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\title[Distribution of Subsequences]
{Metric, fractal dimensional and Baire Results \\ on the \\
Distribution of Sequences and Subsequences}
\author{Martin Goldstern} \email{Martin.Goldstern@tuwien.ac.at}
\thanks{The first author was supported by an Erwin Schr\"odinger
Fellowship from the Austrian Science Foundation (FWF)}
\urladdr{http://info.tuwien.ac.at/goldstern/}
\author{J\"org Schmeling} \email{shmeling@math.fuberlin.de}
\thanks{The second author was partially supported by the Leopoldina
Forschungspreis}
\author{Reinhard Winkler} \email{Reinhard.Winkler@oeaw.ac.at}
\address{ % 2nd address for Winkler and current address for Goldstern
Department of Algebra\\
Wiedner Hauptstr.~810/118\\
Technische Universit\"at\\
A1040 Wien\\
Austria, Europe\\
}
\address{ % address for Goldstern and Schmeling
Fachbereich Mathematik \\
Freie Universit\"at Berlin\\
Arnimallee 3\\
D14195 Berlin
Germany}
\address{ % first address for Winkler
\"Osterreichische Akademie der Wissenschaften\\
Ignaz SeipelPlatz 2\\
A1010 Wien\\
Austria, Europe
}
\subjclass{Primary 11K55; Secondary 28A33}
%11Kxx Probabilistic theory: distribution modulo $1$; metric theory
%
%11K55 Metric theory of other algorithms and expansions; measure
% and Hausdorff dimension, See also {11N99, 28Dxx}
%
%11K99 None of the above but in this section
%
%28A33 Spaces of measures, convergence of measures, See Also {
% 46E27, 60Bxx}
%
%
%60Bxx Probability theory on algebraic and topological structures
%60B10 Convergence of probability measures
%
%28A78 Hausdorff measures
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\begin{document}
\begin{abstract}
Let $X$ be a locally compact metric space. One important object
connected with the distribution behavior of an arbitrary
sequence $\xf$ on~$X$ is the set $M(\xf)$ of limit measures of~$\xf$.
It is defined as the set of accumulation points of the sequence of
the discrete measures induced by~$\xf$.
Using binary representation of reals one gets a natural bijective
correspondence between infinite subsets of the set $\N$ of positive
integers and numbers in the unit interval $I = [0,1)$. Hence to each
sequence $\xf = \xn \in X^{\N}$ and every $a \in I$ there corresponds
a subsequence denoted by~$a\xf$.
We investigate the set $M(a\xf)$ for given $\xf$ with emphasis on the
behavior for ``typical'' $a$ in the sense of Baire category, Lebesgue
measure and Hausdorff dimension. We also treat the special case of
the sequence $(n \alpha) $ mod 1 and its subsequences.
\end{abstract}
\maketitle
% % \include{not}
\section{Introduction}\label{6}
% \pagestyle{headings}
% \markboth{\heute}{\heute}
\subsection*{Notations}
For the whole paper we fix the following abbreviations.
Let $\N$ denote the set of positive integers $1,2,\ldots$ and
let $\R/\Z$ be the one dimensional torus group, identified with
the half open unit interval $I=[0,1)$ in the natural way.
That is, we identify the remainder class $x+\Z$ with the real
number $x  [x]$ (where $[x] $ denotes the
the largest integer $k \in \Z$ with $k \le x$). So the canonical
epimorphism $\R \to \R/\Z $ can be identified with the map $x \mapsto
x  [x] $.
For any set $X $ we write $X^\N $ for the set of all
sequences $\xf=\xn $, $x_n\in X $. Sequences (both infinite and finite) are
always denoted
by boldface letters.
If $\xf=\xn$ is an infinite sequence, we write
$\xf\on N $ for the finite sequence $(x_1,\ldots, x_N) $.
For a finite sequence $(\varepsilon_1, \ldots, \varepsilon_n) $ of
elements of~$X $ we write $[\varepsilon_1,\ldots, \varepsilon_n] $ for
the ``cylinder set''
\[ \{ \xn\in X^\N: x_1=\varepsilon_1, \ldots, x_n=\varepsilon_n\}.
\]
If $X $ is discrete (in particular if $X=\{0,1\}$), then we may
write $C_n(\xf)$ for $[\varepsilon_1, \ldots, \varepsilon_n]$. In
this case the
family $(C_n(\xf))_{n\in\N}$ is a local basis at~$\xf$ in the product
topology on~$X^\N$.
Each number $a \in I$ has
a representation $a = \sum_{n \in \N} a_n2^{n}$ with $a_n \in \{0,1\}$.
This representation is unique if one requires that the set $S_a = \{n
\in \N \  \ a_n=1\}$ is infinite.
(Note that to~$0 \in I$, by identification, there corresponds the
set $S_0 = S_1 = \N$.) Thus the correspondence $a \mapsto S_a$
is a bijection between $I$ and the system of all
infinite subsets of~$\N$ or, equivalently, the set of all strictly
increasing infinite sequences of~integers. If $S_a = \{k_1 < k_2 < \ldots\}$
and $\xf= \xn $, then
we write $a \xf$ for the sequence $(x_{k_n})_{n \in \N} =
(x_i)_{i \in S_a}$. If $\xf = (x_1, \ldots, x_n)$ is a finite
sequence, then $a\xf$ will denote the subsequence $(x_i)_{i \in
\{1,\ldots, n\} \cap S_a}$.
We usually use the Euclidean metric and topology on $[0,1)$,
in particular when we consider measure and Hausdorff dimension. For
arguments dealing with Baire category, however, we identify $[0,1)$
with a subspace of $\N^\N$ and use the Tychonoff topology, which we
call ``the $0$dimensional topology.''
Note that the notions ``meager'' and ``residual'' are the same in the
$0$dimensional topology and the Euclidean topology.
% For a given sequence $a$ we denote by~$C_n(A)$ the cylinder set
% \[
% \{b \in [0,1] \  \ b_1=a_1 \cdots b_n=a_n \}.
% \]
Each $p \in I$ induces the $p$Bernoulli measure $\lambda_p$ on~$I$
which is the Borel measure uniquely
defined by the requirement $\lambda_p(\{a \  \ a_n=1\})=p$ for
all $n \in \N$. If $p \in \{0,1\}$, $\lambda_p$ is concentrated
on~$0=1 \in I$.
Hence the case $0 < p < 1$ is the interesting one and will be assumed
for the rest of the paper.
Let $X$ be a locally compact Hausdorff space with countable topological base.
Thus there is a complete metric $d_X$ inducing the topology.
With $A(\xf)$
we denote the set of accumulation points of~$\xf$. This is the set of
all $y \in X$ such that in each neighborhood $O$ of~$y$ the
relation $x_n \in O$ holds for infinitely many $n \in \N$.
Let now $\M(X)$ denote the set of all nonnegative
Borel measures $\mu$ on~$X$ with $\mu(X)=1$ if $X$ is
compact and $\mu(X)\le 1$ if $X$ is not compact. Let $\K(X)$ denote
the space of all continuous $f: X \to \R$ with compact support
\[
\supp(f) = \overline{\{x \in X \  \ f(x) \neq 0 \}}.
\]
We consider the weak topology on~$\M(X)$ which is the weakest topology
such that all functionals $\mu \mapsto \int_X f d\mu$, $f \in \K(X)$, are
continuous. $\M(X)$ is a compact space with countable topological base,
hence also induced by a metric $d_{\M(X)}$. In most cases it will cause
no confusion to write $d$ for both $d_X$ and~$d_{\M(X)}$.
We will need the fact that $d_{\M(X)}$ can be chosen to be compatible
with convex combinations, i.e.,
\[
d_{\M(X)}(\mu_0, (1t)\mu_0 + t \mu_1) = td_{\M(X)}(\mu_0,\mu_1)
\]
for $0 \le t \le 1$.
(One possibility of defining $d$ is the following: Let $f_n:\R\to
[0,1]$, $n \in \N$, be a countable family spanning a dense subspace
of~$\K(X)$. Then the function $d:\M(X)^2 \to \R$, defined by
\[
d(\mu_1, \mu_2) = \sum_{n \in \N} 2^{n}
\Bigl \int f_n\, d\mu_1  \int f_n \, d\mu_2\Bigr
\]
satisfies the above requirements.)
Each finite sequence $\xf = (x_1,\ldots, x_N)$ induces a probability
measure $\mu_\xf = \frac{1}{N} \sum_{n=1}^N \delta_{x_n}$, where
$\delta_x$ is the Dirac measure concentrate at~$x$. For an infinite
sequence $\xf=\xn$ we let
\[
\mu_{N,\xf} = \mu_{\xf\on N},\qquad
\mu_{N,\xf}(B) = \frac{1}{N}\#\{i\le N: x_i\in B\}
\]
($\# A$ denotes the cardinality of a set~$A$.)
With $M(\xf)$ we denote the
set of all accumulation points of the sequence $(\mu_{N,\xf})_{N \in \N}$
in~$\M(X)$.
It follows from \cite{Si} (for the compact case) or \cite{W2} that
for an arbitrary subset $M \tm \M(X)$ the following holds:
There is a sequence $\xf$
with $M = M(\xf)$ if and only if $M$
is nonempty, closed and connected. $\xf$ is called uniformly
distributed w.r.t.\ $\mu$
($\mu$u.d.) if $M(\xf) = \{ \mu \}$.
If $X$ is a compact group and $\mu$
is the Haar measure on~$X$, we call $\xf$ simply u.d.
In the other extreme case $M(\xf) = \M(X)$, the sequence $\xf$ is called
maldistributed (md.).
Thus we use the notation from \cite{W2} which is a slight modification
of the one in~\cite{My}.
If $A \tm X$ is a closed subset of~$X$, then
the set $\M(A)$ can be considered
as a subspace of~$\M(X)$ in an obvious way: Identify every $\mu\in
\M(A)$ with its extension $\bar\mu\in \M(X)$, defined
by~$\bar \mu(B) = \mu(B \cap A)$ for arbitrary Borel sets $B \tm X$.
Recall that a subset of a topological space is called nowhere dense if
its topological closure has empty interior. A set is of first category
or meager if it is the union of at most countably many nowhere dense sets.
All sets which are not meager are said to be of second category.
A set is called comeager or residual if its complement is meager. Complete
metric spaces (as $I=\R/\Z$,
$X$ and $\M(X)$) are Baire spaces, i.e., nonempty open
subsets are of second category.
\subsection*{Content of the paper}
In section~\ref{5} we investigate the set $M(a\xf)$ for given $\xf$
with emphasis on the behavior for ``typical'' $a$ in the sense of
Baire category, Lebesgue measure and Hausdorff dimension. In
section~\ref{7} we consider sequences depending on an additional
parameter. In section~\ref{8} we focus on the special case of
$n\alpha$sequences. We give Baire results on the typical
distribution of~$(k_n \alpha )_{n\in \N}$ depending on the growth rate
of $(k_n)_{n\in \N}$. For special sequences
$(k_n)_{n\in \N}$ cf.\ also \cite{Be}.
The reader who is interested in connections to
ergodic theory may want to consult the survey paper~\cite{RW}.
The standard textbook on uniform distribution is~\cite{KN}. Many
interesting results and an uptodate bibliography can found in
\cite{DT}.
% % \include{sub}
\section{Subsequences} \label{5}
We fix the general assumption that $X$ be a locally compact
and second countable (hence complete metric) space. Given a
sequence $\xf$ on~$X$ we look, in dependence on~$a$, at its
subsequences $a \xf$
and the corresponding sets $M(a\xf)$. The natural questions on this sets are:
\begin{enumerate}
\item
Which $a$ guarantee $M(a\xf)=M(\xf)$ for arbitrary $\xf$? (density 1,
theorem \ref{9}.)
\item
What can $M(a\xf)$ look like? (Anything reasonable, theorem \ref{1}.)
\item
What is the typical situation in the Baire sense?
(Maldistribution, theorem \ref{2}.)
\item
What is the typical situation in the Lebesgue sense?
($M(a\xf)=M(x)$, theorem \ref{3}.)
\item
What is the (Hausdorff) dimension of all $a$ with $M(a \xf)
\neq M(\xf)$? (Dimension 1, theorem \ref{4}.)
\end{enumerate}
For theorem \ref{9} we recall the notion of~density. For $S \tm \N$
and $n \in \N$ one considers the numbers $d(n,S) = \frac 1n \# S \cap
(0,n]$. Since $0 \le d(n,S) \le 1$ the sequence $(d(n,S))_{n\in\N}$
has an upper and lower limit, the ``upper density $\overline d(S)$''
and the ``lower density $\underline d(S)$.'' If they
coincide, their common value is called the density of~$S$, $d(S)$.
\begin{Theo} \label{9}\
\begin{itemize}
\item[a)] If $S_a$ has density $1$ then $M(a\xf)=M(\xf)$.
\item[b)] If $\overline d(S) < 1$, and $\yf$
is an arbitrary sequence on~$X$, $\# X \ge 2$, then there exists
a sequence $\xf$ such that $a\xf=\yf$ and $M(a\xf)= M(\yf) \not= M(\xf)$.
\end{itemize}
\end{Theo}
\begin{proof}
a): Use the fact
that $ d( n, S_a) \ge 1\varepsilon$ implies $\mu_{N,\xf}(B) 
\mu_{N,a\xf}(B)  \le 2 \varepsilon $ for any set~$B$.
\\
b): First case: $M(\yf)=\M(X)$,
i.e., $\yf$ is maldistributed. Take an arbitrary $x \in X$ and
define $x_n=x$ for all $n \notin S_a$ and $x_{a_n}=y_n$ for $a_n \in S_a$
such that $a\xf=\yf$. Since $\# X \ge 2$ there exists a $\mu \in \M(X)$
with $\mu(\{x\})=0$. Note that
\[
\liminf \mu_{N,\xf}(\{x\}) \ge 1  \overline d > 0.
\]
Thus $\mu \notin M(\xf)$ proving $M(\xf) \neq \M(X) = M(\yf) = M(a\xf)$.
Second case: If there is a $\mu_0 \in \M(X) \setminus M(\yf)$
consider $\N \setminus S_a = \{b_1 < b_2 < \ldots \}$. Let $\zf$
be $\mu_0$u.d.\
and define $x_{b_n}=z_n$ and~$x_{a_n}=y_n$. $M(a\xf)$ is a
nonempty compact set. Hence there is a $\mu_1 \in M(a\xf)$
such that $\e = d(\mu_0, \mu_1) > 0$ takes a the minimal value.
It follows that there is a convex
combination $\mu_t = t\mu_0 + (1t)\mu_1$, $0 < t < 1$, such
that $\mu_t \in M(\xf)$. Here we used $\overline d < 1$.
Since $d(\mu_0, \mu_t) < d(\mu_0,\mu_1)$ we conclude $\mu_t \in
M(\xf) \setminus M(a \xf)$.
\end{proof}
If lower and upper density $\underline d$ and $\overline
d$ of~$S_a$ satisfy $\underline d < \overline d = 1$ general assertions as
in theorem \ref{9}
fail to be true. To illustrate the situation look at the following
examples where we use the notations of theorem \ref{9}:
If $\zf$ is $\mu_0$u.d.\
and $\yf$ is $\mu_1$u.d.\, $\mu_0 \neq \mu_1$,
then in the second case of part b) one
gets $M(\xf) = \{ \mu_t \  \ t \in [\underline d,\overline d] \}$
with $\e = \underline d > 0$.
Thus $M(\xf) = M(a\zf)$. On the other hand, for any $a$
with $\overline d(S_a)=1$, it is possible to construct a maldistributed
sequence $\yf$ such that $a\xf=\yf$ implies $M\xf)=M(\yf)$.
A more detailed
analysis will be the object of a forthcoming paper.
\begin{Theo}\label{1}
Let $\xf \in X^{\N}$ and $M \tm \M(X)$. There exists an~$a \in I$
with $M(a\xf)=M$ if and only if $M$ is closed
and connected with $\emptyset \neq M \tm \M(A(\xf))$.
\end{Theo}
\begin{proof}
First assume $M=M(a\xf)$. Then, by theorem 3.1.1 in \cite{W2}, $M$ is
nonempty, closed and connected. To derive the remaining
relation $M \tm \M(A(\xf))$ it suffices to show that every $x \in
X \setminus A(\xf)$
has a neighborhood $U$ with $\limN \mu_{N,a\xf}(U) = 0$. To show this take
any neighborhood $U$ with compact closure $\overline U$ and with $\overline
U \cap A(\xf) = \emptyset$. Such a $U$ exists since $X$ is
a normal space. If
$x_n \in \overline U$ for infinitely many~$n$, $\overline
U$ would, by compactness, contain an accumulation point
of~$\xf$, contradiction. Thus $x_n \notin U$ for all $n \ge N_0$.
This proves
\[
\limN \mu_{N,a\xf}(U) \le \limN \frac{N_0}N = 0.
\]
For the other direction assume that $M\not=\emptyset $ is closed and
connected.
By \cite[theorem 3.1.2]{W2}, there is sequence $\zf=\yn$ in $A(\xf)$
with $M(\zf)=M$. By the definition of $A(\xf)$ there is a
subsequence $a\xf$ of $\xf$ with $a=(a_1,a_2,\ldots)$ such that $\limn
d(y_n,x_{a_n}) = 0$. This subsequence, obviously, has the required
properties.
\end{proof}
\begin{Theo}\label{2}
$M(a\xf) = \M(A(\xf))$ holds for all $a \in R$ from a
residual set $R \tm I$,
i.e., most subsequences are maldistributed in~$A(\xf)$.
\end{Theo}
\proof
Let $R$ denote the set of all $a$ with $M(a\xf) = \M(A(\xf))$, i.e.,
$a \xf$ is maldistributed in $A(\xf)$:
\[ R = \{ a: M(a\xf) = \M(A(\xf)) \}
\]
The inclusion $\subseteq $
follows from theorem \ref{1}. For the other one we have
to prove that for every $\mu \in \M(A(\xf))$ the set
\[R_{\mu} = \{
a \in I: \mu \in M(a\xf)
\}
\]
is residual.
Since $\M(X)$ is second countable, one may
choose a sequence $(\mu_n)_{n \in \N}$ which is dense in~$\M(A(\xf))$.
We now check that
\[
R = \bigcap_{n \in \N} R_{\mu_n}
\]
is a countable intersection of residual sets, hence residual. In the
above equation the inclusion $\tm$ is trivial. But also the
converse inclusion is clear: Let $a$ be an element from the right hand
side, then
the set $M(a\xf)$ contains the topological closure of
the set of all $\mu_n$ which is $\M(A(\xf))$, hence $a \in R$.
To show that $R_{\mu}$ is residual for $\mu \in \M(A(\xf))$,
just observe
\[
R_{\mu} = \bigcap_{n,k \in \N} O_{n,k}
\]
with the sets $O_{n,k}$ of all $a \in I$ such that
there exists an $n' > n$ with $d(\mu_{n'a\xf},\mu) < \frac 1k$.
Standard arguments show that each $O_{n,k}$ is dense and open.
[$O_{n,k}$ is open since the map $a \mapsto d(\mu_{n,a\xf}, \mu)$ is
continuous (with respect to the $0$dimensional topology on~$I$).
$O_{n,k}$ is dense: Take any $\mu$u.d.\
$\yf = \yn$, $y_n\in
A(\xf)$. Any subsequence $(x_{n_k})_{k\in \N}$ with $d(y_k, x_{n_k})
\to 0$ can be written as $a\xf$ with $a\in O_{n,k}$.]
Thus the countable intersection $R_{\mu}$ of these sets is residual,
and the theorem is proved.
\qed
\begin{Remark} A general principle containing all Baire arguments
of the above type can be found in~\cite{Gr}.
\end{Remark}
% The following theorem is, in some sense, a generalization of results in
% \cite{LT} an other papers.
\begin{Theo}\label{3} $M(a \xf) = M(\xf)$ for $\lambda_p$almost
all~$a$, $0 \tau(\delta)$
the $s$dimensional Hausdorff measure of $A_0$ can be estimated by
%\begin{align*}
%\inf&\left\{\sum_{C_i\in \cal{C}^n} (diam C_i)^s : \cup_i C_i \supset A_{0} ;
% \cal{C}^n\subset\cup_{m\ge n}\cal{C}_m \right\}\\& \le \sum_{n=1}^{\infty}
% 2^{\tau(\delta) n +1} 2^{ s n}< \infty .
%\end{align*}
\[
\lim_{N \to \infty}
\sum_{n=N}^{\infty}
n 2^{n\tau(\delta) } (2^{ n})^2 = 0
\]
Hence $\dim_H(A_0) \le \tau(\delta)$. This completes
the proof of part 1.
\emph{\bf 2. ($\rho(M)=1 \impl \phi(M)=1$)}
By theorem~\ref{9} we have that $M=M(\xf)$. Then theorem~\ref{3} implies
that the set $ \{ a \  \ M(a\xf)=M(\xf) \}$ has full Lebesgue measure and
hence Hausdorff dimension 1.
\emph{\bf 3.($\phi(M)=1 \impl \rho(M)=1$)}
Let us assume that $\rho(M)<1$. In view of the definition of $\rho(M)$ there
is a measure $\mu \in M(\xf)\triangle M$. Let $ \varepsilon =
\frac{1}{10} d(\mu, M)$ if $\mu \in M(\xf)\setminus M$, and
$ \varepsilon = \frac{1}{10} d(\mu, M(\xf))$ if $\mu \in
M\setminus M(\xf)$.
To estimate the Hausdorff dimension of
\[ A_0 := \{ a: M(a\xf)=M\}
\]
we first claim that $a \in A_0$ implies that there are infinitely many
$N\in \N$ such that $ d(\mu_{\xf\on N}, \mu_{a(\xf\on N)}) \le
8\varepsilon$. To check this claim we assume without loss of
generality $\mu \in M (\xf ) \setminus M$ (the other case is
similar). There are infinitely many $N$ such that $d(\mu, \mu_{\xf\on
N}) \le \varepsilon $, and for all large enough $N$ we have
$d(\mu_{a(\xf\on N)}, M(a\xf)) \le \varepsilon$, hence $d(\mu_{\xf\on
N}, \mu_{a(\xf\on N)}) \ge 8 \varepsilon $.
Hence we can find finitely many functions $f_1, \ldots, f_k$ such that
the set $A_0$ is contained in
\[
A_1:=\bigcup_{j=1}^k \{ a: \exists^\infty N\,\,
s(f_j,N)  s(f_j, a, N) \ge 4\varepsilon \}
\]
It is therefore enough to show that for any continuous $f$ and
any~$\varepsilon >0$ the set
$ \{ a: \exists^\infty N\,\, \bigl s(f,N) s(f,N,a) \bigr
\ge 4\varepsilon \}$
has Hausdorff dimension~$<1$.
So fix $f$ and $\varepsilon $. Let $m> 1/\varepsilon$, and define
$T_1, \ldots, T_m$ as in
the proof of~theorem~\ref{3}.
Write $A_{N,{\varepsilon}}$ for the set
\[
A_{N,{\varepsilon}} := \{a: s(f,N)  s(f,N,a)  \ge {4\varepsilon} \}
\]
Note that $a\in A_{N,{\varepsilon}}$
implies that there is some $i\in \{1,\ldots, m\}$ such that
\def\bigdiff#1{
\Bigl
\frac1N \sum_{\two{j\in T_i}{j\le N}} #1 \,\,

\,\,
\frac{1}{\#S_a\cap \{1,\ldots, N\}}
\sum _{\two{j\in T_i\cap S_a}{j\le N}}
#1
\Bigr
}
\[ \bigdiff{f(x_j)} \ge 4{\varepsilon} \frac{\#T_i}{N} \]
Replacing $f(x_j)$ by $\frac {i}{m}$ introduces an error $\le\frac
{1}{m}< \varepsilon $ in each term, hence
\[
\bigdiff{\frac{i}{m}} \ge 3 \varepsilon \frac{\#T_i}{N}
\]
Employing the functions $ \gamma $ and $ \delta$ that we defined in the
proof of theorem~\ref{3} and using $\frac{i}{m}\le 1$ we conclude
that $a \in A_{N, \varepsilon}$ implies
\[ 1  \delta(N,a)\cdot \gamma(i,N,a) \ge3 \varepsilon \qquad \mbox{ for
some $i$}
\]
so $ \delta(N,a) \le 2 \varepsilon $
or $\gamma(i,N,a) \ge \frac 12  \varepsilon $.
Hence
\[ A_1 = \bigcap_{n=1}^\infty \bigcup_{N=n}^\infty A_{N,\varepsilon }
\subseteq
\bigcap_{n=1}^\infty \bigcup_{N=n}^\infty
\{a: \delta(N,a)<2\varepsilon \}
\cup
\bigcap_{n=1}^\infty \bigcup_{N=n}^\infty
\{a: \gamma(i,N,a)< \frac12 \varepsilon \}
\]
We will only estimate
the Hausdorff
dimension of the sets
\[
\bigcap_{n=1}^\infty \bigcup_{N=n}^\infty
\{ a: \gamma(i,N,a) \le \frac12  {\varepsilon} \}
% \qquad
% \mbox{ and }
% \qquad
% \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty
% \{ a: \delta (N,a) \le 2 {4\varepsilon} \}
\]
leaving the second (similar)
estimate to the reader.
As in the first step of the proof of theorem~\ref{3}, let
\[
\C_N({\varepsilon}):= \{ [\varepsilon _1,\ldots, \varepsilon_n]:
\frac{1}{N_i} \sum_{\two{
j \le n}{ j \in T_i}} \varepsilon_j
\le \frac12  {\varepsilon}\}
\qquad \mbox{(where $N_i:= \#(T_i \cap [1,\ldots,N])$),}
\]
and let $C_N(\varepsilon)$ be the union of all the cylinders in~$\C_N(
\varepsilon)$.
Again, $\C_N({\varepsilon})$ contains at most $
\displaystyle
\binomial{N_i}{(\frac 12{\varepsilon}) N_i} = o\bigl(N2^{N\tau(
\frac12  {\varepsilon})}\bigr)$ many cylinders,
% [where $N_i:=\# T_i\cap \{1,\ldots, N\}$]
each of them of diameter $2^{N_i}$. Since $A_1$
is covered by
$ \bigcup_{n=N}^\infty C_n(\varepsilon)$ for any $N\in \N$, we can
as before
bound the the $s$dimensional
Hausdorff measure of $A_1$
for any $s> \tau(\frac12{\varepsilon})$ by
\[
\lim_{N \to \infty} \sum_{n=N}^\infty n2^{n\tau(\frac12{\varepsilon})}2^{ns}
= 0
\]
Hence the Hausdorff dimension of $A_0$ is $\le \tau(\frac12\varepsilon
) < 1$.
%
% Hence, there is
% a $\varepsilon >0$ such that for $a \in \{ a \  \ M(a\xf)=M \}$ we can
% find a continuous function $f\colon X\to [0,1)$ with
% \[
% \limsup_{N \to \infty} \vert s(f,N)  s(f,N,a) \vert > \varepsilon.
% \]
% Following the proof of theorem~\ref{3} we conclude that for
% some $m$ large, a subsequence~$N_k$, $\varepsilon_1 >0$, $\delta >0$ and
% some $i$
% \[
% \gamma(i,N_k,a)\delta(N_k,a)\le 1 \varepsilon
% \]
% and
% \[
% \frac{\#(T_i\cap (0,N_k])}{N_k}\ge\delta.
% \]
% Otherwise, the values of $f$ in the interval $\frac{i1}{m}$ would be obtained
% with the right frequency.
% Let
% \[
% A = \{a \  \ \liminf \gamma(i,N_k,a) \le \frac12  \sqrt{\varepsilon} \}
% \]
% and
% \[
% B = \{a \  \ \liminf \delta(N_k,a) \le 2  \sqrt{\varepsilon} \}.
% \]
% By the above arguments
% \begin{equation}\label{AB}
% \{ a \  \ M(a\xf)=M \} \subset A \cup B.
% \end{equation}
% As in the proof of part 1 we consider special coverings
% of $A$ (respectively $B$).
% Let ${\cal D}_k$ be the set of cylinders $[\epsilon_1 \cdots
% \epsilon_{N_k}]$ such that each cylinder contains at most $(\frac12
% \sqrt{\varepsilon}) \#(T_i\cap (0,N_k])$ 1's in $T_i\cap (0,N_k]$ and is
% arbitrary outside$ T_i\cap (0,N_k]$. There are at
% most $2^{\tau(\frac12 \sqrt{\varepsilon_1}) \delta
% N_k} 2^{(1\delta) N_k}$ of them. Again we can estimate the Hausdorff sum for
% $1>s>1(1\tau(\frac12 \sqrt{\varepsilon_1})\delta)$
% \begin{align*}
% \inf&\left\{\sum_{D_i\in {\cal D}^k} (diam D_i)^s : \cup_i D_i \supset A ;
% {\cal D}^k \subset\cup_{m\ge k} {\cal
% D}_m \right\}\\& \le \sum_{k=1}^{\infty}
% 2^{\tau(\frac12 \sqrt{\varepsilon_1})\delta
% N_k} 2^{(1\delta) N_k} 2^{ s n}< \infty .
% \end{align*}
% This gives part 3.
%
%
%
% We start by proving the following claim:
%
%
%
%
% \begin{Claim}
% Fix a sequence $\xf$ and a function~$f$.
% Then for any ${\varepsilon}>0$ the Hausdorff dimension of the set
% $$ \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty
% \{ a: \gamma(i,N,a) \le \frac12  {\varepsilon} \}
% $$
% is at most ????. A similar estimate for the set
% $ \{ a: \gamma(i,N,a) \le \frac12  {\varepsilon} \}$
% is left to the reader.
%
% \end{Claim}
% \proof As in ???, let
% \[
% \C_N({\varepsilon}):= \{ [\varepsilon _1,\ldots, \varepsilon_n):
% \sum_{\two{
% j \le n}{ j \in T_i}} \varepsilon_j / \# T_i \cap
% \{1,\ldots,N\} \le \frac12  \frac {\varepsilon}4\}
% \]
% Again, $\C_N({\varepsilon})$ contains at most $
% \displaystyle
% \binomial{N_i}{(\frac 12{\varepsilon}) N_i} = o\bigl(N2^{N\tau(
% \frac12  {\varepsilon})}\bigr)$ many cylinders [where $N_i:=\# T_i\cap
% \{1,\ldots, N\}$], each of them of diameter $2^{N_i}$...
%
% ..... . .... .......... .............
%
%
%
%
% \hhh
\emph{\bf 4. ($\phi(M)=0 \impl \rho(M)=0$)}
Let us assume that $\rho(M)=d>0$. In
view of the definition of $\rho(M)$ there is a sequence $a$ with $M(a\xf)=M$
and positive lower density ${\underline d(a)}>d  \delta$. We are going to consider the
subset in $[0,1]$ corresponding to all subsequences of~$a$. For this we
consider for $N \in \N$ the union of cylinder sets $\cJ_N$ defined
by
\[
[\varepsilon_1 \cdots \varepsilon_N] \in \cJ_N
\quad \ifff \quad \varepsilon_j =0
\text{ unless } j=a_k \text{ for } j \in [0,N] \text{ and some } k.
\]
We set
\[
\Lambda_a = \bigcap_{N=1}^\infty \cJ_N = \{b: S_b \subseteq S_a\}.
\]
There is a natural identification of $\Lambda_a$ with the space
$\{0,1\}^\N$ of all $\{0,1\}$sequences.
% Namely, for $\underline\N = \omega_1 \cdots
%\omega_k \cdots \in \Sigma_2$ we set $S_{b(\omega)}=\{k_i\omega_i : k_i\in
%S_a\}$.
We define a measure $\mu_a$ supported on $\Lambda_a$ as the
pushforward of the Bernoulli measure on $\{0,1\}^\N$ generated by the
probability vector $(\frac12,\frac12)$
\[
\mu_a (C_N(b))=
\cases{
\left(\frac12 \right)^{\#(S_a \cap [1,N])} &\text{if $b \in \Lambda_a$}\cr
0 &\text{if $C_N(b) \cap \Lambda_a =\emptyset$}
}
\]
This measure is a probability measure supported on $\Lambda_a$
and has the following property
\[
\liminf_{N \to \infty} \frac{\log \mu_a(C_N(b))}{N \log 2} = \liminf_{N \to \infty} \frac{\#(S_a \cap [0,N])}{N}= {\underline d(a)} >0 \qquad \text{for $\mu_a$a.e. $b$}
\]
Frostman's Lemma (see~\cite[p.221]{Fal}) tells us that
\[
\dimh \mu_a = {\underline d(a)}.
\]
Now we want to apply theorem~\ref{3} to the case $\yf = a\xf$. In
order to do this we identify the subsequences of $a$ with points in the
interval $[0,1)$ in the standard way. Under this identification the
measure $\mu_a$ transforms into the normalized Lebesgue measure
on~$[0,1)$. Hence, theorem~\ref{3} implies
that $\mu_a$a.e.\
subsequence $b$ of $a$ has the same limit behavior
as $a$  i.e., $M(a\xf) = M(b\xf) = M$. By the above considerations
the set of these subsequences $b$ has Hausdorff dimension~${\underline d(a)}$. This
yields $\phi(M) \ge {\underline d(a)} > 0$.
\end{proof}
We actually proved slightly stronger results than stated in
theorem~\ref{HDTheo1}. These results are stated below.
\begin{Cor}\label{HDCor1}
\[\dimh \{b\  \ M(b\xf) = M(a\xf) \} \ge \ud (a).\]
\end{Cor}
\begin{Remark} The estimate in corollary~\ref{HDCor1} is not very
effective for $\ud (a) \le \frac12$. In this case the estimate \[\dimh
\{b\  \ M(b\xf) = M(a\xf) \} \ge \tau(\ud (a)).\] is better. One can
derive this estimate along similar lines as in the proof of part 4 of
the theorem.
\end{Remark}
% For two sequences $S_a$ and $S_b$ we can define a nonsymmetric
% semidistance via
% \[
% \delta'(S_a,S_b)=\lim_{n\to\infty}\frac{\#\{k : k\in S_a\setminus S_b ; k\le
% n\}}{\#\{k : k\in S_a ; k\le n\}}.
% \]
%
% JOERG FRAGEN: SOLL DER FOLGENDE ABSATZ SO BLEIBEN?
%
%
%
% Let $\delta(S_a,S_b):=\delta'(S_a,S_b)+\delta'(S_a,S_b)$ be its
% symmetric semidistance. We can consider the
% equivalence relation $a\sim b$ if
% and only if
% $\delta(S_a,S_b)=0$.
% Then we have shown that
% equivalent sequences have equivalent
% asymptotics: $M(a\xf)=M(b\xf)$ if
% $a\sim b$.
% Moreover, if
% we define an ordering on the equivalence
% classes by~$a\le b$ if
% and only
% if $\delta'(S_b,S_a)=0$ and $M(a\xf)=M(b\xf)$,
% then the above corollary gives an estimate of the
% Hausdorff dimension of maximal chains. To get a complete picture one
% would need to analyze the structure of this ordering, especially to
% estimate the cardinality of maximal chains with given asymptotic
% behavior.
\bigskip
We also want to state another consequence of theorem~\ref{HDTheo1}:
\begin{Theo}\label{4}
Let $\xf$ be an arbitrary sequence on~$X$. Assume that $M(\xf)$
contains a nonatomic measure. Then the set of all $a$
such that $M(a\xf) \neq M(\xf)$ has Hausdorff dimension~$1$.
\end{Theo}
\begin{proof}
By corollary~\ref{DefrhoCor1}
\[
\{a \  \ M(a\xf) \neq M(\xf) \} \supset \{a \  \ \rho (M(a\xf)) < 1 \}.
\]
Let $\mu \in M(\xf)$ be a nonatomic measure. We fix $\varepsilon >0$
and choose an open subset $U \subset X$ with measure $\mu(U)=
\varepsilon$. We
note that such subsets $U$ exist since $\mu$ is nonatomic. Let $a$ be
the subsequence defined by
\[
k \in a \quad \ifff \quad x_k \not \in U.
\]
By the definition of the lower density and of~$M(a\xf)$ this sequence
has a subsequence $a'$ with density $\underline d(a')=1\varepsilon$
and $\mu \not
\in M(a'\xf)$.
By corollary~\ref{HDCor1} we get
\[
\dimh \{b \  \ \rho (M(b\xf)) < 1 \} \ge \dimh \{b \  \ M(b\xf) =
M(a'\xf) \} \ge \ud (a') = 1\varepsilon
\]
Since $\varepsilon$ was arbitrary this concludes the proof.
\end{proof}
% % \include{seq}
\section{Sequences of maps} \label{7}
Let now $\ff$ be a sequence of maps $f_n: X \to Y$.
Thus $\ff: \N \times X \to Y$, $\ff(n,x) = f_n(x)$. For notational
convenience we also write $\ff(x)$ for the sequence $(f_n(x))_{n \in \N}$.
To $\ff$ there corresponds a map $\Phi = \Phi_\ff: I^2 \to
X^{\N}$, $(a,x) \mapsto a \ff(x)$ where $a\ff(x)$ is the
subsequence $(f_{k_n}(x))_{n \in \N}$ of~$(f_n(x))_{n \in \N}$ which
corresponds to~$a$ in the sense described
in the first section. This point of view is similar to that in
\cite{W1}, where several related metric results can be found.
Let $T \tm I \times X$, $a \in I$, $x \in X$, then $_aT = \{x \in X \
 \ (a,x) \in T\}$ and $T_x = \{a \in I \  \ (a,x) \in T\}$ denote
the vertical and horizontal crosssection of~$T$.
Important informations about $T$ and its crosssections
are their Hausdorff dimensions $\dim_H$, their $d$dimensional
Hausdorff measures $m_d$ and their Baire categories as subsets
of $I \times X$ resp.\
$X$ or~$I$. We turn to the Baire aspect.
% LIEBER MARTIN! DER FOLGENDE SATZ IST DIR GEWIDMET. DOCH KEINE EHRE
% OHNE GEGENLEISTUNG: KOENNTEST DU NACHPRUEFEN; OB DER BEWEIS O.K. IST?
% INSBESONDERE BIN ICH MIR NICHT SICHER, OB KURATOWSKI ULAM IN DER
% UNTERSTELLTEN ALLGEMEINHEIT GILT.
\begin{Theo}\label{10}
Let $X,Y$ be complete metric spaces, $Y$ second countable,
and $\ff$ a sequence of
continuous maps $f_n: X \to Y$. Suppose $D \tm Y$ is a dense set such
that for every $y \in D$ the sets
\[
D_y = \bigcup_{n \in \N} f_n^{(1)}(y)
\]
are dense in~$X$ in a strong sense, meaning that every nonempty
open set in~$X$ has an infinite intersection with~$D_y$.
Then the following statements hold: \newline
(i) There is a residual set $R \tm I \times X$ such that $a\ff(x)$
is maldistributed for all $(a,x) \in R$. \newline
(ii) There is a residual set $R \tm I$ such that
for all $a \in R$ the sequence $a\ff(x)$ is
maldistributed for all $x$ from a residual set $R(a) \tm X$.
\newline
(iii) There is a residual set $R \tm X$ such that
for all $x \in X$ $a\ff(x)$ is maldistributed for all $a$ from a
residual set $R(x) \tm I$.
\end{Theo}
\begin{proof}
By the theorem of KuratowskiUlam (cf.~\cite{Ox})
the three statements of the theorem
are equivalent. Hence it suffices to prove (iii). For an arbitrary
nonempty open set $O \tm Y$ define
\[
R(O) = \bigcup_{n \in \N} f_n^{1}(O).
\]
By assumption $R(O)$ is open and dense. Since $Y$ is second countable
it has a countable topological base of nonempty open sets~$O_k$, $k
\in \N$. Since $X$ is a complete metric space this implies
that
\[
R = \bigcap_{k \in \N} R(O_k)
\]
is residual in~$X$. If $x \in R$ and $U$ is a nonempty open set in~$Y$
we have $O_k \tm U$ for some $k$ and,
since $x \in R(O_k)$, $x \in f_n^{1}(O_k)$
for some $n \in \N$, i.e.\
$f_n(x) \in O_k \tm U$.
This proves that $(f_n(x))_{n \in \N}$ is dense in~$Y$,
i.e., $A(\ff(x))=Y$. Thus (iii) follows from theorem \ref{2}.
\end{proof}
\begin{Remark}
To get an analogous metric result [with ``maldistributed'' replaced by
``uniformly distributed''] one has to use Fubini's theorem instead of
KuratowskiUlam's. A sufficient condition on the sequence of
functions for this case is mentioned in Koksma's metric theorem
(cf.~\cite[p.34]{KN}).
\end{Remark}
\noindent
\begin{Elementary examples}
If the maps $f_n$ are expanding in an appropriate sense
the condition of theorem \ref{10} might be fulfilled, maybe
under additional hypotheses on~$X$ and~$Y$. Concrete examples
are: $X= T \tm \R$, $Y=I$, $f_n(x) = a_n x$, $f_n(x) = x^{a_n}$,
$f_n(x) = a_n^x$
etc., or similar sequences on the $s$dimensional torus $X=Y=I^s$.
\end{Elementary examples}
% LIEBER JOERG! AN DIESER STELLE SCHIESST MIR IN DEN SINN, DASS DU
% JA SPEZIALIST DARIN BIST, HAUSDORFFDIMENSIONEN ZU BERECHNEN.
% HIER WIRST DU SICHER LUST HABEN, DEN ABBILDUNGEN GEEIGNETE
% DIFFERENZIERBARKEITEIGENSCHAFTEN UND UNTERE SCHRANKEN IHRER
% ABLEITUNGEN ZU UNTERSTELLEN, UM INTERESSANTE ERGEBNISSE ZU ERHALTEN.
% AN DIESE WERDEN SICH WAHRSCHEINLICH BEISPIELE OBIGER GESTALT ANSCHLIESSEN.
% DESHALB HABE ICH MICH MIT DIESEN BEISPIELEN KURZ GEFASST, WEIL
% ICH DIER NICHT INS HANDWERK PFUSCHEN WOLLTE. ICH BITTE DICH DAHER
% AN DIESER STELLE UM DAS ENTSPRECHENDE THEOREM. SOLLTEN ES GLEICH MEHRERE
% SEIN  NUR ZU!
%
\begin {Examples} The most natural application is to consider subsequences of
ergodic iterationsi.e. sequences $T^{n_k}$ where $T\colon X\to X$ is a
map. The most interesting case
(from the point of view of uniform distributions)
would be if
$T$ is ergodic. This means
\[
\lim_{n\to\infty}\frac1n \sum_{k=0}^{n1}\mu(A\cap T^{k} B)=\mu(A)\mu(B)
\]
for all measurable sets $A$ and~$B$.
This property (for subsequences) obviously implies the above
assumptions for the support of the measure. Unfortunately, this property is
not inherited by subsequences.
(This is one reason why subsequences of~$n\alpha$ are so
interesting.) Therefore it might be better to consider systems which
are topologically mixing. These are maps such that for all pairs of
open sets $U$ and $V$ there is a number $N$ such that for all $n\ge N$
the intersection $U\cap T^N(V)$ is not empty. This condition follows
if for all measurable sets $A$ and $B$
\[
\lim_{n\to\infty}\mu(A\cap T^{n}B)=\mu(A)\mu(B)
\]
holds. Maps satisfying the latter condition are
called mixing. There are many systems
which are known as mixing. For instance any transitive expanding map of
the ntorus is topologically mixing. Also a broad class of systems
decomposes into mixing components. Unfortunately, group translations,
and hence $n\alpha$sequences, are not mixing.
\end {Examples}
% % \include{res}
% oct 14  still not done
% oct 13, 1997
% april 7, 1997
\def\fin{{\rm fin}}
\def\itm#1 {\item[{(#1)}]}
\def\dalpha{\dot{\alpha}}
\def\dbalpha{\dot{\boldsymbol\alpha}}
\def\balpha{\boldsymbol{\alpha}}
\def\dx{\dot{x}}
\section{Results on~$n\alpha$sequences} \label{8}
We now turn to a special case of the sequences $(f_n(x))_{n \in \N}$
discussed in the previous section, namely, the sequence $(n
\alpha)_{n\in \N}$ and its subsequences.
If $a \in [0,1)$, $S_a = \{ k_1 < k_2 < \cdots \,\}$, then we will
identify $a$ with the sequence $(k_1, k_2, \ldots)$. For any $\alpha
\in \R/\Z$ we let $\balpha$ denote the sequence $(\alpha , 2 \alpha ,
3 \alpha , \ldots\,)$, so $a \balpha $ will be the subsequence $(k_1
\alpha , k_2 \alpha, \ldots\,)$.
In this section we will use the ``forcing'' notation, which is a
{\it fa\c{c}on de parler}
that allows us to talk about ``the typical element'' of a
Polish (i.e., complete separable metric) space.
(We use this notation only for proofs, not for the theorems
themselves. The reader who does not want to learn the forcing
calculus will have no difficulty translating our proofs into more
conventional ones.)
\begin{Definition}
\begin{enumerate}
\item Let $X$ be a Polish space. We call the nonempty open
set
of~$X$ ``conditions'' and we view them as giving us information about
the ``typical element $\dx\in p$''. If $p, q$ are conditions with $p
\supseteq q$,
we write $p \le q$ to emphasize that ``$\dx \in q$'' contains
more information than ``$\dx \in p$.''
For any $B \subseteq X$ we write $p \forces B$ (``$p$ forces~$B$'')
as an abbreviation for ``except for a meager set, every element of~$p$ is in~$B$'' or more formally: ``$p \setminus B$ is meager in
$X$''.
\item For each space $X$ we reserve a special variable~$\dot x$ to
denote the ``generic'' or
``typical'' element of~$X$. The ``typical'' real number will be
denoted $\dalpha$.
\\
For any formula (or property) $ \varphi $ we write
$$ p \forces \varphi(\dx)$$
iff $p \forces \{ x: \varphi(x) \}$, i.e., if
``most'' elements
of~$p$ have the property $ \varphi $ (or equivalently, if
most elements $x$
of~$X$ have the property: ``if $x\in p$, then $ \varphi(x)$.'')
\end{enumerate}
\end{Definition}
\begin{Fact}\label{4.1}
If $B$ is a nonmeager Borel set, then there is a condition (i.e, a
nonempty open set) $p$ with $p \forces B$.
\end{Fact}
\begin{proof} All Borel sets have the Baire property
(cf.~\cite[chapter 4]{Ox}).
\end{proof}
\begin{Fact}\label{f}
Assume that $ \varphi(x) $ and $\psi(x)$ are formulas such that the
corresponding sets $\{x: \varphi (x) \}$ and $\{x: \psi (x) \}$ are
Borel sets.
\begin{enumerate}
\item
\label{fcl} % cl=clopen
If $p \Delta q$ is meager, then $p \forces \varphi(\dx) $ iff
$ q \forces \varphi(\dx) $. (We will not need the converse, which is also
true: If ``$p \forces \varphi(\dx) $ implies $q \forces \varphi(\dx) $'' for
every Borel formula $ \varphi(\dx) $, then $ p \setminus q $ is meager.)
\item
\label{fa} % a=and
$p \forces\varphi(\dx) \& \psi(\dx) $ iff: $p \forces \varphi(\dx) $ and
$p \forces \psi(\dx) $.
\item
\label{fn} % n=negation
``$p \forces \varphi(\dx)$'' is false iff there is~$q \ge p$ such
that $q \forces \lnot \varphi(\dx) $. (Apply fact~\ref{4.1} to $p
\setminus \{ x: \varphi(x)\}$.)
\item
\label{fe} % e=existential quantifier
If $ p \forces \exists k\in \N\, \varphi(\dx, k)$
then
there is~$q \ge p$ and $k_0\in \N$ such that
$ q \forces \varphi(\dx, k_0)$.
(This is a thinly disguised version of Baire's theorem.)
\item
\label{fi} % i=inconsistency
It is impossible that $p \forces \varphi(\dx) $ and $p \forces \lnot
\varphi(\dx) $.
\item
\label{fs} % s=stronger
If $p \forces \varphi(\dx) $, $q \ge p$, then also $q \forces
\varphi(\dx) $.
% \item
% \label{fo} % o=open
% For every $p$ there is~$q \ge p$, $q\not=\emptyset $ open, $q
% \setminus p$ meager. (See \cite[???]{Ox})
\end{enumerate}
\end{Fact}
\bigskip
\begin{Theorem} \label{251}
Let ${a} = ({k}_0, {k}_1, {k}_2, \ldots\,)$ be a sequence of natural
numbers.
If\/ $\liminf_n ({k}_{n+1}/{k}_n) > 1 $, then the
set $$ \{ \alpha: {a} \dbalpha \hbox { is u.d. } \} $$
is meager.
\end{Theorem}
This theorem answers a question of \v{S}al\'at \cite{Sa}, who has proved
a similar
result under the assumption
$\liminf_n ({k}_{n+1}/{k}_n) \ge 2 $.
We prepare ourselves for the proof of theorem~\ref{251} with a definition and
a lemma.
\begin{Definition}
We say that a family of functions $(f_1, f_2, \ldots)$ is $
\varepsilon $mixing if: for all intervals~$J$ of length $
\varepsilon $ and all $k\in \N$: $$ \bigcap_{n=1}^k
f^{1}_n(J) \ \not= \ \emptyset $$
More generally we say that $(f_1, f_2, \ldots)$
is $(\varepsilon,{\delta})$mixing if: for all
intervals $J$ of
length $\varepsilon $, and all $k\in N$, and all intervals~$J'$
of length $ {\delta} $:
\[
J' \cap \bigcap_{n=1}^k f^{1}_n(J) \ \not= \ \emptyset
\]
\end{Definition}
\begin{Remark}
Although we work in~$I=[0,1)$, we identify elements in~$I$ with their
equivalence classes in~$\R/\Z$, so an (open) interval can either be of
the form $(a,b)$ or of the form $[0,a) \cup (b,1)$ (for $0\le a \frac{2}{ \varepsilon } {k}_n$ for all~$n$.
\item $ {k}_1 > \frac{\varepsilon } {{\delta} }$.
\end{enumerate}
Then $(f_1, f_2, \ldots\, )$ is $(\varepsilon, {\delta})$mixing.
\end{Lemma}
\begin{proof}
Let~$J$ be an interval of length $ \varepsilon $, $ J' $ an interval
of length $ {\delta} $.
% (We assume $ \varepsilon, {\delta} \in
% (0,1 )$, and by ``interval of length $ {\delta} $ me mean either a set
% $(a, a+ {\delta} )$ with $0 \le a < a+ {\delta} \le 1$ or a set
% $[0, {\delta}  a) \cup (1a, 1)$, with $0 \le a< {\delta} < 1$.
We will show (by induction on~$m$) that
each set $$ J' \cap \bigcap_{i=1}^m f^{1}_i(J)$$
contains in fact an interval $I_m$ of length $ \varepsilon/{k}_m$.
This is clear for~$m=0$, as the length of~$J'$ is $ {\delta} >
\varepsilon/{k}_0$.
Consider~$m>0$.
Note that $f^{1}_m(J) =
\bigcup_{j=1}^{{k}_m} \{\alpha : {k}_m\cdot \alpha  j \in J \}$ is a union
of~${k}_m$ many disjoint intervals, each of length $ \varepsilon /{k}_m$.
By inductive assumption, the set $ J' \cap \bigcap_{i=1}^{m1}
f^{1}_i(J)$
contains an interval $I_{m1} $ of
length $ \varepsilon/{k}_{m1} $.
Since $ \varepsilon / {k}_{m1} > 2/{k}_m$,
we can find a natural number~$j $
such that the interval $$ [\frac{j}{{k}_m}, \frac{ j+1}{{k}_m} ] =
\{ \alpha : {k}_m \cdot \alpha  j \in
[0,1]\}$$
is contained in~$I_{m1} $.
Hence the set
\[I_m:= \{\alpha : {k}_m \cdot \alpha  j \in J \},
\]
an interval of length $ \varepsilon / {k}_m$, is also contained in~$I_m$.
\end{proof}
\bigskip
\noindent {\bf Proof of theorem \ref{251}}
Let $1 < q < \liminf_n ({k}_{n+1}/{k}_n)$. Without loss of generality
we have ${k}_{n+1} / {k}_n>q$ for
{\em all}~$ k$.
Let $\dalpha $ be the ``generic'' real number.
Towards a
contradiction, assume that ``${a} \dbalpha $ is not u.d.'' is not forced,
then by fact~\ref{f}(\ref{fn}) there is a condition~$p$ such
that
\[ p \forces {a} \dbalpha \hbox{ is u.d.}
\]
Pick an integer $c>1$ such that $q ^ c \ge 9 c$, and let $ \varepsilon =
\frac{1}{3c}$. Let $J$ be any interval of length $ \varepsilon $.
So we have $$ p \forces \frac {1}{n} \#\{j
\frac{\varepsilon}{{\delta}} $. Define
for $i=0, 1,2,\ldots, 100 n_0$: $$ f_i(x) = {k}_{(n_0+i)c}x. $$
Note that $ {k}_{(n_0+i+1)c}/ {k}_{(n_0+i)c} \ge q^c \ge 9c > 6c =
2/\varepsilon $.
Hence (by lemma \ref{17}) the functions $f_0, f_1, \ldots$
are $(\varepsilon, {\delta})$mixing, so there is an
interval~$ q \subseteq p $ of positive length such that
\[ q \subseteq \bigcap_{i=0}^{100n_0} f_i^{1}(J) \]
and hence:
\[ q \forces \forall i \in \{0, \ldots, 100n_0\}:
{k}_{(n_0+i)c} \dalpha \in J.\]
But this means that
\[ q \forces \#\{j\le 100n_0: {k}_j \alpha \in J \} \ge
\#\{n_0 c, (n_0+1)c,\ldots, 100n_0c\} \ge 99n_0>\frac{1}{2}(100n_0)\]
a contradiction to~$(*)$.
\begin{Theorem}\label{explicit}
If ${a}=\kn$ with $\lim
k_{n+1}/k_n = \infty$, then the set $\{ \alpha: a\balpha
\mbox{m.d.}\}$ is residual.
\end{Theorem}
\begin{proof} Similar to the above (but easier).
\end{proof}
\bigskip
% DAS FOLGENDE PASST JEDENFALLS NICHT HIERHER. VIELLEICHT KOENNTE MAN
% ES ZU EINER BEMERKUNG KONDENSIEREN UND IN DIE EINLEITUNG VERLEGEN.
Obviously, ``$\liminf$'' is not an appropriate concept
for questions dealing with uniform distribution, since
for example a modification of a very thin subsequence of a sequence of
reals can change the limes inferior without affecting the
distribution behavior of the sequence.
A more appropriate concept
would be the following:
\begin{Definition}
Let $\cJ$ be an ideal of subsets of~$\N$, $x=\xn$ a sequence of real
numbers. We define the lower limit (with respect to~$\cJ$) of this
sequence as follows:
\[ \liminf\nolimits^\cJ x
:= \sup \{ t\in \R \mid \{n:x_n < t \} \in \cJ \},\]
or equivalently,
\[ \liminf\nolimits^\cJ x
:= \sup \{ t\in \R \mid \{n:x_n \le t \} \in \cJ \}.\]
We define $\limsup^\cJ$ dually:
$ \limsup\nolimits^\cJ x
:= \inf \{ t\in \R \mid \{n:x_n \ge t \} \in \cJ \}$.
\end{Definition}
Thus, the usual notion of~$\liminf$ is $\liminf^{\text{fin}}$, where
$\text{fin}$ is the ideal of finite subsets of~$\N$. In our context
the ideal of sets of density~$0$ is the more interesting notion.
For example, we can refine the previous theorem to the following:
\begin{Remark} \label{261}
Let $\cJ$ be the ideal of sets with density 0.
Let ${a} = ( {k}_1, {k}_2, \ldots \,)$, ${k}_1 < {k}_2 < \cdots $.
If $\liminf^\cJ_n ({k}_{n+1}/{k}_n) > 1 $, then the set
\[ \{ \alpha: a \alpha \hbox { is u.d. } \} \]
is meager.
\end{Remark}
We leave the (easy) proof of this fact, as well as the formulation of
similar generalizations for other theorems, to the reader.
%
% \proof Let $1 < q < \liminf^\cJ_n ({k}_{n+1}/{k}_n) > 1 $. Define a
% sequence $(z_n)_{n\in \N}$ by $z_1={k}_1$, $ z_{n+1} =
% \min \{ {k}_j: {k}_j > z_n, {k}_j > q\cdot {k}_z\}$. Then we have
% \[ \#\{ {k}_n:nn$
and a sequence $(k_{n+1}, \ldots, k_m)$ such that
\begin{itemize}
\item $(k_1, \ldots, k_m) \in X_c^\fin$
\item $k_m\ge 2R\cdot m$
\end{itemize}
\item Let~$n$, $R$, $m$ be as above, and assume that
$k_{m+\ell+1}k_{m+\ell} \in [R, 2R]$ for all $\ell\ge 0$.
Then
\begin{itemize}
\itm a for all~$i$: $k_{m+i} \ge R(m+i)$
\itm b $(k_1, k_1, \ldots) \in X_c$
\end{itemize}
\end{enumerate}
\end{Lemma}
\noindent
\begin{proof}[Proof of (1)] %%% ??? !!!
We may assume $n \ge 4R$ [otherwise we start by
replacing $(k_1, \ldots, k_n)$
by~$(k_1, \ldots, k_n, k_n+1, k_n+2,
\ldots, k_n+(4Rn))$]. Let $m= n^3$, and define
\begin{itemize}
\item $k_{n+\ell} = k_n+2\ell$ for $n+\ell\le n^2$.
\item $k_{j n^2 + \ell} = k_{k n^2}+(k+1)\ell$ for $j=1,\ldots,
n1$, $\ell=1,\ldots, n^2$.
\end{itemize}
%
%
We first notice $k_i \ge i $ for all~$i$, hence
\[ \frac{k_{n+\ell+1}}{k_{n+\ell}} \le 1+ \frac{2}{k_{n+\ell}}
\le
1+ \frac{2}{n+\ell} \]
for all $\ell < n^2n$. Next, we see by induction that
\[ k_{\ka n^2} \ge \frac{\ka(\ka+1)}{2} n^2 \ \ \ \
\mbox{for $\ka=1, \ldots, n$} .
\]
so
\begin{align*}
\frac{ k_{\ka n^2 + \ell + 1}} { k_{\ka n^2 + \ell }} &\le
1 + \frac{ \ka+1 } { k_{\ka n^2 + \ell }} =
1 + \frac{ \ka+1 } { k_{\ka n^2} + \ell(\ka+1) } \le \\
&\le
1 + \frac{ \ka+1 } { \frac {1}{2} \ka(\ka+1)n^2 + \ell(\ka+1)} =
1 + \frac{ 2 } { \ka n^2 + 2\ell} \le 1+\frac{2}{\ka n^2 + \ell}.
\end{align*}
Finally, we have $k_{n^3} \ge \frac{n(n+1)}{2} n^2 > n^3\cdot \frac
{n}{2} \ge Rn^3$.
\smallskip
The proof of (2) is trivial.
\end{proof}
\begin{Fact}
\label{continuous}
% \begin{enumerate}
%\item The map $\xf \mapsto M(\xf)$ is continuous (if we use the
% box topology
% on~$X^\N$ and the Hausdorff metric on the closed
% subsets of~$\M(X)$.
%\item Let $U \subseteq \M(X)$ be open, $\mu \in U$. Then the closed
%set~$\{\mu\}$ is an interior point of the set
%$\{ F \subseteq U: F \mbox{ closed }\}$.
%\item
If $\mu_{N,\xf} \to \mu_0$ (weakly, i.e., $\xf$ is $\mu_0$u.d.),
$U \subseteq \M(X)$ an open
set containing~$\mu_0$, then there is $ \varepsilon>0$ such that for
all $\yf \in X^ \N$:
\[ \mbox{ If\/ $ \ \xf  \yf \_\infty < \varepsilon$,
\
then $ M(\yf) \subseteq U$} \]
%\end{enumerate}
\end{Fact}
% \proof MARTIN!!! \qed
\bigskip
\noindent {\bf Proof of theorem \ref{262}:}
Fix an irrational number $ \alpha $, and assume that the set
\[ \{ {\xfora} \in X_{ c }: {\xfora} \balpha \hbox{ is m.d.}\}\]
is not residual. So there is an open set~$p$ such that
\[ p \forces \mbox{ $ \dxf \balpha $ is not m.d.} \]
Now ``$\dxf \balpha $ is not m.d.'' is equivalent to
``$M(\dxf \balpha ) \not= \M(X)$'', so since $M(\dxf \balpha )$ must be
closed this means
\[ p \forces \mbox{there is a basic open set~$U$ such that $U\cap
M(\dxf \balpha ) =\emptyset$.}\]
By \ref{f}(\ref{fe}) can we can find a basic open set $U \subseteq
\M(I)$ and a condition $p' \ge p$ such that
\[ p' \forces U \cap M(\dxforda\balpha) = \emptyset\]
By strengthening $p'$ to a condition $q$ and by shrinking~$U$ we can even
get
\[ q \forces \forall n:
\mu_{n, {\dxforda}\balpha} \notin U\]
Now we will use the fact that $q$ gives only finitely much information
about $\dxforda$.
Let $\mu_0\in U$, and let $\yf = \yn$ be a
sequence with $\mu_{n,\yf}\to \mu_0$.
Choose $ \varepsilon $ as in fact~\ref{continuous}.
The condition $q$ is a cylinder ~$q = [q_1, \ldots, q_n]$.
% basic neighborhood in~$X_c$, defined by $(q_1, \ldots, q_n)\in
% X_c^\fin$.
Without loss of generality % $n \ge k_1$ and
$1/n < \varepsilon $.
Choose $R$ so large that the set $\{ \alpha, 2\alpha, \ldots,
R\alpha\}$ meets every interval of length $\varepsilon $ (mod 1).
By lemma~\ref{71} we may extend the sequence $(q_1, \ldots, q_n)$ to a
sequence $(q_1, \ldots, q_m)$ satisfying $q_m \ge 2R m$.
For all $i > m $ we can now choose
\[q_i \in [q_{i1}+R, q_{i1}+2R]\]
such that $ \alpha q_i \in [y_i\varepsilon, y_i+\varepsilon]$ (mod 1),
and $\qn \in X_c$. Let ${\bf q} = (q_n)_{n\in \N}$.
We have $M({\bf q}) \subseteq
U$, so for large enough $n$ we have $\mu_{n,{\bf q}\balpha} \in U$.
Since trivially $(q_1, \ldots, q_n) \forces \dot {k}_1 = q_1,
\ldots, \dot {k}_n=q_n$, we have
\[ (q_1, \ldots, q_n) \forces \mu_{n,\dxf \balpha} \in U,\]
a contradiction.
% Let $r:= (q_1, \ldots, q_{m^2})$. Then
% \[ r \forces \{ i \le m^2: k_i \alpha \in I \} \supseteq \{ i\le m^2:
% i > m \} \]
% hence
% \[ r \forces card \{ i \le m^2: k_i \alpha \in I \} \ge m^2m >
% m^2(1 \varepsilon ),\]
% a contradiction.
%
\bigskip
We now deal with the (easier) case where $c<2$. It turns out that in
this case the space $X_c$ has a trivial structure: It contains a
countable dense set of isolated points.
\begin{Theorem}\label{263}
Let $c < 2$. Then
for every irrational number $ \alpha $ the set
\[ \{ {a} = \kn \in X_{ c }: {\xfora} \balpha \hbox{ is u.d.}\}\]
contains a countable dense open subset of~$X_{ c}$.
\end{Theorem}
\begin{proof}
Let \[X^*_c:= \{ {\xfora} \in X_c: \exists \ka \,
\exists N\, \forall n\ge N \
k_n = n+\ka\}\]
Clearly $ X^*_c$ is a countable dense subset of~$X_c$, and for any
irrational $ \alpha $ we know that $ {\xfora} \dalpha $ is u.d.\
for
every $ {\xfora}
\in X^*_c$. It remains to show that $X^*_c$ is open.
We will show that all points of~$ X^*_c$ are isolated in~$ X_c$.
Assume that for all $ n> N$ we have $k_n=n+\ka$. Without loss of generality we have
$N\cdot (\frac{2}{c} 1) > \ka$, so~$(n+\ka)\cdot \frac{c}{n} < 2$ for all~$n\ge N$. Let
\[ U:=\{ \yf= \yn\in X_c: y_1=k_1, \ldots, y_N=k_N\}\]
Then $U$ is a basic open neighborhood of~${\xfora}$, and ${\xfora} $
is the only point in~$U$.
\end{proof}
Consider the following question: given $a$, what is
the behavior of~$a \balpha$ for ``typical'' $\alpha$?
\relax From theorem~\ref{2} we know that for a residual set of~$a$'s the
typical behavior of~$a \balpha$ is maldistribution. (An explicit
example of such an~$a$ is mentioned in theorem~\ref{explicit}.)
For some $a$'s (e.g., $a= (1,2,3,\ldots\,)$) the typical behavior is
uniform distribution.
In general, the following dichotomy holds:
\begin{Theorem} For any $a = \kn$ the set
\[
A = \{ \alpha \in \R/\Z: \mbox{$a \balpha$ is u.d.}\}
\]
is either meager or residual.
\end{Theorem}
\begin{proof}
Assume that $A$ is not meager, then there is an interval $[a,b]$
such that $A \cap [a,b]$ is residual in $[a,b]$. Clearly, $ \alpha
\in A$ implies $n \alpha \in A$ for any $n \in \N$. Now for
$n > \frac{1}{ba}$ the set $n\cdot A$ is residual in $n \cdot [a,b]
= \R/\Z$.
\end{proof}
\begin{Open Problem}
Is it true that
\begin{quote}
for all $a$, one of the sets
\[ \{ \alpha: a\balpha \mbox{ is u.d.}\}
\qquad
\{ \alpha: a\balpha \mbox{ is m.d.}\}
\]
is residual?
\end{quote}
The proof of theorem~\ref{251} suggests a negative answer.
\end{Open Problem}
\bigskip
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\end{document}
% LocalWords: Theo Cor Def Mal pt diam cf