measurements. Figure discussed in Section 3.10. Figure 21: The fourteen panels
represent the fourteen observations of IGR J17091-3624 analyzed. The top and
bottom panels show the residuals in terms of $\sigma$ produced when fitting
the spectra with TBabs*(diskbb+powerlaw) (top) and the best reflection model
(bottom), respectively, together with the statistic produced. The NuSTAR FPMA
spectra are shown in blue and the FPMB spectra in red. Figure discussed in
Section 3.11. Figure 22: The left panel shows the posterior distributions
resulting from the MCMC analysis of IGR J17091-3624 for spin, while the right
panel shows the posterior distributions for the inclination of the inner
accretion disk in the model. The different colored lines represent the
fourteen observations analyzed, and the black curves represent the combined
inferred distribution, with the thickness of the lines being proportional to
the ratio of the reflected flux to the total flux in the 3-79 keV band, which
were used as weighting when combining the posterior distributions. The solid
vertical black lines represent the modes of the combined distribution, and the
dashed vertical black lines represent the $1\sigma$ credible intervals of the
measurements. Figure discussed in Section 3.11. Figure 23: The five panels
represent the five observations of GRS 1716-249 analyzed. The top panels show
the unfolded spectra, with the NuSTAR FPMA spectra shown in blue and FPMB
spectra in red. The reported best-fit models are shown by the solid lines,
while the contributions of the relxill components are shown by the dotted
lines and the contributions of the diskbb components are shown through the
dashed lines. The middle and bottom panels show the residuals in terms of
$\sigma$ produced when fitting the spectra with TBabs*(diskbb+powerlaw)
(middle) and the best-performing reflection models (bottom), respectively,
together with the statistic produced. Figure discussed in Section 3.12. Figure
24: The left panel shows the posterior distributions resulting from the MCMC
analysis of GRS 1716-249 for spin, while the right panel shows the posterior
distributions for the inclination of the inner accretion disk in the model.
The different colored lines represent the five observations analyzed, and the
black curves represent the combined inferred distribution, with the thickness
of the lines being proportional to the ratio of the reflected flux to the
total flux in the 3-79 keV band, which were used as weighting when combining
the posterior distributions. The solid vertical black lines represent the
modes of the combined distribution, and the dashed vertical black lines
represent the $1\sigma$ credible intervals of the measurements. Figure
discussed in Section 3.12. Figure 25: The top panels shows the unfolded
spectra of the NuSTAR observations of GRS 1739-278 analyzed, with the NuSTAR
FPMA spectra shown in blue and FPMB spectra in red. The reported best-fit
models are shown by the solid lines, while the contributions of the relxill
components are shown by the dotted lines and the contributions of the diskbb
components are shown through the dashed lines. The middle and bottom panels
show the residuals in terms of $\sigma$ produced when fitting the spectra with
TBabs*(diskbb+powerlaw) (middle) and with the best-performing reflection
models (bottom), respectively, together with the statistic produced. Figure
discussed in Section 3.13. Figure 26: The blue and red curves in the left
panel show the posterior distribution resulting from the MCMC analysis of GRS
1739-278 for spin, while in the right panel they show the posterior
distribution for the inclination of the inner accretion disk in the model. The
black curves represent the distributions inferred using our posterior
combining algorithm on the posterior distributions resulting from the MCMC
runs. The solid vertical black lines represent the modes of the distributions
inferred using the combining algorithm, and the dashed vertical black lines
represent the $1\sigma$ credible intervals of the measurements. Figure
discussed in Section 3.13. Figure 27: The top panel shows the unfolded spectra
of the observation of 1E 1740.7-2942 analyzed, with the NuSTAR FPMA spectra
shown in blue and FPMB spectra in red. The reported best-fit models are shown
by the solid lines, while the contributions of the relxill component are shown
by the dotted lines and the contributions of the diskbb component are shown
through the dashed lines. The middle and bottom panels show the residuals in
terms of $\sigma$ produced when fitting the spectra with
TBabs*(diskbb+powerlaw) (middle) and TBabs*(diskbb+relxill) (bottom),
respectively, together with the statistic produced. Figure discussed in
Section 3.14. Figure 28: The blue curve in the left panel shows the posterior
distribution resulting from the MCMC analysis of 1E 1740.7-2942 for spin,
while in the right panel it shows the posterior distribution for the
inclination of the inner accretion disk in the model. The black curves
represent the distributions inferred using our posterior combining algorithm
on the individual posterior distribution from the MCMC run on ObsID
10002012001. The solid vertical black lines represent the modes of the
distributions inferred using the combining algorithm, and the dashed vertical
black lines represent the $1\sigma$ credible intervals of the measurements.
The red curves show the posterior distributions for spin and inclination
obtained when running the MCMC analysis on fits to the NuSTAR spectra from
ObsID 10002012001 when allowing the power law index to vary freely between the
spectra from the two NuSTAR FPM detectors. Figure discussed in Section 3.14.
Figure 29: Source and background regions used for extracting spectra of T15
(left) and T37 (right). See text in Sections 3.15 and 3.16.  Figure 30: The
top panel shows the unfolded spectra of the observation of T15 analyzed, with
the NuSTAR FPMA spectra shown in blue and FPMB spectra in red. The reported
best-fit models are shown by the solid lines, while the contributions of the
relxill component are shown by the dotted lines and the contributions of the
diskbb component are shown through the dashed lines. The middle and bottom
panels show the residuals in terms of $\sigma$ produced when fitting the
spectra with TBabs*(diskbb+powerlaw) (middle) and
TBabs*(diskbb+relxillD+gauss) (bottom), respectively, together with the
statistic produced. Figure discussed in Section 3.15. Figure 31: The blue
curve in the left panel shows the posterior distribution resulting from the
MCMC analysis of T15 for spin, while in the right panel it shows the posterior
distribution for the inclination of the inner accretion disk in the model. The
black curves represent the distributions inferred using our posterior
combining algorithm on the individual posterior distribution from the MCMC run
on ObsID 90101022002. The solid vertical black lines represent the modes of
the distributions inferred using the combining algorithm, and the dashed
vertical black lines represent the $1\sigma$ credible intervals of the
measurements. Figure discussed in Section 3.15. Figure 32: The top panel shows
the unfolded spectra of the observation of T37 analyzed, with the NuSTAR FPMA
spectra shown in blue and FPMB spectra in red. The reported best-fit models
are shown by the solid lines, while the contributions of the relxill component
are shown by the dotted lines and the contributions of the diskbb component
are shown through the dashed lines. The middle and bottom panels show the
residuals in terms of $\sigma$ produced when fitting the spectra with
TBabs*(diskbb+powerlaw) (middle) and TBabs*(diskbb+relxillCp) (bottom),
respectively, together with the statistic produced. Figure discussed in
Section 3.16. Figure 33: The red curve in the left panel shows the posterior
distribution resulting from the MCMC analysis of T37 for spin, while in the
right panel it shows the posterior distribution for the inclination of the
inner accretion disk in the model. The black curves represent the
distributions inferred using our posterior combining algorithm on the
individual posterior distribution from the MCMC run on ObsID 90201026002. The
solid vertical black lines represent the modes of the distributions inferred
using the combining algorithm, and the dashed vertical black lines represent
the $1\sigma$ credible intervals of the measurements. Figure discussed in
Section 3.16. Figure 34: The top panels shows the unfolded spectra of the
NuSTAR observations of MAXI J1803-298 analyzed, with the NuSTAR FPMA spectra
shown in blue and FPMB spectra in red. The reported best-fit models are shown
by the solid lines, while the contributions of the relxill components are
shown by the dotted lines, the contributions of the diskbb components are
shown through the dashed lines, and the dash-dotted line shows the excess
emission lines required by the data. The middle and bottom panels show the
residuals in terms of $\sigma$ produced when fitting the spectra with
TBabs*(diskbb+powerlaw) (middle) and with the best-performing reflection
models (bottom), respectively, together with the statistic produced. Figure
discussed in Section 3.17. Figure 35: The left panel shows the posterior
distributions resulting from the MCMC analysis of MAXI J1803-298 for spin,
while the right panel shows the posterior distributions for the inclination of
the inner accretion disk in the model. The different colored lines represent
the four observations analyzed, and the black curves represent the combined
inferred distribution, with the thickness of the lines being proportional to
the ratio of the reflected flux to the total flux in the 3-79 keV band, which
were used as weighting when combining the posterior distributions. The solid
vertical black lines represent the modes of the combined distribution, and the
dashed vertical black lines represent the $1\sigma$ credible intervals of the
measurements. Figure discussed in Section 3.17. Figure 36: The three panels
represent the three observations of MAXI J1813-095 analyzed. The top panels
show the unfolded spectra, with the NuSTAR FPMA spectra shown in blue and FPMB
spectra in red. The reported best-fit models are shown by the solid lines,
while the contributions of the relxill components are shown by the dotted
lines and the contributions of the diskbb components are shown through the
dashed lines. The middle and bottom panels show the residuals in terms of
$\sigma$ produced when fitting the spectra with TBabs*(diskbb+powerlaw)
(middle) and the best-performing reflection models (bottom), respectively,
together with the statistic produced. Figure discussed in Section 3.18. Figure
37: The left panel shows the posterior distributions resulting from the MCMC
analysis of MAXI J1813-095 for spin, while the right panel shows the posterior
distributions for the inclination of the inner accretion disk in the model.
The different colored lines represent the three observations analyzed, and the
black curves represent the combined inferred distribution, with the thickness
of the lines being proportional to the ratio of the reflected flux to the
total flux in the 3-79 keV band, which were used as weighting when combining
the posterior distributions. The solid vertical black lines represent the
modes of the combined distribution, and the dashed vertical black lines
represent the $1\sigma$ credible intervals of the measurements. Figure
discussed in Section 3.18. Figure 38: The top panels shows the unfolded
spectra of the NuSTAR observations of MAXI J1848-015 analyzed, with the NuSTAR
FPMA spectra shown in blue and FPMB spectra in red. The reported best-fit
models are shown by the solid lines, while the contributions of the relxill
components are shown by the dotted lines, the contributions of the diskbb
components are shown through the dashed lines, and the dash-dotted line shows
the excess emission lines required by the data. The middle and bottom panels
show the residuals in terms of $\sigma$ produced when fitting the spectra with
TBabs*(diskbb+powerlaw) (middle) and with the best-performing reflection
models (bottom), respectively, together with the statistic produced. Figure
discussed in Section 3.19. Figure 39: The blue and red curves in the left
panel show the posterior distribution resulting from the MCMC analysis of MAXI
J1848-015 for spin, while in the right panel they show the posterior
distribution for the inclination of the inner accretion disk in the model. The
black curves represent the distributions inferred using our posterior
combining algorithm on the posterior distributions resulting from the MCMC
runs. The solid vertical black lines represent the modes of the distributions
inferred using the combining algorithm, and the dashed vertical black lines
represent the $1\sigma$ credible intervals of the measurements. Figure
discussed in Section 3.19. Figure 40: The six panels represent the six
observations of EXO 1846-031 analyzed. The top panels show the unfolded
spectra, with the NuSTAR FPMA spectra shown in blue and FPMB spectra in red.
The reported best-fit models are shown by the solid lines, while the
contributions of the relxill components are shown by the dotted lines and the
contributions of the diskbb components are shown through the dashed lines. The
middle and bottom panels show the residuals in terms of $\sigma$ produced when
fitting the spectra with TBabs*(diskbb+powerlaw) (middle) and the best-
performing reflection models (bottom), respectively, together with the
statistic produced. Figure discussed in Section 3.20. Figure 41: The left
panel shows the posterior distributions resulting from the MCMC analysis of
EXO 1846-031 for spin, while the right panel shows the posterior distributions
for the inclination of the inner accretion disk in the model. The different
colored lines represent the six observations analyzed, and the black curves
represent the combined inferred distribution, with the thickness of the lines
being proportional to the ratio of the reflected flux to the total flux in the
3-79 keV band, which were used as weighting when combining the posterior
distributions. The solid vertical black lines represent the modes of the
combined distribution, and the dashed vertical black lines represent the
$1\sigma$ credible intervals of the measurements. Figure discussed in Section
3.20. Figure 42: The two panels represent the two observations of XTE
J1908+094 analyzed. The top panels show the unfolded spectra, with the NuSTAR
FPMA spectra shown in blue and FPMB spectra in red. The reported best-fit
models are shown by the solid lines, while the contributions of the relxill
components are shown by the dotted lines and the contributions of the diskbb
components are shown through the dashed lines. The middle and bottom panels
show the residuals in terms of $\sigma$ produced when fitting the spectra with
TBabs*(diskbb+powerlaw) (middle) and the best-performing reflection models
(bottom), respectively, together with the statistic produced. Figure discussed
in Section 3.21. Figure 43: The left panel shows the posterior distributions
resulting from the MCMC analysis of XTE J1908+094 for spin, while the right
panel shows the posterior distributions for the inclination of the inner
accretion disk in the model. The different colored lines represent the two
observations analyzed, and the black curves represent the combined inferred
distribution, with the thickness of the lines being proportional to the ratio
of the reflected flux to the total flux in the 3-79 keV band, which were used
as weighting when combining the posterior distributions. The solid vertical
black lines represent the modes of the combined distribution, and the dashed
vertical black lines represent the $1\sigma$ credible intervals of the
measurements. Figure discussed in Section 3.21. Figure 44: The seventeen
panels represent the seventeen observations of GRS 1915+105 analyzed. The top
and bottom panels show the residuals in terms of $\sigma$ produced when
fitting the spectra with TBabs*(diskbb+powerlaw) (top) and the best reflection
model (bottom), respectively, together with the statistic produced. The NuSTAR
FPMA spectra are shown in blue and the FPMB spectra in red. Figure discussed
in Section 3.22. Figure 45: The left panel shows the posterior distributions
resulting from the MCMC analysis of GRS 1915+105 for spin, while the right
panel shows the posterior distributions for the inclination of the inner
accretion disk in the model. The different colored lines represent the
seventeen observations analyzed, and the black curves represent the combined
inferred distribution, with the thickness of the lines being proportional to
the ratio of the reflected flux to the total flux in the 3-79 keV band, which
were used as weighting when combining the posterior distributions. The solid
vertical black lines represent the modes of the combined distribution, and the
dashed vertical black lines represent the $1\sigma$ credible intervals of the
measurements. Figure discussed in Section 3.22. Figure 46: The thirty-four
panels represent the thirty-four observations of Cygnus X-1 analyzed. The top
and bottom panels show the residuals in terms of $\sigma$ produced when
fitting the spectra with TBabs*(diskbb+powerlaw) (top) and the best reflection
model (bottom), respectively, together with the statistic produced. The NuSTAR
FPMA spectra are shown in blue and the FPMB spectra in red. Figure discussed
in Section 3.23. Figure 47: The left panel shows the posterior distributions
resulting from the MCMC analysis of Cygnus X-1 for spin, while the right panel
shows the posterior distributions for the inclination of the inner accretion
disk in the model. The different colored lines represent the observations
analyzed, and the black curves represent the combined inferred distribution,
with the thickness of the lines being proportional to the ratio of the
reflected flux to the total flux in the 3-79 keV band, which were used as
weighting when combining the posterior distributions. The solid vertical black
lines represent the modes of the combined distribution, and the dashed
vertical black lines represent the $1\sigma$ credible intervals of the
measurements. Figure discussed in Section 3.23. Figure 48: The top panels
shows the unfolded spectra of the NuSTAR observations of V404 Cygnus analyzed,
with the NuSTAR FPMA spectra shown in blue and FPMB spectra in red. The
reported best-fit models are shown by the solid lines, while the contributions
of the relxill components are shown by the dotted lines, and the contributions
of the diskbb components are shown through the dashed lines. The middle and
bottom panels show the residuals in terms of $\sigma$ produced when fitting
the spectra with TBabs*(diskbb+powerlaw) (middle) and with the best-performing
reflection models (bottom), respectively, together with the statistic
produced. The flaring state of the source is shown in the left panels, and is
indicated by “$\\_$f,” while the non-flaring state is shown in the right
panels, indicated by “$\\_$n.” Figure discussed in Section 3.24. Figure 49:
The left panel shows the posterior distributions resulting from the MCMC
analysis of V404 Cygnus for spin, while the right panel shows the posterior
distributions for the inclination of the inner accretion disk in the model.
The different colored lines represent the flaring (“$\\_$f”) and non-flaring
(“$\\_$n”) states during the two observations analyzed, and the black curves
represent the combined inferred distribution, with the thickness of the lines
being proportional to the ratio of the reflected flux to the total flux in the
3-79 keV band, which were used as weighting when combining the posterior
distributions. The solid vertical black lines represent the modes of the
combined distribution, and the dashed vertical black lines represent the
$1\sigma$ credible intervals of the measurements. Figure discussed in Section
3.24. Figure 50: Comparison between the new measurements obtained in this
paper and the old values reported in the literature for spin (top-left) and
inclination (top-right). The bottom-left panel shows a comparison between
$1-a_{\rm new}$ and $1-a_{\rm old}$ in logarithmic scale, to better highlight
the differences in measurements where both the initial and final spin are
nearly maximal. All values are taken from Table 1. Figure discussed in Section
4. Figure 51: Top: Change in spin (left) and inclination (right) when compared
to the previously existing measurements. Bottom-left: same as top-left, but
focused on $a_{\rm old}\geq 0.75$, for visual clarity. Bottom-right: change in
spin vs change in inclination. The markers represent the same sources as in
Figure 50, with all values taken from Table 1. Figure discussed in Section 4.


# Superdensity and super-micro-uniformity
in non-integrable flat systems

J. Beck  Department of Mathematics, Rutgers University, Hill Center for the
Mathematical Sciences, Piscataway NJ 08854, USA<EMAIL_ADDRESS>and
W.W.L. Chen  Department of Mathematics and Statistics, Macquarie University,
Sydney NSW 2109, Australia<EMAIL_ADDRESS>
###### Abstract.

We show that on any non-integrable finite polysquare translation surface,
superdensity, an optimal form of time-quantitative density, leads to an
optimal form of time-quantitative uniformity that we call super-micro-
uniformity.

###### Key words and phrases:

geodesics, billiards, density, uniformity

###### 2010 Mathematics Subject Classification:

11K38, 37E35

##  1\. Introduction

Consider billiard in a polygon that satisfies the usual law of reflection in
optics. It is trivial that uniformity of a half-infinite orbit always implies
density, and that the converse is false. However, while density does not in
general imply uniformity, we show here an interesting case when some form of
time-quantitative density implies some form of time-quantitative uniformity.

The purpose of the present paper is to show how superdensity, an optimal form
of time-quantitative density, implies an optimal form of time-quantitative
uniformity that we call super-micro-uniformity. Here super refers to
optimality and micro refers to microscopic scale.

To illustrate the latter, consider the irrational rotation sequence

$\\{k\alpha\\},\quad k=1,2,3,\ldots,$ (1.1)

of fractional parts of $k\alpha$ in the interval $[0,1)$, where $\alpha$ is
irrational. Let $I\subset[0,1)$ be an arbitrary subinterval of length $1/2n$,
and consider the first $n$ elements of the sequence (1.1). Then the expected
number of elements of this $n$-element set in $I$ is clearly $1/2$,
corresponding to $n$ times the length of $I$. On the other hand, the visiting
number $V_{n}(I)$ of $I$, the actual number of elements in $I$ coming from
this $n$-element set is clearly an integer, and so must differ from the
expected number by at least $1/2$. We refer to this as the trivial error.
Indeed, we have same phenomenon if the length of $I$ is $C/n$, where $2C$ is
an odd integer. Here the error is at least $1/2$, and the expected number $C$
is in the constant range.

Given the first $n$ elements of the infinite sequence (1.1), intervals of
length $C/n$ represent test sets in the microscopic scale. Here $C>0$ is a
fixed constant, and $n$ may tend to infinity. The trivial error argument above
implies that in the microscopic scale of $C/n$, we cannot expect perfect local
uniformity in the sense that the ratio of the error term and the expected
number tends to zero as $C$ is fixed and $n$ tends to infinity. To put it
slightly differently, to have perfect local uniformity, it is necessary to
have $C=C(n)\to\infty$ as $n\to\infty$.

It turns out that this necessary condition is sufficient to establish perfect
local uniformity if $\alpha$ is badly approximable. This perfect local
uniformity is what we call super-micro-uniformity. It has the intuitive
meaning that the orbit exhibits uniformity already in the shortest possible
subintervals. We have the following result on super-micro-uniformity of the
irrational rotation sequence generated by a badly approximable $\alpha$.

######  Theorem A.

Let $\alpha$ be a badly approximable real number. For any subinterval
$I\subset[0,1)$, let

$V_{n}(I)=\\{k=1,\ldots,n:\\{k\alpha\\}\in I\\}$

denote the visiting number of $I$ with respect to the first $n$ terms of the
sequence (1.1). Then for every real number $\varepsilon>0$, there exists a
finite threshold $C_{\varepsilon}=C_{\varepsilon}(\alpha)$ such that for any
subinterval $I$ with length $|I|\geqslant C_{\varepsilon}/n$, the inequality

$|V_{n}(I)-n|I||<\varepsilon n|I|$ (1.2)

holds.

The proof of this result is a fairly routine exercise using continued
fractions, so Theorem A is very possibly folklore. However, as we shall
establish a more general result, we briefly outline the ideas here.

First of all, we recall that the convergents $p_{m}/q_{m}$ of $\alpha$ give
excellent rational approximation, in the sense that

$\left|\alpha-\frac{p_{m}}{q_{m}}\right|<\frac{1}{q_{m}q_{m+1}},$

so that

$\left|k\alpha-\frac{kp_{m}}{q_{m}}\right|<\frac{k}{q_{m}q_{m+1}}\leqslant\frac{1}{q_{m+1}},\quad
k=1,\ldots,q_{m}.$

This implies that any segment of $q_{m}$ consecutive terms of the sequence
(1.1) is extremely uniformly distributed in the interval $[0,1)$.

To take advantage of this, it makes sense to look at the Ostrowski
decomposition of integers, using the denominators of the convergents. For
every integer $N$, we can write

$N=\sum_{i=0}^{n}b_{i}q_{i},$

where $n$ is the unique integer satisfying $q_{n}\leqslant N<q_{n+1}$, and the
digits $b_{0},b_{1},\ldots,b_{n}$ satisfy

$\begin{array}[]{ll}b_{0}\in\\{0,1,\ldots,a_{1}-1\\},\vspace{2pt}\\\
b_{i}\in\\{0,1,\ldots,a_{i+1}\\},&i=1,\ldots,n,\vspace{2pt}\\\ b_{i-1}=0\mbox{
if }b_{i}=a_{i+1},&i=1,\ldots,n.\end{array}$

where $a_{1},\ldots,a_{n+1}$ are continued fraction digits of $\alpha$.

Theorem A follows on combining these two ideas in a suitable way.

From the discrete super-micro-uniformity given by (1.2), it is easy to deduce
that every half-infinite geodesic of badly approximable slope $\alpha$ is
super-micro-uniform in the unit torus.

Note that geodesics on the unit torus is the simplest integrable system. If we
consider geodesic flow on an arbitrary finite polysquare translation surface,
then it is typically non-integrable.

######  Theorem 1.

Let $\mathcal{P}$ be a polysquare translation surface with $b$ atomic squares,
and let $\alpha$ be a badly approximable real number. Let $L_{\alpha}(t)$,
$t\geqslant 0$, be a half-infinite geodesic with slope $\alpha$, equipped with
the usual arc-length parametrization. For any positive integer $n$, let
$\mathcal{X}_{n}$ denote the set of the first $n$ intersection points of
$L_{\alpha}(t)$, $t\geqslant 0$, with the vertical edges of $\mathcal{P}$, and
for any subinterval $I$ of any vertical edge of $\mathcal{P}$, let

$V_{n}(I)=|I\cap\mathcal{X}_{n}|$

denote the visiting number of $I$ with respect to $\mathcal{X}_{n}$. Then for
every real number $\varepsilon>0$, there exists a finite threshold
$C_{\varepsilon}=C_{\varepsilon}(\mathcal{P};\alpha)$ such that for any
subinterval $I$ of any vertical edge of $\mathcal{P}$ with length
$|I|\geqslant C_{\varepsilon}/n$, the inequality

$\left|V_{n}(I)-\frac{n|I|}{b}\right|<\varepsilon\frac{n|I|}{b}$

holds. In other words, we have super-micro-uniformity.

The remainder of the paper is devoted to proving this result.

We require a superdensity result in our earlier papers [1, 2]. Let
$\mathcal{P}$ be a polysquare translation surface with $b$ atomic squares, and
let $\alpha$ be a badly approximable real number. Then there exists a finite
superdensity threshold $c_{0}=c_{0}(\mathcal{P};\alpha)$ such that for every
integer $m\geqslant 1$, any geodesic segment of slope $\alpha$ and length
$c_{0}m$ gets $(1/m)$-close to every point of $\mathcal{P}$.

##  2\. Iteration process: step 0

Let $C$ be a constant satisfying $1<C<n$. Let $\mathcal{I}_{n}(\mathcal{P};C)$
denote the collection of any subinterval $I$ of any vertical edge of
$\mathcal{P}$ with length $|I|=C/n$, and let
$I_{0},I_{1}\in\mathcal{I}_{n}(\mathcal{P};C)$ be subintervals satisfying

$V_{n}(I_{0})=\min_{I\in\mathcal{I}_{n}(\mathcal{P};C)}|I\cap\mathcal{X}_{n}|\quad\mbox{and}\quad
V_{n}(I_{1})=\max_{I\in\mathcal{I}_{n}(\mathcal{P};C)}|I\cap\mathcal{X}_{n}|,$

so that $I_{0}$ and $I_{1}$ have respectively the smallest and largest
visiting numbers with respect to $\mathcal{X}_{n}$ among all the subintervals
$I$ under consideration. It is clear that

$|I_{0}\cap\mathcal{X}_{n}|\leqslant\frac{C}{b}\leqslant|I_{1}\cap\mathcal{X}_{n}|.$
(2.1)

Let the real number $\varepsilon$ satisfy $0<\varepsilon<1/2$. We have two
cases:

######  Case A.

We have

$\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{1}\cap\mathcal{X}_{n}|}\geqslant
1-\varepsilon.$ (2.2)

######  Case B.

We have

$\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{1}\cap\mathcal{X}_{n}|}<1-\varepsilon.$
(2.3)

We shall postpone the analysis of Case A to Section 6.

To complete the proof of Theorem 1, we shall show that Case B, where (2.3)
holds, is not possible. Indeed, we shall show that (2.3) leads to a
contradiction. The proof is rather long, and involves a complicated iteration
process, with two possibilities at each step. We shall derive the necessary
contradiction by showing that at some stage of this process, neither
possibility is valid.

We divide the interval $I_{0}$ into $3$ subintervals of equal length, and
denote by $I_{0}(0)$ one of them with the minimum intersection with the set
$\mathcal{X}_{n}$, so that

$\frac{|I_{0}(0)\cap\mathcal{X}_{n}|}{|I_{0}(0)|}\leqslant\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}.$
(2.4)

The $\alpha$-flow moves $I_{0}(0)$ forward. Consider the first time that the
image hits a vertex and splits into $2$ parts, and denote by $I_{0}(1)$ the
larger part. Clearly $I_{0}(1)$ has length

$|I_{0}(1)|\geqslant\frac{|I_{0}(0)|}{2}\geqslant\frac{C}{6n},$

and one of its endpoints is a vertex of $\mathcal{P}$. The $\alpha$-flow moves
$I_{0}(1)$ forward. Consider the first time that the image hits a vertex and
splits into $2$ parts, and denote by $I_{0}(2)$ the larger part. Clearly
$I_{0}(2)$ has length

$|I_{0}(2)|\geqslant\frac{|I_{0}(1)|}{2}\geqslant\frac{C}{12n},$

and one of its endpoints is a vertex of $\mathcal{P}$. The $\alpha$-flow moves
$I_{0}(2)$ forward, and so on. We obtain a sequence $I_{0}(j)$,
$j=0,1,2,3,\ldots,$ of intervals such that the length

$|I_{0}(j)|\geqslant\frac{C}{2^{j}3n},\quad j=0,1,2,3,\ldots,$

and each interval $I_{0}(j)$, $j=1,2,3,\ldots,$ has an endpoint which is a
vertex of $\mathcal{P}$. We next apply the reverse $\alpha$-flow to send each
of these intervals $I_{0}(j)$, $j=1,2,3,\ldots,$ back to a subinterval
$I^{*}_{0}(j)$ of $I_{0}$, giving rise to a nested sequence

$I_{0}\supset I_{0}(0)\supset I^{*}_{0}(1)\supset I^{*}_{0}(2)\supset
I^{*}_{0}(3)\supset\ldots.$

Furthermore, the $\alpha$-flow moves each subinterval $I^{*}_{0}(j)$,
$j=1,2,3,\ldots,$ of $I_{0}$ to $I_{0}(j)$ split-free.

Let $j_{0}$ denote the smallest integer $j=1,2,3,\ldots$ such that
$\operatorname{length}(j_{0})$, the length of the $\alpha$-flow moving
$I^{*}_{0}(j)$ to $I_{0}(j)$, satisfies

$\operatorname{length}(j_{0})>\frac{c_{0}}{|I_{1}|/3}=\frac{3c_{0}n}{C},$
(2.5)

where $c_{0}=c_{0}(\mathcal{P};\alpha)$ is the superdensity threshold.
Superdensity and (2.5) then imply that the $\alpha$-flow moving
$I^{*}_{0}(j_{0})$ forward must intersect the middle-third of the interval
$I_{1}$ before reaching $I_{0}(j_{0})$, and let $I^{*}_{1}(j_{0})$ denote the
image of $I^{*}_{0}(j_{0})$ under the $\alpha$-flow on the vertical edge of
$\mathcal{P}$ that contains $I_{1}$, as shown in Figure 1. Clearly we have

$I^{*}_{1}(j_{0})\subset
I_{1}\quad\mbox{and}\quad\frac{C}{3n}\geqslant|I^{*}_{1}(j_{0})|=|I_{0}(j_{0})|\geqslant\frac{C}{2^{j_{0}}3n}.$
(2.6)

$\begin{array}[]{c}\includegraphics{figure-1.pdf}\\\ \mbox{Figure 1: the
subinterval $I^{*}_{1}(j_{0})$ on the same vertical edge of $\mathcal{P}$ as
$I_{1}$}\end{array}$

It also follows from the definition of $j_{0}$ that
$\operatorname{length}(j_{0}-1)$, the length of the $\alpha$-flow moving
$I^{*}_{0}(j-1)$ to $I_{0}(j-1)$, satisfies

$\operatorname{length}(j_{0}-1)\leqslant\frac{c_{0}}{|I_{1}|/3}=\frac{3c_{0}n}{C}.$
(2.7)

Consider the sequence $q_{1},q_{2},q_{3},\ldots$ of the denominators of the
convergents of $\alpha$. There exists a unique integer $h$ such that

$q_{h}\leqslant\frac{1}{|I_{0}(j_{0}-1)|}<q_{h+1}<(A+1)q_{h},$ (2.8)

where $A$ is an upper bound of the continued fraction digits of $\alpha$.

######  Lemma 1.

Suppose that $q_{h}$ is the denominator of a convergent of $\alpha$ and $I$ is
an interval of real numbers. Consider the equation

$\\{\beta+k\alpha\\}=0,\quad\beta\in I,\quad k=1,\ldots,q_{h}.$

_(i)_ If $I$ has length $|I|\leqslant 1/q_{h}$, then the equation has at most
$3$ solutions $(\beta,k)$.

_(ii)_ If $I$ has length $|I|\geqslant 3/q_{h}$, then the equation has at
least $1$ solution $(\beta,k)$.

###### Proof.

It is well known that the convergent $p_{h}/q_{h}$ of $\alpha$ satisfies

$\left|\alpha-\frac{p_{h}}{q_{h}}\right|<\frac{1}{q_{h}q_{h+1}}.$

It follows that

$\left|k\alpha-\frac{kp_{h}}{q_{h}}\right|<\frac{k}{q_{h}q_{h+1}}\leqslant\frac{1}{q_{h+1}},\quad
k=1,\ldots,q_{h}.$ (2.9)

Consider the mapping

$k\alpha\mapsto\left\\{\frac{kp_{h}}{q_{h}}\right\\}=\frac{\ell(k)}{q_{h}},\quad
k=1,\ldots,q_{h}.$ (2.10)

Since $p_{h}$ and $q_{h}$ are coprime, the numbers $kp_{h}$,
$k=1,\ldots,q_{h}$, form a complete set of residues modulo $q_{h}$, so that

$\\{\ell(k):k=1,\ldots,q_{h}\\}=\\{0,1,\ldots,q_{h}-1\\}.$ (2.11)

Combining (2.9)–(2.11), we conclude that there is a $1$-$1$ correspondence
between the collection of values $k\alpha$, $k=1,\ldots,q_{h}$, and the
collection of intervals

$\displaystyle\mathfrak{I}_{0}$
$\displaystyle=\left[0,\frac{1}{q_{h+1}}\right)\cup\left(1-\frac{1}{q_{h+1}},1\right),$
(2.12) $\displaystyle\mathfrak{I}_{\ell}$
$\displaystyle=\left(\frac{\ell}{q_{h}}-\frac{1}{q_{h+1}},\frac{\ell}{q_{h}}+\frac{1}{q_{h+1}}\right),\quad\ell=1,\ldots,q_{h}-1.$
(2.13)

In particular, we have

$k\alpha\in\mathfrak{I}_{\ell(k)}\bmod{1},\quad k=1,\ldots,q_{h}.$ (2.14)

(i) Suppose that $|I|\leqslant 1/q_{h}$ and $\beta_{0}$ is the lower endpoint
of the interval $I$. It follows from (2.14) that for every $\beta\in I$, we
have

$\\{\beta+k\alpha\\}\in\mathfrak{I}^{*}_{\ell(k)}(\beta_{0})\bmod{1},\quad
k=1,\ldots,q_{h},$

where the intervals $\mathfrak{I}^{*}_{\ell}(\beta_{0})\subset[0,1)$,
$\ell=0,1,\ldots,q_{h}-1$, satisfy

$\displaystyle\mathfrak{I}^{*}_{0}(\beta_{0})$
$\displaystyle=\left[\beta_{0},\frac{1}{q_{h+1}}+\beta_{0}+\frac{1}{q_{h}}\right)\cup\left(1-\frac{1}{q_{h+1}}+\beta_{0},1+\beta_{0}\right)\bmod{1},$
(2.15) $\displaystyle\mathfrak{I}^{*}_{\ell}(\beta_{0})$
$\displaystyle=\left(\frac{\ell}{q_{h}}-\frac{1}{q_{h+1}}+\beta_{0},\frac{\ell}{q_{h}}+\frac{1}{q_{h+1}}+\beta_{0}+\frac{1}{q_{h}}\right)\bmod{1},\quad\ell=1,\ldots,q_{h}-1.$
(2.16)

Note now that each of the $q_{h}$ intervals in (2.15) and (2.16) is of length
less than $3/q_{h}$, so can contain at most $1$ integer. Together they form a
$q_{h}$-cycle through translation by $1/q_{h}$, so that at most $3$ of them
can contain an integer.

(ii) Suppose that $|I|\geqslant 3/q_{h}$ and $\beta_{0}$ is the lower endpoint
of the interval $I$. It follows from (2.14) that

$\\{\beta_{0}+k\alpha\\}\in\mathfrak{I}_{\ell(k)}+\beta_{0},\quad
k=1,\ldots,q_{h},$

where the intervals $\mathfrak{I}_{\ell}+\beta_{0}$,
$\ell=0,1,\ldots,q_{h}-1$, are the translates of the intervals (2.12) and
(2.13) modulo $1$. Consider the collection of intervals

$\mathfrak{I}^{**}_{\ell}(\beta_{0})=\left[\beta_{0}+\frac{\ell+1}{q_{h}},\beta_{0}+\frac{\ell+2}{q_{h}}\right]\bmod{1},\quad\ell=0,\ldots,q_{h}-1.$
(2.17)

It is then not difficult to see that

$\\{\\{\beta+k\alpha\\}:\beta\in
I\\}\supset\mathfrak{I}^{**}_{\ell(k)}(\beta_{0}),\quad k=1,\ldots,q_{h}.$
(2.18)

Note finally that the union of the $q_{h}$ sets in (2.17) is the whole
interval $[0,1)$. It follows from this and (2.18) that $\\{\beta+k\alpha\\}=0$
for some $\beta\in I$ and $k=1,\ldots,q_{h}$. ∎

Let us now focus on the transportation process as the $\alpha$-flow moves the
interval $I^{*}_{0}(j_{0}-1)$ to the interval $I_{0}(j_{0}-1)$. It follows
from the first inequality of (2.8) and Lemma 1(i) that, among any $q_{h}$
consecutive intersections with the vertical edges of $\mathcal{P}$, the flow
hits a vertex at most $3$ times. Now the common length of a geodesic segment
of slope $\alpha$ between consecutive intersections with the vertical edges of
$\mathcal{P}$ is $\sqrt{1+\alpha^{2}}$. In view of the bound (2.7), we clearly
have

$\frac{(j_{0}-1)q_{h}\sqrt{1+\alpha^{2}}}{3}\leqslant\operatorname{length}(j_{0}-1)\leqslant\frac{3c_{0}n}{C}.$
(2.19)

Combining this with the trivial upper bound for $|I_{0}(j_{0}-1)|\leqslant
C/n$ and the second inequality in (2.8), we obtain

$j_{0}-1\leqslant\frac{9c_{0}n}{Cq_{h}\sqrt{1+\alpha^{2}}}\leqslant\frac{9c_{0}n(A+1)|I_{0}(j_{0}-1)|}{C\sqrt{1+\alpha^{2}}}\leqslant\frac{9c_{0}(A+1)}{\sqrt{1+\alpha^{2}}}=c_{1},$
(2.20)

where the constant $c_{1}$ is independent of the parameters $n$, $C$ and
$\varepsilon$. Combining (2.6) and (2.20), we deduce that

$\frac{1}{3}|I_{1}|\geqslant|I^{*}_{1}(j_{0})|\geqslant\frac{c_{2}C}{n}=c_{2}|I_{1}|,$
(2.21)

where the constant

$c_{2}=\frac{1}{2^{c_{1}+1}3}$ (2.22)

is independent of the parameters $n$, $C$ and $\varepsilon$.

Consider again the sequence $q_{1},q_{2},q_{3},\ldots$ of the denominators of
the convergents of $\alpha$. There exists a unique integer $h^{*}$ such that

$q_{h^{*}}\geqslant\frac{3}{|I_{0}(j_{0})|}>q_{h^{*}-1}.$ (2.23)

Combining the continued fraction inequality $q_{h^{*}}<(A+1)q_{h^{*}-1}$ with
the inequality $|I_{0}(j_{0})|\geqslant|I_{0}(j_{0}-1)|/2$ and (2.8), we have

$q_{h^{*}}\leqslant\frac{3(A+1)}{|I_{0}(j_{0})|}\leqslant\frac{6(A+1)}{|I_{0}(j_{0}-1)|}\leqslant
6(A+1)^{2}q_{h}.$ (2.24)

Let us now focus on the transportation process as the $\alpha$-flow moves the
interval $I^{*}_{0}(j_{0})$ to the interval $I_{0}(j_{0})$ via the interval
$I^{*}_{1}(j_{0})$. It follows from the first inequality in (2.23) and Lemma
1(ii) that, among any $q^{h^{*}}$ consecutive intersections with the vertical
edges of $\mathcal{P}$, the flow hits a vertex at least $1$ time. Recall that
the common length of a geodesic segment of slope $\alpha$ between consecutive
intersections with the vertical edges of $\mathcal{P}$ is
$\sqrt{1+\alpha^{2}}$. Thus

$j_{0}q_{h^{*}}\sqrt{1+\alpha^{2}}\geqslant\operatorname{length}(j_{0}).$
(2.25)

Combining (2.19), (2.24) and (2.25), we conclude that for $j_{0}\geqslant 2$,
we have

$\operatorname{length}(j_{0})\leqslant
36(A+1)^{2}\operatorname{length}(j_{0}-1)\leqslant\frac{108(A+1)^{2}c_{0}n}{C}.$
(2.26)

For $j_{0}=1$, combining (2.8), (2.24) and (2.25), we have

$\operatorname{length}(j_{0})\leqslant
6(A+1)^{2}q_{h}\sqrt{1+\alpha^{2}}\leqslant\frac{6(A+1)^{2}\sqrt{1+\alpha^{2}}}{|I_{0}(0)|}=\frac{6(A+1)^{2}n\sqrt{1+\alpha^{2}}}{C}.$

We may assume without loss of generality that $\alpha>1$. Then

$\sqrt{1+\alpha^{2}}<2\alpha<2(a_{0}+1)\leqslant 2(A+1),$

where $a_{0}$ is the first continued fraction digit of $\alpha$. It follows
that for $j_{0}=1$, we have

$\operatorname{length}(j_{0})\leqslant\frac{12(A+1)^{3}n}{C}.$ (2.27)

We may assume that the superdensity threshold $c_{0}\geqslant 1$. Then it
follows from (2.26) and (2.27) that the inequality

$\operatorname{length}(j_{0})\leqslant\frac{108(A+1)^{3}c_{0}n}{C}$ (2.28)

always holds.

We have two cases:

######  Case 1.

We have

$\frac{|I^{*}_{1}(j_{0})\cap\mathcal{X}_{n}|}{|I^{*}_{1}(j_{0})|}\geqslant\left(1-\frac{\varepsilon}{2}\right)\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}.$
(2.29)

######  Case 2.

We have

$\frac{|I^{*}_{1}(j_{0})\cap\mathcal{X}_{n}|}{|I^{*}_{1}(j_{0})|}<\left(1-\frac{\varepsilon}{2}\right)\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}.$
(2.30)

Before we study these cases in detail, we first give some heuristics to
explain the underlying ideas. If Case 1 holds, then we show that the
subinterval $I^{*}_{0}(j_{0})\subset I_{0}$ exhibits a surplus density of
points of $\mathcal{X}_{n}$ compared to $I_{0}$. Removing this subinterval,
the remaining part of $I_{0}$ then exhibits a deficit density of points of
$\mathcal{X}_{n}$ compared to $I_{0}$. If Case 2 holds, then the subinterval
$I^{*}_{1}(j_{0})\subset I_{1}$ exhibits a deficit density of points of
$\mathcal{X}_{n}$ compared to $I_{1}$. Removing this subinterval, the
remaining part of $I_{1}$ then exhibits a surplus density of points of
$\mathcal{X}_{n}$ compared to $I_{1}$. Thus we either obtain a subinterval of
$I_{0}$ that exhibits deficit density of points of $\mathcal{X}_{n}$ compared
to $I_{0}$, or obtain a subinterval of $I_{1}$ that exhibits surplus density
of points of $\mathcal{X}_{n}$ compared to $I_{1}$. We then repeat the
analysis on intervals of length equal to the length of this subinterval, and
this sets up an iteration process. We then show that this iteration has to
terminate after a finite number of steps, and this gives the necessary
contradiction.

Let $I^{\star}$ be an arbitrary subinterval of a vertical edge of
$\mathcal{P}$. Since geodesic flow on $\mathcal{P}$ modulo one is geodesic
flow on the unit torus, and the slope $\alpha$ is badly approximable with
digit upper bound $A$, we have

$|I^{\star}\cap\mathcal{X}_{n}|\leqslant An|I^{\star}|+1.$ (2.31)

###  2.1. Case 1: density decrease

Suppose that the inequality (2.29) holds. To find a lower bound to
$|I^{*}_{0}(j_{0})\cap\mathcal{X}_{n}|$, we consider the transportation
process as the $\alpha$-flow moves the interval $I^{*}_{0}(j_{0})$ to the
interval $I^{*}_{1}(j_{0})$. The length of this flow is at most
$\operatorname{length}(j_{0})$, so the flow intersects the vertical edges of
$\mathcal{P}$ at most

$\frac{\operatorname{length}(j_{0})}{\sqrt{1+\alpha^{2}}}+1<2\operatorname{length}(j_{0})$

times, but these intersection points are not necessarily points of
$\mathcal{X}_{n}$. An analog of (2.31) then says that the number of these
intersection points that are the images of points in $I^{*}_{0}(j_{0})$ under
the $\alpha$-flow is at most

$2A\operatorname{length}(j_{0})|I^{*}_{0}(j_{0})|+1.$

It follows that

$|I^{*}_{0}(j_{0})\cap\mathcal{X}_{n}|\geqslant|I^{*}_{1}(j_{0})\cap\mathcal{X}_{n}|-2A\operatorname{length}(j_{0})|I^{*}_{0}(j_{0})|-1.$
(2.32)

Note that (2.29) gives

$|I^{*}_{1}(j_{0})\cap\mathcal{X}_{n}|\geqslant\left(1-\frac{\varepsilon}{2}\right)|I^{*}_{1}(j_{0})|\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|},$
(2.33)

while (2.28) and $|I_{1}|=C/n$ give

$2A\operatorname{length}(j_{0})\leqslant\frac{216A(A+1)^{3}c_{0}}{|I_{1}|}.$
(2.34)

Combining (2.32)–(2.34) and noting that $|I_{1}\cap\mathcal{X}_{n}|\geqslant
C/b$ and $|I^{*}_{0}(j_{0})|=|I^{*}_{1}(j_{0})|$, we obtain the inequality

$|I^{*}_{0}(j_{0})\cap\mathcal{X}_{n}|\geqslant\left(1-\frac{3\varepsilon}{4}\right)|I^{*}_{0}(j_{0})|\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|},$
(2.35)

provided that each of the last two terms on the right in (2.32) is bounded by

$\frac{\varepsilon}{8}|I^{*}_{0}(j_{0})|\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}.$

This is guaranteed if we ensure that

$C\geqslant\frac{1728A(A+1)^{3}c_{0}b}{\varepsilon}\quad\mbox{and}\quad
C\geqslant\frac{8b}{c_{2}\varepsilon},$ (2.36)

the latter in view of (2.21). Combining (2.3) and (2.35) now leads to the
inequality

$\displaystyle|I^{*}_{0}(j_{0})\cap\mathcal{X}_{n}|$
$\displaystyle\geqslant\left(1-\frac{3\varepsilon}{4}\right)(1-\varepsilon)^{-1}|I^{*}_{0}(j_{0})|\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}$
$\displaystyle>\left(1+\frac{\varepsilon}{4}\right)|I^{*}_{0}(j_{0})|\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|},$
(2.37)

provided that (2.36) holds.

There are at most two subintervals $I_{0,1},I_{0,2}\subset I_{0}$ such that

$I_{0}=I^{*}_{0}(j_{0})\cup I_{0,1}\cup I_{0,2}.$ (2.38)

Removing the interval $I^{*}_{0}(j_{0})$ and combining (2.1) and (2.38), we
obtain

$\displaystyle|I_{0,1}\cap\mathcal{X}_{n}|+|I_{0,2}\cap\mathcal{X}_{n}|=|I_{0}\cap\mathcal{X}_{n}|-|I^{*}_{0}(j_{0})\cap\mathcal{X}_{n}|$
$\displaystyle\quad<|I_{0}\cap\mathcal{X}_{n}|\left(1-\left(1+\frac{\varepsilon}{4}\right)\frac{|I^{*}_{0}(j_{0})|}{|I_{0}|}\right)\leqslant|I_{0}\cap\mathcal{X}_{n}|\left(\frac{|I_{0,1}|+|I_{0,2}|}{|I_{0}|}-\frac{c_{2}\varepsilon}{4}\right),$
(2.39)

noting that (2.21) implies $|I^{*}_{0}(j_{0})|/|I_{0}|\geqslant c_{2}$. There
are two possibilities, either

$\frac{\min\\{|I_{0,1}|,|I_{0,2}|\\}}{|I_{0}|}<\frac{c_{2}\varepsilon}{8},$
(2.40)

or

$\frac{\min\\{|I_{0,1}|,|I_{0,2}|\\}}{|I_{0}|}\geqslant\frac{c_{2}\varepsilon}{8}.$
(2.41)

If (2.40) holds, then we may assume without loss of generality that

$|I_{0,1}|\geqslant|I_{0,2}|,\quad\mbox{so
that}\quad\frac{|I_{0,2}|}{|I_{0}|}<\frac{c_{2}\varepsilon}{8},$

and so it follows from (2.1) that

$|I_{0,1}\cap\mathcal{X}_{n}|\leqslant|I_{0}\cap\mathcal{X}_{n}|\left(\frac{|I_{0,1}|}{|I_{0}|}-\frac{c_{2}\varepsilon}{8}\right).$
(2.42)

On the other hand, if (2.41) holds, then since it follows from (2.1) that

$|I_{0,1}\cap\mathcal{X}_{n}|+|I_{0,2}\cap\mathcal{X}_{n}|\leqslant|I_{0}\cap\mathcal{X}_{n}|\left(\frac{|I_{0,1}|}{|I_{0}|}-\frac{c_{2}\varepsilon}{8}\right)+|I_{0}\cap\mathcal{X}_{n}|\left(\frac{|I_{0,2}|}{|I_{0}|}-\frac{c_{2}\varepsilon}{8}\right),$

we may assume without loss of generality that (2.42) holds again. Note now
that (2.42) leads to the inequality

$|I_{0,1}|\geqslant\frac{c_{2}\varepsilon}{8}|I_{0}|,$

as well as the inequality

$\frac{|I_{0,1}\cap\mathcal{X}_{n}|}{|I_{0,1}|}\leqslant\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}\left(1-\frac{c_{2}\varepsilon}{8}\frac{|I_{0}|}{|I_{0,1}|}\right)\leqslant\left(1-\frac{c_{2}\varepsilon}{8}\right)\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}.$
(2.43)

Thus the switch from $I_{0}$ to $I_{0,1}$ leads to density decrease by a
factor of $1-c_{2}\varepsilon/8$.

The ratio $|I_{0}|/|I_{0,1}|$ is not necessarily an integer. To overcome this
issue, we shall replace $I_{0,1}$ by a suitable subinterval at the expense of
part of the density decrease.

We shall use the following almost trivial observation a number of times.

######  Lemma 2.

Suppose that $I$ is a finite interval of real numbers, $\mathcal{Y}\subset I$
is a finite subset with $m$ elements, and $z$ is a real number satisfying
$0<z\leqslant|I|$.

_(i)_ Then there exists a subinterval $I^{\prime}\subset I$ of length
$|I^{\prime}|=z$ such that

$|I^{\prime}\cap\mathcal{Y}|\leqslant m\frac{z}{|I|-z}.$

_(ii)_ Suppose further that there exists an integer $B\geqslant 1$ such that
every subinterval $I^{\dagger}\subset I$ satisfies

$|I^{\dagger}\cap\mathcal{Y}|\leqslant Bm\frac{|I^{\dagger}|}{|I|}+1.$ (2.44)

Then there exists a subinterval $I^{\prime\prime}\subset I$ of length
$|I^{\prime\prime}|=z$ such that

$|I^{\prime\prime}\cap\mathcal{Y}|\geqslant(m-1)\frac{z}{|I|}-Bm\left(\frac{z}{|I|}\right)^{2}.$

###### Proof.

Write $|I|=kz+w$, where $k\geqslant 1$ is an integer and $0\leqslant w<z$. We
partition the interval $I$ into a union

$I=J_{0}\cup J_{1}\cup\ldots\cup
J_{k},\quad|J_{0}|=w,\quad|J_{1}|=\ldots=|J_{k}|=z.$

(i) Among the intervals $J=J_{1},\ldots,J_{k}$, let $I^{\prime}$ be one for
which $|J\cap\mathcal{Y}|$ is minimal. Observing the inequality

$k=\frac{|I|-w}{z}>\frac{|I|-z}{z},$

we deduce that

$|I^{\prime}\cap\mathcal{Y}|\leqslant\frac{|\mathcal{Y}|}{k}\leqslant
m\frac{z}{|I|-z}.$

(ii) Among the intervals $J=J_{1},\ldots,J_{k}$, let $I^{\prime\prime}$ be one
for which $|J\cap\mathcal{Y}|$ is maximal. Applying (2.44) to the subinterval
$I_{0}$, we have

$|I_{0}\cap\mathcal{Y}|<\frac{Bmz}{|I|}+1.$

Observing this and the inequality

$k=\frac{|I|-w}{z}\leqslant\frac{|I|}{z},$

we deduce that

$|I^{\prime\prime}\cap\mathcal{Y}|\geqslant\frac{|\mathcal{Y}|-|I_{0}\cap\mathcal{Y}|}{k}\geqslant\frac{z}{|I|}\left(m-\frac{Bmz}{|I|}-1\right)=(m-1)\frac{z}{|I|}-Bm\left(\frac{z}{|I|}\right)^{2}.$

This completes the proof. ∎

Let $h_{0}$ be the unique integer satisfying

$h_{0}-1<\frac{16|I_{0}|}{c_{2}\varepsilon|I_{0,1}|}\leqslant h_{0},$ (2.45)

so that

$\frac{|I_{0}|}{h_{0}}\leqslant\frac{c_{2}\varepsilon}{16}|I_{0,1}|\leqslant|I_{0,1}|,$
(2.46)

provided that $\varepsilon$ is sufficiently small.

We now apply Lemma 2(i) with $I=I_{0,1}$,
$\mathcal{Y}=I_{0,1}\cap\mathcal{X}_{n}$ and $z=|I_{0}|/h_{0}$. Then there
exists a subinterval $I_{0}(\star)\subset I_{0,1}$ with $|I_{0}(\star)|=z$
such that

$|I_{0}(\star)\cap\mathcal{X}_{n}|\leqslant|I_{0,1}\cap\mathcal{X}_{n}|\frac{z}{|I_{0,1}|-z}.$

Combining this with the estimate (2.43), we deduce that

$\frac{|I_{0}(\star)\cap\mathcal{X}_{n}|}{|I_{0}(\star)|}\leqslant\frac{|I_{0,1}\cap\mathcal{X}_{n}|}{|I_{0,1}|}\,\frac{|I_{0,1}|}{|I_{0,1}|-z}\leqslant\left(1-\frac{c_{2}\varepsilon}{8}\right)\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}\frac{|I_{0,1}|}{|I_{0,1}|-z}.$
(2.47)

Note from (2.46) that

$\frac{|I_{0,1}|}{|I_{0,1}|-z}\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)^{-1}.$

Combining this with (2.47), we deduce that

$\frac{|I_{0}(\star)\cap\mathcal{X}_{n}|}{|I_{0}(\star)|}\leqslant\left(1-\frac{c_{2}\varepsilon}{8}\right)\left(1-\frac{c_{2}\varepsilon}{16}\right)^{-1}\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}.$
(2.48)

Thus the switch from $I_{0}$ to $I_{0}(\star)$ leads to density decrease by a
factor of $1-c_{2}\varepsilon/16$, with the added benefit that the ratio
$|I_{0}|/|I_{0}(\star)|$ is an integer $h_{0}$.

To obtain a subinterval of $I_{1}$ of the same length as $I_{0}(\star)$, we
next divide $I_{1}$ into $h_{0}$ equal parts, and denote by $I_{1}(\star)$ one
of these subintervals with the maximum intersection with the set
$\mathcal{X}_{n}$. Then

$\frac{|I_{1}(\star)\cap\mathcal{X}_{n}|}{|I_{1}(\star)|}\geqslant\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}.$
(2.49)

Note that

$|I_{1}(\star)|=|I_{0}(\star)|=\frac{C_{1}}{n},\quad
C_{1}=\frac{C}{h_{0}},\quad h_{0}<c_{3}(\varepsilon),$ (2.50)

where the constant $c_{3}(\varepsilon)>0$ is independent of $n$ and $C$.

###  2.2. Case 2: density increase

Suppose that the inequality (2.30) holds. There are at most two subintervals
$I_{1,1},I_{1,2}\subset I_{1}$ such that

$I_{1}=I^{*}_{1}(j_{0})\cup I_{1,1}\cup I_{1,2}.$ (2.51)

Removing the interval $I^{*}_{1}(j_{0})$ and combining (2.30) and (2.51), we
obtain

$\displaystyle|I_{1,1}\cap\mathcal{X}_{n}|+|I_{1,2}\cap\mathcal{X}_{n}|=|I_{1}\cap\mathcal{X}_{n}|-|I^{*}_{1}(j_{0})\cap\mathcal{X}_{n}|$
$\displaystyle\quad>|I_{1}\cap\mathcal{X}_{n}|\left(1-\left(1-\frac{\varepsilon}{2}\right)\frac{|I^{*}_{1}(j_{0})|}{|I_{1}|}\right)\geqslant|I_{1}\cap\mathcal{X}_{n}|\left(\frac{|I_{1,1}|+|I_{1,2}|}{|I_{1}|}+\frac{c_{2}\varepsilon}{2}\right),$
(2.52)

noting that (2.21) implies $|I^{*}_{1}(j_{0})|/|I_{1}|\geqslant c_{2}$. There
are two possibilities, either

$\frac{\min\\{|I_{1,1}|,|I_{1,2}|\\}}{|I_{1}|}<\frac{c_{2}\varepsilon}{5Ab},$
(2.53)

or

$\frac{\min\\{|I_{1,1}|,|I_{1,2}|\\}}{|I_{1}|}\geqslant\frac{c_{2}\varepsilon}{5Ab}.$
(2.54)

If (2.53) holds, then we may assume without loss of generality that

$|I_{1,1}|\geqslant|I_{1,2}|,\quad\mbox{so
that}\quad\frac{|I_{1,1}|}{|I_{1}|}\geqslant\frac{1}{3}\quad\mbox{and}\quad\frac{|I_{1,2}|}{|I_{1}|}<\frac{c_{2}\varepsilon}{5Ab}.$
(2.55)

Combining (2.1), (2.31) and (2.55), we obtain

$|I_{1,2}\cap\mathcal{X}_{n}|\leqslant
An|I_{1,2}|+1\leqslant\frac{c_{2}\varepsilon}{5b}n|I_{1}|+1\leqslant\frac{c_{2}\varepsilon}{4b}n|I_{1}|\leqslant\frac{c_{2}\varepsilon}{4}|I_{1}\cap\mathcal{X}_{n}|,$

provided that $n$ is sufficiently large. Substituting this into (2.2), we
deduce that

$|I_{1,1}\cap\mathcal{X}_{n}|\geqslant|I_{1}\cap\mathcal{X}_{n}|\left(\frac{|I_{1,1}|}{|I_{1}|}+\frac{c_{2}\varepsilon}{4}\right).$
(2.56)

On the other hand, if (2.54) holds, then since it follows from (2.2) that

$|I_{1,1}\cap\mathcal{X}_{n}|+|I_{1,2}\cap\mathcal{X}_{n}|\geqslant|I_{1}\cap\mathcal{X}_{n}|\left(\frac{|I_{1,1}|}{|I_{1}|}+\frac{c_{2}\varepsilon}{4}\right)+|I_{1}\cap\mathcal{X}_{n}|\left(\frac{|I_{1,2}|}{|I_{1}|}+\frac{c_{2}\varepsilon}{4}\right),$

we may assume without loss of generality that (2.56) holds again. Note now
that

$|I_{1,1}|\geqslant\frac{c_{2}\varepsilon}{5Ab}|I_{1}|,$ (2.57)

provided that $\epsilon$ is sufficiently small, and (2.56) leads to the
inequality

$\frac{|I_{1,1}\cap\mathcal{X}_{n}|}{|I_{1,1}|}\geqslant\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}\left(1+\frac{c_{2}\varepsilon}{4}\frac{|I_{1}|}{|I_{1,1}|}\right)\geqslant\left(1+\frac{c_{2}\varepsilon}{4}\right)\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}.$
(2.58)

Thus the switch from $I_{1}$ to $I_{1,1}$ leads to density increase by a
factor $1+c_{2}\varepsilon/4$.

The ratio $|I_{1}|/|I_{1,1}|$ is not necessarily an integer. To overcome this
issue, we shall replace $I_{1,1}$ by a suitable subinterval at the expense of
part of the density increase.

Let $h_{0}$ be the unique integer satisfying

$h_{0}-1<\frac{24Ab|I_{1}|}{c_{2}\varepsilon|I_{1,1}|}\leqslant h_{0},$ (2.59)

so that

$\frac{|I_{1}|}{h_{0}}\leqslant\frac{c_{2}\varepsilon}{24Ab}|I_{1,1}|\leqslant|I_{1,1}|,$
(2.60)

provided that $\epsilon$ is sufficiently small.

We now apply Lemma 2(ii) with $I=I_{1,1}$,
$\mathcal{Y}=I_{1,1}\cap\mathcal{X}_{n}$ and $z=|I_{1}|/h_{0}$. Note that in
view of (2.31), for every subinterval $I^{\dagger}\subset I_{1,1}$, we have

$|I^{\dagger}\cap\mathcal{Y}|=|I^{\dagger}\cap\mathcal{X}_{n}|\leqslant
An|I^{\dagger}|+1.$ (2.61)

On the other hand, it follows from (2.1), (2.58) and $|I_{1}|=C/n$ that

$\frac{|I_{1,1}\cap\mathcal{X}_{n}|}{|I_{1,1}|}\geqslant\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}\geqslant\frac{n}{b}.$
(2.62)

Combining (2.61) and (2.62), we have

$|I^{\dagger}\cap\mathcal{Y}|\leqslant
Ab|I_{1,1}\cap\mathcal{X}_{n}|\frac{|I^{\dagger}|}{|I_{1,1}|}+1,$

so that Lemma 2(ii) is valid with the constant $B=Ab$. It follows that there
exists a subinterval $I_{1}(\star)\subset I_{1,1}$ with $|I_{1}(\star)|=z$
such that

$|I_{1}(\star)\cap\mathcal{X}_{n}|\geqslant(|I_{1,1}\cap\mathcal{X}_{n}|-1)\frac{z}{|I_{1,1}|}-Ab|I_{1,1}\cap\mathcal{X}_{n}|\left(\frac{z}{|I_{1,1}|}\right)^{2}.$

Combining this with (2.60), we have

$\displaystyle\frac{|I_{1}(\star)\cap\mathcal{X}_{n}|}{|I_{1}(\star)|}$
$\displaystyle\geqslant\frac{|I_{1,1}\cap\mathcal{X}_{n}|}{|I_{1,1}|}\left(1-\frac{1}{|I_{1,1}\cap\mathcal{X}_{n}|}-\frac{Ab}{h_{0}}\frac{|I_{1}|}{|I_{1,1}|}\right)$
$\displaystyle\geqslant\frac{|I_{1,1}\cap\mathcal{X}_{n}|}{|I_{1,1}|}\left(1-\frac{1}{|I_{1,1}\cap\mathcal{X}_{n}|}-\frac{c_{2}\varepsilon}{24}\right).$
(2.63)

Next, combining (2.57) and (2.62), and recalling that $|I_{1}|=C/n$, we obtain

$|I_{1,1}\cap\mathcal{X}_{n}|\geqslant\frac{c_{2}C\varepsilon}{5Ab^{2}}.$

We want the bound

$\frac{1}{|I_{1,1}\cap\mathcal{X}_{n}|}\leqslant\frac{c_{2}\varepsilon}{24},$
(2.64)

and this can be guaranteed if we ensure that

$C\geqslant\frac{120Ab^{2}}{(c_{2}\varepsilon)^{2}}.$

Combining (2.2) and (2.64), we now obtain

$\frac{|I_{1}(\star)\cap\mathcal{X}_{n}|}{|I_{1}(\star)|}\geqslant\frac{|I_{1,1}\cap\mathcal{X}_{n}|}{|I_{1,1}|}\left(1-\frac{c_{2}\varepsilon}{12}\right).$

Combining this with (2.58), we deduce that

$\frac{|I_{1}(\star)\cap\mathcal{X}_{n}|}{|I_{1}(\star)|}\geqslant\left(1+\frac{c_{2}\varepsilon}{4}\right)\left(1-\frac{c_{2}\varepsilon}{12}\right)\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}\geqslant\left(1+\frac{c_{2}\varepsilon}{12}\right)\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|},$
(2.65)

provided that $\epsilon$ is sufficiently small. Thus the switch from $I_{1}$
to $I_{1}(\star)$ leads to density increase by a factor
$1+c_{2}\varepsilon/12$, with the added benefit that the ratio
$|I_{1}|/|I_{1}(\star)|$ is an integer $h_{0}$.

To obtain a subinterval of $I_{0}$ of the same length as $I_{1}(\star)$, we
next divide $I_{0}$ into $h_{0}$ equal parts, and denote by $I_{0}(\star)$ one
of these subintervals with the minimum intersection with the set
$\mathcal{X}_{n}$. Then

$\frac{|I_{0}(\star)\cap\mathcal{X}_{n}|}{|I_{0}(\star)|}\leqslant\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}.$
(2.66)

Note that

$|I_{0}(\star)|=|I_{1}(\star)|=\frac{C_{1}}{n},\quad
C_{1}=\frac{C}{h_{0}},\quad h_{0}<c_{3}(\varepsilon),$ (2.67)

where the constant $c_{3}(\varepsilon)>0$ is independent of $n$ and $C$.

##  3\. Iteration process: step 1

Let $\mathcal{I}_{n}(\mathcal{P};C_{1})$ denote the collection of any
subinterval $I$ of any vertical edge of $\mathcal{P}$ with length
$|I|=C_{1}/n$, and let
$I_{0}^{(1)},I_{1}^{(1)}\in\mathcal{I}_{n}(\mathcal{P};C_{1})$ be subintervals
satisfying

$V_{n}(I_{0}^{(1)})=\min_{I\in\mathcal{I}_{n}(\mathcal{P};C_{1})}|I\cap\mathcal{X}_{n}|\quad\mbox{and}\quad
V_{n}(I_{1}^{(1)})=\max_{I\in\mathcal{I}_{n}(\mathcal{P};C_{1})}|I\cap\mathcal{X}_{n}|,$

so that $I_{0}^{(1)}$ and $I_{1}^{(1)}$ have respectively the smallest and
largest visiting numbers with respect to $\mathcal{X}_{n}$ among all the
subintervals $I$ under consideration. It is clear that

$|I_{0}^{(1)}\cap\mathcal{X}_{n}|\leqslant\frac{C_{1}}{b}\leqslant|I_{1}^{(1)}\cap\mathcal{X}_{n}|.$

Furthermore, it either, in Case 1, follows from (2.48)–(2.50) that

$\frac{|I_{0}^{(1)}\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}|}\leqslant\frac{|I_{0}(\star)\cap\mathcal{X}_{n}|}{|I_{0}(\star)|}\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}$
(3.1)

and

$\frac{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}|}\geqslant\frac{|I_{1}(\star)\cap\mathcal{X}_{n}|}{|I_{1}(\star)|}\geqslant\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|},$
(3.2)

or, in Case 2, follows from (2.65)–(2.67) that

$\frac{|I_{0}^{(1)}\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}|}\leqslant\frac{|I_{0}(\star)\cap\mathcal{X}_{n}|}{|I_{0}(\star)|}\leqslant\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}$
(3.3)

and

$\frac{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}|}\geqslant\frac{|I_{1}(\star)\cap\mathcal{X}_{n}|}{|I_{1}(\star)|}\geqslant\left(1+\frac{c_{2}\varepsilon}{12}\right)\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}.$
(3.4)

Combining (2.3) with (3.1) and (3.2), or with (3.3) and (3.4), we obtain

$\frac{|I_{0}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}<1-\varepsilon.$
(3.5)

######  Remark.

Note that (3.5) is the analog of (2.3) and Case B in Step 0. It follows that
if Case B in Step 0 holds, then there is no analog of Case A in Step 1.

We divide the interval $I_{0}^{(1)}$ into $3$ subintervals of equal length,
and denote by $I_{0}^{(1)}(0)$ one of them with the minimum intersection with
the set $\mathcal{X}_{n}$, so that

$\frac{|I_{0}^{(1)}(0)\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}(0)|}\leqslant\frac{|I_{0}^{(1)}\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}|}.$

The $\alpha$-flow moves $I_{0}^{(1)}(0)$ forward. Repeating the argument in
Step 0 between (2.4) and (2.22) with $I_{0},I_{1},C$ replaced by
$I_{0}^{(1)},I_{1}^{(1)},C_{1}$, we obtain a subinterval
$I_{1}^{(1*)}(j_{0}^{(1)})\subset I_{1}^{(1)}$ such that

$\frac{1}{3}|I_{1}^{(1)}|\geqslant|I^{(1*)}_{1}(j_{0}^{(1)})|\geqslant\frac{c_{2}C_{1}}{n}=c_{2}|I_{1}^{(1)}|,$
(3.6)

the analog of (2.21). Note that $j_{0}^{(1)}$ may be different from $j_{0}$,
but it still satisfies the upper bound $c_{1}+1$, where the constant $c_{1}$
is given by (2.20). It then follows from (2.22) that the constant $c_{2}$ in
(3.6) is exactly the same as before.

We have two cases:

######  Case 1.

We have

$\frac{|I^{(1*)}_{1}(j_{0}^{(1)})\cap\mathcal{X}_{n}|}{|I^{(1*)}_{1}(j_{0}^{(1)})|}\geqslant\left(1-\frac{\varepsilon}{2}\right)\frac{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}|}.$
(3.7)

######  Case 2.

We have

$\frac{|I^{(1*)}_{1}(j_{0}^{(1)})\cap\mathcal{X}_{n}|}{|I^{(1*)}_{1}(j_{0}^{(1)})|}<\left(1-\frac{\varepsilon}{2}\right)\frac{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}|}.$
(3.8)

###  3.1. Case 1: density decrease

Suppose that the inequality (3.7) holds. Then an argument analogous to that in
Step 0 between (2.23) and (2.1) now leads to the inequality

$|I^{(1*)}_{0}(j_{0}^{(1)})\cap\mathcal{X}_{n}|\geqslant\left(1-\frac{3\varepsilon}{4}\right)|I^{(1*)}_{0}(j_{0}^{(1)})|\frac{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}|},$
(3.9)

provided that

$C_{1}\geqslant\frac{1728A(A+1)^{3}c_{0}b}{\varepsilon}\quad\mbox{and}\quad
C_{1}\geqslant\frac{8b}{c_{2}\varepsilon}.$

Corresponding to (2.38), there are at most two subintervals
$I_{0,1}^{(1)},I_{0,2}^{(1)}\subset I_{0}^{(1)}$ such that

$I_{0}^{(1)}=I^{(1*)}_{0}(j_{0}^{(1)})\cup I_{0,1}^{(1)}\cup I_{0,2}^{(1)}.$

An argument analogous to that in Step 0 between (2.38) and (2.43) then shows
that, without loss of generality,

$|I_{0,1}^{(1)}|\geqslant\frac{c_{2}\varepsilon}{8}|I_{0}^{(1)}|,$

as well as

$\frac{|I_{0,1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{0,1}^{(1)}|}\leqslant\left(1-\frac{c_{2}\varepsilon}{8}\right)\frac{|I_{0}^{(1)}\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}|}.$

An argument analogous to that in Step 0 between (2.45) and (2.48) then leads
to the existence of a subinterval $I_{0}^{(1)}(\star)\subset I_{0}^{(1)}$
satisfying $|I_{0}^{(1)}|/|I_{0}^{(1)}(\star)|=h_{1}$, where $h_{1}$ is the
unique integer satisfying

$h_{1}-1<\frac{16|I_{0}^{(1)}|}{c_{2}\varepsilon|I_{0,1}^{(1)}|}\leqslant
h_{1},$

such that

$\frac{|I_{0}^{(1)}(\star)\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}(\star)|}\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)\frac{|I_{0}^{(1)}\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}|}.$
(3.10)

To obtain a subinterval of $I_{1}^{(1)}$ of the same length as
$I_{0}^{(1)}(\star)$, we next divide $I_{1}^{(1)}$ into $h_{1}$ equal parts,
and denote by $I_{1}^{(1)}(\star)$ one of these subintervals with the maximum
intersection with the set $\mathcal{X}_{n}$. Then

$\frac{|I_{1}^{(1)}(\star)\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}(\star)|}\geqslant\frac{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}|}.$

Note that

$|I_{1}^{(1)}(\star)|=|I_{0}^{(1)}(\star)|=\frac{C_{2}}{n},\quad
C_{2}=\frac{C_{1}}{h_{1}},\quad h_{1}<c_{3}(\varepsilon),$

where the constant $c_{3}(\varepsilon)>0$ is as in Step 0.

###  3.2. Case 2: density increase

Suppose that the inequality (3.8) holds. Then corresponding to (2.51), there
are at most two subintervals $I_{1,1}^{(1)},I_{1,2}^{(1)}\subset I_{1}^{(1)}$
such that

$I_{1}^{(1)}=I^{(1*)}_{1}(j_{0}^{(1)})\cup I_{1,1}^{(1)}\cup I_{1,2}^{(1)}.$
(3.11)

An argument analogous to that in Step 0 between (2.51) and (2.58) then shows
that, without loss of generality,

$|I_{1,1}^{(1)}|\geqslant\frac{c_{2}\varepsilon}{5Ab}|I_{1}^{(1)}|,$

as well as

$\frac{|I_{1,1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1,1}^{(1)}|}\geqslant\left(1+\frac{c_{2}\varepsilon}{4}\right)\frac{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}|}.$

An argument analogous to that in Step 0 between (2.59) and (2.65) then leads
to the existence of a subinterval $I_{1}^{(1)}(\star)\subset I_{1}^{(1)}$
satisfying $|I_{1}^{(1)}|/|I_{1}^{(1)}(\star)|=h_{1}$, where $h_{1}$ is the
unique integer satisfying

$h_{1}-1<\frac{24Ab|I_{1}^{(1)}|}{c_{2}\varepsilon|I_{1,1}^{(1)}|}\leqslant
h_{1},$

such that

$\frac{|I_{1}^{(1)}(\star)\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}(\star)|}\geqslant\left(1+\frac{c_{2}\varepsilon}{12}\right)\frac{|I_{1}^{(1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(1)}|},$

provided that

$C_{1}\geqslant\frac{120Ab^{2}}{(c_{2}\varepsilon)^{2}}.$ (3.12)

To obtain a subinterval of $I_{0}^{(1)}$ of the same length as
$I_{1}^{(1)}(\star)$, we next divide $I_{0}^{(1)}$ into $h_{1}$ equal parts,
and denote by $I_{0}^{(1)}(\star)$ one of these subintervals with the minimum
intersection with the set $\mathcal{X}_{n}$. Then

$\frac{|I_{0}^{(1)}(\star)\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}(\star)|}\leqslant\frac{|I_{0}^{(1)}\cap\mathcal{X}_{n}|}{|I_{0}^{(1)}|}.$

Note that

$|I_{0}^{(1)}(\star)|=|I_{1}^{(1)}(\star)|=\frac{C_{2}}{n},\quad
C_{2}=\frac{C_{1}}{h_{1}},\quad h_{1}<c_{3}(\varepsilon),$

where the constant $c_{3}(\varepsilon)>0$ is as in Step 0.

##  4\. Iteration process: general step

Let $\mathcal{I}_{n}(\mathcal{P};C_{i})$ denote the collection of any
subinterval $I$ of any vertical edge of $\mathcal{P}$ with length
$|I|=C_{i}/n$, and let
$I_{0}^{(i)},I_{1}^{(i)}\in\mathcal{I}_{n}(\mathcal{P};C_{i})$ be subintervals
satisfying

$V_{n}(I_{0}^{(i)})=\min_{I\in\mathcal{I}_{n}(\mathcal{P};C_{i})}|I\cap\mathcal{X}_{n}|\quad\mbox{and}\quad
V_{n}(I_{1}^{(i)})=\max_{I\in\mathcal{I}_{n}(\mathcal{P};C_{i})}|I\cap\mathcal{X}_{n}|,$

so that $I_{0}^{(i)}$ and $I_{1}^{(i)}$ have respectively the smallest and
largest visiting numbers with respect to $\mathcal{X}_{n}$ among all the
subintervals $I$ under consideration. It is clear that

$|I_{0}^{(i)}\cap\mathcal{X}_{n}|\leqslant\frac{C_{i}}{b}\leqslant|I_{1}^{(i)}\cap\mathcal{X}_{n}|.$

Furthermore, we either, in Case 1 in the previous step and analogous to (3.1)
and (3.2), have

$\frac{|I_{0}^{(i)}\cap\mathcal{X}_{n}|}{|I_{0}^{(i)}|}\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)\frac{|I_{0}^{(i-1)}\cap\mathcal{X}_{n}|}{|I_{0}^{(i-1)}|}$
(4.1)

and

$\frac{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}|}\geqslant\frac{|I_{1}^{(i-1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i-1)}|},$
(4.2)

or, in Case 2 in the previous step and analogous to (3.3) and (3.4), have

$\frac{|I_{0}^{(i)}\cap\mathcal{X}_{n}|}{|I_{0}^{(i)}|}\leqslant\frac{|I_{0}^{(i-1)}\cap\mathcal{X}_{n}|}{|I_{0}^{(i-1)}|}$
(4.3)

and

$\frac{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}|}\geqslant\left(1+\frac{c_{2}\varepsilon}{12}\right)\frac{|I_{1}^{(i-1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i-1)}|}.$
(4.4)

Combining the estimate

$\frac{|I_{0}^{(i-1)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i-1)}\cap\mathcal{X}_{n}|}<1-\varepsilon$

from the previous step with (4.1) and (4.2), or with (4.3) and (4.4), we
obtain

$\frac{|I_{0}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}<1-\varepsilon,$

the analog of (2.3) and (3.5).

On the other hand, iterating (4.1)–(4.4) carefully, we obtain

$\frac{|I_{0}^{(i)}\cap\mathcal{X}_{n}|}{|I_{0}^{(i)}|}\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)^{i_{1}}\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}$
(4.5)

and

$\frac{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}|}\geqslant\left(1+\frac{c_{2}\varepsilon}{12}\right)^{i_{2}}\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|},$
(4.6)

where $i_{1}$ and $i_{2}$ denote respectively the number of times Case 1 and
Case 2 are valid in the previous $i=i_{1}+i_{2}$ steps. Combining (2.1), (4.5)
and (4.6), and recalling that $|I_{0}|=|I_{1}|$ and
$|I_{0}^{(i)}|=|I_{1}^{(i)}|$, we obtain the inequality

$|I_{0}^{(i)}\cap\mathcal{X}_{n}|\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)^{i_{1}}|I_{1}^{(i)}\cap\mathcal{X}_{n}|.$
(4.7)

We divide the interval $I_{0}^{(i)}$ into $3$ subintervals of equal length,
and denote by $I_{0}^{(i)}(0)$ one of them with the minimum intersection with
the set $\mathcal{X}_{n}$. The $\alpha$-flow moves $I_{0}^{(i)}(0)$ forward.
Corresponding to (2.21) and (3.6), we obtain a subinterval
$I_{1}^{(i*)}(j_{0}^{(i)})\subset I_{1}^{(i)}$ such that

$\frac{1}{3}|I_{1}^{(i)}|\geqslant|I^{(i*)}_{1}(j_{0}^{(i)})|\geqslant\frac{c_{2}C_{i}}{n}=c_{2}|I_{1}^{(i)}|,$
(4.8)

where the constant $c_{2}$ in (4.8) is exactly the same as before.

We have two cases:

######  Case 1.

We have

$\frac{|I^{(i*)}_{1}(j_{0}^{(i)})\cap\mathcal{X}_{n}|}{|I^{(i*)}_{1}(j_{0}^{(i)})|}\geqslant\left(1-\frac{\varepsilon}{2}\right)\frac{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}|}.$
(4.9)

######  Case 2.

We have

$\frac{|I^{(i*)}_{1}(j_{0}^{(i)})\cap\mathcal{X}_{n}|}{|I^{(i*)}_{1}(j_{0}^{(i)})|}<\left(1-\frac{\varepsilon}{2}\right)\frac{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}|}.$
(4.10)

###  4.1. Case 1: density decrease

Suppose that the inequality (4.9) holds. Then an argument analogous to that in
Step 0 between (2.23) and (2.48) and that in Step 1 between (3.9) and (3.10)
leads to the existence of a subinterval $I_{0}^{(i)}(\star)\subset
I_{0}^{(i)}$ satisfying $|I_{0}^{(i)}|/|I_{0}^{(i)}(\star)|=h_{i}$, where
$h_{i}$ is an integer and

$\frac{|I_{0}^{(i)}(\star)\cap\mathcal{X}_{n}|}{|I_{0}^{(i)}(\star)|}\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)\frac{|I_{0}^{(i)}\cap\mathcal{X}_{n}|}{|I_{0}^{(i)}|}.$

provided that

$C_{i}\geqslant\frac{1728A(A+1)^{3}c_{0}b}{\varepsilon}\quad\mbox{and}\quad
C_{i}\geqslant\frac{8b}{c_{2}\varepsilon}.$ (4.11)

To obtain a subinterval of $I_{1}^{(i)}$ of the same length as
$I_{0}^{(i)}(\star)$, we next divide $I_{1}^{(i)}$ into $h_{i}$ equal parts,
and denote by $I_{1}^{(i)}(\star)$ one of these subintervals with the maximum
intersection with the set $\mathcal{X}_{n}$. Then

$\frac{|I_{1}^{(i)}(\star)\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}(\star)|}\geqslant\frac{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}|}.$

We have

$|I_{1}^{(i)}(\star)|=|I_{0}^{(i)}(\star)|=\frac{C_{i+1}}{n},\quad
C_{i+1}=\frac{C_{i}}{h_{i}},\quad h_{i}<c_{3}(\varepsilon),$ (4.12)

where the constant $c_{3}(\varepsilon)>0$ is as in Step 0.

###  4.2. Case 2: density increase

Suppose that the inequality (4.10) holds. Then an argument analogous to that
in Step 0 between (2.51) and (2.65) and that in Step 1 between (3.11) and
(3.12) leads to the existence of a subinterval $I_{1}^{(i)}(\star)\subset
I_{1}^{(i)}$ satisfying $|I_{1}^{(i)}|/|I_{1}^{(i)}(\star)|=h_{i}$, where
$h_{i}$ is an integer and

$\frac{|I_{1}^{(i)}(\star)\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}(\star)|}\geqslant\left(1+\frac{c_{2}\varepsilon}{12}\right)\frac{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}|},$

provided that

$C_{i}\geqslant\frac{120Ab^{2}}{(c_{2}\varepsilon)^{2}}.$ (4.13)

To obtain a subinterval of $I_{0}^{(i)}$ of the same length as
$I_{1}^{(i)}(\star)$, we next divide $I_{0}^{(i)}$ into $h_{i}$ equal parts,
and denote by $I_{0}^{(i)}(\star)$ one of these subintervals with the minimum
intersection with the set $\mathcal{X}_{n}$. Then

$\frac{|I_{0}^{(i)}(\star)\cap\mathcal{X}_{n}|}{|I_{0}^{(i)}(\star)|}\leqslant\frac{|I_{0}^{(i)}\cap\mathcal{X}_{n}|}{|I_{0}^{(i)}|}.$

We have

$|I_{0}^{(i)}(\star)|=|I_{1}^{(i)}(\star)|=\frac{C_{i+1}}{n},\quad
C_{i+1}=\frac{C_{i}}{h_{i}},\quad h_{i}<c_{3}(\varepsilon),$ (4.14)

where the constant $c_{3}(\varepsilon)>0$ is as in Step 0.

##  5\. Iteration process: deriving a contradiction

We now attempt to derive the necessary contradiction.

Suppose first that Case 1 holds in Step $i$.

Corresponding to (2.35) and (3.9), we have the inequality

$|I^{(i*)}_{0}(j_{0}^{(i)})\cap\mathcal{X}_{n}|\geqslant\left(1-\frac{3\varepsilon}{4}\right)|I^{(i*)}_{0}(j_{0}^{(i)})|\frac{|I_{1}^{(i)}\cap\mathcal{X}_{n}|}{|I_{1}^{(i)}|},$
(5.1)

provided that

$C_{i}\geqslant\frac{1728A(A+1)^{3}c_{0}b}{\varepsilon}\quad\mbox{and}\quad
C_{i}\geqslant\frac{8b}{c_{2}\varepsilon}.$

Clearly $I_{0}^{(i*)}(j_{0}^{(i)})\subset I_{0}^{(i)}$, so it follows from
(4.7) that

$|I_{0}^{(i*)}(j_{0}^{(i)})\cap\mathcal{X}_{n}|\leqslant\left(1-\frac{c_{2}\varepsilon}{16}\right)^{i_{1}}|I_{1}^{(i)}\cap\mathcal{X}_{n}|.$
(5.2)

On the other hand, combining (4.8) with (5.1) leads to the inequality

$|I^{(i*)}_{0}(j_{0}^{(i)})\cap\mathcal{X}_{n}|\geqslant
c_{2}\left(1-\frac{3\varepsilon}{4}\right)|I_{1}^{(i)}\cap\mathcal{X}_{n}|.$
(5.3)

Clearly (5.2) and (5.3) contradict each other if

$\left(1-\frac{c_{2}\varepsilon}{16}\right)^{i_{1}}<\frac{c_{2}}{2},$

noting that $0<\varepsilon<1/2$. This gives an upper bound
$c_{4}(\varepsilon)$ to $i_{1}$, the number of times that Case 1 holds among
the first $i$ steps.

Suppose next that Case 2 holds in Step $i$.

Combining (2.1) and (4.6), and noting that $|I_{1}|=C/n$, we have

$|I_{1}^{(i)}\cap\mathcal{X}_{n}|\geqslant\left(1+\frac{c_{2}\varepsilon}{12}\right)^{i_{2}}\frac{n}{b}|I_{1}^{(i)}|.$
(5.4)

On the other hand, it follows from (2.31) that

$|I_{1}^{(i)}\cap\mathcal{X}_{n}|\leqslant An|I_{1}^{(i)}|+1.$ (5.5)

Clearly (5.4) and (5.5) contradict each other if

$\left(1+\frac{c_{2}\varepsilon}{12}\right)^{i_{2}}>2Ab,$

provided that $C_{i}\geqslant 1$. This gives an upper bound
$c_{5}(\varepsilon)$ to $i_{2}$, the number of times that Case 2 holds among
the first $i$ steps.

If $i>c_{4}(\varepsilon)+c_{5}(\varepsilon)$, then neither Case 1 nor Case 2
in Step $i$ can be valid. To show that Case B is impossible, it remains to
analyze the various constants in our argument.

Recall that the constants $c_{0}$ and $c_{2}$ depend only on $\mathcal{P}$ and
$\alpha$, and are independent of $n$, $C$ and $\varepsilon$, while the
constant $A$ depends only on $\alpha$, and the constant $b$ depends only on
$\mathcal{P}$. It remains to study the constants $C_{i}$, which must satisfy

$C_{i}=\frac{C}{h_{0}h_{1},\ldots,h_{i-1}}\geqslant\max\left\\{\frac{1728A(A+1)^{3}c_{0}b}{\varepsilon},\frac{8b}{c_{2}\varepsilon},\frac{120Ab^{2}}{(c_{2}\varepsilon)^{2}}\right\\},$
(5.6)

in view of (4.11)–(4.14) and $C_{i}\geqslant 1$. Since
$h_{i}<c_{3}(\varepsilon)$ and the iteration process must stop after at most
$c_{6}(\varepsilon)=c_{4}(\varepsilon)+c_{5}(\varepsilon)$ steps, it follows
that (5.6) is satisfied provided that $C$ is chosen sufficiently large in
terms of $\mathcal{P},\alpha$ and $\varepsilon$.

##  6\. Proof of Theorem 1

We have already shown that Case B leads to a contradiction. To complete the
proof of Theorem 1, it remains to investigate Case A, when the inequality
(2.2) holds. We have the following almost trivial observation.

######  Lemma 3.

Suppose that $J$ is a subinterval of any vertical edge of $\mathcal{P}$ with
length $|J|\geqslant 3C/n$. If (2.2) holds, then

$(1-\varepsilon)\left(\frac{|J|}{|I_{1}|}-3\right)|I_{1}\cap\mathcal{X}_{n}|\leqslant|J\cap\mathcal{X}_{n}|\leqslant\left(\frac{|J|}{|I_{1}|}+3\right)|I_{1}\cap\mathcal{X}_{n}|.$
(6.1)

###### Proof.

Let $k=[n/C]$ denote the integer part of $n/C$. Then we can split any vertical
edge of $\mathcal{P}$ into a union of $k$ special subintervals of length $C/n$
and an extra short interval with length $w$ satisfying $0\leqslant w<C/n$ at
the top end of the vertical edge.

Consider the unique integer $\ell_{0}$ that satisfies the inequalities

$\ell_{0}\leqslant\frac{|J|}{C/n}=\frac{|J|}{|I_{1}|}<\ell_{0}+1.$ (6.2)

Then $J$ contains at least $\ell_{0}-2$ of these special subintervals of
length $C/n$. Combining this observation with (2.2) and the second inequality
in (6.2) leads to the lower bound

$|J\cap\mathcal{X}_{n}|\geqslant(\ell_{0}-2)|I_{0}\cap\mathcal{X}_{n}|>\left(\frac{|J|}{|I_{1}|}-3\right)(1-\varepsilon)|I_{1}\cap\mathcal{X}_{n}|.$

On the other hand, $J$ is covered by $\ell_{0}+2$ special subintervals of
length $C/n$ and the extra short subinterval of length $w$ which is contained
in a subinterval of the vertical edge of length $C/n$. Combining this
observation with the first inequality in (6.2) leads to the upper bound

$|J\cap\mathcal{X}_{n}|\leqslant(\ell_{0}+3)|I_{1}\cap\mathcal{X}_{n}|\leqslant\left(\frac{|J|}{|I_{1}|}+3\right)|I_{1}\cap\mathcal{X}_{n}|.$

This completes the proof. ∎

Let $|J|=3|I_{1}|/\varepsilon=3C/\varepsilon n$. Then

$\displaystyle\left|V_{n}(J)-\frac{n|J|}{b}\right|=\left||J\cap\mathcal{X}_{n}|-\frac{n}{b}\frac{3|I_{1}|}{\varepsilon}\right|$
$\displaystyle\quad\leqslant\left||J\cap\mathcal{X}_{n}|-\frac{3}{\varepsilon}|I_{1}\cap\mathcal{X}_{n}|\right|+\frac{3|I_{1}|}{\varepsilon}\left(\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}-\frac{n}{b}\right).$
(6.3)

With $|J|=3|I_{1}|/\varepsilon$ in (6.1), we have

$\frac{3(1-\varepsilon)^{2}}{\varepsilon}|I_{1}\cap\mathcal{X}_{n}|\leqslant|J\cap\mathcal{X}_{n}|\leqslant\frac{3(1+\varepsilon)}{\varepsilon}|I_{1}\cap\mathcal{X}_{n}|,$

and this implies

$\left||J\cap\mathcal{X}_{n}|-\frac{3}{\varepsilon}|I_{1}\cap\mathcal{X}_{n}|\right|\leqslant
6|I_{1}\cap\mathcal{X}_{n}|.$ (6.4)

On the other hand, it is clear from (2.1) and (2.2) that

$\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}\geqslant\frac{n}{b}\geqslant\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}\geqslant(1-\varepsilon)\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}.$
(6.5)

It then follows from (6.5) that

$\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}-\frac{n}{b}\leqslant\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}-\frac{|I_{0}\cap\mathcal{X}_{n}|}{|I_{0}|}\leqslant\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}-(1-\varepsilon)\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|},$

so that

$\frac{3|I_{1}|}{\varepsilon}\left(\frac{|I_{1}\cap\mathcal{X}_{n}|}{|I_{1}|}-\frac{n}{b}\right)\leqslant
3|I_{1}\cap\mathcal{X}_{n}|.$ (6.6)

It also follows from (6.5) that

$|I_{1}\cap\mathcal{X}_{n}|\leqslant\frac{|I_{1}|}{1-\varepsilon}\frac{n}{b}.$
(6.7)

Substituting (6.4), (6.6) and (6.7) into (6), we conclude that

$\left|V_{n}(J)-\frac{n|J|}{b}\right|\leqslant\frac{3\varepsilon}{1-\varepsilon}\frac{n|J|}{b}.$
(6.8)

Since $n$ and $J$ are arbitrary, the inequality (6.8) proves super-micro-
uniformity with $3\varepsilon(1-\varepsilon)^{-1}$ instead of $\varepsilon$.

## References

  * [1] J. Beck, W.W.L. Chen. Superdensity of non-integrable billiards.
arxiv.org/abs/2104.09089, 25 pp.

  * [2] J. Beck, W.W.L. Chen. Time-quantitative density of non-integrable systems.
arxiv.org/abs/2014.09930, 27 pp.