# Differentiation properties of class ${L}^{1}([0,1]^{2})$ with respect to two
different basis of rectangles

Michihiro Hirayama<EMAIL_ADDRESS>Davit Karagulyan<EMAIL_ADDRESS>
###### Abstract

It is a well known result by Saks [Saks1934] that there exist a function $f\in
L^{1}(\mathbb{R}^{2})$ so that for almost every $(x,y)\in\mathbb{R}^{2}$

$\lim_{\begin{subarray}{c}\mathrm{diam}R\rightarrow 0,\\\ (x,y)\in
R\in\mathcal{R}\end{subarray}}\left|\frac{1}{|R|}\int_{R}f(x,y)\,dxdy\right|=\infty,$

where $\mathcal{R}=\\{[a,b)\times[c,d)\colon a<b,c<d\\}$. In this note we
address the following question: assume we have two different collections of
rectangles; under which conditions there exists a function $f\in
L^{1}(\mathbb{R}^{2})$ so that its integral averages are divergence with
respect to one collection and convergence with respect to another? More
specifically, let $\mathcal{D},\mathcal{C}\subset(0,1]$ and consider
rectangles with side lengths in $\mathcal{D}$ and respectively in
$\mathcal{C}$. We show that if the sets $\mathcal{D}$ and $\mathcal{C}$ are
sufficiently “far” from each other, then such a function can be constructed.
We also show that in the class of positive functions our condition is also
necessary for such a function to exist.

##  1 Introduction

Let $\mathcal{R}=\\{[a,b)\times[c,d)\\}$ be the set of all rectangles with
their sides parallel to the coordinate axis. Given a collection
$\mathcal{C}\subset(0,1]$, let $\mathcal{R}_{\mathcal{C}}\subset\mathcal{R}$
be the collection of all rectangles $[a,b)\times[c,d)$ so that
$b-a\in\mathcal{C}$ and $d-c\in\mathcal{C}$.

######  Definition 1.1.

A family of rectangles $\mathcal{M}\subset\mathcal{R}$ is said to be a basis
of differentiation (or simply a basis), if for any point $z\in\mathbb{R}^{2}$
there exists a sequence of rectangles $R_{k}\in\mathcal{M}$ such that $z\in
R_{k}$, $k\in\mathbb{N}$, and $\mathrm{diam}R_{k}\rightarrow 0$ as
$k\rightarrow\infty$.

Let $\mathcal{C},\mathcal{D}\subset(0,1]$ be two collections. Thus
$\mathcal{R}_{\mathcal{C}}$ and $\mathcal{R}_{\mathcal{D}}$ will be basis of
differentiation if and only if $\liminf\mathcal{C}=0$ and
$\liminf\mathcal{D}=0$. Let $\mathcal{M}\subset\mathcal{R}$ be a
differentiation basis. For any function $f\in
L^{1}\left(\mathbb{R}^{2}\right)$ we define

$\delta_{\mathcal{M}}(z,f)=\limsup_{\begin{subarray}{c}\mathrm{diam}R\rightarrow
0,\\\ z\in
R\in\mathcal{M}\end{subarray}}\left|\frac{1}{|R|}\int_{R}f\,dm-f(z)\right|.$

Here and below, we denote by $m$ or $|\cdot|$ the Lebesgue measure on
$\mathbb{R}^{2}$. The function $f\in L^{1}\left(\mathbb{R}^{2}\right)$ is said
to be differentiable at a point $z\in\mathbb{R}^{2}$ with respect to the basis
$\mathcal{M}$, if $\delta_{\mathcal{M}}(z,f)=0$. Denote

$\mathcal{F}(\mathcal{M})=\left\\{f\in
L^{1}\left(\mathbb{R}^{2}\right)\colon\delta_{\mathcal{M}}(z,f)=0,m\text{-}a.e.\
z\in\mathbb{R}^{2}\right\\}.$

Let $\Phi\colon\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ be a convex function.
Denote by $\Phi(L)\left(\mathbb{R}^{2}\right)$ the class of measurable
functions $f$ defined on $\mathbb{R}^{2}$ such that $\Phi(|f|)\in
L^{1}\left(\mathbb{R}^{2}\right)$. If $\Phi$ satisfies the
$\Delta_{2}$-condition $\Phi(2x)\leq k\Phi(x)$, then
$\Phi(L)\left(\mathbb{R}^{2}\right)$ turns to be an Orlicz space with the norm

$\|f\|_{\Phi}=\inf\left\\{c>0\colon\int_{\mathbb{R}^{2}}\Phi\left(\frac{|f|}{c}\right)\,dm\leq
1\right\\}.$

The following classical theorems of Jessen, Marcinkiewicz, and Zygmund
[Jessen-Marcinkiewicz-Zygmund1935], and Saks [Saks1934] determine the optimal
Orlicz space, which functions have a.e. differentiable integrals with respect
to the entire family of rectangles $\mathcal{R}$ is the space

$L(1+\log L)\left(\mathbb{R}^{2}\right)\subset
L^{1}\left(\mathbb{R}^{2}\right)$

corresponding to the case $\Phi(t)=t\left(1+\log^{+}t\right)$. See also
[deGuzman].

######  Theorem 1.2 (Jessen-Marcinkiewicz-Zygmund [Jessen-Marcinkiewicz-
Zygmund1935]).

If

$f\in L(1+\log L)\left(\mathbb{R}^{2}\right),$

then

$\delta_{\mathcal{R}}(z,f)=0$

for $m$-almost every $z\in\mathbb{R}^{2}$.

######  Theorem 1.3 (Saks [Saks1934]).

If

$\Phi(t)=o(t\log t)\text{ as }t\rightarrow\infty,$

then $\Phi(L)\left(\mathbb{R}^{2}\right)\not\subset\mathcal{F}(\mathcal{R})$.
Moreover, there exists a positive function
$f\in\Phi(L)\left(\mathbb{R}^{2}\right)$ such that
$\delta_{\mathcal{R}}(z,f)=\infty$ everywhere.

In this paper, we are interested in differentiability property of class
$L^{1}([0,1]^{2})$ with respect to two basis $\mathcal{R}_{\mathcal{C}}$ and
$\mathcal{R}_{\mathcal{D}}$. We investigate conditions under which there
exists a function $f\in L^{1}([0,1]^{2})$, so that

$\delta_{\mathcal{R}_{\mathcal{C}}}(z,f)=0\ \text{ for $m$-almost every
}z\in[0,1]^{2}$

and

$\delta_{\mathcal{R}_{\mathcal{D}}}(z,f)\neq 0\ \text{ for $m$-almost every
}z\in[0,1]^{2}.$

To the best of the authors knowledge there is only one result that has some
relation to the problem considered in this note. In [Karagulyan-Karagulyan-
Safaryan2017] the authors study equivalence of differentiation basis of dyadic
rectangles. More specifically consider the basis

$\mathcal{R}^{\text{dyadic
}}=\left\\{\left[\frac{i-1}{2^{n}},\frac{i}{2^{n}}\right)\times\left[\frac{j-1}{2^{m}},\frac{j}{2^{m}}\right)\colon
i,j,n,m\in\mathbb{Z}\right\\}.$ (1)

Let $\Delta=\left\\{\nu_{k}:k=1,2,\ldots\right\\}$ be an increasing sequence
of positive integers. This sequence generates the rare basis
$\mathcal{R}_{\Delta}^{\text{dyadic }}$ of dyadic rectangles of the form (1)
with $n,m\in\Delta$. This kind of bases have also been considered in several
papers [Hagelstein2000, Hare-Stokolos2000, Stokolos2006, Karagulyan2013]. In
[Karagulyan-Karagulyan-Safaryan2017] the authors study under which conditions
the basis $\mathcal{R}^{\text{dyadic }}$ will be equivalent to the basis given
by $\mathcal{R}^{\text{dyadic }}_{\Delta}$. We remark that although in
[Karagulyan-Karagulyan-Safaryan2017] one only considers dyadic rectangles,
this case can be compared with $\mathcal{D}=[0,1]\setminus\mathcal{C}$ in our
case.

Unlike the results considered above, in this note we consider the problem in
full generality, namely we and are interested in all rectangles with their
sides in $\mathcal{C}$ or $\mathcal{D}$ that contain the point of
differentiation.

The paper is self contained and uses some methods from analysis and
probability theory.

##  2 Main results

Intuitively, in order to have divergence and convergence phenomena
simultaneously, the two set $\mathcal{C}$ and $\mathcal{D}$ have to be far
from each other. We now formalize this intuition. For each $x\in\mathcal{D}$,
we define the following two numbers

$\displaystyle\overline{x}$ $\displaystyle=\sup\left\\{a\in\mathcal{C}\colon
a<x\right\\},$ $\displaystyle\underline{x}$
$\displaystyle=\inf\left\\{a\in\mathcal{C}\colon a>x\right\\}.$

The ratios $\overline{x}/x$ and $x/\underline{x}$ denote that distance of $x$
from the set $\mathcal{C}$ from below and above. We prove the following
theorem.

######  Theorem 2.1.

Let $\mathcal{C},\mathcal{D}\subset(0,1]$ and assume

$\liminf_{x\rightarrow
0,x\in\mathcal{D}}\left(\max\left\\{\frac{\overline{x}}{x},\frac{x}{\underline{x}}\right\\}\right)=0.$

Then there exists a function $f\in{L}^{1}([0,1]^{2},m)$ so that for $m$-almost
all $z\in[0,1]^{2}$

$\delta_{\mathcal{R}_{\mathcal{C}}}(z,f)=0,$

and

$\delta_{\mathcal{R}_{\mathcal{D}}}(z,f)=\infty.$

The function that will be constructed in Theorem 2.1 is unbounded both from
above and below. This leads as to the following question

######  Question.

Does there exist a positive function $f$ satisfying the conditions of Theorem
2.1?

We have the following theorem in the opposite direction

######  Theorem 2.2.

Let $\mathcal{C},\mathcal{D}\subset(0,1]$ and assume that

$\liminf_{x\rightarrow
0,x\in\mathcal{D}}\left(\max\left\\{\frac{\overline{x}}{x},\frac{x}{\underline{x}}\right\\}\right)>0.$

Let $f\in L^{1}([0,1]^{2},m)$, with $f\geq 0$ almost surely. If

$\delta_{\mathcal{R}_{\mathcal{C}}}(z,f)=0\ \text{ for $m$-almost every
}z\in[0,1]^{2},$

then

$\delta_{\mathcal{R}_{\mathcal{D}}}(z,f)=0\ \text{ for $m$-almost every
}z\in[0,1]^{2}.$

The proof of Theorem 2.1 is based on the following theorem.

######  Theorem 2.3.

For every $\varepsilon,\delta>0$ and $n>0$, there exist a function $f\in
L^{\infty}([0,1]^{2},m)$, $\eta=\eta(\varepsilon,\delta,n)\in(0,\delta)$ and a
set $Q\subset[0,1]^{2}$, so that

$\|f\|_{L^{1}(m)}\leq 4+o(1),$ (2)

and $|Q|>1-\varepsilon$ such that

  1. (i)

If $z\in Q$ and $z\in A\in\mathcal{R}_{\mathcal{C}}$, then

$\left|\frac{1}{|A|}\int_{A}f\,dm\right|\leq 2.$

  2. (ii)

If $z\in Q$, then there exists $B\in\mathcal{R}_{\mathcal{D}}$, with $z\in B$
and $|B|>\eta$, so that

$\left|\frac{1}{|B|}\int_{B}f\,dm\right|>n.$

  3. (iii)

For every $R\in\mathcal{R}$ with $|R|>\delta$, we have

$\left|\frac{1}{|R|}\int_{R}f\,dm\right|<1.$

##  3 Preliminaries

###  3.1 Auxiliary construction

Let $n\in\mathbb{N}$. Given a decreasing sequence
$b_{1}>b_{2}>\dots>b_{n}>\dots>b_{2n}$, let

$B_{1}=[0,b_{1}]\times[0,b_{2n}],B_{2}=[0,b_{2}]\times[0,b_{2n-1}],\dots,B_{n}=[0,b_{n}]\times[0,b_{n+1}].$

Here the first factor is the height and the second one is the width. Suppose
that the sequence $(b_{j})$ satisfies

$b_{1}b_{2n}\leq b_{2}b_{2n-1}\leq\dots\leq b_{n-1}b_{n+2}\leq b_{n}b_{n+1}.$
(3)

In other words, we suppose that

$|B_{1}|\leq|B_{2}|\leq\dots\leq|B_{n-1}|\leq|B_{n}|.$

For $j=1,\dots,n-1$, let $q_{j}\in\mathbb{N}$ be such that
$q_{j}=\lfloor|B_{j+1}|/|B_{j}|\rfloor$.

Let $\Theta$ be the set of
$(\theta_{n-1},\theta_{n-2},\dots,\theta_{2},\theta_{1})$ with
$\theta_{j}\in\\{0,1,\dots,q_{j}\\}$ satisfying the following condition: if
$\theta_{k}=0$ for some $k$, then $\theta_{l}=0$ for every
$l\in\\{k,k-1,\dots,1\\}$. For each
$\theta=(\theta_{n-1},\theta_{n-2},\dots,\theta_{1})\in\Theta$, let

$|\theta|=\begin{cases}1,&\text{if all }\theta_{j}\neq 0\\\ \max\\{1\leq j\leq
n-1\colon\theta_{j}=0\\}+1,&\text{otherwise}\end{cases}$

and define rectangles $B(\theta)$ as follows. For $\theta\in\Theta$ with
$|\theta|=n$, that is $\theta_{n-1}=0$, we let

$B(\theta)=\left[0,b_{n}\right]\times\left[0,b_{n+1}\right]=B_{n}.$

For $\theta\in\Theta$ with $|\theta|\in\\{1,\dots,n-1\\}$, we let

$B(\theta)=\left[0,b_{|\theta|}\right]\times\left[\sum_{j=|\theta|}^{n-1}(\theta_{j}-1)b_{2n+1-j},\sum_{j=|\theta|}^{n-1}(\theta_{j}-1)b_{2n+1-j}+b_{2n+1-|\theta|}\right],$

where note that $\theta_{n-1}\in\\{1,\dots,q_{n-1}\\}$ as $|\theta|\neq n$.
Note also that for each $\theta\in\Theta$, one has
$|B(\theta)|=b_{|\theta|}b_{2n+1-|\theta|}$.

Define subsets of $[0,1]^{2}$ by

$E=\bigcup_{\theta\in\Theta}B(\theta)\quad\text{and}\quad
F=\bigcup_{\begin{subarray}{c}\theta\in\Theta\colon\\\
|\theta|<n\end{subarray}}B(\theta).$

One has

$\displaystyle E$
$\displaystyle=\bigcup_{j=1}^{n}\bigcup_{|\theta|=j}B(\theta)=\left(\bigcup_{j=1}^{n-1}\bigcup_{|\theta|=j}B(\theta)\right)\cup
B_{n},$ $\displaystyle F$
$\displaystyle=\bigcup_{j=1}^{n-1}\bigcup_{|\theta|=j}B(\theta),$ (4)

where for every $j\in\\{1,\dots,n-1\\}$ there are $q_{n-1}\cdots q_{j}$-many
rectangles $B(\theta)$ of $|\theta|=j$.

###  3.2 Area estimates

######  Lemma 3.1.

Suppose that there is $\lambda\in(0,1)$ such that $b_{k+1}/b_{k}<\lambda$ for
every $k=1,\dots,2n-1$. Then one has

$\frac{1}{n}+\frac{1-\lambda}{n}\sum_{j=1}^{n-1}\frac{q_{n-1}\cdots
q_{j}}{(q_{n-1}+1)\cdots(q_{j}+1)}\leq\frac{|E|}{n|B_{n}|}\leq 1,$ (5)

and

$\frac{1-\lambda}{n-1}\sum_{j=1}^{n-1}\frac{q_{n-1}\cdots
q_{j}}{(q_{n-1}+1)\cdots(q_{j}+1)}\leq\frac{|F|}{(n-1)|B_{n}|}\leq 1.$ (6)

###### Proof.

By construction, one has

$\displaystyle|E|$
$\displaystyle=b_{n}b_{n+1}+q_{n-1}b_{n+2}(b_{n-1}-b_{n})+q_{n-1}q_{n-2}b_{n+3}(b_{n-2}-b_{n-1})$
$\displaystyle\quad+\cdots+q_{n-1}\cdots q_{1}b_{2n}(b_{1}-b_{2})$
$\displaystyle=|B_{n}|+q_{n-1}|B_{n-1}|+q_{n-1}q_{n-2}|B_{n-2}|+\cdots+q_{n-1}\cdots
q_{1}|B_{1}|$
$\displaystyle\quad-(q_{n-1}b_{n+2}b_{n}+q_{n-1}q_{n-2}b_{n+3}b_{n-1}+\cdots+q_{n-1}\cdots
q_{1}b_{2n}b_{2}).$

Since

$q_{n-1}\cdots q_{j+1}q_{j}|B_{j}|\leq q_{n-1}\cdots
q_{j+1}|B_{j+1}|\leq\dots\leq q_{n-1}|B_{n-1}|\leq|B_{n}|$

for every $j=1,\dots,n-1$, it follows that $|E|\leq n|B_{n}|$.

On the other hand, one has

$\displaystyle\frac{|E|}{n|B_{n}|}$
$\displaystyle=\frac{|B_{n}|+\sum_{j=1}^{n-1}q_{n-1}\cdots
q_{j}b_{2n+1-j}(b_{j}-b_{j+1})}{n|B_{n}|}$
$\displaystyle=\frac{1}{n}+\frac{1}{n}\sum_{j=1}^{n-1}\frac{q_{n-1}\cdots
q_{j}b_{2n+1-j}(b_{j}-b_{j+1})}{|B_{n}|}.$

Since

$|B_{n}|\leq(q_{n-1}+1)|B_{n-1}|\leq\dots\leq(q_{n-1}+1)\cdots(q_{j}+1)|B_{j}|$

for each $j=1,\dots,n-1$, it follows that

$\displaystyle\frac{q_{n-1}\cdots q_{j}b_{2n+1-j}(b_{j}-b_{j+1})}{|B_{n}|}$
$\displaystyle\geq\frac{q_{n-1}\cdots
q_{j}}{(q_{n-1}+1)\cdots(q_{j}+1)}\frac{b_{2n+1-j}(b_{j}-b_{j+1})}{b_{j}b_{2n+1-j}}$
$\displaystyle=\frac{q_{n-1}\cdots
q_{j}}{(q_{n-1}+1)\cdots(q_{j}+1)}\left(1-\frac{b_{j}}{b_{j+1}}\right)$
$\displaystyle\geq\frac{q_{n-1}\cdots
q_{j}}{(q_{n-1}+1)\cdots(q_{j}+1)}\left(1-\lambda\right),$

and hence

$\displaystyle\frac{|E|}{n|B_{n}|}$
$\displaystyle\geq\frac{1}{n}+\frac{1-\lambda}{n}\sum_{j=1}^{n-1}\frac{q_{n-1}\cdots
q_{j}}{(q_{n-1}+1)\cdots(q_{j}+1)}.$

We have obtained (5). Since the proof of (6) is same as that of (5), we omit
it. ∎

Let

$\displaystyle K_{E}$
$\displaystyle=\left(\frac{1}{n}+\frac{1-\lambda}{n}\sum_{j=1}^{n-1}\frac{q_{n-1}\cdots
q_{j}}{(q_{n-1}+1)\cdots(q_{j}+1)}\right)^{-1},$ $\displaystyle K_{F}$
$\displaystyle=\left(\frac{1-\lambda}{n-1}\sum_{j=1}^{n-1}\frac{q_{n-1}\cdots
q_{j}}{(q_{n-1}+1)\cdots(q_{j}+1)}\right)^{-1}.$

Then it follows from (5) and (6) in Lemma 3.1 that

$K_{E}^{-1}\frac{n}{n-1}\leq\frac{|E|}{|F|}\leq K_{F}\frac{n}{n-1}.$ (7)

Notice that $K_{E},K_{F}\geq 1$ and they can be arbitrarily close to one by
taking $\lambda\in(0,1)$ small and $(b_{n})$ relevantly.

##  4 Main Lemma

Without loss of generality we can assume that $\mathcal{D}$ is a sequence,
i.e. $\mathcal{D}=\\{b_{n}\\}_{n\geq 1}$. For each $b_{n}\in\mathcal{D}$, we
define

$\displaystyle\overline{a}_{n}$
$\displaystyle=\sup\left\\{a\in\mathcal{C}\colon a<b_{n}\right\\},$
$\displaystyle\underline{a}_{n}$
$\displaystyle=\inf\left\\{a\in\mathcal{C}\colon a>b_{n}\right\\}.$

Recall that we are given $\mathcal{C},\mathcal{D}\subset(0,1]$, where
$\mathcal{D}=\\{b_{n}\\}_{n\in\mathbb{N}}$, and assume

$\liminf_{n\rightarrow\infty}\left(\max\left\\{\frac{\bar{a}_{n}}{b_{n}},\frac{b_{n}}{\underline{a}_{n}}\right\\}\right)=0.$
(8)

Throughout this section, we assume (8). In this section, we prove the
following lemma which will play a fundamental role for the proof of Theorems.

######  Lemma 4.1.

Given $\varepsilon,\delta>0$ and $n\in\mathbb{N}$, there exist sets
$F,D\subset[0,1]^{2}$, a function $h\in L^{1}([0,1]^{2},m)$, and
$\eta=\eta(\varepsilon,\delta,n)\in(0,\delta)$ satisfying
$|D|\leq\varepsilon|F|$ and

$\frac{\|h\|_{L^{1}(m)}}{|F|}=2+o(1),$ (9)

such that for every $z\in[0,1]^{2}$ we have the following.

  1. (i)

If $z\notin D$ and $z\in A\in\mathcal{R}_{\mathcal{C}}$, then

$\left|\frac{1}{|A|}\int_{A}h\,dm\right|\leq 1.$

  2. (ii)

If $z\in F$, then there exists $B\in\mathcal{R}_{\mathcal{D}}$, with $z\in B$
and $|B|>\eta$, so that

$\left|\frac{1}{|B|}\int_{B}h\,dm\right|>n.$

  3. (iii)

For every $R\in\mathcal{R}$ with $|R|>\delta$, we have

$\left|\frac{1}{|R|}\int_{R}h\,dm\right|<1/2.$

###  4.1 Proof of Lemma 4.1

We start with construction of a function and sets in the following two
sections. Then we complete the proof of Lemma 4.1 in Section 4.1.3.

####  4.1.1 Construction of the function

Define $\Theta^{*}=\\{\theta\in\Theta\colon|\theta|=1\\}$. For
$\tau\in(0,b_{2n})$ and $\theta\in\Theta^{*}$, define a subset of $B(\theta)$
by

$B_{\tau}(\theta)=[0,\tau]\times\left[\sum_{j=1}^{n-1}(\theta_{j}-1)b_{2n+1-j},\sum_{j=1}^{n-1}(\theta_{j}-1)b_{2n+1-j}+\tau\right].$

Namely $B_{\tau}(\theta)$ is the square with side lengths $\tau$ at the bottom
left corner of $B(\theta)$ for $\theta\in\Theta^{*}$. Define also
$\sigma\colon\Theta^{*}\to\\{1,-1\\}$ by

$\sigma(\theta)=\begin{cases}+1,\text{ if }\theta_{n-1}\text{ is odd},\\\
-1,\text{ if }\theta_{n-1}\text{ is even}.\end{cases}$

Then define $h\colon[0,1]^{2}\to\mathbb{R}$ by

$h(z)=\begin{cases}\sigma(\theta)\dfrac{n|B_{n}|}{\tau^{2}q_{1}\cdots
q_{n-1}},&\text{if }z\in B_{\tau}(\theta)\text{ for some
}\theta\in\Theta^{*},\\\ 0,&\text{otherwise}.\end{cases}$ (10)

######  Lemma 4.2.

For every $\theta\in\Theta$, one has

$n\leq\frac{1}{|B(\theta)|}\left|\int_{B(\theta)}h\,dm\right|\leq
n\prod_{j=|\theta|}^{n-1}\left(1+\frac{1}{q_{j}}\right).$

###### Proof.

By construction, one has

$\displaystyle\left|\int_{B(\theta)}h\,dm\right|$
$\displaystyle=\frac{n|B_{n}|}{\tau^{2}q_{1}\cdots
q_{n-1}}\times|B(\theta)\cap\text{supp}h|$
$\displaystyle=\frac{n|B_{n}|}{\tau^{2}q_{1}\cdots
q_{n-1}}\times\begin{cases}\tau^{2}q_{1}\cdots q_{|\theta|-1},&|\theta|>1\\\
\tau^{2},&|\theta|=1\end{cases}$
$\displaystyle=\frac{n|B_{n}|}{q_{|\theta|}\cdots q_{n-1}},$

and hence

$\frac{1}{|B(\theta)|}\left|\int_{B(\theta)}h\,dm\right|=\frac{1}{|B(\theta)|}\frac{n|B_{n}|}{q_{|\theta|}\cdots
q_{n-1}}\geq\frac{n|B_{n}|}{|B_{n}|}=n.$

Since
$|B_{n}|\leq(q_{n-1}+1)|B_{n-1}|\leq\cdots\leq(q_{n-1}+1)\cdots(q_{|\theta|}+1)|B(\theta)|$,
one also has

$\frac{1}{|B(\theta)|}\left|\int_{B(\theta)}h\,dm\right|=\frac{1}{|B(\theta)|}\frac{n|B_{n}|}{q_{|\theta|}\cdots
q_{n-1}}\leq n\prod_{j=|\theta|}^{n-1}\left(1+\frac{1}{q_{j}}\right).$

∎

######  Lemma 4.3.

Let $n\in\mathbb{N}$. Then we have

$2\leq\frac{\|h\|_{L^{1}(m)}}{|E|}\leq 2K_{E},$ (11)

and

$2K_{E}^{-1}\frac{n}{n-1}\leq\frac{\|h\|_{L^{1}(m)}}{|F|}\leq
2K_{E}K_{F}\frac{n}{n-1}.$ (12)

###### Proof.

Note first that

$\|h\|_{L^{1}([0,1]^{2},m)}=2\int_{F}|h|\,dm.$

It follows from the construction that

$\displaystyle\int_{F}|h|\,dm=\frac{n|B_{n}|}{\tau^{2}q_{1}\cdots
q_{n-1}}\times|F\cap\text{supp}h|=\frac{n|B_{n}|}{\tau^{2}q_{1}\cdots
q_{n-1}}\times\tau^{2}q_{1}\cdots q_{n-1}=n|B_{n}|,$

and hence Lemma 3.1-(5) implies (11). By using (7), we have (12) from (11). ∎

####  4.1.2 Construction of the exceptional set $D$

Let $h$ be the function of the form (10) defined in Section 4.1.1. In this
section, we will construct a set $D$ while examining the integral averages of
$h$ with respect to $\mathcal{R}_{\mathcal{C}}$ by making use of the condition
(8). Indeed, one uses the following property on the two sets $\mathcal{C}$ and
$\mathcal{D}=\\{b_{j}\\}_{j\in\mathbb{N}}$. The proof is a direct consequence
of the condition (8), and is omitted.

######  Lemma 4.4.

Assume the condition (8). Given $b_{1}>\dots>b_{n}$, and $\lambda_{k}\in(0,1)$
for $k=1,\dots,n$, one can choose $b_{n+1}>\dots>b_{2n}$ such that

$\frac{\overline{a}_{n+k}}{b_{n+k}}\leq\lambda_{k}\frac{b_{n-k+1}}{\overline{a}_{n-k}},$

and

$\frac{b_{n+k}}{\underline{a}_{n+k}}\leq\lambda_{k}\frac{\overline{a}_{n-k+1}}{b_{n-k+1}}$

for $k=1,\dots,n$, where $\overline{a}_{0}=1$ as a convention.

Henceforth, we denote rectangles with side lengths $x$ and $y$ by $A_{xy}$,
where $x$ is the length of vertical side and $y$ is that of the horizontal
one. In other words, we assume that $x$ is the height and $y$ is the width of
our rectangle. Let

$\mathcal{A}_{\mathcal{C}}=\left\\{A_{xy}\subset[0,1]^{2}\colon
A_{xy}\cap\mathrm{supp}h\neq\emptyset,\ x,y\in\mathcal{C}\right\\}.$

We primarily divide our argument into the following cases with respect to the
heights of $A_{xy}\in\mathcal{A}_{\mathcal{C}}$: 1) $x\leq b_{n+1}$, 2)
$x\in(b_{n},b_{1}]$, 3) $x>b_{1}$, and 4) $x\in(b_{n+1},b_{n}]$.

Case 1) $x\leq b_{n+1}$:

By assumption (8), take $b_{n+1}$ so that

$\frac{\overline{a}_{n+1}}{b_{n+1}}\ll b_{n}.$

(Here and below, we will sometimes write $X\ll Y$ if for a given
$\lambda\in(0,1)$ one can take $X$ so small that $X\leq\lambda Y$.) Then for
every $y\in(0,1]$ one has

$xy\leq x\leq\overline{a}_{n+1}\ll b_{n}b_{n+1}=|B_{n}|.$

Let

$D_{1}=\bigcup_{A_{xy}\in\mathcal{A}_{\mathcal{C}}}\left\\{\text{int}A_{xy}\colon(x,y)\in(0,b_{n+1}]\times(0,1]\right\\}.$

Case 2) $b_{n}<x\leq b_{1}$:

There is $r=r(x)\in\\{1,\dots,n-1\\}$ such that

$b_{r+1}<x\leq b_{r}.$

We consider two cases depending on the side $y$.

2-i) The case $y>b_{2n-r}$.

We begin with the case where $A=A_{xy}$ contains
$B(\theta)\in\mathcal{R}_{\mathcal{D}}$ with $|\theta|=r+1$, that is height
$b_{r+1}$ and width $b_{2n-r}$. More precisely, there is $p=p(y)\in\mathbb{N}$
such that $A$ contains $p$-many disjoint copies of $B_{|\theta|}=B_{r+1}$ and
does not contain $(p+1)$-many disjoint copies of it. We then have by Lemma 4.2
that

$\int_{A}h\,dm\leq|B(\theta)|(p+2)n\prod_{j=r+1}^{n-1}\left(1+\frac{1}{q_{j}}\right).$

Since $A$ contains a rectangle of height $x$ and width $pb_{2n-r}$, it follows
that

$\displaystyle\frac{1}{|A|}\int_{A}h\,dm$
$\displaystyle\leq\frac{(p+2)nb_{r+1}b_{2n-r}}{xpb_{2n-r}}\prod_{j=r+1}^{n-1}\left(1+\frac{1}{q_{j}}\right)$
$\displaystyle=n\frac{p+2}{p}\frac{b_{r+1}}{x}\prod_{j=r+1}^{n-1}\left(1+\frac{1}{q_{j}}\right)$
$\displaystyle\leq
n\frac{p+2}{p}\frac{b_{r+1}}{\underline{a}_{r+1}}\prod_{j=r+1}^{n-1}\left(1+\frac{1}{q_{j}}\right).$
(13)

Next, suppose that $A$ contains no rectangles
$B(\theta)\in\mathcal{R}_{\mathcal{D}}$ with $|\theta|=r+1$, that is the case
of $p=0$. One still has

$\displaystyle\frac{1}{|A|}\int_{A}h\,dm\leq\frac{2nb_{r+1}b_{2n-r}}{xy}\prod_{j=r+1}^{n-1}\left(1+\frac{1}{q_{j}}\right)\leq
2n\frac{b_{r+1}}{\underline{a}_{r+1}}\frac{b_{2n-r}}{\underline{a}_{2n-r}}\prod_{j=r+1}^{n-1}\left(1+\frac{1}{q_{j}}\right).$
(14)

2-ii) The case $y\leq b_{2n-r}$.

By definition, one has

$\frac{y}{b_{2n-r}}\leq\frac{\overline{a}_{2n-r}}{b_{2n-r}}\quad\text{and}\quad\frac{b_{r+1}}{x}\geq\frac{b_{r+1}}{\overline{a}_{r}}.$

By applying Lemma 4.4 with $k=n-r$, one obtains

$\frac{\overline{a}_{2n-r}}{b_{2n-r}}\leq\lambda\frac{b_{r+1}}{\overline{a}_{r}},$

and thus

$\frac{y}{b_{2n-r}}\leq\frac{\overline{a}_{2n-r}}{b_{2n-r}}\leq\lambda\frac{b_{r+1}}{\overline{a}_{r}}\leq\lambda\frac{b_{r+1}}{x},$

which means $xy\leq\lambda b_{r+1}b_{2n-r}=|B_{r+1}|$.

Let

$D_{2}=\bigcup_{r=1}^{n-1}\bigcup_{A_{xy}\in\mathcal{A}_{\mathcal{C}}}\left\\{\text{int}A_{xy}\colon(x,y)\in(b_{r+1},b_{r}]\times(b_{2n-r+1},b_{2n-r}]\right\\}.$

Case 3) $x>b_{1}$:

We divide into two cases as follows.

3-i) The case $y>b_{2n}$.

By the same argument as Case 2-i), one can show the convergence of integral
averages of $h$ over $A_{xy}$.

3-ii) The case $y\leq b_{2n}$.

As in Case 2-ii), one obtains

$\frac{y}{b_{2n}}\leq\frac{\overline{a}_{2n}}{b_{2n}}\leq\lambda\frac{b_{1}}{\overline{a}_{0}}\leq\lambda\frac{b_{1}}{x}$

with making use of Lemma 4.4 with $k=n$, and hence $xy\leq\lambda
b_{1}b_{2n}=\lambda|B_{1}|$. Let

$D_{3}=\bigcup_{A_{xy}\in\mathcal{A}_{\mathcal{C}}}\left\\{\text{int}A_{xy}\colon(x,y)\in(b_{1},1]\times(0,b_{2n}]\right\\}.$

Case 4) $b_{n+1}<x\leq b_{n}$:

We divide into two cases as follows.

4-i) The case $y>b_{n+1}$.

For such an $A=A_{xy}\in\mathcal{A}_{\mathcal{C}}$, it follows from the
definition of $h$ (10) that either $\int_{A}h\,dm=0$ or there are at most two
$\theta,\theta^{\prime}\in\Theta$ with $|\theta|=|\theta^{\prime}|=n-1$ such
that

$\left|\int_{A}h\,dm\right|\leq\int_{A}|h|\,dm\leq\int_{B(\theta)}|h|\,dm+\int_{B(\theta^{\prime})}|h|\,dm.$

In the latter case, one has

$\frac{1}{|A|}\left|\int_{A}h\,dm\right|\leq\frac{2}{|A|}\int_{B(\theta)}|h|\,dm\leq
2\frac{|B(\theta)|}{|A|}n\left(1+\frac{1}{q_{n-1}}\right)$

by Lemma 4.2. Since

$|B(\theta)|=b_{|\theta|}b_{2n+1-|\theta|}=b_{n-1}b_{n+2}\leq b_{n+2}$

it follows that

$\displaystyle\frac{1}{|A|}\left|\int_{A}h\,dm\right|\leq
4n\frac{b_{n+2}}{|A|}\leq 4n\frac{b_{n+2}}{b_{n+1}^{2}}.$ (15)

4-ii) The case $y\leq b_{n+1}$.

One has $xy\ll b_{n}b_{n+1}=|B_{n}|$ by (8). Let

$D_{4}=\bigcup_{A_{xy}\in\mathcal{A}_{\mathcal{C}}}\left\\{\text{int}A_{xy}\colon(x,y)\in(b_{n+1},b_{n}]\times(0,b_{n+1}]\right\\}.$

Now we set

$D=D_{1}\cup D_{2}\cup D_{3}\cup D_{4}.$ (16)

Observe here that $D_{4}$ is covered by $D_{2}$, hence one has $D=D_{1}\cup
D_{2}\cup D_{3}$.

######  Lemma 4.5.

Given $\varepsilon\in(0,1)$, one can define $F$ and $D$ such that
$|D|\leq\varepsilon|F|$.

###### Proof.

We see that $D_{1}$ is covered by a rectangle $R$ with height
$2(\overline{a}_{n+1}+\tau q_{n-1})$ and width $1$. Then by the assumption
(8),

$|D_{1}|\leq|R|=2(\overline{a}_{n+1}+\tau q_{n-1})\ll 2(b_{n}b_{n+1}+\tau
q_{n-1})=2(|B_{n}|+\tau q_{n-1}).$

Next, we estimate $|D_{2}|$. As we have seen in Case 2-ii, the area of each
rectangle $A_{xy}$ with $(x,y)\in(b_{r+1},b_{r}]\times(b_{2n-r+1},b_{2n-r}]$
is estimated as

$|A_{xy}|=xy\leq\overline{a}_{r}\overline{a}_{2n-r}\leq\lambda|B_{r+1}|$

for each $r\in\\{1,\dots,n-1\\}$. Here, it follows from Lemma 4.4 with $k=n-r$
that

$\overline{a}_{r}\overline{a}_{2n-r}\leq\lambda b_{r+1}b_{2n-r}$

and

$b_{r+1}b_{2n-r}\leq\lambda\overline{a}_{r+1}\underline{a}_{2n-r}.$

Since $\underline{a}_{2n-r}\leq\overline{a}_{2n-r-1}$ by definition, one has

$\overline{a}_{r}\overline{a}_{2n-r}\leq\lambda^{2}\overline{a}_{r+1}\overline{a}_{2n-r-1}$
(17)

for each $r\in\\{1,\dots,n-1\\}$. Note also that

$\frac{\overline{a}_{r+1}}{\overline{a}_{r}}=\frac{\overline{a}_{r+1}}{b_{r+1}}\frac{b_{r+1}}{\overline{a}_{r}}\leq\frac{\overline{a}_{r+1}}{b_{r+1}}\frac{b_{r+1}}{\underline{a}_{r+1}}\ll
1$ (18)

by (8). By making use of (18) and (17), we can apply the same argument for the
proof of Lemma 3.1 to the sequence
$\\{\overline{a}_{1},\dots,\overline{a}_{n-1},\overline{a}_{n+1},\dots,\overline{a}_{2n-1}\\}$.
As a result, we have

$\displaystyle|D_{2}|$
$\displaystyle\leq(n-1)\left|\cup_{A_{xy}\in\mathcal{A}_{\mathcal{C}}}\left\\{A_{xy}\colon(x,y)\in(b_{r+1},b_{r}]\times(b_{2n-r+1},b_{2n-r}]\right\\}\right|$
$\displaystyle\ll(n-1)\left|\cup_{|\theta|=r+1}B(\theta)\right|\leq(n-1)|B_{n}|.$

In the same way as above, we have $|D_{3}|\ll|B_{n}|$.

Consequently, we have

$|D|\leq|D_{1}|+|D_{2}|+|D_{3}|\ll|B_{n}|+(n-1)|B_{n}|+|B_{n}|\leq\frac{n+1}{n}K_{E}|E|$

by Lemma 3.1-(5), and thus

$|D|\ll K_{E}K_{F}\frac{n+1}{n-1}|F|$

by (7). ∎

####  4.1.3 Proof of Lemma 4.1

######  Lemma 4.6.

Let $R$ be a rectangle with $|R|>4n|B_{n}|$. Then one has

$\frac{1}{|R|}\left|\int_{R}h\,dm\right|<\frac{1}{2}.$

###### Proof.

We have

$\frac{1}{|R|}\left|\int_{R}h\,dm\right|\leq\frac{1}{|R|}\|h\|_{L^{1}(m)}=\frac{2}{|R|}n|B_{n}|<\frac{1}{2}.$

∎

###### Proof of Lemma 4.1.

One can suppose, with the aid of (8), that
$b_{1}>\dots>b_{n}>b_{n+1}>\dots>b_{2n}$ will satisfy

  1. 1.

Condition (3),

  2. 2.

$\dfrac{b_{n}}{\underline{a}_{n}}<\dots<\dfrac{b_{1}}{\underline{a}_{1}}<\dfrac{1}{3n2^{n-1}}$,

  3. 3.

Lemma 4.4,

  4. 4.

$\delta>4nb_{n}b_{n+1}$,

  5. 5.

$b_{n+2}<\dfrac{b_{n+1}^{2}}{4n}$.

For such a sequence, let $F$ and $D$ be the sets and $h$ be the function
defined as (4), (16) and (10), respectively. Then we have Lemma 4.5, and the
property (9) follows from (12). Let $\eta=|B_{1}|$. Then $\eta\in(0,\delta)$,
and the property (ii) follows from Lemma 4.2.

The property (i) follows from the consequences of Cases 2-i), 3-i), and 4-i).
Indeed, we have for (13) that

$\displaystyle\frac{1}{|A|}\int_{A}h\,dm\leq
n\frac{p+2}{p}\frac{b_{r+1}}{\underline{a}_{r+1}}\prod_{j=r+1}^{n-1}\left(1+\frac{1}{q_{j}}\right)\leq
3n\frac{1}{3n2^{n-1}}2^{n-r-1}<1$

for every $r\in\\{1,\dots,n-1\\}$ by the condition 2. Similarly, we have for
(14) that

$\displaystyle\frac{1}{|A|}\int_{A}h\,dm$ $\displaystyle\leq
2n\frac{b_{r+1}}{\underline{a}_{r+1}}\prod_{j=r+1}^{n-1}\left(1+\frac{1}{q_{j}}\right)<1$

for every $r\in\\{1,\dots,n-1\\}$ by the condition 2. For (15), it follows
from by the condition 5 that

$\displaystyle\frac{1}{|A|}\int_{A}h\,dm\leq 4n\frac{b_{n+2}}{b_{n+1}^{2}}<1.$

Lemma 4.6 yields the property (iii). Lemma 4.1 is obtained. ∎

##  5 Proof of Theorem 2.3

###  5.1 Random translations

Let

$\omega=(\alpha,\beta),$

where $\alpha$ and $\beta$ are uniformly distributed on $[0,1]^{2}$. Let
$\omega_{1},\dots,\omega_{N}$ be a set of independent, uniformly distributed
vectors on $[0,1]^{2}$. For $z\in[0,1]^{2}$ and
$\omega=(\omega_{1},\dots,\omega_{N})$, consider the product

$f_{N}(z,\omega)=\prod_{k=1}^{N}\left(1-\mathbbm{1}_{F}(z+\omega_{k})\right).$

Note that $f_{N}(z,\omega)=0$ if and only if
$1-\mathbbm{1}_{F}(z+\omega_{k})=0$ for some $k$, i.e., $z\in F+\omega_{k}$.
Hence $\mathrm{supp}f_{N}(\cdot,\omega)\subset[0,1]^{2}$ is the set which is
not covered by the set

$F_{0}(\omega)=\bigcup_{k=1}^{N}\\{F+\omega_{k}\\}.$

Therefore, the support of $1-f_{N}(\cdot,\omega)$ will be the set that is
covered by the set $F_{0}(\omega)$. Next, consider also

$g_{N}(z,\omega)=\prod_{k=1}^{N}\left(1-\mathbbm{1}_{D}(z+\omega_{k})\right),$

and denote

$D_{0}(\omega)=\bigcup_{k=1}^{N}\\{D+\omega_{k}\\}.$

Note that the support of the function $\left((1-f_{N})\cdot
g_{N}\right)(\cdot,\omega)$ is the set of points that is covered by
$F_{0}(\omega)$ but not by $D_{0}(\omega)$, namely $F_{0}(\omega)\setminus
D_{0}(\omega)$. Denote $Q_{0}(\omega)=F_{0}(\omega)\setminus D_{0}(\omega)$.
For each $\omega=(\omega_{1},\dots,\omega_{N})$, let
$|Q_{0}|(\omega)=|Q_{0}(\omega)|$. Then we have

$|Q_{0}|(\omega)=\int_{[0,1]^{2}}\left((1-f_{N})\cdot
g_{N}\right)(z,\omega)\,dm(z).$

Below, we denote the integration with respect to
$\omega=(\omega_{1},\dots,\omega_{N})$ by $\mathcal{E}$ for notational
simplicity. Namely,

$\mathcal{E}(u)=\int_{([0,1]^{2})^{N}}u(\omega)\,d\omega=\int_{[0,1]^{2}}\dots\int_{[0,1]^{2}}u(\omega_{1},\dots,\omega_{N})\,d\omega_{1}\dots
d\omega_{N}$ (19)

for a function $u$ on $([0,1]^{2})^{N}$.

######  Remark 5.1.

Precisely, the integral (19) above should be written as

$\mathcal{E}(u)=\int_{Y}u(\omega(y))\,dP(y)=\int_{([0,1]^{2})^{N}}u\,dP^{\omega}$

where $(Y,P)$ denotes the underlying probability space on which the $N$-tuple
$\omega=(\omega_{1},\dots,\omega_{N})$ of independent random variables is
defined, and $P^{\omega}$ is the joint distribution. By independency,

$\mathcal{E}(u)=\int_{([0,1]^{2})^{N}}u\,dP^{\omega}=\int_{[0,1]^{2}}\dots\int_{[0,1]^{2}}u\,dP^{\omega_{1}}\dots
dP^{\omega_{N}},$ (20)

and denote (20) by (19) for an abuse of notation.

Henceforth, we construct a function defined on (a neighborhood of) a set
$Q_{0}(\omega)=F_{0}(\omega)\setminus D_{0}(\omega)$ with suitably chosen
$\omega$. Here we have the following lemma on such $\omega$.

######  Lemma 5.2.

Let $N=\lceil 1/|F|\rceil\in\mathbb{N}$. There is an open set
$\Omega\subset([0,1]^{2})^{\lceil 1/|F|\rceil}$ with $|\Omega|>0$ such that
for any $\omega\in\Omega$ we have

$|Q_{0}|(\omega)=|F_{0}\setminus
D_{0}|(\omega)\geq\frac{99}{100}\left(\frac{1}{(2e)^{\varepsilon}}-\frac{1}{e}\right)>0$

for sufficiently small $\varepsilon\in(0,1)$.

In fact, we will prove a more general result which will be of independent
interest. See also Remark 5.6.

######  Proposition 5.3.

Let $A,B\subset[0,1]^{2}$ be sets such that $0<|B|\leq c|A|$ for some $c>0$,
and let $N=\lceil 1/|A|\rceil\in\mathbb{N}$. Suppose that for any
$\omega\in([0,1]^{2})^{N}$ we have $|A\cap(B+\omega)|\leq\varepsilon|A|$ for
some $\varepsilon\in\left(0,\min\\{1,c_{0}\\}\right)$ where
$c_{0}=|B|/|A|\in(0,c]$. Then there exist $\kappa=\kappa(c_{0}|A|)\in(0,1)$
and an open set $\Omega\subset([0,1]^{2})^{N}$ with $|\Omega|>0$ such that for
any $\omega\in\Omega$ we have

$\left|\bigcup_{k=1}^{N}\\{A+\omega_{k}\\}\setminus\bigcup_{k=1}^{N}\\{B+\omega_{k}\\}\right|>\frac{99}{100}\left(\frac{1}{(\kappa^{-1}e)^{c_{0}}}-\frac{1}{e^{(1+c_{0}-\varepsilon)}}\right).$

Note that $\kappa=\kappa(c_{0}|A|)\in(0,1)$ in Proposition 5.3 can be taken
such that $\kappa\to 1$ as $c_{0}|A|\to 0$. Once Proposition 5.3 is obtained,
we have Lemma 5.2.

###### Proof of Lemma 5.2.

Take $A=F$ and $B=D$ in Proposition 5.3. One can take $c=\varepsilon$ by Lemma
4.5. Hence Proposition 5.3 yields the result with $\kappa=1/2$. ∎

We postpone the proof of Proposition 5.3 for a short while, and proceed the
argument. Henceforth, we let $N=\lceil 1/|F|\rceil$. Fix sufficiently small
$\varepsilon\in(0,1)$, and set

$\chi=\frac{99}{100}\left(\frac{1}{(2e)^{\varepsilon}}-\frac{1}{e}\right)>0.$

As a result, we obtain the following lemma.

######  Lemma 5.4.

Let $\delta>0$ and $n\in\mathbb{N}$. There exist a function $h_{0}\in
L^{1}([0,1]^{2},m)$ with

$\|h_{0}\|_{L^{1}}\leq 4K_{E}K_{F}\frac{n}{n-1},$

a set $Q_{0}\subset[0,1]^{2}$ with $|Q_{0}|>\chi$, and
$\eta=\eta(\varepsilon,\delta,n)\in(0,\delta)$ such that we have the
following.

  1. (i)

If $z\in Q_{0}$ and $z\in A\in\mathcal{R}_{\mathcal{C}}$, then

$\left|\frac{1}{|A|}\int_{A}h_{0}\,dm\right|\leq 2.$

  2. (ii)

If $z\in Q_{0}$, then there exists $B\in\mathcal{R}_{\mathcal{D}}$ with $B\ni
z$ and $|B|>\eta$ such that

$\left|\frac{1}{|B|}\int_{B}h_{0}\,dm\right|>n.$

  3. (iii)

For every $R\in\mathcal{R}$ with $|R|>\delta$, we have

$\left|\frac{1}{|R|}\int_{R}h\,dm\right|<1.$

###### Proof.

Let $\Omega$ be the set as in Lemma 5.2. Define

$\Omega_{0}=\left\\{\omega\in\Omega\colon|\omega_{i}-\omega_{j}|\notin\\{b_{1},\dots,b_{n}\\}\right\\}.$

We see $|\Omega_{0}|=|\Omega|>0$.

Since $\omega\mapsto Q_{0}(\omega)=F_{0}(\omega)\setminus D_{0}(\omega)$ is
continuous with respect to the Hausdorff metric on the space of all closed
subsets of $[0,1]^{2}$, there exists
$\omega_{0}=((\omega_{0})_{1},\dots,(\omega_{0})_{N})\in\Omega_{0}$ such that
for any rectangle $A=A_{xy}$ with sides $x,y\in\mathcal{C}$ containing $z\in
Q_{0}(\omega_{0})$, there will be cancellations for all but at most two
rectangles from $\mathcal{R}_{D}$, say $B=B(\theta)+(\omega_{0})_{i}$ and
$B^{\prime}=B(\theta^{\prime})+(\omega_{0})_{j}$ for some
$\theta,\theta^{\prime}\in\Theta$ and $i,j\in\\{1,\dots,N\\}$. Namely, let

$h_{0}=h_{F+(\omega_{0})_{1}}+\cdots+h_{F+(\omega_{0})_{N}},$

then it follows that either $\int_{A}h_{0}\,dm=0$ or

$\displaystyle\left|\frac{1}{|A|}\int_{A}h_{0}\,dm\right|$
$\displaystyle\leq\left|\frac{1}{|A|}\int_{A\cap
B}h_{F+(\omega_{0})_{i}}\,dm(z)\right|+\left|\frac{1}{|A|}\int_{A\cap
B^{\prime}}h_{F+(\omega_{0})_{j}}\,dm\right|\leq 2$

by Lemma 4.1-(i). Hence the first property follows.

The argument same as above shows the third property by Lemma 4.1-(iii).

Since $\omega_{0}\in\Omega_{0}$, Lemma 4.1-(ii) yields the property (ii) by
taking $\tau>0$ small if necessary. (Recall that $\tau>0$ determines the
height of the support of $h_{0}$.)

One has

$\displaystyle\|h_{0}\|_{L^{1}(m)}$
$\displaystyle\leq\left\|h_{F+(\omega_{0})_{1}}\right\|_{L^{1}(m)}+\cdots+\left\|h_{F+(\omega_{0})_{N}}\right\|_{L^{1}(m)}$
$\displaystyle\leq 2K_{E}K_{F}|F|\frac{n}{n-1}\left(\frac{1}{|F|}+1\right)\leq
4K_{E}K_{F}\frac{n}{n-1}$

by (12) and (7) with noting $N=\lceil 1/|F|\rceil$. Hence Lemma holds for
$Q_{0}=Q_{0}(\omega_{0})$ and $h_{0}$ defined as above. ∎

####  5.1.1 Proof of Proposition 5.3

In this section, we denote $X=[0,1]^{2}$ for notational simplicity. Given
$\omega=(\omega_{1},\dots,\omega_{N})\in X^{N}$, denote

$A_{0}(\omega)=\bigcup_{k=1}^{N}\\{A+\omega_{k}\\}\quad\text{and}\quad
B_{0}(\omega)=\bigcup_{k=1}^{N}\\{B+\omega_{k}\\}.$

For each $\omega=(\omega_{1},\dots,\omega_{N})$, let $|A_{0}\setminus
B_{0}|(\omega)=|A_{0}(\omega)\setminus B_{0}(\omega)|$.

######  Lemma 5.5.

We have

$\mathcal{E}\left(|A_{0}\setminus
B_{0}|\right)\geq\left(1-c_{0}|A|\right)^{N}-\left(1-(1+c_{0}-\varepsilon)|A|\right)^{N}.$
(21)

###### Proof.

We see

$\displaystyle\mathcal{E}\left(|A_{0}\setminus B_{0}|\right)$
$\displaystyle=\int_{X^{N}}|A_{0}\setminus B_{0}|(\omega)\,d\omega$
$\displaystyle=\mathcal{E}\left(\int_{X}b_{N}(z,\omega)\,dm(z)\right)-\mathcal{E}\left(\int_{X}(a_{N}\cdot
b_{N})(z,\omega)\,dm(z)\right),$ (22)

where

$a_{N}(z,\omega)=\prod_{k=1}^{N}\left(1-\mathbbm{1}_{A}(z+\omega_{k})\right)\quad\text{and}\quad
b_{N}(z,\omega)=\prod_{k=1}^{N}\left(1-\mathbbm{1}_{B}(z+\omega_{k})\right).$

For the first term in the right-hand side of (5.1.1), by the Fubini theorem,

$\displaystyle\mathcal{E}\left(\int_{X}b_{N}(z,\omega)\,dm(z)\right)$
$\displaystyle=\mathcal{E}\left(\int_{X}\prod_{k=1}^{N}(1-\mathbbm{1}_{B}(z+\omega_{k}))\,dm(z)\right)$
$\displaystyle=\int_{X}\mathcal{E}\left(\prod_{k=1}^{N}(1-\mathbbm{1}_{B}(z+\omega_{k}))\right)\,dm(z).$

Since $|B|=c_{0}|A|$ by assumption, we obtain

$\mathcal{E}\left(\prod_{k=1}^{N}\left(1-\mathbbm{1}_{B}(z+\omega_{k})\right)\right)=(1-|B|)^{N}=(1-c_{0}|A|)^{N}$

for every $z\in X$, and thus

$\mathcal{E}\left(\int_{X}b_{N}(z,\omega)\,dm(z)\right)=(1-c_{0}|A|)^{N}.$
(23)

Next, for the second term in the right-hand side of (5.1.1), by the Fubini
theorem again, one has

$\mathcal{E}\left(\int_{X}(a_{N}\cdot
b_{N})(z,\omega)\,dm(z)\right)=\int_{X}\mathcal{E}(a_{N}\cdot
b_{N})(z)\,dm(z),$

where

$\displaystyle\mathcal{E}(a_{N}\cdot b_{N})(z)$
$\displaystyle=\mathcal{E}\left(\left(\prod_{k=1}^{N}\left(1-\mathbbm{1}_{A}(z+\omega_{k})\right)\right)\prod_{k=1}^{N}\left(1-\mathbbm{1}_{B}(z+\omega_{k})\right)\right)$
$\displaystyle=\mathcal{E}\left(\prod_{k=1}^{N}\left(1-\mathbbm{1}_{A}(z+\omega_{k})-\mathbbm{1}_{B}(z+\omega_{k})+\mathbbm{1}_{A\cap
B}(z+\omega_{k})\right)\right)$ $\displaystyle=\left(1-|A|-|B|+|A\cap
B|\right)^{N}$
$\displaystyle\leq\left(1-|A|-c_{0}|A|+\varepsilon|A|\right)^{N}$ (24)

for every $z\in[0,1]^{2}$. Hence we have

$\mathcal{E}\left(\int_{X}(a_{N}\cdot
b_{N})(z,\omega)\,dm(z)\right)\leq\left(1-(1+c_{0}-\varepsilon)|A|\right)^{N}$
(25)

Lemma 5.5 follows from (5.1.1), (23) and (25). ∎

###### Proof of Proposition 5.3.

Since $1/|A|\leq N<(1+|A|)/|A|$, it follows from Lemma 5.5 that

$\left(1-c_{0}|A|\right)^{N}\geq\left(1-c_{0}|A|\right)^{(1+|A|)/|A|}\geq\left(\kappa\cdot
e^{-1}\right)^{c_{0}}$

for some $\kappa=\kappa(c_{0}|A|)\in(0,1)$ such that $\kappa\to 1$ as
$c_{0}|A|\to 0$, and

$\left(1-(1+c_{0}-\varepsilon)|A|\right)^{N}\leq\left(1-(1+c_{0}-\varepsilon)|A|\right)^{1/|A|}\leq
e^{-(1+c_{0}-\varepsilon)}.$

Hence, by Lemma 5.5, we arrive at

$\mathcal{E}\left(|A_{0}\setminus B_{0}|\right)=\int_{X^{N}}|A_{0}\setminus
B_{0}|(\omega)\,d\omega\geq\frac{1}{(\kappa^{-1}e)^{c_{0}}}-\frac{1}{e^{(1+c_{0}-\varepsilon)}}.$
(26)

Since $\omega\mapsto|A_{0}\setminus B_{0}|(\omega)$ is a non-negative and
continuous function, it follows from (26) that there is an open set
$\Omega\subset X^{N}=X^{\lceil 1/|A|\rceil}$ with $|\Omega|>0$ such that for
any $\omega\in\Omega$ we have

$|A_{0}\setminus
B_{0}|(\omega)\geq\frac{99}{100}\left(\frac{1}{(\kappa^{-1}e)^{c_{0}}}-\frac{1}{e^{(1+c_{0}-\varepsilon)}}\right).$

Proposition is obtained. ∎

######  Remark 5.6.

Proposition 5.3 concerns the case where $c_{0}>0$ (or $c$) is rather large. If
$c>0$ is small, this is exactly the case of Lemma 5.2, then we have a better
upper bound

$\mathcal{E}(a_{N}\cdot b_{N})(z)\leq\left(1-|A|\right)^{N}$

in stead of (24), and thus

$\mathcal{E}\left(|A_{0}\setminus
B_{0}|\right)\geq\left(1-c|A|\right)^{N}-\left(1-|A|\right)^{N}$

in stead of (21).

###  5.2 Proof of Theorem 2.3

###### Proof.

Denote $X_{0}=[0,1]^{2}$, and let $h_{0}\in L^{1}(X_{0})$ and
$Q_{0}=F_{0}(\omega_{0})\setminus D_{0}(\omega_{0})\subset X_{0}$ be as in
Lemma 5.4. Hence $|Q_{0}|>\chi$ by Lemma 5.4. Let $h_{0}^{*}=|Q_{0}|h_{0}$.
Then $\|h_{0}^{*}\|_{L^{1}(m)}\leq(4+o(1))|Q_{0}|$. Since we fix such an
$\omega_{0}\in\Omega_{0}$, we will omit $\omega_{0}$ and write, say $D_{0}$
instead of $D_{0}(\omega_{0})$ for notational simplicity. By Lemma 4.5, one
has

$|D_{0}|\leq|D|\left(\frac{1}{|F|}+1\right)<\varepsilon_{0}=\frac{\varepsilon}{2^{3}}.$

Here and below, we let $\varepsilon_{k}=\varepsilon/2^{k+3}$ for
$k\in\mathbb{N}\cup\\{0\\}$.

Let

$Y_{1}=X_{0}\setminus\left(Q_{0}\sqcup
D_{0}\right)=X_{0}\setminus\left(F_{0}\cup D_{0}\right).$

Since the boundary of the set $Q_{0}\sqcup D_{0}$ consists of straight lines,
we can partition $Y_{1}$ into squares of small size. Precisely, there exist
$a_{1}=a_{1}(\varepsilon)\in(0,1)$ and a finite family of squares $S_{1,a}$
with side lengths $a\in[a_{1},1)$ such that

$\left|Y_{1}\setminus\bigsqcup_{a\geq a_{1}}S_{1,a}\right|<\varepsilon_{0}.$

(Note that the disjoint union includes squares of same side lengths.) Let

$X_{1}=\bigsqcup_{a\geq a_{1}}S_{1,a}.$

Inside each square $S_{1,a}$, we can find a function $h_{1,a}\in L^{1}(X_{0})$
and a subset $Q_{1,a}=F_{1,a}\setminus D_{1,a}\subset S_{1,a}$ for which all
the properties of Lemma 5.4 hold, and hence

$\left|\bigsqcup_{a\geq a_{1}}Q_{1,a}\right|>\chi\left|\bigsqcup_{a\geq
a_{1}}S_{1,a}\right|=\chi|X_{1}|.$

Define

$Q_{1}^{*}=\bigsqcup_{a\geq a_{1}}Q_{1,a}\quad\text{ and }\quad
D_{1}^{*}=\bigsqcup_{a\geq a_{1}}D_{1,a}.$

Then $Q_{1}^{*},D_{1}^{*}\subset X_{1}$, and one has

$|D_{1}^{*}|<\varepsilon_{1}.$

Notice that each support $\mathrm{supp}h_{1,a}$ consists of finitely many
horizontal line segments. Hence we may and do assume that the supports
$\mathrm{supp}h_{1,a}$ in different squares $S_{1,a}$ never lay on the same
horizontal line. We may do assume further that no two such line segments, one
from $\mathrm{supp}h_{0}^{*}$ and the other from any of
$\mathrm{supp}f_{1,a}$, lay on the same horizontal line. In short, one can
assume that any two horizontal line segments in
$\mathrm{supp}h_{0}^{*}\sqcup\\{\mathrm{supp}h_{1,a}\colon a\geq a_{1}\\}$
have a “vertical gap”. Let

$h_{1}^{*}=\sum_{a\geq a_{1}}|Q_{1,a}|h_{1,a}\in L^{1}(X_{0}).$

Then, by construction, we still have all the properties of Lemma 5.4 for
$Q_{0}\sqcup Q_{1}^{*}$ and $h_{0}^{*}+h_{1}^{*}$ by the same argument of the
proof for Lemma 5.4. Here each function $h_{1,a}$ is multiplied by the area
$|Q_{1,a}|$ just to have $\|h_{1}^{*}\|_{L^{1}(m)}\leq(4+o(1))|Q_{1}^{*}|$.
Hence it follows that
$\|h_{0}^{*}+h_{1}^{*}\|_{L^{1}(m)}\leq(4+o(1))(|Q_{0}|+|Q_{1}^{*}|)$.

Next, define

$Y_{2}=X_{1}\setminus\left(Q_{1}^{*}\sqcup D_{1}^{*}\right),$

and repeat the procedure described above. Namely, there exist
$a_{2}=a_{2}(\varepsilon)\in(0,a_{1})$ and a finite family of squares
$S_{2,a}$ with side lengths $a\in[a_{2},a_{1})$ with

$\left|Y_{2}\setminus\bigsqcup_{a\in[a_{2},a_{1})}S_{2,a}\right|<\varepsilon_{1}$

such that for each $S_{2,a}$, there are $h_{2,a}\in L^{1}(X_{0})$ and
$Q_{2,a}=F_{2,a}\setminus D_{2,a}\subset S_{2,a}$ for which all the properties
of Lemma 5.4 hold such that

$\left|\bigsqcup_{a\in[a_{2},a_{1})}Q_{2,a}\right|>\chi\left|\bigsqcup_{a\in[a_{2},a_{1})}S_{2,a}\right|\quad\text{
and }\quad\left|\bigsqcup_{a\in[a_{2},a_{1})}D_{2,a}\right|<\varepsilon_{2}.$

Define

$\displaystyle X_{2}=\bigsqcup_{a\in[a_{2},a_{1})}S_{2,a},\quad
Q_{2}^{*}=\bigsqcup_{a\in[a_{2},a_{1})}Q_{2,a},\quad
D_{2}^{*}=\bigsqcup_{a\in[a_{2},a_{1})}D_{2,a}.$

Note that one has $Q_{2}^{*},D_{2}^{*}\subset X_{2}$, and
$|D_{2}^{*}|<\varepsilon_{2}$. We let

$h_{2}^{*}=\sum_{a\in[a_{2},a_{1})}|Q_{2,a}|h_{2,a}\in L^{1}(X_{0}),$

where $\mathrm{supp}h_{2,a}$ are placed such that any two horizontal line
segments in
$\mathrm{supp}h_{0}^{*}\sqcup\mathrm{supp}h_{1}^{*}\sqcup\mathrm{supp}h_{2}^{*}$
have a vertical gap. Hence, all the properties of Lemma 5.4 still hold for
$Q_{0}\sqcup Q_{1}^{*}\sqcup Q_{2}^{*}$ and $h_{0}^{*}+h_{1}^{*}+h_{2}^{*}$.
Notice that
$\|h_{0}^{*}+h_{1}^{*}+h_{2}^{*}\|_{L^{1}(m)}\leq(4+o(1))(|Q_{0}|+|Q_{1}^{*}|+|Q_{2}^{*}|)$
holds.

Once $X_{k}$, and $Q_{k}^{*},D_{k}^{*}\subset X_{k}$ are defined, by letting

$Y_{k+1}=X_{k}\setminus\left(Q_{k}^{*}\sqcup D_{k}^{*}\right),$

one can define $X_{k+1}\subset Y_{k+1}$, and $Q_{k+1}^{*},D_{k+1}^{*}\subset
X_{k+1}$ with

$\left|Y_{k+1}\setminus
X_{k+1}\right|<\varepsilon_{k},\quad\left|Q_{k+1}^{*}\right|>\chi\left|X_{k+1}\right|,\quad\quad|D_{k+1}^{*}|<\varepsilon_{k+1}$

and a function $h_{k+1}^{*}\in L^{1}(X_{0})$, with
$\|h_{k+1}^{*}\|_{L^{1}(m)}\leq(4+o(1))|Q_{k+1}^{*}|$, having the vertical gap
property among
$\mathrm{supp}h_{0}^{*}\sqcup\mathrm{supp}h_{1}^{*}\sqcup\dots\sqcup\mathrm{supp}h_{k+1}^{*}$
such that all the properties of Lemma 5.4 hold for $Q_{0}\sqcup
Q_{1}^{*}\sqcup\dots\sqcup Q_{k+1}^{*}$ and
$h_{0}^{*}+h_{1}^{*}+\dots+h_{k+1}^{*}$.

Now, we show that the area of $\bigsqcup_{k=0}^{n-1}Q_{k}^{*}$, where
$Q_{0}^{*}=Q_{0}$, can be arbitrarily close to one.

######  Lemma 5.7.

One can take $n\in\mathbb{N}$ so large that

$\left|\bigsqcup_{k=0}^{n-1}Q_{k}^{*}\right|>1-\varepsilon.$

###### Proof.

One has

$|Q_{0}^{*}|=|Q_{0}|=|X_{0}|-|D_{0}|-|Y_{1}|>1-\varepsilon_{0}-(|X_{1}|+\varepsilon_{0})=1-|X_{1}|-2\varepsilon_{0},$

and

$\displaystyle|Q_{0}^{*}\sqcup Q_{1}^{*}|=|Q_{0}^{*}|+|Q_{1}^{*}|$
$\displaystyle>(1-|X_{1}|-2\varepsilon_{0})+(|X_{1}|-|D_{1}^{*}|-|Y_{2}|)$
$\displaystyle>1-2\varepsilon_{0}-\varepsilon_{1}-|Y_{2}|$
$\displaystyle>1-2\varepsilon_{0}-\varepsilon_{1}-(|X_{2}|+\varepsilon_{1})$
$\displaystyle>1-|X_{2}|-2\varepsilon_{0}-2\varepsilon_{1}.$

Hence by induction, one obtain

$\left|\bigsqcup_{k=0}^{n-1}Q_{k}^{*}\right|>1-|X_{n}|-2\sum_{k=0}^{n-1}\varepsilon_{k}=1-|X_{n}|-2\sum_{k=0}^{n-1}\frac{\varepsilon}{2^{k+3}}>1-|X_{n}|-\frac{\varepsilon}{2}.$

Next, one sees that $|X_{n}|$ strictly decreases to 0 as $n$ grows. Indeed,

$\displaystyle|X_{n+1}|\leq|Y_{n+1}|$
$\displaystyle=|X_{n}|-|Q_{n}^{*}|-|D_{n}^{*}|$
$\displaystyle<|X_{n}|-|Q_{n}^{*}|$ $\displaystyle<|X_{n}|-\chi|X_{n}|$
$\displaystyle=(1-\chi)|X_{n}|<\dots<(1-\chi)^{n+1}|X_{0}|=(1-\chi)^{n+1}.$

Take $n\in\mathbb{N}$ so large that $(1-\chi)^{n}<\varepsilon/2$. Then it
follows that

$\left|\bigsqcup_{k=0}^{n-1}Q_{k}^{*}\right|>1-|X_{n}|-\frac{\varepsilon}{2}>1-\varepsilon.$

Lemma is obtained. ∎

Take an $n\in\mathbb{N}$ so large as in Lemma 5.7. Letting
$Q=\bigsqcup_{k=0}^{n-1}Q_{k}^{*}$ and
$f=h_{0}^{*}+h_{1}^{*}+\dots+h_{n-1}^{*}$ yields the Theorem 2.3. ∎

##  6 Proof of Theorem 2.1

###  6.1 Setup

To prove Theorem 2.1, we use Theorem 2.3 recursively to find sequences of sets
$\\{Q_{n}\\}_{n\geq 1}$ and functions $\\{f_{n}\\}_{n\geq 1}$ as follows. Let
$\varepsilon_{1}=1/2$, $\delta_{1}=1/2$, and $a_{1}=1$. Then there are
$Q_{1}\subset[0,1]^{2}$ with $|Q_{1}|>1-\varepsilon_{1}$ and $f_{1}\in
L^{\infty}(m)$ with $\|f_{1}\|_{L^{1}(m)}\leq 2$ and
$\eta_{1}\in(0,\delta_{1})$ such that

  1. (i1)

If $z\in Q_{1}$ and $z\in A\in\mathcal{R}_{\mathcal{C}}$, then

$\left|\frac{1}{|A|}\int_{A}f_{1}\,dm\right|\leq 2.$

  2. (ii1)

If $z\in Q_{1}$, then there exists $B\in\mathcal{R}_{\mathcal{D}}$, with $z\in
B$ and $|B|>\eta_{1}$, so that

$\left|\frac{1}{|B|}\int_{B}f_{1}\,dm\right|>a_{1}(=1).$

  3. (iii1)

For every $R\in\mathcal{R}$ with $|R|>\delta_{1}$, we have

$\left|\frac{1}{|R|}\int_{R}f_{1}\,dm\right|<1.$

Suppose that $Q_{n-1}\subset[0,1]^{2}$, $f_{n-1}\in L^{\infty}(m)$ and
$\eta_{n-1}>0$ are defined such that the corresponding properties (in-1),
(iin-1), (iiin-1) hold. Let $\varepsilon_{n}=1/(n+1)^{2}$,
$\delta_{n}=\eta_{n-1}$, and

$a_{n}=\left(2\sup_{k\leq n-1}\|f_{k}\|_{L^{\infty}(m)}+n\right)n^{2}.$

Then there are $Q_{n}\subset[0,1]^{2}$ with $|Q_{n}|>1-\varepsilon_{n}$ and
$f_{n}\in L^{\infty}(m)$ with $\|f_{n}\|_{L^{1}(m)}\leq 2$ and
$\eta_{n}\in(0,\delta_{n})$ such that

  1. (in)

If $z\in Q_{n}$ and $z\in A\in\mathcal{R}_{\mathcal{C}}$, then

$\left|\frac{1}{|A|}\int_{A}f_{n}\,dm\right|\leq 2.$

  2. (iin)

If $z\in Q_{n}$, then there exists $B\in\mathcal{R}_{\mathcal{D}}$, with $z\in
B$ and $|B|>\eta_{n}$, so that

$\left|\frac{1}{|B|}\int_{B}f_{n}\,dm\right|>a_{n}.$

  3. (iiin)

For every $R\in\mathcal{R}$ with $|R|>\delta_{n}$, we have

$\left|\frac{1}{|R|}\int_{R}f_{n}\,dm\right|<1.$

Notice that $\delta_{n}>\eta_{n}=\delta_{n+1}>\eta_{n+1}=\delta_{n+2}$ by
construction, and thus both $\eta_{n}$ and $\delta_{n}$ are decreasing
sequences in particular.

###  6.2 Proof of Theorem 2.1

###### Proof of Theorem 2.1.

Define

$f=\sum_{n=1}^{\infty}\frac{1}{n^{2}}f_{n}.$

Note that by property (2) of Theorem 2.3, we have

$\left\|\sum_{n=1}^{\infty}\frac{1}{n^{2}}f_{n}\right\|_{L^{1}(m)}\leq\sum_{n=1}^{\infty}\frac{1}{n^{2}}\|f_{n}\|_{L^{1}(m)}<\infty.$

Hence $f$ is well defined and $f\in L^{1}(m)$. By the Borel-Cantelli lemma, we
also have that

$m\left(\liminf_{n\to\infty}Q_{n}\right)=1.$

Since $f_{k}\in L^{\infty}(m)$, then by Theorem 1.2 we have that for all
$k\in\mathbb{N}$, there is $\Gamma_{k}\subset[0,1]^{2}$ with $m(\Gamma_{k})=1$
such that

$\lim_{\begin{subarray}{c}\mathrm{diam}R\to 0,\\\ z\in
R\in\mathcal{R}\end{subarray}}\frac{1}{|R|}\int_{R}f_{k}\,dm=f_{k}(z)$ (27)

for every $z\in\Gamma_{k}$. It follows from (27) and Theorem 2.3-(i) that for
every $z\in Q_{k}\cap\Gamma_{k}$ we have

$|f_{k}(z)|\leq 2.$ (28)

Denote $\Gamma_{\infty}=\bigcap_{k=1}^{\infty}\Gamma_{k}$ and

$\Lambda=\left(\liminf_{n\to\infty}Q_{n}\right)\cap\Gamma_{\infty}.$

Clearly $m(\Lambda)=1$.

First, we prove convergence, namely
$\delta_{\mathcal{R}_{\mathcal{C}}}(z,f)=0$ for $m$-almost every
$z\in[0,1]^{2}$. Let $z\in\Lambda$. Hence there exists $M=M_{z}\in\mathbb{N}$
so that $z\in Q_{n}$ for all $n\geq M$. Then for every
$A\in\mathcal{R}_{\mathcal{C}}$ with $A\ni z$ we have

$\displaystyle\left|\frac{1}{|A|}\int_{A}f\,dm-f(z)\right|$
$\displaystyle=\left|\sum_{n=1}^{\infty}\frac{1}{n^{2}}\frac{1}{|A|}\int_{A}f_{n}\,dm-\sum_{n=1}^{\infty}\frac{f_{n}(z)}{n^{2}}\right|$
$\displaystyle\leq\left|\sum_{n=1}^{M}\frac{1}{n^{2}}\frac{1}{|A|}\int_{A}f_{n}\,dm-\sum_{n=1}^{M}\frac{f_{n}(z)}{n^{2}}\right|$
(29)
$\displaystyle\quad+\sum_{n=M+1}^{\infty}\frac{1}{n^{2}}\left|\frac{1}{|A|}\int_{A}f_{n}\,dm\right|+\sum_{n=M+1}^{\infty}\frac{|f_{n}(z)|}{n^{2}}.$
(30)

Since $z\in Q_{n}\cap\Gamma_{\infty}$ for all $n\geq M$, it follows from (28)
that $|f_{n}(z)|\leq 2$ for all $n\geq M$. Note also that

$\left|\frac{1}{|A|}\int_{A}f_{n}\,dm\right|\leq 2$

by the property (in). Thus for the last two terms (30), we have

$\sum_{n=M+1}^{\infty}\frac{1}{n^{2}}\left|\frac{1}{|A|}\int_{A}f_{n}\,dm\right|+\sum_{n=M+1}^{\infty}\frac{|f_{n}(z)|}{n^{2}}\leq\sum_{n=M}^{\infty}\frac{4}{n^{2}}<\frac{4}{M-1}.$

For the first term (29), since $x\in\Gamma_{\infty}$, then we have by (27)
that

$\lim_{\begin{subarray}{c}\mathrm{diam}A\to 0,\\\ z\in
A\in\mathcal{R}_{\mathcal{C}}\end{subarray}}\left|\sum_{n=1}^{M}\frac{1}{n^{2}}\frac{1}{|A|}\int_{A}f_{n}\,dm-\sum_{n=1}^{M}\frac{f_{n}(z)}{n^{2}}\right|=0.$

Consequently, it follows that

$\lim_{\begin{subarray}{c}\mathrm{diam}A\to 0,\\\ z\in
A\in\mathcal{R}_{\mathcal{C}}\end{subarray}}\left|\sum_{n=1}^{\infty}\frac{1}{n^{2}}\frac{1}{|A|}\int_{A}f_{n}\,dm-\sum_{n=1}^{\infty}\frac{f_{n}(z)}{n^{2}}\right|=0.$

In other words $\delta_{\mathcal{R}_{\mathcal{C}}}(z,f)=0$ for $m$-almost
every $z\in[0,1]^{2}$.

We now prove divergence. Let $z\in\Lambda$. Hence there exists
$M_{z}\in\mathbb{N}$ so that $z\in Q_{n}$ for all $n\geq M$. Take $N>M_{z}$.
Then by the property (iiN), there exists $B\in\mathcal{R}_{\mathcal{D}}$, with
$z\in B$ and $|B|>\eta_{N}$ such that

$\left|\frac{1}{|B|}\int_{B}f_{N}\,dm\right|>a_{N}.$

Hence for such a rectangle $B\in\mathcal{R}_{\mathcal{D}}$, we have

$\displaystyle\left|\frac{1}{|B|}\int_{B}f\,dm\right|$
$\displaystyle=\left|\frac{1}{|B|}\int_{B}\sum_{n=1}^{\infty}\frac{1}{n^{2}}f_{n}\,dm\right|$
$\displaystyle>\left|\frac{1}{|B|}\int_{B}\frac{1}{N^{2}}f_{N}\,dm\right|-\left|\sum_{n\neq
N}^{\infty}\frac{1}{n^{2}}\frac{1}{|B|}\int_{B}f_{n}\,dm\right|$
$\displaystyle>\frac{a_{N}}{N^{2}}-\left|\sum_{n\neq
N}^{\infty}\frac{1}{n^{2}}\frac{1}{|B|}\int_{B}f_{n}\,dm\right|.$

Here we have

$\displaystyle\sum_{n=1}^{N-1}\frac{1}{n^{2}}\frac{1}{|B|}\int_{B}|f_{n}|\,dm\leq\left(\sup_{n\leq
N-1}\|f_{n}\|_{L^{\infty}(m)}\right)\cdot\sum_{n=1}^{N-1}\frac{1}{n^{2}}\leq
2\sup_{n\leq N-1}\|f_{n}\|_{L^{\infty}(m)}.$

Notice that we have $|B|>\delta_{n}$ for all $n>N$ since
$|B|>\eta_{N}=\delta_{N+1}$ and $\delta_{k}$ is a decreasing sequence in $k$.
Thus by the property (iiin) with $n>N$, we have

$\left|\frac{1}{|B|}\int_{B}f_{n}\,dm\right|\leq 1$

for all $n>N$, and hence

$\left|\sum_{n>N}^{\infty}\frac{1}{n^{2}}\frac{1}{|B|}\int_{B}f_{n}\,dm\right|\leq\sum_{n=N+1}^{\infty}\frac{1}{n^{2}}<\frac{1}{N}.$

It follows that

$\left|\frac{1}{|B|}\int_{B}f\,dm\right|>\frac{a_{N}}{N^{2}}-2\sup_{n\leq
N-1}\|f_{n}\|_{L^{\infty}(m)}-\frac{1}{N}=N-\frac{1}{N}.$

Consequently, we have

$\left|\frac{1}{|B|}\int_{B}f\,dm-f(z)\right|\geq\left|\frac{1}{|B|}\int_{B}f\,dm\right|-|f(z)|>N-\frac{1}{N}-|f(z)|.$

Letting $N\to\infty$, the divergence property
$\delta_{\mathcal{R}_{\mathcal{D}}}(z,f)=\infty$ holds for $m$-almost every
$z\in[0,1]^{2}$. ∎

##  7 Proof of Theorem 2.2

###### Proof.

By assumption, one can take $c\in(0,1)$ so small that

$1\geq\max\left\\{\frac{\overline{a}}{a},\frac{a}{\underline{a}}\right\\}>c>0$

holds for every sufficiently small $a\in\mathcal{D}$. Suppose
$\delta_{\mathcal{R}_{\mathcal{C}}}(z,f)=0$ for $m$-almost every
$z\in\mathbb{R}^{2}$. For a given $\delta>0$ consider the set

$E_{\delta}=\left\\{z\in\mathbb{R}^{2}\colon\sup_{\begin{subarray}{c}{\rm
diam}A<\delta,\\\ z\in
A\in\mathcal{R}_{\mathcal{C}}\end{subarray}}\left|\frac{1}{|A|}\int_{A}f\,dm-f(z)\right|<1\right\\}.$

We have $|E_{\delta}|>0$ for $\delta>0$ small enough. Then for the
characteristic function $\mathbbm{1}_{E_{\delta}}$ of $E_{\delta}$, we will
have by Theorem 1.2, that almost all $z\in E_{\delta}$ are Lebesgue, namely
for almost every $z\in E_{\delta}$ we have

$\lim_{\begin{subarray}{c}{\rm diam}R\rightarrow 0,\\\ z\in
R\in\mathcal{R}\end{subarray}}\frac{|R\cap E_{\delta}|}{|R|}=1.$

Take $B\in\mathcal{R}_{\mathcal{D}}$ with $z\in B$ and ${\rm diam}B$ so small
such that

$\frac{|B\cap E_{\delta}|}{|B|}>1-c^{2}.$ (31)

We now wish to cover the rectangle $B$ with rectangles of sides
$x,y\in\mathcal{C}$. More specifically, we have the following lemma.

######  Lemma 7.1.

There is a collection $\\{A_{k}\in\mathcal{R}_{\mathcal{C}}\\}_{q=1}^{n}$ for
some $n\leq\left(\frac{1}{c}+1\right)^{2}$ such that

  1. (i)

$B\subset\bigcup_{q=1}^{n}A_{q}$,

  2. (ii)

$|B\cap A_{q}|\geq c^{2}|B|$ for every $q\in\\{1,\dots,n\\}$,

  3. (iii)

$\dfrac{|A_{q}|}{|B|}\leq\dfrac{1}{c^{2}}$ for every $q\in\\{1,\dots,n\\}$.

###### Proof.

Let $B=[s,s+a]\times[t,t+b]$ for some small $a,b\in\mathcal{D}$ and some
$(s,t)\in\mathbb{R}^{2}$. Then by assumption, there exist $x,y\in\mathcal{C}$
such that

$(1a)\quad 1\geq\frac{x}{a}>c\quad\text{or}\quad(2a)\quad 1\geq\frac{a}{x}>c,$

and

$(1b)\quad 1\geq\frac{y}{b}>c\quad\text{or}\quad(2b)\quad 1\geq\frac{b}{y}>c.$

Hence, consider the following four cases. For the case where $x\in\mathcal{C}$
satisfies $(2a)$ and $y\in\mathcal{C}$ does $(2b)$, by taking

$A_{1}=[s,s+x]\times[t,t+y],$

we have $|B\cap A_{1}|=|B|$ and

$\frac{|A_{1}|}{|B|}=\frac{xy}{ab}<\frac{1}{c^{2}}.$

For the case where $x\in\mathcal{C}$ satisfies $(2a)$ and $y\in\mathcal{C}$
does $(1b)$, we consider

$A_{\ell}=\left[s,s+x\right]\times\left[t+(\ell-1)y,t+\ell y\right]$

for $\ell\in\\{1,\dots,\lfloor b/y\rfloor\\}$, and also

$A_{\lfloor b/y\rfloor+1}=\left[s,s+x\right]\times\left[t+b-y,t+b\right]$

to cover up. Then we have

$\frac{|B\cap A_{\ell}|}{|B|}=\frac{ay}{ab}>c,$

and

$\frac{|A_{\ell}|}{|B|}=\frac{xy}{ab}\leq\frac{1}{c}$

for every $\ell\in\\{1,\dots,\lfloor b/y\rfloor+1\\}$. The case where
$x\in\mathcal{C}$ satisfies $(1a)$ and $y\in\mathcal{C}$ does $(2b)$ is
similar, hence omit this case.

For the case where $x\in\mathcal{C}$ satisfies $(1a)$ and $y\in\mathcal{C}$
does $(1b)$, we define

$A_{k,\ell}=\left[s+(k-1)x,s+kx\right]\times\left[t+(\ell-1)y,t+\ell y\right]$

for $k\in\\{1,\dots,\lfloor a/x\rfloor\\}$ and $\ell\in\\{1,\dots,\lfloor
b/y\rfloor\\}$. Define also

$A_{\lfloor
a/x\rfloor+1,\ell}=\left[s+a-x,s+a\right]\times\left[t+(\ell-1)y,t+\ell
y\right]$

for $\ell\in\\{1,\dots,\lfloor b/y\rfloor\\}$,

$A_{k,\lfloor
b/y\rfloor+1}=\left[s+(k-1)x,s+kx\right]\times\left[t+b-y,t+b\right]$

for $k\in\\{1,\dots,\lfloor a/x\rfloor\\}$, and

$A_{\lfloor a/x\rfloor+1,\lfloor
b/y\rfloor+1}=\left[s+a-x,s+a\right]\times\left[t+b-y,t+b\right]$

Denote the whole $\\{A_{k.\ell}\\}$ by $\\{A_{q}\\}_{q=1}^{n}$, where we have

$n\leq\left(\left\lfloor\frac{a}{x}\right\rfloor+1\right)\left(\left\lfloor\frac{b}{y}\right\rfloor+1\right)\leq\left(\frac{1}{c}+1\right)^{2}.$

We also have

$\frac{|B\cap A_{q}|}{|B|}=\frac{xy}{ab}>c^{2},$

and

$\frac{|A_{q}|}{|B|}=\frac{xy}{ab}\leq 1$

for every $q\in\\{1,\dots,n\\}$.

In all cases above, we have a collection
$\\{A_{q}\in\mathcal{R}_{\mathcal{C}}\\}_{q=1}^{n}$ such that

$B\subset\bigcup_{q=1}^{n}A_{q}.$

Lemma is obtained. ∎

It follows from (31) and Lemma 7.1-(iii) that $E_{\delta}\cap
A_{q}\neq\emptyset$ for all $q\in\\{1,\dots,n\\}$. Hence

$\left|\frac{1}{|A_{q}|}\int_{A_{q}}f\,dm\right|\leq 1+f(z)$

for all $q\in\\{1,\dots,n\\}$. Thus we have by Lemma 7.1 that

$\displaystyle\int_{B}f\,dm\leq\sum_{q=1}^{n}\int_{A_{q}}f\,dm\leq\left(1+f(z)\right)\sum_{q=1}^{n}|A_{q}|\leq\left(1+f(z)\right)\left(\frac{1}{c}+1\right)^{2}\frac{1}{c^{2}}|B|,$

and which implies

$\frac{1}{|B|}\int_{B}f\,dm\leq\left(\frac{1}{c}+1\right)^{2}\frac{1}{c^{2}}\left(1+f(z)\right).$

Consequently, we have

$\limsup_{\begin{subarray}{c}{\rm diam}B\rightarrow 0,\\\ z\in
B\in\mathcal{R}_{\mathcal{D}}\end{subarray}}\frac{1}{|B|}\int_{B}f\,dm<\infty.$

Thus, for almost every Lebesgue points $z\in E_{\delta}$ we have that the
upper differential is bounded. Then, due to a theorem of Besicovitch
[Besicovitch1935], it follows that $\delta_{\mathcal{R}_{\mathcal{D}}}(z,f)=0$
for $m$-almost every $z\in\mathbb{R}^{2}$. ∎

## Acknowledgments

The first author was supported by Japan Society for the Promotion of Science
(JSPS) KAKENHI Grant Number 19K03558. The second author is supported by the
Knut and Alice Wallenberg foundation (KAW).

## References

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