# The dimension of the set of $\psi$-badly approximable points in all ambient
dimensions; on a question of Beresnevich and Velani.

Henna Koivusalo
(Bristol)  Jason Levesley
(York)  Benjamin Ward111B. W. gratefully acknowledges support from the EPSRC
research grant (EP/W522430/1), and Australian Research Council Discovery grant
(No. 200100994).
(La Trobe)  Xintian Zhang
(Bristol)

###### Abstract

Let $\psi:{\mathbb{N}}\to[0,\infty)$, $\psi(q)=q^{-(1+\tau)}$ and let
$\psi$-badly approximable points be those vectors in ${\mathbb{R}}^{d}$ that
are $\psi$-well approximable, but not $c\psi$-well approximable for
arbitrarily small constants $c>0$. We establish that the $\psi$-badly
approximable points have the Hausdorff dimension of the $\psi$-well
approximable points, the dimension taking the value $(d+1)/(\tau+1)$ familiar
from theorems of Besicovitch and Jarník. The method of proof is an entirely
new take on the Mass Transference Principle by Beresnevich and Velani (Annals,
2006); namely, we use the colloquially named ‘delayed pruning’ to construct a
sufficiently large $\liminf$ set and combine this with ideas inspired by the
proof of the Mass Transference Principle to find a large $\limsup$ subset of
the $\liminf$ set. Our results are a generalisation of some $1$-dimensional
results due to Bugeaud and Moreira (Acta Arith, 2011), but our method of proof
is nothing alike.

##  1 Introduction

Throughout let $\psi:{\mathbb{N}}\to(0,\infty)$ be a monotonic decreasing
function. We will use the notation $\psi_{\tau}$ when $\psi(q)=q^{-\tau}$ for
$\tau\in(0,\infty)$. The set of $\psi$-well approximable points, denoted by
$\mathcal{W}_{d}(\psi)$, is defined to be the set

$\mathcal{W}_{d}(\psi):=\left\\{x\in[0,1]^{d}:\max_{1\leq i\leq
d}\left|x_{i}-\frac{p_{i}}{q}\right|<\frac{\psi(q)}{q}\quad\text{for
i.m.}\,\,\,(p,q)\in{\mathbb{Z}}^{d}\times{\mathbb{N}}\right\\}\,,$

where i.m. denotes infinitely many. The main object of interest in this
article is the Hausdorff dimension of the $\psi$-badly approximable points;
those points of $\mathcal{W}_{d}(\psi)$ for which the approximation by $\psi$
cannot be improved by an arbitrarily small constant. More precisely,

$\mathbf{Bad}_{d}(\psi):=\mathcal{W}_{d}(\psi)\backslash\bigcap_{k=1}^{\infty}\mathcal{W}_{d}\left(\frac{1}{k}\psi\right).$

In [8, Question 1] Beresnevich and Velani asked for the Hausdorff dimension of
$\mathbf{Bad}_{d}(\psi)$ and conjectured that the Hausdorff dimension was
equal to the dimension of $\mathcal{W}_{d}(\psi)$. We prove this conjecture
true for a certain class of approximation functions.

We recall the definition of Hausdorff measure and dimension. (For more details
on such notions see for example [29, 16].) For $s\geq 0$ the Hausdorff
$s$-measure of a set $F\subset{\mathbb{R}}^{d}$ is defined as

${\cal H}^{s}(F):=\lim_{\rho\to
0^{+}}\inf\left\\{\sum_{i=1}^{\infty}|B_{i}|^{s}:F\subset\bigcup_{i=1}^{\infty}B_{i}\quad\text{
and }\quad|B_{i}|<\rho\,\,\forall\,\,i\right\\}\,$

where $|\cdot|$ denotes the diameter of a set. The Hausdorff dimension of set
$F\subset{\mathbb{R}}^{d}$ is defined as

$\operatorname{\dim_{H}}F:=\inf\left\\{s\geq 0:{\cal H}^{s}(F)=0\right\\}\,.$

In the case when $s=d$, the ambient dimension of our setting, Hausdorff
$s$-measure coincides (up to a constant) with Lebesgue measure which we denote
by $\lambda_{d}$. The literature surrounding the metric properties of
$\mathbf{Bad}_{d}(\psi)$ is extensive, and we only highlight some of the main
results below.

For approximation function $\psi_{1/d}$, the set
$\mathbf{Bad}_{d}(\psi_{1/d})=\mathbf{Bad}_{d}$ is the classical set of badly
approximable points. As a consequence of a landmark theorem of Khintchine [25]
we know that $\lambda_{d}(\mathbf{Bad}_{d})=0$ (This can be proven by a range
of methods, see [4] for an overview of techniques). It is then natural to ask,
what is the Hausdorff dimension of $\mathbf{Bad}_{d}$. Jarník [21] and Schmidt
[31] showed that despite the set being null, it has full Hausdorff dimension
in the cases where $d=1$ and $d>1$, respectively. That is

$\operatorname{\dim_{H}}\mathbf{Bad}_{d}=d\,.$

In the direction of $\psi$-well approximable points, Besicovitch [10] and
Jarník [22] independently proved that whenever $\tau>\frac{1}{d}$

$\operatorname{\dim_{H}}\mathcal{W}_{d}(\psi_{\tau})=\frac{d+1}{\tau+1}\,.$

Dodson extended this result to the case of general approximation functions
$\psi$ [15, Theorem 2]. As
$\mathbf{Bad}_{d}(\psi)\subset\mathcal{W}_{d}(\psi)$, these results give
immediate upper bounds for the Hausdorff dimension of
$\mathbf{Bad}_{d}(\psi)$. As is often the case, establishing the corresponding
lower bound is the main obstacle in determining the exact dimension.

The following refinement of $\mathcal{W}_{d}(\psi)$ and
$\mathbf{Bad}_{d}(\psi)$, first introduced in [5], is useful as they can be
used to define sets whose points satisfy more delicate approximation
properties. Let $\psi,\phi:{\mathbb{N}}\to(0,\infty)$ be monotonic decreasing
functions with $\psi(q)>\phi(q)$ for all $q\in{\mathbb{N}}$ and denote

$\displaystyle D_{d}(\psi,\phi)$
$\displaystyle:=\left\\{x\in[0,1]^{d}:\underset{\stackrel{{\scriptstyle\text{for
all $(p,q)\in{\mathbb{Z}}^{d}\times{\mathbb{N}}$}}}{{\text{with $q$
sufficiently
large}}}}{\underbrace{\,\quad\phi(q)/q\,\,\leq\quad\,}}\max_{1\leq i\leq
d}\left|x_{i}-\frac{p_{i}}{q}\right|\underset{\text{for i.m.
$(p,q)\in{\mathbb{Z}}^{d}\times{\mathbb{N}}$
}}{\underbrace{\,\quad<\,\,\psi(q)/q\quad\,}}\right\\}$
$\displaystyle=\mathcal{W}_{d}(\psi)\backslash\mathcal{W}_{d}(\phi).$

With these sets defined, the set of $\psi$-badly approximable points can now
be alternatively written as

$\mathbf{Bad}_{d}(\psi)=\bigcup_{k\in{\mathbb{N}}}D_{d}\left(\psi,\frac{1}{k}\psi\right),$

and the set of exact approximation order $\psi$ is

$\mathbf{Exact}_{d}(\psi):=\bigcap_{k\in{\mathbb{N}}}D_{d}\left(\psi,\left(1-\frac{1}{k}\right)\psi\right)=\mathcal{W}_{d}(\psi)\backslash\bigcup_{k=1}^{\infty}\mathcal{W}_{d}\left(\left(1-\frac{1}{k}\right)\psi\right).$

Trivially
$\mathbf{Exact}_{d}(\psi)\subset\mathbf{Bad}_{d}(\psi)\subset\mathcal{W}_{d}(\psi)$
and so $\mathbf{Exact}_{d}(\psi)$ can be used for lower bounds of dimension.

One of the first results on the set $\mathbf{Exact}_{1}(\psi)$ was given by
Jarník [23] who, using the theory of continued fractions, proved that
$\mathbf{Exact}_{1}(\psi)\not=\emptyset$ for functions $\psi(q)=o(q^{-1})$.
Without the condition $\psi(q)=o(q^{-1})$ the theory becomes significantly
more complicated. In particular, a classical result of Hurwitz tells us that
$\mathcal{W}_{1}\left(\frac{1}{\sqrt{5}}\psi_{1}\right)=[0,1]$ and so

$\mathbf{Exact}_{1}(\psi_{1})\subset\mathcal{W}_{1}(\psi_{1})\backslash\mathcal{W}_{1}\left(\frac{1}{\sqrt{5}}\psi_{1}\right)=\emptyset\,.$

Conversely, if $\psi(q)=cq^{-1}$ for some $c>0$ small it is not necessarily
true that $\mathbf{Exact}_{1}(\psi)=\emptyset$. In fact, for $c\to 0$, Moreira
showed that the Hausdorff dimension of $\mathbf{Exact}_{1}(\psi)$ tends to one
[28, Theorem 2]. Nevertheless, throughout we will suppose the approximation
function satisfies $\psi(q)=o\left(q^{-1/d}\right)$.

Since the result of Jarník there has been gradual progress towards
establishing the Hausdorff dimension of $\mathbf{Exact}_{1}(\psi)$. In [18]
Güting proved that

$\operatorname{\dim_{H}}\left(\bigcup_{k\in{\mathbb{N}}}D_{1}\left(\psi_{\tau},\psi_{\tau+\frac{1}{k}}\right)\right)=\operatorname{\dim_{H}}\mathcal{W}_{1}(\psi_{\tau})=\frac{2}{1+\tau},$

for $\tau>1$. In [5] Beresnevich, Dickinson and Velani improved upon Güting’s
result by calculating the Hausdorff measure at the dimension and showed that

${\cal
H}^{\frac{2}{1+\tau}}\left(\bigcup_{k\in{\mathbb{N}}}D_{1}\left(\psi_{\tau},\psi_{\tau+\frac{1}{k}}\right)\right)=\infty\,.$

Furthermore they considered approximation functions with ”logarithmic error”
and showed that for $\tau>\frac{1}{d}$

$\operatorname{\dim_{H}}\left(\bigcup_{k\in{\mathbb{N}}}D_{d}\left(\psi_{\tau},\,q\mapsto
q^{-\tau}(\log q)^{-\frac{1}{k}}\right)\right)=\frac{d+1}{\tau+1}.$

This was established by comparing the Hausdorff measures of the two
approximation sets $\mathcal{W}_{d}(\psi_{\tau})$ and
$\mathcal{W}_{d}\left(q\mapsto q^{-\tau}(\log q)^{-\varepsilon}\right)$ for
small $\varepsilon>0$. However the technique used here cannot be applied in
the ”constant error” case as the Hausdorff measure of the two sets would be
the same. In a series of papers Bugeaud [12, 13] and Bugeaud and Moreira [14]
proved a complete result in one dimension222Indeed, this result of Bugeaud and
Moreira doesn’t only hold for $\psi_{\tau}$ but also for general approximation
functions $\psi$. We state this special case for simplicity as it is the
relevant statement for the purposes of the current work., showing that

$\operatorname{\dim_{H}}\mathbf{Exact}_{1}(\psi_{\tau})=\frac{2}{1+\tau}.$

It is then a consequence of the inclusions
$\mathcal{W}_{1}(\psi)\supset\mathbf{Bad}_{1}(\psi)\supset\mathbf{Exact}_{1}(\psi)$
that

$\operatorname{\dim_{H}}\mathbf{Bad}_{1}(\psi_{\tau})=\frac{2}{1+\tau}\,,$

Furthermore, in [12, Theorem 2] Bugeaud also showed that

${\cal H}^{\frac{2}{1+\tau}}(\mathbf{Bad}_{1}(\psi_{\tau}))=\infty\,.$

The series of papers by Bugeaud and Moreira rely on results from the theory of
continued fractions. Higher dimensional versions of continued fractions have
been extensively studied, see for example [11, 32] and [9, 24] for more recent
approaches. However, trying to apply arguments from $1$-dimension to higher
dimensions often fails and in particular, the methods of Bugeaud and Moreira
do not seem to be applicable to the study of $\mathbf{Bad}_{d}(\psi_{\tau})$.

In the current work we compute the Hausdorff dimension
$\mathbf{Bad}_{d}(\psi_{\tau})$ for any dimension $d$. That is, we prove the
following;

######  Theorem 1.

Let $\psi_{\tau}(q)=q^{-\tau}$ with $\tau>\frac{1}{d}$, and let
$B\subset[0,1]^{d}$ be any open ball. Then

$\operatorname{\dim_{H}}\mathbf{Bad}_{d}(\psi_{\tau})\cap
B=\frac{d+1}{1+\tau}.$

The proof has been inspired by the proof of the Mass Transference Principle
(MTP) of Beresnevich and Velani [7]. The MTP is a powerful technique,
introduced to Diophantine approximation in 2006, when it was used to prove a
Hausdorff measure version of the Duffin-Schaeffer conjecture. The original MTP
states, loosely speaking, that if a limsup set of balls has full Lebesgue
measure, then the limsup set of the same balls shrunk in a controlled way
satisfies a dimension lower bound. The MTP has since been adapted, e.g. for
other measures, and other shapes [1, 26, 33, 34]. However, in studying
$\psi$-badly approximable points, we are facing a set-up not covered by the
many generalisations: we need a Hausdorff dimension estimate for the $\limsup$
set given by balls centred outside of the zero measured set in which we would
like the dimension lower bounds to hold! The proof we present relies on a few
key ideas: We emulate $\mathbf{Bad}_{d}(\psi_{\tau})$ by a set of almost full
measure (see § 3), and for ‘positive’ approximation we use balls centred at
rationals that are not too far away from this set (see § 4). These allow us to
approach the dimension bound in the spirit of the proof of MTP, although there
are some further geometric obstacles to overcome (see § 5).

Our main result is actually a consequence of the following result, which we
also state for comparison to the series of results of Moreira and Bugeaud on
dimension of $\mathbf{Exact}_{1}(\psi_{\tau})$.

######  Theorem 2.

Let $\psi_{\tau}$ and $B\subset[0,1]^{d}$ be as above with $\tau>\frac{1}{d}$.
Then there exists constant $0<C<1$ dependent only on $B$, $\tau$ and $d$ such
that

$\operatorname{\dim_{H}}D_{d}\left(\psi_{\tau},C\psi_{\tau}\right)\cap
B=\frac{d+1}{1+\tau}.$

######  Remark 1.

Note that Theorem 2 immediately implies Theorem 1, since
$D_{d}\left(\psi_{\tau},C\psi_{\tau}\right)\subset\mathbf{Bad}_{d}(\psi_{\tau})$.
The constant $C>0$ is implicit in our proofs and hasn’t been optimised.
However, we cannot make it arbitrary close to $1$, and so we cannot prove any
result on $\mathbf{Exact}_{d}(\psi_{\tau})$. Very recently Fregoli has proven
a restricted higher dimensional version of the results of Bugeaud and Moreira
[12, 13, 14] for $n\geq 3$ and $\tau>1$. The technique used by Fregoli took
their results and lifted it to higher dimensions by a clever observation.
Briefly, for $\boldsymbol{x}=(x_{1},\boldsymbol{y})\in{\mathbb{R}}^{d}$ if
$x_{1}\in\mathbf{Exact}_{1}(\psi_{\tau})$ and $\boldsymbol{y}\in
W_{d-1}(\psi_{\tau})$ then $\boldsymbol{x}\in\mathbf{Exact}_{d}(\psi_{\tau})$.
The dimension result can then be proven via the result of Bugeaud and Moreira
and a Theorem of Jarník on fibres. See [17] for more details, we only wish to
point out that our technique is significantly different. For other notions of
size it was recently shown by Schleischitz that for certain functions $\psi$
the set $\mathbf{Exact}_{d}(\psi)$ has full packing dimension, see [30,
Theorem 3.6 & Corollary 7].

######  Remark 2.

For this set of approximation functions the restriction that
$\tau>\frac{1}{d}$ is appropriate. If $\tau=\frac{1}{d}$ then as previously
mentioned $\mathbf{Bad}_{d}(\psi_{\tau})=\mathbf{Bad}_{d}$, that is, the
classical set of $d$-dimensional badly approximable points which has full
Hausdorff dimension [31]. If $\tau<\frac{1}{d}$ then
$\mathbf{Bad}_{d}(\psi_{\tau})=\emptyset$ since
$\mathcal{W}_{d}(c\psi_{\tau})\supseteq\mathcal{W}_{d}(\psi_{1/d})=[0,1]^{d}$
for any constant $c>0$ and so $\mathbf{Bad}_{d}(\psi_{\tau})\subset
W_{d}(\psi_{\tau})\backslash[0,1]^{d}=\emptyset$.

######  Remark 3.

In the current work we haven’t considered general approximation functions
$\psi$. Analogous to the results of Bugeaud and Moreira [12, 13, 14], we
suspect that for monotonic decreasing $\psi(q)$ of order $o(q^{-\frac{1}{d}})$

$\operatorname{\dim_{H}}\mathbf{Bad}_{d}(\psi)=\operatorname{\dim_{H}}\mathcal{W}_{d}(\psi),$

where the Hausdorff dimension of
$\operatorname{\dim_{H}}\mathcal{W}_{d}(\psi)$ is known [15]. We believe that
it may be possible to adapt our argument to deduce this result in the case
$\sum_{q=1}^{\infty}\psi(q)^{d}<\infty$. The case where
$\sum_{q=1}^{\infty}\psi(q)^{d}=\infty$ appears significantly harder.

######  Remark 4.

The theory of exact approximation has also been studied in a range of other
settings. ZL Zhang proved a version of Bugeaud and Moreiras’ result in the
setting of the field of formal series [35], and He & Xiong [19] have proven an
analogue in the setting of approximation by complex rational numbers. Both
papers appeal to results from the theory of continued fractions in their
respective setting to prove their main theorem. Very recently Bandi, Ghosh and
Nandi [3] have proven a general theory in the setting of exact approximation.
In their paper they obtain a result on exact approximation sets in hyperbolic
metric spaces. However, as far as we can see the $d$-dimensional real result
presented here cannot be deduced from the general theory presented in their
paper.

The rest of the paper is organised as follows. In the following section we
give an outline of the proof of Theorem 1. In § 3 we find a subset ${\cal
C}^{\tau}(N)\subset\mathbf{Bad}_{d}(\psi_{\tau})$ of large measure, as
quantified in § 3.1. In § 4 we define the subset of rationals $Q(N,\tau)$ that
we consider for positive approximation on ${\cal C}^{\tau}(N)$, and show that
this subcollection of rationals is not too small. In § 5 we present a
construction of a Cantor set and a mass distribution, following the general
outline of the proof of Mass Transference Principle, but relying on a new
geometric approximation Lemma 7.

##  2 Outline of the proof of Theorem 1

Before going into the proof of Theorem 1 we give an overview of the
methodology used.

Firstly, since
$\mathbf{Bad}_{d}(\psi_{\tau})\subseteq\mathcal{W}_{d}(\psi_{\tau})$ the upper
bound follows from the Jarník-Besicovitch Theorem. The main substance of the
proof is in establishing a corresponding lower bound. To accomplish this we
construct a Cantor subset ${\cal D}(B)$ of $\mathbf{Bad}_{d}(\psi_{\tau})\cap
B$ where $B$ is any arbitrary open ball in $(0,1)^{d}$. The construction of
${\cal D}(B)$ can be split into three main steps.

Step 1: We construct a set that avoids ‘dangerous balls’ i.e. parts of the
unit cube that lie ‘too close’ to rational points. The name ‘dangerous balls’
is taken from the analogous set that is removed when constructing a Cantor
subset of badly approximable points, see for example [6, §7]. In this setting,
given $N\in{\mathbb{N}}$ and $\tau>\frac{1}{d}$, we construct a set ${\cal
C}^{\tau}(N)$ such that

${\cal
C}^{\tau}(N)\cap\bigcup_{\frac{p}{q}\in{\mathbb{Q}}^{d}\cap[0,1]^{d}}B\left(\frac{p}{q},c_{N}q^{-1-\tau}\right)=\emptyset\,,$

for some constant $c_{N}>0$, where $B(x,r)=\\{y\in{\mathbb{R}}^{d}:\lVert
x-y\rVert<r\\}$ denotes the open ball with centre $x\in{\mathbb{R}}^{d}$ of
radius $r>0$ and $\|\cdot\|$ is the standard supremum norm in
$\mathbb{R}^{d}$. This set, and properties of ${\cal C}^{\tau}(N)$ that will
be required later in the proof of the main result, are dealt with in §3.

Step 2: We then construct a $\limsup$ subset of $\mathcal{W}_{d}(\psi_{\tau})$
such that each ball in the $\limsup$ set retains a positive proportion of its
mass when intersected with ${\cal C}^{\tau}(N)$. We achieve this by
constructing a set $Q(N,\tau)\subset{\mathbb{Q}}^{d}$ such that for any
$\frac{p}{q}\in Q(N,\tau)$ the inequality

$\lambda_{d}\left({\cal C}^{\tau}(N)\cap
B\left(\frac{p}{q},q^{-1-\tau}\right)\right)\geq\kappa\lambda_{d}\left(B\left(\frac{p}{q},q^{-1-\tau}\right)\right)\,$

holds for some fixed constant $\kappa>0$ independent of $\frac{p}{q}$. This is
established in the course of proving Lemma 5.

Step 3: The final part of the argument is to construct a Cantor subset and a
mass distribution, inspired by the Mass Transference Principle (a now standard
tool for proving metric results on $\limsup$ sets, see [2]). A key step in the
construction of the Cantor set ${\cal D}(B)$ is to show that for every ball
$\tilde{B}=B\left(\frac{p}{q},q^{-1-\tau}\right)$ with
$\frac{p}{q}\in{\mathbb{Q}}$ there exists a finite disjoint collection of
balls of the form
$B\left(\frac{p^{\prime}}{q^{\prime}},q^{\prime-1-\tau}\right)$ with
$\frac{p^{\prime}}{q^{\prime}}\in{\mathbb{Q}}$ that are contained in
$\tilde{B}$ and intersect a positive proportion of ${\cal C}^{\tau}(N)$. A
crucial lemma to ensure this happens is Lemma 7 of §5. This statement is the
counterpart of the $K_{G,B}$ lemma from the proof of the MTP [7].

This methodology is sufficient to prove Theorem 2 which in turn implies
Theorem 1.

##  3 Step 1: The construction of ${\cal C}^{\tau}(N)$

${\cal C}^{\tau}(N)$ The following ‘Simplex lemma’ is a crucial element in the
construction of ${\cal C}^{\tau}(N)$. The proof can be found in [27].

######  Lemma 1.

Let $d\geq 1$ be an integer and $Q\in{\mathbb{N}}$. Let
$E\subset{\mathbb{R}}^{d}$ be a convex set of $d$-dimensional Lebesgue measure

$\lambda_{d}(E)\leq(d!)^{-1}Q^{-(d+1)}.$

Suppose $E$ contains $d+1$ rational points with denominator $1\leq q\leq Q$.
Then these rational points lie on some hyperplane of ${\mathbb{R}}^{d}$.

By splitting up the construction of ${\cal C}^{\tau}(N)$ into suitable levels,
this result tells us that regions containing ‘dangerous balls’ will look like
thickened $(d-1)$-dimensional hyperplanes in $[0,1]^{d}$.

Fix some $N\in{\mathbb{N}}$. (We will need to take this $N$ large enough along
the course of the proofs, but it will eventually be fixed.) Construct ${\cal
C}^{\tau}(N)$ as follows:

  1. 1.

Split $[0,1]^{d}$ into $t^{d}$ cubes of sidelength $t^{-1}$ for some
$t\in{\mathbb{N}}$ where $t$ is chosen to satisfy $t>(d!)^{1/d}$. Now split
each of these cubes into $N^{d+1}$ cubes $I$ each with side-length
$t^{-1}N^{-\frac{d+1}{d}}$ and volume $t^{-d}N^{-(d+1)}$. Let $C_{1}$ denote
the set of cubes constructed in this way. Note there are $t^{d}N^{(d+1)}$
cubes in total.

  2. 2.

Within each cube $I$ we want to eventually remove the rational points
$(\frac{p_{1}}{q},\dots,\frac{p_{d}}{q})$ with $q$ bounded by $1\leq q<N$.
Apply Lemma 1 to each cube in $I\in C_{1}$ to establish that any such rational
points are contained in some hyperplane $L$ which we choose to be the minimal
such affine subspace. That is, if there is only a single point $\frac{p}{q}\in
I$ then $L$ is a point, if there are two points
$\frac{p}{q},\frac{p^{\prime}}{q^{\prime}}\in I$ then $L$ is the unique line
between $\frac{p}{q}$ and $\frac{p^{\prime}}{q^{\prime}}$ intersected with $I$
and so on. Lemma 1 tells us that at worst $L$ could be a $d$-dimensional
affine hyperplane. For simplicity we will call any such $L$ a hyperplane. Let
${\cal L}_{1}$ denote the collection of all hyperplanes $L$ constructed in
this manner for each cube in $C_{1}$. Remember these sets for later.

  3. 3.

Repeat this process; Assume that levels up to $n-1$ have been constructed. At
the $n$th level split each cube $I\in C_{n-1}$ into $N^{d+1}$ cubes each with
side length $t^{-1}N^{-(n-1)\frac{d+1}{d}}\times
N^{-\frac{d+1}{d}}=t^{-1}N^{-n\frac{d+1}{d}}$ and volume $t^{-d}N^{-n(d+1)}$.
Let $C_{n}$ denote all cube constructed in this layer. For a cube $I\in C_{j}$
for $j=1,\dots,n-1$ let $C_{n}(I)$ denote all cubes in $C_{n}$ which are
contained in $I$. Apply Lemma 1 to each $I^{\prime}\in C_{n}$ to find a
hyperplane $L$ containing all rational points
$(\frac{p_{1}}{q},\dots,\frac{p_{d}}{q})\in I^{\prime}$ with $1\leq q<N^{n}$.
Let ${\cal L}_{n}$ denote the collection of all level $n$ hyperplanes $L$
constructed in this way. For a cube $I\in C_{j}$ for $j=1,\dots,n-1$, let
${\cal L}_{n}(I)$ denote the set of all hyperplanes that correspond to cubes
$I^{\prime}\in C_{n}(I)$. Observe that a hyperplane $L\in{\cal L}_{n}$
contains all the rational points with denominators

$N^{n-1}\leq q<N^{n}.$

  4. 4.

Removal of the hyperplanes: Recall we eventually want to remove all dangerous
balls $B\left(\tfrac{p}{q},cq^{-(1+\tau)}\right)$ for some constant $c>0$.
Choose

$c_{N}=t^{-1}N^{-u(1+\tau)}$

with $u\in{\mathbb{N}}$ constant such that

$u>3>2+\frac{(d+1)(d-1)}{(1+\tau)d^{2}}.$ (1)

Then

$\bigcup_{N^{n-1}\leq
q<N^{n}}B\left(\frac{p}{q},c_{N}q^{-(1+\tau)}\right)\subset\bigcup_{N^{n-1}\leq
q<N^{n}}B\left(\frac{p}{q},t^{-1}N^{-(n-1+u)(1+\tau)}\right).$

At layer

$\ell(1):=\left\lfloor u\frac{(1+\tau)d}{d+1}\right\rfloor$

remove all cubes $I\in C_{\ell(1)}$ that intersect the
$t^{-1}N^{-u(1+\tau)}$-thickening of all hyperplanes contained in ${\cal
L}_{1}$. That is, for a hyperplane $L\in{\cal L}_{1}$ constructed in the cube
$I^{\prime}\in C_{1}$ and

$\delta(1)=t^{-1}N^{-u(1+\tau)}\,,$

let

$L^{\delta(1)}:=\left\\{\mathbf{x}\in
I^{\prime}:\operatorname{dist}(\mathbf{x},L)=\inf_{(a_{1},\dots,a_{d})\in
L}\left\\{\max_{1\leq i\leq d}|x_{i}-a_{i}|\right\\}\leq\delta(1)\right\\}\,.$
(2)

Then, denoting ${\cal S}_{\ell(1)}$ as the first level surviving set of cubes
in $C_{\ell(1)}$, we have

${\cal S}_{\ell(1)}:=\left\\{I\in C_{\ell(1)}:\,\,I\cap\bigcup_{L\in{\cal
L}_{1}}L^{\delta(1)}=\emptyset\right\\}.$

  5. 5.

Iterative removal of hyperplanes: We follow the same step as the construction
of the first layer of surviving cubes, with the small exception that we only
remove hyperplanes that still contain ’significant’ rational points after the
previous level pruning. That is, at layer

$\ell(n):=\left\lfloor(n-1+u)\frac{(1+\tau)d}{d+1}\right\rfloor$ (3)

remove all cubes $I\in C_{\ell(n)}$ that intersect the
$t^{-1}N^{-(n-1+u)(1+\tau)}$-thickening of all hyperplanes contained in ${\cal
L}_{n}^{*}$, where

${\cal L}_{n}^{*}:=\left\\{L\in{\cal L}_{n}:\,\,\exists\frac{p}{q}\in L\text{
with }\begin{cases}i)\quad N^{n-1}\leq q<N^{n}\,,\\\ ii)\quad
B\left(\tfrac{p}{q},\delta(n)\right)\cap\bigcup\limits_{I\in{\cal
S}_{\ell(n-1)}}I\neq\emptyset\,.\end{cases}\right\\}.$

That is, for a hyperplane $L\in{\cal L}_{n}^{*}$ and

$\delta(n)=t^{-1}N^{-(n-1+u)(1+\tau)}\,,$

let

$L^{\delta(n)}:=\left\\{\mathbf{x}\in
I^{\prime}:\operatorname{dist}(\mathbf{x},L)=\inf_{(a_{1},\dots,a_{d})\in
L}\left\\{\max_{1\leq i\leq d}|x_{i}-a_{i}|\right\\}\leq\delta(n)\right\\}\,.$
(4)

Then, denoting ${\cal S}_{\ell(n)}$ as the surviving set of cubes in
$C_{\ell(n)}$, we have

${\cal S}_{\ell(n)}:=\left\\{I\in C_{\ell(n)}:\,\,I\cap\bigcup_{L\in{\cal
L}_{n}^{*}}L^{\delta(n)}=\emptyset\quad\text{ \&
}\quad\exists\,\,I^{\prime}\in{\cal S}_{\ell(n-1)}\text{ s.t. }I\subset
I^{\prime}\right\\}\,.$

For ease of notation in later stages let

$\widehat{L^{\delta(n)}}:=\left\\{I\in C_{\ell(n)}:I\cap
L^{\delta(n)}\neq\emptyset\right\\}\,,$

so in particular

${\cal S}_{\ell(n)}=\left(\bigcup_{I\in{\cal
S}_{\ell(n-1)}}C_{\ell(n)}(I)\right)\backslash\left(\bigcup_{L\in{\cal
L}^{*}_{n}}\widehat{L^{\delta(n)}}\right).$

For any cube $I\in C_{j}$ with $j=1,\dots,\ell(n)-1$ let ${\cal
S}_{\ell(n)}(I)$ denote the set of cubes in ${\cal S}_{\ell(n)}$ that are
contained in $I$.

  6. 6.

Let

${\cal C}^{\tau}(N)=\bigcap_{n\in{\mathbb{N}}}\bigcup_{I\in{\cal
S}_{\ell(n)}}I.$

Observe that our constructed set ${\cal C}^{\tau}(N)$ does indeed avoid all
dangerous balls. We have the following statement

######  Lemma 2.

${\cal
C}^{\tau}(N)\cap\bigcup_{\frac{p}{q}\in{\mathbb{Q}}^{d}\cap[0,1]^{d}}B\left(\frac{p}{q},c_{N}q^{-(1+\tau)}\right)=\emptyset\,.$

###### Proof.

Suppose there exists $\frac{p}{q}\in{\mathbb{Q}}^{d}$ such that

$B\left(\frac{p}{q},c_{N}q^{-(1+\tau)}\right)\cap{\cal
C}^{\tau}(N)\neq\emptyset.$

We show this to be false. Suppose $N^{k-1}\leq q<N^{k}$ for
$k\in{\mathbb{N}}$. Then

$c_{N}q^{-(1+\tau)}\leq
c_{N}N^{-(k-1)(1+\tau)}=t^{-1}N^{-(k-1+u)(1+\tau)}=\delta(k)\,.$ (5)

By the Simplex Lemma there exists hyperplane
$L=L\left(\tfrac{p}{q}\right)\in{\cal L}_{k}$ containing $\frac{p}{q}$. If
$L\left(\tfrac{p}{q}\right)\in{\cal L}^{*}_{k}$ then by construction we have
that

$L\left(\tfrac{p}{q}\right)^{\delta(n)}\cap{\cal C}^{\tau}(N)=\emptyset\,,$

and so by (5)

$B\left(\frac{p}{q},c_{N}q^{-(1+\tau)}\right)\cap{\cal C}^{\tau}(N)\subseteq
B\left(\frac{p}{q},\delta(k)\right)\cap{\cal C}^{\tau}(N)=\emptyset.$

Thus we must have $L\left(\tfrac{p}{q}\right)\not\in{\cal L}^{*}_{k}$. We know
that $L\left(\tfrac{p}{q}\right)$ contains a rational with denominator
satisfying $i)$ (the condition appearing in ${\cal L}^{*}_{k}$), thus since
$L\left(\tfrac{p}{q}\right)\not\in{\cal L}^{*}_{k}$ it must hold that $ii)$
fails. That is

$B\left(\frac{p}{q},\delta(k)\right)\cap\bigcup_{I\in{\cal
S}_{\ell(k-1)}}I=\emptyset\,.$

Since ${\cal C}^{\tau}(N)$ is a nested Cantor set we have that

$\bigcup_{I\in{\cal S}_{\ell(k-1)}}I\supseteq{\cal C}^{\tau}(N)\,,$

and so, again, we are forced to conclude that

$B\left(\frac{p}{q},\delta(k)\right)\cap{\cal C}^{\tau}(N)=\emptyset.$

Appealing to (5) we obtain that
$B\left(\tfrac{p}{q},c_{N}q^{-(1+\tau)}\right)$ has empty intersection with
${\cal C}^{\tau}(N)$. This exhausts all possibilities for $\tfrac{p}{q}$ and
so contradicts our initial assumption, thus ${\cal C}^{\tau}(N)$ does indeed
avoid all dangerous neighbourhoods of rational points. ∎

###  3.1 Lebesgue measure of ${\cal C}^{\tau}(N)$

The following measure theoretic properties on ${\cal C}^{\tau}(N)$ will be
required later. The first of these statements tells us that for large enough
$N\in{\mathbb{N}}$ a significant proportion of $[0,1]^{d}$ is contained in
${\cal C}^{\tau}(N)$.

######  Lemma 3.

Suppose $\tau>\frac{1}{d}$. Then for any $\varepsilon>0$ there exists
$N\in{\mathbb{N}}$ such that

$\lambda_{d}({\cal C}^{\tau}(N))\geq 1-\varepsilon.$

Precisely,

$\lambda_{d}({\cal C}^{\tau}(N))\geq
1-3t^{-d}\sum_{n=1}^{\infty}N^{\frac{(d+1)}{d}(n-\ell(n))}.$

###### Proof.

Observe that

$\lambda_{d}({\cal C}^{\tau}(N))\geq
1-\lambda_{d}\left(\bigcup_{n\in{\mathbb{N}}}\left\\{I\in
C_{\ell(n)}:I\cap\bigcup_{L\in{\cal
L}_{n}}L^{\delta(n)}\neq\emptyset\right\\}\right).$

While we may be removing significantly less cubes (recall in later layers we
only neighbourhoods of hyperplanes in the reduced set ${\cal L}_{n}^{*}$) the
calculation below is sufficient.

Note that $L$ is a hyperplane and the fact that the hyperplane is contained in
a cube $I\in C_{n}$ of side length $t^{-1}N^{-n\frac{d+1}{d}}$. Hence the
thickened hyperplane $L^{\delta(n)}$ (recall (4)) intersects at most

$3N^{(\ell(n)-n)\frac{(d+1)(d-1)}{d}}$

cubes $I\in C_{\ell(n)}$. Hence

$\displaystyle\lambda_{d}\left(\bigcup_{n\in{\mathbb{N}}}\left\\{I\in
C_{\ell(n)}:I\cap\bigcup_{L\in{\cal
L}_{n}}L^{\delta(n)}\neq\emptyset\right\\}\right)$
$\displaystyle\leq\sum_{n=1}^{\infty}\sum_{L\in{\cal
L}_{n}}\lambda_{d}\left(\left\\{I\in C_{\ell(n)}:I\cap
L^{\delta(n)}\neq\emptyset\right\\}\right),$
$\displaystyle\leq\sum_{n=1}^{\infty}N^{n(d+1)}3N^{(\ell(n)-n)\frac{(d+1)(d-1)}{d}}\lambda_{d}(I),$
$\displaystyle\leq
3t^{-d}\sum_{n=1}^{\infty}N^{n(d+1)+(\ell(n)-n)\frac{(d+1)(d-1)}{d}-\ell(n)(d+1)},$
$\displaystyle\leq
3t^{-d}\sum_{n=1}^{\infty}N^{\frac{(d+1)}{d}(n-\ell(n))}\,=\varepsilon_{N}.$
(6)

Thus, providing $\tau>\frac{1}{d}$ (and so $\ell(n)-n>\rho n$ for some
$\rho>0$), the above summation is convergent, and for a suitably large choice
of $N$ we have that

$\lambda_{d}({\cal C}^{\tau}(N))\geq 1-\varepsilon_{N}.$

with $\varepsilon_{N}\to 0$ as $N\to\infty$. ∎

The following lemma gives us an even stronger statement than Lemma 3, namely
that every surviving cube in the construction ${\cal C}^{\tau}(N)$ retains
much of its mass when later layers are removed.

######  Lemma 4.

Suppose $\tau>\frac{1}{d}$. Let $n,N\in{\mathbb{N}}$. Then for all $I\in{\cal
S}_{\ell(n)}$,

$\lambda_{d}(I\cap{\cal
C}^{\tau}(N))\geq\left(1-3\left(\sum_{k=n+1}^{\ell(n)}N^{(\ell(n)-\ell(k))\frac{d+1}{d}}+\sum_{k=\ell(n)+1}^{\infty}N^{(k-\ell(k))\frac{d+1}{d}}\right)\right)\lambda_{d}(I).$

######  Remark 5.

Note the constant bound given here is larger than the constant proven in Lemma
3 for all $n\in{\mathbb{N}}$. So we could take $\varepsilon_{N}$ (given by
(3.1)) to be the universal constant for all surviving cube
$I\in\bigcup_{n\in{\mathbb{N}}}{\cal S}_{\ell(n)}$.

###### Proof.

The proof is similar to that of Lemma 3. Firstly, since $I\in{\cal
S}_{\ell(n)}$ we clearly have that

$I\not\in\bigcup_{k=1}^{n}\left\\{I^{\prime}\in
C_{\ell(k)}:I^{\prime}\cap\bigcup_{L\in{\cal
L}_{k}}L^{\delta(k)}\neq\emptyset\right\\}.$

and so

$\lambda_{d}\left(I\cap\bigcup_{k=1}^{n}\bigcup_{I^{\prime}\in{\cal
S}_{\ell(k)}}I^{\prime}\right)=\lambda_{d}(I).$

Hence

$\lambda_{d}(I\cap{\cal
C}^{\tau}(N))\geq\lambda_{d}(I)-\lambda_{d}\left(\bigcup_{k=n}^{\infty}\left\\{I^{\prime}\in
C_{\ell(k)}(I):I^{\prime}\cap\bigcup_{L\in{\cal
L}_{k}}L^{\delta(k)}\neq\emptyset\right\\}\right).$

To count the number of hyperplanes from $\bigcup_{k>n}\bigcup_{L\in{\cal
L}_{k}}L$ contained in $I$ observe that

$\\#\left\\{L\in{\cal L}_{k}:I\cap L^{\delta(k)}\neq\emptyset\right\\}\leq
1\quad\quad n+1\leq k\leq\ell(n),$ (7)

since $I$ intersects one unique cube $I^{\prime}\in C_{k}$ for $n+1\leq
k\leq\ell(n)$ (in this case $I$ is actually contained in such cube). And that

$\\#\left\\{L\in{\cal L}_{k}:I\cap L^{\delta(k)}\neq\emptyset\right\\}\leq
N^{(k-\ell(n))(d+1)}\quad\quad k\geq\ell(n)+1,$ (8)

since we are now looking at the number of cubes $I^{\prime}\in C_{k}(I)$ that
generate hyperplanes.

Now we consider the number of subcubes $I^{\prime}\in C_{\ell(k)}(I)$ that
could intersect with a thickened hyperplane $L\in{\cal L}_{k}$. We have that

$\\#\\{I^{\prime}\in C_{\ell(k)}(I):I^{\prime}\cap
L^{\delta(k)}\neq\emptyset\\}\leq
3N^{(\ell(k)-\ell(n))\frac{(d+1)(d-1)}{d}}\quad\quad\text{ if }L\in{\cal
L}_{k}\text{ for }n+1\leq k\leq\ell(n),$ (9)

since there are at most $N^{(\ell(k)-\ell(n))(d+1)}$ $\ell(k)$-level cubes in
$I$. Also,

$\\#\\{I^{\prime}\in C_{\ell(k)}(I):I^{\prime}\cap
L^{\delta(k)}\neq\emptyset\\}\leq
3N^{(\ell(k)-k)\frac{(d+1)(d-1)}{d}}\quad\quad\text{ if $L\in{\cal L}_{k}$ for
$k\geq\ell(n)+1$},$ (10)

since each hyperplane of interest is now contained in some cube in $C_{k}$.

Bring these together we have that

$\displaystyle\lambda_{d}\left(\bigcup_{k\in{\mathbb{N}}_{>n}}\left\\{I^{\prime}\in
C_{\ell(k)}(I):I^{\prime}\cap\bigcup_{L\in{\cal
L}_{k}}L^{\delta(k)}\neq\emptyset\right\\}\right)$
$\displaystyle\leq\sum_{k=n+1}^{\infty}\,\,\,\sum_{L\in{\cal L}_{k}:L\cap
I\neq\emptyset}\lambda_{d}\left(\left\\{I^{\prime}\in
C_{\ell(k)}(I):I^{\prime}\cap L^{\delta(k)}\neq\emptyset\right\\}\right),$
$\displaystyle\overset{\eqref{hyperplanes_1}-\eqref{hyperplane_intersect_2}}{\leq}\sum_{k=n+1}^{\ell(n)}3t^{-d}N^{(\ell(k)-\ell(n))\frac{(d+1)(d-1)}{d}}N^{-\ell(k)(d+1)}\quad+$
$\displaystyle\sum_{k=\ell(n)+1}^{\infty}3t^{-d}N^{(k-\ell(n))(d+1)}N^{(\ell(k)-k)\frac{(d+1)(d-1)}{d}}N^{-\ell(k)(d+1)}$
$\displaystyle=t^{-d}N^{-\ell(n)(d+1)}3\left(\sum_{k=n+1}^{\ell(n)}N^{(\ell(n)-\ell(k))\frac{d+1}{d}}+\sum_{k=\ell(n)+1}^{\infty}N^{(k-\ell(k))\frac{d+1}{d}}\right).$

And so we have that for $I\in{\cal S}_{\ell(n)}$

$\lambda_{d}(I\cap{\cal
C}^{\tau}(N))\geq\left(1-3\left(\sum_{k=n+1}^{\ell(n)}N^{(\ell(n)-\ell(k))\frac{d+1}{d}}+\sum_{k=\ell(n)+1}^{\infty}N^{(k-\ell(k))\frac{d+1}{d}}\right)\right)\lambda_{d}(I).$

∎

##  4 Step 2: A suitable subset of $\mathcal{W}_{d}(\psi_{\tau})$

Recall we want to construct a subset of $\mathcal{W}_{d}(\psi_{\tau})$ such
that any ball in the corresponding $\limsup$ set retains enough mass when
intersected with ${\cal C}^{\tau}(N)$. To this end we construct the following
subset of ${\mathbb{Q}}^{d}$. Define

$Q(N,\tau):=\left\\{\frac{p}{q}\in{\mathbb{Q}}^{d}:\exists\,L\in{\cal
L}^{*}_{n}\,\text{ with }\,\frac{p}{q}\in L\quad\text{ and
}\begin{cases}i)\,N^{n-1}\leq q\leq N^{n}\,,\\\
ii)\,B\left(\tfrac{p}{q},\delta(n)\right)\cap\bigcup_{I\in{\cal
S}_{\ell(n-1)}}I\neq\emptyset\,.\end{cases}\right\\}\,.$

We say $\tfrac{p}{q}$ is a leading rational of a hyperplane $L\in{\cal L}_{n}$
if $L$ is the hyperplane on which $\tfrac{p}{q}$ lies while satisfying the
above conditions.

Essentially, for each hyperplane removed in the construction of ${\cal
C}^{\tau}(N)$ we associate a set of rational points $\tfrac{p}{q}$ and
significant intersection with surviving cubes of the previous layer.

Note a few important properties of leading rationals:

  1. 1.

Every hyperplane $L\in\bigcup_{n\in{\mathbb{N}}}{\cal L}^{*}_{n}$ contains at
least one leading rational. This is clear from how we define the sets ${\cal
L}^{*}_{n}$ and $Q(N,\tau)$. Namely, if $L$ does not contain a leading
rational, then it would not have been removed.

  2. 2.

If $\tfrac{p}{q}$ is a leading rational of a hyperplane $L\in{\cal L}^{*}_{n}$
that was constructed in cube $I\in C_{n}$ then

$B\left(\tfrac{p}{q},2q^{-1-\frac{1}{d}}\right)\supseteq I\,.$

To see this observe that trivially $\tfrac{p}{q}\in I$, that the sidelenght of
$I$ is $N^{-n(1+\tfrac{1}{d})}$, and that $N^{n-1}\leq q\leq N^{n}$.

  3. 3.

If $\tfrac{p}{q}$ is not a leading rational then there exists a hyperplane
$L\in\bigcup_{n\in{\mathbb{N}}}{\cal L}^{*}_{k}$, say $L\in{\cal L}^{*}_{n}$,
for which $\tfrac{p}{q}\in\widehat{L^{\delta(k)}}$ and any leading rational of
$L$, say $\tfrac{r}{s}$, has $s<q$. This may not be so obvious. To see this,
suppose $N^{n-1}<q\leq N^{n}$. Then $\tfrac{p}{q}$ must, by the simplex lemma,
lie on a hyperplane of ${\cal L}_{n}$. If $\tfrac{p}{q}$ is not a leading
rational of said hyperplane, then it must lie in some previous level removed
strip. Any leading rational, say $\frac{r}{s}$, of a hyperplane from a
previously removed layer must, by definition, have denominator $s<N^{n-1}<q$.

Let

$\mathcal{W}_{d}(Q(N,\tau),\psi):=\left\\{\mathbf{x}\in[0,1]^{d}:\max_{1\leq
i\leq d}\left|x_{i}-\frac{p_{i}}{q}\right|<\frac{\psi(q)}{q}\quad\text{ for
i.m. }\,\,\frac{p}{q}\in Q(N,\tau)\right\\}.$

The following two lemmas are crucial properties of the set $Q(N,\tau)$.

######  Lemma 5.

Let $\psi(q)=3q^{-\frac{1}{d}}$ and suppose that $\tau>\frac{1}{d}$. Then

$\mathcal{W}_{d}(Q(N,\tau),\psi)\supseteq{\cal C}^{\tau}(N)\,.$

###### Proof.

Take any $\mathbf{x}\in{\cal C}^{\tau}(N)$. By Dirichlet’s Theorem
$\mathbf{x}\in\mathcal{W}_{d}(\psi)$. Let $A(\mathbf{x})$ be the set of all
those rational points for which

$\mathbf{x}\in B\left(\frac{p}{q},q^{-1-\frac{1}{d}}\right).$

If the cardinality of $A(\mathbf{x})\cap Q(N,\tau)$ is infinite, then clearly
$\mathbf{x}\in\mathcal{W}_{d}(Q(N,\tau),\psi)$, so assume $A(\mathbf{x})\cap
Q(N,\tau)$ is finite. Let $\frac{p}{q}\in A(\mathbf{x})\backslash Q(N,\tau)$.
Let $N^{m-1}<q\leq N^{m}$. Since $\tfrac{p}{q}$ is not a leading rational, by
$3.$ there exists some hyperplane $L$, say $L\in{\cal L}^{*}_{n}$ with $n<m$,
such that $\tfrac{p}{q}\in\widehat{L^{\delta(n)}}$. Let $\tfrac{r}{s}$ be a
leading rational of $L$. Then by property $3.$ of leading rational $s<q$, and
by the triangle inequality

$\left|\mathbf{x}-\frac{r}{s}\right|\leq\left|\mathbf{x}-\frac{p}{q}\right|+\left|\frac{p}{q}-\frac{r}{s}\right|<3s^{-1-\tfrac{1}{d}}.$
(11)

If the sequence of rational points in $A(\mathbf{x})$ can be associated to
infinitely many different leading rationals, then we are done by the above
argument. We now prove that there must be hyperplanes $L\in{\cal L}_{n}^{*}$
from arbitrarily high levels $n$ associated to the points in the sequence
$A(\mathbf{x})$. That is, they are leading rationals on these hyperplanes
themselves, or we find a nearby leading rational by the above argument.

Assume to the contrary that the sequence $A(\mathbf{x})$ is associated to
finitely many hyperplanes. Suppose $m$ is the largest level in which this
finite sequence of hyperplanes appear. By definition of $\mathbf{x}\in{\cal
C}^{\tau}(N)$ we have that

$\mathbf{x}\in[0,1]^{d}\backslash\left(\bigcup_{j\leq m}\bigcup_{L\in{\cal
L}^{*}_{j}}\widehat{L^{\delta(j)}}\right)\,,$

which is an open set, and so there exists some $\varepsilon>0$ such that

$d\left(\mathbf{x},\bigcup_{j\leq m}\bigcup_{L\in{\cal
L}^{*}_{j}}\widehat{L^{\delta(j)}}\right)>\varepsilon\,.$

However, for any $\frac{u}{v}\in A(\mathbf{x})$ with $v>\varepsilon^{-1}$ we
have that

$\left|\mathbf{x}-\frac{u}{v}\right|<v^{-1-\tfrac{1}{d}}<\varepsilon\,,$

and so

$\frac{u}{v}\not\in\bigcup_{j\leq m}\bigcup_{L\in{\cal
L}^{*}_{j}}\widehat{L^{\delta(j)}}\,,$

contradicting the finiteness of the sequence of associated hyperplanes. Thus
we have an infinite sequence of associated hyperplanes to $A(\mathbf{x})$,
which by (11) we can associate to an infinite sequence of rational points
which sufficiently approximate $\mathbf{x}$ so that
$\mathbf{x}\in\mathcal{W}_{d}(Q(N,\tau),\psi)$ as required.

∎

Lastly, the following lemma on the measure of balls in the construction of
$\mathcal{W}_{d}(Q(N,\tau),\psi_{\tau})$ will be required later. It
essentially tells us that each ball $B\left(\frac{p}{q},q^{-1-\tau}\right)$
with $\frac{p}{q}\in Q(N,\tau)$ contains a surviving cube that covers a
constant proportion of $B\left(\frac{p}{q},q^{-1-\tau}\right)$ independent of
$\frac{p}{q}$.

######  Lemma 6.

Let $B(\tau)=B\left(\tfrac{p}{q},q^{-1-\tau}\right)$ for some $\frac{p}{q}\in
Q(N,\tau)$. Then there exists $I\in\bigcup_{n\in{\mathbb{N}}}{\cal
S}_{\ell(n)}$ with $I\subset B(\tau)$ and

$\lambda_{d}(B(\tau))\leq C\lambda_{d}(I),$

with $C>0$ independent of $\frac{p}{q}$.

###### Proof.

Suppose that $N^{n-1}\leq q<N^{n}$. Since $\frac{p}{q}\in Q(N,\tau)$ we have
that

$B\left(\frac{p}{q},\delta(n)\right)\cap\bigcup_{I\in{\cal
S}_{\ell(n-1)}}I\neq\emptyset\,.$

Let $I^{*}\in{\cal S}_{\ell(n-1)}$ be some cube with positive intersection
with $B\left(\tfrac{p}{q},\delta(n)\right)$. Then, by considering the
sidelength of $I^{*}$ we have that

$B\left(\frac{p}{q},3\max\left\\{t^{-1}N^{-\ell(n-1)\left(1+\tfrac{1}{d}\right)},\delta(n)\right\\}\right)\supseteq
I^{*}\,.$

Now observe that

$\displaystyle 3t^{-1}N^{-\ell(n-1)\left(1+\frac{1}{d}\right)}$
$\displaystyle\leq
3t^{-1}N^{-\left\lfloor(n-2+u)\frac{(1+\tau)d}{(d+1)}\right\rfloor\frac{d+1}{d}}$
$\displaystyle\leq
3t^{-1}N^{-(n-1+u)\frac{(1+\tau)d}{(d+1)}\frac{d+1}{d}}\,\,\,\,\,\,(=3\delta(n))$
$\displaystyle=3t^{-1}N^{-(u-1)(1+\tau)}N^{-n(1+\tau)}$
$\displaystyle<q^{-(1+\tau)}\,.$

Hence

$B\left(\frac{p}{q},q^{-(1+\tau)}\right)\supseteq
B\left(\frac{p}{q},3\max\left\\{t^{-1}N^{-\ell(n-1)(1+\tfrac{1}{d})},\delta(n)\right\\}\right)\supseteq
I^{*}\,.$

It remains to see that

$\displaystyle\lambda_{d}(I^{*})$ $\displaystyle=t^{-d}N^{-\ell(n)(d+1)}$
$\displaystyle\geq t^{-d}N^{-(n-2+u)(1+\tau)d}$ $\displaystyle\geq
t^{-d}N^{-(u-1)(1+\tau)d}q^{-(1+\tau)d}$
$\displaystyle=2^{-d}t^{-d}N^{-(u-1)(1+\tau)d}\lambda_{d}\left(B\left(\frac{p}{q},q^{-(1+\tau)}\right)\right)\,,$

and so taking $C=2^{d}t^{d}N^{(u-1)(1+\tau)d}$ in the lemma completes the
proof. ∎

##  5 Step 3: Construction of a Cantor subset of
$\mathbf{Bad}_{d}(\psi_{\tau})$

The following lemma is crucial in the construction of our Cantor subset of
$\mathbf{Bad}_{d}(\psi_{\tau})$.

######  Lemma 7 ($T_{G,I}$ Lemma).

Let $\\{B_{i}\\}_{i\in{\mathbb{N}}}$ be a sequence of balls in
${\mathbb{R}}^{d}$ with $r(B_{i})\to 0$ as $i\to\infty$. Suppose that there
exists some constant $C>0$ such that

$\lambda_{d}\left(I\cap\limsup_{i\to\infty}B_{i}\right)\geq C\lambda_{d}(I)$
(12)

for any $I\in\bigcup_{n\in{\mathbb{N}}}{\cal S}_{\ell(n)}$. Then for any
$I\in\bigcup_{n\in{\mathbb{N}}}{\cal S}_{\ell(n)}$ and any $G>1$ there is a
finite subcollection

$T_{G,I}\subset\\{B_{i}:i\geq G\\}$

such that the balls are disjoint, lie inside $I$, and

$\lambda_{d}\left(\bigcup_{\tilde{B}\in
T_{G,I}}\tilde{B}\right)\geq\kappa_{1}\lambda_{d}(I),$

with $\kappa_{1}=C5^{-d}4^{-1}$.

######  Remark 6.

Note that Lemma 7 is applicable to our setting with

$\\{B_{i}\\}_{i\in{\mathbb{N}}}=\left\\{B\left(\frac{p}{q},2q^{-1-\frac{1}{d}}\right)\right\\}_{\frac{p}{q}\in
Q(N,\tau)}\,,$

i.e.
$\limsup_{i\to\infty}B_{i}=\mathcal{W}_{d}\left(Q(N,\tau),2\psi_{1/d}\right)$,
since

$\displaystyle\lambda_{d}\left(I\cap\mathcal{W}_{d}\left(Q(N,\tau),2\psi_{1/d}\right)\right)$
$\displaystyle\overset{\text{
Lemma~{}\ref{Q_approximations}}}{\geq}\lambda_{d}\left(I\cap{\cal
C}^{\tau}(N)\cap\mathcal{W}_{d}(\psi_{1/d})\right)$
$\displaystyle=\lambda_{d}\left(I\cap{\cal C}^{\tau}(N)\right)$
$\displaystyle\overset{\text{
Lemma~{}\ref{measure_cC_cubes}}}{\geq}(1-\varepsilon_{N})\lambda_{d}(I).$

Note that we can order the collection of balls in a decreasing order.

###### Proof.

Suppose that $I\in{\cal S}_{\ell(n)}$ and let ${\cal S}_{\ell(n+1)}^{*}(I)$
denote the set of cubes $I^{\prime}\in{\cal S}_{\ell(n+1)}$ contained in the
interior of $I$ (remove the edge cubes). Observe that

$\lambda_{d}\left(\bigcup_{I^{\prime}\in{\cal
S}_{\ell(n+1)}^{*}(I)}I^{\prime}\right)\geq\frac{1}{2}\lambda_{d}(I)$ (13)

since we are removing at most $2^{d}+1$ hyperplanes worth of cubes which (for
sufficiently large $N\in{\mathbb{N}}$) is small relative to the size of $I$.

Let

${\cal G}:=\left\\{B_{i}:B_{i}\cap\bigcup_{I^{\prime}\in{\cal
S}_{\ell(n+1)}^{*}(I)}I^{\prime}\neq\emptyset\,,\,\,i\geq G\right\\}.$

We may assume $r(B_{i})$ is monotonically decreasing (since we can order
$B_{i}$ however we want) and so for sufficiently large $i$ ($i\geq i_{0}$ such
that $r(B_{i_{0}})<\frac{1}{2}N^{-\ell(n+1)\frac{d+1}{d}}$) any ball
$B_{i}\in{\cal G}$ is contained in $I$. By the $5r$-covering lemma (see for
example [20, Theorem 1.2] ) there exists a disjoint subcollection ${\cal
G}^{\prime}\subseteq{\cal G}$ such that

$\bigcup_{B_{i}\in{\cal G}}B_{i}\subset\bigcup_{B_{i}\in{\cal
G}^{\prime}}5B_{i}$

and the balls in ${\cal G}^{\prime}$ are disjoint. It follows that

$\displaystyle\lambda_{d}\left(\bigcup_{B_{i}\in{\cal
G}^{\prime}}5B_{i}\right)$
$\displaystyle\geq\lambda_{d}\left(\bigcup_{I^{\prime}\in{\cal
S}_{\ell(n+1)}^{*}(I)}I^{\prime}\cap\limsup_{i\to\infty}B_{i}\right)$
$\displaystyle=\sum_{I^{\prime}\in{\cal
S}_{\ell(n+1)}^{*}(I)}\lambda_{d}\left(I^{\prime}\cap\limsup_{i\to\infty}B_{i}\right)$
$\displaystyle\overset{\eqref{partial_measure}}{\geq}C\sum_{I^{\prime}\in{\cal
S}_{\ell(n+1)}^{*}(I)}\lambda_{d}(I^{\prime})$
$\displaystyle\overset{\eqref{eq1.1}}{\geq}\frac{C}{2}\lambda_{d}(I).$

Furthermore, since ${\cal G}^{\prime}$ is a disjoint collection of balls we
have that

$\displaystyle\lambda_{d}\left(\bigcup_{B_{i}\in{\cal
G}^{\prime}}5B_{i}\right)$ $\displaystyle\leq\sum_{B_{i}\in{\cal
G}^{\prime}}5^{d}\lambda_{d}(B_{i})$
$\displaystyle=5^{d}\lambda_{d}\left(\bigcup_{B_{i}\in{\cal
G}^{\prime}}B_{i}\right).$

Hence

$\lambda_{d}\left(\bigcup_{B_{i}\in{\cal
G}^{\prime}}B_{i}\right)\geq\frac{C}{5^{d}2}\lambda_{d}(I).$

Since the balls of ${\cal G}^{\prime}$ are disjoint and $r(B_{i})\to 0$ as
$i\to\infty$ we have that

$\lambda_{d}\left(\bigcup_{B_{i}\in{\cal G}^{\prime}:i\geq j}B_{i}\right)\to
0\quad\text{ as }\quad j\to\infty.$

Hence there exists $j_{0}\geq G$ such that

$\lambda_{d}\left(\bigcup_{B_{i}\in{\cal G}^{\prime}:i\leq
j_{0}}B_{i}\right)\geq\frac{C}{5^{d}4}\lambda_{d}(I).$

Letting

$T_{G,I}=\\{B_{i}\in{\cal G}^{\prime}:i\leq j_{0}\\}$

completes the proof. ∎

###  5.1 Constructing ${\cal D}(B)$

Armed with Lemma 7 we now proceed with the construction of a Cantor subset
${\cal D}(B)$ of

${\cal C}^{\tau}(N)\cap W_{d}(Q(N,\tau),\psi_{\tau})\cap B\,$

for any ball $B\subset[0,1]^{d}$. A sketch of the construction is as follows:

  1. i)

Firstly given a ball $B$ we choose a suitable $N\in{\mathbb{N}}$ such that $B$
has non-empty intersection with ${\cal C}^{\tau}(N)$. (We will choose $N$
larger later when necessary.)

  2. ii)

To a surviving cube $I\subset B$ we then apply Lemma 7 to obtain a disjoint
collection of balls from
$\left\\{B\left(\frac{p}{q},2q^{-1-\frac{1}{d}}\right)\right\\}_{\frac{p}{q}\in
Q(N,\tau)}$ contained in $I$ that cover a positive proportion of $I$.

  3. iii)

We then shrink each of these balls i.e.

$B\left(\frac{p}{q},2q^{-1-\frac{1}{d}}\right)\mapsto
B\left(\frac{p}{q},q^{-1-\tau}\right)\,.$

The collection of these shrunken balls completes the first layer.

  4. iv)

We repeat the same process as above. That is, for each shrunken ball from the
first layer we apply Lemma 6 to find a surviving cube $I^{\prime}$ of
$C^{\tau}(N)$. We then choose suitable $G^{\prime}>0$ and apply Lemma 7 to
each surviving cube $I^{\prime}$, and then shrink each of the balls in the set
$T_{G^{\prime},I^{\prime}}$. This collection of balls then becomes the second
layer. We continue inductively.

Let us now present the full details of the construction. Fix
$B\subset[0,1]^{d}$. We construct ${\cal D}(B)$ as follows:

  1. 1.

Fix $N\in{\mathbb{N}}$ such that

$N\geq r(B)^{\frac{d}{d+1}}\quad\&\quad
3t^{-d}\sum_{n=1}^{\infty}N^{\frac{(d+1)}{d}(n-\ell(n))}=\varepsilon_{N}<\frac{1}{4}\lambda_{d}(B)\,.$
(14)

This ensures that

$\lambda_{d}(B\cap{\cal C}^{\tau}(N))\geq\frac{3}{4}\lambda_{d}(B),$

Furthermore, assume $N$ is large enough such that there exists $I_{0}\in{\cal
S}_{\ell(1)}$ with $I_{0}\subset B$. Set $E_{0}=I_{0}$.

  2. 2.

Apply Lemma 7 to $I_{0}$ with $G_{0}\geq 1$ chose sufficiently large such that

$\left.\begin{array}[]{c}r(B_{i})<t^{-1}N^{-\ell(n_{0})\frac{d+1}{d}}\\\
4r(B_{i}(\tau))<r(B_{i})\end{array}\right\\}\quad\forall\,\,i\geq G_{0}.$ (15)

Let $T_{G_{0},I_{0}}$ denote the set of balls $B_{i}$ from Lemma 7.

  3. 3.

Shrinking: For each $B_{i}\in T_{G_{0},I_{0}}$ shrink the balls

$B_{i}=B\left(\frac{p}{q},2q^{-\frac{d+1}{d}}\right)\mapsto
B\left(\frac{p}{q},q^{-(1+\tau)}\right)=B_{i}(\tau),$

and let the first layer

$E_{1}=T_{G_{0},I_{0}}^{\tau}=\left\\{B_{i}(\tau):B_{i}\in
T_{G_{0},I_{0}}\right\\}.$

  4. 4.

Induction: Suppose $E_{n-1}$ has been constructed. For each $B_{i}(\tau)\in
E_{n-1}$ we proceed as follows.

    1. a)

nesting: By Lemma 6 there exists some $I\in\bigcup_{n}{\cal S}_{\ell(n)}$, say
$I(i)$, with $I(i)\subseteq B_{i}(\tau)$ and

$\lambda_{d}(B_{i}(\tau))\leq C(N,d,u,\tau)\lambda_{d}(I(i)).$

For $I(i)$ related to $B_{i}(\tau)$ in this way say $I(i)\sim B_{i}(\tau)$.

    2. b)

covering: Choose $G_{n}>G_{n-1}$ such that

$4r(B_{j})<\min\\{r(I(k)):I(k)\sim B_{k}(\tau)\in
E_{n-1}\\}\quad\forall\,\,j\geq G_{n}.$

(Note that $r(B_{j})\to 0$ with $j$.) Again, note this is only a lower bound,
$G_{n}$ can be chosen as large as we want, and indeed, we will increase it if
necessary in the proof of Lemma 8. Apply Lemma 7 to $I(i)$ with $G_{n}$ chosen
as above. Let $T_{G_{n},I(i)}$ be the set of balls from Lemma 7.

    3. c)

shrinking: For each $B_{i}\in T_{G_{n},I(i)}$ shrink the ball and let

$T_{G_{n},I(i)}^{\tau}=\left\\{B_{j}(\tau):B_{j}\in T_{G_{n},I(i)}\right\\}.$

Define the $n$th layer to be

$E_{n}=\left\\{B_{j}(\tau):B_{j}(\tau)\in\bigcup_{I(k)\sim B_{k}(\tau)\in
E_{n-1}}T^{\tau}_{G_{n},I(k)}\right\\}.$

Let

${\cal D}(B)=\bigcap_{n\in{\mathbb{N}}}\bigcup_{B(\tau)\in E_{n}}B(\tau)$

This completes the construction of ${\cal D}(B)$.
Note that ${\cal D}(B)\subset
D_{d}(\psi_{\tau},c_{N}\psi_{\tau})\subset\mathbf{Bad}_{d}(\psi_{\tau})$ since
for any $\mathbf{x}\in{\cal D}(B)$ we have an associated sequence

$I_{0}\supset B_{i_{1}}(\tau)\supset I(i_{1})\supset B_{i_{2}}(\tau)\supset
I(i_{2})\dots$

such that $\mathbf{x}$ is contained in the limit set. So $\mathbf{x}\in{\cal
C}^{\tau}(N)\cap W(Q(N,\tau),\psi_{\tau})\subset
D_{d}(\psi_{\tau},c_{N}\psi_{\tau})\subset\mathbf{Bad}_{d}(\psi_{\tau})$.

######  Remark 7.

The methodology of the proof given here follows closely to the usual proof of
the Mass Transference Principle. The key additional step in the construction
is the nesting step, which we require to ensure ${\cal D}(B)\subset{\cal
C}^{\tau}(N)$, and replacing the $K_{G,B}$-lemma [7, Lemma 5] by $T_{G,I}$
Lemma 7 above.

###  5.2 Constructing a measure on ${\cal D}(B)$

To obtain a lower bound to the Hausdorff dimension of ${\cal D}(B)$ and hence
to $\mathbf{Bad}_{d}(\psi_{\tau})$, we aim to use the mass distribution
principle [16, Proposition 4.2]. Namely, we need to construct a measure $\mu$
supported on ${\cal D}(B)$ such that measures of balls are in good control.

To that end, construct a mass distribution $\mu$ on ${\cal D}(B)$ as follows:
Set $\mu(I_{0})=1$, then for each $B_{i}(\tau)\in E_{1}$ define

$\mu(B_{i}(\tau))=\frac{\lambda_{d}(B_{i})}{\sum_{B_{j}\in
T_{G_{0},I_{0}}}\lambda_{d}(B_{j})}\times\mu(I_{0}).$

Note

$\sum_{B_{i}(\tau)\in E_{1}}\mu(B_{i}(\tau))=\mu(I_{0}).$

Now inductively suppose the mass has been distributed over balls in $E_{n-1}$.
For each $B_{i}(\tau)\in E_{n}$ there exists, by construction, a unique
$B_{k}(\tau)\in E_{n-1}$ such that $B_{i}(\tau)\subset B_{k}(\tau)$. Let
$I(k)\sim B_{k}(\tau)$ and define

$\mu(B_{i}(\tau))=\frac{\lambda_{d}(B_{i})}{\sum_{B_{j}\in
T_{G_{n},I(k)}}\lambda_{d}(B_{j})}\times\mu(B_{k}(\tau)).$

Again the mass is preserved. By Proposition 1.7 of [16] this extends to a
measure $\mu$ with support contained in ${\cal D}(B)$ and for general subset
$F\subset[0,1]^{d}$

$\mu(F)=\inf\left\\{\sum_{i}\mu(B_{i}(\tau)):F\cap{\cal
D}(B)\subset\bigcup_{i}B_{i}(\tau)\,\text{ with
}\,B_{i}(\tau)\in\bigcup_{n}E_{n}\right\\}.$

Recall that $s=(1+d)/(1+\tau)$.

######  Lemma 8.

Let $\epsilon>0$. Let $F$ be a ball contained in $B$. Then there is a choice
of suitable $(G_{n})$ in steps 4 of the construction of ${\cal D}(B)$ such
that

$\mu(F)\leq r(F)^{s-\epsilon}.$

###### Proof.

Case 1: $B_{i}(\tau)\in E_{n}$.

Assume first that $F=B_{i}(\tau)\in E_{1}$, a ball of centre
$\tfrac{p}{q}\in{\mathbb{Q}}^{d}$ and radius $q^{-(1+\tau)}$. Then

$\displaystyle\mu(B_{i}(\tau))$
$\displaystyle=\frac{\lambda_{d}(B_{i})}{\sum_{B_{j}\in
T_{G_{0},I_{0}}}\lambda_{d}(B_{j})}\times 1$
$\displaystyle\overset{\text{Lemma~{}\ref{TGI}}}{\leq}\lambda_{d}(B_{i})\frac{1}{\kappa_{1}\lambda_{d}(I_{0})}$
$\displaystyle\leq\frac{1}{\kappa_{1}\lambda_{d}(I_{0})}2^{2d}q^{-(d+1)}$
$\displaystyle=\frac{2^{2d}}{\kappa_{1}\lambda_{d}(I_{0})}q^{-(1+\tau)\frac{d+1}{1+\tau}}$
$\displaystyle=\frac{2^{2d}}{\kappa_{1}\lambda_{d}(I_{0})}r(B_{i}(\tau))^{s}.$

For balls $B_{i}(\tau)\in E_{n}$ centred at some
$\tfrac{p}{q}\in{\mathbb{Q}}^{d}$ with $n>1$ there exists $B_{k}(\tau)\in
E_{n-1}$ with $B_{i}(\tau)\subset B_{k}(\tau)$. Let

$\displaystyle\mu(B_{i}(\tau))$
$\displaystyle=\frac{\lambda_{d}(B_{i})}{\sum_{B_{j}\in
T_{G_{0},I(k)}}\lambda_{d}(B_{j})}\times\mu(B_{k}(\tau))$
$\displaystyle\overset{\text{Lemma~{}\ref{TGI}}}{\leq}\lambda_{d}(B_{i})\frac{\mu(B_{k}(\tau))}{\kappa_{1}\lambda_{d}(I(k))}$
$\displaystyle\leq\frac{\mu(B_{k}(\tau))}{\kappa_{1}\lambda_{d}(I(k))}2^{2d}q^{-(d+1)}$
$\displaystyle=\frac{2^{2d}}{\kappa_{1}}r(B_{i}(\tau))^{s}\frac{\mu(B_{k}(\tau))}{\lambda_{d}(I(k))}$
$\displaystyle\overset{\text{Lemma~{}\ref{I in
B(t)}}}{\leq}\frac{2^{2d}}{\kappa_{1}C(N,d,u,\tau)}r(B_{i}(\tau))^{s}\frac{\mu(B_{k}(\tau))}{\lambda_{d}(B_{k}(\tau))}$

The value of $G_{n}$ can be chosen as large as we like, and hence for any
$\epsilon>0$ we can suppose it was large enough to guarantee

$r(B_{i}(\tau))^{-\epsilon}>\frac{\mu(B_{k}(\tau))}{\lambda_{d}(B_{k}(\tau))}.$
(16)

Note that this can be done in a uniform manner since for any $B_{i}(\tau)\in
E_{n}$ we have $i>G_{n}$ and the set $E_{n-1}$ is finite. Hence,

$\mu(B_{i}(\tau))\leq C^{\prime}(N,d,u,\tau)r(B_{i}(\tau))^{s-\epsilon}.$

Observe the constant is independent of the layer which $B_{i}(\tau)$ is from.

Case 2: $F$ general ball.

Now fix $F$, a ball contained in $B$. We prove the general statement that for
every $\epsilon>0$

$\mu(F)\ll_{N,d,u,\tau}r(F)^{s-\epsilon}.$ (17)

Firstly note that if $F\cap{\cal D}(B)=\emptyset$ then $\mu(F)=0$ so the
statement is trivially satisfied. Hence without loss of generality $F\cap{\cal
D}(B)\neq\emptyset$. Since $F$ has non-empty intersection with ${\cal D}(B)$
at least one ball in each layer $E_{n}$ must intersect $F$. If there is
exactly one ball $B_{i_{n}}(\tau)$ per layer $E_{n}$ (that intersects $F$)
then note that

$\mu(B_{i_{n}}(\tau))\to 0\quad\text{ as }\,n\to\infty$

and so (17) holds, since there exists some $n\in{\mathbb{N}}$ such that

$\mu(B_{i_{n}}(\tau))\ll r(F)^{s-\epsilon}.$

Hence assume there exists $n_{F}\in{\mathbb{N}}$ such that at least two balls,
say $B_{i}(\tau),B_{j}(\tau)\in E_{n_{F}}$ have non-empty intersection with
$F$. Furthermore $n_{F}$ is the first layer in which this happens.

Note $B_{i}(\tau),B_{j}(\tau)$ belong to the same ball $B_{k}(\tau)\in
E_{n_{F}-1}$ (otherwise the $n_{F}-1$ layer has non-empty intersection with at
least two balls, contradicting the choice of $n_{F}$). If

$r(F)\geq r(B_{k}(\tau)),$

then

$\mu(F)\leq\underset{B_{i}(\tau)\cap F\neq\emptyset}{\sum_{B_{i}(\tau)\in
E_{n_{F}-1}:}}\mu(B_{i}(\tau))=\mu(B_{k}(\tau)).$

We can then use Case 1 of the proof to estimate $\mu(B_{k}(\tau))$ from above
and obtain

$\mu(F)\leq C^{\prime}(N,d,u,\tau)r(B_{k}(\tau))^{s-\epsilon}\leq
C^{\prime}(N,d,u,\tau)r(F)^{s-\epsilon}$

as required.

So assume

$r(F)<r(B_{k}(\tau)).$ (18)

We have that

$\lambda_{d}(5F)\geq\sum_{B_{i}:B_{i}(\tau)\in{\cal F}}\lambda_{d}(B_{i}),$
(19)

where

${\cal F}=\left\\{B_{i}(\tau)\in T_{G_{n_{F}},I(k)}^{\tau}:B_{i}(\tau)\cap
F\neq\emptyset\right\\}.$

See [7, Lemma 7], setting $A=F$ and $M=B_{i}(\tau)$, $cM=B_{i}$ for
justification of the above statement. Hence

$\displaystyle\mu(F)$ $\displaystyle\leq\sum_{B_{i}(\tau)\in{\cal
F}}\mu(B_{i}(\tau))$ $\displaystyle\leq\sum_{B_{i}:B_{i}(\tau)\in{\cal
F}}\frac{\lambda_{d}(B_{i})}{\sum_{B_{j}\in
T_{G_{n_{F}},I(k)}}\lambda_{d}(B_{j})}\times\mu(B_{k}(\tau))$
$\displaystyle\overset{\text{Lemma~{}\ref{TGI}}}{\leq}\frac{1}{\kappa_{1}}\sum_{B_{i}:B_{i}(\tau)\in{\cal
F}}\lambda_{d}(B_{i})\times\frac{\mu(B_{i}(\tau))}{\lambda_{d}(I(k))}$
$\displaystyle\overset{\text{Lemma~{}\ref{I in
B(t)}}}{\leq}\frac{1}{\kappa_{1}C(N,d,u,\tau)}\sum_{B_{i}:B_{i}(\tau)\in{\cal
F}}\lambda_{d}(B_{i})\times\frac{\mu(B_{i}(\tau))}{\lambda_{d}(B_{k}(\tau))}$
$\displaystyle\overset{\eqref{F
count}}{\leq}\frac{5^{d}}{\kappa_{1}C(N,d,u,\tau)}\lambda_{d}(F)\frac{\mu(B_{i}(\tau))}{\lambda_{d}(B_{k}(\tau))}.$

Finally, recalling that $d>s$ we continue with

$\displaystyle\mu(F)$
$\displaystyle\overset{(d>s)}{\leq}\frac{10^{d}}{\kappa_{1}C(N,d,u,\tau)}r(F)^{s}\frac{\mu(B_{i}(\tau))}{\lambda_{d}(B_{k}(\tau))}$
$\displaystyle\leq\frac{10^{d}}{\kappa_{1}C(N,d,u,\tau)}r(F)^{s-\epsilon},$

so (17) holds, where the last line follows on the observation that

$r(F)^{-\epsilon}\overset{\eqref{assumption2}}{>}r(B_{k}(\tau))^{-\epsilon}\overset{\eqref{assumption1}}{>}\frac{\mu(B_{i}(\tau))}{\lambda_{d}(B_{k}(\tau))}.$

∎

We can now finish the proof of Theorem 2. By the mass distribution principle,
see for example [16, Proposition 4.2], we have that

$\operatorname{\dim_{H}}{\cal D}(B)\geq s-\varepsilon.$

Since $\varepsilon>0$ is arbitrary the lower bound dimension result of Theorem
2, and thus Theorem 1, follows.

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