# Arbitrary finite intersections of doubling measures and applications

Theresa C. Anderson  (Anderson) Department of Mathematics, Carnegie Mellon
University  ,  Elisa Bellah  (Bellah) Department of Mathematics, Carnegie
Mellon University  ,  Zoe Markman  (Markman) Department of Mathematics,
Swarthmore College  ,  Teresa Pollard  (Pollard) Department of Mathematics,
New York University  and  Josh Zeitlin  (Zeitlin) Department of Mathematics,
Yale University

###### Abstract.

Using a wide array of machinery from diverse fields across mathematics, we
provide a construction of a measure on the real line which is doubling on all
$n$-adic intervals for any finite list of $n\in\mathbb{N}$, yet not doubling
overall. In particular, we extend previous results in the area, where only two
coprime numbers $n$ were allowed, by using substantially new ideas. In
addition, we provide several nontrivial applications to reverse Hölder
weights, $A_{p}$ weights, Hardy spaces, BMO and VMO function classes, and
connect our results with key principles and conjectures across number theory.

##  1\. Introduction

A longstanding folkloric question in harmonic analysis is to construct a
measure $\mu$ on the real line that is $n$-adic doubling for every $n$, yet
not doubling overall (see Section 2 for definitions). Analogous, and similarly
longstanding, analogues of this question exist for other ubiquitous objects in
analysis, such as the reverse Hölder weight classes, Muckenhoupt $A_{p}$
weight classes, bounded mean oscillation (BMO) function classes, Hardy spaces
($H^{1}$) and the vanishing mean oscillation (VMO) function class. These
classes have been extensively studied and we refer the reader to [8], [7],
[12], [11], [15], and [18], among others, for the requisite background.

Outside of unpublished lecture notes of Peter Jones from 1978 showing that
$BMO_{2}\cap BMO_{3}\neq BMO$, the authors are aware of little progress toward
these questions until 2019. (The only reference to these notes that we are
aware of in the literature come from a paper of Krantz [14]; a sketch of
Jones’s results is given, but details are sparse.) Then in 2019, Boylan, Mills
and Ward explicitly constructed a measure on $[0,1]$ that was dyadic and
triadic doubling, yet not doubling overall. This involved a careful re-
weighting procedure and technical calculations; moreover the result relied on
number theory that did not generalize beyond the primes $2$ and $3$
(corresponding to the _bases_ $2$ and $3$ for dyadic and triadic intervals).
In 2022, Anderson and Hu were able to extend the Boylan-Mills-Ward
construction to any distinct prime bases $p$ and $q$, which required several
new ideas. Notably, the number theory behind the bases, though elementary, was
significantly more intricate, along with the geometry behind selecting
intervals on which to re-weight the measure. Additionally, a new procedure to
avoid the specific calculations of [6] was developed. Anderson, Travesset and
Veltri extended this construction to coprime bases, which only involved
changing the number theoretic part of the construction [4]. However, there was
still a significant gap in resolving the folkloric question about measures.
Here, we tighten this gap as well as provide a wide-range of new applications
for all the weight and function classes mentioned above. Namely, we are able
to extend the results of [2] to any finite set of _arbitrary_ bases (which
includes any two bases, greatly generalizing [4]). Moreover, our proofs use a
novel interplay of number theory, topology, group theory, measure theory,
geometry, harmonic and functional analysis. Finally we make precise an analogy
to the _Hasse principle_ from arithmetic geometry in this harmonic analytic
setting.

Our first contribution is to extend the Anderson-Hu construction to any two
bases, which relies in part on lemmas from [2] that are generalized herein,
rather than the more technical machinery from [4]. However, the main
contribution is not from the more general elementary number theory developed
here, but on the interval selection used to re-weight the measure. A key part
of [2] involved showing that specific distinguished points $Y(J)$ on $p$-adic
intervals $J$ and $Z(I)$ on $q$-adic intervals $I$ were extremely close. That
is, for any $\varepsilon>0$ we have

(1.1) $|Y(J)-Z(I)|<\varepsilon|I|.$

While this inequality can be shown to hold for coprime bases, it does not
extend to the general two bases case; our insight is to replace $Y(J)$ with a
revolving distinguished point $\zeta$ that may change from interval to
interval, yet still satisfies the closeness relationship. Moreover, our new
interval selection and re-weighting procedure automatically generalizes to a
finite list of multiples of the larger base. This variant shows that there is
a dyadic, triadic and $6$-adic doubling measure that is not doubling, as an
example, complimenting results in [3], [15], [17], [20] [21]. In attempting to
extend this construction to three arbitrary bases we discovered an observation
connected with _far numbers_ that illustrates the limit of this construction
and justifies the need for a new one. A detailed number theoretic explanation
that our construction would be difficult to generalize farther is given in the
appendix.

This leads to our second main contribution – constructing a measure that is
$n$-adic doubling for any finite list of $n$, yet not overall. This result
generalizes all previous described work, while using a significantly different
argument.

######  Theorem 1.1.

There exists an infinite family of measures that are simultaneously
$n_{i}$-adic doubling for any finite list of positive integers
$n_{1},\dots,n_{k}$, but not doubling overall.

The new process involves firstly using the whole real line, instead of just
$[0,1]$ to select distinguished dyadic intervals on which to reweight our
measure (though later on we are able to compactify our results). This process
stems from relaxing the restriction (1.1) by adding absolute values. Namely,
let $\alpha\in\mathbb{N}$ and $\varepsilon<2^{-2\alpha}$. We show that inside
$[\alpha,\alpha+1]$ there exists some arbitrarily large $x\in\mathbb{N}$ such
that

$\left|\frac{1}{2^{x}}-\frac{1}{n_{i}^{[x\log_{n_{i}}2]}}\right|<\frac{\varepsilon}{2^{x}}$

for all $n_{i}$ in our finite list, where the brackets denotes the closest
integer function. In this new construction we are choosing the sizes of each
interval simultaneously, as opposed to [2] where they are chosen
independently. The proof of the existence of such an $x$ involves an intricate
interplay of topology, number theory and geometry, and has deep connections
with Schanuel’s conjecture (an outstanding open problem in transcendental
number theory [5]). Finally, we make partial progress toward the infinite base
case and show the significant obstacles that would need to be overcome for
this case to be true.

Both of these constructions lead to our third contribution of applications to
weight and function classes. Notably, we are able to weave previous progress
initiated in [2] with both functional analysis and our new constructions to
show that any finite intersection of corresponding weight and function classes
is never the whole class. We show

######  Theorem 1.2.

Let $S\subset\mathbb{N}$ be finite. Then we have:

  1. (1)

$\bigcap_{n\in S}RH_{r}^{n}\neq RH_{r}$

  2. (2)

$\bigcap_{n\in S}A_{p}^{n}\neq A_{p}$

  3. (3)

$\bigcap_{n\in S}BMO_{n}\neq BMO$

  4. (4)

$\sum_{n\in S}H^{1}_{n}\neq H^{1}$

  5. (5)

$\bigcap_{n\in S}VMO_{n}\neq VMO$

Precise definitions of all classes are given in Section 4. All these results
hold over both $\mathbb{R}$ and $\mathbb{T}=[0,1]$. In particular, these
results hold for sums and intersections involving two arbitrary natural
numbers.

These results are different from, and complimentary to, the array of results
in [15], where the authors show that the intersection (or sum) of two
generalized dyadic function classes is indeed the full function class (see
also [1] for discussion and extension of their results). Thus, our results
illustrate the subtle difference among generalized $n$-adic grids for a single
$n$ and intersections of standard $n$-adic grids. Finally, the proofs revolve
around a construction stemming from our measure, which is modified and
interwoven with key relationships between the weight and function classes,
both specifically and in a general functional analysis sense.

We write toward a broad audience and do not assume any deep familiarity with
the techniques in [2], but rather introduce tools and concepts as we go.
Several times we simultaneously give details of intermediate steps in [2]
while generalizing them, while other times we opt for a non-technical
description of steps which are essentially identical here. Armed with this big
picture, the interested reader can then refer to [2] for more details. We
include several remarks to put the wide array of results and techniques in
context, as well as to provide intuition.

This paper is organized as follows. In Section 2 we simultaneously generalize
and streamline the argument in [2] as to apply to a wider range of cases,
including the general two bases case. The next section, Section 3, begins with
a connection to _far numbers_ , indicating that our general construction in
the previous section does not generalize. From there we outline our new
construction to get Theorem 1.1. Finally, we outline partial progress toward
the infinite base case and the obstacles to overcome; as well as briefly
discuss the relationship with this line of work and the _Hasse principle_. The
final section, Section 4, contains all the applications to weight and function
classes, proving Theorem 1.2.

###  1.1. Acknowledgements

This research, and authors Anderson, Markman, Pollard and Zeitlin, were
partially supported by NSF DMS-2231990 and NSF CAREER DMS-2237937 (T.C.
Anderson). In addition, Markman, Pollard and Zeitlin benefited from funding
from Carnegie Mellon’s SUAMI program. The authors also thank David Cruz-Uribe
for providing several references.

##  2\. Extension of the Anderson-Hu Construction

Our main goal in this section will be to extend the Anderson-Hu construction
to the general two bases case. We first recall several definitions and set
some notation.

######  Definition 2.1.

A _doubling measure_ $\mu$ is a measure for which there exists a positive
absolute constant $C$ such that for every interval $I\subset\mathbb{R}$ we
have

$\mu(2I)\leq C\mu(I),$

where $2I$ is the interval which shares the same midpoint of $I$ and twice the
length of $I$. Equivalently, $\mu$ is doubling if and only if there is a
positive absolute constant $C$ so that

$\frac{1}{C}\leq\frac{\mu(I_{1})}{\mu(I_{2})}\leq C$

for any two adjacent intervals $I_{1}$ and $I_{2}$ of equal length.

######  Definition 2.2.

For a positive integer $n$, the _standard $n$-adic system_ $\mathcal{D}(n)$ is
the collection of $n$-adic intervals in $\mathbb{R}$ of the form

(2.1) $I=\left[\frac{k-1}{n^{m}},\frac{k}{n^{m}}\right),\quad
m,k\in\mathbb{Z}.$

For an $n$-adic interval $I$ defined as in (2.1), the _$n$ -adic children_ of
$I$ are given by

$I_{j}=\left[\frac{k-1}{n^{m}}+\frac{j-1}{n^{m+1}},\frac{k-1}{n^{m}}+\frac{j}{n^{m+1}}\right),\quad
1\leq j\leq n.$

We say that $n$-adic intervals $J_{1}$ and $J_{2}$ are _$n$ -adic siblings_ if
they are the children of some common $n$-adic interval $I$. Note that an
arbitrary pair of $n$-adic intervals are not necessarily $n$-adic siblings.

######  Definition 2.3.

A measure $\mu$ is an _n-adic doubling measure_ if there exists a positive
absolute constant $C$ so that

$\frac{1}{C}\leq\frac{\mu(J_{1})}{\mu(J_{2})}\leq C$

for any two $n$-adic siblings $J_{1}$ and $J_{2}$.

It should be noted that any doubling measure is automatically $n$-adic
doubling for every $n\in\mathbb{N}$. Furthermore, it can be quickly checked
that a measure is $n$-adically doubling if and only if it is $n^{k}$-adically
doubling for every positive integer $k$. Further details on doubling and
related concepts can be found in [2].

Finally, we set some notation. Given an $n$-adic interval $I$ as in (2.1), we
write $Y(I)$ be the right endpoint of $I_{1}$, that is

(2.2) $Y(I)=\frac{k-1}{n^{m}}+\frac{1}{n^{m+1}},$

and $Z(I)$ be the left endpoint of $I_{n}$, that is

(2.3) $Z(I)=\frac{k-1}{n^{m}}+\frac{n-1}{n^{m+1}}.$

as illustrated in Figure 1.

$\frac{k-1}{n^{m}}$$\frac{k}{n^{m}}$$I_{1}$$I_{2}$$\cdots$$I_{n}$$Y(I)$$Z(I)$
Figure 1. Illustration of distinguished points $Y(I)$ and $Z(I)$

Our main result of this section extends the Anderson-Hu construction to show
the following.

######  Theorem 2.4.

There exists an infinite family of measures that are both $u$-adic and
$v$-adic doubling for any pair of positive integers $u$ and $v$, but not
doubling overall.

Our key insight in proving Theorem 2 is a new interval selection procedure to
that in [2], where the intervals that are selected will be given a modified
measure, with all other intervals given Lebesgue measure. The modified measure
is an iterative reweighting, which is unchanged from the construction of
Anderson-Hu (having its genesis in [6]). Therefore, before describing our new
contributions, we begin with a description of the re-weighting procedure from
[2], in nontechnical terms, so that the reader can build intuition without
having to read [2] and also convince themselves that this procedure will give
the desired results on our selected intervals.

###  2.1. Description of the Anderson-Hu construction

In [2], the authors construct a measure which is $p$-adic and $q$-adic
doubling for distinct primes $p$ and $q$ but not doubling overall. To do this,
they select an infinite family of $q$-adic intervals $I^{\alpha}$ indexed by
parameters $\alpha\in\mathbb{N}$. Then they re-weight the Lebesgue measure on
the $q$-adic children of $I^{\alpha}$ to the right and left of the
distinguished point $Z(I^{\alpha})$, as defined in (2.3).

We give a description of the reweighting procedure here. Choose any $a<1$ and
$b>1$ satisfying $a(q-1)+b=q$. We reweight each interval $I^{\alpha}$
iteratively in $2\alpha$ steps as follows. For convenience, set
$I=I^{\alpha}$. First, redistribute the Lebesgue measure on the $q$ children
of $I$ by setting

$\mu(I_{q})=b|I_{q}|=\frac{b|I|}{q}\text{ and
}\mu(I_{j})=a|I_{j}|=\frac{a|I|}{q}$

for $j=1,\dots,q-1$. Next, set $H^{(1)}:=I_{q-1}$ and $G^{(1)}:=I_{q}$ and
redistribute the measure to the children of $H^{(1)}$ and $G^{(1)}$ as in the
previous step. That is, set

$\mu(H^{(1)}_{q})=\frac{b\mu(H^{(1)})}{q}\text{ and
}\mu(H^{(1)}_{j})=\frac{a\mu(H^{(1)})}{q}$

and similarly

$\mu(G^{(1)}_{q})=\frac{b\mu(G^{(1)})}{q}\text{ and
}\mu(G^{(1)}_{j})=\frac{a\mu(G^{(1)})}{q}$

for $j=1,\dots,q-1$. Next, set $H^{(2)}:=H^{(1)}_{q}$ and
$G^{(2)}:=G^{(1)}_{1}$ and repeat the reweighting procedure on $H^{(2)}$ and
$G^{(2)}$ from the previous step. We then iterate in this way for $\alpha$
steps. That is, we set

$\mu(H^{(k)}_{q})=\frac{b\mu(H^{(k)})}{q}\text{ and
}\mu(H^{(k)}_{j})=\frac{a\mu(H^{(k)})}{q}$
$\mu(G^{(k)}_{q})=\frac{b\mu(G^{(k)})}{q}\text{ and
}\mu(G^{(1)}_{j})=\frac{a\mu(G^{(k)})}{q}$

for $j=1,\dots,q-1$ and then define $H^{(k+1)}:=H^{(k)}_{q}$ and
$G^{(k+1)}:=G^{(k)}_{1}$ for $k=1,\dots,\alpha$ as illustrated in the figure
below

$\textZeta$$H^{(2\alpha)}$$E^{(2\alpha-1)}$$E^{(2\alpha-2)}$$E^{(2\alpha-3)}$$1$$q-1$$q(q-1)$$q^{2}(q-1)$$a^{\alpha}b^{\alpha}$$a^{\alpha+1}b^{\alpha-1}$$a^{\alpha+1}b^{\alpha-2}$$a^{\alpha+1}b^{\alpha-3}$$\textZeta$$H^{(\alpha)}$$E^{(\alpha)}$$E^{(\alpha-1)}$$a^{\alpha+1}$$q^{\alpha}$$q^{\alpha}(q-2)$$q^{\alpha}(q-1)$$q^{\alpha-1}(q-1)$$ba^{\alpha-1}$$a^{\alpha}$$q^{\alpha+1}$$q^{\alpha+1}(q-2)$$q^{\alpha+1}(q-1)$$E^{(\alpha-2)}$$ba^{\alpha-2}$$a^{\alpha-1}$
Figure 2. Illustrated is $\mu$ on the left hand side of $\textZeta$.
Weightings are listed in red and lengths in blue.

Observe that the first $\alpha$ steps of this procedure gives

$\mu(H^{(\alpha)})=\frac{a^{\alpha}|I|}{q^{\alpha}}\text{ and
}\mu(G^{(\alpha)})=\frac{b^{\alpha}|I|}{q^{\alpha}},$

as calculated in [2]. Note further that the intervals $H^{(\alpha)}$ and
$G^{(\alpha)}$ are adjacent and of equal length. Hence, if we perform this
process on infinitely many intervals $I^{\alpha}$ indexed by infinitely many
positive integers $\alpha$, we see that there is no universal constant
bounding the ratio of $\mu(G^{(\alpha)})$ and $\mu(H^{(\alpha)})$ for all
$\alpha$, thus breaking the doubling property as defined in 2.1. However, re-
weighting the $q$-adic intervals in this way preserves $q$-adic doubling as
the ratio of the measures of any $q$-adic siblings is either $1,$
$\frac{a}{b}$ or $\frac{b}{a}$. Note here that $G^{(\alpha)}$ and
$H^{(\alpha)}$ are not $q$-adic siblings, but they are adjacent $q$-adic
intervals of equal length.

After $\alpha$ steps, we now have a measure which is $q$-adic doubling, but
not doubling overall. To allow for a measure which is simultaneously $p$-adic
doubling, we reverse the above procedure on each interval $I^{\alpha}$ for an
additional $\alpha$ steps. That is, for $k=\alpha+1,\dots,2\alpha$ we set

$\mu(H^{(k)}_{q})=\frac{a\mu(H^{(k)})}{q}\text{ and
}\mu(H^{(k)}_{j})=\frac{b\mu(H^{(k)})}{q}$
$\mu(G^{(k)}_{q})=\frac{b\mu(G^{(k-1)})}{q}\text{ and
}\mu(G^{(k)}_{j})=\frac{a\mu(G^{(k)})}{q}.$

Observe that these additional $\alpha$ steps do not have any effect on the
$q$-adic doubling (as the ratio of $q$-adic intervals are still either $1$,
$\frac{a}{b}$ or $\frac{b}{a}$) or the failure of doubling overall (since the
measures of $H^{(\alpha)}$ and $G^{(\alpha)}$ are unchanged) but it does help
to guarantee $p$-adic doubling. In [2], the authors select intervals
$I^{\alpha}$ so that the difference between $Y(J^{\alpha})$ and
$Z(I^{\alpha})$ is suitably bounded, where $J^{\alpha}$ is the smallest
$p$-adic interval containing $I^{\alpha}$. The reverse weighting procedure
from steps $\alpha+1$ to $2\alpha$ guarantees $p$-adic doubling due to the
_exhaustion procedure_ in [2], where the closeness of $Y(J^{\alpha})$ and
$Z(I^{\alpha})$ is key: essentially on these scales the $p$-adic interval must
not span too many $q$-adic children with differing weights.

The existence of a family of intervals $I^{\alpha}$ on $[0,1]$ so that the
difference between $Y(J^{\alpha})$ and $Z(I^{\alpha})$ is suitably bounded,
and thus the existence of the desired measure, is guaranteed by the following
Proposition.

######  Proposition 2.5 (Proposition 3.5 in [2]).

Given any interval $\widetilde{J}\subseteq[0,1]$ and any $\varepsilon>0,$
there exists a $q$-adic interval $I\subset\widetilde{J}$ such that

$0<Y(J)-Z(I)\leq\varepsilon|I|$

where $J$ is the smallest $p$-adic interval containing $I.$

The existence of such a pair of intervals in Proposition 3.5 relies on the
number theory explored in [2] which guarantees that given some sufficiently
nice $p$-adic point of the form $\frac{k}{p^{a}}$ there exists an infinite
sequence of points $\frac{j}{q^{b}}$ such that

$\frac{k}{p^{a}}-\frac{j}{q^{b}}=\frac{1}{p^{a}q^{b}}.$

If we let $J=\left[\frac{k-1}{p^{a}},\frac{k-1+p}{p^{a}}\right)$ be a $p$-adic
interval then we can choose a large enough $b$ such that

$I=\left[\frac{j-q+1}{q^{b}},\frac{j+1}{q^{b}}\right)\subseteq J$

and $J$ is the largest $p$-adic interval containing $I.$ This framework was
extended in [4] to allow for such a sequence of $m$ and $n$-adic points with
$\gcd(m,n)=1$. Extending and simplifying this framework is thus our starting
point.

###  2.2. Extending the construction

Before proving our main result of this section, we first show how a simple
generalization of [2] can extend the construction to the $q$ and $q^{n}p^{m}$
bases case, where $p,q$ are distinct primes. Beginning here allows us to
recall the techniques of [2] which we will alter to prove Theorem 2.4. Due to
the discussion above, the existence of a measure which is $q$-adic and
$q^{n}p^{m}$-adic doubling, for distinct primes $q$ and $p$, follows
immediately from the construction of the measure outlined in Subsection 2.1 on
the selected intervals $I^{(\alpha)}$ and the following proposition.

######  Proposition 2.6.

Let $p$ and $q$ be primes with $p>q$. Given any interval
$\tilde{J}\subset[0,1]$ and any $\varepsilon>0$, there exists a $q$-adic
interval $I\subset\tilde{J}$ such that $0<Y(J)-Z(I)\leq\varepsilon|I|$, where
$J$ is the smallest $p^{m}q^{n}$-adic interval that contains $I$, for any
fixed positive integers $m$ and $n$.

###### Proof.

Fix integers $m,n$, interval $\tilde{J}\subset[0,1]$, and $\varepsilon>0$, and
let $J^{\prime}$ be the largest $q$-adic interval contained in $\tilde{J}$.
Say that $J^{\prime}$ has length $\dfrac{1}{q^{m^{\prime}}}$. As in [2] let
$m(p,q)$ denote the smallest positive integer satisfying

$q^{p-1}\not\equiv 1(\text{mod}\,p^{m(p,q)+1})$

and let $O_{\ell}(p,q)$ denote the order of $q^{p-1}$ in
$(\mathbb{Z}/p^{\ell}\mathbb{Z})^{\times}$ for any nonnegative integer $\ell$.
By Proposition 2.1 of [2], when $\ell$ is sufficiently large there exists a
constant $C(p,q)$ so that

$\frac{O_{\ell}(p,q)}{p^{\ell-1}}=\frac{1}{p^{C(p,q)}}.$

Choose a large enough integer $m_{1}$ so that all of the following hold

$m_{1}>\max\left\\{\frac{m(p,q)}{q-1},m^{\prime},\frac{m^{\prime}+C(p,q)+1}{p-1}\right\\},$

$m\mid m_{1}$ and $\dfrac{1}{p^{m_{1}(q-1)}}<\varepsilon$. Next, choose a
positive integer $k$ with

$k\equiv 1(\bmod(p^{m}q^{n})^{C(p,q)+1})$

so that for $m_{1}$ large enough we have $k/(p^{m}q^{n})^{s}\in J^{\prime}$
where $s={m_{1}(q-1)}/{m}$, and let

$\displaystyle
J=\left[\frac{k-1}{(p^{m}q^{n})^{s}},\frac{k-1+p^{m}q^{n}}{(p^{m}q^{n})^{s}}\right].$

Observe that $J$ is a $p^{m}q^{n}$-adic interval of length
$1/(p^{m}q^{n})^{s-1}$. Indeed, if we write

$k=\kappa\cdot(p^{m}q^{n})^{C(p,q)+1}+1$

for some $\kappa\in\mathbb{Z}$, then

$\frac{k-1}{(p^{m}q^{n})^{s}}=\frac{\kappa\cdot(p^{m}q^{n})^{C(p,q)+1}+1-1}{(p^{m}q^{n})^{s}}=\frac{\kappa\cdot(p^{m}q^{n})^{C(p,q)}}{(p^{m}q^{n})^{s-1}}$

and

$\frac{k-1+p^{m}q^{n}}{(p^{m}q^{n})^{s}}=\frac{\kappa\cdot(p^{m}q^{n})^{C(p,q)+1}+1-1+p^{m}q^{n}}{(p^{m}q^{n})^{s}}=\frac{\kappa\cdot(p^{m}q^{n})^{C(p,q)}+1}{(p^{m}q^{n})^{s-1}}$

.

Also note that that

$Y(J)=\frac{k}{(p^{m}q^{n})^{s}}\in J^{\prime}$

by choice of $k$. Since we chose $p>q$ and $m_{1}$ large enough, we see that
$Y(J)$ is in the interior of $J^{\prime}$. So, if we choose $m_{1}$
sufficiently large, we can make the interval $J$ arbitrarily small to ensure
that $J\subseteq J^{\prime}$. By Proposition 2.3 of [2], there are infinitely
many pairs $m_{2},j\in\mathbb{N}$ such that

$\frac{k}{p^{m_{1}(q-1)}}-\frac{j}{q^{m_{2}(p-1)}}=\frac{1}{p^{m_{1}(q-1)}q^{m_{2}(p-1)}}$

with $j\equiv-1(\bmod\,q)$. We now choose such an $m_{2}$ subject to the
additional constraint that

$q^{m_{2}(p-1)}>10q\cdot p^{m_{1}(q-1)},$

and set

$I=\left[\frac{j+1-q}{q^{ns+m_{2}(p-1)}},\frac{j+1}{q^{ns+m_{2}(p-1)}}\right).$

By our choice of $j\equiv-1(\bmod\,q)$ we can see that $I$ is a $q$-adic
interval. Furthermore, we have

$Z(I)=\frac{j}{q^{ns+m_{2}(p-1)}}.$

We are left to show that our intervals $I$ and $J$ satisfy the following
conditions.

  1. (1)

$Y(J)>Z(I)$

To see this, we have

$\displaystyle Y(J)-Z(I)$
$\displaystyle=\frac{1}{q^{ns}}\left(\frac{k}{p^{ms}}-\frac{j}{q^{m_{2}(p-1)}}\right)$
$\displaystyle=\frac{1}{q^{ns}}\left(\frac{k}{p^{m_{1}(q-1)}}-\frac{j}{q^{m_{2}(p-1)}}\right)$
$\displaystyle=\frac{1}{q^{ns}}\left(\frac{1}{p^{m_{1}(q-1)}q^{m_{2}(p-1)}}\right),$

which is greater than $0$ by construction.

  2. (2)

$I\subset J$

Let $l(I)$ denote the left most endpoint of $I$. Then we have

$\displaystyle|[l(I),Y(J)]|$ $\displaystyle=|[l(I),Z(I)]|+|[Z(I),Y(J)]|$
$\displaystyle=\frac{q-1}{q^{ns+m_{2}(p-1)}}+\frac{1}{p^{m_{1}(q-1)}q^{m_{2}(p-1)}}$
$\displaystyle=\frac{1}{q^{ns+m_{2}(p-1)}}\left[q+\left(\frac{1}{p^{m_{1}(q-1)}}-1\right)\right]$
$\displaystyle<\frac{q}{q^{ns+m_{2}(p-1)}}$
$\displaystyle<\frac{q}{10q^{ns+1}p^{ms}}$
$\displaystyle=\frac{1}{10(p^{m}q^{n})^{s}}<|[l(J),Y(J)]|=\frac{1}{(p^{m}q^{n})^{s}}$

  3. (3)

$J$ is the smallest $pq$-adic interval containing $I$.

Note that $Y(J)$ is again an interior point of $I$, as

$Y(J)-Z(I)=\frac{1}{q^{ns}}\left(\frac{1}{p^{m_{1}(q-1)}q^{m_{2}(p-1)}}\right)\leq\frac{1}{q^{ns+m_{2}(p-1)}}=|[Z(I),r(I)]|,$

where $r(I)$ denotes the right endpoint of $I$. Furthermore, all other
$pq$-adic intervals with side-length less than or equal to $|J|$ are either
disjoint from $Y(J)$ or contain $Y(J)$ as an endpoint.

  4. (4)

$Y(J)-Z(I)<\varepsilon|I|$

This follows from our choice of $m_{1}$, that is:

$\displaystyle Y(J)-Z(I)$
$\displaystyle=\frac{1}{q^{ns}}\left(\frac{1}{p^{m_{1}(q-1)}q^{m_{2}(p-1)}}\right)$
$\displaystyle<\frac{1}{q^{ns}}\left(\frac{\varepsilon
q}{q^{m_{2}(p-1)}}\right)$ $\displaystyle=\varepsilon|I|.$

By the same methodology used in [2] this completes the proof. ∎

###  2.3. Proof of Theorem 2.4

In order to prove our main result of this section, we take advantage of the
insight that the methods used in [2] to show the original measure is $p$-adic
doubling (i.e. the exhaustion procedure) only require (1.1) for _some_ $Y$ and
$Z$. That is, that the endpoint of some child of a $p$-adic interval can be
made arbitrarily close to some “distinguished point” on the selected $q$-adic
intervals.

In [2], these distinguished points are chosen to be $Y(J)$ and $Z(I)$. Given
$u,v\in\mathbb{N}$ with $u>v$, in our extensions we will keep $Z(I)$ as the
“distinguished point” on every $v$-adic interval $I$, but replace $Y(J)$ by a
“revolving” interior endpoint. As may be expected, this makes the initial
framework much more amenable to generalization.

We first generalize the number-theoretic framework developed by [2] to its
full power, and afterwards we will detail the changes which must be made to
their interval selection procedure in order to complete our extensions.

######  Proposition 2.7.

(Generalization of Proposition 2.1 in [2]) Let $p$ be prime and $u=p^{k}$ for
an integer $k\geq 1$, and let $v$ be any integer with $\gcd(u,v)=1$. Let
further $O_{m}(u,v)$ denote the order of $v^{\varphi(u)}$ in
$\left(\mathbb{Z}\big{/}\left(u^{m}\mathbb{Z}\right)\right)^{*}$ for each
$m\geq 1$, where $\varphi(u)$ denotes the Euler-totient function. Then there
exists some integer $C(u,v)\geq 0$, such that

$\frac{O_{m}(u,v)}{u^{m-1}}=\frac{1}{u^{C(u,v)}}.$

when $m$ is sufficiently large.

###### Proof.

Let $m(u,v)$ be the smallest integer such that

$v^{\varphi(u)}\not\equiv 1\ (\textrm{mod}\ u^{m(u,v)+1}).$

This implies that there exists some $N_{0}\in\\{1,2,\dots,m(u,v)\\}$ such that

(2.4) $\left(v^{\varphi(u)}\right)^{u^{N_{0}}}\equiv 1\ (\textrm{mod}\
u^{m(u,v)+1}),$

since by Euler’s theorem (applied to $v$), it is always true that

$\left(v^{\varphi(u)}\right)^{u^{m(u,v)}}\equiv 1\ (\textrm{mod}\
u^{m(u,v)+1}).$

In slightly more detail, recall that if $u=p^{k}$ then
$\varphi(u)=p^{k-1}(p-1)$. We see

$u^{N_{0}}\varphi(u)=p^{kN_{0}+k-1}(p-1),$

and as $p^{km(u,v)+k-1}(p-1)=\varphi(p^{km(u,v)+k})$ we see that such an
$N_{0}$ must indeed exist. Without loss of generality, we assume that the
$N_{0}$ fixed above is the smallest, namely

(2.5) $\left(v^{\varphi(u)}\right)^{u^{N_{0}-1}}\not\equiv 1\ (\textrm{mod}\
u^{m(u,v)+1}).$

We claim that for any $\ell\geq 0$, we have

(2.6) $\left(v^{\varphi(u)}\right)^{u^{N_{0}+\ell-1}}\not\equiv 1\
(\textrm{mod}\ u^{m(u,v)+\ell+1}).$

We prove the claim by induction. The case when $\ell=0$ is exactly (2.5).
Assume (2.6) holds when $\ell=k$, that is

(2.7) $\left(v^{\varphi(u)}\right)^{u^{N_{0}+k-1}}\not\equiv 1\ (\textrm{mod}\
u^{m(u,v)+k+1}),$

By raising both sides of equation ((2.4)) to $u^{k-1}$ we have

(2.8) $\left(v^{\varphi(u)}\right)^{u^{N_{0}+k-1}}\equiv 1\ (\textrm{mod}\
u^{m(u,v)+k}),$

which, together with (2.7), implies that we can write

$\left(v^{\varphi(u)}\right)^{u^{N_{0}+k-1}}=u^{m(u,v)+k}\cdot s+1,$

where $u\nmid s$.

Taking the $u$-th power on both sides of the above equation, we have

$\displaystyle\left(v^{\varphi(u)}\right)^{u^{N_{0}+k}}$ $\displaystyle\equiv$
$\displaystyle\left(\left(v^{\varphi(u)}\right)^{u^{N_{0}+k-1}}\right)^{u}$
$\displaystyle\equiv$ $\displaystyle\left(u^{m(u,v)+k}\cdot s+1\right)^{u}$
$\displaystyle\equiv$ $\displaystyle u^{m(u,v)+k+1}\cdot s+1$
$\displaystyle\not\equiv$ $\displaystyle 1\quad(\textrm{mod}\
u^{m(u,v)+k+2});$

in the last line above, we use the fact that $u\nmid s$. In particular, we
have used that $m(u,v)+k\geq 1$, to show (via the binomial theorem)
$2(m(u,v)+k\geq m(u,v)+k+1$. Therefore, (2.6) is proved.

With this, for any $\ell\geq 0$ we have

$(v^{\varphi(u)})^{u^{N_{0}+\ell-1}}\neq 1(\bmod u^{m(u,v)+\ell+1}),$

which implies that

$O_{m(u,v)+\ell+1}(u,v)=u^{N_{0}+\ell}.$

Rewriting $m$ as $m(u,v)+(m-m(u,v)-1)+1=m(u,v)+\ell+1$, as in [2] this yields

$O_{m}(u,v)=u^{-(m(u,v)-N_{0})},$

and setting $C(u,v)=(m(u,v)-N_{0})$ gives $u^{-C(u,v)}$, as desired. ∎

After the next Proposition, we will have the necessary tools to generalize the
interval selection procedure, namely, Theorem 3.4 and Proposition 3.5 of [2].

######  Proposition 2.8.

(Generalization of Proposition 2.3 in [2]) Let $p$ be prime, and $u=p^{k}$ for
an integer $k\geq 1$. Let $v$ be coprime with $C(u,v)$ and $m(u,v)$ be defined
as above. Then for any $m_{1}>\frac{m(u,v)}{\varphi(v)}$ and

(2.9) $\displaystyle k$ $\displaystyle\in$
$\displaystyle\left\\{1,1+u^{C(u,v)+1},1+2u^{C(u,v)+1},\dots,u^{m_{1}\varphi(v)}-u^{C(u,v)+1}+1\right\\}$
$\displaystyle=$
$\displaystyle\left\\{x\in\left[1,u^{m_{1}\varphi(v)}\right]:x\equiv 1\
\left(\textrm{mod}\ u^{C(u,v)+1}\right)\right\\},$

there exists infinitely many pairs $j$ and $m_{2}$, where
$m_{2}\in\mathbb{N}$, and

(2.10) $\displaystyle j$ $\displaystyle\in$
$\displaystyle\left\\{v-1,2v-1,\dots,v^{m_{2}\varphi(u)}-1\right\\}$
$\displaystyle=$
$\displaystyle\left\\{y\in\left[1,v^{m_{2}\varphi(u)}\right]:y\equiv-1\
(\textrm{mod}\ v)\right\\},$

such that

(2.11)
$\frac{k}{u^{m_{1}\varphi(v)}}-\frac{j}{v^{m_{2}\varphi(u)}}=\frac{1}{u^{m_{1}\varphi(v)}v^{m_{2}\varphi(u)}}.$

###### Proof.

Note that (2.11) is equivalent to find infinitely many pairs $m_{2}$ and $j$
which satisfies (2.10) for the equation

(2.12) $kv^{m_{2}\varphi(u)}-ju^{m_{1}\varphi(v)}=1,$

where $m_{1}>\frac{m(u,v)}{\varphi(v)}$ and $k$ satisfies (2.9).

To begin with, we note that if (2.12) holds, then $j$ automatically satisfies
(2.11), which follows by taking the modulus $v$ on both sides of (2.12) and
Euler’s theorem.

Therefore, it suffices for us to solve (2.12) for infinitely many pairs
$m_{2}$ and $j$. Taking modulus $u^{m_{1}\varphi(v)}$ on both sides of (2.12),
we see that it suffices to solve

(2.13) $kv^{m_{2}\varphi(u)}\equiv 1\quad\left(\textrm{mod}\
u^{m_{1}\varphi(v)}\right),$

where

$k\in\left\\{x\in\left[1,u^{m_{1}\varphi(v)}\right]:x\equiv 1\
\left(\textrm{mod}\ u^{C(u,v)+1}\right)\right\\}.$

Denote

$G_{m_{1}}(u,v):=\left\\{x\in\left[1,u^{m_{1}\varphi(v)}\right]:x\equiv 1\
\left(\textrm{mod}\ u^{C(u,v)+1}\right)\right\\}.$

The solubility of (2.14) will again follow if we show that

  1. (a)

The set $G_{m_{1}}(u,v)$ is a subgroup of
$\left(\mathbb{Z}/\left(u^{m_{1}\varphi(v)}\right)\mathbb{Z}\right)^{*}$;

  2. (b)

$v^{\varphi(u)}$ is a generator of the group $G_{m_{1}}(u,v)$.

We now show (a) and (b). To begin, we note that

(2.14) $v^{\varphi(u)}\equiv 1\left(\textrm{mod}\ u^{C(u,v)+1}\right),$

such that $v^{\varphi(u)}\in G_{m_{1}}(u,v)$. This is because $m(u,v)$ is the
smallest integer such that

$v^{\varphi(u)}\not\equiv 1\ (\textrm{mod}\ u^{m(u,v)+1})$

and $C(u,v)=m(u,v)-N_{0}$ for some $N_{0}\in\\{1,\dots,m(u,v)\\}$. Next, it
follows from (2.14) that

$\left(v^{\varphi(u)}\right)^{\ell}\in G_{m_{1}}(u,v),\quad\forall\ell\geq 1,$

and hence

$\left\langle v^{\varphi(u)}\right\rangle\subseteq G_{m_{1}}(u,v),$

.

Therefore, both assertions $(a)$ and $(b)$ will follow if we can show

$\left\langle v^{\varphi(u)}\right\rangle=G_{m_{1}}(u,v).$

This equality follows from the fact that

$O_{m_{1}\varphi(v)}=u^{m_{1}\varphi(v)-C(u,v)-1}=\left|G_{m_{1}}(u,v)\right|$

which again hinges on Proposition 2.7. ∎

We can now move on to generalizing the interval selection. Given the previous
results, let $\\{pc_{1},...,pc_{k}\\}$ be a finite set of multiples of some
prime $p$, where we have $c_{i}\in\mathbb{N}$ for all $i\in[1,k]$. Now, given
some $v$ coprime to $p$, we can always find $n\in\mathbb{N}$ such that
$v<p^{n}=u$, and proceed to build our measure off of $v,u$. Also note that
from this point we will work with the finite set of $n^{th}$ multiples
$\\{(pc_{1})^{n},...,(pc_{k})^{n}\\}=\\{uc_{1}^{\prime},...,uc_{k}^{\prime}\\}$
(such that $c_{i}^{\prime}=c_{i}^{n}$). As mentioned earlier, this will give
us an equivalent result.

######  Theorem 2.9 (Generalization of Theorem 3.4 in [2]).

There exists a collection of $v$-adic intervals
$\\{I^{\alpha_{\ell}}_{\ell}\\}_{\ell\geq 1}$ on $[0,1)$, where
$\alpha_{\ell}\geq 1$ is a positive integer associated to $\ell$, such that

  1. (1)

The collection of $pc_{i}$-adic intervals $\\{J_{\ell}^{i}\\}_{\ell\geq 1}$ is
pairwise disjoint and contained in $[0,1)$, where $J_{\ell}^{i}$ is the
smallest $pc_{i}$-adic interval that contains $I_{\ell}^{\alpha_{\ell}}$ for
all $i\in[1,k]$. In particular, the collection
$\\{I_{\ell}^{\alpha_{\ell}}\\}_{\ell\geq 1}$ is also pairwise disjoint;

  2. (2)

For each $\alpha\geq 1,\alpha\in\mathbb{N}$, there are only finitely many
$\ell\geq 1$, such that $\alpha_{\ell}=\alpha$;

  3. (3)

For each $\ell\geq 1$,

(2.15)
$0<\zeta\left(J_{\ell}^{i}\right)-Z\left(I_{\ell}^{\alpha_{\ell}}\right)\leq
q^{-100\alpha_{\ell}}\left|I_{\ell}^{\alpha_{\ell}}\right|.$

where $\zeta(J_{\ell}^{i})$ is an interior endpoint of some child of
$J_{\ell}^{i}$. Note that since $J_{\ell}^{i}$ is the smallest $pc_{i}$-adic
interval which contains $I_{\ell}^{\alpha_{\ell}}$, condition (2.15) in
particular guarantees that the right endpoint of $I_{\ell}^{\alpha_{\ell}}$ is
to the right of $\zeta(J^{\ell})$.

######  Remark 2.10.

As mentioned above, the measure created via this method will always also be
$v$ and $p$-adic doubling, and if we let $c_{1}=1$ such that $J_{\ell}^{1}$ is
our largest $p$-adic interval containing $I_{\ell}^{\alpha_{\ell}}$, then
$\zeta(J_{\ell}^{1})$ will always equal $Y(J_{\ell}^{1})$.

As before this theorem’s proof relies on a generalized Proposition 3.5 of [2]
(Proposition 2.11 below); however, while in the original construction the work
was essentially done after the proof of Proposition 3.5, we have to take more
care. After the proof of the following proposition we will go into more detail
regarding how our intervals are to be selected.

######  Proposition 2.11 (Generalized Proposition 3.5).

Given any interval $\tilde{J}\subset[0,1]$ and any $\varepsilon>0$, there
exists a $v$-adic interval $I\subset\tilde{J}$ such that

$0<\zeta\left(J^{i}\right)-Z\left(I\right)\leq\varepsilon|I|,$

for all integer $i\in[1,k]$, where $J^{i}$ is the smallest $pc_{i}$-adic
interval that contains $I$.

######  Remark 2.12.

Note in particular that we do not require $J^{i}\subseteq\tilde{J}$ (or even
$J^{i}\subseteq[0,1]$) – however we will necessarily have $I\subset\tilde{J}$
and $I\subset J^{i}$.

###### Proof.

We start with fixing an interval $\widetilde{J}\subset[0,1]$ and some
$\varepsilon>0$, and we let $J^{\prime}$ be the largest $v$-adic interval
which is contained in $\widetilde{J}$ with sidelength $1/v^{t_{1}^{\prime}}$.
We choose
$t_{1}>\max\left\\{\frac{m(u,v)}{\varphi(v)},t_{1}^{\prime}\right\\}$ and
$\frac{1}{u^{t_{1}\varphi(v)}}<\varepsilon v$, and we choose

$k\equiv 1(\bmod u^{C(u,v)+1}),$

and note that choosing $t_{1}$ large enough we can find a $k$ such that
$\frac{k}{u^{t_{1}\varphi(v)}}\in J^{\prime}$. Fix such a pair of $t_{1}$ and
$k$, and let

$J:=\left[\frac{k-1}{u^{t_{1}\varphi(v)}},\frac{k+u-1}{u^{t_{1}\varphi(v)}}\right].$

We then have

$Y(J)=\frac{k}{u^{t_{1}\varphi(v)}}$

and $J\subseteq J^{\prime}\subseteq\widetilde{J}$ due to the choice of $t_{1}$
and the fact $u=p^{n}>v$.

Note that

$\frac{k}{u^{t_{1}\varphi(v)}}=\frac{kc_{i}^{\prime
t_{1}\varphi(v)}}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}}$

is also the endpoint of some $uc_{i}^{\prime}$-adic interval. It follows,
writing

$\frac{kc_{i}^{\prime t_{1}\varphi(v)}}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}}$

in lowest terms, that there must exist some $y\in[1,uc_{i}^{\prime}-1]$ such
that $\frac{kc_{i}^{\prime
t_{1}\varphi(v)}}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}}$ is the right endpoint
of the $y^{th}$ child of some $uc_{i}^{\prime}$-adic interval $J^{i}$, denoted
$\zeta(J^{i})$ – for example if $\frac{kc_{i}^{\prime
t_{1}\varphi(v)}}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}}$ is in lowest terms
(which occurs when $u,c_{i}^{\prime}$ are coprime as $u\nmid k$), then
$\zeta(J^{i})$ will be the right endpoint of the $y^{th}$ child of

$J^{i}=\left[\frac{kc_{i}^{\prime
t_{1}\varphi(v)}-y}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}},\frac{kc_{i}^{\prime
t_{1}\varphi(v)}-y+uc_{i}^{\prime}}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}}\right).$

Henceforth we will let $J^{i}$ denote the smallest $uc_{i}^{\prime}$-adic
interval such that $\zeta(J^{i})$ is the endpoint of the $y^{th}$ child.

By Proposition 2.8, there exists infinitely many pairs $t_{2}\in\mathbb{N}$
and

$j\equiv-1(\bmod v)$

such that

(2.16)
$\frac{k}{u^{t_{1}\varphi(v)}}-\frac{j}{v^{t_{2}\varphi(u)}}=\frac{1}{u^{t_{1}\varphi(v)}v^{t_{2}\varphi(u)}}.$

We choose such a pair $t_{2}$ and $j$, with $t_{2}$ sufficiently large such
that

(2.17)
$\frac{v-1}{v^{t_{2}\varphi(u)}}+\frac{1}{u^{t_{1}\varphi(v)}v^{t_{2}\varphi(u)}}<\inf_{i\in[1,k]}\left\\{\frac{1}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}}\right\\}$

and let

$I:=\left[\frac{j+1-v}{v^{t_{2}\varphi(u)}},\frac{j+1}{v^{t_{2}\varphi(u)}}\right].$

such that

$\textZeta(I)=\frac{j}{v^{t_{2}\varphi(u)}}.$

.

We now proceed to check items (1) through (4) in the proof of Proposition 3.5
of [2], and doing so complete the proof.

  1. (1)

The inequality $\zeta(J^{i})>Z(I)$ follows because $\zeta(J^{i})=Y(J)>Z(I)$
for all $i\in[i,k]$.

  2. (2)

To show $I\subset J^{i}$, consider

$\displaystyle|[l(I),\zeta(J^{i})]|$
$\displaystyle=|[l(I),Z(I)]|+|[Z(I),\zeta(J^{i})]|$
$\displaystyle=\frac{v-1}{v^{t_{2}\varphi(u)}}+\frac{1}{u^{t_{1}\varphi(v)}v^{t_{2}(c-1)}}$
$\displaystyle<\frac{1}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}}$
$\displaystyle\leq|[l(J^{i}),\zeta(J^{i})]|$

by our choice of $t_{2}$. Explicitly, this tells us that $l(I)>l(J^{i})$ and
$Z(J^{i})\geq\zeta(J^{i})>Z(I)$ which gives $r(J^{i})>r(I)$, so we do have
that $I\subset J^{i}$ for all $i$.

  3. (3)

As before $I$ contains $\zeta(J^{i})$ as an interior point (this is obvious
considering just $I,J$), so indeed $J^{i}$ must be the smallest
$uc_{i}^{\prime}$-adic interval containing $I$.

  4. (4)

The inequality $\zeta(J^{i})-Z(I)<\varepsilon|I|$ again follows exactly as
before by our choice of $t_{1}$, as

$\zeta(J^{i})-Z(I)=\frac{k}{u^{t_{1}\varphi(v)}}-\frac{j}{v^{t_{2}\varphi(u)}}=\frac{1}{u^{t_{1}\varphi(v)}v^{t_{2}\varphi(u)}}<\frac{\varepsilon
v}{v^{t_{2}\varphi(u)}}=\varepsilon|I|.$

∎

With the proof complete, we can detail the additional complications that arise
when applying the above result to prove Theorem 2.9.

To begin, consider $uc_{i}^{\prime}$ such that $u,c_{i}$ are coprime, where

$J^{i}=\left[\frac{kc_{i}^{\prime
t_{1}\varphi(v)}-y}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}},\frac{kc_{i}^{\prime
t_{1}\varphi(v)}-y+uc_{i}^{\prime}}{(uc_{i}^{\prime})^{t_{1}\varphi(v)}}\right).$

Recalling that $\zeta(J^{i})=Y(J)$, we see that it is possible that such a
$J^{i}$ could possibly extend to the left of $l(J)$ (for example this may
happen if $\zeta(J^{i})=Z(J^{i})$).

In the application of Proposition 3.5 to Theorem 3.4 in [2], the authors began
by choosing $p$-adic intervals, and then selecting $J_{\ell}$ and
$I^{\alpha_{\ell}}_{\ell}$ within each of these intervals. We also want to
begin by choosing $p^{n}$-adic intervals, but if we choose our intervals
carelessly (not paying attention to the initial $\tilde{J}$ in which our
interval selection begins) it seems possible that one of our $J^{i}_{\ell}$
could intersect some $J^{j}_{\ell^{\prime}}$ and the corresponding
$I_{\ell^{\prime}}$, which would be problematic.

To avoid this, we note that for $u,c_{i}^{\prime}$ coprime $J^{i}$ will always
have total length less than $J$; thus it suffices to choose $u$-adic intervals
$J$ at least twice their length apart. In other words, if we choose to find
$J_{\ell}$ and $I_{\ell}$ within a $u$-adic $\tilde{J}$ of length $u^{-d}$, we
would want our next $\tilde{J^{\prime}}$ to be at least $u^{-d}$ units right
of $r(\tilde{J})$. As we can make all of our intervals arbitrarily small, this
can clearly be accomplished (an alternate way to look at it is that, given our
largest $\tilde{J}$ has length $u^{-d}$, $d$ can clearly be chosen such that
$2u^{-d}+2u^{-2d}+2u^{-3d}+\cdots=\frac{2}{u^{d}-1}$ becomes arbitrarily
small)

Now we deal with the case where $u,c_{i}^{\prime}$ are not coprime. As alluded
to earlier, the trouble here is that we a priori don’t have any bounds on the
size of $J^{i}$ chosen within an arbitrary $\tilde{J}$. However, if we choose
an $uc_{i}^{\prime}$-adic point contained strictly inside an
$uc_{i}^{\prime}$-adic interval $\tilde{J}_{i}$, then this
$uc_{i}^{\prime}$-adic must be the endpoint of some descendant of
$\tilde{J}_{i}$.

With this insight, we choose our intervals as follows:

  1. (1)

choose a sequence of $u$-adic intervals $\tilde{J_{\ell}}$, spaced apart as
described above

  2. (2)

begin with the least $c_{i}^{\prime}$ not coprime to $u$, and choose some
$uc_{i}^{\prime}$-adic interval contained within each $\tilde{J_{\ell}}$ (for
all $\ell$) – denote this interval $\tilde{J^{i}_{\ell}}$

  3. (3)

repeat this process for the next largest $c_{j}^{\prime}$ not coprime to $u$,
choosing some $uc_{j}^{\prime}$-adic interval contained within each
$\tilde{J_{\ell}^{i}}$ denoted $\tilde{J_{\ell}^{j}}$

  4. (4)

iterate steps (2) and (3) until there are no more $c_{i}^{\prime}$ which are
not coprime to $u$

  5. (5)

let $c_{w}^{\prime}$ denote the largest $c_{i}^{\prime}$ not coprime to $u$,
such that we end up with infinitely many disjoint $c_{w}^{\prime}$-adic
intervals $\tilde{J_{\ell}^{w}}$

Choosing $t_{1}$ large enough, we can always find
$Y(J_{\ell})=\zeta(J^{i}_{\ell})$ contained within each of our
$\tilde{J_{\ell}^{w}}$, and by the reasoning given above, all of our
$uc_{i}^{\prime}$-adic intervals $J^{i}_{\ell}$ will be disjoint, and each
will be the largest $uc_{i}^{\prime}$-adic interval containing our $v$-adic
interval $I_{\ell}$, as desired.

From the comments in the introduction in [2] and from a careful reading of
those results, we verify that we have generalized all steps needed to prove
Theorem 2.4. Indeed, the construction of the measure by re-weighting the
chosen intervals is the same as [2] (and as described above) once we have
selected our intervals $I^{(\alpha)}$. Therefore, the proofs that this measure
constructed on the chosen $I^{(\alpha)}$ is not doubling but is $u$ and
$v$-adic doubling remain carry through without change.

##  3\. A new construction for any finite list of bases: interplay of number
theory, topology, and analysis

We begin by giving some intuition as to why the construction given above fails
to admit a generalization to arbitrary finite lists of natural numbers. While
details are outlined in the appendix, essentially it relies on the properties
of the _far-numbers_ as discussed in [2] and [3].

######  Definition 3.1.

(Definition 2.7 in [2]) A real number $\delta$ is _$n$ -far_ if the distance
from $\delta$ to each given rational of the form $k/n^{m}$ is at least some
fixed multiple of $1/n^{m}$, where $m\geq 0$, $k\in\mathbb{Z}$. That is, if
there exists $C>0$ such that

(3.1) $\left|\delta-\frac{k}{n^{m}}\right|\geq\frac{C}{n^{m}},\quad\forall
m\geq 0,k\in\mathbb{Z}.$

where $C$ depends on $\delta$ but is independent of $m$ and $k$.

With this definition in mind, we now attempt to extend the framework of [2] in
the setting of primes $p$ and $q$, to the setting of three distinct primes.
Naively, given primes $p,q,$ and $r$ one might try to extend the construction
in [2] by finding a $p$-adic interval $J_{p}$ and a $q$-adic interval $J_{q}$
which are the largest $p$ and $q$-adic intervals (respectively) containing
some $r$-adic interval $I_{r}$ and simultaneously satisfy

$Y(J_{p})-Z(I_{r})\leq\varepsilon|I_{r}|$

and

$Y(J_{q})-Z(I_{r})\leq\varepsilon|I_{r}|$

for some given arbitrarily small $\varepsilon>0$. Observe that all $p$-adic
points contained in $(0,1)$ are $q$-far, and similarly all $q$-adic points
contained in $(0,1)$ are $p$-far. More concretely, if we have
$Y(J_{p})=\frac{k}{p^{n}}$ and $Y(J_{q})=\frac{j}{q^{m}}$ for some
$k,j,m,n\in\mathbb{N}$, then it can be easily verified that
$|Y(J_{p})-Y(J_{q})|\geq\frac{1}{p^{n}q^{m}}$ (see [1]). The triangle
inequality then yields

$|Y(J_{p})-Y(J_{q})|\leq|Y(J_{p})-Z(I_{r})|+|Y(J_{q})-Z(I_{r})|\leq
2\varepsilon|I_{r}|,$

such that $\frac{1}{p^{n}q^{m}}\leq 2\varepsilon|I_{r}|$ for any arbitrarily
small $\varepsilon$. This would imply that at least one of
$\frac{1}{q^{m}}=\frac{|J_{q}|}{q}$ or $\frac{1}{p^{n}}=\frac{|J_{p}|}{p}$ was
of order much smaller than $|I_{r}|=\frac{1}{r^{k}}$ for some
$k\in\mathbb{N}$. Recalling that we would want $J_{p},J_{q}$ to both contain
$I_{r}$ this seems inherently problematic, and motivates us to try and find a
new approach in order to select intervals which are sufficiently “close” for
multiple primes. This new approach is outlined throughout the rest of this
section.

######  Theorem 3.2.

Let $\\{n_{1},\ldots,n_{k}\\}\subseteq\mathbb{N}$ be a finite set of natural
numbers which are not powers of $2.$ Then, there exists an infinite family of
measures which are $n_{i}$-adic doubling for all $i$ but not doubling overall.

Note that for technical reasons, we will always exclude the set of bases that
only consists of $1\in\mathbb{N}$, otherwise we would tautologically claim the
existence of a measure that is both non-doubling and doubling.

######  Remark 3.3.

It suffices to assume that no $n$ is a power of $2$ because we will construct
a dyadic doubling measure, which is equivalent to a $2^{n}$-adic doubling
measure for each $n\in\mathbb{N}.$ We choose to make it dyadic doubling only
out of convenience to guarantee the proper containment relations from the
intervals that will be constructed..

By a careful reading of [2], in particular, Sections $5$ and $6$ (and as
discussed in Section 2 here), it suffices to find an infinite family of
disjoint intervals $I^{(\alpha)}$ (with $\alpha\in\mathbb{N}$), where each
contains a dyadic interval $I_{2}^{(\alpha)}$ satisfying (1.1). Namely, for
all $1\leq i\leq m$ we want that

$|Y(I_{n_{i}}^{(\alpha)})-Z(I_{2}^{(\alpha)})|<2^{-100\alpha}|I_{2}^{(\alpha)}|$

where $I_{n_{i}}^{(\alpha)}$ is the smallest $n_{i}$-adic interval in
$I^{(\alpha)}$ containing $I_{2}^{(\alpha)}.$ We will construct our measure
through dyadic intervals and apply the weighting procedure from Section $4$ of
[2] (and outlined in Section 2 here) to these intervals. Then, we get a
measure which is dyadic doubling and $n_{i}$-adic doubling but not doubling
overall. However, as related above, guaranteeing the closeness relationship is
problematic. Therefore, we will permit a key relaxation of (1.1) by allowing

$|Y(I_{n_{i}}^{(\alpha)})-Z(I_{2}^{(\alpha)})|<2^{-2\alpha}|I_{2}^{(\alpha)}|$

rather than trying to get

$0<Y(I_{n_{i}}^{(\alpha)})-Z(I_{2}^{(\alpha)})<2^{-2\alpha}|I_{2}^{(\alpha)}|.$

As discussed in Section $5$ and $6$ in [2], this bound on the absolute value
is all that is really needed. Indeed, the same proof holds just with us having
to switch directions in running through the outlined exhaustion procedure; the
authors in [2] choose the specific interior points to be in the given
orientation for number theoretic purposes. Our replacement for this stems from
the closeness relationship below.

######  Lemma 3.4.

Let $\\{n_{1},\ldots,n_{k}\\}\subseteq\mathbb{N},$ $\alpha\in\mathbb{N}$ and
$\varepsilon=2^{-100\alpha}>0.$ Then, there exists infinitely many
$x\in\mathbb{N}$ such that for all $1\leq i\leq k$ that

(3.2)
$\left|\frac{1}{2^{x}}-\frac{1}{n_{i}^{\left[x\log_{n_{i}}(2)\right]}}\right|<\varepsilon\,\frac{1}{2^{x}}.$

The significance of the $\alpha$ here is that in the measure construction we
will choose our $x$ such that $\varepsilon=2^{-100\alpha}<<2^{-2\alpha}$ just
as done in [2].

######  Remark 3.5.

Upon showing the existence of such an $x$ we can construct dyadic intervals
$I_{2}^{\alpha}$ in $[\alpha,\alpha+1]$ such that if $J_{i}^{\alpha}$ is the
smallest $n_{i}$-adic interval containing $I_{2}^{\alpha}$ then
$|Y(J_{i}^{\alpha})-Z(I_{2}^{\alpha})|<\varepsilon|I_{2}^{\alpha}|$ which is
sufficient to guarantee the existence of a measure which is dyadic doubling
and $n_{i}$-adic doubling for each $n_{i}$. Alternatively, we could construct
this measure on any $n$-adic intervals where $n$ is at most the smallest of
our list of natural numbers with only minor modifications in the proofs below.

###### Proof.

Re-arranging Equation 3.2 is equivalent to finding values of $x$ so that

$\left|1-\frac{2^{x}}{n_{i}^{\left[x\log_{n_{i}}(2)\right]}}\right|<\varepsilon$

for all $i$. Note that the orbit of rationally independent points
$(\theta_{1},\ldots,\theta_{m})\in\mathbb{T}^{m}$ under multiplication by an
integer is dense in $\mathbb{T}^{m}.$ It is an open problem as to whether

$S=\\{\log_{n_{i}}(2)|1\leq i\leq k\\}$

is a rationally independent set – this is equivalent to asking whether
$\\{\frac{1}{\log n_{i}}|1\leq i\leq k\\}$ is rationally independent, which
the Schaunel conjecture (if true) would imply. Hence we will consider two
cases without casting any aspersions on the validity of either case.

If $S$ is rationally independent, then the orbit of these points is dense in
$\mathbb{T}^{k}.$ If the set $S$ is rationally dependent, then the orbit under
the $\mathbb{Z}$-action will be dense in some linear subspace of
$\mathbb{T}^{k}$ (as proven below in Proposition 3.6).

Thus in either case, for $\frac{1}{2}>\delta>0$, there exist infinitely many
$x$ such that $|\\{x\log_{n_{i}}2\\}|\leq\delta$ and given $\varepsilon$, we
can pick $\delta$ small enough such that $|1-n_{i}^{\pm\delta}|<\varepsilon.$
For any such $x$ we then get that

$\left|1-\frac{2^{x}}{n_{i}^{\left[x\log_{n_{i}}2\right]}}\right|\leq\left|1-n_{i}^{\pm\delta}\right|<\varepsilon.\qed$

######  Proposition 3.6.

Let $\\{\theta_{1},\ldots,\theta_{k}\\}\subseteq\mathbb{R}\setminus\mathbb{Q}$
be rationally independent. Then, the orbit of $(\theta_{1},\ldots,\theta_{k})$
under the action of $\mathbb{Z}$ is dense in some linear subspace of
$\mathbb{T}^{k}.$

###### Proof.

Because $\theta_{1},\ldots,\theta_{k}$ are rationally independent there is
some minimal $\mathbb{Q}$-spanning set. Without loss of generality say

$\\{\theta_{1},\ldots,\theta_{k}\\}\subseteq\mathrm{Span}_{\mathbb{Q}}\\{\theta_{1},\ldots,\theta_{j}\\}$

with $j<k$ and $\theta_{1},\ldots,\theta_{j}$ rationally independent. Consider
the linear subspace

$V:=\mathrm{Span}_{\mathbb{R}}\left\\{\begin{pmatrix}\theta_{1}\\\ 0\\\ 0\\\
\vdots\\\ 0\\\ \theta_{j+1}\\\ \vdots\\\
\theta_{k}\end{pmatrix},\begin{pmatrix}0\\\ \theta_{2}\\\ 0\\\ \vdots\\\ 0\\\
\theta_{j+1}\\\ \vdots\\\ \theta_{k}\end{pmatrix},\ldots,\begin{pmatrix}0\\\
0\\\ 0\\\ \vdots\\\ \theta_{j}\\\ \theta_{j+1}\\\ \vdots\\\
\theta_{k}\end{pmatrix},\right\\}/\mathbb{Z}^{k}\subseteq\mathbb{T}^{k}.$

We aim to show that the orbit of $(\theta_{1},\ldots,\theta_{k})$ is dense in
this $j$-dimensional subspace $V\subseteq\mathbb{T}^{k}.$ Take
$(x_{1},\ldots,x_{k})\in V$ and $\varepsilon>0.$ Since
$(\theta_{1},\ldots,\theta_{j})$ is rationally independent, the
$\mathbb{Z}$-orbit is dense in $\mathbb{T}^{j}.$ Hence, there is some
$x\in\mathbb{Z}$ such that for all $1\leq i\leq j$ we have
$|x_{i}-x\theta_{i}|<\frac{\varepsilon}{C}$ where

$C=\max_{j+1\leq\ell\leq
k}\left\\{\sum_{i=1}^{j}|r_{i}|:\theta_{\ell}=r_{1}\theta_{1}+\cdots+r_{j}\theta_{j}\right\\}.$

This constant is well defined because $\theta_{1},\ldots,\theta_{j}$ are
rationally independent, so for each $\theta_{\ell}$ with $\ell>j$, there are
unique $r_{1},\ldots,r_{j}$ such that
$\theta_{\ell}=r_{1}\theta_{1}+\cdots+r_{j}\theta_{j}.$ Then, for all
$j+1\leq\ell\leq k$ we get that

$x_{\ell}=r_{1}x_{1}+\cdots+r_{j}x_{j}$

and

$\theta_{\ell}=r_{1}\theta_{1}+\cdots+r_{j}\theta_{j}.$

This gives

$|x_{\ell}-x\theta_{\ell}|=\sum_{i=1}^{j}|r_{i}||x_{i}-x\theta_{i}|<\frac{\varepsilon}{C}(\sum_{i=1}^{j}|r_{i}|)<\varepsilon.$

Hence, for each $(x_{1},\ldots,x_{k})\in V$ we have an $x\in\mathbb{N}$ such
that

$\|(x\theta_{1},\ldots,x\theta_{k})-(x_{1},\ldots,x_{k})\|_{\infty}<\varepsilon;$

thus the $\mathbb{Z}$-orbit of $(\theta_{1},\ldots,\theta_{k})$ is dense in
$\mathbb{T}^{k}.$ ∎

######  Remark 3.7.

If the Schanuel conjecture [5] were true then $\\{\log_{n_{i}}(2)|1\leq i\leq
k\\}$ would be rationally independent and thus we could guarantee that the
orbit would be dense in the torus. Hence, we would be able to choose points in
the orbit in any neighborhood of $0.$ In particular we could guarantee that
$\\{x\log_{n_{i}}(2)\\}<\frac{1}{2}$ and arbitrarily close to $0.$ we could
then choose $x$ such that $[x\log_{n_{i}}2]=\lfloor x\log_{n_{i}}(2)\rfloor$
giving

$\frac{1}{2^{x}}<\frac{1}{n_{i}^{[x\log_{n_{i}}(2)]}}$

for all $n_{i}$. this would guarantee that the intervals we construct would
have $Z(I_{2}^{(\alpha)})$ would be to the left of the
$Y(I_{n_{i}}^{(\alpha)})$ for each $n_{i}$ which would mimic the construction
in [2]. This alignment would give a nice interpretation as to why the original
number theoretic construction of [2] does not carry over in this setting.
Explicit details related to this are given in the appendix. The relationship
to the Schanuel conjecture connects many areas in our setting.

Now, we prove Theorem 1.1.

###### Proof of Theorem 1.1.

We begin by selecting the collection of intervals on which to alter our
measure. For each $\alpha\in\mathbb{N}$ find a requisite $x$ as constructed
Lemma 3.4 for $\varepsilon:=2^{-100\alpha}.$ Then, define

$I_{2}^{(\alpha)}=\left[\alpha,\alpha+\frac{1}{2^{x-1}}\right)$

and

$I_{n_{i}}^{(\alpha)}=\left[\alpha,\alpha+\frac{1}{n_{i}^{[x\log_{n_{i}}(2)]-1}}\right).$

This gives

$|Y(I_{n_{i}}^{(\alpha)})-Z(I_{2}^{(\alpha)})|\leq\varepsilon|I_{2}^{(\alpha)}|.$

Hence it suffices to show that for all $n_{i}$ we have the containment
$I_{2}^{(\alpha)}\subseteq I_{n_{i}}^{(\alpha)}$ as the proximity of
$Y(I_{n_{i}}^{(\alpha)})$ and $Z(I_{2}^{(\alpha)})$ guarantees that this is
the smallest such interval containing it. We enlose a picture of what nested
$n$-adic intervals would look like for $n=2,3,5,6$ in Figure 3. Set the
following notation: for an interval $I=[a,b)\subset\mathbb{R}$, let $l(I)=a$
and $r(I)=b$ denote the left and right-most endpoints of the closure of $I$,
respectively. Containment follows from looking at the relative position of the
endpoints. Indeed, $l(I_{2}^{(\alpha)})=l(I_{n_{i}}^{(\alpha)})$ and we will
show that

$r(I_{n_{i}}^{(\alpha)})=\alpha+n_{i}y$

is to the right of

$r(I_{2}^{(\alpha)})=\alpha+2z$

where $y=\frac{1}{n_{i}^{[x\log_{n_{i}}(2)]}}$ and $z=\frac{1}{2^{x}}.$
Because $n_{i}>2$ we have that

$r(I_{n_{i}}^{(\alpha)})-r(I_{2}^{(\alpha)})=(n_{i}-2)y+2(y-z)>0$

as $2(y-z)>-2\varepsilon z$ and $\varepsilon$ is much smaller than
$n_{i}-2\geq 1$.

As discussed earlier, one can follow the procedure in Sections $4,5$ and $6$
of [2] to get the main result. In particular, the proof of $n$-adic doubling
for each $n$ is essentially the same as in [2] but with some of the details
slightly changed. First off, the “trivial” cases of finding the ratio for
$\frac{\mu(J_{j_{1}})}{\mu(J_{j_{2}})}$ for children of the $n$-adic interval
$J$ containing $I_{2}^{(\alpha)}$ are the exact same in this case.

In the other cases, the proof is slightly different. This is because
$Y(I_{n}^{(\alpha)})$ may be to the right or the left of
$Z(I_{2}^{(\alpha)}).$ We run through these cases.

Suppose that $J=I_{n}^{(\alpha)}$ for some $n\in\mathbb{N}$ and large enough
$\alpha.$ Let $J_{1},\ldots,J_{n}$ denote the $n$-adic children of $J.$ First,
suppose that $Y(J)>Z(I)$ where $I=I_{2}^{(\alpha)}.$ In that case we get that

$\frac{\mu(J_{j_{1}})}{\mu(J_{j_{2}})}=1\;\forall
j_{1},j_{2}\in\\{3,\ldots,n\\}$

as on those intervals we’ll just get the Lebesgue measure as they are outside
of $I.$

Then, for $J_{i}$ for $i=1,2$ we get

$\frac{a|I|}{2}\leq\mu(J_{1})\leq|I|$

and

$\frac{b|I|}{4}\leq\mu(J_{2})\leq 2|I|.$

In the case where $Y(J)<Z(I)$ we get

$\frac{\mu(J_{j_{1}})}{\mu(J_{j_{2}})}=1\;\forall
j_{1},j_{2}\in\\{4,\ldots,n\\}$

as on those intervals we’ll just get the Lebesgue measure as they are outside
of $I.$ Then, for $J_{i}$ for $i=1,2,3$ we get

$\frac{a|I|}{4}\leq\mu(J_{1})\leq\frac{a|I|}{2}$

and

$\frac{b|I|}{4}\leq\mu(J_{2})\leq|I|$

and

$\frac{a|I|}{4}\leq\mu(J_{3})\leq\frac{3|I|}{2}.$

The case discussed above is the only one that is essentially different (and
only in a small way). When it comes to the exhaustion procedure most of the
details are then identical with the exception of replacing $p$ by $n$. Thus
there exists a family of measures which is $n_{i}$-adic doubling for each
$n_{i}$ but not doubling overall. ∎

$0$$Y(I_{3})$$Z(I_{2})$$Y(I_{5})$$Y(I_{6})$ Figure 3.
$I_{2},I_{3},\ldots,I_{n}$. Not to scale.

Now, with an eye towards applications, we will compactify the construction of
this measure. That way we will get applications to weight and function classes
over compact sets.

Here is the adjusted procedure. Given an initial $\alpha$ take an $x$ such
that

$\left|\frac{1}{2^{x}}-\frac{1}{n^{[x\log_{n}2]}}\right|<\frac{2^{-100\alpha}}{2^{x}}.$

Then, we choose

$I_{2}^{(\alpha)}=\left[0,\frac{1}{2^{x-1}}\right)\text{ and
}I_{n}^{(\alpha)}=\left[0,\frac{1}{n^{[x\log_{n}2]-1}}\right).$

We keep the same re-weighting procedure on $I_{2}^{(\alpha)}$ but with the
exception of re-weighting the first step as follows: give the interval
$[0,\frac{1}{2^{x+1}}]$ the weight $1$, the interval
$[\frac{1}{2^{x+1}},\frac{1}{2^{x}}]$ with the weight $a$, and
$[\frac{1}{2^{x}},\frac{1}{2^{x-1}}]$ the weight $b-1.$ Then for each
successive iteration go back to the original re-weighting procedure of
multiplying by $b$ on the right half and $a$ on the left half. Then for any
successive $\alpha$ take $x$ such that

$\left|\frac{1}{2^{x}}-\frac{1}{n^{[x\log_{n}2]}}\right|<\frac{2^{-100\alpha}}{2^{x}},$

with $x$ sufficiently larger than the $x$ associated to the previous $\alpha$.
This can be done since there are infinitely many $x$ of this form associated
to this $\alpha.$ This re-weighting still provides us with a measure as we
still get

$\displaystyle\frac{1}{2^{x-1}}$ $\displaystyle=\mu([0,\frac{1}{2^{x-1}}])$
$\displaystyle=\mu([0,\frac{1}{2^{x+1}}])+\mu([\frac{1}{2^{x+1}},\frac{1}{2^{x}}])+\mu([\frac{1}{2^{x}},\frac{1}{2^{x+1}}])$
$\displaystyle=1(\frac{1}{2^{x+1}})+(a)(\frac{1}{2^{x+1}})+(b-1)(\frac{1}{2^{x}})=\frac{1}{2^{x-1}}.$

Repeating the same procedure then compactifies the measure to $[0,1]$, which
is $n$-adic doubling for all $n$ in our list but not doubling overall.

###  3.1. Obstacles to extending to an infinite set of bases

It remains an open question to construct a measure that is doubling on an
infinite set of bases, yet not doubling overall. Here we show that while parts
of our construction generalize to the infinite base setting, there are still
significant obstacles to overcome to prove the folkloric conjecture. In
particular, in attempting to solve this question we run into a key obstacle.
This obstacle turns out to be more general and we shed insight into what a
construction to overcome this would need to surmount. Nevertheless, we show
that at least our interval selection generalizes to the case of infinitely
many $n\in\mathbb{N}.$ That is, we can find some $x\in\mathbb{N}$ such that

$\left|\frac{1}{2^{x}}-\frac{1}{n^{[x\log_{n}2]}}\right|<\frac{\varepsilon}{2^{x}}$

for all $n\in\mathbb{N}$.

We begin by using the fact that the infinite torus $\mathbb{T}^{\infty}$ is
metrizable, and thus is endowed with the triangle inequality, to extend the
interval selection above. This would accommodate an infinite base set. Though
standard, for completeness we show the metrizability of $\mathbb{T}^{\infty}$
in the appendix.

######  Lemma 3.8.

Let $\varepsilon>0.$ Then, there exists infinitely many $x\in\mathbb{N}$ such
that for all $n\in\mathbb{N}$ which is not a power of $2$ we have

$\left|\frac{1}{2^{x}}-\frac{1}{n^{[x\log_{n}2]}}\right|<\frac{\varepsilon}{2^{x}}.$

###### Proof.

Consider

$z=(\log_{3}(2),\log_{5}(2),\log_{6}(2),\ldots)\in\mathbb{T}^{\omega}$

where $z_{i}$ is $\log_{n_{i}}(2)$ where $n_{i}$ is the $i^{\text{th}}$
natural number which is not a power of $2.$ By Proposition 5.1 we have that in
any finite sub-torus $0$ is a limit point of the $\mathbb{Z}$-orbit of $z.$
Hence, by Proposition 5.2 we have that $0$ is a limit point of the
$\mathbb{Z}$-orbit of $z$ in $\mathbb{T}^{\omega}.$ Hence, there exists
infinitely many $x\in\mathbb{N}$ such that
$|\\{x\log_{n}2\\}|<\delta<\frac{1}{2}$ for each $n$ such that

$\left|1-\frac{2^{x}}{n^{[x\log_{n}2]}}\right|=|1-n^{\delta}|<\varepsilon$

so

$\left|\frac{1}{2^{x}}-\frac{1}{n^{[x\log_{n}2]}}\right|<\frac{\varepsilon}{2^{x}}.$

In the case where $n$ is a power of $2$ then there are infinitely many $x$
such that we get $0.$ ∎

######  Remark 3.9.

While the interval construction works for an infinite list, the issue that
remains is the proof of $n$-adic doubling. This is because the constants
chosen to ensure doubling cannot be made uniform under the given exhaustion
procedure in [2]. In particular, all of the constants for $n$-adic doubling
depend on $N=\lfloor\log_{2}n\rfloor+1$ which becomes arbitrarily large, so we
cannot “exhaust” all $n$-adic children (in the language of [2]) for infinitely
many $n$ uniformly. This indicates a logarithmic growth in the doubling
constant, depdendent on the base $n$. Making this constant uniform does not
seem possible under this construction.

###  3.2. Relationship to Hasse principle

Finally, we comment about an analogy of our progress toward a Hasse principle.
In general terms, the Hasse principle states that if a system of equations has
a solution $\mathrm{mod}\,p$ for all primes $p$ and a real solution (which can
be viewed as $\mathrm{mod}\,p_{\infty}$, or the prime at infinity), then it
has an integer solution. In [2], we introduced an analogue of the Hasse
principle with the “solution $\mathrm{mod}\,p$” being replaced by “$p$-adic
doubling” and an “integer solution” being replaced by “doubling”. Thus, being
able to construct a measure that is doubling for all primes yet not overall
would be an analogue of the Hasse principle in this harmonic analysis setting.
Here we amplify this analogy by additionally connecting the existence of a
real solution to our setting. In our setup, the real solution corresponds to
the measure being doubling on any interval around $0$. Since $0$ is never an
interior point for any $n$-adic interval, capturing the behavior of intervals
containing $0$ cannot be done by the $n$-adic framework and needs to be
considered separately, just as real solutions do for the Hasse principle.
Indeed, the failure of doubling for our examples stems from intervals not
including $0$, so all of our measures on $\mathbb{R}$ satisfy doubling “at
$0$”. Interestingly, in the next section we show that for the function class
$VMO$, the infinite intersection of $n$-adic VMO is never equal to the full
VMO; however in this example, the failure occurs around the point $0$ (and
thus this example would not show the failure of the Hasse principle, since
there would be no “real solution”). Either finding a new construction or
showing one cannot exist to demonstrate or break the Hasse principle in this
setting would be an exciting development.

##  4\. Applications

We are now ready to show a variety of applications to weight and function
classes. We begin with definitions and short remarks, segway into an
observation about the VMO function class involving infinite intersections, and
closing with the proof of Theorem 1.2.

Let $w_{\mu}$ be the weight associated to the measure $\mu$ in Section 4, that
is

$\mu(I)=\int_{I}w_{\mu}dx\text{, for any interval $I$}.$

######  Definition 4.1.

Let $1<r<\infty$. We say $w\in RH_{r}$, the _r-reverse Hölder class_ , if

$(\fint_{I}w^{r})^{\frac{1}{r}}\leq C\fint_{I}w$

for all intervals $I$, where $C$ is an absolute constant. Moreover, we say
$w\in RH_{1}$ if $w\in RH_{r}$ for some $r>1$, that is

$RH_{1}=\bigcup_{r>1}RH_{r}.$

######  Definition 4.2.

Let $1<r<\infty$. We say $w\in RH_{r}^{p}$ if

$(\fint_{P}w^{r})^{\frac{1}{r}}\leq C\fint_{P}w$

for all $p$-adic intervals $P$, where $C$ is an absolute constant and $w$ is
$p$-adic doubling.

######  Definition 4.3.

Let $1<r<\infty$. We say a weight $w\in A_{r}$, the _Muckenhoupt $A_{r}$
class_ if

$\sup_{I}(\fint_{I}w(x)dx)(\fint_{I}w(x)^{\frac{-1}{r-1}}dx)^{r-1}<\infty$

where the supremum is taken over all intervals $I$. Moreover, we say $w\in
A_{\infty}$ if $w\in A_{r}$ for some $r>1$, that is

$A_{\infty}=\bigcup_{r>1}A_{r}.$

######  Definition 4.4.

Let $1<r<\infty$. We say a weight $w\in A_{r}^{p}$ if

$\sup_{P}(\fint_{P}w(x)dx)(\fint_{P}w(x)^{\frac{-1}{r-1}}dx)^{r-1}<\infty$

where the supremum is taken over all $p$-adic intervals $P$. Moreover, we say
$w\in A_{\infty}$ if $w\in A_{r}^{p}$ for some $r>1$, that is

$A_{\infty}^{p}=\bigcup_{r>1}A_{r}^{p}.$

######  Definition 4.5.

We say a function is in $BMO$ if and only if

$\displaystyle||f||_{BMO}\coloneqq\sup_{I}\fint_{I}|f-\fint_{I}f|<\infty$

where $I$ is any interval and $\fint_{I}$ gives the average over $I$ with
respect to Lebesgue measure. A function is similarly in $BMO_{p}$ if and only
if it satisfies the above condition when restricted to $p$-adic intervals $I$.

The class $VMO$, or functions of vanishing mean oscillation, was introduced by
Sarason on the torus [18]. However, Coifman and Weiss [9] modified Saranson’s
definition when they extended it to the real line so as to have a duality
relationship with the Hardy space $H^{1}$ (see discussion below). We list the
definition on the real line first.

######  Definition 4.6.

A function $f$ in $BMO(\mathbb{R})$ is said to be in $VMO(\mathbb{R})$ if it
satisfies the following conditions:

  1. (1)

$\lim_{\delta\rightarrow 0}\sup_{I:|I|<\delta}\fint_{I}|f-\fint_{I}f|=0$;

  2. (2)

$\lim_{N\rightarrow\infty}\sup_{I:|I|>N}\fint_{I}|f-\fint_{I}f|=0$; and

  3. (3)

$\lim_{R\rightarrow\infty}\sup_{I:I\cap
B(0,R)=\emptyset}\fint_{I}|f-\fint_{I}f|=0$.

where $B(0,R)$ is the ball centered at zero of radius $R$.

With this definition, $VMO$ is the closure in the $BMO$ norm of the class
$C_{0}$ (continuous functions with compact support). The definition on the
torus is the same, but without the second and third conditions.

######  Definition 4.7.

A function $f$ in $BMO_{p}(\mathbb{R})$ is said to be in $VMO_{p}(\mathbb{R})$
if it satisfies the following conditions:

  1. (1)

$\lim_{\delta\rightarrow 0}\sup_{I:|I|<\delta}\fint_{I}|f-\fint_{I}f|=0$;

  2. (2)

$\lim_{N\rightarrow\infty}\sup_{I:|I|>N}\fint_{I}|f-\fint_{I}f|=0$; and

  3. (3)

$\lim_{R\rightarrow\infty}\sup_{I:I\cap
B(0,R)=\emptyset}\fint_{I}|f-\fint_{I}f|=0$,

where $I$ is always a $p$-adic interval.

We will also work with the Hardy sapces $H^{1}$ and $n$-adic Hardy spaces
$H^{1}_{n}$. The classical Hardy space $H^{1}$ is a subspace of $L^{1}$, and
has several useful and equivalent definitions. We refer the reader to [13] for
a description of these spaces. Since we will not be using the particulars of
the definitions and only duality in our proofs, we will content ourselves with
using the duality relationships as a definition. Note that with Hardy spaces
we use sums instead of intersections; this can be seen as stemming from the
role of the maximal functions in the definitions.

We will make use of the following duality relationships shortly.

######  Proposition 4.8.

The dual of $H^{1}$ is $BMO$, and the dual of $VMO$ is $H^{1}$. That is

  1. (1)

$H^{1*}=BMO$

  2. (2)

$VMO^{*}=H^{1}$.

These hold true both on $\mathbb{R}$ and $\mathbb{T}$. They also hold in the
$n$-adic setting.

The proof of $H^{1}$ duality is due to Fefferman-Stein [10] and VMO duality is
due to Coifman-Weiss [9].

We first show, via a simple construction, that the _infinite_ intersection of
the $n$-adic $VMO$ spaces on the real line is never $VMO$. The structure of
$VMO$ permits such a claim, showing how different it is from the other
function classes.

######  Theorem 4.9.

$\bigcap_{n\in\mathbb{N}}VMO_{n}\neq VMO$

###### Proof.

Define $f$ such that:

$\displaystyle f(x)=\begin{cases}0,x<0\\\ 1,x\geq 0\end{cases}$

We observe $0$ is an endpoint of all $n$-adic intervals. So, when restricting
to $n$-adic intervals, $f$ will be constant and thus $VMO_{n}$ for all
$n\in\mathbb{N}$. We show $\lim_{\delta\rightarrow
0}\sup_{I:|I|<\delta}\fint_{I}|f-\fint_{I}f|>0$. Fix $\delta>0$ to be
arbitrarily small. Then let $J$ be the interval $(-r,r)$ such that
$|J|<\delta$. Noting that $\fint_{J}f=\frac{1}{2r}\int_{J}f=\frac{1}{2}$, we
compute:

$\displaystyle\sup_{I:|I|<\delta}\fint_{I}|f-\fint_{I}f|$
$\displaystyle\geq\fint_{J}|f-\fint_{J}f|=\frac{\int_{(-r,0)}|0-\frac{1}{2}|+\int_{(0,r)}|1-\frac{1}{2}|}{2r}=\frac{1}{2}.$

Since this estimate holds as $\delta\to 0$, then $f\notin VMO$.

∎

######  Remark 4.10.

The exact same proof also shows that for any finite set $S\subseteq\mathbb{N}$
that

$\cap_{n\in S}VMO_{n}\neq VMO.$

Now we can prove Theorem 1.2. This proof includes an alternative proof of the
$VMO$ result in the above remark and actually much more, since it uses a much
more expansive framework. Indeed this proof ties all the weight and function
cases together with the measures constructed earlier via a functional analysis
framework.

###### Proof of Theorem 1.2.

We construct the function $w_{\mu}$ to be the weight associated with our
measure $\mu$, defined at the top of this section. Then
$w_{\mu}\in\bigcup_{n\in S}RH_{r}^{n}$, and $w_{\mu}\not\in RH_{r}$, by the
remarks in the introduction and the fact that the construction of the measure
is the same as in [2]. Note that the proof of this result for two distinct
primes appears in the final section of [2], including carefully computing
bounds on the reverse Hölder constants. The computations therein only rely on
the construction of the measure, and the exhaustion procedure, and as these
remain unchanged, the proof also follows through in this setting. We leave the
details to the interested reader. Thus, $RH_{r}^{n}\not\subseteq\bigcup_{n\in
S}RH_{r}^{n}$.

We select our $r$ such that $\max_{n\in S}{1-\frac{\ln n}{\ln 2}}<r<\infty$.
By eplacing all instances of $r$ in the reverse Hölder weights case with
$\frac{-1}{r-1}$, we get an identical argument, such that
$A_{r}\not\subseteq\bigcap_{n\in S}A_{r}^{n}$. See [2] for details.

We then apply the fact that for any $A_{\infty}$ weight $w$, $\log|w|\in BMO$
(the same holds in the $n$-adic case). Let $f(x)=\log w_{\mu}(x)$. Since
$w_{\mu}\in\bigcap_{n\in S}A_{r}^{n}$, we know $f\in\bigcap_{n\in S}BMO_{n}$.
We then apply an identical argument to [4] to observe that $w_{\mu}\not\in
BMO$. Observe that our measure construction differs slightly from [4], but not
in a way that substantially alters our proof. In particular, we get that
$||f||_{BMO}\geq\frac{\alpha}{4}\log\frac{b}{a}$. Since $\alpha$ can be made
to be arbitrarily large, we get that $||f||_{BMO}$ is unbounded. Also in a
similar manner as [4], we see that $f\notin BMO_{n}$ for any $n\in S$.

Now to show the Hardy space result we use duality. A standard theorem from
functional analysis says that if $X,X_{1},\dots,X_{k},Y,Y_{1},\dots,Y_{k}$ are
Banach spaces with $X^{*}=Y$ and $X_{i}^{*}=Y_{i}$, and if
$Y\neq\cap_{i}Y_{i}$, then $X\neq\sum_{i}X_{i}$. We apply this result first
with $X=H^{1}$, $X_{i}=H^{1}_{i}$, $Y=BMO$ and $Y_{i}=BMO_{i}$ to conclude
that

$\sum_{i:n_{i}\in S}H^{1}_{n_{i}}\neq H^{1}.$

Note that the functional analysis result also holds under the following
modification: if $Y\neq\sum_{i}Y_{i}$, then $X\neq\cap_{i}X_{i}$. We then
apply this functional analysis result for a second time with $X=VMO$,
$X_{i}=VMO_{i}$, $Y=H^{1}$, and $Y_{i}=H^{1}_{i}$ to conclude that

$\sum_{i:n_{i}\in S}VMO_{n_{i}}\neq VMO.$

These results hold on $\mathbb{R}$ and $\mathbb{T}$ since duality is the same
in both cases, and we gave a compact version of our measure, originally
defined on $\mathbb{R}$, in Section 3. ∎

##  5\. Appendix: Description of necessity of new construction and
intermediate lemmas

###  5.1. Number theory behind need for a new construction in Section 3

Here we show, given points $k_{i}/p_{i}^{m_{i}}$ for primes
$p_{1},\ldots,p_{n}$ satisfying the criterion established in [2], there is
only a finite number of $q$-adic points ${j}/{q^{n}}$ such that

$\frac{k_{i}}{p_{i}^{m_{i}}}-\frac{j}{q^{n}}=\frac{1}{p_{i}^{m_{i}}q^{n}}$

for all $i.$ In [2], it was shown that there are infinitely many $q$-adic
points $\frac{j}{q^{n}}$ satisfying the inequality above for a single prime
$p_{i}$, which guaranteed the existence of at least one $q$-adic interval
sitting completely inside of the chosen $p$-adic interval. As the existence of
such a $q$-adic interval was necessary for the construction of the measure in
[2], the number theory process developed in [2] and [4] would be difficult to
generalize to more than two bases.

The main criterion for interval selection in [2] is that inside any interval
$\widetilde{J}$ and a chosen $\varepsilon>0$ we can choose a $q$-adic interval
$I\subset\widetilde{J}$ such that $0<Y(J)-Z(I)<\varepsilon|I|$ where $J$ is
the smallest $p$-adic interval containing $I.$ Now, in order to generalize the
construction of this measure for finitely many primes (and then finitely many
natural numbers altogether) we want to be able to select a $q$-adic interval
$I$ inside any $\widetilde{J}$ such that $0<Y(J_{i})-Z(I)<\varepsilon|I|$ with
$J_{i}$ denoting the largest $p_{i}$-adic interval containing $I$ for any
collection of primes $p_{i}>q.$ In Proposition 3.5 in [2], the method for
constructing these intervals is to start with a $p$-adic interval $J$ defined
by $Y(J)=\frac{k}{p^{m}}$ and $k\equiv q^{l(p-1)}\mod p^{m}.$ The authors
assume $k\equiv q^{l(p-1)}\mod p^{m}$ for the convenience of applying their
number theoretic results. From there, they construct an infinite sequence of
$n$ and a corresponding $j$ for each $n$ such that

$\frac{k}{p^{m}}-\frac{j}{q^{n}}=\frac{1}{p^{m}q^{n}}$

so by choosing $\frac{1}{p^{m}}<\varepsilon q$ and $n$ they get that

$\left[\frac{j-q+1}{q^{n}},\frac{j+1}{q^{n}}\right)\subseteq\left[\frac{k-1}{p^{m}},\frac{k-1+p}{p^{m}}\right).$

The difficulty in extending this number theoretic procedure is then due to the
need to find infinitely many solutions $n$ to a system of equations. That is,
if given

$Y(J_{1})=\frac{k_{1}}{p_{1}^{m_{1}}}\text{ and
}Y(J_{2})=\frac{k_{2}}{p_{2}^{m_{2}}}$

with $k_{1}\equiv q^{l_{1}(p_{1}-1)(p_{2}-1)}\mod p_{1}^{m_{1}}$ and
$k_{2}\equiv q^{l_{2}(p_{1}-1)(p_{2}-1)}\mod p_{2}^{m_{2}}$ (the proper
analogue to the $k\equiv q^{l(p-1)}$ criterion) then if

(5.1)
$\frac{k_{1}}{p_{1}^{m_{1}}}-\frac{j}{q^{n}}=\frac{1}{p_{1}^{m_{1}}q^{n}}$

and

(5.2)
$\frac{k_{2}}{p_{2}^{m_{2}}}-\frac{j}{q^{n}}=\frac{1}{p_{2}^{m_{2}}q^{n}}$

solving for $j$ gives

$\frac{k_{2}q^{n}-1}{p_{2}^{m_{2}}}=\frac{k_{1}q^{n}-1}{p_{1}^{m_{1}}}$

so

$\left(\frac{k_{2}}{p_{2}^{m_{2}}}-\frac{k_{1}}{p_{1}^{m_{1}}}\right)q^{n}=\frac{1}{p_{2}^{m_{2}}}-\frac{1}{p_{1}^{m_{1}}}$

and thus

$n=\log_{q}\left(\frac{p_{1}^{m_{1}}-p_{2}^{m_{2}}}{k_{2}p_{1}^{m_{1}}-k_{1}p_{2}^{m_{2}}}\right).$

Since $q,p_{1},p_{2}$ are fixed primes, and the remaining variables on the
right-hand side of the equation above are fixed by the intervals $J_{1}$ and
$J_{2}$, there is at most one integer solution $n$ satisfying the equations
5.1 and 5.2. From our discussion above, we see that the argument in [2] cannot
be directly extended to more than two prime bases.

###  5.2. Topological lemmas

######  Proposition 5.1.

Let $\mathbb{T}^{\infty}:=\mathbb{T}^{\omega}$ denote the countable direct
product of the torus $\mathbb{R}/\mathbb{Z}.$ Then $\mathbb{T}^{\omega}$ is
metrizable and in particular the induced product topology is equivalent to the
topology generated by the metric

$d(x,y)=\sum_{k=1}^{\infty}\frac{|x_{k}-y_{k}|}{2^{k}}.$

###### Proof.

This metric is likely well-known – we found it in [19]. There the fact that
the open balls from this metric are equivalent to the product topology on
$\mathbb{T}^{\omega}$ is presented without proof. For completeness, we give a
proof.

First, let $x\in\mathbb{T}^{\omega}$ and $\varepsilon>0$ and take the open
ball $B_{\varepsilon}(x)$ under the metric topology. We show that we can find
some open set under the product topology which fits inside
$B_{\varepsilon}(x).$, by taking the open intervals inside the first $n$ tori
such that $\frac{n+1}{2^{n}}<\varepsilon$ with the open ball
$|z_{i}-x_{i}|<\frac{2^{i}\varepsilon}{n+1}.$

Next, take some open set under the product topology. These are going to be a
countable union of finite products of open sets as these form a sub-basis of
the product topology. Hence, it is sufficient to consider finite products of
open sets. That is, we want to show that given some $U=U_{i_{1}}\times\cdots
U_{i_{k}}$ where $U_{i_{j}}$ is an open set of the $i_{j}^{\text{th}}$ copy of
$\mathbb{T}$ we can find an open ball $B_{\varepsilon}(x)\subseteq U.$ Suppose
without loss of generality that
$U=B_{\varepsilon_{1}}(x_{1})\times\cdots\times B_{\varepsilon_{k}}(x_{k}).$
Then, choose $y\in U$ and $\delta>0$ small enough such that if
$|z_{i}-y_{i}|<\frac{2^{i}\delta}{k+1}$ then $|z_{i}-x_{i}|<\varepsilon_{i}$
for all $i.$ Hence, $B_{\delta}(y)\subseteq U.$ ∎

######  Proposition 5.2.

Suppose $S\subseteq\mathbb{T}^{\omega}$ has the property that $0$ is a limit
point of $S$ inside any finite torus. Then, $0\in\overline{S}.$

###### Proof.

Consider $B_{\varepsilon}(0),$ the open ball of radius $\varepsilon$ under
$d.$ We show that there are infinitely many points of $S$ inside
$B_{\varepsilon}(0).$

Take $n$ such that $\frac{n+1}{2^{n}}<\varepsilon.$ Then, by the hypothesis
there exists infinitely many points of $z\in S$ such that
$|z_{i}|<2^{i}\frac{\varepsilon}{n+1}$ for $1\leq i\leq n$ as $S$ is dense in
any finite torus. Thus, for the chosen $z$ we get

$d(z,0)<\sum_{i=1}^{n}2^{i}\frac{\varepsilon}{2^{i}(n+1)}+\sum_{i=n+1}^{\infty}\frac{|z_{i}|}{2^{i}}<n\frac{\varepsilon}{n+1}+\sum_{i=n+1}^{\infty}\frac{|z_{i}|}{2^{i}}$
$<n\frac{\varepsilon}{n+1}+\frac{1}{2^{n}}<n\frac{\varepsilon}{n+1}+\frac{\varepsilon}{n+1}=\varepsilon.$

Hence we have infinitely many points inside this ball so therefore we have
that $0$ is a limit point of $S$ inside the infinite torus. ∎

######  Remark 5.3.

In effect all we are using in these two lemmas is the existence of a metric.
We do not need to explicitly use the one given, only the triangle inequality.

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