with those values of $\lambda$ such that
$1 + h_1 +\frac{d}{8} \le \alpha_1 \, q'\, \lambda \le
h_1 +\frac{3d}{8} + 1$, then we could modify the argument very
slightly to show that with a different choice of $C(d)$ we can assume
that all the intervals which occur in the definition of a set of
type $(r, c, \epsilon)$ occur to the right. We will need this
remark later. Thus, it only remains to prove the lemma in the case
$r = 1$.
This is very simple. If $L_1$ is the interval $(h, h+d)$, we set
$C(d)$ equal to $\frac{d}{4}$, and $B(d)$ equal to
$\frac{20}{d} + 1$. Finally for $\epsilon > 0$ we set
$N$ equal to ${\rm Max}\, \left(\frac{2}{\epsilon},B + 1\right)$.
Then if $s = \frac{p}{n}$, $s < 1$, we have as above for no
$k \le B(d)$ is $ks \equiv 0\;({\rm modulo}\; 1)$, so that
there exists an integer $k\le B(d)$, such that
$ks = \alpha + E$, where $E$ is an integer and
$\alpha< \frac{d}{20}$. Also $E < B+1$. We have
$\alpha = \frac{kpEn}{n} = \frac{n'}{n}$, and $n' \ge N$
since $k$ and $E$ do not exceed $N$. Let $v$ be any integer.
There exists a consecutive sequence of integers $\lambda$
satisfying both $\frac{v}{\alpha}\le \lambda\le \frac{v+1}{\alpha}$
and $h\le \lambda\alpha \le h+d\;({\rm modulo}\; 1)$ numbering
at least $\frac{d}{2\alpha}$ terms. If we set $q = k$, intervals
of the form
$\left[\frac{\lambda}{n}, \frac{\lambda+1}{n}\right]$
will be of the desired type, and the total length of these
intervals will e at least as great as
$\frac{d}{2\alpha n}= \frac{d}{2n'}$. The distance from
the midpoint of this block of intervals to any point in the
interval
$\left[\frac{v}{\alpha n}, \frac{v+1}{\alpha n}\right]$
is at most $\frac{2}{n'}$, using again the fact that $n > n'$.
Thus we see that our choice of $C$ and $B$ satisfy the conditions
of the lemma, since for some $v$ every point is contained in
an interval of the form
$\left[\frac{v}{\alpha n}, \frac{v+1}{\alpha n}\right]$.
\medskip\noindent
{\bf Theorem 5.} Let $n_j^i$, $j=1,\ldots, r$ be a normal
sequence of $r$tuples. Let $L_1,\ldots, L_r$ be the
intervals $[h_j, h_j + d]$. Then there exists a constant
$C(d)$ depending only on $d$, such that if $\epsilon > 0$,
there exists an integer $N$ such that for $i > N$ the set
$S_i = \{x\, \; n_j^i x\in L_j\; {\rm mod}\; 1, \ {\rm all}\ j\}$
is a set of type $(r, C(d), \epsilon)$.
\medskip\noindent
{\bf Proof.} We shall drop the superscript in $n_j^i$ without risk
of confusion. Let $n_1 > n_2 > \cdots > n_r$. If for $j= 2,\ldots, r$
and for some integer $k$, we have
$$
h_j  \frac{3}{4}\, d\, \frac{n_j}{n_1}
\le
\frac{h_1 + k}{n_1}\, n_j
\le
h_j + d  \frac{n_j}{n_1}\, d\ ({\rm modulo}\; 1)\,,
\leqno(55)
$$
then the interval
$A_k = \left[\frac{k+h_1}{n_1} + \frac{3}{4}\, \frac{d}{n_1}\,,
\frac{k+h_1}{n_1} + \frac{d}{n_1}\right]$
will belong to the set $S$. We see this because if $x$ lies in
this interval then $n_1 x$ lies in
$\left[h_1 + \frac{3}{4}\, d, h_1 + d\right]\;({\rm modulo}\; 1)$,
and $n_j x$ for $j\ne 1$ lies in
$[h_j, h_j + d]\; {\rm modulo}\; 1$.
We notice that the intervals $A_k$ are of length at least
$\frac{d}{4n_1}$, and that every point of the interval
$\left[\frac{k}{n_1}, \frac{k+1}{n_1}\right]$
is no more than $\frac{2}{n_1}$ distant from its midpoint.
This means that the interval
$\left[\frac{k}{n_1}, \frac{k+1}{n_1}\right]$
is of type $\left(1, \frac{d}{8}, \frac{2}{n_1}\right)$
relative to $S$. The condition on $k$ in $(55)$ says that
$k\, \frac{n_j}{n_1}$ lies in a certain interval modulo~1 of length
at least $\frac{3}{4}\, d$, remembering that $n_1> n_j$. We may thus
apply our lemma to the numbers $\frac{n_j}{n_1}$. We then conclude
that there exists an integer $q \le B(d)$, such that the intervals
of the form
$\left[\frac{k}{n_1}, \frac{k+1}{n_1}\right]$
where $qk$ satisfies $(55)$, is of type $(r1, C, \epsilon)$.
This follows because the hypotheses of the lemma will be satisfied
for any $N$ if we go out far enough in the sequence of $r$tuples,
since the sequence is normal. Therefore we may say that the
intervals
$\left[\frac{qk}{n_1}, \frac{qk+1}{n_1}\right]$
are of type $\left(1, \frac{d}{8}, \frac{2}{n_1}\right)$
relative to $S$. Now, if a set $S$ is of type
$(r, C,\epsilon)$ then the set $qS$ is easily seen to be of
type $(r, C, q\epsilon)$. Thus the intervals
$\left[\frac{qk}{n_1}, \frac{qk+1}{n_1}\right]$
are of type $(r1, C, q\epsilon)$. Each such interval is of
type $\left(1 , \frac{d}{8q}, \frac{q+2}{n_1}\right)$
relative to $S$ since the interval
$\left[\frac{qk}{n_1}, \frac{qk+1}{n_1}\right]$
is of type
$\left(1 , \frac{d}{8q}, \frac{2}{n_1}\right)$.
Thus if we insure that $n_1$ is sufficiently large we will
have that $S$ is of type $(r, C', \epsilon)$,
where $C'$ is a new constant which depends only on $d$.
Thus the theorem is proved. Now we have the main theorem of
this section:
\medskip\noindent
{\bf Theorem 6.} There exists an $H^r$ set which is not contained
in a countable union of $H^{r1}$ sets.
\medskip\noindent
{\bf Proof.} We remark first that every $H^{r1}$ set is trivially
also an $H^r$ set. For if a give $H^{r1}$ set is defined relative
to the intervals $L_1,\ldots, L_{r1}$ and the sequence
$n_j^i$, $1\le j \le r1$, then it is also defined relative to the
sequence $n_j^i$, $1\le j \le r$ where $n_r^i$ is taken to be a
sufficiently rapidly increasing sequence so that the sequence
of $r$tuples is still normal, and $L_r$ is taken to be
the interval $[0, 1]$.
We shall now define a set $S$, which satisfies the statement of
the theorem. Let $a_1^k < a_2^k \cdots < a_r^k$ be a sequence
of increasing positive integers such that
$$
a_2^k  a_1^k \to \infty, \ldots,a_r^ka_{r1}^k \to \infty
\leqno(56)
$$
as $k$ tends to infinity. We define the set $S$ as the set of all
$x$ such that for each $k$, not all the points
$$
3^{a_1^k}\, x, 3^{a_2^k}\, x, \ldots, 3^{a_r^k}\, x
$$
lie in the open interval $\left[\frac{1}{3}, \frac{2}{3}\right]$.
Clearly $S$ is an $H^{r1}$ set, so that by an application of
the Baire category theorem we need only prove the following:
If $x\in S$, and $I$ is an interval containing $x$, no
$H^{r1}$ set can contain $S\cap I$.
Let $x$ and $I$ be such a point and interval. Every number $z$
can be expanded in the ternary system,\footnote{In original
``$\infty$'' missing.}
$$
z = \sum_{k=1}^\infty \frac{\theta(k)}{3^k}
\leqno(57)
$$
where $\theta(k)$ takes only the values $0, 1, 2$. Unless the expansion
of $z$ ends in all zeroes or all twos, then saying that
$3^n\, z$ lies in the interval
$\left[\frac{1}{3}, \frac{2}{3}\right]$,
is equivalent to asserting that
$\theta(n+1) = 1$. So excluding the exceptional numbers mentioned
of which there are only a countable number, to say that
$z$ lies in $S$ means exactly that for no $k$ do we have
$$
\theta(a_1^k + 1) = \theta(a_2^k + 1) = \ldots = \theta(a_r^k + 1) = 1\,.
\leqno(58)
$$
Even if the number $z$ does terminate in zeroes then the above
condition implies that $z$ is in the set $S$. Therefore if we
modify our original $x$ by setting all $\theta(k) = 0$ for
$k$ sufficiently large, we may still assume that it lies in
$S\cap I$, and that it terminates in zeroes. Assume further
that $x$ is the midpoint of $I$, and that $I$ is of length
$2\delta$. Let $T$ be an $H^{r1}$ set which contains $S\cap I$.
Then by our theorem, we know that there exists a constant
$C$, such that the complement of $T$, which we denote by $U$,
is a set of type $(r1, C, \epsilon)$ for arbitrary $\epsilon$.
Let $n$ be an integer having certain properties which we shall
specify later on, and also having the property that for
$k\ge a_1^n$, $\theta (k) = 0$. Since $U$ is of type
$(r1,C, \epsilon)$, there exists an interval $I_1$ with
midpoint $m_1$ and length $\delta_1$ such that
$$
xm_1< \epsilon\,, \quad
\frac{\delta_1}{xm_1} > C\,,
\leqno(59)
$$
and $I_1$ is an interval of type $(r2, C, \epsilon)$ with respect
to $U$. By the remark which was made in the course of the proof
of the lemma, we can
assume that $I_1$ lies to the right of $x$. By choosing $\epsilon$
small enough we can insure %%%???
that the interval $I_1$ lies completely within our original
interval $I$. As a matter of fact, we can choose $\epsilon$
so small that each one of the $r1$ intervals which we shall
choose all lie in the original interval $I$. Now since
$m_1 > x$, it must agree with $x$ in its ternary expansion
up to the place
$$
\left[\frac{\logxm_1}{\log 3}\right]  1\,.
\leqno(60)
$$
Now if we assume that
$$
r\in < 3^{a_1^n  2}\,, %%??
\leqno(61)
$$
this place will be one beyond which $x$ terminates in zeroes.
Define $z_1$ to be the number which agrees with $m_1$ up to
the place
$$
\left[\frac{\log \frac{1}{2}}{\log 3}\right] + 1\,,
$$
beyond which $z_1$ terminates in zeroes. Let us call the range
of places between
$$
\left[\frac{\logxm_1}{\log 3}\right]  1
\quad
\left[\frac{\log \frac{1}{2}}{\log 3}\right] + 1
$$
$R_1$. Now because $I_1$ is of length $\delta_1$, $z_1$ lies in
the interval $I_1$. Also because of $(59)$, the number of places
in $R_1$ is at most some bounded quantity which depends only on $C$.
Now there must exist an interval $I_2$ with midpoint $m_2$ and
of length $\delta_2$, which is of type $(r2,C, \epsilon)$ with
respect to $U$ and such that
$$
\frac{\delta_2}{z_1m_2} > C, \quad
z_1m_2 < \epsilon\,.
\leqno(62)
$$
Again we may assume that\footnote{Original contains here
``$m_2\ z_1\cdot z_1$ and $m_2$''.}
$z_1$ and $m_2$ %%%??
will then agree in their ternary expansions up to the place
$$
\left[\frac{\logz_1m_2}{\log 3}\right]  1 \,.
$$
Define $z_2$ to agree with $m_2$ up to the
place%\footnote{Original contains ``$\log \frac{2}{2}$''.}
$$
\left[\frac{\log \frac{2}{2}}{\log 3} \right]+ 1
$$
and be zero beyond that. Then again, the number $z_2$ lies in
the interval $I_2$, and this number agrees with the number $x$
in all but possibly two ranges of places $R_1$ and $R_2$,
the second being defined as all places lying between
$$
\left[\frac{\logz_1m_2}{\log 3}\right]  1
\quad {\rm and}\; \
\left[\frac{\log \frac{2}{2}}{\log 3}\right] + 1\,.
$$
Because of $(62)$, both ranges have at most a bounded number
of places in them. Proceeding in this manner, we eventually
obtain a number $z$, which lies in $U$, and which agrees
with $x$ in all but $r1$ ranges of places each of length
bounded by a number depending only on $C$. Now if $n$ were
chosen so large that the difference\footnote{Original
contains typo ``$a_j^n \ {}_1$''.}
$a_{j+1}^n  a_j^n$ was always greater than this constant,
for all $j$, it would follow that this number $z$ would lie in
$S$, since the conditions $(58)$ could only occur for at most
$r1$ of the $a_j^n$. Hence $z$ is in $S$ and simultaneously
in the complement of $T$, so that we have a contradiction. Therefore
the theorem is proved.
\bigskip\noindent
{\bf 5.} It is not a priori obvious from the definition that an
$H^r$ set for $r> 1$ is a set of measure zero. Of course, once we
know that such a set is a set of uniqueness then it follows that
it must be a set of measure zero. For $r=1$ it is clear that
an $H^1$ set is a set of measure zero, for we constantly remove
from the unit interval a set of fixed measure which is more and
more evenly dispersed, so that we decimate any fixed in the set.
For $r > 1$ the situation is clarified by the following theorem:
\medskip\noindent
{\bf Theorem 7.} Let $n_j^k$, $1\le j \le r$ be a sequence
of $r$tuples such that for any $r$ integers $a_j$, not all
zero, the sum $\sum_{j=1}^r a_j n_j^k$ is zero for only a finite
number of values of $k$. Then for $f_j$, $1\le j\le r$, bounded
measurable functions, we have
$$
\int_0^1 f_1(n_1^kx)\, f_2(n_2^kx) \ldots f(n_r^k x) dx
\to
\prod_{j=1}^r \int_0^1 f_j(x) dx\,.
\leqno(63)
$$
\medskip\noindent
{\bf Proof.} Assume first that $f_j$ are finite exponential polynomials.
Then the expression on the left of $(63)$ will consist of the
product of the constant terms of $f_j$ if $k$ is sufficiently large.
This is so because of the condition on $n_j^k$. This is exactly what
the right side is, so that the theorem is true in this case.
In the general case, we may approximate each $f_j$ by an
exponential polynomial in the $L^1$ norm, such that the maximum
of each polynomial is not more than the maximum of the
corresponding $f_j$. Then it follows that the right and left
hand sides of $(63)$ approach the corresponding expressions
for the $f_j$ as the approximation becomes closer. Here we must
use the fact that $nx$ is a measurepreserving transformation for
all $n$. Hence the theorem is proved in this case also.
\medskip
Now, assume that $n_j^k$ are a normal sequence of $r$tuples.
This means precisely that the $r+1$tuples
$1,n_1^k, \ldots,n_r^k$ satisfy the condition of our theorem.
Thus if $I$ is an arbitrary interval and $L_1,\ldots, L_r$
are intervals of length $\delta$ each, then the set
$S_k= \{x\, \; x\in I, \ \, n_j^k x \in L_j\, , \ {\rm all}\ j\}$
has measure approaching $\delta^r\, I$. This we see by applying
the theorem to the characteristic functions of these intervals.
Thus if $E$ is an $H^r$ set defined by the normal sequence
$n_j^k$ and the intervals $L_j$, then if $U$ is any open set
containing $E$, for some $k$ each interval in $U$ is decimated
by the corresponding $S_k$ to the extent that $\delta^r$ times
the measure of $U$ lies in the complement of $E$. This clearly
implies that $E$ is of measure zero.
\chapter{}
In this section we shall prove a theorem concerning multiple
trigonometric series. In the theory of trigonometric series
in one variable, the famous CantorLebesgue theorem states
that if a trigonometric series converges on a set of
positive measure, then the coefficients must tend to zero.
No complete analogue of this theorem can exist, since we
must first specify a particular method of summation when
we are dealing with the case of several variables.
Shapiro in [9] has obtained results which would give an
almost complete answer to the question of the uniqueness of
multiple series, if some kind of analogue could be proved.
It appears that no results along this line have been published.
The theorem which we shall prove is much weaker than the
corresponding CantorLebesgue theorem. The terms of a multiple
series are in one to one correspondence with the set of all
$r$tuples of integers. A method of summation is described
by a sequence $E_n$, of finite sets of $r$tuples of integers,
such that $E_n$ is contained in $E_{n+1}$ and the union of
$E_n$ consists of all $r$tuples. At the $n^{{\rm th}}$ stage
of the summation we consider the sum of all terms corresponding
to $r$tuples contained in $E_n$. Then we have the following
definition.
\medskip\noindent
{\bf Definition.} A method of summation $E_n$, for a trigonometric
series in $r$ variables, is said to be regular if and only if
there exists a constant $K$ such that for every lattice point
$a=(a_1,\ldots,a_r)$, there exists $n_0$ such that $a$
belongs to $E_{n_0}$ and the maximum of the absolute value of
all coordinates of all lattice points in $E_{n_0}$ is smaller
than $K\, \displaystyle\mathop{{\rm Max}}_{1\le i \le r} a_i$.
\medskip
It is clear that the usual methods of spherical and cubical summation
are regular. Now we have our theorem.
\medskip\noindent
{\bf Theorem.} Let
$$
\sum_{j_1,\ldots,j_r=\infty}^\infty c_{j_1\ldots j_r}
e^{i\, (j_1x_1+\ldots + j_r x_r)}
\leqno(1)
$$
be a trigonometric series in $r$ variables, which converges
almost everywhere by a regular method of summation. Then for
all $\gamma > 1$, there exists $b > 0$, such that
$$
c_{j_1\ldots j_r} \le b\, \gamma^{j}
\quad {\rm where} \; \
j = \displaystyle\mathop{{\rm Max}}_{1\le i \le r} j_i\,.
\leqno(2)
$$
\medskip\noindent
{\bf Proof.} The variables $x_1,\ldots, x_r$ range over the
reals modulo~$2\pi$, which we consider as being canonically
identified with the reals modulo~1.
By applying Egoroff's theorem to the partial sums
$$
s_n = \sum_{j\in E_n} C_j\, e^{i\, j\cdot x}\,,
$$
where $j = (j_1,\ldots, j_r)$, $x = (x_1,\ldots, x_r)$, and
$j\cdot x = j_1 x_1 + \ldots + j_r x_r$, we find that for any
$\epsilon > 0$, there exists a closed set $G$, of measure
greater than $1\epsilon$, such that $s_n$ are uniformly
bounded in absolute value by some constant $B$, for all $x$
in $G$. Let $a = (a_1,\ldots, a_r)$ be a lattice point. Let
$E_{n_0}$ be the set which corresponds to it in the definition
of regularity. Set
$N = 2 \displaystyle \mathop{{\rm Max}}_{{\scriptstyle j\in E_n}\atop
{\scriptstyle 1\le i \le r}}
j_i + 2$.
Then $N \le K \displaystyle \mathop{{\rm Max}}_{1\le i \le r} a_i$.
The set $G$ can be visualized as being contained in the
$r$fold direct product $T^r$ of the unit circle.
Because the measure of $G$ is more than $1\epsilon$,
it follows that there is a set $G'$, in the space
$T^{r1}$ of measure greater than $1\epsilon^{1/2}$
such that if $(x_2,\ldots, x_r)$ belongs to $G'$, the
set of all $x_1$ such that $(x_1,x_2,\ldots, x_r)$
belongs to $G$ is of measure also greater than
$1 \epsilon ^{1/2}$. We now need the following simple lemma.
\medskip\noindent
{\bf Lemma.} Let $S$ be a set on the unit circle of measure
greater than $\delta$. Then for any integer $k > 0$, there
exist $k$ points in $S$, $z_1,\ldots , z_k$, such that the
distance between any two of them is greater than $\frac{\delta}{k}$.
\medskip\noindent
{\bf Proof.} Let $z_1$ be any point in $S$. Then for any $\eta > 0$,
let $z_2$ be a point in $S$, if it exists, such that
$z_2 > z_1 +\frac{\delta}{k}$ and the measure of the set of
all points in $S$ lying in the interval $[z_1,z_2]$ is
smaller than $\frac{\delta}{k} +\eta$. Here $[z_1, z_2]$
denotes the set of all points lying to the right of $z_1$
and to the left of $z_2$. Choose $z_3$ to be a point in $S$,
if it exists, which has the same relationship to $z_2$, as
$z_2$ had to $z_1$. Now assume that the process terminates
after $m$ steps, that is, no $z_{m+1}$ can be found having the
appropriate property. This means that the only elements of
$S$ lying in the interval $[z_m, z_1]$ are contained
in $\left[z_m, z_m +\frac{\delta}{k}\right]$. Thus it follows
that the measure of $S$ is smaller than
$m\, \left(\frac{\delta}{k} +\eta\right)$, unless $m=k$. This shows
that $z_1,\ldots, z_k$ can be so chosen in $S$ that they are placed
cyclically around the circle, and the distance between two
consecutive points is greater than $\frac{\delta}{k}$.
This prove the lemma.
\medskip
Now consider $s_{n_0}\, (x_1,\ldots,x_r)$ as a function
of $x_1$ alone, holding $x_2,\ldots,x_r$ fixed.
By multiplying $s_{n_0}\, (x_1,\ldots, x_n)$ by a
suitable power of $e^{ix}$, we obtain a trigonometric
polynomial
$\bar s_{n_0}\, (x_1,\ldots,x_r)$
which contains only positive powers of $e^{ix_1}$ and is of
degree smaller than $N$. Set $\zeta = e^{x_1}$, so that
$\bar s_{n_0}$ can be regarded as a polynomial of degree
smaller than $N$ in the complex variable $\zeta$. For each
$(x_2,\ldots, x_n)$ in $G'$, $\bar s_{n_0}$ is still bounded in
absolute value by $B$ on the points $(x_1,\ldots,x_n)$ where
$x_1$ belongs to a set $S$ of measure greater than
$1\epsilon^{1/2}$. Thus by the lemma there are $N$ points
$z_1,\ldots, z_N$ in $S$, such that the distance between
any two of them is greater than
$\lambda = \frac{1\epsilon^{1/2}}{N}$.
Set $\zeta_k = e^{iz_k}$. Now, if we apply the Lagrange interpolation
formula to $\bar s_{n_0}$, it follows that
$$
\bar s_{n_0} (\zeta)
=
\sum_{k=1}^N \bar s_{n_0} (\zeta_k)\,
\frac{\prod_{j\ne k}\, (\zeta  \zeta_j)}
{\prod_{j\ne k}\, (\zeta_k  \zeta_j)}\,.
\leqno(3)
$$
The denominator of each term in $(3)$ obviously exceeds the
corresponding product where the $\zeta_j$ are $m^{{\rm th}}$
roots of unity, and $m = \left[\frac{1}{\lambda}\right]+1$.
Now, the product $\prod_{\theta}(1\theta)$, where the product
is extended over all the $m^{{\rm th}}$ roots of unity different
from $1$, equals $m$. On the other hand, in the denominator which
occurs in $(3)$, only $N1$ terms occur in the product. Thus
there are $mN$ terms which are additional, and each one of
these terms is bounded in absolute value by 2. Hence we obtain
$$
\prod_{j\ne k} (\zeta_k\zeta_j) \ge \frac{m}{2^{mN}}
\ge \frac{N}{2^{\delta N}}
\leqno(4)
$$
where $\delta$ is a quantity which goes to zero with $\epsilon$.
Now we estimate the numerators which occur in $(3)$. We have
$$
\left\prod_{j\ne k}\, (\zeta  \zeta_j)\, \right
\le
\left\prod_\theta (1 \theta)\, \right\,,
\leqno(5)
$$
where the product is now extended over those $N1$ $m^{{\rm th}}$
roots of unity which are furthest from $1$. The same product
extended over the remaining roots is bounded from below by
$$
\left\prod_{k=1}^{\delta\, m} (1e^{\frac{2\pi i k}{m}})^2\, \right
\ge \frac{1}{2^{2\, \delta\, m}}\,
\left(\frac{1}{m}\cdot \frac{2}{m}\ldots \frac{[\delta\, m]}{m}\right)^2\,,
\leqno(6)
$$
where $\delta$ as before tends to zero with $\epsilon$. The right
hand side of $(6)$ can be estimated by Stirling's formula. It follows
easily that the right side of $(6)$ is greater than
$2^{\delta\, N}$, where $\delta$ here is again a quantity which
tends to zero with $\epsilon$. Hence it follows from $(5)$ that
$$
\left\prod_{j\ne k}\, (\zeta  \zeta_j)\, \right
\le N\, 2^{\delta\, N}\,.
\leqno(7)
$$
Now $\bar s_{n_0}(\zeta_j)\le B$, so that from $(3)$ it follows
that $\bar s_{n_0}(\zeta)$ is bounded by $b\, \gamma^N$, where
$b$ depends only on $B$, and $\gamma$ is a number which tends to
one as $\epsilon$ tends to zero. Thus it follows that the
coefficients of $\bar s_{n_0}(\zeta)$ are bounded by
$b\, \gamma^N$. These coefficients are functions of
$x_2,\ldots, x_n$, and are bounded by $b\, \gamma^N$
whenever $(x_2, \ldots, x_n)$ lies in $G'$, which is a
set of measure greater than $1 \epsilon^{1/2}$. Now if we
apply the same argument as above to each of these functions
we find that they in turn are bounded by $b_1\, \gamma_1^N$
where $\gamma_1$ also tends to one as $\epsilon$ tends to
zero. If we apply this process $r$ times we eventually obtain
the result that the coefficient $c_a \le b\, \gamma^N$
for suitable choices of $b$ and $\gamma$. Since
$N \le K \displaystyle \mathop{{\rm Max}}_{1\le i \le r} a_i$,
it follows that the theorem holds after a suitable redefinition
of $\gamma$ and $b$.
\bigskip\noindent
{\bf 2.} Because our theorem allows the coefficients to grow quite
rapidly, it may be of some interest to construct an example of a
series where the coefficients grow at a reasonably rapid pace.
However, we cannot prove that our theorem is a best possible result.
Given any function $\omega(n)$ which tends to zero as $n$
tends to infinity, we shall now show that there exists a double
trigonometric series with the property that it converges almost
everywhere, and yet for some sequence of coefficients $m_k$ and
$n_k$ we have $C_{m_k,n_k} > \omega(N_k)\, N_k$ where
$N_k = {\rm Max}\, (m_k, n_k)$. Let $K_n(t)$ denote the
Fejer kernel, that is,
$$
K_n(t) = \frac{1}{2\, (n+1)}\,
\left(\frac{\sin(n+1)\frac{1}{2}\, t}{\sin \frac{1}{2}\, t}\right)^2\,.
\leqno(8)
$$
Then constant coefficient of $K_n(t)$ is 1, and $K_n(t)$ is a
trigonometric polynomial of degree $n$. Let $n_k$ be an
increasing sequence of integers, and $d_k$ positive numbers
such that $\sum_{k=1}^\infty \frac{d_k}{n_k}$ converges and
$d_k > \omega(n_k)\, n_k$. Then it is clear that the series
$\sum_{k=1}^\infty d_k\, K_{n_k}(x)$ converges absolutely at
all points except for $x=0$. Hence it follows that the
corresponding double series,
$$
\sum_{k=1}^\infty d_k\, e^{i n_ky} \, K_{n_k}(x)
\leqno(9)
$$
converges at all points where $x\ne 0$, if we sum the series
by the method of square summation. More precisely, we take as
the $n^{{\rm th}}$ partial sum all those terms of degree not
greater than $n$ in either $x$ or $y$. On the other hand it is
quite clear that the coefficients tend to infinity rapidly as
was indicated.
\chapter{}
\noindent
{\bf 1.} The first example of a trigonometric series which converges
to zero almost everywhere, and which is not identically zero, was
given by Mensov. He constructed a measure on a perfect set of
Lebesgue measure zero and considered its FourierStieltjes series.
Riemann's localization theorem then tells us that such a series
converges to zero on the complement of the support of the
measure provided only that the coefficients of this series
tend to zero. An exposition of Mensov's result can be found either
in [2] or [13]. The particular set of multiplicity, or Mset,
which is thus constructed is very similar to the usual Cantor
set. The Cantor set is constructed by removing from the unit
interval the middle third. From each of the remaining segments
one removes the middle third and so on. What remains is precisely
the Cantor set. If at each stage, instead of using the fraction
one third, we use the number $\alpha_i$, where $0<\alpha_i<1$, we
obtain a more general class of sets. Mensov's example is exactly
when the $\alpha_i$ tend to zero, while $\sum \alpha_i =\infty$.
Salem [8], investigated the case of equal $\alpha_i$ and proved
the remarkable theorem that unless this common ratio belongs to
a certain denumerable class of algebraic numbers, the set is
an Mset, whereas in the contrary case it is a Uset.
In his proof he used a formula for the FourierStieltjes coefficients
of a certain measure which originated with Carleman. This formula
is valid in the case of Mensov's example, but the proofs of Mensov's
result referred to above do not use this explicit formula.
Therefore, it seems of interest to give a proof which unifies the
treatment of these two cases, and seems to be conceptually much
clearer. We begin by deriving the formula which was referred to.
\medskip
Let $\xi_i$ be a sequence of real numbers such that $0< \xi_i<\frac{1}{2}$.
Let $\mu_i$ be the measure which assigns mass $\frac{1}{2}$ to the
point $0$ and mass $\frac{1}{2}$ to the point
$\xi_1\ldots \xi_{i1}(1\xi_i)$. Now set
$\nu_j = \mu_1 \star \mu_2 \ldots \star \mu_j$ where $\star$
denotes convolution. Set $\alpha_i = 12\xi_i$, and let
$S_j$ be the collection of intervals which remain after
$j$ dissections as described above, using the ratios $\alpha_i$.
Then clearly $S_j$ consists of $2^j$ intervals and the measure
$\nu_j$ assigns mass $\frac{1}{2^j}$ to the left hand endpoint
of each of these intervals. As $j$ tends to infinity,
$\nu_j$ tends weakly to a measure $\nu$ which has its support
on the set $S=\cap_{j=1}^\infty S_j$. This measure has been
constructed entirely analogously to the ordinary Cantor function.
Now let $C_n$ be the Fourier coefficients of $\nu$, that is,
$$
C_n = \int_0^1 e^{2\pi i n x}\, d\nu(x)\,.
\leqno(1)
$$
Since $\nu$ is an infinite convolution, $C_n$ is the infinite
product of the corresponding coefficients for each of the measures
$\mu_k$. So\footnote{In original ``$(1)^n$'' missing.}
$$
\leqalignno{
C_n &=
\prod_{k=1}^\infty \left(\frac{1+e^{2\pi i n \xi_1\ldots \xi_{k1}
(1\xi_k)}}{2}\right)
&(2)
\cr\noalign{\vskip 5pt}
&=\prod_{k=1}^\infty e^{\pi i n \xi_1\ldots \xi_{k1}(1\xi_k)}\,
\prod_{k=1}^\infty \cos\pi n \xi_1\ldots \xi_{k1}(1\xi_k)
&
\cr\noalign{\vskip 5pt}
&= (1)^n\, \prod_{k=1}^\infty \cos\pi n \xi_1\ldots \xi_{k1}(1\xi_k)\,,
&(3)
\cr
}
$$
Since the first factor is obviously $(1)^n$. Now under the assumption
that $\xi_k$ tends to $\frac{1}{2}$, we shall prove that the
product $(3)$ tends to zero as $n$ tends to infinity.
Set $\theta_k = \xi_1\ldots \xi_{k1}(1\xi_k)$. Then
clearly $\frac{\theta_{k+1}}{\theta_k}$ tends to $1/2$.
In the case where ${\xi_i} = \frac{1}{2}$, then clearly
the measure $\nu_j$ consists of $2^j$ equally spaced masses
of mass $\frac{1}{2^j}$ each. The Fourier coefficients
of this measure are easily computed and we may express
the result in the following equation.
$$
\cos x\, \cos\frac{x}{2}\ldots \cos\frac{x}{2^k}
=
\frac{\sin 2x}{2^{k+1}\, \sin\frac{x}{2^k}}\,.
\leqno(4)
$$
This formula may also be easily verified by induction on $k$.
Now if we assume that $\frac{x}{2^k} < \frac{1}{10}$, it
follows that
$$
\left \cos x\, \cos\frac{x}{2}\ldots \cos\frac{x}{2^k} \right
\le \frac{1}{x}\,.
\leqno(5)
$$
Let $A > 0$ and $m$ an integer such that $\frac{3A}{2^m} < \frac{1}{10}$.
Then the expression
$$
\left\prod_{\ell = 0}^m \cos \frac{x}{2^\ell}
 \prod_{\ell = 0}^m \cos \frac{\theta_{k+\ell}}{\theta_k}\, x
\right
\leqno(6)
$$
where $x$ lies between $A$ and $3A$ approaches $0$ uniformly in
$x$ as $k$ tends to infinity and $m$ is kept fixed. This is true
because the quantity $\frac{\theta_{k+\ell}}{\theta_k}$ approaches
$\frac{1}{2}$ for fixed $\ell$ as $k$ tends to infinity.
Now if $n$ is large enough there is a value of $k$ such that
$A 2\, m$,}
\cr
}
$$
where $C$ is some absolute constant.
Assume that the measure $\mu$ has Fourier coefficients $C_n$, and
$\sum_{n=\infty}^\infty C_n\le B$. Then the
measure $\mu(\xi^{1}(x))$ is a measure whose Fourier coefficients
we denote by $\tilde C_m$. We have then
$$
\tilde C_m =
\int_0^1 e^{2\pi i m x}\, d\mu(\xi^{1}(x))
=
\int_0^1 e^{2\pi i m \xi(x)}\, d\mu(x)\,.
\leqno(10)
$$
Now the Fourier series for $e^{2\pi i m \xi(x)}$ converges absolutely,
so that in $(10)$ we may integrate term by term to obtain
$$
\tilde C_m = \sum_n \lambda_{n, m} C_m
=
\sum_{n \ge 2 \, m} \lambda_{n, m} C_n
+
\sum_{n < 2 \, m} \lambda_{n, m} C_n\,.
\leqno(11)
$$
The first sum in $(11)$ is less than $\alpha_1\, B\, m^{1/2}$,
while the second is less than $\frac{\alpha_2}{m}\, V$, where
$V$ is the total variation of $\mu$ and $\alpha_1$ and
$\alpha_2$ are constants.
Let $\Omega(N)$ be an arbitrary positive function tending monotonically
to infinity. We now propose to construct a set of measures
$\mu_k$ with the following properties:
\smallskip
a) the total variation of $\mu_k$ is not greater than $k$,
\smallskip
b) if $\mu_k$ has the Fourier series
$\sum_{n=\infty} ^\infty d_{n, k}\, e^{2\pi i n x}$,
then
$$
\sum_{n=\infty}^\infty d_{n, k} \le \omega(k)\,,
\leqno(12)
$$
where $\omega(k)$ tends monotonically to infinity.
\smallskip
c) For each $k$, let $p = p(k)$ be a positive integer such that
$\frac{\alpha_2\, k}{q} < \frac{\Omega(q)}{q^{1/2}}$
for all $q \ge p$, and $p(k+1) > p(k)$. We also demand that
$\omega(k)$ satisfy $\alpha_1\, \omega(k) < \Omega(p)$.
This can be achieved by choosing $p$ large enough. Set
$$
\tilde d_{m, k} = \sum_n \lambda_{n, m}\, d_{n, k}\,.
\leqno(13)
$$
Then $(9)$, $(11)$ and a) imply that there exists $M$ and
$\epsilon$, such that
$$
d_{n, k}  d_{n, k+1} < \epsilon
\quad
{\rm for}\ \; n< M
\leqno(14)
$$
implies
$$
\tilde d_{m, k}  \tilde d_{m, k+1}
<
\frac{\Omega(m)}{2^k\, m^{1/2}}
\quad
{\rm for}\ \;
m \le p(k+1)\,.
\leqno(15)
$$
This is so because in the second summand of $(11)$,
$\sum_{n> 2\, m} \lambda_{n, m}$
convergent sequence %%???
while $d_{n, k}$ are bounded by $k$. We now assume that $\mu_k$
satisfy condition $(14)$.
\smallskip
Under these conditions let $m$ be an integer, $k_1$ such that
$p(k_1) \le m\le p(k_1+1)$. Then by $(11)$,
$$
\tilde d_{m, k_1}
\le
\frac{\alpha_1\, \omega(k_1)}{m^{1/2}}
+\frac{\alpha_2\, k_1}{m} < \frac{2\, \Omega(m)}{m^{1/2}}\,.
\leqno(16)
$$
On the other hand, $(15)$ implies that for all $k> k_1$ we have
$$
\tilde d_{m, k_1}  \tilde d_{m, k}
<
\frac{\Omega(m)}{m^{1/2}}
\leqno(17)
$$
so that $\tilde d_m =
\displaystyle \mathop{{\rm Lim}}_{k\to \infty} \tilde d_{m, k}$
is such that
$$
\tilde d_m <
\frac{3\, \Omega(m)}{m^{1/2}}\,.
\leqno(18)
$$
We also observe that since $(16)$ holds for $m> p$, it
follows that
$\tilde d_{m, k} \le \frac{3\, \Omega(m)}{m^{1/2}}$.
In particular $\tilde d_{m, k}$ are uniformly bounded
if\footnote{In original ``$\Omega$'' missing.}
$\Omega(N)$ tends to infinity slowly enough.
Now we shall proceed to construct a sequence of sets $S_k$. let
$d_1,d_2,\ldots$ be a sequence of integers tending to infinity.
Let $S_1$ be the union of all intervals of the form
$\left[\frac{r}{d_1}, \frac{r}{d_1} + \frac{1}{4\, d_1}\right]$
for some integer $r$. Then the measure of $S_1$ is onefourth.
$S_1$ has the property that the set $S_1+S_1$, which means all
points of the form $x+y$ where $x$ and $y$ belong to $S_1$, is
the union of all intervals of the form
$\left[\frac{r}{d_1}, \frac{r}{d_1} + \frac{1}{2\, d_1}\right]$
and is of measure $\frac{1}{2}$. $S_1$ thus consists of the unit
interval with $d_1$ intervals removed. Let $S_2$ consist of all the
intervals of $S_1$, from each one of which $d_2$ intervals have been
removed in precisely the same manner. Then the measure of $S_2$
will be $\frac{1}{16}$ while that of $S_2+S_2$ will be $\frac{1}{4}$.
We proceed in this manner with $d_3, d_4,\ldots$, to construct
the sets $S_k$. The measure of $S_k$ is $\frac{1}{4^k}$ while that
of $S_k+S_k$ is $\frac{1}{2^k}$.
Now measures $\nu_k$ will be defined inductively. Each interval of
$S_k$ has the property that $\nu_k$ assigns to the first half of
it a multiple of Lebesgue measure and to the second half another
multiple of Lebesgue measure. Now each half interval of $S_k$
contains certain intervals of $S_{k+1}$. $\nu_{k+1}$ assigns
to each of these intervals identical distributions of mass
so that the sum of the measures of these intervals is the same
as that of the original half interval. However this is done in
such a manner that the {\it total\/} variation is
$\frac{k+1}{k}$ times that of the original interval. We first
observe that $\nu_{k+1}$ will have any finite number of its
Fourier coefficients arbitrarily close to the corresponding
coefficients of $\nu_k$ provided $d_{k+1}$ is chosen large enough.
For, as $d_{k+1}$ approaches infinity it is obvious that
$\nu_{k+1}$ approaches $\nu_k$ weakly. In particular it follows
that if $F_{k+1}(x)$ and $F_k(x)$, defined as the integrals of
$\nu_{k+1}$ and $\nu_k$ respectively, i.e.,
$$
F_{k+1}(x) = \int_0^x d\nu_{k+1}\,,
\quad
F_{k}(x) = \int_0^x d\nu_{k}\,,
$$
then $F_{k+1}(x)  F_k(x)$ can be made uniformly small. We can
thus assume that all the $F_k(x)$ are uniformly bounded. This
remark will be needed presently.
The quantity
$\displaystyle\mathop{{\rm Max}}_{\delta> 0}\, \delta\, \eta(\delta)$
also does not depend upon the choice of the sequence $d_k$.
This is so because this quantity is the maximum ``density''
of the measure and only depends upon the Lebesgue measure of
$S_k$. Now set $\mu_k = \nu_k\star \nu_k$. Then by the lemma we have
$$
\sum_{n = N}^N d_{n, k} \le A\cdot V\cdot N\cdot
\eta\left(\frac{1}{N}\right)\,.
\leqno(19)
$$
Thus it follows that by choosing $d_k$ large enough we can be certain
that $\mu_k$ satisfy all the properties mentioned above.
The support of $\mu_k$ is contained in $S_k+S_k$. The intersection
of these sets is a closed set of measure zero. Now I claim that
the series $\sum_n \tilde d_n e^{2\pi i n x}$ converges to
zero everywhere outside the transform of this set by the
transformation $\xi(x)$. For it is certainly true that the series
$\sum_n \tilde d_{n, k} e^{2\pi i n x}$ converges to zero outside the
transform of $S_k+S_k$. Now the assertion follows from the following
lemma:
\medskip\noindent
{\bf Lemma.} If the sequence of trigonometrical
series
$\sum_n \tilde d_{n, k} e^{2\pi i n x}$ converges to zero
in an open interval, and $d_{n, k}$ are uniformly bounded, if
$\displaystyle\tilde d_n =
\mathop{{\rm Lim}}_{k\to \infty}\tilde d_{n, k}$
exist, then
$\sum_n \tilde d_n e^{2\pi i n x}$ converges to zero in that
interval, provided
$\displaystyle\mathop{{\rm Lim}}_{k\to \infty}\tilde d_n = 0$.
\medskip\noindent
{\bf Proof.} This is a trivial consequence of formal multiplication.
If $\gamma(x)$ converges rapidly, is nonzero at a point of the
interval, zero outside it, then the formal product of $\gamma(x)$
with each of the series
$\sum_n \tilde d_{n, k} e^{2\pi i n x}$ is identically zero.
The hypotheses imply that the product of $\gamma(x)$ and
$\sum_n \tilde d_n e^{2\pi i n x}$ is zero. Therefore the
conclusion of the lemma follows from wellknown theorems
concerning formal multiplication.
Since $\xi(x)$ is piecewise differentiable it follows that
$\sum_n \tilde d_n e^{2\pi i n x}$ converges almost
everywhere to zero. It remains to prove that
$\sum_n \tilde d_n e^{2\pi i n x}$ is not a FourierStieltjes
series. Let $F_k(x)$ be the integral of the measure
$\nu_k(x)$, that is, $\nu_k(x)= dF_k(x)$. All the functions
$F_k(x)$ are uniformly bounded provided we take the sequence
$d_k$ growing quickly enough. The functions $F_k(x)$ converge
boundedly to a function $F(x)$ of infinite variation.
The Fourier series of $F(x)$ is the integrated series
of $\sum_n \tilde d_n e^{2\pi i n x}$.
Since $F(x)$ is of infinite variation it follows that
$\sum_n \tilde d_n e^{2\pi i n x}$ is not a FourierStieltjes
series.
Thus we have shown that there exist series which converge to zero
almost everywhere and which are not FourierStieltjes series.
Furthermore the coefficients can be made smaller than
$\frac{\Omega(m)}{m^{1/2}}$ where $\Omega(m)$ is any
function tending to infinity.
\bigskip\noindent
{\bf 3.} In the preceding section we quoted a theorem of Wiener
concerning the Fourier series of measures. In this section we shall
present a generalization of the theorem to several dimensions by
means of a rather simple proof which seems more lucid than the
one in [13]. Also a generalization in a different direction will
be given.
\medskip\noindent
{\bf Theorem.} Let $\mu$ be a measure on the $n$dimensional
torus $T^n$, with Fourier series
$$
d\mu\sim \sum_j C_j \, e^{2\pi i \, (j\cdot x)}
\leqno(20)
$$
where $j$ ranges over all $n$dimensional lattice points
$j = (j_1,\ldots, j_n)$, $x = (x_1,\ldots, x_n)$
and $j\cdot x = j_1x_1+\ldots + j_n x_n$. Let $S_p$ be an
expanding sequence of rectangles in the space of all lattice
points, and $S_p$ denote the number of lattice points contained
in $S_p$. Then
$$
\mathop{{\rm Lim}}_{p\to \infty} \frac{1}{S_p}\,
\sum_{j\in S_p} C_j^2
\leqno(21)
$$
exists and equals $\sum_Q \mu(Q)^2$ where $Q$ ranges
over all points having nonzero mass.
\medskip\noindent
{\bf Proof.} We assume $S_p$ is the set of all lattice points
$j$, such that $m_{p, k}\le j_k \le m_{p, k}$,
$1\le k\le n$. We assume that $m_{p, k}$ tends to infinity
as $p$ tends to infinity. Set
$$
\eqalign{
f_p(x_1,\ldots, x_n)
&= \frac{1}{S_p} \sum_{j\in S_p} e^{2\pi i \, (j\cdot x)}
\cr\noalign{\vskip 5pt}
&= \prod_{k=1}^n \frac{1}{2\, m_{p, k} + 1}
\sum_{m_{p, k}\le j_k \le m_{p, k}} e^{2\pi i j_k x_k}\,.
\cr
}
\leqno(22)
$$
It is clear that $f_p \le 1$, and since $f_p$ is a product
of Dirichlet kernels, it follows from the formula for the sum
of a geometric series that $f_p$ converges to 0 everywhere
except at $(0,0,\ldots,0)$ where it is 1.
It is also clear that
$$
\frac{1}{S_p} \sum_{j\in S_p} C_j^2
=
\mathop{\int\!\!\int}_{T^n\times T^n}\, f_p(xy)\, d\mu(x)d\bar \mu(y)\,,
\leqno(23)
$$
as may be seen by direct substitution. By the Lebesgue monotone
convergence theorem it follows that the limit of $(21)$
exists and equals
$$
\mathop{\int\!\!\int}_{T^n\times T^n}\, \delta(xy)\, d\mu(x)d\bar \mu(y)\,,
\leqno(24)
$$
where $\delta(x)$ is zero for $x\ne 0$, $\delta(0) = 1$. This
quantity by Fubini's theorem is clearly $\sum_Q \mu(Q)^2$.
Now we shall restrict ourselves to the case of one variable.
Let $\mu$ be a measure which has no point masses, i.e.,
a continuous measure. If $d\mu$ has the Fourier series
$\sum_n C_n\, e^{2\pi i n x}$, let us examine the quantity
$\frac{1}{N}\, \sum_{k=1}^N C_{n_k}^2$,
where $n_k$ is some increasing sequence of integers. Then
this expression is equal to
$$
\frac{1}{N}\,
\mathop{\int\!\!\int}_{T^1\times T^1}\, f_N(xy)\, d\mu(x)d\bar \mu(y)\,,
\leqno(25)
$$
where
$$
F_N(x) = \frac{1}{N}\, \sum_{k=1}^N e^{2\pi i n_k x}\,.
$$
As before the $f_N(x)$ are unformly bounded by 1 in
absolute value. If the $n_k$ are a polynomial sequence,
i.e., $n_k$ is a polynomial in $k$, then wellknown
results of Weyl [11], tell us that $f_N(x)$ approaches
zero for all irrational values of $x$. Hence $(25)$
approaches $(24)$ where $\delta(x)$ is now a function
which is zero at all irrational points. Since there are only
a countable number of rational points and $\mu$ is assumed
continuous we have the following result:
\medskip\noindent
{\bf Theorem.} If $\mu$ is a continuous measure, then
$$
\frac{1}{N} \, \sum_{k=1}^N C_{n_k}^2
$$
approaches zero if $n_k$ is a polynomial sequence.
\chapter{}
Green's theorem in two dimensions says that if $C$ is a simple
closed curve bounding the region $Q$, if $A(x, y)$ and $B(x, y)$
are continuous functions having derivatives, then under
suitable further conditions we have,
$$
\int_C A d + B dy
=
\mathop{\int\!\!\int}_Q \left(\frac{\partial B}{\partial x}
\frac{\partial A}{\partial y}\right)\, dxdy\,,
\leqno(1)
$$
where the line integral is taken in a positive sense around
the curve $C$. In [3], Bochner investigated under which
conditions $(1)$ holds. There it was shown that if $A$ and
$B$ have certain regularity properties and if the integrand
on the right of $(1)$ behaves well, then $(1)$ does hold.
Here we shall prove $(1)$ under what may be regarded as the
weakest possible hypotheses. This question was also treated
by Shapiro in [10], and though he assumes certain regularity
of $A$ and $B$, namely, the existence of the differential,
he allows certain exceptional sets which we cannot allow. The
proof of our theorem is modeled after the proof of the LoomanMensov
theorem as contained, for example, in [6]. We will not deal
with the topological difficulties involved so that our theorem
will only treat the case in which $Q$ is a rectangle.
\medskip\noindent
{\bf Theorem.} Let $A(x, y)$ and $B(x, y)$ be two functions
defined on the rectangle $Q$, and continuous on the closure of $Q$.
Assume further that the partial derivatives
$$
\frac{\partial A}{\partial x}\;\
\frac{\partial A}{\partial y}\;\
\frac{\partial B}{\partial x}\;\
\frac{\partial B}{\partial y}
$$
exist everywhere in the interior of $Q$, except perhaps at a
countable number of points. If
$\frac{\partial B}{\partial x}\frac{\partial A}{\partial y}$
is Lebesgue integrable in the rectangle $Q$, then $(1)$ holds.
\medskip\noindent
{\bf Proof.} We first need a lemma which is contained in [6].
In the following, the word ``rectangle'' means the direct
product of two intervals.
\medskip\noindent
{\bf Lemma.} Let $w(x, y)$ be a function defined in a square
$Q$, such that $\frac{\partial w}{\partial x}$
and $\frac{\partial w}{\partial y}$ exist almost everywhere
in $Q$. Let $F$ be a closed, nonempty set in $Q$, and
$N$ a finite constant such that
$$
\eqalign{
&w(x, y + k) w(x, y)\le N\, k\,,
\cr\noalign{\vskip 5pt}
&w(x+h, y ) w(x, y)\le N\, h\,,
\cr
}
\leqno(2)
$$
whenever $(x, y)$ belongs to $F$ and $(x+h, y)$ and
$(x, y+k)$ belong to $Q$. Let $J$ be the smallest rectangle
containing $F$ and assume $J$ is the product of $(a_1,b_1)$
and $(a_2, b_2)$. Then
$$
\eqalign{
&\left\mathop{\int\!\!\int}_{F}\, \frac{\partial w}{\partial x}
dx dy

\int_{a_2}^{b_2} [w(b_1, y)  w(a_1, y)]dy\, \right
\le 5\, N \cdot QF\,,
\cr\noalign{\vskip 5pt}
&\left\mathop{\int\!\!\int}_{F}\, \frac{\partial w}{\partial y}
dx dy

\int_{a_1}^{b_1} [w(x, b_2)  w(x, a_2)]dx\, \right
\le 5\, N \cdot QF\,.
\cr
}
\leqno(3)
$$
\medskip
It is clearly enough to prove the theorem for all rectangles
$Q'$ properly contained in $Q$, for, in this case, we can
approximate $Q$ from the interior by a sequence of such rectangles,
and for each of which $(1)$ holds. Now by the Lebesgue integrability
of $\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}$
the right side approaches the corresponding integral over $Q$,
and by the continuity of $A$ and $B$ so does the left side. Hence
we may assume in the original statement of the theorem that
$A$ and $B$ are actually defined in a neighborhood of $Q$
where they are continuous and have derivatives at all but a
countable number of points. Now, let $E$ be the set of all
points $P$ in $Q$, such that $(1)$ holds for integrations taken
over all rectangles in a sufficiently small neighborhood of $P$.
We shall show that $E$ is all of $Q$. Let $F = QE$. Since
$E$ is obviously open, $F$ is closed. Let $H_n$ be the set of
all points $(x, y)$ in $Q$, such that
$\left\frac{A(x+h, y)  A(x, y)}{h}\right$,
$\left\frac{A(x, y+k)  A(x, y)}{h}\right$,
$\left\frac{B(x+h, y)  B(x, y)}{h}\right$,
$\left\frac{B(x, y+k)  B(x, y)}{h}\right$
are all bounded by $n$ whenever $h\le \frac{1}{n}$,
$k\le \frac{1}{n}$, and all the quantities involved are defined.
Clearly $Q$, with a countable number of exceptions, is the union
of all these closed sets $H_n$. Therefore by the Baire category
theorem, since $F$ is also closed, either $F$ contains an isolated
point in the interior of $Q$, or there is some square $I$, in
the interior of $Q$, such that $I\cap F$ is nonempty and is contained
in $H_N$ for some $N$. If a rectangle lies completely in $E$, then
the HeineBorel theorem shows that $(1)$ holds for it. Hence
we see that isolated points of $F$ cannot occur and so the second
alternative holds. Then the conditions of the lemma hold and we have
$$
\left\int_J A dx + B dy 
\mathop{\int\!\!\int}_{J\cap F}
\left(\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}\right)
dx dy \, \right \le 10\, N\cdot JF\,.
\leqno(4)
$$
Here $J$ is the smallest rectangle containing $I\cap F$ and
$\partial J$ denotes the boundary of $J$. The set $IJ$ is a
finite union of rectangles each of which can be approximated
from the interior by rectangles wholly contained in $E$.
Hence $(1)$ holds for the set $IJ$, where the line integral is
taken around its boundary in the positive sense. Thus we have
$$
\int_{\partial(IJ)} Adx + Bdy
=
\mathop{\int\!\!\int}_{I  J}
\left(\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}\right)
dxdy\,.
\leqno(5)
$$
So by $(4)$ we have
$$
\left
\int_{\partial I} Adx + Bdy

\mathop{\int\!\!\int}_{(J\cap F)\cup (Ij)}
\left(\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}\right)
dxdy\, \right
\le 10\, N\cdot JF\,.
\leqno(6)
$$
From $(6)$ it follows that
$$
\left
\int_{\partial I} Adx + Bdy \right
\le 10\, N\cdot I +
\mathop{\int\!\!\int}_{I}
\left\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}\right
dxdy\,.
\leqno(7)
$$
Now $(7)$ holds equally well for any square $I'$ contained in $I$.
Thus the set function which assings to every square $I'$ the
quantity\footnote{In original the $'$ is missing.}
$\int_{\partial I'} Adx + B dy$ is dominated by an
absolutely continuous measure and hence extends to an absolutely
continuous measure defined on all Borel sets. Thus, it is given
by the indefinite integral of some function. If we can then
show that the derivative of this measure in the sense of averages
taken over smaller and smaller squares is equal to
$\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}$
almost everywhere, then we will know that $(1)$ holds for all rectangles
in $I$ and hence that $I\cap F$ is empty, which is a contradiction.
How the derivative of the measure at almost all points not in $F$
is clearly
$\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}$.
This is merely the theorem concerning the differentiation of
indefinite integrals, since at such points $(1)$ does hold.
On the other hand, if $P$ is a point of density of $F$, then
for\footnote{In original ``a'' missing.} a sufficiently small
square $I$ around $P$, we have that $\frac{IF}{I}$
is arbitrarily small. Thus from $(6)$ it follows that
$\frac{1}{I}\int_{\partial I} Adx + Bdy$
approaches
$
\frac{1}{I}\,
\displaystyle \mathop{\int\!\!\int}_{(J\cap F)\cup (Ij)}
\left(\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}\right)
dxdy
$.
This last quantity approaches the derivative of the integral
taken with respect to the sets $(J\cap F)\cup (IJ)$. These
are a regular sequence of sets in the sense of [6] p.~106,
since $\frac{(J\cap F)\cup (IJ)}{I}$ tends to 1,
and so this derivative is equal to
$\frac{\partial B}{\partial x}  \frac{\partial A}{\partial y}$
at almost all points of $F$. Thus the measure is the desired
indefinite integral and so $(1)$ holds at all points $P$.
Now by the HeineBorel theorem it follows that $(1)$ holds
for the rectangle $Q$ itself.
\smallskip
We may easily generalize this result to any number of dimensions.
\chapter*{Bibliography}
\begin{enumerate}
\smallskip\item
S. Banach, Theorie des operations lineaires, (Warsaw, 1932).
\smallskip\item
N. Bary, The uniqueness problem in the representation of functions
by trigonometric series, Amer. Math. Soc. Trans. No. 52, (1951).
\smallskip\item
S. Bochner, GreenGoursat Theorem, Math. Zeit, 63, 230242, (1955).
\smallskip\item
J. Marcinkiewicz and A. Zygmund, Two theorems on trigonometrical
series, Mat. Sbornik, N.S. (44), 733738 (1937).
\smallskip\item
I.I. PyatetskiiShapiro, The uniqueness problem of the
representation of functions by trigonometric series,
Maskov, Gos. Univ. Ucenye Zapiski, Matematika 5 (1952).
\smallskip\item
S. Saks, The theory of the integral, (Warsaw, 1937).
\smallskip\item
R. Salem, On singular monotonic functions of Cantor type,
J. Math. Physics 21, 6982 (1942).
\smallskip\item
R. Salem, Sets of uniqueness and sets of multiplicity, Trans. Amer.
Math. Soc. 54, 218288.
\smallskip\item
V.L. Shapiro, Uniqueness of multiple trigonometric series,
Annals of Math. 66, 467480 (1957).
\smallskip\item
V.L. Shapiro, On Green's Theorem, J. London Math. Soc., 32,
261269 (1957).
\smallskip\item
H. Weyl, Uber die Gleichverteilung von Zahlen mod Eins, Math.
Annalen, 77, 313352 (1916).
\smallskip\item
N. Wiener and A. Wintner, FourierStieltjes transforms and
singular infinite convolutions, Amer. J. Math. 60, 513522 (1938).
\smallskip\item
A. Zygmund, Trigonometrical Series, (Warsaw, 1935).
\end{enumerate}
\chapter*{Notes}
Paul Cohen's thesis has remained unpublished, except for the results
of Chapter~4 (the shortest chapter) which appeared in [Coh59].
However, the rest of the thesis also presents interesting results, and
in fact, some of these were reproduced by Ash and Welland in [AWe72],
see also [AWa97]. Apart from this, it appears that the results of the
thesis have largely gone unnoticed. The main references on the
uniqueness questions are the books of Bary [Ba64],
Salem [Sal63], Meyer [Mey72],
Zygmund [Zy79], Kechris and Louveau [KL87], and
Kahane and Salem [KS94].
Even though a number of the results stated in these works were
proved in Paul Cohen's thesis, it is not cited in the
references. Therefore, it seemed appropriate that this work should
finally be more widely available. I have included a short list of
references in order to indicate the progress made in the last 40
years.
The most substantial reference is the book of Kechris and Louveau
[KL87] which also gives a complete update on the progress done on the
topics of Chapter~1. As noted above, some of the results of Chapter~2
are reproved in [AWe72]. The general reference for Chapter~3 is the
updated version of the book of Kahane and Salem [KS94] which
summarizes progress until 1994.
\section*{Chapter 1}
\noindent
{\bf Section 1.}
This section simplifies and extends some results previously obtained
by PiatetskiShapiro [PS52]. The notation used here is not
standard, and the notations of Kahane and Salem [KS94]
have been adopted. The class $W$ is written $A$. The class of
sequences $a_n$ such that ${\rm Lim}\, a_n = 0$ is now referred to as
the class of {\it pseudofunctions.}
At the end of the section it is stated that there are no known
sets which have an ordinal $> 1$ associated to them. This
ordinal, introduced by PiatetskiShapiro, is now known as the
{\it rank\/} of the set of uniqueness. A survey of results
is given by Kechris and Louveau [KL87, Chapter~V]. For example,
it follows from a theorem of Banach [KL87, p.~156] that
the rank is $<\omega_1$ and McGehee [MG68]
has shown that the rank is in unbounded in $\omega_1$.
\medskip\noindent {\bf Section 2.} This section uses the results of
PiatetskiShapiro to give a spimple proof of a special case of a
theorem of Zygmund and Marcinkiewicz. The proof should be compared
with the one given by Kahane and Salem [KS94, p.~61], see also [KL87,
p.~71]. The general proof, i.e., not limited to closed sets, is given
by N.~Bary [Ba64, p.~364].
\medskip\noindent
{\bf Section 3.} The classes $H^n$ studied here
are now denoted $H^{(n)}$. Moreover, the notation
$n_i^j$ used in this section should be $n_i^{(j)}$
in order to be consistent with the rest of the text.
\medskip\noindent
{\bf Section 4.}
Another explicit construction of an $H^{(n)}$ set which is not
a countable union of $H^{(n1)}$ sets is given by
N.~Bary [Ba64, p.~382].
\section*{Chapter 2}
This chapter presents a generalization of the CantorLebesgue theorem
to multiple trigonometric series. Recall that this theorem states that
if a trigonometric series converges on a set of positive measure then
its coefficients must go to zero. In this chapter, Paul Cohen proves
that if a multiple trigonometric series converges almost everywhere
for a ``regular'' method of summation, e.g., circular or square, then
the coefficients grow more slowly than exponential. Moreover,
the other result in this chapter shows that an analogue of
the CantorLebesgue theorem cannot generalize directly, as
an example is given of a double trigonometric series which
converges almost everywhere using square convergence, but
whose coefficients do not go to zero.
The first result may appear to be somewhat weak, but in fact it was
recently shown by Ash and Wang [AWa97] that Cohen's result is optimal
for square summation, i.e., for any function $\varphi(n)$ which goes
to infinity slower than exponentially, there is a square convergent
trigonometric series which has coefficients which grow like
$\varphi(n)$.
In the case of spherical summation, much progress has been made.
The CantorLebesgue theorem does generalize in this case as
was shown by Cooke [Co71] and Zygmund [Zy72] in dimension 2,
and by B.~Connes (in slightly less general form)
in dimensions greater than 2 [BC76].
Due to a result of Shapiro [Sh57], Cooke's result immediately
proved the uniqueness of double trigonometric results for
circular summation. A recent advance was made by Bourgain
[Bo76] who extended this uniqueness result to all dimensions.
Similarly, uniqueness
results were also obtained for ``unrestricted rectangular
convergence'' by Ash, Freiling, and Rinne [AFR93].
The case of square convergence still remains open.
\section*{Chapter 3}
In the first section, a simple proof is given that
Mensov's original example is a set of multiplicity.
The following comment is implicit in the proof, but
is added here for clarity:
In order to ensure that the resulting set has measure
zero, one just have that $\prod (2 \xi_k) = 0$,
while $\xi_k < 1/2$ for all $k$, i.e.,
$\sum \alpha_i = \infty$ as in Mensov's original construction.
\section*{Bibliography}
\begin{description}
\item[{\rm [AFR93]}]
J.M. Ash, C. Freiling, and D. Rinne,
{\sl Uniqueness of rectangularly convergent trigonometric series,}
Ann. of Math. {\bf 137} (1993), 145166.
\item[{\rm [AWe72]}]
J.M. Ash and G.V. Welland,
{\sl Convergence, uniqueness, and summability of multiple
trigonometric series,}
Trans. Amer. Math. Soc. {\bf 163} (1972), 401436.
\item[{\rm [AWa97]}]
J.M. Ash and G. Wang, {\sl A survey of uniqueness
questions in multiple trigonometric series,}
in ``Harmonic analysis and nonlinear differential equations,''
Contemporary Mathematics {\bf 208}, Amer. Math. Soc., Providence,
RI, 1997, p.~3571.
\item[{\rm [Ba64]}]
N.K. Bary, {\sl A treatise on trigonometric series, 2 Vols.,}
Pergamon Press, Macmillan, New York, 1964.
\item[{\rm [Bo96]}]
J. Bourgain, {\sl Spherical summation and uniqueness of
multiple trigonometric series,} Internat. Math. Res. Notices
(1996), 93107.
\item[{\rm [Coh59]}]
P.J. Cohen, {\sl On Green's theorem,}
Proc. Amer. Math. Soc. {\bf 10} (1959), 109112.
\item[{\rm [BC76]}]
B. Connes, {\sl Sur les coefficients des s\'eries
trigonom\'etriques convergentes sph\'eriquement,}
C.R. Acad. Sci. Paris, S\'er. A {\bf 283} (1976), 159161.
\item[{\rm [Co71]}]
R. Cooke, {\sl A CantorLebesgue theorem in two dimensions,}
Proc. Amer. Math. Soc. {\bf 30} (1971), 547550.
\item[{\rm [KS94]}]
J.P. Kahane and R. Salem, {\sl Ensembles parfaits et
s\'eries trigonom\'etriques,} Nouvelle \'edition revue et
augment\'ee, Hermann, Paris, 1994.
\item[{\rm [KL87]}]
A.S. Kechris and A. Louveau,
{\sl Descriptive Set Theory and the Structure of Sets of
Uniqueness,} London Math. Soc. Lecture Note Series {\bf 128},
Cambridge University Press, Cambridge, 1987.
\item[{\rm [Ly95]}]
R. Lyons, {\sl Seventy years of Rajchman measures,}
Proc. of Conference in Honor of JeanPierre Kahane
(Orsay, 1993),
J. Fourier Anal. Appl., Special issue (1995), 363377.
\item[{\rm [MG68]}]
O.C. McGehee, {\sl A proof of a statement of Banach on
the weak$*$ topology,} Mich. Math. J. {\bf 15} (1968),
135140.
\item[{\rm [Mey72]}]
Y. Meyer, {\sl Algebraic numbers and harmonic analysis,}
NorthHolland, Elsevier, New York, 1972.
\item[{\rm [PS52]}]
I. PiatetskiShapiro,
{\sl On the problem of uniqueness of
a function in a trigonometric series,} (Russian)
Moskov. Gos. Univ. Uw. c. Zap. Mat. {\bf 155(5)} (1952), 5472.
\item[{\rm [PS54]}]
I. PiatetskiShapiro,
{\sl Supplement to the work ``On the problem of uniqueness of
a function in a trigonometric series,''} (Russian)
Moskov. Gos. Univ. Uw. c. Zap. Mat. {\bf 165(7)} (1954), 7997.
\item[{\rm [Sal63]}]
R. Salem, {\sl Algebraic numbers and Fourier analysis,}
D.C.~Heath, Boston, 1963.
\item[{\rm [Sh57]}]
V.L. Shapiro, {\sl Uniqueness of multiple trigonometric series,}
Ann. of Math. {\bf 66} (1957), 467480.
\item[{\rm [Zy72]}]
A. Zygmund, {\sl A CantorLebesgue theorem for double
trigonometric series,} Studia Math. {\bf 43} (1972), 173178.
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A. Zygmund, {\sl Trigonometric Series, 2nd Edition, 2 Vols.,}
Cambridge University Press, Cambridge, 1979.
\end{description}
\end{document}