# Subsets of virtually nilpotent groups with the SBM property

Ryan Burkhart and Isaac Goldbring  Department of Mathematics
University of California, Irvine, 340 Rowland Hall (Bldg.# 400), Irvine, CA
92697-3875<EMAIL_ADDRESS>http://www.math.uci.edu/ isaac

###### Abstract.

We extend Leth’s notion of subsets of the integers satisfying the Standard
interval measure (SIM) property to the class of virtually nilpotent groups and
name the corresponding property the Standard ball measure (SBM) property. In
order to do this, we define a natural measure on closed balls in asymptotic
cones associated to such groups and show that this measure satisfies the
Lebesgue density theorem. We then prove analogs of various properties known to
hold for SIM sets in this broader context, occasionally assuming extra
properties of the group, such as the small spheres property and the small gaps
property.

The results of this paper are taken from Burkhart’s doctoral dissertation,
written under Goldbring’s supervision.

Goldbring was partially supported by NSF grant DMS-2054477.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Some nonstandard analysis
    2. 2.2 Growth rates for finitely generated groups
    3. 2.3 Asymptotic cones for finitely generated groups
  3. 3 A measure on the unit ball of asymptotic cones
    1. 3.1 Introducing the measure
    2. 3.2 The Lebesgue Density Property
    3. 3.3 Consequences of the small sphere property
  4. 4 The BM property for internal sets
    1. 4.1 The Ball Measure Property
    2. 4.2 Properties of BM sets
  5. 5 The SBM property for standard sets
    1. 5.1 SBM Sets and Their Properties
    2. 5.2 Supra-SBM Sets
    3. 5.3 Musings on dependence on generators

##  1\. Introduction

In the 1980s, Steven Leth introduced a property of subsets of the natural
numbers that he called the SIM (short for standard interval-measure) property
[12]. The property is motivated by ideas from nonstandard analysis and,
roughly speaking, demands that there be a connection between the _internal_
notion of small gap sizes on hyperfinite intervals and the _external_ notion
of the (normalized) standard part of the set having large Lebesgue measure.
More precisely, given an infinite, hyperfinite interval
$I=[y,z]\subseteq{}^{*}\mathbb{N}$, one can consider the map
$\operatorname{st}_{I}:I\to[0,1]$ given by
$\operatorname{st}_{I}(x):=\operatorname{st}\left(\frac{x-y}{z-y}\right)$. It
can be shown that if an internal subset $A$ of $I$ is such that
$\operatorname{st}_{I}(A)$ has large Lebesgue measure, then all gaps of $A$ on
$I$ have small size (that is, small internal cardinality) compared to the size
of $I$. The SIM property is motivated by asking that the converse relation
hold, uniformly over all infinite, hyperfinite subintervals of $I$.

In the same paper, Leth showed that subsets of the natural numbers with the
SIM property satisfy the conclusion of a theorem of Steward and Tijdeman [19]
known to hold for sets with positive Banach density: if
$A_{1},\ldots,A_{n}\subseteq\mathbb{N}$ have the SIM property, then
$D(A_{1})\cap\cdots\cap D(A_{n})$ is syndetic, where, for any subset
$A\subseteq\mathbb{N}$, $D(A)$ is the set of those $n\in\mathbb{N}$ which may
be written as $a_{i}-a_{j}$ for infinitely many pairs $(a_{i},a_{j})\in
A^{2}$.

In joint work of Leth and the second author, further properties of SIM sets
were established. For example, Leth had shown that being a SIM set is not
simply a notion of size and it became apparent that it would be sensible to
consider the class of supra-SIM sets, that is, the class of sets containing
SIM sets. In [7], it was shown that the class of supra-SIM sets is a partition
regular class, whence is combinatorially natural. Moreover, they were able to
prove a version of Jin’s theorem for sumsets of SIM sets, namely the sum of
any two SIM sets is piecewise syndetic [10] (Jin’s theorem had the same
conclusion but with $A$ and $B$ assumed to have positive Banach density), as
well as a version of Nathanson’s theorem from [14], namely if $A$ is SIM, then
for any $n\in\mathbb{N}$, there are $B,C\subseteq\mathbb{N}$ with $B$ infinite
and $|C|=n$ for which $B+C\subseteq A$. (Nathanson’s result had the same
conclusion under the assumption that $A$ had positive Banach density.)
Nathanson’s theorem was partial progress on a sumset conjecture of Erdos,
which asked if $A$ has positive lower density, must $A$ contain the sum of two
infinite sets? This conjecture was recently resolved in the affirmative, under
the weaker assumption of positive Banach density, by Donaldson, Moreira, and
Richter [13]. Whether or not the theorem holds under the assumption that $A$
is simply a SIM set is an interesting open question.

It is relatively routine to generalize the definition of SIM set from subsets
of $\mathbb{N}$ to subsets of $\mathbb{Z}$. By replacing intervals by cubes,
one can easily generalize the definition further to subsets of
$\mathbb{Z}^{n}$ and thus to subsets of any finitely generated abelian group.
However, generalizing the definition beyond the finitely generated abelian
case is not entirely clear as many of the arguments rely heavily on the
“geometry” of intervals and cubes.

One idea of how to expand the class of groups to which this concept is
meaningful is to replace cubes by balls in the Cayley graph (with respect to
some given finite set of generators). The natural quotient map from an
infinite, hyperfinite ball to the asymptotic cone of the group with respect to
the given set of generators and some infinite hypernatural number is then a
suitable replacement for the map $\operatorname{st}_{I}$ considered above.
However, many of the results mentioned above rely heavily on the Lebesgue
density theorem for Lebesgue measure on $[0,1]$. Since the Lebesgue measure is
the pushforward of the Loeb measure associated to the hyperfinite counting
measure on $I$ via $\operatorname{st}_{I}$, it is thus natural in our extended
setting to ask that the Lebesgue density theorem hold for the pushforward of
the Loeb measure associated to the hyperfinite counting measure on infinite,
hyperfinite balls in the Cayley graph. We show that, assuming that the group
has polynomial growth, this pushforward measure does indeed satisfy the
Lebesgue density theorem. By Gromov’s celebrated theorem [8], the assumption
of polynomial growth is equivalent to the assumption that the group is
virtually nilpotent.

In this paper, we show how to extend the definition of SIM sets to the class
of finitely generated virtually nilpotent groups, obtaining a class of sets
that we call SBM sets, where SBM is short for the standard ball measure
property. However, to obtain results that parallel the results discussed
above, we sometimes need to assume that our groups satisfy a property we call
the small sphere property, a property known to hold in all nilpotent groups of
class 2 and conjectured to hold for all nilpotent groups. For some of our
results, a third property, which we call the small gap property, is also
assumed. With these assumptions in hand, we prove direct analogs of the
aformentioned results known to hold for SIM sets (with the SBM analog of Jin’s
theorem being the only exception).

While it is satisfying to extend the class of SIM sets to a fairly large class
of nonabelian groups, it would be desirable to extend the notion even further.
In particular, it would be interesting to see if this notion could be suitably
extended to the class of amenable groups, perhaps using well-behaved Folner
sequences as replacements for balls in the Cayley graph. Ideally one could
even extend the notion past amenable groups. For example, how to extend the
notion to finitely generated free groups is also not clear to us at present.

We assume the reader is familiar with the basics of nonstandard analysis. The
monograph [5] contains a complete introduction and is aimed at an audience
interested in combinatorial applications. Many of the proofs appearing later
in this paper are based on the arguments given in Chapter 16 of that
monograph. Nevertheless, some basic notation and information on certain
nonstandard constructions, such as hyperfinite products in groups and the Loeb
measure construction, are given in the next section.

##  2\. Preliminaries

###  2.1. Some nonstandard analysis

In this subsection, we only recall a few facts from nonstandard analysis that
we use throughout the paper. We first remind the reader that, for any set $X$,
letting $\mathcal{P}(X)$ denote the powerset of $X$, we can naturally identify
the nonstandard extension ${}^{*}\mathcal{P}(X)$ of $\mathcal{P}(X)$ with a
subset of $\mathcal{P}(X^{*})$; in this way, elements of
${}^{*}\mathcal{P}(X)$ will be called internal subsets of ${}^{*}X$; subsets
of ${}^{*}X$ that are not internal are called external. In a similar way, if
$\mathcal{P}_{f}(X)$ denotes the finite subsets of $X$, then elements of
${}^{*}\mathcal{P}_{f}(X)$ are special internal subsets of ${}^{*}X$ called
hyperfinite subsets of ${}^{*}X$. By transfer, for each hyperfinite subset $A$
of ${}^{*}X$, there is a unique element $N$ of ${}^{*}\mathbb{N}$ for which
there is an internal bijection $f:A\to[1,N]:=\\{M\in{}^{*}\mathbb{N}\ :\ 1\leq
M\leq N\\}$; we refer to this $N$ as the internal cardinality of $A$, denoted
by $|A|$. Note also that, by transfer, an internal subset of a hyperfinite set
is once again hyperfinite.

Given any hyperfinite set $A$, there is a natural finitely-additive
probability measure $\mu$ on the algebra (but not $\sigma$-algebra!) of
internal subsets of $A$ given by
$\mu(B):=\operatorname{st}\left(\frac{|B|}{|A|}\right)$. It is a standard fact
that the hypotheses of the Carathéodory extension theorem hold in this
context, whence we get an extension of $\mu$, denoted $\mu_{L}$ and called the
Loeb measure, defined on a canonical $\sigma$-algebra of subsets of $A$
containing the internal subsets, called the $\sigma$-algebra of Loeb
measurable sets. An example to keep in mind is the following: if
$N\in{}^{*}\mathbb{N}\setminus\mathbb{N}$ and
$A=\\{0,\frac{1}{N},\ldots,\frac{N-1}{N},1\\}$ and
$\operatorname{st}:A\to[0,1]$ is the usual standard part map, then the
pushforward of the Loeb measure space on $A$ via $\operatorname{st}$ is the
usual Lebesgue measure on $[0,1]$.

We also remind the reader of the notion of hyperfinite products in groups.
Suppose that $G$ is an arbitrary group. Let $G_{fs}$ denote the set of finite
sequences from $G$ and let $p:G_{fs}\to G$ denote the product map
$p((g_{1},\ldots,g_{n})):=g_{1}\cdots g_{n}$. We can thus consider the
nonstandard extension of ${}^{*}G_{fs}$, which is the set of hyperfinite
sequences from ${}^{*}G$; we suggestively write elements of ${}^{*}G_{fs}$ as
$(g_{1},\ldots,g_{N})$ for some element $N\in{}^{*}\mathbb{N}$. Similarly, we
consider the nonstandard extension $p:{}^{*}G_{fs}\to{}^{*}G$ of $p$ and
suggestively write $g_{1}\cdots g_{N}$ instead of $p((g_{1},\ldots,g_{N}))$.

###  2.2. Growth rates for finitely generated groups

From now on, throughout this paper, $G$ will always denote a finitely
generated group with finite generating set $S$. For simplicity, we will always
assume that our generating set $S$ is symmetric, that is, contains the
identity and is closed under inverses.

The word metric on $G$ with respect to $S$ is the metric

$d_{S}(x,y)=\min\\{n\in{\mathbb{N}}:x^{-1}y=s_{1}s_{2}\cdot\cdot\cdot
s_{n}\text{ for some }s_{1},...,s_{n}\in S\\}.$

Note that $d_{S}$ is left-invariant, that is, for all $x,y,g\in G$,
$d_{S}(x,y)=d_{S}(gx,gy)$. Also notice that, for ${\mathbb{Z}}$ with the
standard symmetric generating set $S=\\{1,-1\\},$ we have that $d_{S}$ is
simply the usual notion of distance in ${\mathbb{Z}}$.

For any $g\in G$ and any $n\in{\mathbb{N}}$, define the closed ball of radius
$n$ and center $g$ to be $B(g,n)=\\{h\in G:|g^{-1}h|\leq n\\}$, where we also
allow $B(g,0)=\\{g\\}$. In what follows, we will often omit the word “closed”
and simply refer to this as a “ball’. The open ball of radius $n$ with center
$g$ is then defined as $U(g,n)=\\{h\in G:|g^{-1}h|<n\\}=B(g,n-1)$, allowing us
to define the sphere of radius $n$ with center $g$ by $S(g,n)=B(g,n)\backslash
B(g,n-1)$. Note that we can (and will) also allow real (instead of natural)
number radii, noting that spheres with non-integer radii are empty.

For $g\in G$, set the word length of $g$ with respect to $S$ to be
$|g|_{S}=d_{S}(g,e)$, where $e$ denotes the identity of the group. For
$n\in{\mathbb{N}}$, set

$\alpha_{n}=\alpha_{n,S}=|\\{g\in G:|g|_{S}\leq n\\}|=|B(g,e)|.$

Note that we have the trivial upper bound $\alpha_{n}\leq|S|^{n}$. However, in
many groups, $\alpha_{n}$ will grow much more slowly than this exponential
upper bound. For example, if $G$ is abelian, then it is readily verified that
$\alpha_{n}<(n+1)^{|S|}$ for large enough $n$. This is an example of the
following phenomenon:

######  Definition 2.1.

We say that $G$ has polynomial growth if there is $d\in\mathbb{N}$ and $C>0$
such that $\alpha_{n}\leq Cn^{d}$ for all $n\in\mathbb{N}$. The degree of
polynomial growth of a group is defined to be the minimum degree $d$ for which
the above inequality holds for some positive constant $C$.

Since any two word metrics on $G$ are bi-Lipschitz equivalent, the notion of
polynomial growth is indeed independent of the choice of generating set.

In many cases, it will be useful to have both upper and lower bounds for
$\alpha_{n}$.

######  Theorem 2.2.

[1] If $G$ has polynomial growth with degree $d$, then there exists a constant
$c\in{\mathbb{N}}$ such that $\frac{1}{c}n^{d}\leq\alpha_{n}\leq cn^{d}$ for
all $n\in{\mathbb{N}}$.

It was shown by Wolf in [21] that all virtually nilpotent groups are of
polynomial growth. Thirteen years later, Gromov proved the converse in [8],
thereby completely characterizing groups with polynomial growth.

######  Theorem 2.3 (Gromov [8]).

The finitely generated groups with polynomial growth are precisely the
finitely generated groups which are virtually nilpotent.

By Gromov’s theorem, in the sequel, we use the terms polynomial growth and
virtually nilpotent interchangeably.

We extend the above notions to the nonstandard setting as follows. The metric
$d_{S}$ on $G$ extends to an internal metric, also denoted $d_{S}$, on
${}^{*}G$, which, by transfer, satisfies, for all $x,y\in^{*}G$:

$d_{S}(x,y)=\min\\{N\in\text{ }^{*}{\mathbb{N}}:x^{-1}y=s_{1}s_{2}\cdots
s_{N}\text{ for some }s_{1},...,s_{N}\in S\\}.$

Here, $(s_{1},s_{2},\ldots,s_{N})$ ranges over hyperfinite sequences from $S$
and $s_{1}s_{2}\cdots s_{N}$ is the corresponding hyperfinite product, as
introduced in the previous subsection. Similarly, for any $g\in{}^{*}G$ and
any $R\in{}^{*}\mathbb{R}^{>0}$, we have notions of internally closed balls
$B(g,R)$, internally open balls $U(g,R)$, and internal spheres $S(g,R)$.

We note for future reference one nonstandard consequence of the polynomial
growth assumption.

######  Lemma 2.4.

Suppose that $G$ has polynomial growth. Let $M\leq N$ be infinite elements of
${}^{*}{\mathbb{N}}$. Then $\frac{\alpha_{M}}{\alpha_{N}}\approx
0\iff\frac{M}{N}\approx 0$.

###### Proof.

Let $d$ be the degree of polynomial growth of $G$, as witnessed by the
constant $c$. First assume that $\frac{M}{N}\approx 0$. Then since
$\alpha_{M}\leq cM^{d}$ and $\alpha_{N}\geq\frac{1}{c}N^{d}$, we get that
$\frac{\alpha_{M}}{\alpha_{N}}\leq\frac{c^{2}M^{d}}{N^{d}}=c^{2}(\frac{M}{N})^{d}$,
which is infinitesimal.

For the other direction, assume $\frac{\alpha_{M}}{\alpha_{N}}\approx 0$.
Since $\alpha_{M}\geq\frac{1}{c}M^{d}$ and $\alpha_{N}\leq cN^{d}$, this gives
us that $\frac{M^{d}}{c^{2}N^{d}}\leq\frac{\alpha_{M}}{\alpha_{N}}\approx 0$.
Since $\frac{1}{c^{2}}$ is an element of ${\mathbb{R}}$, this means
$(\frac{M}{N})^{d}\approx 0$ and thus $\frac{M}{N}\approx 0$. ∎

###  2.3. Asymptotic cones for finitely generated groups

In this section, we recall the asymptotic cone construction for finitely
generated groups, using the approach pioneered by van den Dries and Wilkie [4]
via nonstandard analysis. (While asymptotic cones can be defined for any
metric space, we will only need the construction for finitely generated groups
equipped with their word metric.)

Throughout this subsection, we fix a finitely generated group $G$ with finite
(symmetric) generating set $S$. Given $N\in$
${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$, we define the relation $\sim_{N}$
on ${}^{*}G$ by setting $x\sim_{N}y$ if and only if $\frac{d_{S}(x,y)}{N}$ is
infinitesimal. The following properties of $\sim_{N}$ are routine to verify:

######  Lemma 2.5.

Fix $M,N\in{}^{*}\mathbb{N}\setminus\mathbb{N}$ and $x,y,z\in{}^{*}G$.

  1. (1)

$\sim_{N}$ is an equivalence relation on ${}^{*}G$.

  2. (2)

Let $S^{\prime}$ be another finite generating set for $G$, with corresponding
word metric $d_{S^{\prime}}$ and equivalence relation $\sim_{N}^{\prime}$.
Then $\sim_{N}$ and $\sim_{N}^{\prime}$ are the same relation on ${}^{*}G$.

  3. (3)

If $\frac{M}{N}$ is neither infinite nor infinitesimal, than $\sim_{M}$ and
$\sim_{N}$ are the same equivalence relation.

  4. (4)

If $x\sim_{N}y$, then $\frac{|d_{S}(x,z)-d_{S}(y,z)|}{N}$ is infinitesimal.

  5. (5)

We have $x\sim_{N}y\iff y\in B(x,M)$ for some $M\in$ ${}^{*}{\mathbb{N}}$ such
that $\frac{M}{N}\approx 0$.

For $x\in^{*}G$ and $N\in{}^{*}\mathbb{N}\setminus\mathbb{N}$, we let
$\operatorname{eq}_{N}(x)$ denote the equivalence class of $x$ with respect to
$\sim_{N}$. Items (2) and (3) in Lemma 2.5 imply that these equivalence
classes are independent of the choice of generating set for $G$ and the choice
of representative for the archimedean class of $N$. Given $E\subseteq G$, we
set $\operatorname{eq}_{N}(E)=\\{\operatorname{eq}_{N}(x)\ :\ x\in E\\}$ for
the set of equivalence classes of elements of $E$.

In the sequel, to simplify notation, we often overline a variable name to
indicate that it is an equivalence class and similarly for sets of equivalence
classes. Consequently, for $\overline{x}\in\operatorname{eq}_{N}(^{*}G)$, we
have that $\overline{x}=eq^{-1}_{N}(\overline{x})=\\{g\in$ ${}^{*}G$ :
$\operatorname{eq}_{N}(g)=\overline{x}\\}$. Implicit in this notation is that
$x$ itself is a representative of the equivalence class $\overline{x}$, that
is, $\operatorname{eq}_{N}(x)=\overline{x}$. (One drawback of this notation is
that we suppress mention of $N$; we thus only use this notation when $N$ has
been fixed ahead of time.)

######  Definition 2.6.

The asymptotic cone $C(G,g,N)$ of $G$ with base point $g\in$ ${}^{*}G$ and
scaling factor $N\in$ ${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$ is defined as

$\left\\{\operatorname{eq}_{N}(x):x\in{}^{*}G\text{ and
}\frac{d_{S}(x,g)}{N}\text{ is finite}\right\\}.$

By Lemma 2.5(4), the definition above is indeed independent of choice of
representative. The asymptotic cone can be equipped with the metric $d_{N}$
given by

$d_{N}\left(\operatorname{eq}_{N}(x),\operatorname{eq}_{N}(y)\right)=\operatorname{st}\left(\frac{d_{S}(x,y)}{N}\right).$

Note that this standard part is indeed defined as
$\frac{d_{S}(x,y)}{N}\leq\frac{d_{S}(x,g)}{N}+\frac{d_{S}(y,g)}{N}$ and both
terms on the right hand side of this inequality are finite by assumption.
Furthermore, Lemma 2.5(4) implies that this definition is indeed independent
of the choice of representatives for equivalence classes.

An easy saturation argument shows that $C(G,g,N)$ is a complete metric space.
Moreover, the asymptotic cone is also a geodesic metric space. (See [6] for
proofs of both of these facts.)

We have obscured mention of the generating set $S$ in the above notation. The
reason for that is the fact that any two word metrics for $G$ are bi-Lipschitz
equivalent remains true for the asymptotic cones with respect to these
generating sets (using the same base point and scaling factor $N$), whence the
construction is essentially independent of the choice of generator. The map

$\operatorname{eq}_{N}(x)\mapsto\operatorname{eq}_{N}(gx):C(G,e,N)\to
C(G,g,N)$

is an isometry, whence the asymptotic cone (with respect to a given $N$) is
also essentially independent of base point. That being said, the homeomorphism
type of the asymptotic cone can very much depend on the choice of $N$; see,
for example, [11]. However, for the groups we will be interested in, the
choice of $N$ will be irrelevant.

######  Example 2.7.

All asymptotic cones of $\mathbb{Z}$ with respect to the generating set
$\\{1,-1\\}$ are isometric to $\mathbb{R}$ with its usual distance. Indeed,
working for simplicity with the base point $0$, we see that the map

$\operatorname{eq}_{N}(x)\mapsto\operatorname{st}(\frac{x}{N}):C(\mathbb{Z},0,N)\to\mathbb{R}$

is an isometry.

The result in the previous example is a special case of a more general result
of Pansu [15], namely that when $G$ is virtually nilpotent, the isometry type
of the asymptotic cones of $G$ (with respect to a given generating set) is
independent of the choice of $N\in{}^{*}\mathbb{N}\setminus\mathbb{N}$.

The following theorem will be of great use to us going forward:

######  Theorem 2.8.

If $X$ is the metric space of an asymptotic cone defined via a finitely
generated group $G$, then the following are equivalent:

  1. (1)

$X$ is proper (that is, closed balls are compact).

  2. (2)

$X$ is separable.

  3. (3)

$G$ is virtually nilpotent.

In particular, some asymptotic cone of $G$ is proper (resp. separable) if and
only if they all are.

###### Proof sketch.

(1) implies (2) holds for any metric space; the converse was shown by Sisto in
[18].

(1) implies (3) was shown by Sapir [16], building on work by Hrushovski [9].
Conversely, if $G$ is nilpotent, then some asymptotic cone of $G$ is proper by
the work of van den Dries and Wilkie [4]; that all such cones must be proper
follows from the aforementioned result of Pansu [15]. ∎

##  3\. A measure on the unit ball of asymptotic cones

###  3.1. Introducing the measure

Throughout this section, $G$ is a finitely generated group with finite
(symmetric) generating set $S$.

For $N\in$ ${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$ and $g\in$ ${}^{*}G$, we
let $\mu_{g,N}$ denote the Loeb measure on the hyperfinite set $B(g,N)$ and
let ${\mathcal{L}}_{g,N}$ denote the $\mu_{g,N}$-measurable subsets of
$B(g,N)$. Let $\mathcal{A}_{g,N}$ denote the set of subsets $\overline{A}$ of
$\operatorname{eq}_{N}\big{(}B(g,N)\big{)}$ such that
$\overline{A}\in\mathcal{A}_{g,N}$ if and only if
$\operatorname{eq}_{N}^{-1}(\overline{A})\cap B(g,N)\in{\mathcal{L}}_{g,N}$.
For $\overline{A}\in\mathcal{A}_{g,N}$, set
$\lambda_{g,N}(\overline{A})=\mu_{g,N}\big{(}\operatorname{eq}_{N}^{-1}(\overline{A})\cap
B(g,N)\big{)}$. In other words, $\mathcal{A}_{g,N}$ and $\lambda_{g,N}$ are
the pushforwards of ${\mathcal{L}}_{g,N}$ and $\mu_{g,N}$ by
$\operatorname{eq}_{N}$.

Suitably adapting the example from Subsection 2.1, letting $G={\mathbb{Z}}$,
$g=0$, and $N$ be any element of ${}^{*}{\mathbb{N}}\setminus\mathbb{N}$,
$\lambda_{g,N}$ is the Lebesgue measure on $[-1,1]$ and $\mathcal{A}_{g,N}$ is
the collection of Lebesgue measurable subsets of $[-1,1]$.

Note that, in the metric space $(C(G,g,N),d_{N})$,
$\operatorname{eq}_{N}\big{(}B(g,N)\big{)}$ is the closed ball of radius $1$
centered at the equivalence class $\overline{g}=\operatorname{eq}_{N}(g)$. In
what follows, we will write $B_{N}(\overline{g},r)$ to refer to
$\operatorname{eq}_{N}\big{(}B(g,rN)\big{)}$, which is the closed ball of
radius $r\in{\mathbb{R}}$ with center $\overline{g}$ in the asymptotic cone
$C(G,g,N)$. Note that this is independent of the choice of representative of
$\overline{g}$. We also consider the open ball of radius $r$ and center
$\overline{g}$, denoted $U_{N}(\overline{g},r)$, and note that
$U_{N}(\overline{g},r)=\bigcup_{n\in{\mathbb{N}}}\operatorname{eq}_{N}\left(B\big{(}g,r(1-\frac{1}{n})N\big{)}\right)$.
Finally, we also consider the sphere
$S_{N}(\overline{g},r)=B_{N}(\overline{g},r)\backslash U_{N}(\overline{g},r)$.

######  Proposition 3.1.

For any $g\in$ ${}^{*}G$ and any $N\in$
${}^{*}{\mathbb{N}}\setminus\mathbb{N}$, if $A\subseteq$ ${}^{*}G$ is
internal, then $\operatorname{eq}_{N}\big{(}A\cap B(g,N)\big{)}$ is
$\lambda_{g,N}$-measurable.

###### Proof.

Set $E=A\cap B(g,N)$. It suffices to establish that

$\operatorname{eq}_{N}^{-1}\left(\operatorname{eq}_{N}(E)\right)=\bigcap_{n=1}^{\infty}\left\\{x\in
B(g,N):x\in B\left(y,\frac{N}{n}\right)\text{ for some }y\in E\right\\}.$

Towards this end, first assume
$x\in\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}(E)\big{)}$. Then
there is $y\in E$ such that $x\in B(y,M)$ for some $M\in^{*}\mathbb{N}$ with
$\frac{M}{N}\approx 0$. Thus $x\in B(y,\frac{N}{n})$ for all
$n\in{\mathbb{N}}$, as desired.

Conversely, assume that $x$ is in the intersection on the right side of the
above display. Then for any $n\in{\mathbb{N}}$, there is $a_{n}\in E$ such
that $x\in B(a_{n},\frac{N}{n})$. Note that $\\{M\in$ ${}^{*}{\mathbb{N}}:x\in
B(a,M)$ for some $a\in E\\}$ is internal since $E$ is internal, and thus has a
minimum $M^{\prime}$. By the above observation, $M^{\prime}<\frac{N}{n}$ for
all $n\in{\mathbb{N}}$, meaning $\frac{M^{\prime}}{N}\approx 0$. Taking $a\in
E$ such that $x\in B(a,M^{\prime})$, we see that
$\operatorname{eq}_{N}(x)=\operatorname{eq}_{N}(a)$ and thus
$x\in\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}(E)\big{)}$. ∎

######  Corollary 3.2.

If $B_{N}(\overline{g},1)$ is separable (that is, if $G$ is virtually
nilpotent), then the Borel subsets of $B_{N}(\overline{g},1)$ are
$\lambda_{g,N}$-measurable.

###### Proof.

First note that, since $B_{N}(\overline{g},1)$ is separable, the Borel subsets
form the smallest $\sigma$-algebra containing all open balls, whence it
suffices to show that open balls are $\lambda_{g,N}$-measurable. By
Proposition 3.1, the closed ball $B_{N}(\overline{h},r)$ is
$\lambda_{g,N}$-measurable as
$B_{N}(\overline{h},r)=\operatorname{eq}_{N}(B(h,rN)\cap B(g,N))$. Using the
fact that
$U_{N}(\overline{h},r)=\bigcup_{n\in{\mathbb{N}}}B_{N}\big{(}\overline{h},(1-\frac{1}{n})r\big{)}$,
we see that all open balls are $\lambda_{g,N}$-measurable as well. ∎

From now on, we will restrict ourselves to the virtually nilpotent case and
will only be concerned with the behavior of $\lambda_{g,N}$ on the Borel
subsets of $B_{N}(\overline{g},1)$; we consequently restrict our attention to
that a priori smaller $\sigma$-algebra.

###  3.2. The Lebesgue Density Property

One property that will come up frequently in our later proofs is whether or
not $\lambda_{g,N}$ satisfies a version of the Lebesgue density theorem, which
we formulate in the following way.

######  Definition 3.3.

Let $\mu$ be an outer measure on the metric space $X$. We say that $(X,\mu)$
satisfies the Lebesgue Density Theorem (LDT) if, for all $\mu$-measurable
$A\subseteq X$, we have that

$\lim_{r\to 0^{+}}\frac{\mu\big{(}A\cap
B(x,r)\big{)}}{\mu\big{(}B(x,r)\big{)}}=1$

for $\mu$-almost all $x$ in $A$.

In order to put ourselves in the context of the previous definition, we define
the outer measure $\lambda^{0}_{g,N}$ on $B_{N}(\overline{g},1)$ by

$\lambda^{0}_{g,N}(A)=\inf\\{\lambda_{g,N}(B):A\subseteq B,B\text{ a Borel
set}\\}.$

This is an outer measure which agrees with $\lambda_{g,N}$ on Borel sets and
whose measurable sets include the Borel sets.

The goal of this subsection will be to establish the following theorem:

######  Theorem 3.4.

Let $G$ be a finitely generated virtually nilpotent group. Then for all
infinite $N\in$ ${}^{*}{\mathbb{N}}$ and all $g\in$ ${}^{*}G$, we have that
$\lambda_{g,N}^{0}$ satisfies the Lebesgue Density Theorem.

If we let $G={\mathbb{Z}}$ with the standard generating set, we have already
discussed the fact that, for any $g\in{}^{*}{\mathbb{Z}}$, $\lambda_{g,N}$ is
the Lebesgue measure on $B_{N}(\overline{g},1)\cong[-1,1]$. Thus, the previous
theorem generalizes the basic fact that the Lebesgue measure on $[-1,1]$
satisfies the conclusion of the Lebesgue Density Theorem.

Our proof of Theorem 3.4 will use a more general result about the validity of
the LDT for outer measures satsisfying a certain list of properties.

######  Definition 3.5.

Suppose that $\mu$ is an outer measure on the metric space $X$. We say that
$\mu$:

  1. (1)

is Borel regular if all Borel sets are $\mu$-measurable and, for all
$A\subseteq X$, we have that $\mu(A)=\mu(B)$ for some Borel set $B$;

  2. (2)

is open $\sigma$-finite if $X$ can be covered by countably many open sets of
finite measure;

  3. (3)

has the symmetric Vitali property (SVP) if, for any $E\subseteq X$ with
$\mu(E)<\infty$ and any collection $\mathcal{B}$ of closed balls with centers
in $E$ satisfying

$\inf\\{r\in{\mathbb{R}}^{+}:B(x,r)\in\mathcal{B}\\}=0$

for every $x\in E$, there is a countable, pairwise disjoint subcollection
$\mathcal{B}^{\prime}=\\{B_{n}:n\in{\mathbb{N}}\\}\subseteq\mathcal{B}$ of
balls such that $\mu(E\backslash\bigcup_{n\in{\mathbb{N}}}B_{n})=0$;

  4. (4)

has the doubling property if there exists a constant $C$ with
$\mu\big{(}B(x,r)\big{)}\leq C\mu\big{(}B(x,\frac{r}{2})\big{)}$ for all $x\in
X$ and $r\in{\mathbb{R}}^{+}$.

The following lemma appears as [17, Remark 3.11].

######  Lemma 3.6.

If $X$ is separable and $\mu$ is a Borel regular outer measure on X with the
doubling property, then $\mu$ has the Symmetric Vitali Property.

The following is [17, Theorem 3.16]:

######  Theorem 3.7.

Let $\mu$ be an outer measure on the metric space $X$ that is Borel regular,
open $\sigma$-finite, and has the symmetric Vitali property. Then $(X,\mu)$
satisfies the Lebesgue Density Theorem.

We are now ready for the proof of the main result of this subsection:

###### Proof of Theorem 3.4.

The fact that $\lambda_{g,N}^{0}$ is Borel regular and open $\sigma$-finite is
clear. It remains to verify the symmetric Vitali property. Using Lemma 3.6 and
the fact that $B_{N}(\overline{g},1)$ is separable (by the assumption that $G$
is virtually nilpotent), it suffices to establish that $\lambda_{g,N}^{0}$ has
the doubling property.

First recall that, by transfer, there is $c>0$ and $d\in\mathbb{N}$ such that

$\frac{N^{d}}{c}\leq|B(g,N)|\leq cN^{d}.$

Claim: For $\overline{x}\in B_{N}(\overline{g},1)$ and $r\in(0,2)$, we have:

$\lambda_{g,N}\left(B_{N}\big{(}\overline{x},\frac{r}{2}\big{)}\right)\geq\frac{\frac{1}{c}(\frac{r}{16}N)^{d}}{|B(g,N)|}.$

Proof of claim:

There are four cases to consider:

Case 1: Assume $B_{N}(\overline{x},\frac{r}{2})\subseteq
B_{N}(\overline{g},1)$. Then

$\lambda_{g,N}\left(B_{N}\big{(}\overline{x},\frac{r}{2}\big{)}\right)\geq\frac{\big{|}B(x,\frac{r}{4}N)\big{|}}{|B(g,N)|}\geq\frac{\frac{1}{c}(\frac{r}{4}N)^{d}}{|B(g,N)|}\geq\frac{\frac{1}{c}(\frac{r}{16}N)^{d}}{|B(g,N)|}.$

In the remaining three cases, we assume $B_{N}(\overline{x},\frac{r}{2})$ is
not a subset of $B_{N}(\overline{g},1)$ and set
$a:=d_{N}(\overline{x},\overline{g})>1-\frac{r}{2}$.

Case 2: Assume $r\leq 1$. Then, using the fact that the asymptotic cone is a
geodesic metric space, we can take a $\overline{z}$ such that
$d_{N}(\overline{x},\overline{z})=\frac{r}{4}$ and
$d_{N}(\overline{z},\overline{g})=a-\frac{r}{4}>0$, whence
$B_{N}(\overline{z},\frac{r}{4})\subseteq B_{N}(\overline{g},1)\cap
B_{N}(\overline{x},\frac{r}{2})$. Thus, we can use the inequality from the
first case, replacing $r$ with $\frac{r}{2}$, giving us

$\lambda_{g,N}\left(B_{N}\big{(}\overline{x},\frac{r}{2}\big{)}\right)\geq\lambda_{g,N}\left(B_{N}\big{(}\overline{z},\frac{r}{4}\big{)}\right)\geq\frac{\big{|}B(z,\frac{r}{8}N)\big{|}}{|B(g,N)|}\geq\frac{\frac{1}{c}(\frac{r}{16}N)^{d}}{|B(g,N)}.$

In the last last two cases, we assume $r>1$.

Case 3: Assume $a\geq\frac{1}{2}$. Then we can take $\overline{z}$ with
$d_{N}(\overline{x},\overline{z})=\frac{1}{4}$ and
$d_{N}(\overline{z},\overline{g})=a-\frac{1}{4}>0$, whence
$B_{N}(\overline{z},\frac{1}{4})\subseteq B_{N}(\overline{x},\frac{r}{2})\cap
B_{N}(\overline{g},1)$. Thus, we can use the inequality from the first case
with $r=\frac{1}{2}$, giving us

$\lambda_{g,N}\bigg{(}B_{N}\Big{(}\overline{x},\frac{r}{2}\Big{)}\bigg{)}\geq\lambda_{g,N}\left(B_{N}\Big{(}\overline{z},\frac{1}{4}\Big{)}\right)\geq\frac{\big{|}B(z,\frac{1}{8}N)\big{|}}{|B(g,N)|}\geq\frac{\frac{1}{c}(\frac{1}{8}N)^{d}}{|B(g,N)|}\geq\frac{\frac{1}{c}(\frac{r}{16}N)^{d}}{|B(g,N)|}.$

Case 4: $a<\frac{1}{2}$. Then there exists $\overline{z}$ with
$d_{N}(\overline{x},\overline{z})=\frac{1}{4}$, so that
$d_{N}(\overline{z},\overline{g})<\frac{3}{4}$. Thus
$B_{N}(\overline{z},\frac{1}{4})$ is again contained in both of our balls, and
so we can use the same inequality from case three. The claim is now proven.

Next note that, for any $r>0$, we have

$\lambda_{g,N}\big{(}B_{N}(\overline{x},r)\big{)}\leq\frac{|B(x,2rN)|}{|B(g,N)|}\leq\frac{c(2rN)^{d}}{|B(g,N)|}.$

Combining this observation with the claim gives, for all $r\in(0,2)$,

$\lambda_{g,N}\big{(}B_{N}(\overline{x},r)\big{)}\leq\frac{c(2rN)^{d}}{|B(g,N)|}=\frac{c^{2}(32)^{d}\frac{1}{c}(\frac{r}{16}N)^{d}}{|B(g,N)|}\leq
c^{2}(32)^{d}\lambda_{g,N}\left(B_{N}\big{(}\overline{x},\frac{r}{2}\big{)}\right),$

and so the doubling property holds with $C=c^{2}(32)^{d}$ (as
$\lambda_{g,N}^{0}$ agrees with $\lambda_{g,N}$ on Borel sets). ∎

######  Remark 3.8.

Besides showing that there is a well-defined isometry type of asymptotic cones
for a given virtually nilpotent group, Pansu [15] also showed that this
asymptotic cone can be equipped with a natural Lie group structure. (See [2]
for a nice description of this construction.) In particular, these asymptotic
cones possess an (essentially) unique Haar measure; we leave it for future
work to compare the restriction of this Haar measure to the unit ball of the
asymptotic cone and the measure on the unit ball we have been considering in
this paper. (We thank Alessandro Sisto for pointing us to this fact.)

###  3.3. Consequences of the small sphere property

While the assumption of polynomial growth will prove useful going forward (as
has already been evidenced by the validity of the LDT for the measure we have
considered on closed balls in asymptotic cones established in the previous
subsection), it does not seem to be quite powerful enough to prove all of our
results below. Indeed, we will want spheres in asymptotic cones to behave much
as they do in euclidean space, namely that a sphere of radius $r$ of dimension
(that is, degree of polynomial growth) $d$ has size on the order of $r^{d-1}$.
We formalize this desire into the following definition:

######  Definition 3.9.

A finitely generated group $G$ with symmetric generating set $S$ has the Small
Sphere (SS) Property if there exists some constant $c^{\prime}$ such that, for
any $n\in{\mathbb{N}}$ and $g\in G$, we have that $|S(g,n)|\leq
c^{\prime}n^{d-1}$.

As defined, the SS property is a property that may or may not hold for a given
group _with respect to a given generating set_ ; it is unclear to us if this
property is independent of generating set.

The SS property has been shown to hold in nilpotent groups of nilpotency class
2 (such as the discrete Heisenberg group), regardless of the generating set,
by Stoll in [20]. Indeed, Stoll was able to show that this growth rate of
spheres is sharp in this case, that is, there exist nilpotent groups of
nilpotecy class 2 where the size of spheres of radius $n$ is precisely on the
order of $n^{d-1}$.

In [2], Breuillard and Le Donne made progress on finding a growth rate for
spheres in arbitrary virtually nilpotent groups. They proved that for a
nilpotent group of nilpotency class $r$, the size of the spheres of radius $n$
is bounded by a growth function on the order of $n^{d-\beta_{r}}$, where
$\beta_{r}=\frac{2}{3r}$. However, they mention that this growth rate is
unlikely to be sharp. Indeed, it has been conjectured that spheres in all
virtually nilpotent groups satisfy the SS property. Breuillard and Le Donne
refer to this as a “folklore conjecture” and no further progress on this
conjecture seems to have been made at this time.

Nevertheless, we will concern ourselves primarily with groups with the SS
property (with respect to some generating set), as this will allow us to prove
many of the nice properties of SIM subsets of ${\mathbb{Z}}$ in our more
general context. One of the main consequences this assumption gives us is the
following:

######  Lemma 3.10.

Suppose that $G$ is a finitely generated virtually nilpotent group with the SS
property. Fix $N\in$ ${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$ and $g\in$
${}^{*}G$. Then $\lambda_{g,N}\big{(}S_{N}(\overline{x},r)\cap
B_{N}(\overline{g},1)\big{)}=0$ for all $r\in{\mathbb{R}}^{+}$ and all $x\in$
${}^{*}G$.

###### Proof.

For $\delta>0$, let $B_{\delta}=\bigcup_{n\in I_{\delta}}S(x,n)$, where
$I_{\delta}\subseteq$ ${}^{*}{\mathbb{N}}$ is the infinite, hyperfinite
interval
$\big{[}\lfloor(1-\delta)rN\rfloor+1,\lfloor(1+\delta)rN\rfloor\big{]}$. Note
that $B_{\delta}$ is internal and $S_{N}(\overline{x},r)\subseteq B_{\delta}$,
whence $\lambda_{g,N}\big{(}S_{N}(\overline{x},r)\cap
B_{N}(\overline{g},1)\big{)}\leq\frac{|B_{\delta}\cap
B(g,N)|}{|B(x,N)|}\leq\frac{|B_{\delta}|}{|B(x,N)|}$. Assume that $c$ is the
constant witnessing the polynomial growth of $G$ with degree $d$, and
$c^{\prime}$ is the corresponding constant for the SS property. Then we have
that $|I_{\delta}|\leq 2\delta rN$, and $|S(x,M)|\leq
c^{\prime}\big{(}(1+\delta)rN\big{)}^{d-1}$ for all $M\in I_{\delta}$. Thus,
$|B_{\delta}|\leq(2\delta
rN)\big{(}c^{\prime}\left((1+\delta)rN\right)^{d-1}\big{)}$ and
$|B(x,N)|\geq\frac{1}{c}N^{d}$. Therefore,

$\lambda_{g,N}\big{(}S_{N}(\overline{x},r)\cap
B_{N}(\overline{g},1)\big{)}\leq\frac{2c^{\prime}(1+\delta)^{d-1}\delta
r^{d}N^{d}}{\frac{1}{c}N^{d}}=2cc^{\prime}(1+\delta)^{d-1}\delta r^{d}$

for all $\delta>0$. Letting $\delta$ tend to $0$, we see that
$\lambda_{g,N}\big{(}S_{N}(\overline{x},r)\cap
B_{N}(\overline{g},1)\big{)}=0$, as desired. ∎

One nice consequence of Lemma 3.10 is that the measure of balls becomes
exactly what we would expect them to be, as long as they are of reasonable
size.

######  Corollary 3.11.

Suppose that $G$ is a finitely generated virtually nilpotent group with the SS
property. Then for any $N\in\mathbb{}^{*}N\setminus\mathbb{N}$,
$r\in\mathbb{R}^{{}^{+}}$, and $g,h\in$ ${}^{*}G$, we have that

$\lambda_{g,N}\Big{(}\operatorname{eq}_{N}\big{(}B(h,rN)\cap
B(g,N)\big{)}\Big{)}=\operatorname{st}\left(\frac{\big{|}B(h,rN)\cap
B(g,N)\big{|}}{|B(g,N)|}\right).$

###### Proof.

By Lemma 3.10, we know that
$\lambda_{g,N}\left(\operatorname{eq}_{N}\big{(}S(h,rN)\big{)}\cap
B_{N}(\overline{g},1)\right)=0$ and
$\lambda_{g,N}\left(\operatorname{eq}_{N}\big{(}S(g,N)\big{)}\right)=0$ (as
$\operatorname{eq}_{N}\big{(}S(h,rN)\big{)}\subseteq S_{N}(\overline{h},r)$
and $\operatorname{eq}_{N}\big{(}S(g,N)\big{)}\subseteq
S_{N}(\overline{g},1)$). Therefore, the left hand side of the display
appearing in the corollary equals

$\lambda_{g,N}\Big{(}\operatorname{eq}_{N}\big{(}B(h,rN)\cap
B(g,N)\big{)}\backslash\big{(}\operatorname{eq}_{N}\big{(}S(h,rN)\big{)}\cup\operatorname{eq}_{N}\big{(}S(g,N)\big{)}\big{)}\Big{)}.$

Notice that
$\operatorname{eq}_{N}^{-1}\Big{(}\operatorname{eq}_{N}\big{(}B(h,rN)\cap
B(g,N)\big{)}\backslash\left(\operatorname{eq}_{N}\big{(}S(h,rN)\big{)}\cup\operatorname{eq}_{N}\big{(}S(g,N)\big{)}\right)\Big{)}$
is a subset of both $B(h,rN)$ and $B(g,N)$, since the only equivalence classes
left would be those that were some appreciable distance from $S(h,rN)$ and
$S(g,N)$. Consequently, we have

$\lambda_{g,N}\Big{(}\operatorname{eq}_{N}\big{(}B(h,rN)\cap
B(g,N)\big{)}\Big{)}\leq\mu_{g,N}\big{(}B(h,rN)\cap B(g,N)\big{)}.$

On the other hand, $B(h,rN)\cap B(g,N)$ is a subset of
$\operatorname{eq}_{N}^{-1}\left(\operatorname{eq}_{N}\big{(}B(h,rN)\cap
B(g,N)\big{)}\right)$, meaning
$\lambda_{g,N}\left(\operatorname{eq}_{N}\big{(}B(h,rN)\cap
B(g,N)\big{)}\right)\geq\mu_{g,N}\big{(}B(h,rN)\cap B(g,N)\big{)}$. Combining
these bounds, we see that the left-hand side of the display in the corollary
equals

$\mu_{N}\big{(}B(h,rN)\cap
B(g,N)\big{)}=\operatorname{st}\left(\frac{\big{|}B(h,rN)\cap
B(g,N)\big{|}}{|B(g,N)|}\right),$

as desired. ∎

Using the same techniques, we prove the following lemmas hold in virtually
nilpotent groups with the SS property. The first lemma tells us that we can
restrict internal sets to our given ball before or after moving to the
equivalence classes without changing the measure.

######  Lemma 3.12.

Suppose that $G$ is a finitely generated virtually nilpotent group with the SS
property. Then for any $g\in$ ${}^{*}G$, any $N\in$
${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$, and any internal $A\subseteq$
${}^{*}G$, we have

$\lambda_{g,N}\big{(}\operatorname{eq}_{N}(A)\cap\operatorname{eq}_{N}(B(g,N))\big{)}=\lambda_{g,N}\left(\operatorname{eq}_{N}\big{(}A\cap
B(g,N)\big{)}\right).$

###### Proof.

Since $\operatorname{eq}_{N}\big{(}A\cap
B(g,N)\big{)}\subseteq\operatorname{eq}_{N}(A)\cap\operatorname{eq}_{N}(B(g,N))$,
the left-hand side is at least the right-hand side. To show equality, note
that

$\big{(}\operatorname{eq}_{N}(A)\cap\operatorname{eq}_{N}(B(g,N))\big{)}\backslash\operatorname{eq}_{N}\big{(}A\cap
B(g,N)\big{)}\subseteq S_{N}(\overline{g},1),$

as any equivalence class in $\operatorname{eq}_{N}(A)$ and
$\operatorname{eq}_{N}(B(g,N))$ that is not represented by an element of
$A\cap B(g,N)$ must contain elements outside of $B(g,N)$. Thus, by Lemma 3.10,

$\lambda_{g,N}\big{(}\operatorname{eq}_{N}(A)\cap\operatorname{eq}_{N}(B(g,N))\big{)}=\lambda_{g,N}\left(\big{(}\operatorname{eq}_{N}(A)\cap\operatorname{eq}_{N}(B(g,N))\big{)}\backslash
S_{N}(\overline{g},1)\right)=$
$=\lambda_{g,N}\left(\operatorname{eq}_{N}\big{(}A\cap
B(g,N)\big{)}\right).\text{ \qed}$

This next lemma gives us an easy way to convert the measure from one ball to a
superball or subball.

######  Lemma 3.13.

Suppose that $G$ is a finitely generated virtually nilpotent group with the SS
property. Then for any $g,g^{\prime}\in$ ${}^{*}G$, any $N\in$
${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$, any $r\in(0,2)$ such that
$B(g^{\prime},rN)\subseteq B(g,N)$, and any internal $A\subseteq$ ${}^{*}G$,
we have

$\lambda_{g,N}\Big{(}\operatorname{eq}_{N}\big{(}A\cap
B(g^{\prime},rN)\big{)}\Big{)}=\operatorname{st}\left(\frac{|B(g^{\prime},rN)|}{|B(g,N)|}\right)\lambda_{g^{\prime},rN}\Big{(}\operatorname{eq}_{N}\big{(}A\cap
B(g^{\prime},rN)\big{)}\Big{)}.$

###### Proof.

By definition, whenever
$\overline{B}\subseteq\operatorname{eq}_{N}(^{*}G)=\operatorname{eq}_{rN}(^{*}G)$
is $\lambda_{g,N}$ or $\lambda_{g^{\prime},rN}$ measurable, respectively, we
have
$\lambda_{g,N}(\overline{B})=\mu_{g,N}\big{(}\operatorname{eq}_{N}^{-1}(\overline{B})\cap
B(g,N)\big{)}$ and
$\lambda_{g^{\prime},rN}(\overline{B})=\mu_{g^{\prime},rN}\big{(}\operatorname{eq}_{N}^{-1}(\overline{B})\cap
B(g^{\prime},rN)\big{)}$. Next observe that

$\Big{(}\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}\big{(}B(g^{\prime},rN)\big{)}\big{)}\cap
B(g,N)\Big{)}\Big{\backslash}\Big{(}\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}\big{(}B(g^{\prime},rN)\big{)}\big{)}\cap
B(g^{\prime},rN)\Big{)}$

is contained in $\operatorname{eq}_{N}^{-1}(S_{N}(\overline{g^{\prime}},r))$,
which has $\mu_{g,N}$-measure $0$ by Lemma 3.10, and whence the restriction to
$B(g^{\prime},rN)$ does not affect the measure. The lemma follows from noting
that the following four quantities are equal, where the infima in the second
and third bullets are over internal subsets $C$ of ${}^{*}G$:

  * •

$\mu_{g,N}\Big{(}\operatorname{eq}_{N}^{-1}\Big{(}\operatorname{eq}_{N}\big{(}A\cap
B(g^{\prime},rN)\big{)}\Big{)}\cap B(g,N)\Big{)}$

  * •

$\inf\left\\{\operatorname{st}\left(\frac{|C|}{|B(g,N)|}\right):\operatorname{eq}_{N}^{-1}\Big{(}\operatorname{eq}_{N}\big{(}A\cap
B(g^{\prime},rN)\big{)}\Big{)}\cap B(g,N)\subseteq C\right\\}$

  * •

$\inf\left\\{\operatorname{st}\left(\frac{|C|}{|B(g^{\prime},rN)|}\cdot\frac{|B(g^{\prime},rN)|}{|B(g,N)|}\right):\operatorname{eq}_{N}^{-1}\Big{(}\operatorname{eq}_{N}\big{(}A\cap
B(g^{\prime},rN)\big{)}\Big{)}\cap B(g^{\prime},rN)\subseteq C\right\\}$

  * •

$\operatorname{st}\left(\frac{|B(g^{\prime},rN)|}{|B(g,N)|}\right)\mu_{g^{\prime},rN}\Big{(}\operatorname{eq}_{N}^{-1}\Big{(}\operatorname{eq}_{N}\big{(}A\cap
B(g^{\prime},rN)\big{)}\Big{)}\cap B(g^{\prime},rN)\Big{)}.$ ∎

##  4\. The BM property for internal sets

###  4.1. The Ball Measure Property

Throughout this section, we fix a finitely generated group $G$ with symmetric
generating set $S$.

######  Definition 4.1.

Given a ball $I=B(g,N)$, with $g\in$ ${}^{*}G$ and $N\in$
${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$, and internal $A\subseteq$
${}^{*}G$, we define the following two quantities:

  * •

$\lambda_{I}(A)=\lambda_{g,N}\big{(}\operatorname{eq}_{N}(A\cap I)\big{)}$ .

  * •

$g_{A}(I)=\max\left\\{\frac{|B(x,M)|}{|I|}:B(x,M)\subseteq I,\text{
}B(x,M)\cap A=\emptyset\right\\}$.

Using the notation from the previous definition, note that $\lambda_{I}(A)$
can equivalently be expressed as
$\lambda_{e,N}\left(\operatorname{eq}_{N}\big{(}g^{-1}A\cap
B(e,N)\big{)}\right)$. We will sometimes refer to $\lambda_{I}(A)$ as the
"measure" of $A$ in $I$. The quantity $g_{A}(I)$ is a measure of the size of
gaps of $A$ in $I$ and notice that the maximum appearing in the definition
indeed exists as $\big{\\{}M\in$ ${}^{*}{\mathbb{N}}:B(x,M)\subseteq I,$
$B(x,M)\cap A=\emptyset\text{ for some }x\in I\big{\\}}$ is an internal subset
of $[1,2N]$. Just like the corresponding properties in the original definition
of the IM property, $\lambda_{I}(A)$ is an external property of $A$ whereas
$g_{A}(I)$ is an internal property. As in the case of the integers, without
any further assumptions, there is always one relationship between these two
quantities, assuming the group is virtually nilpotent and satisfies the SS
property:

######  Lemma 4.2.

Suppose that $G$ is a virtually nilpotent group satisfying the SS property.
Let $I=B(g,N)$ be an infinite, hyperfinite ball and $A$ an internal subset of
$I$. For any $\varepsilon>0$, if $\lambda_{I}(A)\geq 1-\varepsilon$, then
$g_{A}(I)\leq\varepsilon$.

###### Proof.

Assume that $g_{A}(I)>\varepsilon$. Then there is some infinite hyperfinite
ball $J\subseteq I$ such that $\operatorname{st}(\frac{|J|}{|I|})>\varepsilon$
and $A\cap J=\emptyset$. Let $\partial J$ be the sphere that makes up the
boundary of $J$. Observe that
$\operatorname{eq}_{N}(A)\cap\big{(}\operatorname{eq}_{N}(J)\backslash\operatorname{eq}_{N}(\partial
J)\big{)}=\emptyset$. Thus, using Lemma 3.10 and Corollary 3.11, we have that
the following quantities coincide:

  * •

$\lambda_{I}(A)$

  * •

$\lambda_{g,N}\big{(}\operatorname{eq}_{N}(A)\big{)}$

  * •

$1-\lambda_{g,N}\big{(}\operatorname{eq}_{N}(J)\backslash\operatorname{eq}_{N}(\partial
J)\big{)}$

  * •

$1-\lambda_{g,N}\big{(}\operatorname{eq}_{N}(J)\big{)}+\lambda_{g,N}\big{(}\operatorname{eq}_{N}(\partial
J)\big{)}$

  * •

$1-\lambda_{g,N}\big{(}\operatorname{eq}_{N}(J)\big{)}$

  * •

$1-\operatorname{st}\left(\frac{|J|}{|I|}\right)$.

Since the last quantity is strictly less than $1-\varepsilon$, the lemma is
proven. ∎

Notice that even this proof of such a basic relationship (seemingly) requires
the use of the SS property. The definition of the BM property seeks to find a
connection between these quantities in the other direction:

######  Definition 4.3.

Given an internal set $A\subseteq$ ${}^{*}G$ and infinite hyperfinite ball
$I$, we say that $A$ has the ball measure property (or BM property for short)
on $I$ if: for every $\varepsilon>0$, there is $\delta>0$ such that, for all
infinite hyperfinite balls $J\subseteq I$ with $g_{A}(J)\leq\delta$, we have
$\lambda_{J}(A)\geq 1-\varepsilon$.

For simplicity, we often just say $A$ is BM on $I$. Note that if
$G={\mathbb{Z}}$ with generating set $S=\\{-1,1\\}$, then the notion of the BM
property reduces to the definition of the IM property on intervals of odd
length which, since a single element has no effect on either $g_{A}(I)$ or
$\lambda_{I}(A)$, can be easily extended to include all intervals. Thus, the
previous definition truly is a generalization of the IM property.

While we have defined this concept for all finitely generated groups, given
that the basic fact Lemma 4.2 used the assumptions of being virtually
nilpotent and having the SS property, it seems that to establish a nice theory
of sets with the BM property, these assumptions will need to be present
throughout. Consequently, we henceforth assume that $G$ is of polynomial
growth and has the SS property.

###  4.2. Properties of BM sets

Suppose that $A$ has the BM property on $I$. Let $\delta(A,I,\varepsilon)$ be
the supremum of the $\delta$’s that witness $A$ being BM on $I$ for the given
$\varepsilon$. As in the IM case, we want to avoid the situation where $A$ is
BM on $I$ simply because there is some $\delta>0$ with $g_{A}(J)>\delta$ for
all infinite balls $J\subseteq I$. In this spirit, we establish the following
proposition:.

######  Proposition 4.4.

Let $A\subseteq$ ${}^{*}G$ be an internal set and let $I$ be an infinite
hyperfinite ball such that $A$ is BM on $I$. Then the following statements are
equivalent:

  1. (i)

For every $\delta>0$, there is an infinite subball $J_{\delta}$ of $I$ such
that $g_{A}(J_{\delta})<\delta$.

  2. (ii)

There is an infinite subball $J$ of $I$ such that $\lambda_{J}(A)>0$.

###### Proof.

First assume that (i) holds. Then there is a subball $J$ of $I$ such that
$g_{A}(J)<\delta(A,I,\frac{1}{2})$. By definition, this means that
$\lambda_{J}(A)\geq 1-\frac{1}{2}=\frac{1}{2}>0$.

For the other direction, assume that $J=B(g,N)$ is an infinite subball of $I$
such that $\lambda_{J}(A)>0$, meaning
$\lambda_{g,N}\big{(}\operatorname{eq}_{N}(A\cap J)\big{)}>0$. By the LDT and
Corollary 3.11, for any $\delta>0$, we can find a subball
$J^{\prime}=B(g^{\prime},N^{\prime})\subseteq J$ such that

$1-\delta<\frac{\lambda_{g,N}\big{(}\operatorname{eq}_{N}(A\cap
J)\cap\operatorname{eq}_{N}(J^{\prime})\big{)}}{\lambda_{g,N}(\operatorname{eq}_{N}(J^{\prime}))}=\operatorname{st}\left(\frac{|J|}{|J^{\prime}|}\right)\lambda_{g,N}\big{(}\operatorname{eq}_{N}(A\cap
J)\cap\operatorname{eq}_{N}(J^{\prime})\big{)}.$

By Lemma 3.12 and Lemma 3.13, we have that the right-hand side of the previous
display equals

$\operatorname{st}\left(\frac{|J|}{|J^{\prime}|}\right)\lambda_{g,N}\big{(}\operatorname{eq}_{N}(A\cap
J^{\prime})\big{)}=\lambda_{g^{\prime},N^{\prime}}\big{(}\operatorname{eq}_{N}(A\cap
J^{\prime})\big{)}.$

Thus, by Lemma 4.2, we see that $g_{A}(J^{\prime})<\delta$. Letting
$J_{\delta}=J^{\prime}$, we have that (i) holds. ∎

Using the same terminology as in [5], we say that $A$ has the enhanced BM
property on $I$ if $A$ is BM on $I$ and $\lambda_{I}(A)>0$.

We can now establish the “internal partition regularity” of the BM property:

######  Theorem 4.5.

Let $A$ be an internal subset of ${}^{*}G$ with the enhanced BM property on an
infinite hyperfinite ball $I$. Assume that $A\cap I=B_{1}\cup\cdots\cup B_{n}$
with each $B_{i}$ internal. Then there is an $i\in\\{1,\ldots,n\\}$ and an
infinite subball $J\subseteq I$ such that $B_{i}$ has the enhanced BM property
on $J$.

###### Proof.

We prove this theorem by induction, the base case $n=1$ being trivial. Now
assume that $n>1$ and the result is true for partitions of size $n-1$. Let
$A\cap I=B_{1}\cup\cdots\cup B_{n}$ be an internal partition. If there is an
$i\in\\{1,\ldots,n\\}$ and an infinite subball $J\subseteq I$ such that
$B_{i}\cap J=\emptyset$ and $\lambda_{J}(A)>0$, then we can use the assumption
on $J$ and the $n-1$ remaining $B_{j}$’s to obtain $j$ and
$J^{\prime}\subseteq J\subseteq I$ that satisfies the conclusion of the
theorem. Thus, we may assume going forward that this is not the case, that is,
whenever $\lambda_{J}(A)>0$, we have that $B_{i}\cap J\neq\emptyset$ for all
$i=1,\ldots,n$. We will show that this implies that each $B_{i}$ is BM on $I$.
Since $\lambda_{I}(B_{i})$ must be positive for some $i\in\\{1,\ldots,n\\}$,
this $B_{i}$ wil be as desired.

Fix $i$ and set $B=B_{i}$. Fix $\epsilon>0$ and let $J=B(g,N)$ be an infinite
subball of $I$ with $g_{B}(J)\leq\delta(A,I,\varepsilon)$. (If no such $J$
exists then $B$ has the BM property on $I$ and we are done.) Since $B\subseteq
A$, we know that $g_{A}(J)\leq g_{B}(J)\leq\delta(A,I,\varepsilon)$. By
definition, this gives us that $\lambda_{J}(A)\geq 1-\varepsilon$. Now take
$B_{N}(\overline{x},r)\subseteq
B_{N}(\overline{g},1)\backslash\operatorname{eq}_{N}(B)$, so that $B(x,rN)\cap
B=\emptyset$. Our assumption then gives us that $\lambda_{B(x,rN)}(A)=0$, and
so $\lambda_{J}\big{(}A\cap B(x,rN)\big{)}=0$. Thus,
$\lambda_{J}(B)=\lambda_{J}(A)\geq 1-\varepsilon$, meaning $B$ has the $BM$
property on $I$. ∎

We next proceed to establish a result about “difference sets” for which we
will need the following notion:

######  Definition 4.6.

Let $A_{1},...,A_{n}$ be internal subsets of ${}^{*}G$ and let
$I_{1},...,I_{n}$ be hyperfinite balls.

  * •

A $\delta$-configuration with respect to these sets is a sequence of subballs
$J_{1},...,J_{n}$ of $I_{1},...,I_{n}$ respectively such that all $J_{i}$ have
the same radius and such that $g_{A_{i}}(J_{i})\leq\delta$ for all
$i=1,\ldots,n$. We will refer to the shared radius of $J_{1},...,J_{n}$ as the
radius of the configuration.

  * •

If $J_{1},...,J_{n}$ is a $\delta$-configuration with respect to
$A_{1},...,A_{n}$ and $I_{1},...,I_{n}$, then a $\delta$-subconfiguration of
$J_{1},\ldots,J_{n}$ is a $\delta$-configuration with respect to
$A_{1},...,A_{n}$ and $J_{1},...,J_{n}$.

  * •

A strong $\delta$-subconfiguration is a $\delta$-subconfiguration
$K_{1},...,K_{n}$ of $J_{1},...,J_{n}$ for which there exists a $c\in$
${}^{*}G$ such that, writing $J_{i}=B(a_{i},N)$, we have that
$K_{i}=B(a_{i}c,M)$.

######  Theorem 4.7.

Let $A_{1},...,A_{n}$ be internal subsets of ${}^{*}G$ which are BM on the
balls $I_{1},...,I_{n}$ respectively. Take $\varepsilon>0$ such that
$\varepsilon<\frac{1}{n+1}$, and $\delta>0$ such that
$\delta<\min_{i=1,...,n}\delta(A_{i},I_{i},\varepsilon)$. Then there is a
$w\in{\mathbb{N}}$ such that any $\delta$-configuration has a strong
$\delta$-subconfiguration of radius at most $w$.

###### Proof.

Let $A_{1},...,A_{n}$, $I_{1},...,I_{n}$, $\varepsilon$, and $\delta$ be as
above. Take $J_{i}=B(a_{i},R)$ to be a $\delta$-configuration, with $R\in$
${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$. Then, by our choice of $\delta$,
we have that $\lambda_{J_{i}}(A_{i})\geq 1-\frac{1}{n+1}$, meaning
$\lambda_{a_{i}^{-1}J_{i}}(a_{i}^{-1}A_{i})\geq 1-\frac{1}{n+1}$. Note that
$a_{i}^{-1}J_{i}=B(e,R)$. Therefore, each $a_{i}^{-1}A_{i}$ has measure at
least $1-\frac{1}{n+1}$ on $B(e,R)$, whence

$\lambda_{e,R}\left(\bigcap_{i=1}^{n}\operatorname{eq}_{R}\big{(}a_{i}^{-1}A_{i}\cap
B(e,R)\big{)}\right)>0.$

By the Lebesgue density theorem,
$\bigcap_{i=1}^{n}\operatorname{eq}_{R}\big{(}a_{i}^{-1}A_{i}\cap
B(e,R)\big{)}$ has a point of density $\overline{b}$. Thus, there must be an
$r\in(0,1)$ such that

$\dfrac{\lambda_{e,R}\left(\Big{(}\bigcap_{i=1}^{n}\operatorname{eq}_{R}\big{(}a_{i}^{-1}A_{i}\cap
B(e,R)\big{)}\Big{)}\cap
B_{R}(\overline{b},r)\right)}{\lambda_{e,R}\big{(}B_{R}(\overline{b},r)\big{)}}\geq
1-\delta.$

Let $K_{i}=B\big{(}a_{i}b,\lfloor rR\rfloor\big{)}$. By the previous display,
we have that $g_{A_{i}}(K_{i})\leq\delta$ for all $i=1,\ldots,n$. Thus the
$K_{i}$ form a strong $\delta$-subconfiguration with radius $rR<R$.

The import of the previous discussion is that every $\delta$-configuration
with infinite radius has a _proper_ strong $\delta$-subconfiguration. Let
${\mathcal{C}}$ denote the (internal) set of $\delta$-configurations of
$A_{1},...,A_{n},I_{1},...,I_{n}$ and let $f:{\mathcal{C}}\to$
${}^{*}{\mathbb{N}}$ be the internal function defined by sending any
$\delta$-configuration $J_{1},...,J_{n}$ to the minimal radius of a strong
$\delta$-subconfiguration of $J_{1},...J_{n}$ (which exists by transfer). We
have thus shown that $f({\mathcal{C}})\subseteq{\mathbb{N}}$ whence, by
overflow, $f({\mathcal{C}})\subseteq[0,w]$ for some $w\in{\mathbb{N}}$. Thus,
this $w$ satisfies the conditions of the theorem. ∎

The previous lemma has consequences for difference sets. To state these
consequences, we establish some terminology and remind the reader about
syndetic sets in arbitrary groups.

######  Definition 4.8.

For any $A\subseteq$ ${}^{*}G$, we define

$D(A)=\\{g\in G:g=a^{-1}b\text{ for infinitely many pairs }a,b\in A\\}.$

For any group $G$, recall that a subset $A$ of $G$ is called syndetic if there
is a finite subset $F\subseteq G$ such that $AF=G$. Equivalently, $A$ is
syndetic if there exists $n\in{\mathbb{N}}$ such that $B(g,n)\cap
A\neq\emptyset$ for all $g\in G$. Note that if $H\leq G$ is a subgroup, then
$H$ is syndetic if and only if $H$ has finite index in $G$.

We can now state the following lemma, which we will use in the next section.

######  Lemma 4.9.

Suppose that $A$ has the enhanced BM property on some infinite, hyperfinite
ball $I$. Then there is some $r\in{\mathbb{N}}$ such that, for every $g\in G$,
there is $z\in B(e,r)$ with $x^{-1}y\in\\{h^{-1}ghz:h\in$ ${}^{*}G\\}$ for
infinitely many pairs $x,y\in A$.

###### Proof.

Let $A_{1}=A_{2}=A$ and $I_{1}=I_{2}=I=B(v,N)$. By Theorem 4.7, there is
$w\in{\mathbb{N}}$ such that every $\delta$-configuration $B(a_{i},R)$ has
some $c\in$ ${}^{*}G$ with the sequence $B(a_{i}c,w)$ containing a strong
$\delta$-subconfiguration of $B(a_{i},R)$. In particular, this implies that
$B(a_{i}c,w)\cap A\neq\emptyset$.

Fix $g\in G$. Since $\lambda_{v,N}\big{(}\operatorname{eq}_{N}(A\cap
I)\big{)}>0$, we can find countably many distinct points of density of
$\operatorname{eq}_{N}(A\cap I)$, thereby giving us pairwise disjoint balls
$B_{N}(\overline{a_{i,\varepsilon}},b_{\varepsilon})\subseteq\operatorname{eq}_{N}(A\cap
I)$ with
$\lambda_{a_{i,\varepsilon},b_{\varepsilon}N}\big{(}\operatorname{eq}_{N}(A\cap
I)\cap B_{N}(\overline{a_{i,\varepsilon}},b_{\varepsilon})\big{)}\geq
1-\varepsilon$ for all $i\in{\mathbb{N}}$. Thus
$g_{A}\big{(}B(a_{i,\varepsilon},b_{\varepsilon}N)\big{)}\leq\varepsilon$.
Since the gap function is internal and we can find such balls for any
$\varepsilon>0$, there must be countably many pairwise disjoint balls
$J_{i}=B(a_{i},R)$ with $g_{A}(J_{i})\approx 0$ for all $i\in\mathbb{N}$.
Since $g\in G$, it is also true that $g_{A}\big{(}B(a_{i}g,R)\big{)}\approx
0$, as the elements of $B(a_{i}g,R)$ that are not in $J_{i}$ are all contained
in one of $S(a_{i},R+1),...,S(a_{i},R+|g|)$, all of which have infinitesimal
size when compared to $J_{i}$ by the SS property. Consequently, the
$J_{i},B(a_{i}g,R)$ forms a $\delta$-configuration of
$A_{1},A_{2},I_{1},I_{2}$ for any $\delta$ and any $i$. Thus, by our choice of
$w$, for every $i\in{\mathbb{N}}$, there is a $c_{i}\in$ ${}^{*}G$ such that

$A\cap B(a_{i}c_{i},w)\neq\emptyset\text{ and }A\cap
B(a_{i}gc_{i},w)\neq\emptyset.$

If $x_{i}\in A\cap B(a_{i}c_{i},w)$ and $y_{i}\in A\cap B(a_{i}gc_{i},w)$,
then $x_{i}=a_{i}c_{i}x$ and $y_{i}=a_{i}gc_{i}y$ with $|x|,|y|\leq w$. Since

$x_{i}^{-1}y_{i}=x^{-1}c_{i}^{-1}a_{i}^{-1}a_{i}gc_{i}y=x^{-1}c_{i}^{-1}gc_{i}x(x^{-1}y),$

we see that $x_{i}^{-1}y_{i}\in B(h^{-1}gh,2w)$ for some $h\in$ ${}^{*}G$,
meaning there must be a $z\in B(e,2w)$ such that infinitely many $i$ satisfy
$x_{i}^{-1}y_{i}\in\\{h^{-1}ghz:h\in$ ${}^{*}G\\}$. ∎

As a corollary, we get the following results.

######  Corollary 4.10.

If $A$ has the enhanced BM property on some infinite, hyperfinite ball $I$,
then there is some $r\in{\mathbb{N}}$ such that $B(g,r)\cap D(A)\neq\emptyset$
for any $g\in Z(G)$.

###### Proof.

Let $r$ be as in Lemma 4.9. Fix $g\in Z(G)$. By Lemma 4.9, there is $z\in
B(e,r)$ such that there are infinitely many $x,y\in A$ with
$x^{-1}y\in\\{h^{-1}ghz:h\in$ ${}^{*}G\\}=\\{gz\\}$. In other words, $gz\in
D(A)$. Since $z\in B(e,r)$, $gz\in B(g,r)$, finishing the proof. ∎

Thus, we see that sets with the enhanced BM property are, in some sense,
"syndetic on the center of $G$".

######  Corollary 4.11.

If $A$ has the enhanced BM property on some infinite, hyperfinite ball $I$ and
$Z(G)$ has finite index in $G$, then $D(A)$ is syndetic in $G$.

##  5\. The SBM property for standard sets

###  5.1. SBM Sets and Their Properties

We now move to the setting of standard sets, maintaining our standing
assumptions that $G$ is a finitely generated virtually nilpotent group with
the SS property.

######  Definition 5.1.

We say that a set $A\subseteq G$ has the standard BM property (or SBM property
for short) if:

  1. (1)

${}^{*}A$ is BM on every infinite, hyperfinite subball of ${}^{*}G$, and

  2. (2)

${}^{*}A$ has the enhanced BM property on some infinite hyperfinite subball of
${}^{*}G$.

Note that for $G={\mathbb{Z}}$ with the standard generating set
$S=\\{1,-1\\}$, we recover the definition of SIM set.

Since the SBM property is a property of standard sets, it is possible to give
an equivalent definition using only standard notions (as Leth did for subsets
of the integers with the SIM property in [12]). We merely establish the
notation for stating this equivalent reformulation, leaving the proof of the
theorem to the reader. Since this standard reformulation is not terribly
enlightening, the reader choosing to skip the verification will lose nothing
in what follows.

Following the notation in [5], for $A\subseteq G$ and
$0<\delta<\varepsilon<1$, let the function
$F_{\delta,\varepsilon,A}:{\mathbb{N}}\to{\mathbb{N}}\cup\\{\infty\\}$ be
defined as follows. First, if $g_{A}(I)>\delta$ for every subball of $G$ with
radius $\geq n$, set $F_{\delta,\varepsilon,A}(n)=0$. Otherwise, let
$F_{\delta,\varepsilon,A}(n)$ be the minimum $k$ such that there is a subball
$I$ of $G$ with radius $\geq n$ and $g_{A}(I)\leq\delta$ for which there are
disjoint subballs $I_{1},...,I_{k}\subseteq I$ with $I_{i}\cap A=\emptyset$
for all $i$ and $\sum_{i=1}^{k}|I_{i}|\geq\varepsilon|I|$. Finally, if no such
$k$ exists, we let $F_{\delta,\varepsilon,A}(n)=\infty$. Note that either
there exists an $n\in{\mathbb{N}}$ such that $F_{\delta,\varepsilon,A}(m)=0$
for all $m\geq n$ or else $F_{\delta,\varepsilon,A}$ is an increasing
function.

######  Theorem 5.2.

$A$ has the SBM property if and only if: for all $\varepsilon>0$, there is
$\delta>0$ such that $\lim_{n\to\infty}F_{\delta,\varepsilon,A}(n)=\infty$.

We now turn to a discussion of examples and non-examples of sets with the SBM
property. First, an easy example:

######  Example 5.3.

If $A\subseteq G$ is syndetic, then $A$ has the SBM property.

###### Proof.

Assume $A\subseteq G$ is syndetic. Then there is $n\in{\mathbb{N}}$ such that
$B(g,n)\cap A\neq\emptyset$ for all $g\in G$. By transfer, this will also be
true of ${}^{*}A$ for all $g\in$ ${}^{*}G$. Thus, $g_{A}(I)\approx 0$ and
$\lambda_{I}(A)=1$ for all infinite, hyperfinite balls $I\subseteq$ ${}^{*}G$,
whence $A$ is SBM. ∎

In order to proceed further and to see that the SBM property behaves in a way
that mimics the SIM property, we will ask that our groups satisfy one further
property.

######  Definition 5.4.

We say that the finitely generated group $G$ has the small gap property (or SG
property for short) if, for any $r\in(0,1)$, there is $\delta>0$, such that
$\lim_{n\to\infty}\eta_{n}(G,\delta)>r,$ where

$\eta_{n}(G,\delta)=\min\\{\rho(A,B)\ :\ g_{B}(A)<\delta,A,B\text{ subballs of
}G,|A|>n\\}$

and

$\rho(A,B)=\left\\{\frac{|C|}{|A|}:C\subseteq A\cap B\text{ is a
subball}\right\\}.$

In other words, if $G$ has the SG property and we know that the gap size
between two balls $A$ and $B$ is small, then as long as $A$ is large enough,
this also tells us that there is a ball in the intersection that is close in
size to $A$.

It is unclear at this time what groups satisfy this property and whether or
not this property is dependent on the generating set. Certainly,
${\mathbb{Z}}^{d}$ will have this property with its standard generators, but
we have yet to verify any non-abelian examples. That being said, it seems
possible that the following holds.

######  Conjecture 5.5.

Let $G$ be a finitely generated, virtually nilpotent group with the SS
property. Then $G$ has the SG property.

From this point forward, besides assuming that our groups are virtually
nilpotent and have the SS property, we also assume that our groups have the SG
property, and will make clear when this is used. One consequence of the SG
property is that an infinite ball will always have the BM property on any
other infinite ball, formalized in the following lemma.

######  Lemma 5.6.

If $A$ and $B$ are infinite, hyperfinite balls in ${}^{*}G$, then $B$ has the
BM property on $A$.

###### Proof.

Fix $\varepsilon\in(0,1)$ and let $\delta>0$ be as in the definition of the SG
property for $r=1-\varepsilon$. Assume that the infinite hyperfinite subball
$A^{\prime}\subseteq A$ is such that $g_{B}(A^{\prime})<\delta$. Then the SG
property and transfer tell us that there is a ball $C\subseteq A^{\prime}\cap
B$ with $\operatorname{st}\big{(}\frac{|C|}{|A^{\prime}|}\big{)}\geq
1-\epsilon$. By Corollary 3.11,

$\lambda_{A^{\prime}}(B)\geq\lambda_{A^{\prime}}(C)=\operatorname{st}\left(\frac{|C|}{|A^{\prime}|}\right)>1-\varepsilon.$

Thus, $B$ has the SBM property on $A$. ∎

The following lemma is another consequence of the SG property. First, we need
to recall the generalization of piecewise syndeticity to arbitrary groups:
$B\subseteq G$ is piecewise syndetic if there is a finite set $F\subseteq G$
such that, for all finite $T\subseteq G$, $gT\subseteq BF$ for some $g\in G$.

######  Theorem 5.7.

If $B$ is piecewise syndetic, then there is $A\subseteq B$ with the SBM
property.

We defer the proof of Theorem 5.7 for later, as it will follow immediately
from Theorem 5.11 below. Theorem 5.7 implies that many sets contain SBM sets.
However, containing an SBM set is not enough to say that a set is SBM. In
fact, the following holds.

######  Lemma 5.8.

Let $A\subseteq G$ be a non-syndetic set. Then, there is $B\supseteq A$ such
that $B$ is not SBM.

###### Proof.

For every $n\in{\mathbb{N}}$, let $x_{n}\in G$ be such that
$B(x_{n},n^{2})\cap A=\emptyset$. Then set

$B=A\cup\bigcup_{n\in{\mathbb{N}}}\bigcup_{0\leq k\leq n}S(x_{n},kn).$

Assume that $c$ is a constant witnessing the polynomial growth of $G$ of
degree $d$. Fix an infinite $N\in$ ${}^{*}{\mathbb{N}}$. Fix $\delta>0$ and
take $m\in{\mathbb{N}}$ such that $m^{d}>\frac{c^{2}}{\delta}$. Let
$I=B(x_{N},N^{2})$, and consider $J=B(x_{N},mN)$. By the definition of $B$, we
have

$g_{{}^{*}B}(J)\leq\frac{|B(x_{N},N)|}{|B(x_{N},mN)|}\leq\frac{cN^{d}}{\frac{1}{c}m^{d}N^{d}}=\frac{c^{2}}{m^{d}}<\delta.$

On the other hand, ${}^{*}B\cap J=\bigcup_{0\leq k\leq m}S(x_{N},kN)$ is a
finite union of $\lambda_{x_{N},mN}$-measure 0 sets by the SS property. Thus,
for every $\delta>0$, we can find a $J\subseteq I$ with
$g_{{}^{*}B}(J)<\delta$ and $\lambda_{J}($ ${}^{*}B)=0$. Therefore ${}^{*}B$
is not BM on $I$ and thus $B$ is not SBM. ∎

Note that this lemma only made use of the SS property, and did not require the
SG property. However, assuming we do have the SG property, we can combine the
previous lemma with Theorem 5.7 to conclude that every set that is piecewise
syndetic but not syndetic contains an SBM set, but is also contained in a set
that is not SBM.

We end this subsection with one other fact about SBM sets.

######  Corollary 5.9.

If $A$ has the SBM property, then there is an $r\in{\mathbb{N}}$ such that for
all $x\in Z(G)$, $B(x,r)\cap D(A)\neq\emptyset$.

###### Proof.

This follows immediately from Corollary 4.10 and the fact that $D(A)=D($
${}^{*}A)$. ∎

###  5.2. Supra-SBM Sets

We stated in the previous section that all piecewise syndetic sets contain an
SBM set, and that SBM sets are not closed under supersets. The following
definition is thus natural:

######  Definition 5.10.

A set $A\subseteq G$ is supra-SBM if there is a $B\subseteq A$ such that $B$
has the SBM property.

With this definition (and making use of the SG property), we prove the
following, generalizing a result of Goldbring and Leth [7, Theorem 3.4].

######  Theorem 5.11.

Let $A\subseteq G$ be such that ${}^{*}A$ has the enhanced BM property on some
ball $I$. Then $A$ is supra-SBM.

###### Proof.

Let $d$ be the degree of polynomial growth of $G$ and let $c$ be the constant
witnessing this growth. Fix $k\in{\mathbb{N}}$. Let $\delta(k)$ be as in the
definition of the small gap property for
$r=\min\big{\\{}\frac{1}{5^{d}c^{2}},(1-\frac{1}{k})\big{\\}}$. Fix
$\delta_{k}<\min\big{(}\delta($ ${}^{*}A,I,\frac{1}{k}),\delta(k)\big{)}$. For
any $k,n\in{\mathbb{N}}$, if we take $J\subseteq I$ an infinite subball such
that $g_{{}^{*}A}(J)<\delta_{k}$, then $\lambda_{J}($ ${}^{*}A)\geq
1-\frac{1}{k}$ by the definition of $\delta_{k}$. Therefore, if we take $n$
disjoint subballs $J_{1},...,J_{n}$ of $J$ with $J_{i}\text{ }\cap$
${}^{*}A=\emptyset$ for each $i$, we have
$\sum_{i=1}^{n}\frac{|J_{i}|}{|J|}<\frac{1}{k}$. In other words, the sum of
$n$ disjoint gaps of ${}^{*}A$ in $J$ have sizes adding to less than
$\frac{|J|}{k}$. This statement is internal and is true for all infinite $J$,
so by underflow, there must be some $M_{n,k}\in{\mathbb{N}}$ such that
whenever $|J|>M_{n,k}$, then $n$ disjoint gaps of ${}^{*}A$ on $J$ add up to
less than $\frac{|J|}{k}$. Also, since $\lambda_{I}($ ${}^{*}A)>0$, for every
$n\in{\mathbb{N}}$ there exists an infinite subball $J$ of $I$ with
$g_{{}^{*}A}(J)<\frac{1}{n}$.

Since there is an infinite ball $I$ with all of these internal properties on
${}^{*}A$, by transfer, we can define a sequence of pairwise disjoint balls
$I_{n}$ in $G$ with the following properties:

  * •

Letting $B(e,r_{n})$ be the smallest ball around the identiy which contains
$I_{1},...,I_{n-1}$, we have that $I_{n}\cap B(e,nr_{n})=\emptyset$.

  * •

$I_{n}$ has a subball $J$ with radius at least $n$ such that
$g_{A}(J)<\frac{1}{n}$.

  * •

For all $k\leq n$ and all $J\subseteq I_{n}$, if $m<n$, $|J|>M_{m,k}$, and
$g_{A}(J)<\delta_{k}$, then at least $m+1$ disjoint gaps of $A$ on $J$ are
necessary for the size of the gaps to add up to $\frac{|J|}{k}$.

Let $B=\bigcup_{n\in{\mathbb{N}}}(A\cap I_{n})$. We show that $B$ has the SBM
property.

Let $H$ be an infinite, hyperfinite subball of ${}^{*}G$. We show that
${}^{*}B$ has the BM property on $H$ as witnessed by the function
$\delta_{H}(\varepsilon)=\frac{1}{5^{d}c^{2}}\delta_{k}$ for any
$k>\frac{2}{\epsilon}$. To see this, take $J\subseteq H$ such that
$g_{{}^{*}B}(J)<\frac{1}{5^{d}c^{2}}\delta_{k}$, and assume that $I_{M}$ is
the ball with largest index which intersects $J$. Note that every $I_{K}$ with
$K<M$ satisfies $I_{K}\subseteq B(e,r_{K})$, whereas every element of $I_{M}$
is outside of $B(e,Kr_{K})$. Therefore, the radius of $J$ is at least
$\frac{(K-1)r_{K}}{2}$, and so $\frac{|B(e,r_{K})|}{|J|}$ is infinitesimal as
$\frac{2r_{K}}{(K-1)r_{K}}$ is infinitesimal. Thus, all elements of $I_{K}$
with $K<M$ are in the equivalence class of the identity modulo $\sim_{M}$.

As constructed, ${}^{*}B\cap H\subseteq B(e,r_{M})\cup I_{M}$. We know that
$\frac{|B(e,r_{M})|}{|J|}$ is infinitesimal and that $g_{B(e,r_{M})\cup
I_{M}}(J)\leq g_{{}^{*}B}(J)\leq\frac{1}{5^{d}c^{2}}\delta_{k}$. Assume,
towards a contradiction, that $g_{I_{M}}(J)>\delta_{k}$. Then there is a
closed ball $C\subseteq J$ such that $C\cap I_{M}=\emptyset$ and
$|C|>\delta_{k}|J|$. Let $r_{C}$ be the radius of $C$ and $x$ be the center of
$C$. Take $y\in C$ such that $d(x,y)=r_{C}$. Let $x^{\prime}$ and $y^{\prime}$
be such that $d(x,x^{\prime})=d(y,y^{\prime})=\lfloor\frac{r_{C}}{4}\rfloor$
and $d(x,y^{\prime})=d(y,x^{\prime})=r_{C}-\lfloor\frac{r_{C}}{4}\rfloor$. In
other words, let $x^{\prime}$ and $y^{\prime}$ be points along some path of
length $r_{C}$ from $x$ to $y$ where $x^{\prime}$ is approximately
$\frac{1}{4}$ of the way from $x$ to $y$ and $y^{\prime}$ is approximately
$\frac{3}{4}$ of the way from $x$ to $y$. Since $d(x,y)=r_{C}$, we must have
that $d(x^{\prime},y^{\prime})\geq\frac{r_{C}}{2}$. This tells us that the
balls $B(x^{\prime},\frac{r_{C}}{5})$ and $B(y^{\prime},\frac{r_{C}}{5})$ are
disjoint and contained in $C$. Indeed, if $a$ is in the ball around
$x^{\prime}$ and $b$ is in the ball around $y^{\prime}$, then
$d(a,b)\geq\frac{r_{C}}{10}$. Finally, note that if $B(e,r_{M})\cap
B(x^{\prime},\frac{r_{C}}{5})\neq\emptyset$, then $B(e,r_{M})\cap
B(y^{\prime},\frac{r_{C}}{5})=\emptyset$. Thus

$g_{B(e,r_{M})\cup
I_{M}}(J)\geq\frac{1}{c|J|}\left(\frac{r_{C}}{5}\right)^{d}\geq\frac{|C|}{|J|c^{2}5^{d}}>\frac{\delta_{k}}{c^{2}5^{d}},$

contradicting our original inequality. Thus, we must have that
$g_{I_{M}}(J)\leq\delta_{k}$.

Using the SG property and our definition of $\delta_{k}$, we know that there
is a ball $J^{\prime}\subseteq I_{M}\cap J$ such that
$\frac{|J^{\prime}|}{|J|}>\frac{1}{5^{d}c^{2}}$. This gives us that

$g_{{}^{*}B}(J^{\prime})\leq
g_{{}^{*}B}(J)\left(\frac{|J|}{|J^{\prime}|}\right)\leq\left(\frac{1}{5^{d}c^{2}}\delta_{k}\right)(5^{d}c^{2})=\delta_{k}.$

So $g_{{}^{*}A}(J^{\prime})=g_{{}^{*}B}(J^{\prime})\leq\delta_{k}$. Since
$|J^{\prime}|$ is infinite, it is larger than $M_{m,k}$ for all
$m\in{\mathbb{N}}$. Therefore, the cardinalities of $m$ disjoint gaps of
${}^{*}A$ (and therefore ${}^{*}B$) on $J^{\prime}$ can not add to
$\frac{|J^{\prime}|}{k}$ for any $m\in{\mathbb{N}}$.

Assume now, towards a contradiction, that
$\lambda_{J^{\prime}}(^{*}B)<1-\frac{1}{k}$. Let
$J^{\prime}=B(g^{\prime},N^{\prime})$, so that
$\lambda_{J^{\prime}}(^{*}B)=\lambda_{g^{\prime},N^{\prime}}\big{(}\operatorname{eq}_{N^{\prime}}(^{*}B\cap
J^{\prime})\big{)}$. We know that $\operatorname{eq}_{N^{\prime}}(^{*}B\cap
J^{\prime})$ is a closed set, so that
$\overline{J_{c}}=\operatorname{eq}_{N^{\prime}}(J^{\prime})\backslash\operatorname{eq}_{N^{\prime}}(^{*}B\cap
J^{\prime})$ is an open set with
$\lambda_{g^{\prime},N^{\prime}}(\overline{J_{c}})>\frac{1}{k}$. Since
$\overline{J_{c}}$ is open, for every $\overline{x}\in\overline{J_{c}}$, we
can find an $n_{x}\in{\mathbb{N}}$ such that
$B_{N^{\prime}}(\overline{x},\frac{1}{n_{x}})\subseteq\overline{J_{c}}$. Let
$\mathcal{B}=\bigcup_{\overline{x}\in\overline{J_{c}}}\\{B_{N^{\prime}}(\overline{x},\frac{1}{n}):n\geq
n_{x}\\}$. By the symmetric Vitali property, there is a countable pairwise
disjoint subcollection
$\mathcal{B}^{\prime}=\big{\\{}\overline{B_{n}}:n\in{\mathbb{N}}\big{\\}}$ of
$\mathcal{B}$ such that
$\lambda_{g^{\prime},N^{\prime}}\big{(}\overline{J_{c}}\backslash\bigcup_{n\in{\mathbb{N}}}\overline{B_{n}}\big{)}=0$,
whence
$\lambda_{g^{\prime},N^{\prime}}\big{(}\bigcup_{n\in{\mathbb{N}}}\overline{B_{n}}\big{)}=\lambda_{g^{\prime},N^{\prime}}(\overline{J_{c}})>\frac{1}{k}$.
By countable additivity, we have that
$\sum_{n=1}^{\infty}\lambda_{g^{\prime},N^{\prime}}(\overline{B_{n}})>\frac{1}{k}$.
Thus, there must be some $m\in{\mathbb{N}}$ such that
$\sum_{n=1}^{m}\lambda_{g^{\prime},N^{\prime}}(\overline{B_{n}})>\frac{1}{k}$.
Because the $\overline{B_{n}}$ are closed subballs of the open set
$\overline{J_{c}}$ and $\overline{J_{c}}\cap\operatorname{eq}_{N^{\prime}}($
${}^{*}B)=\emptyset$, we know that
$\operatorname{eq}_{N^{\prime}}^{-1}(\overline{B_{n}})\cap$
${}^{*}B=\emptyset$. Consequently, by overflow, for each $n\leq m$ we can find
a closed ball $C_{n}\subseteq J^{\prime}$ such that
$\operatorname{eq}_{N^{\prime}}^{-1}(\overline{B_{n}})\subseteq C_{n}$ and
$C_{n}\cap\text{}^{*}B=\emptyset$. This gives us $m$ gaps of ${}^{*}B$ in
$J^{\prime}$ such that

$\frac{|\bigcup_{n\leq m}C_{n}|}{|J^{\prime}|}\geq\sum_{n\leq
m}\lambda_{g^{\prime},N^{\prime}}(\overline{B_{n}})>\frac{1}{k}.$

This contradicts the fact that no finite number of gaps of ${}^{*}B$ can add
up to $\frac{|J^{\prime}|}{k}$. Thus, $\lambda_{J^{\prime}}($ ${}^{*}B)\geq
1-\frac{1}{k}$.

Finally, $\lambda_{J}($ ${}^{*}B)\geq\lambda_{J}($ ${}^{*}B\cap
J^{\prime})=\operatorname{st}\left(\frac{|J^{\prime}|}{|J|}\right)\lambda_{J^{\prime}}($
${}^{*}B)\geq(1-\frac{1}{k})^{2}$. We conclude that
$\lambda_{J}(^{*}B)>1-\frac{2}{k}$, as desired.

It remains to show that ${}^{*}B$ has the enhanced BM property on some
infinite ball. To see this, note that, for any $N\in$
${}^{*}{\mathbb{N}}\backslash{\mathbb{N}}$, our construction yields that
$I_{N}$ has a subball $J$ with radius at least $N$ and
$g_{{}^{*}A}(J)\leq\delta_{N}<\frac{1}{N}\approx 0$. Since
$g_{{}^{*}B}(J)=g_{{}^{*}A}(J)$, $\lambda_{J}($ ${}^{*}B)=1$, and so ${}^{*}B$
has the enhanced BM property on $J$. ∎

Using this theorem and the definition of SBM set, we are able to reformulate
the definition of supra-SBM sets in the following way.

######  Corollary 5.12.

$A\subseteq G$ is supra-SBM if and only if there is a $B\subseteq A$ and an
infinite, hyperfinite ball $I\subseteq$ ${}^{*}G$ such that ${}^{*}B$ has the
enhanced BM property on $I$.

###### Proof.

First, assume $A$ is supra-SBM. By definition, there is $B\subseteq A$ with
$B$ SBM. This $B$ is as desired by the definition of the SBM property. For the
other direction, if $B\subseteq A$ and ${}^{*}B$ has the enhanced BM property
on $I$, then by Theorem 5.11 $B$ is supra-SBM, and thus so is $A$. ∎

With Theorem 5.11, the proof that all piecewise syndetic sets are supra-SBM
becomes apparent:

###### Proof of Theorem 5.7.

Assume $A\subseteq G$ is piecewise syndetic. Then there is a
$d\in{\mathbb{N}}$ such that, for every $n\in{\mathbb{N}}$, there exists a
$g\in G$ with $A$ having gaps of size at most $d$ on $B(g,n)$. Thus, by
transfer, there is an infinite, hyperfinite ball $I\subseteq$ ${}^{*}G$ such
that ${}^{*}A$ has gaps of size at most $d$ on $I$. Therefore, for every
$J\subseteq I$, $g_{{}^{*}A}(J)\approx 0$ and $\lambda_{J}(^{*}A)=1$. By
Theorem 5.11, ${}^{*}A$ has the enhanced BM property on this $I$ and thus $A$
is supra-SBM. ∎

Theorem 5.11 also has the following nice Ramsey-theoretic consequence:

######  Corollary 5.13.

Being supra-SBM is a partition regular property of subsets of $G$.

###### Proof.

Let $A\subseteq G$ be a supra-SBM set and suppose $A=B_{1}\cup\cdots\cup
B_{n}$. Then there is a subset $B$ of $A$ that is SBM and $B=(B\cap
B_{1})\cup\cdots\cup(B\cap B_{n})$. Thus ${}^{*}B=$ ${}^{*}(B\cap
B_{1})\cup\cdots\cup$ ${}^{*}(B\cap B_{n})$, and ${}^{*}B$ has the enhanced BM
property on some infinite hyperfinite ball $I$. By Theorem 4.5, there is an
$i\in\\{1,\ldots,n\\}$ and an infinite subball $J\subseteq I$ such that
${}^{*}(B\cap B_{i})\cap J$ has the enhanced BM property on $J$. Theorem 5.11
implies that $B\cap B_{i}$ is supra-SBM, whence so is $B_{i}$. ∎

The last result about supra-SBM sets which we will generalize from the theory
of supra-SIM sets is Nathanson’s theorem, as discussed in the introduction.
Proving this is achieved through repeated use of the following lemma.

######  Lemma 5.14.

Suppose that $B\subseteq G$ is supra-SBM. Then there is an $r\in{\mathbb{N}}$
such that, for every $t\in Z(G)$, there is an element $t^{\prime}\in B(t,r)$
for which $B\cap Bt^{\prime}$ is supra-SBM.

###### Proof.

Assume that $A\subseteq B$ is SBM and ${}^{*}A$ has the enhanced BM property
on the infinite, hyperfinite ball $I$. Let $w\in{\mathbb{N}}$ be as in Theorem
4.7 for $A_{1}=A_{2}=$ ${}^{*}A$ and $I_{1}=I_{2}=I$ (for some $\epsilon>0$
and the corresponding $\delta$). As a reminder, this implies that whenever
$J_{i}=B(a_{i},N)\subseteq I$, $i=1,2$, are such that
$g_{{}^{*}A}(J_{i})\leq\delta$, there is some $c\in G$ such that ${}^{*}A\cap
B(a_{i}c,w)\not=\emptyset$.

Now, fix $t\in Z(G)$ and set

$B_{t}=\bigcup_{k\in B(e,2w)}t\text{}^{*}Ak.$

Claim: If $J$ is an infinite, hyperfinite subball of $I$ such that
$\lambda_{J}(^{*}A)>0$, then

${}^{*}A\cap B_{t}\cap J\neq\emptyset.$

Proof of the Claim: By the Lebesgue density theorem, there is $a_{1}\in J$ and
$R\in$ ${}^{*}{\mathbb{N}}$ such that $\lambda_{B(a_{1},R)}(^{*}A)>1-\delta$.
By Lemma 4.2, $g_{{}^{*}A}(B(a_{1},R))<\delta$. Note that because
$d(a_{1},a_{1}t^{-1})=|t^{-1}|$, the symmetric difference $B(a_{1},R)\triangle
B(a_{1}t^{-1},R)$ is contained in a union of spheres:

$B(a_{1},R)\triangle
B(a_{1}t^{-1},R)\subseteq\bigcup_{i=0}^{|t^{-1}|}S(a_{1},R-i)\cup
S(a_{1}t^{-1},R-i).$

Since $t^{-1}\in G$, by the SS property, this is a finite union of measure 0
sets. Thus, we can restrict ${}^{*}A$ to $B(a_{1},R)\cap B(a_{1}t^{-1},R)$
without changing our measure. In other words,

$\lambda_{B(a_{1},R)}(^{*}A)=\lambda_{B(a_{1},R)}(^{*}A\cap B(a_{1},R)\cap
B(a_{1}t^{-1}R))=$ $=\lambda_{B(a_{1}t^{-1},R)}(^{*}A\cap B(a_{1},R)\cap
B(a_{1}t^{-1}R))=\lambda_{B(a_{1}t^{-1},R)}(^{*}A).$

Furthermore, since $t^{-1}\in Z(G)$, we have that

$\lambda_{B(t^{-1}a_{1},R)}(^{*}A)=\lambda_{B(a_{1}t^{-1},R)}(^{*}A)=\lambda_{B(a_{1},R)}(^{*}A)>1-\delta.$

Thus $g_{{}^{*}A}(B(t^{-1}a_{1},R))<\delta$. Therefore, we can apply Theorem
4.7 to $J_{1}=B(a_{1},R)$ and $J_{2}=B(t^{-1}a_{1},R)$, yielding $c\in G$ such
that ${}^{*}A\cap B(a_{1}c,w)\not=\emptyset$ and ${}^{*}A\cap
B(t^{-1}a_{1}c,w)\not=\emptyset$. Note that

${}^{*}A\cap B(t^{-1}a_{1}c,w)\neq\emptyset\iff(t\text{ }^{*}A)\cap
B(a_{1}c,w)\neq\emptyset.$

Let $d\in$ ${}^{*}A\cap B(a_{1}c,w)\subseteq$ ${}^{*}A\cap J$. Then note that
$B(a_{1}c,w)\subseteq B(d,2w)$ and therefore $B(d,2w)=dB(e,2w)$ contains an
element $x$ of $t$ ${}^{*}A$. Thus, there is $k^{-1}\in B(e,2w)$ such that
$dk^{-1}=x$. This $k$ is also contained in $B(e,2w)$, and so $d\in t$
${}^{*}Ak\subseteq B_{t}$, finishing the proof of the claim.

The claim implies that, for any infinite hyperfinite subball $J\subseteq I$,
we have

$\lambda_{J}(^{*}A\cap B_{t})=\lambda_{J}(^{*}A),$

as every infinite subball of $J$ either satisfies $\lambda_{J}(^{*}A)=0$ or
contains an element of ${}^{*}A\cap B_{t}$. Thus, ${}^{*}A\cap B_{t}$ has the
enhanced BM property on $I$. Since $B_{t}$ is a finite union of internal sets,
Theorem 4.5 implies that there is $k\in B(e,2w)$ such that ${}^{*}A\cap
t\text{ }^{*}Ak$ has the enhanced BM property on some infinite subball of $I$.
It follows that $A\cap tAk$ is supra-SBM, whence, so is $B\cap tBk$. Since
$t\in Z(G)$, $B\cap tBk=B\cap Btk$. Thus, for any $t\in Z(G)$, there is
$t^{\prime}=tk\in B(t,2w)$ such that $B\cap Bt^{\prime}$ is supra-SBM. ∎

With this result in place, we prove Nathanson’s theorem for supra-SBM sets
under the additional assumption that the group has infinite center.

######  Theorem 5.15.

If $Z(G)$ is infinite and $A\subseteq G$ is supra-SBM, then for every
$n\in{\mathbb{N}}$, there are $B,C\subseteq G$ such that $B$ is infinite,
$|C|=n$, and $BC\subseteq A$.

###### Proof.

First, note that since $Z(G)$ is infinite, for any supra-SBM set $X$, Lemma
5.14 implies there must be infinitely many $t$ such that $X\cap Xt$ is supra-
SBM. For $n=1$, apply Lemma 5.14 to $A$, yielding $t_{1}\in G$ such that
$B_{1}=A\cap At_{1}$ is supra-SBM, and, in particular, is infinite. Letting
$C_{1}=\\{t_{1}^{-1}\\}$, we have our result. Now, arguing inductively, fix
$m\in{\mathbb{N}}$ and assume we have constructed $B_{m},C_{m}\subseteq G$
such that $B_{m}\subseteq A$ is supra-SBM,
$C_{m}=\\{t_{1}^{-1},...,t_{m}^{-1}\\}$, and $B_{m}C_{m}\subseteq A$. Then,
applying Lemma 5.14 to $B_{m}$, we find $t_{m+1}\in G$ such that
$B_{m+1}=B_{m}\cap B_{m}t_{m+1}$ is supra-SBM and $t_{m+1}^{-1}\not\in C_{m}$.
Letting $C_{m+1}=C_{m}\cup\\{t_{m+1}^{-1}\\}$, we see that this $B_{m+1}$ and
$C_{m+1}$ are as desired. ∎

It is shown in [3] that all finitely generated nilpotent groups have infinite
center. In particular, since Theorem 5.15 uses the SS property but not the SG
property, it applies to finitely generated groups of nilpotency class 2, such
as the discrete Heisenberg group.

###  5.3. Musings on dependence on generators

As a closing remark, it behooves us to make at least some mention of the role
of generating set for the concepts defined in this paper. At the present time,
we have been unable to verify whether or not these concepts are independent of
generating set. One naïve way around this would be to redefine our properties
in the following way.

######  Definition 5.16.

We say that a set $A\subseteq G$ has the weak SBM (WSBM) property if there
exists a finite, symmetric generating set $S$ of $G$ such that $G$ has the SS
and SG properties with respect to this generating set and $A$ is SBM with
respect to $S$. Similary, we say that a set $A\subseteq G$ has the supra-WSBM
property if there is $B\subseteq A$ such that $B$ has the WSBM property.

While these definitions are not very illuminating, they do provide us with a
generator independent class of sets that satisfy all of the important
properties of SBM and supra-SBM sets respectively in virtually nilpotent
groups satisfying the SS and SG properties with respect to some generating
set.

It is natural to ask: what sets are or are not WSBM or supra-WSBM? In order to
partially answer this question, we need the following:

######  Lemma 5.17.

Fix an internal set $A\subseteq$ ${}^{*}G$. Let $S$ and $S^{\prime}$ be finite
generating sets for the virtually nilpotent group $G$, with corresponding word
metrics $d_{S}$ and $d_{S^{\prime}}$, which in turn induce measures
$\lambda_{g,N}$ and $\lambda_{g,N}^{\prime}$ on the corresponding $d_{S}$ and
$d_{S^{\prime}}$ balls respectively. Assume $G$ has the SS property with
respect to $S^{\prime}$. Then the existence of an infinite hyperfinite $d_{S}$
ball $I$ such that $\lambda_{I}(A)>0$ implies the existence of an infinite
hyperfinite $d_{S^{\prime}}$ ball $I^{\prime}$ such that
$\lambda_{I^{\prime}}^{\prime}(A)>0$.

###### Proof.

Let $K\in{\mathbb{N}}$ be such that the identity map on $G$ is a $K$-bi-
Lipschitz map between $(G,d_{S})$ and $(G,d_{S^{\prime}})$. Let $d$ be the
degree of polynomial growth of $G$. Assume that $I$ is an infinite,
hyperfinite $d_{S}$ subball of $G$ such that $\lambda_{I}(A)>0$. By the
Lebesgue density theorem, for any $\epsilon>0$, we can find an infinite
$d_{S}$ subball $J=B(g,N)\subseteq I$ with $\lambda_{J}(A)>1-\epsilon$. We
know that $J$ contains the $d_{S^{\prime}}$ ball
$I^{\prime}=B^{\prime}(g,\lfloor\frac{N}{K}\rfloor)$. Furthermore, there exist
constants $c$ and $c^{\prime}$ such that $|J|\leq cN^{d}$ and
$|I^{\prime}|\geq\frac{c^{\prime}}{K^{d}}N^{d}$. Thus, we see that
$\lambda_{J}(I^{\prime})\geq\frac{c^{\prime}}{cK}$.

Let $\epsilon=\frac{c^{\prime}}{2cK}$. Then, since $\lambda_{J}(A)>1-\epsilon$
and $\lambda_{J}(I^{\prime})>\epsilon$, we must have that $\lambda_{J}(A\cap
I^{\prime})>0$. To conclude the proof, we show that
$\lambda_{I^{\prime}}^{\prime}(A)>0$. Assume, towards a contradiction, that
$\lambda_{I^{\prime}}^{\prime}(A)=0$. In other words,

$\lambda_{g,\lfloor\frac{N}{K}\rfloor}^{\prime}\big{(}\operatorname{eq}_{N}(A\cap
I^{\prime})\big{)}=\mu_{g,\lfloor\frac{N}{K}\rfloor}^{\prime}\big{(}\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}(A\cap
I^{\prime})\big{)}\cap I^{\prime}\big{)}=0.$

Using the SS property for $G$ with respect to $S^{\prime}$, we have that

$\mu_{g,\lfloor\frac{N}{K}\rfloor}^{\prime}\big{(}\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}(A\cap
I^{\prime})\big{)}\big{)}=\mu_{g,\lfloor\frac{N}{K}\rfloor}^{\prime}\big{(}\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}(A\cap
I^{\prime})\big{)}\cap I^{\prime}\big{)}=0.$

Thus, for any $\delta>0$, there must be an internal set $B$ such that
$\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}(A\cap
I^{\prime})\big{)}\subseteq B$ and
$\operatorname{st}\big{(}\frac{|B|}{|I^{\prime}|}\big{)}<\delta$. Since $B$
will also contain
$\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}(A\cap
I^{\prime})\big{)}\cap J$, $B$ serves to witness

$\lambda_{J}(A\cap
I^{\prime})=\mu_{g,N}\big{(}\operatorname{eq}_{N}^{-1}\big{(}\operatorname{eq}_{N}(A\cap
I^{\prime})\big{)}\cap
J\big{)}\big{)}\leq\operatorname{st}\Big{(}\frac{|B|}{|J|}\Big{)}\leq\operatorname{st}\Big{(}\frac{|B|}{|I^{\prime}|}\Big{)}<\delta.$

Letting $\delta\to 0$, we see that $\lambda_{J}(A\cap I^{\prime})=0$, a
contradiction. ∎

Lemma 5.17 shows that many sets in groups with the SS property with respect to
some generating set $S$ are not supra-WSBM.

######  Corollary 5.18.

Let $G$ be a virtually nilpotent group and let $S$ be a finite generating set
for $G$ such that $G$ has the SS property with respect to $S$. Let $A\subseteq
G$ be such that $\lambda_{I}(^{*}A)=0$ for all infinite hyperfinite $d_{S}$
balls $I\subseteq$ ${}^{*}G$. Then $A$ is not supra-WSBM.

###### Proof.

Fix a finite generating set $S^{\prime}$ for $G$ with corresponding measure
$\lambda_{I}^{\prime}$, and let $A$ be as in the statement of the corollary.
The contrapositive of Lemma 5.17 gives us that $\lambda_{I}^{\prime}(^{*}A)=0$
for all infinite hyperfinite $d_{S^{\prime}}$ balls $I$. Therefore, given
$B\subseteq A$, we have that $\lambda_{I}^{\prime}(^{*}B)=0$ for all infinite
hyperfinite $d_{S^{\prime}}$ balls $I$. Thus ${}^{*}B$ does not have the
enhanced SBM property on any infinite hyperfinite $d_{S^{\prime}}$ ball. Since
$S^{\prime}$ is an arbitrary finite generating set for $G$, this yields the
desired conclusion. ∎

######  Example 5.19.

Let $G$ be a finitely generated virtually nilpotent group, and let $S$ be a
finite generating set for $G$ such that $G$ has the SS property with respect
to $S$. Let $A=\\{a_{n}\\}_{n\in{\mathbb{N}}}\subseteq G$ be such that
$a_{1}\neq e$ and $|a_{n+1}|\geq n|a_{n}|$. Then $A$ is not supra-SSBM.

###### Proof.

Let $I=B(g,R)\subseteq$ ${}^{*}G$ be an infinite hyperfinite ball. By
Corollary 5.18, it suffices to show that $\lambda_{I}(^{*}A)=0$. By transfer,
$|a_{N+1}|\geq N|a_{N}|$ for all $N\in{}^{*}\mathbb{N}$. Consequently, for all
$M\leq N$, we have that

$d_{S}(a_{N+1},a_{M})\geq|a_{N+1}|-|a_{M}|\geq|a_{N+1}|-|a_{N}|\geq(N-1)|a_{N}|.$

This also implies that $\\{a_{1},...,a_{N}\\}\subseteq B(e,|a_{N}|)$. Let
$a_{M_{1}},...,a_{M_{K}}$ be all the elements of ${}^{*}A$ contained in $I$,
with $M_{1}<\cdots<M_{K}$. If there are no such elements, then clearly
$\lambda_{I}(^{*}A)=0$. Indeed, we may assume that $K$ is infinite, as
otherwise $\lambda_{I}(^{*}A)=0$ as the $\lambda_{g,N}$-measure of finitely
many equivalence classes is always 0.

If $\operatorname{eq}_{R}(a_{M_{K}})=\operatorname{eq}_{R}(a_{M_{K-1}})$, then
$\dfrac{(M_{k}-2)|a_{M_{K}-1}|}{R}\approx 0$, and thus
$\operatorname{eq}_{R}(a_{M_{1}})=\cdots=\operatorname{eq}_{R}(a_{M_{K}})$.
This again gives us that $\lambda_{I}(^{*}A)=0$.

Finally, assume that
$\operatorname{eq}_{R}(a_{M_{K}})\neq\operatorname{eq}_{R}(a_{M_{K-1}})$.
Since $M_{K-1}\leq M_{K}-1$, we know that

$d_{S}(a_{M_{K}},a_{M_{K-1}})\geq((M_{K}-1)-1)|a_{M_{K}-1}|\geq(M_{K}-2)|a_{M_{K-1}}|.$

Thus, in order for $a_{M_{K}}$ and $a_{M_{K-1}}$ to both be in $B(g,R)$, we
must have that

$2R\geq d_{S}(a_{M_{K}},g)+d_{S}(a_{M_{K-1}},g)\geq
d_{S}(a_{M_{K}},a_{M_{K-1}})\geq(M_{K}-2)|a_{M_{K-1}}|.$

Rearranging this inequality, we find

$\frac{|a_{M_{K-1}}|}{R}\leq\frac{2}{M_{K}-2}\approx 0.$

Thus, $\operatorname{eq}_{R}(B(e,|a_{M_{K-1}}|))$ is a single equivalence
class, meaning once again that $\lambda_{I}(^{*}A)=0$ as the measure of two
equivalence classes. ∎

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