# Semicontinuity of Structure for Small Sumsets in Compact Abelian Groups

John T. Griesmer

###### Abstract

We study pairs of subsets $A,B$ of a compact abelian group $G$ where the
sumset $A+B:=\\{a+b:a\in A,b\in B\\}$ is small. Let $m$ and $m_{*}$ be Haar
measure and inner Haar measure on $G$, respectively. Given $\varepsilon>0$, we
classify all pairs $A,B$ of Haar measurable subsets of $G$ satisfying
$m(A),m(B)>\varepsilon$ and $m_{*}(A+B)\leq m(A)+m(B)+\delta$ where
$\delta=\delta(\varepsilon)$ is small. We also study the case where the
$\delta$-popular sumset $A+_{\delta}B:=\\{t\in G:m(A\cap(t-B))>\delta\\}$ is
small. We prove that for all $\varepsilon>0$, there is a $\delta>0$ such that
if $A$ and $B$ are subsets of a compact abelian group $G$ having
$m(A),m(B)>\varepsilon$ and $m(A+_{\delta}B)\leq m(A)+m(B)+\delta$, then there
are sets $S,T\subseteq G$ such that $m(A\triangle S)+m(B\triangle
T)<\varepsilon$ and $m(S+T)\leq m(S)+m(T)$. Appealing to known results, the
latter inequality yields strong structural information on $S$ and $T$, and
therefore on $A$ and $B$.

title = Semicontinuity of Structure for Small Sumsets in Compact Abelian
Groups, author = John T. Griesmer, plaintextauthor = John T. Griesmer,
runningtitle = Semicontinuity for small sumsets, keywords = sumset, inverse
theorem, ultraproduct, year=2019, number=18, received=27 August 2018,
revised=20 May 2019, published=29 November 2019, doi=10.19086/da.11089,

[classification=text]

##  1 Sumsets in compact abelian groups

If $G$ is an abelian group and $A$ and $B$ are subsets of $G$, let $A+B$
denote the _sumset_ $\\{a+b:a\in A,b\in B\\}$ and $A-B$ denote the _difference
set_ $\\{a-b:a\in A,b\in B\\}$. If $t\in G$ write $A+t$ for the _translate_
$\\{a+t:a\in A\\}$. Let $-A$ denote the set $\\{-a:a\in A\\}$. In this article
all topological groups are assumed to be Hausdorff, and all finite groups are
endowed with the discrete topology.

If $G$ is a compact abelian group with Haar probability measure $m$, let
$m_{*}$ denote the corresponding _inner Haar measure_ :
$m_{*}(A)=\sup\\{m(C):C\subseteq A,C\text{ is compact}\\}$. We say that a set
$A\subseteq G$ is _Haar measurable_ (or _$m$ -measurable_) if it lies in the
completion of the Borel $\sigma$-algebra of $G$ with respect to $m$. We are
forced to consider inner Haar measure, as the sumset of two $m$-measurable
sets can fail to be $m$-measurable. The structure of Haar measurable sets
$A,B\subseteq G$ satisfying $m(A),m(B)>0$ and

$m_{*}(A+B)\leq m(A)+m(B)$ (1.1)

is well understood; see §3 for a brief (and incomplete) summary or [19, 22,
23, 29, 44] for comprehensive exposition.

For $\delta\geq 0$, let $A+_{\delta}B:=\\{t\in G:m(A\cap(t-B))>\delta\\}$ be
the _$\delta$ -popular sumset_. The function $t\mapsto m(A\cap(t-B))$ is
continuous, so $A+_{\delta}B$ is an open subset of $A+B$. Thus the hypothesis
$m(A+_{\delta}B)\leq m(A)+m(B)+\delta$ is weaker than Inequality (1.1). Our
first result is the following theorem, which says that for very small
$\delta$, pairs of sets satisfying this weaker inequality resemble pairs
satisfying (1.1). This resolves the special case of Conjecture 5.1 in [44]
with the additional assumption that $G$ is abelian. Here $A\triangle B$
denotes the symmetric difference $(A\setminus B)\cup(B\setminus A)$.

######  Theorem 1.1.

For all $\varepsilon>0$ there exists $\delta>0$ such that for every compact
abelian group $G$ with Haar probability measure $m$ and all $m$-measurable
sets $A,B\subseteq G$ with $m(A),m(B)>\varepsilon$ and $m(A+_{\delta}B)\leq
m(A)+m(B)+\delta$, there exist $m$-measurable sets $S,T\subseteq G$ such that
$m(A\triangle S)+m(B\triangle T)<\varepsilon$, $S+T$ is $m$-measurable, and
$m(S+T)\leq m(S)+m(T)$.

Note: $\delta$ depends only on $\varepsilon$ and not on $G$ in Theorem 1.1.

Theorem 1.1 generalizes Theorem 1.5 of [44], which imposes the additional
assumption that $G$ is connected. A quantitative version of this result in the
case where $G$ is a cyclic group of prime order is proved in [32]. See also
[33] for a version in $\mathbb{Z}$, and more recently [41]. As a byproduct of
the proof of Theorem 1.1, we provide in Theorem 1.15 a classification of sets
satisfying $m_{*}(A+B)\leq m(A)+m(B)+\delta$, where $\delta$ is very small
compared to $m(A)$ and $m(B)$. This generalizes Theorem 1.3 of [44], which
assumes connectedness of $G$. Theorems 1.1 and 1.15 are new for every infinite
compact disconnected abelian group, such as
$\mathbb{T}\times(\mathbb{Z}/N\mathbb{Z})$.

Before stating Theorem 1.15 we introduce some terminology and notation.

######  Notation 1.2.

Let $(X,\mu)$ be a measure space and $A,B\subseteq X$. We write $A\sim_{\mu}B$
if $\mu(A\triangle B)=0$ and write $A\subset_{\mu}B$ if $\mu(A\setminus B)=0$.
If $f$ and $g$ are functions on $X$, we write $f\equiv_{\mu}g$ if $f(x)=g(x)$
for $\mu$-almost every $x$.

For any given set $X$, we will consider only one measure $\mu$ on $X$, and
hence only one $\sigma$-algebra of measurable sets; we will not write the
$\sigma$-algebra explicitly.

If $G$ is an abelian group, a measure $\mu$ on $G$ is _translation invariant_
if for every $\mu$-measurable set $A\subseteq G$ and every $t\in G$, $A+t$ is
$\mu$-measurable and $\mu(A+t)=\mu(A)$. For such a measure and $f,g\in
L^{2}(\mu)$, we define the _$\mu$ -convolution_ $f*_{\mu}g$ by
$f*_{\mu}g(x):=\int f(t)g(x-t)\,d\mu(t)$. In the sequel the measure $\mu$ may
be understood from context, and we may write $f*g$ in place of $f*_{\mu}g$.
Note that if $A:=\\{x:f(x)>0\\}$ and $B:=\\{x:g(x)>0\\}$, then
$\\{x:f*_{\mu}g(x)>0\\}\subseteq A+B$. Furthermore, the $\delta$-popular
sumset defined above can be written in terms of convolution as
$A+_{\delta}B=\\{x\in G:1_{A}*_{m}1_{B}(x)>\delta\\}$. Theorem 1.1 thereby
yields information on the supports of $f$ and $g$ under a strong assumption on
a level set of $f*g$. For connected groups, the recent article [9] proves more
quantitative and detailed results on the relationship between $f$, $g$, and
$f*g$.

If $H$ is a compact abelian group and $\pi:G\to H$ is a homomorphism, we say
that $\pi$ is _$\mu$ -measurable_ if $\pi^{-1}(A)$ is $\mu$-measurable for
every open set $A\subseteq H$. If $m$ is Haar probability measure on $H$, we
say that $\pi$ _preserves $\mu$_ if $\mu(\pi^{-1}(A))=m(A)$ for every
$m$-measurable set $A\subseteq H$. Note that $\pi$ preserves $\mu$ if and only
if $\int f\circ\pi\,d\mu=\int f\,dm$ for all $f\in L^{1}(m)$.

If the measures $\mu$ and $m$ are clear from context, we may say that “$\pi$
is measure preserving” instead of “$\pi$ preserves $\mu$”.

######  Lemma 1.3.

Let $G$ be an abelian group and $\mu$ a translation invariant probability
measure on $G$. If $H$ is a compact abelian group with Haar probability
measure $m$ and $\pi:G\to H$ is a surjective $\mu$-measurable homomorphism,
then $\pi$ preserves $\mu$. Furthermore, if $f,g\in L^{2}(m)$ then
$(f\circ\pi)*_{\mu}(g\circ\pi)=(f*_{m}g)\circ\pi$.

###### Proof.

Define a measure $\eta$ on the $\sigma$-algebra of Borel subsets of $H$ by
$\eta(A):=\mu(\pi^{-1}(A))$. It is easy to check that $\eta$ is a translation
invariant probability measure, so uniqueness of Haar measure implies
$\eta(A)=m(A)$ for all $m$-measurable subsets of $H$. Thus $\pi$ preserves
$\mu$. The last assertion is straightforward to verify from the definition of
convolution and the fact that $\pi$ preserves $\mu$. ∎

Our results relate subsets of a given compact abelian group $G$ to subsets of
quotients of $G$ which are isomorphic to $\mathbb{T}$ or to a finite group;
the following definition helps describe such sets.

######  Definition 1.4 (Bohr intervals and cyclic progressions).

Let $\mathbb{T}:=\mathbb{R}/\mathbb{Z}$ with the usual topology. The elements
of $\mathbb{T}$ are $a+\mathbb{Z}$, where $a\in\mathbb{R}$, so every subset of
$\mathbb{T}$ may be identified with a set $E+\mathbb{Z}$, where
$E\subseteq[0,1]\subseteq\mathbb{R}$. An _interval_ in $\mathbb{T}$ is a set
of the form $[a,b]+\mathbb{Z}$, where $a,b\in\mathbb{R}$ and $0<b-a\leq 1$. We
will abuse notation and write $[a,b]$ for the set $[a,b]+\mathbb{Z}$ contained
in $\mathbb{T}$. Unless otherwise specified, the intervals we consider are
closed, as indicated in the definition.

Let $\lambda$ denote Haar probability measure on $\mathbb{T}$. Note that if
$I$ and $J$ are intervals in $\mathbb{T}$ then $I+J$ is an interval, and
$\lambda(I+J)=\min\\{1,\lambda(I)+\lambda(J)\\}$.

Let $G$ be an abelian group and $\mu$ a translation invariant probability
measure on $G$. We say that $A\subseteq G$ is a _Bohr interval_ if there is a
surjective $\mu$-measurable homomorphism $\tau:G\to\mathbb{T}$ and an interval
$I\subseteq\mathbb{T}$ such that $A=\tau^{-1}(I)$.

If $A,B\subseteq G$ are Bohr intervals, we say that $A$ and $B$ are _parallel_
if there is a single surjective $\mu$-measurable homomorphism
$\tau:G\to\mathbb{T}$ and intervals $I,J\subseteq\mathbb{T}$ such that
$A=\tau^{-1}(I)$ and $B=\tau^{-1}(J)$.

If $N\in\mathbb{N}$, we say that $A\subseteq G$ is an _$N$ -cyclic
progression_111This terminology is not standard, but it satisfies our desire
to include the cardinality of the group $\mathbb{Z}/N\mathbb{Z}$ in the
terminology, as a bound on $N$ is stated in Theorem 1.15. if there is a
surjective $\mu$-measurable homomorphism $\tau:G\to\mathbb{Z}/N\mathbb{Z}$ and
an arithmetic progression $P\subseteq\mathbb{Z}/N\mathbb{Z}$ of common
difference $1$ such that $A=\tau^{-1}(P)$.

If $A$ and $B$ are $N$-cyclic progressions, we say that $A$ and $B$ are
_parallel_ if there is a single surjective $\mu$-measurable homomorphism
$\tau:G\to\mathbb{Z}/N\mathbb{Z}$ such that $A=\tau^{-1}(P)$ and
$B=\tau^{-1}(Q)$ for arithmetic progressions
$P,Q\subseteq\mathbb{Z}/N\mathbb{Z}$ of common difference $1$.

If $K\leq G$ is a $\mu$-measurable finite index subgroup of $G$,
$C^{\prime}\subseteq K$ is a Bohr interval in $K$, and $a\in G$, we say that
$a+C^{\prime}$ is a _relative Bohr interval_ in $G$. A _relative $N$-cyclic
progression_ is defined similarly.

######  Remark 1.5.

An $N$-cyclic progression is not an arithmetic progression in $G$, unless $G$
itself is a cyclic group of order $N$.

######  Notation 1.6.

We write $C_{N}$ for the subgroup
$\\{\frac{j}{N}:j\in\mathbb{Z}\\}\subseteq\mathbb{T}$, so that $C_{N}$ is
isomorphic to $\mathbb{Z}/N\mathbb{Z}$. If $\tau:G\to\mathbb{T}$ is a
$\mu$-measurable homomorphism such that $\tau(G)=C_{N}$ and
$I\subseteq\mathbb{T}$ is an interval, then $\tau^{-1}(I)$ is an $N$-cyclic
progression. The $N$-cyclic progressions we encounter in our proofs will have
this form, and the next lemma estimates their measure.

######  Lemma 1.7.

If $I\subseteq\mathbb{T}$ is an interval, then $N\lambda(I)-1\leq|C_{N}\cap
I|\leq N\lambda(I)+1$. Consequently, if $\tau:G\to C_{N}$ is a surjective
$\mu$-measurable homomorphism, then $\tilde{I}:=\tau^{-1}(I)$ is an $N$-cyclic
progression satisfying
$\lambda(I)-\frac{1}{N}\leq\mu(\tilde{I})\leq\lambda(I)+\frac{1}{N}$.

###### Proof.

Haar measure on $C_{N}$ is normalized counting measure, so by Lemma 1.3 it
suffices to estimate the cardinality of $I\cap C_{N}$. To do this we identify
$I\cap C_{N}$ with a set of the form
$[a,b]\cap\\{\frac{j}{N}:j\in\mathbb{Z}\\}$ contained in $\mathbb{R}$, where
$b-a=\lambda(I)$. The last intersection has the same cardinality as
$[Na,Nb]\cap\mathbb{Z}$, so we use the estimate
$d-c-1\leq|[c,d]\cap\mathbb{Z}|\leq d-c+1$, which holds for all $c\leq
d\in\mathbb{R}$. Thus $N(b-a)-1\leq|I\cap C_{N}|\leq N(b-a)+1$, which is
equivalent to the estimate stated in the lemma. ∎

######  Remark 1.8.

If $A$ and $B$ are parallel Bohr intervals such that $A=~{}\tau^{-1}(I)$ and
$B=\tau^{-1}(J)$ where $I,J$ are intervals in $\mathbb{T}$, then Lemma 1.3
implies $\mu(A)=\lambda(I)$, $\mu(B)=\lambda(J)$, and
$\mu(A+B)=\min\\{1,\mu(A)+\mu(B)\\}$. Similarly, if $A$ and $B$ are parallel
$N$-cyclic progressions then $\mu(A+B)=\min\\{1,\mu(A)+\mu(B)-\frac{1}{N}\\}$.

######  Remark 1.9.

Pairs of parallel Bohr intervals are called “parallel Bohr sets” in [44].
These form a special case of the _Sturmian pairs_ considered in [4, 5], which
deal with the abelian and nonabelian settings.

######  Example 1.10 (Bohr intervals).

Let $I\subseteq\mathbb{T}$ be the interval corresponding to $[0,\frac{1}{3}]$
and define $\tau:\mathbb{T}\to\mathbb{T}$ by $\tau(x)=2x$. Then
$\tilde{I}:=\tau^{-1}(I)$ is a Bohr interval. Here $\tilde{I}$ corresponds to
$[0,\frac{1}{6}]\cup[\frac{1}{2},\frac{2}{3}]$, so a Bohr interval in
$\mathbb{T}$ is not necessarily an interval.

Let $I\subseteq\mathbb{T}$ be the interval corresponding to $[0,\frac{1}{3}]$
and define a homomorphism $\tau:\mathbb{T}\times\mathbb{T}\to\mathbb{T}$ by
$\tau(x,y)=x-2y$. Then
$\tilde{I}:=\tau^{-1}(I)=\\{(x,y)\in\mathbb{T}\times\mathbb{T}:x-2y\in I\\}$
is a Bohr interval in $\mathbb{T}\times\mathbb{T}$.

######  Example 1.11 ($N$-cyclic progressions).

If $N,a\in\mathbb{N}$, with $a\leq N$, then the residues $\\{0,1,\dots,a-1\\}$
in $\mathbb{Z}/N\mathbb{Z}$ form an $N$-cyclic progression. If $p$ is prime
and $A\subseteq\mathbb{Z}/p\mathbb{Z}$ is an arithmetic progression, i.e. a
set of the form $\\{a,a+d,a+2d,\dots,a+kd\\}$, where
$a,d\in\mathbb{Z}/p\mathbb{Z}$, $d\neq 0$, $k\in\mathbb{N}$, then $A$ is a
$p$-cyclic progression.

For the following definitions, fix an abelian group $G$ with a translation
invariant probability measure $\mu$, and let $\mu_{*}$ be the associated inner
measure.

######  Definition 1.12 (Periodicity).

Given a subgroup $K\leq G$, a set $A\subseteq G$ is a union of cosets of $K$
if and only if $A=A+K$. The latter equation will abbreviate the assertion “$A$
is a union of cosets of $K$”. We may call such a set $A$ _periodic with
respect to_ $K$.

If $A\sim_{\mu}A+K$, we say that $A$ is _essentially periodic with respect to_
$K$. When $A$ is essentially periodic with respect to $K$, then $A$ is
$\mu$-similar to a union of cosets of $K$, but the converse does not hold. For
example, if $G$ is an infinite compact abelian group with Haar probability
measure $\mu$, $K\leq G$ is a proper open subgroup, $c\notin K$, and
$A=K\cup\\{c\\}$, then $A\sim_{\mu}K$ but $A\nsim_{\mu}A+K$.

The following definition describes sets which are close to being periodic. It
is stronger than merely insisting that a given set has small symmetric
difference with a periodic set.

######  Definition 1.13.

If $K\leq G$ is a $\mu$-measurable subgroup with $\mu(K)>0$ and
$\varepsilon\geq 0$, a set $A\subseteq G$ is _$\varepsilon$ -periodic with
respect to $K$_ if $\mu(A+K)-\mu_{*}(A)\leq\varepsilon\mu(K)$.

The term “quasi-periodic” is introduced in [22] and [23] to describe pairs of
sets satisfying $|A+B|\leq|A|+|B|$. Here is the natural generalization of the
term to a group with a translation invariant measure.

######  Definition 1.14 (Quasi-periodicity).

Let $K\leq G$ be a $\mu$-measurable subgroup with $\mu(K)>0$. We say that
$A\subseteq G$ is _quasi-periodic_ with respect to $K$ if $A$ can be
partitioned into two sets $A_{1}$ and $A_{0}$, where $A_{0}\neq\varnothing$,
$(A_{1}+K)\cap A_{0}=\varnothing$, $A_{1}\sim_{\mu}A_{1}+K$, and $A_{0}$ is
contained in a coset of $K$. For such $A_{1},A_{0}$, the expression $A_{1}\cup
A_{0}$ is called a _quasi-periodic decomposition of $A$ with respect to $K$_,
or simply a _quasi-periodic decomposition_ , if the group $K$ is clear from
context.

If $\varepsilon>0$, we say that $A\subseteq G$ is _$\varepsilon$ -quasi
periodic with respect to $K$_ if $A=A_{1}\cup A_{0}$, where
$A_{0}\neq\varnothing$, $(A_{1}+K)\cap A_{0}=\varnothing$, $A_{1}$ is
$\varepsilon$-periodic with respect to $K$, and $A_{0}$ is contained in a
coset of $K$.

We allow $A_{1}$ to be empty in the definitions of “quasi-periodic” and
“$\varepsilon$-quasi-periodic”, but we do not allow $A_{0}$ to be empty.

If we say that “$A=A_{1}\cup A_{0}$ is a quasi-periodic (or
$\varepsilon$-quasi-periodic) decomposition of $A$ with respect to $K$”, then
$A_{1}$ and $A_{0}$ satisfy the conditions listed in Definition 1.14.

For a compact abelian group $G$ with Haar measure $m$, Theorem 1.15 provides
detailed structural information about pairs of sets $A,B\subseteq G$ when
$\delta$ is very small and $m_{*}(A+B)\leq m(A)+m(B)+\delta$. Our methods do
not provide a specific $\delta$ for which our conclusions hold. Note that an
$m$-measurable subgroup $K\leq G$ has finite index if and only if $K$ is
compact and open, if and only if $m(K)>0$.

######  Theorem 1.15.

For all $\varepsilon>0$, there exists $\delta>0$ and $d\in\mathbb{N}$ such
that if $G$ is a compact abelian group with Haar probability measure $m$ and
$A,B\subseteq G$ have $m(A)>\varepsilon$, $m(B)>\varepsilon$, and
$m_{*}(A+B)\leq m(A)+m(B)+\delta$, then there is a compact open subgroup
$K\leq G$ (possibly $K=G$) having index at most $d$ such that exactly one of
the following holds.

  1. (I)

$A+B$ is $\varepsilon$-periodic with respect to $K$ and $m(A+B+K)\leq
m(A+K)+m(B+K)$.

  2. (II)

$A+B$ is not $\varepsilon$-periodic with respect to $K$, while $A$ and $B$
have $\varepsilon$-quasi-periodic decompositions $A=A_{1}\cup A_{0}$,
$B=B_{1}\cup B_{0}$ with respect to $K$, where at least one of $A_{1},B_{1}$
is nonempty and

$m_{*}(A_{0}+B_{0})\leq m(A_{0})+m(B_{0})+\varepsilon m(K),$

or

  3. (III)

$m_{*}(A+B)<(1-\varepsilon)m(K)$, and there are
$A^{\prime},B^{\prime}\subseteq K$ such that $m(A^{\prime})<m(A)+\varepsilon$,
$m(B^{\prime})<m(B)+\varepsilon$ and $a,b\in G$ such that $A\subseteq
a+A^{\prime}$ and $B\subseteq b+B^{\prime}$, where

    1. (III.1)

$A^{\prime},B^{\prime}$ are parallel Bohr intervals in $K$, or

    2. (III.2)

$A^{\prime},B^{\prime}$ are parallel $N$-cyclic progressions in $K$ for some
$N>d$.

Note: $\delta$ and $d$ are independent of the group $G$.

######  Remark 1.16.

Theorem 1.15 does not assume $m_{*}(A+B)<1-\varepsilon$ or even
$m_{*}(A+B)<1$; the possibility $m_{*}(A+B)\approx 1$ is accounted for in
conclusion (I), where the group $K$ may be equal to $G$. See §13.1 for
elaboration.

Connected groups have no proper open subgroups, so the special case of Theorem
1.15 where $G$ is connected recovers the following result from [44].

######  Theorem 1.17 ([44], Theorem 1.3).

For all $\varepsilon>0$, there exists $\delta>0$ such that if $G$ is a
connected compact abelian group and $A,B\subseteq G$ have $m(A)>\varepsilon$,
$m(B)>\varepsilon$, $m(A)+m(B)<1-\varepsilon$ and $m_{*}(A+B)\leq
m(A)+m(B)+\delta$, then there are parallel Bohr intervals
$A^{\prime},B^{\prime}\subseteq G$ such that $A\subseteq A^{\prime}$,
$B\subseteq B^{\prime}$, and $m(A^{\prime}\setminus A)+m(B^{\prime}\setminus
B)<\varepsilon$.

The following corollary of Theorem 1.15 is a weak version of Theorem 21.8 of
[24] and Theorem 1.3 of [40]. We obtain it by specializing Theorem 1.15 to the
case where $G$ is a cyclic group of prime order and adding the hypothesis
$m(A)+m(B)<1-\varepsilon$, which rules out conclusion (I).

######  Corollary 1.18.

For all $\varepsilon>0$, there exists $N\in\mathbb{N}$, $\delta>0$, such that
for every prime $p>N$ and all sets $A,B\subseteq\mathbb{Z}/p\mathbb{Z}$
satisfying $|A|,|B|\geq\varepsilon p$, $|A|+|B|<(1-\varepsilon)p$, and
$|A+B|\leq|A|+|B|+\delta p$, there are arithmetic progressions $I\supseteq A$,
$J\supseteq B$ in $\mathbb{Z}/p\mathbb{Z}$ both having common difference
$k\neq 0$ such that $|I|\leq(1+\varepsilon)|A|$ and
$|J|\leq(1+\varepsilon)|B|$.

Corollary 1.18 follows immediately from Theorem 1.15: when $\varepsilon>0$ is
fixed, $\delta$ and $d$ are as in the conclusion, and $p>d$, there is no
proper subgroup of $\mathbb{Z}/p\mathbb{Z}$ having index at most $d$, so only
conclusion (III.2) can hold, and the subgroup $K$ therein is equal to $G$.

Theorem 1.15 is new for every infinite disconnected compact abelian group,
such as $\mathbb{T}\times(\mathbb{Z}/N\mathbb{Z})$. The special case of
Theorem 1.15 where $G$ is an arbitrary finite abelian group is apparently also
new.

###  1.19 Acknowledgement

Two anonymous referees for https://discreteanalysisjournal.com/_Discrete
Analysis_ contributed many corrections and improvements to this article.

##  2 Outline of proofs

We prove Theorems 1.1 and 1.15 in §12 and §11, respectively. The proofs are
carried out in three parts, which we outline here. We assume familiarity with
ultraproducts and Loeb measure; the reader can refer to Appendix A for a
synopsis and references. We now fix a nonprincipal ultrafilter $\mathcal{U}$
on $\mathbb{N}$.

Part 1. Classify all pairs of $\mu$-measurable sets $A,B\subseteq\mathbf{G}$
satisfying $\mu(A)>0$, $\mu(B)>0$, and $\mu_{*}(A+B)\leq\mu(A)+\mu(B)$, where
$\mathbf{G}$ is an ultraproduct $\prod_{n\to\mathcal{U}}G_{n}$ of compact
abelian groups $G_{n}$ with Haar probability measure $m_{n}$, and $\mu$,
$\mu_{*}$ are the corresponding Loeb measure and inner Loeb measure on
$\mathbf{G}$. This is done in Propositions 10.1 and 10.2, whose proofs we
outline here.

To prove Propositions 10.1 and 10.2, we use Corollary 5.5 to model an
arbitrary sumset in $\mathbf{G}$ by a sumset in a compact quotient of
$\mathbf{G}$. We then apply known inverse theorems in compact abelian groups
to these models and transfer the information to the original group
$\mathbf{G}$. Given $\mu$-measurable sets $A,B\subseteq\mathbf{G}$, Corollary
5.5 provides:

  1. $\bullet$

a compact abelian group $G$ with Haar probability measure $m$,

  2. $\bullet$

a surjective $\mu$-measurable (and measure preserving) homomorphism
$\pi:\mathbf{G}\to G$, and

  3. $\bullet$

Borel sets $C,D\subseteq G$ such that $C+D$ is Borel,

$A\subset_{\mu}A^{\prime}:=\pi^{-1}(C),\quad
B\subset_{\mu}B^{\prime}:=\pi^{-1}(D),\quad\text{and}\quad
A^{\prime}+B^{\prime}\subset_{\mu}A+B.$ (2.1)

Assuming $A,B\subseteq\mathbf{G}$ satisfy $\mu_{*}(A+B)\leq\mu(A)+\mu(B)$, the
containments in (2.1) and the fact that $\pi$ preserves $\mu$ imply
$m(C+D)\leq m(C)+m(D)$. This inequality allows us to apply Theorems 3.2 and
3.5 to classify $C$ and $D$. Some ad hoc arguments then lead to the desired
classification of $A$ and $B$, based on the structure of $C$ and $D$ and the
hypothesis $\mu_{*}(A+B)\leq\mu(A)+\mu(B)$. Some of these arguments are
carried out in §9, which generalizes some lemmas from [19] while providing
more efficient proofs.

Part 2. Prove that if an _internal_ set
$\mathbf{A}=\prod_{n\to\mathcal{U}}A_{n}$ in $\mathbf{G}$ is highly structured
(i.e. $\mathbf{A}$ is quasi-periodic, or a Bohr interval, etc.), then the
constituent sets $A_{n}$ are themselves highly structured. For example, if
$\mathbf{A}$ is a $\mu$-measurable internal finite index subgroup of
$\mathbf{G}$, we will see that for $\mathcal{U}$-many $n$, the set $A_{n}$ is
also a finite index subgroup of $G_{n}$ with the same index as $\mathbf{A}$.
This is done in §6 and §8.

Parts 1 and 2 both rely on Lemma 4.6 and Proposition 4.14, two of the main
results of [43]. Lemma 4.6 characterizes the $\mu$-measurable homomorphisms
$\chi:\mathbf{G}\to\mathcal{S}^{1}$, and Proposition 4.14 exploits this
characterization to find compact quotients of $\mathbf{G}$ which are useful
for studying convolutions and sumsets.

Part 3. Prove Theorem 1.15 by contradiction, following the strategy of [44].
Assuming Theorem 1.15 is false, then for some $\varepsilon>0$ and each
$n\in\mathbb{N}$ there is a compact abelian group $G_{n}$ with Haar
probability measure $m_{n}$ and sets $A_{n},B_{n}\subseteq G_{n}$ with
$m_{n}(A_{n}),m_{n}(B_{n})>\varepsilon$, while $A_{n}+B_{n}$ has inner Haar
measure at most $m_{n}(A_{n})+m_{n}(B_{n})+\frac{1}{n}$, and none of the
conclusions (I)-(III) in Theorem 1.15 are satisfied by $A_{n},B_{n}$, and a
compact open subgroup $K_{n}\leq G_{n}$ of index at most $n$. We consider the
ultraproduct $\mathbf{G}=\prod_{n\to\mathcal{U}}G_{n}$ with Loeb measure $\mu$
based on $m_{n}$ and the internal sets
$\mathbf{A}:=\prod_{n\to\mathcal{U}}A_{n}$ and
$\mathbf{B}:=\prod_{n\to\mathcal{U}}B_{n}\subseteq\mathbf{G}$. The definition
of Loeb measure implies $\mu(\mathbf{A}),\mu(\mathbf{B})\geq\varepsilon$, and
$\mu_{*}(\mathbf{A}+\mathbf{B})\leq\mu(\mathbf{A})+\mu(\mathbf{B})$, so we may
appeal to Propositions 10.1 and 10.2 to identify the structure of $\mathbf{A}$
and $\mathbf{B}$. The results of §6 and §8 then derive the desired structure
of $A_{n}$ and $B_{n}$. This produces a fixed $d\in\mathbb{N}$ and compact
open subgroups $K_{n}\leq G_{n}$ of index at most $d$ such that $A_{n},B_{n}$,
and $K_{n}$ satisfy the conclusion of Theorem 1.15 for $\mathcal{U}$-many $n$,
contradicting the assumption to the contrary.

In §7 we provide a separate proof of Theorem 1.17, where the main ideas of our
appoach appear and connectedness prevents the profusion of cases seen in the
general setting.

The proof of Theorem 1.1 follows an outline similar to the proof of Theorem
1.15: we assume, to get a contradiction, that for some $\varepsilon>0$ and
each $n\in\mathbb{N}$ there are $A_{n},B_{n}\subseteq G_{n}$ such that
$m_{n}(A_{n}),m_{n}(B_{n})>\varepsilon$ and
$m_{n}(A_{n}+_{\frac{1}{n}}B_{n})<m_{n}(A_{n})+m_{n}(B_{n})+\frac{1}{n}$,
while $A_{n},B_{n}$ do not satisfy the conclusion of Theorem 1.15. In the
ultraproduct $\mathbf{G}$, this produces a pair $\mathbf{A},\mathbf{B}$ where
$\mu(\mathbf{A}+_{0}\mathbf{B})\leq\mu(\mathbf{A})+\mu(\mathbf{B})$.
Proposition 5.1, Lemma 5.3, and some ad hoc arguments then provide sets
$A^{\prime}$ and $B^{\prime}$ such that $A^{\prime}\sim_{\mu}\mathbf{A}$ and
$B^{\prime}\sim_{\mu}\mathbf{B}$ while
$\mu(A^{\prime}+B^{\prime})\leq\mu(A^{\prime})+\mu(B^{\prime})$. Lemma 12.3, a
consequence of Propositions 10.1 and 10.2, then provides sets $S_{n}$,
$T_{n}\subseteq G_{n}$ having $m_{n}(S_{n}+T_{n})\leq
m_{n}(S_{n})+m_{n}(T_{n})$ and $m_{n}(A_{n}\triangle
S_{n})+m_{n}(B_{n}\triangle T_{n})<\varepsilon$, contradicting our assumption
to the contrary.

##  3 Summary of known inverse theorems

We summarize some known inverse theorems for sumsets in a locally compact
abelian (LCA) group $G$. Our proofs proceed by transferring these results to
ultraproducts of compact abelian groups.

######  Definition 3.1.

Let $G$ be an abelian group and $A\subseteq G$. The group $H(A):=\\{g\in
G:A+g=A\\}$ is called the _stabilizer_ of $A$.

Note that $A+H(A)=A$, and $H(A)$ is the largest set $S$ such that $A+S=A$.
Consequently, if $H(A)$ is open, then $A$ is open as well.

The next three results are due to M. Kneser.

######  Theorem 3.2 ([29], Satz 1).

Let $G$ be a locally compact abelian group with Haar measure $m$. If
$A,B\subseteq G$ are $m$-measurable sets satisfing $m_{*}(A+B)<m(A)+m(B)$ then
$H:=H(A+B)$ is compact and open (so $A+B$ is clopen and therefore measurable),
and

$m(A+B)=m(A+H)+m(B+H)-m(H).$ (3.1)

Equation (3.1) plays a crucial role in our arguments; some of its consequences
are developed in §9.

If $G$ is a connected LCA group then $G$ has no proper open subgroup, so the
next corollary follows from Theorem 3.2.

######  Corollary 3.3.

If $G$ is a connected LCA group with Haar measure $m$ and $A,B\subseteq G$ are
$m$-measurable sets, then $m_{*}(A+B)\geq\min\\{m(G),m(A)+m(B)\\}$.

When $G$ is compact, the following theorem classifies the pairs of $m$-nonnull
sets where equality holds in Corollary 3.3. We use terminology from Definition
1.4 to state it.

######  Theorem 3.4 ([29], Satz 2).

If $G$ is a connected compact abelian group with Haar measure $m$ and
$A,B\subseteq G$ are $m$-measurable sets satisfying $m(A),m(B)>0$ and

$m_{*}(A+B)=m(A)+m(B)<1,$

then there are parallel Bohr intervals $A^{\prime},B^{\prime}\subseteq G$ with
$m(A^{\prime})=m(A)$, $m(B^{\prime})=m(B)$, and $A\subseteq A^{\prime}$,
$B\subseteq B^{\prime}$.

The next result extends Theorem 3.4 to disconnected groups. It is a
consequence of Theorem 1.3 in [19], classifying the pairs of subsets $A,B$ of
a compact abelian group where $m_{*}(A+B)=m(A)+m(B)$. We could apply those
results from [19] directly, but we find it easier to use the following
theorem, which presents the same result in a more convenient organization. Its
derivation from the results of [19] is explained in Remark 3.7. Here we use
terminology and notation from §1.

######  Theorem 3.5.

Let $G$ be a compact abelian group with Haar probability measure $m$ and let
$A,B\subseteq G$ be $m$-measurable sets with $m(A),m(B)>0$ and
$m_{*}(A+B)=m(A)+m(B)$. Then there is a compact open subgroup $K\leq G$
(possibly $K=G$) such that exactly one of the following holds.

  1. (I)

$A+B\sim_{m}A+B+K$,

  2. (II)

$A+B\nsim_{m}A+B+K$, and $A$ and $B$ have quasi-periodic decompositions
$A=A_{1}\cup A_{0}$, $B=B_{1}\cup B_{0}$ with respect to $K$, where at least
one of $A_{1}$ or $B_{1}$ is nonempty and
$m_{*}(A_{0}+B_{0})=m(A_{0})+m(B_{0})$, or

  3. (III)

$A+B\nsim_{m}A+B+K$, and there are parallel Bohr intervals
$A^{\prime},B^{\prime}\subseteq K$ having $m(A^{\prime})=m(A)$,
$m(B^{\prime})=m(B)$ and $a,b\in G$ such that $A\subseteq a+A^{\prime}$ and
$B\subseteq b+B^{\prime}$.

While the possibilities (I)-(III) are mutually exclusive for a given subgroup
$K$, a pair $A,B$ may satisfy (I) with a given group $K$ and satisfy (II) with
a different subgroup $K^{\prime}$ in place of $K$.

######  Remark 3.6.

The equation $m_{*}(A_{0}+B_{0})=m(A_{0})+m(B_{0})$ in conclusion (II) leads
to Corollary 10.15, which seems indispensable to our approach.

######  Remark 3.7.

Theorem 1.3 of [19] has the same hypothesis as Theorem 3.5 here. To see how
Theorem 3.5 follows, we list here the four conclusions (P), (E), (K), and (QP)
from [19] and explain how each implies one of (I)-(III) in Theorem 3.5.

  1. (P)

There is a compact open subgroup $K\leq G$ with $A\sim_{m}A+K$ and
$B\sim_{m}B+K$.

If $A$ and $B$ satisfy (P) then $A$ and $B$ satisfy (I) in Theorem 3.5. To see
this, note that if $A\sim_{m}A+K$, then for every coset $K^{\prime}$ of $K$ we
have $A\cap K^{\prime}\sim_{m}K^{\prime}$ or $A\cap K^{\prime}=\varnothing$.
Thus $A+B\sim_{m}A+B+K$, meaning (I) is satisfied in Theorem 3.5.

  2. (E)

There are measurable sets $A^{\prime}\supseteq A$ and $B^{\prime}\supseteq B$
such that

$m(A^{\prime})+m(B^{\prime})>m(A)+m(B),\textit{ and
}m_{*}(A^{\prime}+B^{\prime})=m_{*}(A+B).$

In this case $m_{*}(A^{\prime}+B^{\prime})<m(A^{\prime})+m(B^{\prime})$, so
Theorem 3.2 implies the stabilizer $K:=H(A^{\prime}+B^{\prime})$ is compact
and open. The containment $A+B\subseteq A^{\prime}+B^{\prime}$ then implies
$A+B\sim_{m}A+B+K$, meaning $A,B$, and $K$ satisfy (I) in Theorem 3.5.

  3. (K)

There is a compact open subgroup $K\leq G$, parallel Bohr intervals
$\tilde{I},\tilde{J}\subseteq G$, and $a,b\in G$ such that $A\subseteq
a+\tilde{I}$, $B\subseteq b+\tilde{J}$, while $m(A)=m(\tilde{I})$,
$m(B)=m(\tilde{J})$.

If a pair $A,B$ satisfies (K) and does not satisfy conclusion (I) of Theorem
3.5, then it is easy to verify that $A,B$ satisfies conclusion (III).

  4. (QP)

There is a compact open subgroup $K\leq G$ and partitions $A=A_{1}\cup A_{0}$,
$B=B_{1}\cup B_{0}$ such that $A_{0}\neq\varnothing$, $B_{0}\neq\varnothing$,
at least one of $A_{1}\neq\varnothing$, $B_{1}\neq\varnothing$, and

    1. (QP.1)

$A_{1}+K\sim_{m}A_{1}$, $B_{1}+K\sim_{m}B_{1}$, while $A_{0}$ and $B_{0}$ are
each contained in a coset of $K$ and $(A_{1}+K)\cap A_{0}=(B_{1}+K)\cap
B_{0}=\varnothing$;

    2. (QP.2)

$A_{0}+B_{0}+K$ is a unique expression element of $A+B+K$ in $G/K$;

    3. (QP.3)

$m_{*}(A_{0}+B_{0})=m(A_{0})+m(B_{0})$.

If $A,B$ satisfies (QP) and $A+B\nsim_{m}A+B+K$, then it is easy to check that
$A,B$ satisfies conclusion (II) of Theorem 3.5 here.

######  Remark 3.8.

The pairs of subsets of a discrete abelian group satisfying $|A+B|<|A|+|B|$
are classified in [27] (see [22] for exposition). The pairs satisfying
$|A+B|=|A|+|B|$ are classified in [23]. These results can be combined with
Theorem 1.15 to yield additional detail, as the inequality $m(A+B+K)\leq
m(A+K)+m(B+K)$ in conclusion (I) reduces the study of the pair $A+K,B+K$ to
the study of sets $A^{\prime},B^{\prime}\subseteq G/K$ satisfying
$|A^{\prime}+B^{\prime}|\leq|A^{\prime}|+|B^{\prime}|$.

##  4 Characters of ultraproduct groups

We will use ultraproducts and Loeb measure throughout the remainder of this
article; see Appendix A for a summary and references. Here we briefly
summarize notation.

If $\mathcal{U}$ is a nonprincipal ultrafilter on $\mathbb{N}$ and
$(X_{n})_{n\in\mathbb{N}}$ is a sequence of sets, we write
$\prod_{n\to\mathcal{U}}X_{n}$ for the corresponding ultraproduct, which we
denote by $\mathbf{X}$. Ultraproducts and internal sets will be denoted with
boldface letters. If $Y$ is a compact Hausdorff topological space,
$(y_{n})_{n\in\mathbb{N}}$ is a sequence of elements of $Y$, and $\mathcal{U}$
is an ultrafilter on $\mathbb{N}$, we write $\lim_{n\to\mathcal{U}}y_{n}$ for
the unique element $y\in Y$ such that for every neighborhood $V$ of $y$,
$\\{n\in\mathbb{N}:y_{n}\in V\\}\in\mathcal{U}$. For a sequence of functions
$f_{n}:X_{n}\to Y$, we denote by $\lim_{n\to\mathcal{U}}f_{n}$ the function
from $\prod_{n\to\mathcal{U}}X_{n}$ to $Y$ defined by
$f(\mathbf{x})=\lim_{n\to\mathcal{U}}f_{n}(x_{n})$, where
$(x_{n})_{n\in\mathbb{N}}$ is a representative of $\mathbf{x}$.

Note that some authors use “$\lim_{n\to\mathcal{U}}$” for what we call
“$\prod_{n\to\mathcal{U}}$”.

For this section, fix a sequence of compact abelian groups $G_{n}$ with Haar
probability measure $m_{n}$ and a nonprincipal ultrafilter $\mathcal{U}$ on
$\mathbb{N}$. Let $\mathbf{G}$ be the ultraproduct
$\prod_{n\to\mathcal{U}}G_{n}$ and $\mu$ the Loeb measure corresponding to
$(m_{n})_{n\in\mathbb{N}}$. As usual $\mathcal{S}^{1}$ denotes the circle
group $\\{z\in\mathbb{C}:|z|=1\\}$ with the group operation of multiplication
and the usual topology.

We assume familiarity with basic harmonic analysis on compact abelian groups,
as introduced in [12] or [36]. If $G$ is a compact abelian group,
$\widehat{G}$ denotes its Pontryagin dual, the group of continuous
homomorphisms $\chi:G\to\mathcal{S}^{1}$ with the discrete topology.

######  Definition 4.1.

Call $f:\mathbf{G}\to\mathbb{C}$ a _strong character_ of $\mathbf{G}$ if there
are characters $\chi_{n}\in\widehat{G}_{n}$ such that
$f=\lim_{n\to\mathcal{U}}\chi_{n}$. Equivalently, $f$ is a strong character of
$\mathbf{G}$ if $f=\leftidx^{\circ}(\prod_{n\to\mathcal{U}}\chi_{n})$, where
$\chi_{n}\in\widehat{G}_{n}$ for $\mathcal{U}$-many $n$. Every strong
character is $\mu$-measurable by Part (ii) of Proposition A.4. One can easily
verify that every strong character is a homomorphism from $\mathbf{G}$ to
$\mathcal{S}^{1}$. Write $\widehat{\mathbf{G}}$ for the set of strong
characters of $\mathbf{G}$, and note that $\widehat{\mathbf{G}}$ is a group
under pointwise multiplication.222We do not refer to a topology on
$\mathbf{G}$ when we define $\widehat{\mathbf{G}}$, in contrast to the dual
$\widehat{G}$ of a locally compact abelian group $G$, which (by definition)
grants $\widehat{G}$ the topology of uniform convergence on compact subsets of
$G$.

A _weak character_ of $\mathbf{G}$ is a $\mu$-measurable homomorphism
$\rho:\mathbf{G}\to\mathcal{S}^{1}$. Of course every strong character is also
weak character, and Lemma 4.6 says that every weak character of $\mathbf{G}$
is a strong character of $\mathbf{G}$, motivating our choice of the notation
$\widehat{\mathbf{G}}$.

######  Definition 4.2.

If $G$ is a group, $f$ is a function on $G$, and $t\in G$, we write $f_{t}$
for the _translate_ of $f$ by $t$: the function defined by $f_{t}(x):=f(x-t)$.

######  Definition 4.3.

Let $\mathcal{F}(\mathbf{G})$ be the uniform closure of the linear span of the
strong characters of $\mathbf{G}$. Note that $\mathcal{F}(\mathbf{G})$ is
closed under pointwise multiplication and complex conjugation, and so is an
algebra of functions. Following [43] we call $\mathcal{F}(\mathbf{G})$ the
_Fourier algebra_ of $\mathbf{G}$.

######  Definition 4.4.

Let $L^{2}(\mu)_{wm}$ be the set of $f\in L^{2}(\mu)$ such that

$\int\Bigl{|}\int
f\bar{g}_{\mathbf{t}}\,d\mu\Bigr{|}^{2}\,d\mu(\mathbf{t})=0\quad\text{for all
}g\in L^{2}(\mu).$

Equivalently, $f\in L^{2}(\mu)_{wm}$ if $f*g\equiv_{\mu}0$ for all $g\in
L^{2}(\mu)$.

To see the equivalence asserted in the definition, observe that $\int
f\bar{g}_{\mathbf{t}}\,d\mu=f*h(\mathbf{t})$, where
$h(\mathbf{x}):=\overline{g(-\mathbf{x})}$, so $f\in L^{2}(\mu)_{wm}$ if and
only if $f*g\equiv_{\mu}0$ for all $g\in L^{2}(\mu)$. The subscript “ _wm_ ”
is to indicate an analogy with weak mixing in dynamics.

The next lemma is Lemma 6.3 in [43]. Readers familiar with [14] or [15] will
recognize it as a limiting version of an equivalence between two definitions
of uniformity. Here we write $f\perp\mathcal{F}(\mathbf{G})$ to mean $\int
f\bar{g}\,d\mu=0$ for every $g\in\mathcal{F}(\mathbf{G})$.

######  Lemma 4.5.

Let $f:\mathbf{G}\to\mathbb{C}$ be bounded and $\mu$-measurable. Then
$f\perp\mathcal{F}(\mathbf{G})$ if and only if $f\in L^{2}(\mu)_{wm}$.

The next lemma is part of Proposition 6.1 in [43].

######  Lemma 4.6.

If $\chi:\mathbf{G}\to\mathcal{S}^{1}$ is a $\mu$-measurable homomorphism,
then $\chi$ is a strong character of $\mathbf{G}$. Consequently, if
$\tau:\mathbf{G}\to\mathbb{T}$ is a $\mu$-measurable homomorphism, then there
is a sequence of continuous homomorphisms $\tau_{n}:G_{n}\to\mathbb{T}$ such
that $\tau=\lim_{n\to\mathcal{U}}\tau_{n}$.

The second assertion in Lemma 4.6 follows from the first as $\mathbb{T}$ is
isomorphic to $\mathcal{S}^{1}$, and with this isomorphism characters of a
compact group $G$ correspond to continuous homomorphisms from $G$ to
$\mathbb{T}$.

We say a character $\chi$ of a group $G$ is _trivial_ if $\chi(g)=1$ for all
$g\in G$.

######  Lemma 4.7.

If $\chi_{n}\in\widehat{G}_{n}$ for all $n$, then
$\chi:=\lim_{n\to\mathcal{U}}\chi_{n}$ is trivial if and only if $\chi_{n}$ is
trivial for $\mathcal{U}$-many $n$.

###### Proof.

If $\chi_{n}$ is trivial for $\mathcal{U}$-many $n$, then clearly $\chi$ is
trivial. To prove the converse, assume $\chi_{n}$ is nontrivial for
$\mathcal{U}$-many $n$. For these $n$, the image of $\chi_{n}$ is a nontrivial
subgroup of $\mathcal{S}^{1}$. Thus there is an $x_{n}\in G_{n}$ such that the
real part of $\chi_{n}(x_{n})$ is not positive, meaning
$|\chi_{n}(x_{n})-1|\geq\sqrt{2}$. It follows that
$\lim_{n\to\mathcal{U}}\chi_{n}(x_{n})\neq 1$, so $\chi$ is nontrivial. ∎

######  Corollary 4.8.

The group $\widehat{\mathbf{G}}$ is isomorphic to the group
$\prod_{n\to\mathcal{U}}\widehat{G}_{n}$.

Note: here we are only considering the objects as groups, so the isomorphism
is simply a group isomorphism.

###### Proof.

By Lemma 4.6, a surjective homomorphism
$\Phi:\prod_{n\to\mathcal{U}}\widehat{G}_{n}\to\widehat{\mathbf{G}}$ may be
defined by $\Phi(\bm{\chi}):=\lim_{n\to\mathcal{U}}\chi_{n}$, where
$(\chi_{n})_{n\in\mathbb{N}}$ is a representative of $\bm{\chi}$. Lemma 4.7
implies that the kernel of $\Phi$ consists only of a single element, so $\Phi$
is one-to-one. Thus $\Phi$ is an isomorphism. ∎

Corollary 4.8 allows us to identify the ultraproducts $\mathbf{G}$ of compact
groups where $\widehat{\mathbf{G}}$ is torsion free.

######  Lemma 4.9.

For a fixed $k$, the group $\widehat{\mathbf{G}}$ has an element of order $k$
if and only if for $\mathcal{U}$-many $n$, $\widehat{G}_{n}$ has an element of
order $k$. Consequently, $\widehat{\mathbf{G}}$ is torsion free if and only if
for all $k\in\mathbb{N}\setminus\\{1\\}$ and $\mathcal{U}$-many $n$,
$\widehat{G}_{n}$ does not have an element of order $k$.

###### Proof.

This follows immediately from the definition of
$\prod_{n\to\mathcal{U}}\widehat{G}_{n}$ and Corollary 4.8. ∎

######  Corollary 4.10.

Let $H$ be a finite group and $\tau:\mathbf{G}\to H$ a surjective
$\mu$-measurable homomorphism. Then there are continuous homomorphisms
$\tau_{n}:G_{n}\to H$ such that $\tau=\lim_{n\to\mathcal{U}}\tau_{n}$. If
$K_{n}=\ker\tau_{n}$, then for $\mathcal{U}$-many $n$, we have
$G_{n}/K_{n}\cong H$. Furthermore $\ker\tau=\prod_{n\to\mathcal{U}}K_{n}$.

Corollary 4.10 says that every $\mu$-measurable finite index subgroup of
$\mathbf{G}$ is internal.

###### Proof.

Every finite abelian group is a product of finite cyclic groups, so it
suffices to prove the special case of the corollary where $H$ is a finite
cyclic group. In this case we consider $H$ as the subgroup of
$\mathcal{S}^{1}$ generated by $\exp(2\pi i/N)$, where $N$ is the order of
$H$. Note that $H$ is the unique subgroup of $\mathcal{S}^{1}$ having order
$N$, and that if $z\in\mathcal{S}^{1}$ satisfies $z^{N}=1$, then $z\in H$. We
consider $\tau$ as a homomorphism into $\mathcal{S}^{1}$ with image $H$. Lemma
4.6 then implies $\tau=\lim_{n\to\mathcal{U}}\chi_{n}$ for some
$\chi_{n}\in\widehat{G}_{n}$. Let $\tau^{N}$ be the homomorphism defined by
$\tau^{N}(\mathbf{g}):=\tau(\mathbf{g})^{N}$, and similarly define
$\chi_{n}^{N}.$ Continuity of the map $z\mapsto z^{N}$ implies
$\tau^{N}=\lim_{n\to\mathcal{U}}\chi_{n}^{N}$, and $\tau^{N}$ is the trivial
homomorphism, as the image of $\tau$ is generated by an element of order $N$.
Lemma 4.7 then implies $\chi_{n}^{N}$ is trivial for $\mathcal{U}$-many $n$,
meaning $\chi_{n}(G_{n})\subseteq H$ for $\mathcal{U}$-many $n$. The
hypothesis $\tau(\mathbf{G})=H$ then implies $|\tau_{n}(G_{n})|=|H|$ for
$\mathcal{U}$-many $n$, and therefore $\tau_{n}(G_{n})=H$ for
$\mathcal{U}$-many $n$. Thus $G_{n}/K_{n}\cong H$ for $\mathcal{U}$-many $n$.

To see that $\ker\tau=\prod_{n\to\mathcal{U}}K_{n}$, let
$\mathbf{g}\in\mathbf{G}$, choose a representative $(g_{n})_{n\in\mathbb{N}}$
of $\mathbf{g}$, and note that the finiteness of $H$ implies
$\lim_{n\to\mathcal{U}}\tau_{n}(g_{n})=0$ if and only if $\tau_{n}(g_{n})=0$
for $\mathcal{U}$-many $n$. ∎

The next definition provides a useful splitting of $L^{2}(\mu)$.

######  Definition 4.11.

Let $L^{2}(\mu)_{c}$ denote the closure of $\mathcal{F}(\mathbf{G})$ in
$L^{2}(\mu)$. If $f\in L^{2}(\mu)$, write $f^{(c)}$ for the orthogonal
projection of $f$ onto $L^{2}(\mu)_{c}$, and $f^{(wm)}$ for $f-f^{(c)}$. Thus
$f^{(wm)}\perp\mathcal{F}(\mathbf{G})$. Of course $f^{(c)}$ and $f^{(wm)}$ are
defined only up to equality on $\mu$-null sets, but for our purposes this
ambiguity can be resolved by choosing a genuine function to represent
$f^{(c)}$.

The “$c$” in the notation above is to indicate that the translates
$\\{f_{\mathbf{t}}:\mathbf{t}\in\mathbf{G}\\}$ of a bounded function form a
precompact subset of $L^{2}(\mu)$ if and only if $f\in L^{2}(\mu)_{c}$.

The following lemma is a collection of common facts about the orthogonal
projection of a function onto a space spanned by an algebra of functions, but
they are often not stated explicitly for our setting.

######  Lemma 4.12.

Let $f\in L^{2}(\mu)$.

  1. (i)

If $f(\mathbf{x})\geq 0$ for $\mu$-almost every $\mathbf{x}$, then
$f^{(c)}(\mathbf{x})\geq 0$ for $\mu$-almost every $\mathbf{x}$.

  2. (ii)

If $f:\mathbf{G}\to[0,1]$, then $f^{(c)}(\mathbf{x})\in[0,1]$ for $\mu$-almost
every $\mathbf{x}$.

  3. (iii)

$\int f^{(c)}\,d\mu=\int f\,d\mu$.

  4. (iv)

If $A\subseteq\mathbf{G}$ is $\mu$-measurable, then
$A\subset_{\mu}\\{\mathbf{x}:1_{A}^{(c)}(\mathbf{x})>0\\}$.

  5. (v)

If $f:\mathbf{G}\to[0,1]$ then
$\\{\mathbf{x}:f(\mathbf{x})>0\\}\subset_{\mu}\\{\mathbf{x}:f^{(c)}(\mathbf{x})>0\\}$.

###### Proof.

(i) Suppose $f$ is a nonnegative element of $L^{2}(\mu)$. We will prove that
there is a sequence of nonnegative elements of $\mathcal{F}(\mathbf{G})$
converging in $L^{2}(\mu)$ to $f^{(c)}$, which implies $f^{(c)}$ is
$\mu$-almost everywhere nonnegative. Let $h_{n}$ be any sequence of elements
of $\mathcal{F}(\mathbf{G})$ converging to $f^{(c)}$ in $L^{2}(\mu)$. Since
$\mathcal{F}(\mathbf{G})$ is a uniformly closed algebra of functions
containing the constant functions $0$ and $1_{\mathbf{G}}$, we have that for
all $h\in\mathcal{F}(\mathbf{G})$,
$h^{+}:=\sup\\{h,0\\}\in\mathcal{F}(\mathbf{G})$, by Lemma 8 in Chapter 6 of
[6]. Since $f$ is nonnegative we have
$\|f-h_{n}^{+}\|_{L^{2}(\mu)}\leq\|f-h_{n}\|_{L^{2}(\mu)}$ for each $n$, so
$h_{n}^{+}$ converges to $f^{(c)}$, as $f^{(c)}$ is the unique element $h$ in
$L^{2}(\mu)_{c}$ which minimizes $\|f-h\|_{L^{2}(\mu)}$.

Part (ii) follows from Part (i) and the nonnegativity of the functions
$1_{\mathbf{G}}-f$ and $f$.

Part (iii) is a consequence of the fact that $\mathcal{F}(\mathbf{G})$
contains the constant functions on $\mathbf{G}$.

To prove Part (iv), we use the identity $\int f\cdot g^{(c)}\,d\mu=\int
f^{(c)}\cdot g^{(c)}\,d\mu$, which follows immediately from the fact that the
map $f\mapsto f^{(c)}$ is an orthogonal projection. Let
$E=A\setminus\\{\mathbf{x}:1_{A}^{(c)}(\mathbf{x})>0\\}$. Then
$1_{E}1_{A}^{(c)}=0$, so that $\int 1_{E}^{(c)}1_{A}^{(c)}\,d\mu=\int
1_{E}1_{A}^{(c)}\,d\mu=0$. Now $0\leq 1_{E}\leq 1_{A}\leq 1_{\mathbf{G}}$, so
Part (i) implies $0\leq 1_{E}^{(c)}\leq 1_{A}^{(c)}$. Combining this last
inequality with the integral $\int 1_{E}^{(c)}1_{A}^{(c)}\,d\mu=0$, we get
that $\int 1_{E}^{(c)}\,d\mu=0$, which by Part (iii) implies $\mu(E)=0$.

Part (v) follows from Part (iv), the linearity of the map $f\mapsto f^{(c)}$,
and bounding $f$ below by linear combinations of characteristic functions
supported on $\\{x:f(x)>0\\}$. ∎

######  Lemma 4.13.

If $f,g:\mathbf{G}\to\mathbb{C}$ are bounded $\mu$-measurable functions, then
$f*g\equiv_{\mu}f^{(c)}*g^{(c)}$.

###### Proof.

Write $f$ as $f^{(c)}+f^{(wm)}$. Lemma 4.12 implies $f^{(c)}$ is bounded, so
Lemma 4.5 implies $f^{(wm)}$ is a bounded element of $L^{2}(\mu)_{wm}$, and we
get that $f^{(wm)}*g\equiv_{\mu}0$ by Definition 4.4. Thus
$f*g\equiv_{\mu}f^{(c)}*g$, and by symmetry we get that
$f^{(c)}*g\equiv_{\mu}f^{(c)}*g^{(c)}$. ∎

The conclusion of Lemma 4.13 cannot be improved to assert
$f*g=f^{(c)}*g^{(c)}$. See Remark 5.2 for elaboration.

The following proposition is part of Theorem 6 of [43], cosmetically modified
to deal with two functions on $\mathbf{G}$ instead of one.

######  Proposition 4.14.

If $f,g\in L^{2}(\mu)_{c}$ are bounded functions, then there is a compact
metrizable abelian group $G$, a surjective measure preserving quotient map
$\pi:\mathbf{G}\to G$, and bounded Borel functions $\tilde{f},\tilde{g}$ on
$G$ such that $f\equiv_{\mu}\tilde{f}\circ\pi$ and
$g\equiv_{\mu}\tilde{g}\circ\pi$.

One can prove Proposition 4.14 by imitating the proof in [43], with the
following minor modification: where that proof considers the set of characters
$\chi$ appearing in the Fourier series of $f$, to prove Proposition 4.14
consider instead the set of characters appearing in the Fourier series of $f$
or the Fourier series of $g$ (or both).

##  5 Reducing to a compact quotient of $\mathbf{G}$G

Here we collect some results allowing us to study sumsets in an ultraproduct
$\mathbf{G}$ of compact groups by studying related sumsets in a compact
quotient of $\mathbf{G}$.

For this section, fix a sequence of compact abelian groups $G_{n}$ with Haar
probability measure $m_{n}$ and a nonprincipal ultrafilter $\mathcal{U}$ on
$\mathbb{N}$. Let $\mathbf{G}$ denote the ultraproduct
$\prod_{n\to\mathcal{U}}G_{n}$ and $\mu$ be the Loeb measure corresponding to
$(m_{n})_{n\in\mathbb{N}}$. We continue to use the notation
$\widehat{\mathbf{G}}$ of Definition 4.1: the elements of
$\widehat{\mathbf{G}}$ are the functions $\chi:\mathbf{G}\to\mathcal{S}^{1}$
of the form $\chi=\lim_{n\to\mathcal{U}}\chi_{n}$, where
$\chi_{n}\in\widehat{G}_{n}$ for each $n$; in particular the elements of
$\widehat{\mathbf{G}}$ are $\mu$-measurable.

The main results of this section are Proposition 5.1 and its consequences,
Lemma 5.4 and Corollary 5.5. Here we abbreviate the set $\\{x:f(x)>0\\}$ as
$\\{f>0\\}$.

######  Proposition 5.1.

Let $f,g:\mathbf{G}\to[0,1]$ be $\mu$-measurable. Then there is a compact
metrizable quotient $G$ of $\mathbf{G}$ with Haar measure $m$, $\mu$-measure
preserving quotient map $\pi:\mathbf{G}\to G$, and Borel functions
$\tilde{f},\tilde{g}:G\to[0,1]$ such that

$\displaystyle\\{f>0\\}\subset_{\mu}\\{\tilde{f}\circ\pi>0\\},\quad\\{g>0\\}\subset_{\mu}\\{\tilde{g}\circ\pi>0\\},and$
(5.1) $\displaystyle f*_{\mu}g\equiv_{\mu}(\tilde{f}*_{m}\tilde{g})\circ\pi.$
(5.2)

###### Proof.

Let $f,g:\mathbf{G}\to[0,1]$ be $\mu$-measurable functions. Lemma 4.12 implies
$f^{(c)}$ and $g^{(c)}$ are bounded, so Proposition 4.14 provides a compact
metrizable group $G$, a $\mu$-measure preserving quotient map
$\pi:\mathbf{G}\to G$, and Borel functions
$\tilde{f},\tilde{g}:G\to\mathbb{C}$ such that

$f^{(c)}\equiv_{\mu}\tilde{f}\circ\pi,\quad
g^{(c)}\equiv_{\mu}\tilde{g}\circ\pi.$ (5.3)

We will show that $\tilde{f}$ and $\tilde{g}$ satisfy the conclusion of
Proposition 5.1.

Lemma 4.12 implies $0\leq f^{(c)}(\mathbf{x}),g^{(c)}(\mathbf{x})\leq 1$ for
$\mu$-a.e. $\mathbf{x}$, so $0\leq\tilde{f}(x),\tilde{g}(x)\leq 1$ for
$m$-a.e. $x$. Lemma 4.12 also implies $\\{f>0\\}\subset_{\mu}\\{f^{(c)}>0\\}$
and $\\{g>0\\}\subset_{\mu}\\{g^{(c)}>0\\}$. Since
$f^{(c)}\equiv_{\mu}\tilde{f}\circ\pi$, we get the essential containments
stated in (5.1).

To prove Equation (5.2), first note that Lemma 4.13 implies
$f*_{\mu}g\equiv_{\mu}f^{(c)}*_{\mu}g^{(c)}$. It therefore suffices to show
that $f^{(c)}*_{\mu}g^{(c)}\equiv_{\mu}(\tilde{f}*_{m}\tilde{g})\circ\pi$, and
this follows immediately from Lemma 1.3. ∎

######  Remark 5.2.

To be clear, Proposition 5.1 asserts that
$(f*g)(\mathbf{x})=(\tilde{f}*\tilde{g})(\pi(\mathbf{x}))$ for $\mu$-almost
every $\mathbf{x}\in\mathbf{G}$. This cannot be improved to replace
“$\mu$-almost every $\mathbf{x}$” with “every $\mathbf{x}$”. Examples
exhibiting the obstruction are discussed in §13.2.6.

The next lemma lets us pass between convolutions and sumsets; the main
application will take an estimate on the size of a sumset in $\mathbf{G}$ and
obtain a similar estimate on the size of a related sumset in a compact
quotient of $\mathbf{G}$.

######  Lemma 5.3.

Let $G$ be a compact metrizable abelian group with Haar probability measure
$m$. If $f,g:G\to[0,1]$ are $m$-measurable functions let $C_{0}:=\\{x\in
G:f(x)>0\\}$ and $D_{0}:=\\{x\in G:g(x)>0\\}$. Then there are Borel sets
$C\subseteq C_{0}$, $D\subseteq D_{0}$ such that $m(C)=m(C_{0})$,
$m(D)=m(D_{0})$, $C+D$ is Borel, and $C+D\subseteq\\{x\in G:f*g(x)>0\\}$.

For a proof of Lemma 5.3, see Lemma 2.13 of [18]. When $G=\mathbb{T}^{d}$ for
some $d\in\mathbb{N}$, we can take $C$ and $D$ to be countable unions of
compact subsets of the points of Lebesgue density of $\\{x:f(x)>0\\}$ and
$\\{x:g(x)>0\\}$, respectively.

######  Lemma 5.4.

Let $f,g:\mathbf{G}\to[0,1]$ be $\mu$-measurable functions and
$E:=\\{\mathbf{x}\in\mathbf{G}:f*_{\mu}g(\mathbf{x})>0\\}$. If
$A:=\\{\mathbf{x}\in\mathbf{G}:f(\mathbf{x})>0\\}$ and
$B:=\\{\mathbf{x}\in\mathbf{G}:g(\mathbf{x})>0\\}$, there is a compact
quotient group $G$ with Haar probability measure $m$, a $\mu$-measure
preserving quotient map $\pi:\mathbf{G}\to G$, and Borel sets $C,D\subseteq G$
such that $A\subset_{\mu}A^{\prime}:=\pi^{-1}(C)$,
$B\subset_{\mu}B^{\prime}:=\pi^{-1}(D)$ and
$A^{\prime}+B^{\prime}\subset_{\mu}E$. Furthermore $C+D$ is Borel and
$m(C+D)\leq\mu(E)$.

###### Proof.

Let $G$, $\pi:\mathbf{G}\to G$, and $\tilde{f},\tilde{g}:G\to[0,1]$ be as in
Proposition 5.1. Consider the level sets $C_{0}:=\\{t\in G:\tilde{f}(t)>0\\}$,
$D_{0}:=\\{t\in G:\tilde{g}(t)>0\\}$, and apply Lemma 5.3 to choose Borel sets
$C\subseteq C_{0}$, $D\subseteq D_{0}$ having $C\sim_{m}C_{0}$,
$D\sim_{m}D_{0}$ such that $C+D$ is Borel and

$C+D\subseteq V:=\\{t\in G:\tilde{f}*_{m}\tilde{g}(t)>0\\}.$ (5.4)

Let $A^{\prime}:=\pi^{-1}(C)$ and $B^{\prime}:=\pi^{-1}(D)$. Now Proposition
5.1 implies $(\tilde{f}*_{m}\tilde{g})\circ\pi\equiv_{\mu}f*_{\mu}g$, so
$E\sim_{\mu}\pi^{-1}(V)$, and (5.4) implies
$A^{\prime}+B^{\prime}\subset_{\mu}E$.

The essential containment $A\subset_{\mu}A^{\prime}$ follows from the
similarity
$A^{\prime}\sim_{\mu}\\{\mathbf{x}\in\mathbf{G}:\tilde{f}\circ\pi(\mathbf{x})>0\\}$
and Proposition 5.1. Likewise $B\subset_{\mu}B^{\prime}$. The inequality
$m(C+D)\leq\mu(E)$ follows from the containments
$\pi^{-1}(C+D)=A^{\prime}+B^{\prime}\subset_{\mu}E$, since $\pi$ preserves
$\mu$. ∎

Specializing Lemma 5.4 to the case where $f=1_{A}$ and $g=1_{B}$ are
characteristic functions of sets and observing that
$A+_{0}B:=\\{\mathbf{x}\in\mathbf{G}:1_{A}*1_{B}(\mathbf{x})>0\\}\subseteq
A+B$, we obtain the following corollary.

######  Corollary 5.5.

If $A,B\subseteq\mathbf{G}$ are $\mu$-measurable sets then there is a compact
quotient $G$ with Haar measure $m$, a $\mu$-measure preserving quotient map
$\pi:\mathbf{G}\to G$, and Borel sets $C,D\subseteq G$ such that

  1. (i)

$A\subset_{\mu}A^{\prime}:=\pi^{-1}(C)$,
$B\subset_{\mu}B^{\prime}:=\pi^{-1}(D)$ and
$A^{\prime}+B^{\prime}\subset_{\mu}A+_{0}B\subseteq A+B$. Furthermore $C+D$ is
Borel (so that $A^{\prime}+B^{\prime}$ is $\mu$-measurable) and
$m(C+D)\leq\mu_{*}(A+B)$.

  2. (ii)

If $A^{\prime\prime}\sim_{\mu}A$ and $B^{\prime\prime}\sim_{\mu}B$, then
$A^{\prime}+B^{\prime}\subset_{\mu}A^{\prime\prime}+B^{\prime\prime}$.

###### Proof.

Part (i) follows immediately from Lemma 5.4. Part (ii) follows from Part (i)
and the observation that $A^{\prime\prime}+_{0}B^{\prime\prime}=A+_{0}B$. ∎

##  6 Bohr intervals and sumsets

For this section fix a sequence of compact abelian groups $G_{n}$ with Haar
probability measure $m_{n}$ and a nonprincipal ultrafilter $\mathcal{U}$ on
$\mathbb{N}$. Let $\mathbf{G}$ be the ultraproduct
$\prod_{n\to\mathcal{U}}G_{n}$ and $\mu$ the corresponding Loeb measure. Write
$\widehat{\mathbf{G}}$ for the group of strong characters of $\mathbf{G}$, as
defined in §4. Lemma 6.1 and Corollary 6.2 characterize the internal Bohr
intervals in $\mathbf{G}$ in terms of their constituent sets. Lemma 6.4 and
Corollary 6.5 show that certain sets related to Bohr intervals are themselves
Bohr intervals.

Recall (Definition 1.4) that if $G$ is a group and $\mu$ is a translation
invariant probability measure on $G$, a Bohr interval in $G$ is a set of the
form $\tau^{-1}(I)$, where $\tau:G\to\mathbb{T}$ is a surjective
$\mu$-measurable homomorphism and $I\subseteq\mathbb{T}$ is an interval.
Furthermore, we consider only closed intervals in $\mathbb{T}$.

Here we characterize those sequences of sets $A_{n}\subseteq G_{n}$ such that
$\prod_{n\to\mathcal{U}}A_{n}$ is a Bohr interval in $\mathbf{G}$; we will see
that this forces the $A_{n}$ to be contained in Bohr intervals or $N$-cyclic
progressions (Definition 1.4) not much larger than $A_{n}$. We use the
notation $C_{N}$ (Notation 1.6) and continue to write $\lambda$ for Lebesgue
measure ($=$ normalized Haar measure) on $\mathbb{T}$.

######  Lemma 6.1.

For each $n\in\mathbb{N}$, let $\tau_{n}:G_{n}\to\mathbb{T}$ be a continuous
homomorphism, and define $\tau:\mathbf{G}\to\mathbb{T}$ as
$\lim_{n\to\mathcal{U}}\tau_{n}$. Assume that $\tau(\mathbf{G})=\mathbb{T}$.
Then

  1. (i)

Either

    1. $\cdot$

$\tau_{n}(G_{n})=\mathbb{T}$ for $\mathcal{U}$-many $n$, or

    2. $\cdot$

$\tau_{n}(G_{n})=C_{N_{n}}$ for $\mathcal{U}$-many $n$, where
$\lim_{n\to\mathcal{U}}N_{n}=\infty$.

  2. (ii)

If $I\subseteq\mathbb{T}$ is an interval let
$\tilde{I}_{n}:=\tau_{n}^{-1}(I)$. Then either $\tilde{I}_{n}$ is a Bohr
interval of $m_{n}$-measure $\lambda(I)$ for $\mathcal{U}$-many $n$, or
$\tilde{I}_{n}$ is an $N_{n}$-cyclic progression of $m_{n}$-measure at most
$\lambda(I)+\frac{1}{N_{n}}$, where $\lim_{n\to\mathcal{U}}N_{n}=\infty$.

  3. (iii)

If $\mathbf{A}=\prod_{n\to\mathcal{U}}A_{n}$ is an internal set and
$\mathbf{A}\subseteq\tau^{-1}(I)$, then there are intervals
$I_{n}^{\prime}\subseteq\mathbb{T}$ such that
$\lim_{n\to\mathcal{U}}\lambda(I_{n}^{\prime})\leq\lambda(I)$ and
$A_{n}\subseteq\tau_{n}^{-1}(I_{n}^{\prime})$ for $\mathcal{U}$-many $n$.

###### Proof.

(i) The continuity of $\tau_{n}$ implies that $\tau_{n}(G_{n})$ is either
$\mathbb{T}$ or a finite cyclic subgroup of $\mathbb{T}$, as these are the
only compact subgroups of $\mathbb{T}$. The hypothesis that $\tau$ is
surjective implies that for all $x\in\mathbb{T}$, there is a sequence of
elements $g_{n}\in G_{n}$ such that $\lim_{n\to\mathcal{U}}\tau_{n}(g_{n})=x$.
Thus if $\tau_{n}(G_{n})\neq\mathbb{T}$ for $\mathcal{U}$-many $n$, then
$\tau_{n}(G_{n})=C_{N_{n}}$, where $\lim_{n\to\mathcal{U}}N_{n}=\infty$.

(ii) If $\tau_{n}(G_{n})=\mathbb{T}$, then $\tau_{n}^{-1}(I)$ is a Bohr
interval of $m_{n}$-measure $\lambda(I)$, by definition. If
$\tau_{n}(G_{n})\neq\mathbb{T}$ for $\mathcal{U}$-many $n$, then Part (i)
implies that $\tau_{n}(G_{n})=C_{N_{n}}$ for $\mathcal{U}$-many $n$, and
$\lim_{n\to\mathcal{U}}N_{n}=\infty$. For such $n$, $\tilde{I}_{n}$ is an
$N$-cyclic progression, and the estimate of $m_{n}(\tilde{I}_{n})$ then
follows from Lemma 1.7.

(iii) The hypothesis $\mathbf{A}\subseteq\tau^{-1}(I)$ is equivalent to the
condition $\lim_{n\to\mathcal{U}}\tau_{n}(a_{n})\in I$ whenever $a_{n}\in
A_{n}$. Let $J\subseteq\mathbb{T}$ be an open interval containing $I$ and
$\bar{J}$ its closure. Observe that $\tau_{n}(A_{n})\subseteq\bar{J}$ for
$\mathcal{U}$-many $n$: assuming otherwise, there is a sequence $a_{n}\in
A_{n}$ such that $\tau_{n}(a_{n})\notin\bar{J}$ for $\mathcal{U}$-many $n$,
contradicting the condition $\lim_{n\to\mathcal{U}}\tau_{n}(a_{n})\in I$. If
we let $I_{n}^{\prime}$ be the smallest closed interval containing
$\tau_{n}(A_{n})$, we then have $I_{n}^{\prime}\subseteq\bar{J}$ for
$\mathcal{U}$-many $n.$ Thus
$L:=\lim_{n\to\mathcal{U}}\lambda(I_{n}^{\prime})\leq\lambda(\bar{J})$. Since
$J$ is an arbitrary open interval containing the interval $I$, this implies
$L\leq\lambda(I)$, so the $I_{n}^{\prime}$ are the desired intervals. ∎

######  Corollary 6.2.

If
$\mathbf{A}=\prod_{n\to\mathcal{U}}A_{n},\mathbf{B}=\prod_{n\to\mathcal{U}}B_{n}\subseteq\mathbf{G}$
are internal sets and $\tilde{I},\tilde{J}\subseteq\mathbf{G}$ are parallel
Bohr intervals such that $\mathbf{A}\subseteq\tilde{I}$,
$\mathbf{B}\subseteq\tilde{J}$, and $\mu(\tilde{I})=\mu(\mathbf{A})$,
$\mu(\tilde{J})=\mu(\mathbf{B})$ then there are
$\tilde{I}_{n},\tilde{J}_{n}\subseteq G_{n}$ such that

$A_{n}\subseteq\tilde{I}_{n},\quad
B_{n}\subseteq\tilde{J}_{n},\quad\lim_{n\to\mathcal{U}}m_{n}(\tilde{I}_{n}\setminus
A_{n})+m_{n}(\tilde{J}_{n}\setminus B_{n})=0,$ (6.1)

and $\tilde{I}_{n}$ and $\tilde{J}_{n}$ are either parallel Bohr intervals or
parallel $N_{n}$-cyclic progressions, where
$\lim_{n\to\mathcal{U}}N_{n}=\infty$.

###### Proof.

By the definition of “parallel Bohr intervals”, we can write $\tilde{I}$ and
$\tilde{J}$ as $\tau^{-1}(I)$ and $\tau^{-1}(J)$ for some surjective
$\mu$-measurable homomorphism $\tau:\mathbf{G}\to\mathbb{T}$ and intervals
$I,J$ contained in $\mathbb{T}$. Lemma 4.6 implies
$\tau=\lim_{n\to\mathcal{U}}\tau_{n}$ for some continuous homomorphisms
$\tau_{n}:G_{n}\to\mathbb{T}$. Parts (ii) and (iii) of Lemma 6.1 then
guarantee the existence of Bohr intervals or $N$-cyclic progressions
$\tilde{I}_{n},\tilde{J}_{n}$ satisfying (6.1). ∎

We often need to infer information about a group element $b$ based on the fact
that a translate $A+b$ is contained in a certain Bohr interval. This possible
under strong hypotheses, according to Lemma 6.4. The following definition
helps abbreviate the statement.

######  Definition 6.3.

Let $G$ be a group and $A,B\subseteq G$. Define

$\bar{A}_{B}:=\\{g\in G:g+B\subseteq A+B\\}.$

If $\nu$ is a measure on $G$, let $\bar{A}_{B,\nu}:=\\{g\in
G:g+B\subset_{\nu}A+B\\}$.

The following lemma is elementary and well known in the special case where $G$
is the torus $\mathbb{T}$.

######  Lemma 6.4.

If $G$ is an abelian group with translation invariant probability measure
$\nu$ and $A,B\subseteq G$ are parallel Bohr intervals with $\nu(A)+\nu(B)<1$,
then $\bar{A}_{B}=\bar{A}_{B,\nu}=A$ and $\bar{B}_{A}=\bar{B}_{A,\nu}=B$.

We omit the proof. The general case follows from the case where $G$ is
$\mathbb{T}$, since the homomorphism $\tau:G\to\mathbb{T}$ in the definition
of “Bohr interval” is always surjective and measure preserving.

######  Corollary 6.5.

With $G$ as in Lemma 6.4, let $A,B\subseteq G$ satisfy
$\nu_{*}(A+B)\leq\nu(A)+\nu(B)$, and suppose $\tilde{I},\tilde{J}\subseteq G$
are parallel Bohr intervals such that $\nu(\tilde{I})+\nu(\tilde{J})<1$,
$A\sim_{\nu}\tilde{I}$, and $B\sim_{\nu}\tilde{J}$. Then $A\subseteq\tilde{I}$
and $B\subseteq\tilde{J}$.

###### Proof.

By Lemma 6.4, to prove $A\subseteq\tilde{I}$ it suffices to show that for all
$a\in A$, $a+\tilde{J}\subset_{\nu}\tilde{I}+\tilde{J}$. Note that the
similarities $A\sim_{\nu}\tilde{I}$ and $B\sim_{\nu}\tilde{J}$ imply
$1_{A}*1_{B}=1_{\tilde{I}}*1_{\tilde{J}}$, and the level set
$E:=\\{x:1_{\tilde{I}}*1_{\tilde{J}}(x)>0\\}$ satisfies
$E\sim_{\nu}\tilde{I}+\tilde{J}$. Hence $\tilde{I}+\tilde{J}\subset_{\nu}A+B$,
so the hypothesis $\nu_{*}(A+B)\leq\nu(A)+\nu(B)$ implies
$\nu_{*}(A+B)=\nu(\tilde{I}+\tilde{J})$. In particular if $C$ is a
$\nu$-measurable subset of $A+B$, then $C\subset_{\nu}\tilde{I}+\tilde{J}$.
Thus $a+B\subset_{\nu}\tilde{I}+\tilde{J}$, so the similarity
$B\sim_{\nu}\tilde{J}$ implies $a+\tilde{J}\subset_{\nu}\tilde{I}+\tilde{J}$
for all $a\in A$, as desired. This establishes $A\subseteq\tilde{I}$, and the
containment $B\subseteq\tilde{J}$ follows by symmetry. ∎

##  7 Proof of Theorem 1.17

We prove Theorem 1.17 at the end of this section. The main ingredient of the
proof is Proposition 7.1, the natural analogue of Theorem 3.4 in the setting
of ultraproducts of connected compact groups. However, we will not assume in
Proposition 7.1 that the constituent groups $G_{n}$ are connected; we merely
assume that $\widehat{\mathbf{G}}$ is torsion free. In fact
$\widehat{\mathbf{G}}$ is torsion free whenever the $G_{n}$ are connected, but
this is so for other ultraproducts, such as
$\prod_{n\to\mathcal{U}}\mathbb{Z}/p_{n}\mathbb{Z}$, where $p_{n}$ is the
$n^{\text{th}}$ prime.

Fix a sequence of compact abelian groups $G_{n}$ with Haar probability measure
$m_{n}$ and a nonprincipal ultrafilter $\mathcal{U}$ on $\mathbb{N}$. Let
$\mathbf{G}=\prod_{n\to\mathcal{U}}G_{n}$ with corresponding Loeb measure
$\mu$. Let $\widehat{\mathbf{G}}$ denote the group of strong characters of
$\mathbf{G}$, as in §4. Recall the term “Bohr interval” from Definition 1.4.

######  Proposition 7.1.

Let $\mathbf{G}$ be as above and assume $\widehat{\mathbf{G}}$ is torsion
free. If $A$, $B\subseteq\mathbf{G}$ are $\mu$-measurable sets satisfying
$\mu(A),\mu(B)>0$ and $\mu_{*}(A+B)\leq\mu(A)+\mu(B)<1$, then there are
parallel Bohr intervals $\tilde{I},\tilde{J}\subseteq\mathbf{G}$ such that
$A\subseteq\tilde{I}$, $B\subseteq\tilde{J}$, and $\mu(A)=\mu(\tilde{I})$,
$\mu(B)=\mu(\tilde{J})$.

###### Proof.

By Corollary 6.5, it suffices to find Bohr intervals
$\tilde{I},\tilde{J}\subseteq\mathbf{G}$ such that $A\sim_{\mu}\tilde{I}$ and
$B\sim_{\mu}\tilde{J}$.

First we show that every compact quotient $G$ of $\mathbf{G}$ with
$\mu$-measurable quotient map $\pi:\mathbf{G}\to G$ is connected. To see this,
note that the map $\pi^{*}:\widehat{G}\to\widehat{\mathbf{G}}$,
$\pi^{*}(\chi):=\chi\circ\pi$ is an injective homomorphism, so that
$\widehat{G}$ is isomorphic to a subgroup of $\widehat{\mathbf{G}}$. Hence,
$\widehat{G}$ is torsion free, and we conclude that $G$ is connected by
Theorem 2.5.6(c) of [36].

Now apply Corollary 5.5 to find a compact metrizable quotient $G$ of
$\mathbf{G}$ and a $\mu$-measurable quotient map $\pi$ and Borel sets
$C,D\subseteq G$, $A^{\prime}:=\pi^{-1}(C)$, $B^{\prime}:=\pi^{-1}(D)$, such
that $C+D$ is Borel, $A^{\prime}+B^{\prime}\subset_{\mu}A+B$, and
$A\subset_{\mu}A^{\prime}$, $B\subset_{\mu}B^{\prime}$. Then
$\mu(A^{\prime})\geq\mu(A)$ and $\mu(B^{\prime})\geq\mu(B)$, so

$\displaystyle m(C+D)=\mu(A^{\prime}+B^{\prime})$
$\displaystyle\leq\mu_{*}(A+B)$
$\displaystyle\leq\mu(A)+\mu(B)\leq\mu(A^{\prime})+\mu(B^{\prime})=m(C)+m(D).$

Then $m(C+D)\leq m(C)+m(D)$ and $m(C+D)<1$, so Corollary 3.3 and the
connectedness of $G$ imply $m(C+D)=m(C)+m(D)$. The inequalities displayed
above are therefore all equalities, and in particular
$\mu(A)+\mu(B)=\mu(A^{\prime})+\mu(B^{\prime})$. The containments
$A\subset_{\mu}A^{\prime}$ and $B\subset_{\mu}B^{\prime}$ then imply
$\mu(A)=\mu(A^{\prime})$ and $\mu(B)=\mu(B^{\prime})$, so

$A\sim_{\mu}A^{\prime}\quad\text{and}\quad B\sim_{\mu}B^{\prime}.$ (7.1)

Theorem 3.4 implies that there are parallel Bohr intervals
$I^{\prime},J^{\prime}\subseteq G$ such that $C\subseteq I^{\prime}$,
$D\subseteq J^{\prime}$ and $m(C)=m(I^{\prime})$, $m(D)=m(J^{\prime})$. Then
$\tilde{I}:=\pi^{-1}(I^{\prime})$ and $\tilde{J}:=\pi^{-1}(J^{\prime})$ are
parallel Bohr intervals such that $A^{\prime}\subseteq\tilde{I}$,
$B^{\prime}\subseteq\tilde{J}$, and $\mu(A^{\prime})=\mu(\tilde{I})$,
$\mu(B^{\prime})=\mu(\tilde{J})$. The similarities in (7.1) then imply
$A\sim_{\mu}\tilde{I}$ and $B\sim_{\mu}\tilde{J}$, as desired. ∎

###### Proof of Theorem 1.17.

Suppose, to get a contradiction, that Theorem 1.17 fails for a given
$\varepsilon>0$. Then for all sufficiently large $n$, there is a connected
compact abelian group $G_{n}$ with Haar probability measure $m_{n}$, inner
Haar measure $m_{n*}$, and sets $A_{n},B_{n}\subseteq G_{n}$ having
$m_{n}(A_{n}),m_{n}(B_{n})>\varepsilon$ such that

  1. (a.1)

$m_{n*}(A_{n}+B_{n})\leq
m_{n}(A_{n})+m_{n}(B_{n})+\frac{1}{n}<1-\frac{\varepsilon}{2}$, and

  2. (a.2)

for every pair of parallel Bohr intervals $\tilde{I}_{n}$,
$\tilde{J}_{n}\subseteq G_{n}$ having
$m_{n}(\tilde{I}_{n})<m(A_{n})+\frac{\varepsilon}{2}$,
$m_{n}(\tilde{J}_{n})<m(B_{n})+\frac{\varepsilon}{2}$, we have
$A_{n}\nsubseteq\tilde{I}_{n}$ or $B_{n}\nsubseteq\tilde{J}_{n}$.

Let $\mathcal{U}$ be a nonprincipal ultrafilter on $\mathbb{N}$, and form the
ultraproduct $\mathbf{G}=\prod_{n\to\mathcal{U}}G_{n}$ with Loeb measure $\mu$
corresponding to $m_{n}$. Consider the internal sets
$\mathbf{A}=\prod_{n\to\mathcal{U}}A_{n}$,
$\mathbf{B}=\prod_{n\to\mathcal{U}}B_{n}\subseteq\mathbf{G}$. Then

$\mu(\mathbf{A})=\lim_{n\to\mathcal{U}}m_{n}(A_{n})\geq\varepsilon,\
\mu(\mathbf{B})=\lim_{n\to\mathcal{U}}m_{n}(B_{n})\geq\varepsilon,$

and $\mu_{*}(\mathbf{A}+\mathbf{B})=\lim_{n\to\mathcal{U}}m_{n*}(A_{n}+B_{n})$
by Lemma A.6, so
$\mu_{*}(\mathbf{A}+\mathbf{B})\leq\mu(\mathbf{A})+\mu(\mathbf{B})~{}<~{}1$.

Observe that $\widehat{\mathbf{G}}$ is torsion free by Lemma 4.9, as each
$\widehat{G}_{n}$ is torsion free due to the connectedness of $G_{n}$.
Proposition 7.1 implies that there are parallel Bohr intervals
$\tilde{I},\tilde{J}\subseteq\mathbf{G}$ with $\mathbf{A}\subseteq\tilde{I}$,
$\mathbf{B}\subseteq\tilde{J}$, and $\mu(\mathbf{A})=\mu(\tilde{I})$,
$\mu(\mathbf{B})=\mu(\tilde{J})$. Corollary 6.2 and the connectedness of
$G_{n}$ then imply that for $\mathcal{U}$-many $n$, there are parallel Bohr
intervals $\tilde{I}_{n},\tilde{J}_{n}\subseteq\mathbb{T}$ containing $A_{n}$
and $B_{n}$, respectively, such that $m_{n}(\tilde{I}_{n}\setminus
A_{n})+m_{n}(\tilde{J}_{n}\setminus B_{n})<\frac{\varepsilon}{2}$. These
inequalities and containments contradict assumption (a.2). ∎

##  8 Periodic and quasi-periodic sets in ultraproducts

Here we study periodic and quasi-periodic subsets of ultraproducts
(Definitions 1.12 and 1.14). Let $(G_{n})_{n\in\mathbb{N}}$ be a sequence of
compact abelian groups with Haar probability measure $m_{n}$ and inner Haar
measure $m_{n*}$. Let $\mathcal{U}$ be a nonprincipal ultrafilter on
$\mathbb{N}$, and fix the ultraproduct
$\mathbf{G}=\prod_{n\to\mathcal{U}}G_{n}$ with Loeb measure $\mu$
corresponding to $m_{n}$. Let $\mu_{*}$ denote the inner measure associated to
$\mu$. If $K\leq\mathbf{G}$ is a $\mu$-measurable finite index subgroup,
Corollary 4.10 implies $K$ is internal: there is a sequence of compact open
subgroups $K_{n}\leq G_{n}$ such that
$K=\mathbf{K}:=\prod_{n\to\mathcal{U}}K_{n}$. With a view toward the proofs of
Theorems 1.1 and 1.15, we investigate how decompositions of an internal set
$\mathbf{A}$ by cosets of $\mathbf{K}$ are related to coset decompositions of
the constituent sets $A_{n}$ by cosets of $K_{n}$.

For the remainder of the section, fix coset representatives
$\mathbf{x}^{(1)},\dots,\mathbf{x}^{(k)}$ for $\mathbf{K}$, so that
$\mathbf{G}$ is the disjoint union
$\bigcup_{j=1}^{k}(\mathbf{x}^{(j)}+\mathbf{K})$. For each $j\leq k$ and
$n\in\mathbb{N}$, choose $x_{n}^{(j)}\in G_{n}$ so that
$(x_{n}^{(j)})_{n\in\mathbb{N}}$ represents $\mathbf{x}^{(j)}$ (in the sense
of §A.2). We write $\mathbf{K}^{(j)}$ for $\mathbf{x}^{(j)}+\mathbf{K}$ and
write $K^{(j)}_{n}$ for $x^{(j)}_{n}+K_{n}$.

######  Lemma 8.1.

Suppose $\mathbf{G}$, $\mathbf{K}$, $\mathbf{K}^{(j)}$, and $K_{n}^{(j)}$ are
as defined in the preceding paragraph. Let
$\mathbf{A}=\prod_{n\to\mathcal{U}}A_{n}\subseteq\mathbf{G}$ be an internal
set. Then the following hold for $\mathcal{U}$-many $n$. To be clear, this
means that there is a $U\in\mathcal{U}$ such that for all $n\in U$, the
following are true.

  1. (i)

The map $\phi_{n}:G_{n}/K_{n}\to\mathbf{G}/\mathbf{K}$ defined by
$\phi_{n}(K_{n}^{(j)})=\mathbf{K}^{(j)}$ is an isomorphism.

  2. (ii)

$A_{n}\cap K_{n}^{(j)}=\varnothing$ iff
$\mathbf{A}\cap\mathbf{K}^{(j)}=\varnothing$, while $A_{n}\cap
K_{n}^{(j)}=K_{n}^{(j)}$ iff
$\mathbf{A}\cap\mathbf{K}^{(j)}=\mathbf{K}^{(j)}$.

  3. (iii)

$\phi_{n}(A_{n}+K_{n})=\mathbf{A}+\mathbf{K}$; consequently
$\mu(\mathbf{A}+\mathbf{K})=m_{n}(A_{n}+K_{n})$.

Furthermore,

  1. (iv)

For all $j$, $\lim_{n\to\mathcal{U}}m_{n*}(A_{n}\cap
K_{n}^{(j)})=\mu_{*}(\mathbf{A}\cap\mathbf{K}^{(j)})$; if the sets $A_{n}$ are
$m_{n}$-measurable we have $\lim_{n\to\mathcal{U}}m_{n}(A_{n}\cap
K_{n}^{(j)})=\mu(\mathbf{A}\cap\mathbf{K}^{(j)})$.

###### Proof.

To prove (i) first observe that by Corollary 4.10 we have $|G_{n}/K_{n}|=k$
for $\mathcal{U}$-many $n$. Thus it suffices to prove that
$\Phi:=\prod_{n\to\mathcal{U}}\phi_{n}$ is an isomorphism from
$\prod_{n\to\mathcal{U}}(G_{n}/K_{n})$ to $\mathbf{G}/\mathbf{K}$. That $\Phi$
is a homomorphism follows from the identity
$(x_{n}^{(i)}+x_{n}^{(j)})\sim_{\mathcal{U}}(x_{n}^{(i)})+(x_{n}^{(j)})$ for
each $i,j\leq k$. The injectivity of $\Phi$ follows from the fact that if
$x_{n}^{(i)}-x_{n}^{(j)}\in K_{n}$ for $\mathcal{U}$-many $n$, then
$\mathbf{x}^{(i)}-\mathbf{x}^{(j)}\in\mathbf{K}$, and surjectivity then
follows from the equality of cardinalities of the domain and codomain of
$\Phi$. So $\Phi$ is an isomorphism, and hence $\phi_{n}$ is an isomorphism
for $\mathcal{U}$-many $n$.

Part (ii) follows from the definition of the ultraproduct and our choice of
$(x_{n}^{(j)})_{n\in\mathbb{N}}$. Part (iii) is an immediate consequence of
Part (ii). Part (iv) follows from the definition of Loeb measure and Lemma
A.6. ∎

Before stating the next lemma, we remark that if
$\mathbf{A}\subseteq\mathbf{G}$ is internal, $\mathbf{K}$ is an internal
subgroup, and $\mathbf{A}=A_{1}\cup A_{0}$ is a quasi-periodic decomposition
of $\mathbf{A}$ with respect to $\mathbf{K}$, then the sets $A_{1}$ and
$A_{0}$ are internal: $A_{0}$ is the intersection of $\mathbf{A}$ with a coset
of $\mathbf{K}$, while $A_{1}$ is the difference $\mathbf{A}\setminus A_{0}$.

######  Lemma 8.2.

Suppose $\mathbf{G},\mathbf{K}$, $\mathbf{A}$, $K_{n}$, and $\phi_{n}$ are as
in Lemma 8.1, and $\mathbf{A}$ has a quasi-periodic decomposition
$\mathbf{A}=\mathbf{A}_{1}\cup\mathbf{A}_{0}$ with respect to $\mathbf{K}$. If
$\varepsilon>0$ then for $\mathcal{U}$-many $n$, $A_{n}$ has an
$\varepsilon$-quasi-periodic decomposition $A_{n}=A_{n,1}\cup A_{n,0}$ with
respect to $K_{n}$, such that
$\phi_{n}(A_{n,l}+K_{n})=\mathbf{A}_{l}+\mathbf{K}$ for $l=0,1$.

###### Proof.

Write $\mathbf{A}_{l}$ as $\prod_{n\to\mathcal{U}}A_{n,l}$ for $l=0,1$. Parts
(i) and (ii) of Lemma 8.1 imply that for $\mathcal{U}$-many $n$, $A_{n,0}$ is
contained in a coset of $K_{n}$, $A_{n,0}+K_{n}$ is disjoint from $A_{n,1}$,
and $m_{n}(A_{n}+K_{n})=\mu(\mathbf{A}+\mathbf{K})$. Part (iv) then implies
$\lim_{n\to\mathcal{U}}m_{n}(A_{n,1}+K_{n})-m_{n*}(A_{n,1})=0$, and in
particular $m_{n}(A_{n,1}+K_{n})-m_{n*}(A_{n,1})<\varepsilon\mu(K_{n})$ for
$\mathcal{U}$-many $n$. ∎

######  Lemma 8.3.

With $\mathbf{K}$ and $K_{n}$ as in the preceding lemmas, let
$\mathbf{A}=\prod_{n\to\mathcal{U}}A_{n}$ and
$\mathbf{B}=\prod_{n\to\mathcal{U}}B_{n}\subseteq\mathbf{G}$ be internal sets
such that $\mathbf{A}+\mathbf{B}=\mathbf{A}+\mathbf{B}+\mathbf{K}$. Then

  1. (i)

$A_{n}+B_{n}=A_{n}+B_{n}+K_{n}$ for $\mathcal{U}$-many $n$.

  2. (ii)

If, in addition,
$\mu(\mathbf{A}+\mathbf{B})=\mu(\mathbf{A}+\mathbf{K})+\mu(\mathbf{B}+\mathbf{K})-\mu(\mathbf{K})$,
then

$m_{n}(A_{n}+B_{n})=m_{n}(A_{n}+K_{n})+m_{n}(B_{n}+K_{n})-m_{n}(K_{n})\quad\text{for
$\mathcal{U}$-many $n$.}$

###### Proof.

Parts (i) and (ii) follow from Lemma 8.1 and the fact that Haar measure on
$G_{n}/K_{n}$ and Loeb measure on $\mathbf{G}/\mathbf{K}$ are both normalized
counting measure. ∎

##  9 Reducing to tame critical pairs

Here we collect some technical lemmas for §10, where the majority of our
proofs are carried out. These are mainly consequences of the _conclusion_ of
Theorem 3.2. The results of this section do not require countable additivity
of the relevant measures, so we work in a slightly more general setting than
in the other sections.

Call a measure $\nu$ on a group $G$ _symmetric_ if $\nu(-A)=\nu(A)$ for all
$\nu$-measurable sets $A$. Note that Haar measure $m$ on a compact abelian
group is symmetric, so the associated Loeb measure on an ultraproduct of
compact abelian groups is also symmetric.

For the remainder of the section, we fix an abelian group $G$ and a _finitely_
additive (or countably additive), symmetric, translation invariant measure
$\nu$ on $G$. We do not assume that $\nu(G)$ is finite. We continue denote by
$\nu_{*}$ the inner measure associated to $\nu$:
$\nu_{*}(A)=\sup\\{\nu(C):C\subseteq A\text{ is }\nu\text{-measurable}\\}$.

######  Lemma 9.1.

If $A,B\subseteq G$ have finite measure and satisfy $\nu(A)+\nu(B)>\nu(G)$,
then $A+B=G$. If $K\leq G$ is a $\nu$-measurable subgroup such that
$0<\nu(K)<\infty$ and $\nu((a+K)\cap A)+\nu((b+K)\cap B)>\nu(K)$ for some
$a\in A,b\in B$, then $a+b+K\subseteq A+B$.

###### Proof.

If $\nu(A)+\nu(B)>\nu(G)$ then for all $t\in G$, we have $\nu(A\cap(t-B))>0$,
and in particular $t\in A+B$. Thus $A+B=G$. To prove the second assertion,
define the measure $\nu_{K}$ as $\nu$ restricted to $K$. Then
$\nu_{K}((A-a)\cap K)+\nu_{K}((B-b)\cap K)>\nu(K)$, so by the first part of
the lemma we have $K\subseteq A+B-a-b$, meaning $a+b+K\subseteq A+B$. ∎

######  Definition 9.2.

If $K$ is a $\nu$-measurable finite index subgroup of $G$ and $A\subseteq G$,
we say that $A$ is _$K$ -solid_ if for all $g\in G$, either $\nu((g+K)\cap
A)>0$ or $(g+K)\cap A=\varnothing$.

######  Lemma 9.3.

Let $K\leq G$ be a $\nu$-measurable finite index subgroup of $G$, $A\subseteq
G$, and let $B\subseteq G$ be $K$-solid. If $A+B\sim_{\nu}A+B+K$ and $g\in G$,
then $g+B\subset_{\nu}A+B+K$ if and only if $g+B+K\subseteq A+B+K$.

In the notation of Definition 6.3, the conclusion of the lemma says
$\bar{A}_{B,\nu}=\bar{A}_{B+K}$.

###### Proof.

The inclusion $\bar{A}_{B+K}\subseteq\bar{A}_{B,\nu}$ follows from the
similarity $A+B\sim_{\nu}A+B+K$. To prove the reverse inclusion, let
$g\in\bar{A}_{B,\nu}$, so that $g+B\subset_{\nu}A+B$. If $g+B\nsubseteq
A+B+K$, then for some coset $h+K$ disjoint from $A+B+K$,
$(g+B)\cap(h+K)\neq\varnothing$, so the $K$-solidity of $B$ implies
$\nu((g+B)\cap(h+K))>0$, and we get that $g+B\not\subset_{\nu}A+B+K$, and
therefore $g\notin\bar{A}_{B,\nu}$. So we conclude that $g+B\subseteq A+B+K$,
meaning $g\in\bar{A}_{B+K}$. ∎

In the next definition we use the term “stabilizer” from Definition 3.1.

######  Definition 9.4.

We say that two $\nu$-measurable subsets $A,B$ of $G$ form a _critical pair_
if $\nu_{*}(A+B)<\nu(A)+\nu(B)$. We say they form a _tame critical pair_ if,
in addition, $A+B$ is $\nu$-measurable, the stabilizer $H:=H(A+B)$ is
$\nu$-measurable, and

$\nu(A+B)=\nu(A+H)+\nu(B+H)-\nu(H).$ (9.1)

In other words, $A,B$ is a tame critical pair if it satisfies the conclusion
of Theorem 3.2.

Observe that the criticality of $A,B$ and Equation (9.1) imply $\nu(H)>0$.

The following lemmas provide useful information for tame pairs and related
pairs of subsets of $G$. The first of these uses notation from Definition 6.3
and the following.

######  Notation 9.5.

If $H\leq G$ is a subgroup and $g\in G$, we write $H_{g}$ for the coset $g+H$.

######  Lemma 9.6.

Let $A,B\subseteq G$ be a tame critical pair and $H:=H(A+B)$ the stabilizer of
$A+B$. Then

  1. (i)

for all $a\in A$, $b\in B$

$\begin{split}\nu(A\cap H_{a})+\nu(A+B)\geq\nu(A)+\nu(B),\\\ \nu(B\cap
H_{b})+\nu(A+B)\geq\nu(A)+\nu(B),\end{split}$ (9.2)

In particular, $A$ and $B$ are $H$-solid.

  2. (ii)

We have

$\bar{A}_{B,\nu}=\bar{A}_{B}=A+H,\ \bar{B}_{A,\nu}=\bar{B}_{A}=B+H.$ (9.3)

  3. (iii)

Furthermore, if $g\in G$ and $A+g\nsubseteq A+B$, then
$\nu((A+g)\cup(A+B))\geq\nu(A)+\nu(B).$

###### Proof.

To prove the first inequality in Part (i), fix $a\in A$. Let
$A^{\prime}:=A\cap H_{a}$, $A^{\prime\prime}:=A\setminus A^{\prime}$, so that
$A^{\prime\prime}\cap H_{a}=\varnothing$ and
$\nu(A)=\nu(A^{\prime})+\nu(A^{\prime\prime})$. Then $A+H$ is the disjoint
union $(A^{\prime\prime}+H)\cup H_{a}$, so
$\nu(A+H)-\nu(H)=\nu(A^{\prime\prime}+H)$. Replacing $\nu(A+H)-\nu(H)$ with
$\nu(A^{\prime\prime}+H)$ in Equation (9.1) and adding $\nu(A^{\prime})$ to
both sides results in the following:

$\nu(A^{\prime})+\nu(A+B)=\nu(A^{\prime})+\nu(A^{\prime\prime}+H)+\nu(B+H)\geq\nu(A^{\prime})+\nu(A^{\prime\prime})+\nu(B)=\nu(A)+\nu(B),$

meaning $\nu(A^{\prime})+\nu(A+B)\geq\nu(A)+\nu(B)$. This is the desired
inequality. The second inequality in Part (i) follows by symmetry. Note that
the criticality of $A,B$ and (9.2) imply that $A$ and $B$ are $H$-solid.

To prove Part (ii) we first establish $\bar{A}_{B,\nu}=\bar{A}_{B}$. Note that
the tameness of $A+B$ implies $A+B$ is a union of cosets of $H$, so the
$H$-solidity of $B$ and Lemma 9.3 imply $\bar{A}_{B,\nu}=\bar{A}_{B}$. To
prove that $\bar{A}_{B,\nu}=A+H$, we reduce the problem to the special case
where $\nu$ is counting measure. Lemma 9.3 implies
$\bar{A}_{B,\nu}=\bar{A}_{B+H}$, and the last set is clearly a union of cosets
of $H$. So all sets we are currently considering in the proof are unions of
cosets of $H$, and we may work in the quotient $G/H$. We consider $G/H$ as a
discrete group with counting measure, so we may apply Theorem 3.2 in this
setting. Replacing $A$ with $A+H$ and $B$ with $B+H$ in $G/H$, the problem is
to prove that $\bar{A}_{B}=A$, assuming $|A+B|=|A|+|B|-1$ and the stabilizer
$H$ of $A+B$ is trivial. We clearly have $A\subseteq\bar{A}_{B}$, so we will
prove the reverse inclusion. Assume, to get a contradiction, that there is a
$c\notin A$ such that $c+B\subseteq A+B$. Setting $\tilde{A}=A\cup\\{c\\}$, we
get that $\tilde{A}+B=A+B$, so $H(\tilde{A}+B)=H(A+B)$. Our choice of
$\tilde{A}$ then implies $|\tilde{A}+B|=|A+B|=|\tilde{A}|+|B|-2$, which is
strictly less than $|\tilde{A}+H|+|B+H|-|H|$, as $|H|=1$. This contradicts
Inequality (3.1) in Theorem 3.2 applied to $\tilde{A},B$. Thus
$\bar{A}_{B}=A$.

Part (iii) is a consequence of Part (i) and the fact that $A+B$ is a union of
cosets of $H$: under the hypothesis of (iii) we have that
$(A+g)\setminus(A+B)$ contains a set of the form $(A\cap H_{a})+g$ which is
disjoint from $A+B$. ∎

######  Lemma 9.7.

Suppose $H$ is a subgroup of $G$ having $0<\nu(H)<\infty$ and $A,B\subseteq G$
satisfy

$\nu(A+H)+\nu(B+H)-\nu(H)<\nu(A)+\nu(B).$ (9.4)

Then for all $a\in A$, $b\in B$, we have $\nu(A\cap H_{a})+\nu(B\cap
H_{b})>\nu(H)$.

###### Proof.

Inequality (9.4) can be written as $\nu(A+H)-\nu(A)+\nu(B+H)-\nu(B)<\nu(H)$,
meaning

$\nu((A+H)\setminus A)+\nu((B+H)\setminus B)<\nu(H).$ (9.5)

Now $A+H=\bigcup_{a\in A}H_{a}$ and $B+H=\bigcup_{b\in B}H_{b}$, so (9.5)
implies $\nu(H_{a}\setminus A)+\nu(H_{b}\setminus B)<\nu(H)$ for all $a,b\in
H$. The last inequality can be written as $\nu(H)-\nu(A\cap
H_{a})+\nu(H)-\nu(B\cap H_{b})<\nu(H)$, which can be rearranged to yield the
desired conclusion. ∎

The next two lemmas form a major portion of the proof of Proposition 10.1,
generalizing Theorem 3.2 to ultraproducts of compact abelian groups.

######  Lemma 9.8.

Suppose $A,B\subseteq G$ is a critical pair and $A^{\prime},B^{\prime}$ is a
tame critical pair such that $A\subseteq A^{\prime}$, $B\subseteq B^{\prime}$,
and $\nu_{*}(A+B)=\nu(A^{\prime}+B^{\prime})$. Let
$H=H(A^{\prime}+B^{\prime})$. Then for all $a\in A^{\prime},b\in B^{\prime}$,
we have

$\nu(A\cap H_{a})+\nu(B\cap H_{b})>\nu(H).$ (9.6)

Consequently,

$A+H=A^{\prime}+H,\quad B+H=B^{\prime}+H,\quad A+B=A^{\prime}+B^{\prime}+H,$
(9.7)

and $A,B$ is a tame critical pair with $H(A+B)=H$.

###### Proof.

The proof of Inequality (9.6) is similar to the proof of Lemma 9.7. We have

$\nu(A)+\nu(B)>\nu_{*}(A+B)=\nu(A^{\prime}+B^{\prime})=\nu(A^{\prime}+H)+\nu(B^{\prime}+H)-\nu(H),$

so $\nu(A^{\prime}+H)-\nu(A)+\nu(B^{\prime}+H)-\nu(B)<\nu(H)$. Since
$A\subseteq A^{\prime}+H$ and $B\subseteq B^{\prime}+H$, this inequality
implies $\nu(H_{a}\setminus A)+\nu(H_{b}\setminus B)<\nu(H)$ for all $a\in
A^{\prime},b\in B^{\prime}$. As in the proof of Lemma 9.7, the latter
inequality can be rearranged to yield (9.6).

Now Inequality (9.6) implies that $A\cap H_{a}\neq\varnothing$ for all $a\in
A^{\prime}$. Together with the assumption $A\subseteq A^{\prime}$ this implies
$A+H=A^{\prime}+H$, and by symmetry we get $B+H=B^{\prime}+H$. The equation
$A+B+H=A^{\prime}+B^{\prime}+H$ follows from these equations, and the equation
$A+B=A^{\prime}+B^{\prime}+H$ then follows from (9.6) and Lemma 9.1. We
therefore have $H(A+B)=H(A^{\prime}+B^{\prime}+H)$, while the definition of
$H$ implies $A^{\prime}+B^{\prime}=A^{\prime}+B^{\prime}+H$, and we may
conclude $H(A+B)=H(A^{\prime}+B^{\prime})$. The tameness of $A,B$ then follows
from the tameness of $A^{\prime},B^{\prime}$ and the equations in (9.7). ∎

######  Lemma 9.9.

Suppose $A,B\subseteq G$ is a critical pair and $A^{\prime},B^{\prime}$ is a
tame critical pair with $A^{\prime}\subseteq A,B^{\prime}\subseteq B$, and
$\nu(A)=\nu(A^{\prime})$, $\nu(B)=\nu(B^{\prime})$. Then $A,B$ is a tame
critical pair with $H(A+B)=H(A^{\prime}+B^{\prime})$.

###### Proof.

We must prove that $H:=H(A^{\prime}+B^{\prime})$ is the stabilizer of $A+B$
and and that Equation (9.1) is satisfied. To prove this, we will show that
$A+B=A^{\prime}+B^{\prime}$, $A+H=A^{\prime}+H$, and $B+H=B^{\prime}+H$.

We first show that

$A^{\prime}+b\subset_{\nu}A^{\prime}+B^{\prime}\quad\text{ for all }b\in B.$
(9.8)

To prove (9.8) we assume otherwise and derive a contradiction by proving that
$\nu_{*}(A+B)\geq\nu(A)+\nu(B)$, contrary to the hypothesis. So we assume
there is a $b\in B$ such that
$C:=(A^{\prime}+b)\setminus(A^{\prime}+B^{\prime})$ satisfies $\nu(C)>0$. Then
Part (iii) of Lemma 9.6 implies
$\nu((A^{\prime}+b)\cup(A^{\prime}+B^{\prime}))\geq\nu(A^{\prime})+\nu(B^{\prime})$.
The containments in the hypothesis imply
$(A^{\prime}+b)\cup(A^{\prime}+B^{\prime})\subseteq A+B$, so
$\nu_{*}(A+B)\geq\nu(A^{\prime})+\nu(B^{\prime})=\nu(A)+\nu(B)$, which is the
desired contradiction.

Having proved (9.8), Part (ii) of Lemma 9.6 and (9.8) together imply
$B\subseteq B^{\prime}+H$. Likewise we have $A\subseteq A^{\prime}+H$. The
containments $A^{\prime}\subseteq A$ and $B^{\prime}\subseteq B$ then imply
$A+H=A^{\prime}+H$ and $B+H=B^{\prime}+H$, and the equation
$A^{\prime}+B^{\prime}=A^{\prime}+B^{\prime}+H$ implies
$A+B=A^{\prime}+B^{\prime}+H$. Finally, the tameness of
$A^{\prime},B^{\prime}$ implies
$\nu(A+B)=\nu(A^{\prime}+B^{\prime})=\nu(A+H)+\nu(B+H)-\nu(H)$, as desired. ∎

The next definition is useful in dealing with quasi-periodic pairs (Definition
1.14).

######  Definition 9.10.

If $H$ is a subgroup of $G$, $A,B\subseteq G$, and $a\in A,b\in B$, we say
that $a+b+H$ is a _unique expression element of $A+B+H$ in $G/H$_ if
$a^{\prime}\in A$, $b^{\prime}\in B$ and $a^{\prime}+b^{\prime}+H=a+b+H$ imply
$a^{\prime}\in a+H$ and $b^{\prime}\in b+H$.

The next lemma deals with a case arising frequently in the proofs of our main
results.

######  Lemma 9.11.

Assume $A,B\subseteq G$ satisfy $\nu_{*}(A+B)=\nu(A)+\nu(B)$,
$B^{\prime}\subseteq B$ satisfies $\nu(B^{\prime})=\nu(B)$, and $A,B^{\prime}$
is a tame critical pair in $G$. Let $H=H(A+B^{\prime})$. Then

  1. (i)

$(A+B+H)\setminus(A+B^{\prime})$ is a coset of $H$,

  2. (ii)

$B^{\prime}\sim_{\nu}B^{\prime}+H$,

  3. (iii)

$A$ has a quasi-periodic decomposition $A_{1}\cup A_{0}$ with respect to $H$,
where

$\nu(A_{0})=\nu_{*}((A+B)\setminus(A+B^{\prime})).$

  4. (iv)

If $A+B\nsim_{\nu}A+B+H$, then setting $B_{1}:=B\cap(B^{\prime}+H)$,
$B_{0}:=B\setminus B_{1}$ yields a quasi-periodic decomposition of $B$ with
respect to $H$ where $\nu(B_{0})=0$, and $A_{0}+B_{0}+H$ is a unique
expression element of $A+B+H$. Furthermore,
$\nu_{*}(A_{0}+B_{0})=\nu(A_{0})+\nu(B_{0})$.

###### Proof.

Throughout this proof we let $C$ denote $(A+B+H)\setminus(A+B^{\prime})$. We
first observe that $C$ is nonempty, as $A+B^{\prime}$ is a union of cosets of
$H$ and is a proper subset of $A+B$.

We prove Part (i) by contradiction: assume that $C$ has nonempty intersection
with at least two cosets of $H$. We may then choose $a_{1},a_{2}\in A$ and
$b_{1},b_{2}\in B$ such that $a_{1}+b_{1}$ and $a_{2}+b_{2}$ lie in $C$ and do
not occupy the same coset of $H$. Let $A_{1}=A\cap(a_{1}+H)$,
$A_{2}=A\cap(a_{2}+H)$. Then $A_{1}+b_{1}$, $A_{2}+b_{2}$, and $A+B^{\prime}$
are mutually disjoint subsets of $A+B$, while Lemma 9.6 (i) implies
$\nu(A_{i}+b_{i})+\nu(A+B^{\prime})\geq\nu(A)+\nu(B^{\prime})$ for each $i$.
The last inequality, the criticality of $A,B^{\prime}$, and the disjointness
of $C$ from $A+B^{\prime}$ imply
$\nu_{*}(A+B)>\nu(A)+\nu(B^{\prime})=\nu(A)+\nu(B)$, contradicting our
assumption to the contrary.

To prove Parts (ii) and (iii), choose $a\in A$ and $b\in B$ such that
$a+b\in(A+B)\setminus(A+B^{\prime})$. Let $A_{0}:=A\cap(a+H)$, so that Lemma
9.6 (i) implies $\nu(A_{0})+\nu(A+B^{\prime})\geq\nu(A)+\nu(B^{\prime})$,
while the containment $(A_{0}+b)\cup(A+B^{\prime})\subseteq A+B$ and the
hypothesis on $\nu(A+B)$ imply the reverse inequality. Thus

$\nu(A_{0})+\nu(A+B^{\prime})=\nu(A)+\nu(B^{\prime}),$ (9.9)

and the assumption that $A,B^{\prime}$ is a tame critical pair allows us to
use Equation (9.1) with $B^{\prime}$ in place of $B$. Combining that equation
with (9.9) and rearranging, we obtain

$\nu(A+H)-\nu(A)+\nu(B^{\prime}+H)-\nu(B^{\prime})=\nu(H)-\nu(A_{0}).$ (9.10)

Since $A_{0}$ is contained in a coset of $H$, the right hand side above is
equal to $\nu((A_{0}+H)\setminus A_{0})$. So we can rewrite (9.10) as

$\nu((A+H)\setminus A)+\nu((B^{\prime}+H)\setminus
B^{\prime})=\nu((A_{0}+H)\setminus A_{0}).$ (9.11)

Since $(A_{0}+H)\setminus A_{0}\subseteq(A+H)\setminus A$, we have
$\nu((A+H)\setminus A)\geq\nu((A_{0}+H)\setminus A_{0})$, and (9.11) then
implies

$\nu(A+H)-\nu(A)=\nu(H)-\nu(A_{0})$ (9.12)

and $\nu(B^{\prime}+H)-\nu(B^{\prime})=0$. The last equation implies (ii).
Setting $A_{1}=A\setminus A_{0}$, Equation (9.12) implies
$A_{1}\sim_{\nu}A_{1}+H$. Thus $A_{1}\cup A_{0}$ is the desired quasi-periodic
decomposition of $A$. To complete the proof of (iii), note that the right hand
side of Equation (9.9) is $\nu_{*}(A+B)$, so that equation can be rewritten as
$\nu(A_{0})=\nu_{*}(A+B)-\nu(A+B^{\prime})$.

We prove Part (iv) using the following claim.

######  Claim.

The additional hypothesis (iv) implies $C+H$ is a unique expression element of
$(A+H)+(B+H)$ in $G/H$.

To prove the Claim, we will show that $C+H$ is disjoint from $A_{1}+B$,
meaning that if $a\in A,b\in B$ satisfy $a+b+H=C+H$, then $a\in A_{0}+H$ and
$b\in C-A_{0}+H$. Assume, to get a contradiction, that $C+H$ is not disjoint
from $A_{1}+B$. Choose $a_{1}\in A_{1}$ and $b\in B$ such that $a_{1}+b\in
C+H$. Since $A_{1}\sim_{\nu}A_{1}+H$, we conclude that $C+H\subset_{\nu}A+B$.
Together with Part (i) and the equation $A+B^{\prime}=A+B^{\prime}+H$, this
implies $A+B\sim_{\nu}A+B+H$, contradicting the hypothesis in (iv). This
proves the Claim.

Now let $B_{1}$ be as in the statement of Part (iv), and let $A_{0}$ be as
defined in the proof of Part (iii). In particular $A_{0}+b_{0}\subseteq C+H$
for some $b_{0}\in B$. Let $B_{0}=B\cap(b_{0}+H)$. We will show that
$B=B_{1}\cup B_{0}$, which by Part (ii) yields the desired quasi-periodic
decomposition. If $A+b\subseteq A+B^{\prime}$, we will show that $b\in
B^{\prime}+H$, and otherwise we will show that $b\in B_{0}$. Under the first
assumption, we apply Part (ii) of Lemma 9.6 to $A,B^{\prime}$ and conclude
$b\in B^{\prime}+H$. If $A+b\nsubseteq A+B^{\prime}$, the containment
$A+B\subseteq C\cup(A+B^{\prime})$ implies $(A+b)\cap C\neq\varnothing$. Then
there is an $a\in A$ such that $a+b\in C$. The Claim then implies $a\in
A_{0}$, so $b\in B_{0}$. The last equation in the conclusion of (iv) follows
from (iii) and the fact that $\nu(B_{0})=0$. ∎

The following lemma is used only in the proof of Lemma 12.3.

######  Lemma 9.12.

Let $K$ be a $\nu$-measurable subgroup of $G$ with $0<\nu(K)<\infty$. Suppose
$C,D\subseteq G$ are periodic with respect to $K$, and
$\nu(C+D)=\nu(C)+\nu(D)-\nu(K)$. Assume further that $C_{0},D_{0}$ are cosets
of $K$ contained in $C$ and $D$, respectively, and $C_{0}+D_{0}$ is a unique
expression element of $C+D+K$ in $G/K$. Let $A_{1}=C\setminus C_{0}$,
$B_{1}=D\setminus D_{0}$, and suppose $A_{0}\subseteq C_{0}$ and
$B_{0}\subseteq D_{0}$ are nonempty and satisfy
$\nu_{*}(A_{0}+B_{0})\leq\nu(A_{0})+\nu(B_{0})$. If $A=A_{1}\cup A_{0}$ and
$B=B_{1}\cup B_{0}$, then $\nu_{*}(A+B)\leq\nu(A)+\nu(B)$.

###### Proof.

Let $E=(A+B_{1})\cup(A_{1}+B)$. The definitions of $A$ and $B$ and the fact
that $C_{0}+D_{0}$ is a unique expression element of $C+D+K$ then imply
$E=(C+D)\setminus(C_{0}+D_{0})$, so $\nu(E)=\nu(C+D)-\nu(K)$. Note that
$\nu(C+D)-\nu(K)$ simplifies to

$\nu(C)-\nu(K)+\nu(D)-\nu(K)=\nu(A_{1})+\nu(B_{1}).$

Then

$\nu_{*}(A+B)=\nu(E)+\nu_{*}(A_{0}+B_{0})\leq\nu(A_{1})+\nu(A_{0})+\nu(B_{1})+\nu(B_{0})=\nu(A)+\nu(B),$

as desired. ∎

##  10 Lifting Theorems 3.2 and 3.5 to ultraproducts

The main results of this section are Propositions 10.1 and 10.2, the natural
generalizations of Theorems 3.2 and 3.5 to ultraproducts of compact abelian
groups.

For this section fix a sequence of compact abelian groups $G_{n}$ with Haar
probability measure $m_{n}$ and a nonprincipal ultrafilter $\mathcal{U}$ on
$\mathbb{N}$. Let $\mathbf{G}$ be the ultraproduct
$\prod_{n\to\mathcal{U}}G_{n}$ and $\mu$ the Loeb measure on $\mathbf{G}$
corresponding to $(m_{n})_{n\in\mathbb{N}}$. Let $\mu_{*}$ be the inner
measure on $\mathbf{G}$ corresponding to $\mu$. Recall Definition 3.1: the
stabilizer of a set $C\subseteq\mathbf{G}$ is the subgroup

$H(C):=\\{g\in\mathbf{G}:C+g=C\\}.$

######  Proposition 10.1.

If $A,B\subseteq\mathbf{G}$ are $\mu$-measurable sets with
$\mu_{*}(A+B)<\mu(A)+\mu(B)$, then the stabilizer $\mathbf{H}:=H(A+B)$ is an
internal $\mu$-measurable finite index subgroup of $\mathbf{G}$ and

$\mu(A+B)=\mu(A+\mathbf{H})+\mu(B+\mathbf{H})-\mu(\mathbf{H}).$ (10.1)

The proof of Proposition 10.1 is presented after the proof of Lemma 10.6.

In the next proposition we use the terms “Bohr interval” and “quasi-periodic”,
from Definitions 1.4 and 1.14.

######  Proposition 10.2.

Let $A,B\subseteq\mathbf{G}$ be $\mu$-measurable sets satisfying $\mu(A)>0$,
$\mu(B)>0$, and $\mu_{*}(A+B)=\mu(A)+\mu(B)$. Then there is an internal
$\mu$-measurable finite index subgroup $\mathbf{K}\leq\mathbf{G}$ such that
exactly one of the following holds.

  1. (I)

$A+B\sim_{\mu}A+B+\mathbf{K}$.

  2. (II)

$A+B\nsim_{\mu}A+B+\mathbf{K}$ and $A$ and $B$ have quasi-periodic
decompositions $A=A_{1}\cup A_{0}$, $B=B_{1}\cup B_{0}$ with respect to
$\mathbf{K}$ where at least one of $A_{1}$, $B_{1}$ is nonempty and
$\mu_{*}(A_{0}+B_{0})=\mu(A_{0})+\mu(B_{0})$.

  3. (III)

$A+B\nsim_{\mu}A+B+\mathbf{K}$ and there are parallel Bohr intervals
$A^{\prime},B^{\prime}\subseteq\mathbf{K}$ and
$\mathbf{a},\mathbf{b}\in\mathbf{G}$ such that $\mu(A)=\mu(A^{\prime})$,
$\mu(B)=\mu(B^{\prime})$, and $A\subseteq\mathbf{a}+A^{\prime}$,
$B\subseteq\mathbf{b}+B^{\prime}$.

The proof of Proposition 10.2 is presented following the proof of Lemma 10.13.

######  Remark 10.3.

The condition $A+B\nsim_{\mu}A+B+\mathbf{K}$ in conclusion (II) above implies
that $A_{0}+B_{0}+\mathbf{K}$ is a unique expression element of
$A+B+\mathbf{K}$ in $\mathbf{G}/\mathbf{K}$ (Definition 9.10): assuming
otherwise we would have $A+B\sim_{\mu}A+B+\mathbf{K}$. Consequently,

$\displaystyle\mu_{*}((A+B)\setminus(A_{0}+B_{0}))$
$\displaystyle=\mu_{*}(A+B)-\mu_{*}(A_{0}+B_{0})$
$\displaystyle=\mu(A)+\mu(B)-\mu(A_{0})-\mu(B_{0})$
$\displaystyle=\mu(A_{1})+\mu(B_{1}).$

Thus if $A,B$ satisfies conclusion (II) of Proposition 10.2, then

$\mu_{*}((A+B)\setminus(A_{0}+B_{0}))=\mu(A_{1})+\mu(B_{1}).$ (10.2)

We thereby deduce the following corollary (cf. Corollary 3.1 of [19]).

######  Corollary 10.4.

If $G$, $A$, and $B$ are as in Proposition 10.2 and $A,B$, and $\mathbf{K}$
satisfy conclusion (II) therein, then
$\mu(A+B+\mathbf{K})=\mu(A+\mathbf{K})+\mu(B+\mathbf{K})-\mu(\mathbf{K}).$

###### Proof.

Note that the quasi-periodic decompositions of $A$ and $B$ with respect to
$\mathbf{K}$ yield $\mu(A+\mathbf{K})=\mu(A_{1})+\mu(\mathbf{K})$, and
$\mu(B+\mathbf{K})=\mu(B_{1})+\mu(\mathbf{K})$. Adding these and rearranging
yields

$\mu(A_{1})+\mu(B_{1})+\mu(\mathbf{K})=\mu(A+\mathbf{K})+\mu(B+\mathbf{K})-\mu(\mathbf{K}).$
(10.3)

Equation (10.2) and the fact that $A_{0}+B_{0}$ is a unique expression element
of $A+B+\mathbf{K}$ in $\mathbf{G}/\mathbf{K}$ implies
$\mu(A+B+\mathbf{K})=\mu(A_{1})+\mu(B_{1})+\mu(\mathbf{K})$, whereby (10.3)
provides the desired simplification. ∎

The remainder of this section is dedicated to the proofs of Propositions 10.1
and 10.2. The next definitions introduce some useful terminology for these
proofs.

For the following definitions let $G$ be an abelian group and $\nu$ a finitely
additive translation invariant measure on $G$. We call a pair of subsets
$A,B\subseteq G$ _critical_ (as in §9) if $\nu_{*}(A+B)<\nu(A)+\nu(B)$, or
_subcritical_ if $\nu_{*}(A+B)=\nu(A)+\nu(B)$.

We continue to use the term “tame” (Definition 9.4) when considering critical
pairs. For subcritical pairs we extend the definition as follows.

######  Definition 10.5 (Tameness and type).

A pair of sets $A,B\subseteq\mathbf{G}$ is a _tame subcritical pair_ if
$\nu_{*}(A+B)=\nu(A)+\nu(B)$ and there is a $\nu$-measurable finite index
subgroup $K\leq G$ such that $A,B$, and $K$ satisfy one of the conclusions
(I)-(III) of Proposition 10.2. If $A,B$, and $K$ satisfy conclusion (I), we
say that the triple $A,B,K$ _has type (I)_ , and similarly for the other
conclusions. We say that $A,B$ has type (I) if there exists a finite index
subgroup $K\leq G$ such that $A,B,K$ has type (I), and similarly for the other
types.

When the ambient group is compact, the terms “tame” and “type” can be applied,
simply replacing Proposition 10.2 with Theorem 3.5 in the definition.

######  Lemma 10.6.

Let $G$ be a compact quotient of $\mathbf{G}$ with $\mu$-measurable quotient
map $\pi:\mathbf{G}\to G$. If $C,D\subseteq G$ is a subcritical pair (critical
pair) in $G$, then $\pi^{-1}(C)$, $\pi^{-1}(D)$ is a tame subcritical pair
(tame critical pair) in $\mathbf{G}$.

###### Proof.

We only prove the assertion for critical pairs. The assertion for subcritical
pairs follows from a similar routine analysis, using Theorem 3.5 in place of
Theorem 3.2. Let $m$ denote Haar measure on $G$.

Assume $C,D$ is a critical pair in $G$. Let $A=\pi^{-1}(C)$, $B=\pi^{-1}(D)$,
so the $\mu$-measurability of $\pi$ and Lemma 1.3 imply $\mu(A)=m(C)$,
$\mu(B)=m(D)$, and

$\mu(A+B)=m(C+D)<m(C)+m(D)=\mu(A)+\mu(B),$

so $A,B$ is a critical pair. Theorem 3.2 guarantees the existence of a compact
open (and therefore finite index) subgroup $H\leq G$ such that $C+D=C+D+H$ and
$m(C+D)=m(C+H)+m(D+H)-m(H)$. The quotient $G/H$ is finite, so if $\rho:G\to
G/H$ denotes the quotient map, then $\tau:=\rho\circ\pi$ is a $\mu$-measurable
homomorphism from $\mathbf{G}$ onto the finite group $G/H$. Corollary 4.10
implies the kernel $\mathbf{H}$ of $\tau$ is an internal $\mu$-measurable
finite index subgroup of $\mathbf{G}$. Now $\mathbf{H}$ is the stabilizer of
$A+B$ and the equation
$\mu(A+B)=\mu(A+\mathbf{H})+\mu(B+\mathbf{H})-\mu(\mathbf{H})$ follows as
$\pi$ preserves $\mu$. Thus $A,B$ is a tame critical pair in $\mathbf{G}$. ∎

Perhaps there is a direct proof of Proposition 10.1 imitating the proof of
Theorem 3.2 in [29], but we do not pursue this approach. Instead we lift the
result from the compact setting using an argument from [26], where an analogue
of Theorem 3.2 is proved for upper Banach density in $\mathbb{Z}$. For this
argument we require Lemmas 9.8 and 9.9.

###### Proof of Proposition 10.1.

Consider $\mu$-measurable sets $A,B\subseteq\mathbf{G}$ satisfying
$\mu_{*}(A+B)<\mu(A)+\mu(B)$. Our goal, in the terminology of Definition 9.4,
is to prove that $A,B$ is a tame critical pair. Apply Corollary 5.5 to find a
compact abelian quotient $G$ of $\mathbf{G}$ with Haar probability measure
$m$, a $\mu$-measure preserving quotient map $\pi:\mathbf{G}\to G$, and sets
$C,D\subseteq G$, such that
$A\subset_{\mu}A^{\prime}:=\pi^{-1}(C),B\subset_{\mu}B^{\prime}:=\pi^{-1}(D)$,
$A^{\prime}+B^{\prime}$ is $\mu$-measurable and
$A^{\prime}+B^{\prime}\subset_{\mu}A+B$. Then

$\displaystyle m(C+D)=\mu(A^{\prime}+B^{\prime})$
$\displaystyle\leq\mu_{*}(A+B)<\mu(A)+\mu(B)\leq\mu(A^{\prime})+\mu(B^{\prime})=m(C)+m(D).$

Thus $m(C+D)<m(C)+m(D)$, so $C,D$ is a critical pair in $G$. Then Theorem 3.2
implies $C,D$ is a tame critical pair, so Lemma 10.6 implies
$A^{\prime},B^{\prime}$ is a tame critical pair. Let $A^{\prime\prime}=A\cap
A^{\prime}$ and $B^{\prime\prime}=B\cap B^{\prime}$. We will prove that the
pair $A^{\prime\prime},B^{\prime\prime}$ is tame. First, observe that the
essential containment $A\subset_{\mu}A^{\prime}$ implies
$A^{\prime\prime}\sim_{\mu}A$, and likewise $B^{\prime\prime}\sim_{\mu}B$. We
now show that
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})=\mu(A^{\prime}+B^{\prime})$. Note
that $A^{\prime\prime}+B^{\prime\prime}\subseteq A^{\prime}+B^{\prime}$, so
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})\leq\mu(A^{\prime}+B^{\prime})$.
The reverse inequality follows from Part (ii) of Corollary 5.5 and the
similarities noted above.

We have shown that
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})=\mu(A^{\prime}+B^{\prime})$, while
the inequality
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})<\mu(A^{\prime\prime})+\mu(B^{\prime\prime})$
follows from the similarities $A^{\prime\prime}\sim_{\mu}A$ and
$B^{\prime\prime}\sim_{\mu}B$, the containments $A^{\prime\prime}\subseteq A$,
$B^{\prime\prime}\subseteq B$, and the fact that $A,B$ is a critical pair. We
may therefore apply Lemma 9.8 to conclude that
$A^{\prime\prime},B^{\prime\prime}$ is a tame critical pair. Now Lemma 9.9
implies $A,B$ is a tame critical pair, completing the proof of Proposition
10.1. ∎

The next definition introduces some pairs of sets for the proof of Proposition
10.2.

######  Definition 10.7.

A subcritical pair of sets $A,B\subseteq\mathbf{G}$ is

  1. $\bullet$

_extendible_ if there are $A^{\prime}\supseteq A$, $B^{\prime}\supseteq B$
such that $\mu(A^{\prime})+\mu(B^{\prime})>\mu(A)+\mu(B)$, while
$\mu_{*}(A^{\prime}+B^{\prime})=\mu_{*}(A+B)$,

  2. $\bullet$

_reducible_ if there are sets $A^{\prime\prime}\subseteq A$,
$B^{\prime\prime}\subseteq B$ such that $\mu(A^{\prime\prime})=\mu(A)$,
$\mu(B^{\prime\prime})=\mu(B)$, and
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})<\mu_{*}(A+B)$. In this case we say
$A,B$ _reduces to_ $A^{\prime\prime},B^{\prime\prime}$.

If the pair $A,B$ is not extendible, we say it is _nonextendible_ , and if it
is not reducible, we say that it is _irreducible_.

In subsequent proofs it will be convenient to assume that a pair $A,B$ is
nonextendible and irreducible. The following three lemmas allow us to do so,
by proving that if a subcritical pair is extendible or reducible, then it is a
tame subcritical pair of type (I) or type (II) (in the terminology of
Definition 10.5).

######  Lemma 10.8.

If $A,B\subseteq\mathbf{G}$ satisfy $\mu(A),\mu(B)>0$,
$\mu_{*}(A+B)=\mu(A)+\mu(B)$, and $A,B$ is extendible, then $A,B$ has type
(I).

###### Proof.

If $A,B$ is extendible, then there exist sets
$A^{\prime},B^{\prime}\subseteq\mathbf{G}$ such that $A\subseteq A^{\prime}$,
$B\subseteq B^{\prime}$, $\mu(A^{\prime})+\mu(B^{\prime})>\mu(A)+\mu(B)$, and
$\mu_{*}(A^{\prime}+B^{\prime})=\mu_{*}(A+B)$. Thus
$\mu_{*}(A^{\prime}+B^{\prime})<\mu(A^{\prime})+\mu(B^{\prime})$. Then
Proposition 10.1 implies that there is an internal $\mu$-measurable finite
index subgroup $\mathbf{H}\leq\mathbf{G}$ such that
$A^{\prime}+B^{\prime}=A^{\prime}+B^{\prime}+\mathbf{H}$. The assumption
$\mu_{*}(A^{\prime}+B^{\prime})=\mu_{*}(A+B)$ and the containment
$A+B\subseteq A^{\prime}+B^{\prime}$ now imply
$A+B\sim_{\mu}A^{\prime}+B^{\prime}+\mathbf{H}$, so
$A+B\sim_{\mu}A+B+\mathbf{H}$, meaning $A,B$ has type (I). ∎

######  Lemma 10.9.

Suppose that $A,B\subseteq G$ satisfies $\mu_{*}(A+B)=\mu(A)+\mu(B)$, and
$A,B$ reduces to a tame critical pair $A^{\prime},B^{\prime}$. Let
$\mathbf{H}=H(A^{\prime}+B^{\prime})$, and assume that
$A+B\nsim_{\mu}A+B+\mathbf{H}$. Then $A$ and $B$ have quasi-periodic
decompositions with respect to $\mathbf{H}$ where at least one of
$\mu(A_{0})=0$ or $\mu(B_{0})=0$, and

$\mu_{*}(A_{0}+B_{0})=\mu(A_{0})+\mu(B_{0}).$ (10.4)

###### Proof.

We consider cases based on $\mu_{*}(A+B^{\prime})$ and
$\mu_{*}(A^{\prime}+B)$.

Case 1: $\mu_{*}(A+B^{\prime})<\mu_{*}(A+B)$ or
$\mu_{*}(A^{\prime}+B)<\mu_{*}(A+B)$. By symmetry we need only consider the
first inequality. In this case Proposition 10.1 implies $A,B^{\prime}$ is a
tame critical pair. Lemma 9.1 and Lemma 9.7 now imply
$A+B^{\prime}=A^{\prime}+B^{\prime}$, so
$H(A+B^{\prime})=H(A^{\prime}+B^{\prime})$. Part (iv) of Lemma 9.11 then
implies the desired conclusion.

Case 2: Neither $\mu_{*}(A+B^{\prime})<\mu_{*}(A+B)$ nor
$\mu_{*}(A^{\prime}+B)<\mu_{*}(A+B)$. In this case we have
$\mu_{*}(A^{\prime}+B)=\mu_{*}(A+B^{\prime})=\mu_{*}(A+B)$. We apply Lemma
9.11 with $A^{\prime}$ in place of $A$, and thus obtain a quasi-periodic
decomposition $B=B_{1}\cup B_{0}$ with respect to $\mathbf{H}$ where
$\mu(B_{0})=0$ and $B_{1}=B\cap(B^{\prime}+\mathbf{H})$. Similarly we apply
Lemma 9.11 with $B^{\prime}$ in place of $A$ and $A$ in place of $B$. We
thereby obtain a quasi-periodic decomposition $A=A_{1}\cup A_{0}$ with respect
to $\mathbf{H}$ such that $\mu(A_{0})=0$. The assumption that
$A+B\nsim_{\mu}A+B+\mathbf{H}$ then implies $A_{0}+B_{0}+\mathbf{H}$ is a
unique expression element of $A+B+\mathbf{H}$. In particular,
$A_{0}+B_{0}\subseteq D:=(A+B)\setminus(A+B_{1})$. Now our assumption that
$\mu_{*}(A+B)=\mu_{*}(A+B^{\prime})$ implies $\mu_{*}(A+B_{1})=\mu_{*}(A+B)$.
Then $\mu_{*}(D)=0$, so that $\mu_{*}(A_{0}+B_{0})=0$, which implies (10.4). ∎

######  Lemma 10.10.

If $A,B\subseteq\mathbf{G}$ has $\mu_{*}(A+B)=\mu(A)+\mu(B)$ and $A,B$ is
reducible and does not satisfy conclusion (I) in Proposition 10.2, then $A,B$
satisfies conclusion (II) of Proposition 10.2 with at least one of
$\mu(A_{0})=0$ or $\mu(B_{0})=0$.

###### Proof.

Suppose the pair $A,B$ is as above and $A,B$ reduces to
$A^{\prime},B^{\prime}$, where $\mu_{*}(A^{\prime}+B^{\prime})<\mu(A)+\mu(B)$.
Let $\mathbf{H}=H(A^{\prime}+B^{\prime})$, so that $\mathbf{H}$ is an internal
$\mu$-measurable finite index subgroup of $\mathbf{G}$, by Proposition 10.1.
Our hypothesis implies $A+B\nsim_{\mu}A+B+\mathbf{H}$, so Lemma 10.9 yields
the desired conclusion. ∎

The next lemma uses notation from Definition 6.3.

######  Lemma 10.11.

Let $A,B\subseteq\mathbf{G}$ be $\mu$-measurable sets having $\mu(A),\mu(B)>0$
such that $\mu_{*}(A+B)\leq\mu(A)+\mu(B)$, and assume the stabilizer
$\mathbf{H}$ of $A+B$ is a $\mu$-measurable finite index subgroup of
$\mathbf{G}$. Then either $\bar{A}_{B}=A+\mathbf{H}$, or
$\bar{A}_{B}=(A+\mathbf{H})\cup(\mathbf{g}+\mathbf{H})$ for some
$\mathbf{g}\in\mathbf{G}\setminus(A+\mathbf{H})$.

###### Proof.

Since $\mathbf{H}$ is the stabilizer of $A+B$, we have
$A+\mathbf{H}\subseteq\bar{A}_{B}$, and we aim to establish the reverse
containment. If $\bar{A}_{B}\neq A+\mathbf{H}$, then there is a
$\mathbf{g}\in\mathbf{G}\setminus(A+\mathbf{H})$ with $\mathbf{g}+B\subseteq
A+B$. Setting $C:=A\cup(\mathbf{g}+\mathbf{H})$, we have
$\mu(C)=\mu(A)+\mu(\mathbf{H})>\mu(A)$, while $C+B=A+B$, so
$\mu(C+B)=\mu(A+B)<\mu(C)+\mu(B)$. Furthermore $H(C+B)=H(A+B)$, so Proposition
10.1 implies the hypotheses of Lemma 9.6 are satisfied with $C$ in place of
$A$. Then $\bar{C}_{B}=C+\mathbf{H}$. Now $\bar{C}_{B}=\bar{A}_{B}$, so
$\bar{A}_{B}=C+\mathbf{H}=(A+\mathbf{H})\cup(\mathbf{g}+\mathbf{H})$.∎

The next lemma deals with a case arising in the proof of Proposition 10.2.

######  Lemma 10.12.

Suppose $A,B\subseteq\mathbf{G}$ have $\mu(A),\mu(B)>0$, and
$\mu_{*}(A+B)=\mu(A)+\mu(B)$. Furthermore assume there are sets
$A^{\prime}\subseteq A$ and $B^{\prime}\subseteq B$ and an internal
$\mu$-measurable finite index subgroup $\mathbf{H}\leq\mathbf{G}$ such that

  1. $\bullet$

$A^{\prime}\sim_{\mu}A,B^{\prime}\sim_{\mu}B$, and
$\mu_{*}(A^{\prime}+B^{\prime})=\mu(A^{\prime})+\mu(B^{\prime})$ (so
$\mu_{*}(A^{\prime}+B^{\prime})=\mu_{*}(A+B)$),

  2. $\bullet$

$A^{\prime}$ and $B^{\prime}$ are both $\mathbf{H}$-solid (Definition 9.2),

  3. $\bullet$

$A^{\prime}+B^{\prime}\sim_{\mu}A^{\prime}+B^{\prime}+\mathbf{H}$, and
$\mathbf{H}$ is the stabilizer of $A^{\prime}+B^{\prime}+\mathbf{H}$.

Then $A,B$ satisfies conclusion (I) or conclusion (II) of Proposition 10.2.

###### Proof.

First note that for all $a\in A$,
$a+B^{\prime}\subset_{\mu}A^{\prime}+B^{\prime}$. This follows from the
assumption $\mu_{*}(A^{\prime}+B^{\prime})=\mu_{*}(A+B)$ and the containment
$A+B^{\prime}\subseteq A+B$. Thus
$A\subseteq\bar{A^{\prime}}_{B^{\prime},\mu}$ in the notaion of Definition
6.3. Furthermore, the $\mathbf{H}$-solidity of $B^{\prime}$ implies, via Lemma
9.3, that
$\bar{A^{\prime}}_{B^{\prime},\mu}=\bar{A^{\prime}}_{B^{\prime}+\mathbf{H}}$.
We consider two cases.

Case 1:
$\mu(A^{\prime}+\mathbf{H})+\mu(B^{\prime}+\mathbf{H})>\mu(A^{\prime})+\mu(B^{\prime})$.
In this case we will prove that $A\subseteq A^{\prime}+\mathbf{H}$ and
$B\subseteq B^{\prime}+\mathbf{H}$. Let $C=A^{\prime}+\mathbf{H}$,
$D=B^{\prime}+\mathbf{H}$. Then $\mu(C+D)<\mu(C)+\mu(D)$, so we may apply
Proposition 10.1 to the pair $C,D$. Our hypothesis on $A^{\prime},B^{\prime}$
implies that $\mathbf{H}$ is the stabilizer of $C+D$. Then Lemma 9.6 implies
$\bar{C}_{D}=C+\mathbf{H}$. In other words: if
$a+B^{\prime}+\mathbf{H}\subseteq A^{\prime}+B^{\prime}+\mathbf{H}$, then
$a\in A^{\prime}+\mathbf{H}$. So we have shown that
$\bar{A^{\prime}}_{B^{\prime}+\mathbf{H}}=A^{\prime}+\mathbf{H}$, and the
previously established identities and containments imply $A\subseteq
A^{\prime}+\mathbf{H}$. The containment $B\subseteq B^{\prime}+\mathbf{H}$
follows by symmetry.

Now $A\subseteq C$ and $B\subseteq D$, while $\mu_{*}(C+D)=\mu_{*}(A+B)$. The
identity $C+D=C+D+\mathbf{H}$ and containment $A+B\subseteq C+D$ then imply
$A+B\sim_{\mu}A+B+\mathbf{H}$. Thus $A,B$ satisfy conclusion (I) of
Proposition 10.2.

Case 2: $A^{\prime}+\mathbf{H}\sim_{\mu}A^{\prime}$ and
$B^{\prime}+\mathbf{H}\sim_{\mu}B^{\prime}$. In this case Lemma 9.1 implies
$A^{\prime}+B^{\prime}=A^{\prime}+B^{\prime}+\mathbf{H}$, so $\mathbf{H}$ is
the stabilizer of $A^{\prime}+B^{\prime}$. Then
$\bar{A^{\prime}}_{B^{\prime},\mu}=\bar{A^{\prime}}_{B^{\prime}}=\bar{A^{\prime}}_{B^{\prime}+\mathbf{H}}$,
so that $A\subseteq\bar{A^{\prime}}_{B^{\prime}+\mathbf{H}}$. Lemma 10.11
implies $\bar{A^{\prime}}_{B^{\prime}+\mathbf{H}}$ is either
$A^{\prime}+\mathbf{H}$, or
$(A^{\prime}+\mathbf{H})\cup(\mathbf{g}+\mathbf{H})$ for some
$\mathbf{g}\notin A^{\prime}+\mathbf{H}$. We therefore have the following
disjunction:

$A\subseteq A^{\prime}+\mathbf{H},$ (10.5)

or

$A\subseteq(A^{\prime}+\mathbf{H})\cup(\mathbf{g}+\mathbf{H})\text{ for some
}\mathbf{g}\notin A^{\prime}+\mathbf{H}.$ (10.6)

If one of the containments $A\subseteq A^{\prime}+\mathbf{H}$, $B\subseteq
B^{\prime}+\mathbf{H}$, or $A+B\subseteq A^{\prime}+B^{\prime}+\mathbf{H}$
holds, then we are done: the similarities $A\sim_{\mu}A^{\prime}$ and
$B\sim_{\mu}B^{\prime}$ would imply $A+B\sim_{\mu}A+B+\mathbf{H}$, so that $A$
and $B$ satisfy Conclusion (I) of Proposition 10.2. We therefore assume these
containments fail, so that (10.6) holds, and likewise
$B\subseteq(B^{\prime}+\mathbf{H})\cup(\mathbf{g}+\mathbf{H})$ for some
$\mathbf{g}\notin B^{\prime}+\mathbf{H}$. We aim to produce quasi-periodic
decompositions $A=A_{1}\cup A_{0}$, $B=B_{1}\cup B_{0}$ satisfying conclusion
(II) of Proposition 10.2. Based on the containment (10.6), we set
$A_{1}:=A\cap(A^{\prime}+\mathbf{H})$ and $A_{0}:=A\setminus A_{1}$. Then
$A_{1}\cup A_{0}$ is a quasi-periodic decomposition of $A$. By symmetry we
produce a quasi-periodic decomposition of $B$ as $B_{1}\cup B_{0}$, where
$B_{0}=B\setminus(B^{\prime}+\mathbf{H})$. Since we are assuming
$A+B\nsubseteq A^{\prime}+B+\mathbf{H}$, we have that
$A_{0}+B_{0}\subseteq(A+B)\setminus(A^{\prime}+B^{\prime})$. The hypothesis
$\mu_{*}(A+B)=\mu_{*}(A^{\prime}+B^{\prime})$ then implies
$\mu_{*}(A_{0}+B_{0})=0$, whence $\mu(A_{0})=\mu(B_{0})=0$. Thus $A=A_{1}\cup
A_{0}$ and $B=B_{1}\cup B_{0}$ are decompositions of $A$ and $B$ satisfying
conclusion (II) of Proposition 10.2. ∎

The next lemma forms the majority of our proof of Proposition 10.2. It says
that if $\mu_{*}(A+B)=\mu(A)+\mu(B)$,
$\mu_{*}(A^{\prime}+B^{\prime})=\mu(A^{\prime})+\mu(B^{\prime})$,
$A^{\prime}\sim_{\mu}A$, $B^{\prime}\sim_{\mu}B$, and the pair
$A^{\prime},B^{\prime}$ satisfies the conclusion of Proposition 10.2, then so
does the pair $A,B$.

######  Lemma 10.13.

Let $A,B\subseteq\mathbf{G}$ have $\mu(A)$, $\mu(B)>0$ and

$\mu_{*}(A+B)=\mu(A)+\mu(B).$

If $A^{\prime},B^{\prime}\subseteq\mathbf{G}$ is a tame subcritical pair such
that $A\sim_{\mu}A^{\prime}$ and $B\sim_{\mu}B^{\prime}$, then $A,B$ is also a
tame subcritical pair.

Note: in Lemma 10.13 we do not assume $A^{\prime}+B^{\prime}\sim_{\mu}A+B$.

###### Proof.

By Lemmas 10.8 and 10.10, we may assume $A,B$ is irreducible and
nonextendible.

Since $A^{\prime},B^{\prime}$ is a tame subcritical pair, there is a
$\mu$-measurable finite index subgroup $\mathbf{K}^{\prime}\leq\mathbf{G}$
such that $A^{\prime},B^{\prime},\mathbf{K}^{\prime}$ satisfies one of the
conclusions of Proposition 10.2. Our goal is to prove that there is a
$\mu$-measurable finite index subgroup $\mathbf{K}$ (not necessarily equal to
$\mathbf{K}^{\prime}$) such that the triple $A,B,\mathbf{K}$ satisfies one of
the conclusions of Proposition 10.2. In the terminology of Definition 10.5, we
want to show that $A,B,\mathbf{K}$ has type (I), type (II), or type (III).

The additional hypothesis that $A,B$ is irreducible implies

$\mu_{*}((A^{\prime}+B^{\prime})\triangle(A+B))=0$ (10.7)

since otherwise setting $A^{\prime\prime}=A\cap A^{\prime}$ and
$B^{\prime\prime}=B\cap B^{\prime}$ would yield a pair with
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})<\mu(A)+\mu(B)$ and
$A\sim_{\mu}A^{\prime\prime}$, $B\sim_{\mu}B^{\prime\prime}$.

We consider three cases.

Case 1: $A^{\prime}\subseteq A,B^{\prime}\subseteq B$. We consider several
subcases based on the type of $A^{\prime},B^{\prime},\mathbf{K}^{\prime}$.

Subcase 1.1: $A^{\prime},B^{\prime},\mathbf{K}^{\prime}$ has type (I). Here
$A^{\prime}+B^{\prime}\sim_{\mu}A^{\prime}+B^{\prime}+\mathbf{K}^{\prime}$.
Then the stabilizer $\mathbf{H}$ of
$A^{\prime}+B^{\prime}+\mathbf{K}^{\prime}$ contains $\mathbf{K}^{\prime}$,
which is an internal $\mu$-measurable finite index subgroup of $\mathbf{G}$.
It follows that $\mathbf{H}$ is also such a subgroup, and
$A^{\prime}+B^{\prime}\sim_{\mu}A^{\prime}+B^{\prime}+\mathbf{H}$. We will
derive one of two possibilities: either $A+B\sim_{\mu}A+B+\mathbf{H}$, or $A$
and $B$ have quasi-periodic decompositions with respect to $\mathbf{H}$ such
that $A,B$, and $\mathbf{H}$ satisfy conclusion (II) of Proposition 10.2. Note
that $A^{\prime},B^{\prime}$ and $\mathbf{H}$ satisfy all of the hypotheses of
Lemma 10.12, except possibly the assumption that $A^{\prime}$ and $B^{\prime}$
are $\mathbf{H}$-solid. Let $A^{\prime\prime}\subseteq A^{\prime}$ and
$B^{\prime\prime}\subseteq B^{\prime}$ be $\mathbf{H}$-solid sets with
$\mu(A^{\prime\prime})=\mu(A^{\prime})$ and
$\mu(B^{\prime\prime})=\mu(B^{\prime})$. We will show that
$A^{\prime\prime},B^{\prime\prime}$ and $\mathbf{H}$ do satisfy the hypothesis
of Lemma 10.12. To see this, note that the irreducibility of $A,B$ and the
containments $A^{\prime\prime}\subseteq A$, $B^{\prime\prime}\subseteq B$
imply $\mu_{*}(A^{\prime\prime}+B^{\prime\prime})=\mu_{*}(A+B)$, and therefore
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})=\mu_{*}(A^{\prime}+B^{\prime})$.
It follows that
$A^{\prime\prime}+B^{\prime\prime}\sim_{\mu}A^{\prime\prime}+B^{\prime\prime}+\mathbf{H}$,
and we have verified the hypotheses of Lemma 10.12 with $A^{\prime\prime}$ and
$B^{\prime\prime}$ in place of $A^{\prime}$ and $B^{\prime}$. We may now
conclude that $A,B,\mathbf{H}$ has type (I) or type (II).

Subcase 1.2: $A^{\prime},B^{\prime},\mathbf{K}^{\prime}$ has type (II). In
this case
$A^{\prime}+B^{\prime}\nsim_{\mu}A^{\prime}+B^{\prime}+\mathbf{K}^{\prime}$,
and there are quasi-periodic decompositions $A^{\prime}=A_{1}^{\prime}\cup
A_{0}^{\prime}$, $B^{\prime}=B_{1}^{\prime}\cup B_{0}^{\prime}$ with respect
to $\mathbf{K}^{\prime}$ such that
$\mu_{*}(A_{0}^{\prime}+B_{0}^{\prime})=\mu(A_{0}^{\prime})+\mu(B_{0}^{\prime})$,
and at least one of $A_{1}^{\prime},B_{1}^{\prime}$ is nonempty. Without loss
of generality we assume $A_{1}^{\prime}\neq\varnothing$. We consider two
further subcases.

Subcase 1.2.1: $\mu(A_{0}^{\prime})>0$ and $\mu(B_{0}^{\prime})>0$. Here we
will show that $A,B,\mathbf{K}^{\prime}$ has type (II). We may assume
$A^{\prime},B^{\prime}$ is nonextendible, as otherwise Lemma 10.8 implies
$A^{\prime}+B^{\prime}\sim_{\mu}A^{\prime}+B^{\prime}+\mathbf{H}$ for some
$\mu$-measurable finite index subgroup $\mathbf{H}$ and we could argue as in
Subcase 1.1. Define $A_{0}:=A\cap(A_{0}^{\prime}+\mathbf{K}^{\prime})$ and
$A_{1}:=A\setminus A_{0}$. We will show that
$A_{1}\sim_{\mu}A_{1}+\mathbf{K}^{\prime}$. To do so, we will show that for
all $a\in A_{1}$, $a+\mathbf{K}^{\prime}\subset_{\mu}A_{1}$. Fix $a\in A_{1}$.
First observe that irreducibility of $A,B$ implies
$a+B^{\prime}\subset_{\mu}A^{\prime}+B^{\prime}$.

######  Claim.

With the preceding choice of $a$, we have
$a+B^{\prime}\subset_{\mu}(A_{1}^{\prime}+B^{\prime})\cup(A^{\prime}+B_{1}^{\prime})$.

To prove the Claim, assume otherwise. Then
$(a+B^{\prime})\cap(A_{0}^{\prime}+B_{0}^{\prime}+\mathbf{K}^{\prime})\neq\varnothing$,
so there is a $b\in B^{\prime}$ such that $a+b\in
A_{0}+B_{0}+\mathbf{K}^{\prime}$. Since $a\notin A_{0}+\mathbf{K}^{\prime}$,
$b\notin B_{0}^{\prime}$, so $b\in B_{1}^{\prime}$. Since
$B_{1}^{\prime}\sim_{\mu}B_{1}^{\prime}+\mathbf{K}^{\prime}$, we then have
$A_{0}^{\prime}+B_{0}^{\prime}+\mathbf{K}^{\prime}\subset_{\mu}a+B^{\prime}$,
and the observation preceding the Claim then implies
$A_{0}^{\prime}+B_{0}^{\prime}+\mathbf{K}^{\prime}\subset_{\mu}A^{\prime}+B^{\prime}$.
But then
$A^{\prime}+B^{\prime}\sim_{\mu}A^{\prime}+B^{\prime}+\mathbf{K}^{\prime}$,
contradicting our present hypothesis on $A^{\prime},B^{\prime}$ and
establishing the Claim.

The similarities $A_{1}^{\prime}\sim_{\mu}A_{1}^{\prime}+\mathbf{K}^{\prime}$
and $B_{1}^{\prime}\sim_{\mu}B_{1}+\mathbf{K}^{\prime}$, the
$\mathbf{K}^{\prime}$-solidity of $B^{\prime}$, and the Claim yield the
essential containment
$a+B^{\prime}+\mathbf{K}^{\prime}\subset_{\mu}A^{\prime}+B^{\prime}$. Then
$a+\mathbf{K}^{\prime}\subset_{\mu}A^{\prime}$ by the nonextendibility of
$A^{\prime},B^{\prime}$: otherwise setting
$C:=A^{\prime}\cup(a+\mathbf{K}^{\prime})$ we would have
$\mu(C)>\mu(A^{\prime})$ and
$\mu_{*}(C+B^{\prime})=\mu_{*}(A^{\prime}+B^{\prime})$.

We have shown that for all $a\in A_{1}$,
$a+\mathbf{K}^{\prime}\subset_{\mu}A^{\prime}$. The similarity
$A\sim_{\mu}A^{\prime}$ then implies
$A_{1}\sim_{\mu}A_{1}+\mathbf{K}^{\prime}$. Likewise we obtain
$B_{1}\sim_{\mu}B_{1}+\mathbf{K}^{\prime}$.

We have shown that $A_{1}\cup A_{0}$ and $B_{1}\cup B_{0}$ are quasi-periodic
decompositions of $A$ and $B$ with respect to $\mathbf{K}^{\prime}$. To show
$\mu_{*}(A_{0}+B_{0})=\mu(A_{0})+\mu(B_{0})$, observe that
$A_{0}^{\prime}+B_{0}^{\prime}$ is disjoint from
$(A_{1}^{\prime}+B^{\prime})\cup(A^{\prime}+B_{1}^{\prime})$; otherwise we
would have
$A^{\prime}+B^{\prime}\sim_{\mu}A^{\prime}+B^{\prime}+\mathbf{K}^{\prime}$.
Irreducibility of $A,B$ implies $\mu_{*}(A+B)=\mu_{*}(A^{\prime}+B^{\prime})$,
so the desired equation follows from the similarities
$A_{0}\sim_{\mu}A_{0}^{\prime}$, $B_{0}\sim_{\mu}B_{0}^{\prime}$ and the
assumption that
$\mu_{*}(A_{0}^{\prime}+B_{0}^{\prime})=\mu(A_{0}^{\prime})+\mu(B_{0}^{\prime})$.

Subcase 1.2.2: $\mu(A_{0}^{\prime})=0$ or $\mu(B_{0}^{\prime})=0$. Assume,
without loss of generality, that $\mu(A_{0}^{\prime})=0$. In this case let
$A^{\prime\prime}:=A^{\prime}\setminus A_{0}^{\prime}$. Then
$A^{\prime\prime}\sim_{\mu}A^{\prime\prime}+\mathbf{K}^{\prime}$, so
$A^{\prime\prime}+B^{\prime}\sim_{\mu}A^{\prime\prime}+B^{\prime}+\mathbf{K}^{\prime}$.
Irreducibility of $A,B$ implies
$\mu_{*}(A^{\prime\prime}+B^{\prime})=\mu_{*}(A+B)=\mu(A^{\prime\prime})+\mu(B^{\prime})$,
so Subcase 1.1 applies with the pair $A^{\prime\prime},B^{\prime}$ in place of
the pair $A^{\prime},B^{\prime}$. We conclude that $A,B$ has type (I) or type
(II).

Subcase 1.3: $A^{\prime},B^{\prime},\mathbf{K}^{\prime}$ has type (III). By
translating $A^{\prime}$ and $B^{\prime}$, we may assume $A^{\prime}$ and
$B^{\prime}$ are contained in parallel Bohr intervals
$\tilde{I},\tilde{J}\subseteq\mathbf{K}^{\prime}$, respectively, with
$\mu(\tilde{I})=\mu(A^{\prime})$, $\mu(\tilde{J})=\mu(B^{\prime})$. Thus
$A\sim_{\mu}\tilde{I}$ and $B\sim_{\mu}\tilde{J}$, whence Corollary 6.5
implies $A,B$ has type (III).

Case 2: $A\subseteq A^{\prime},B\subseteq B^{\prime}$. In this case, if
$A^{\prime},B^{\prime},\mathbf{K}^{\prime}$ has type (I) or type (III), then
we immediately conclude that $A,B$ also has type (I) or type (III). If
$A^{\prime},B^{\prime},\mathbf{K}^{\prime}$ has type (II), the containments
$A+B\subseteq A^{\prime}+B^{\prime}\subseteq
A^{\prime}+B^{\prime}+\mathbf{K}^{\prime}$ and the similarities
$A\sim_{\mu}A^{\prime}$, $B\sim_{\mu}B^{\prime}$ easily imply that $A,B$ has
type (I) or type (II).

Case 3: the general case. Let $A^{\prime\prime}:=A\cap A^{\prime}$,
$B^{\prime\prime}:=B\cap B^{\prime}$, so that $A\sim_{\mu}A^{\prime\prime}$,
$B\sim_{\mu}B^{\prime\prime}$. If
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})<\mu(A^{\prime\prime})+\mu(B^{\prime\prime})$,
then $A,B$ is reducible, contrary to our assumptions. If
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})=\mu(A^{\prime\prime})+\mu(B^{\prime\prime})$,
then we may apply the result from Case 2 with
$A^{\prime\prime},B^{\prime\prime}$ in place of $A,B$ to conclude that
$A^{\prime\prime},B^{\prime\prime}$ is a tame pair. Then Case 1 with
$A^{\prime\prime},B^{\prime\prime}$ in place of $A^{\prime},B^{\prime}$
implies $A,B$ is a tame pair. ∎

###### Proof of Proposition 10.2.

With the terminology introduced in Definition 10.5, Proposition 10.2 can be
stated as “If $A,B\subseteq\mathbf{G}$ satisfy $\mu_{*}(A+B)=\mu(A)+\mu(B)$,
then $A,B$ is a tame pair”.

Assume $A,B\subseteq\mathbf{G}$ satisfy $\mu(A),\mu(B)>0$ and
$\mu_{*}(A+B)=\mu(A)+\mu(B)$. By Lemmas 10.8 and 10.10, we may assume that
$A,B$ is nonextendible and irreducible. Apply Corollary 5.5 to find a compact
metrizable quotient $G$ of $\mathbf{G}$ with $\mu$-measure preserving quotient
map $\pi:\mathbf{G}\to G$ and Borel sets $C,D\subseteq G$ such that $C+D$ is
Borel while $A^{\prime}:=\pi^{-1}(C)$, $B^{\prime}:=\pi^{-1}(D)$ satisfy
$A\subset_{\mu}A^{\prime}$, $B\subset_{\mu}B^{\prime}$, and
$A^{\prime}+B^{\prime}\subset_{\mu}A+B$. Let $A^{\prime\prime}=A\cap
A^{\prime}$ and $B^{\prime\prime}=B\cap B^{\prime}$, so that
$A^{\prime\prime}\sim_{\mu}A$ and $B^{\prime\prime}\sim_{\mu}B$. The
irreducibility of $A,B$ then implies
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})=\mu_{*}(A+B)$, and hence
$\mu_{*}(A^{\prime\prime}+B^{\prime\prime})=\mu(A^{\prime\prime})+\mu(B^{\prime\prime})$.
We now consider two cases.

Case 1: $A^{\prime\prime}\nsim_{\mu}A^{\prime}$ or
$B^{\prime\prime}\nsim_{\mu}B^{\prime}$. In this case
$A^{\prime\prime},B^{\prime\prime}$ is extendible, as
$A^{\prime\prime}\subseteq A^{\prime}$, $B^{\prime\prime}\subseteq
B^{\prime}$, and
$\mu(A^{\prime}+B^{\prime})\leq\mu_{*}(A+B)=\mu_{*}(A^{\prime\prime}+B^{\prime\prime})$.
Then Lemma 10.8 implies $A^{\prime\prime},B^{\prime\prime}$ is a tame pair.
Lemma 10.13 and the similarities $A^{\prime\prime}\sim_{\mu}A$,
$B^{\prime\prime}\sim_{\mu}B$ then imply $A,B$ is tame.

Case 2: $A^{\prime\prime}\sim_{\mu}A^{\prime}$ and
$B^{\prime\prime}\sim_{\mu}B^{\prime}$. In this case we also have
$A\sim_{\mu}A^{\prime}$ and $B\sim_{\mu}B^{\prime}$. Now the containments
$A^{\prime\prime}+B^{\prime\prime}\subseteq
A^{\prime}+B^{\prime}\subset_{\mu}A+B$ and irreducibility of $A,B$ imply
$\mu(A^{\prime}+B^{\prime})=\mu_{*}(A+B)=\mu(A)+\mu(B)$, and therefore
$\mu(A^{\prime}+B^{\prime})=\mu(A^{\prime})+\mu(B^{\prime}).$ Then
$m(C+D)=m(C)+m(D)$, and Theorem 3.5 implies $C,D$ is a tame pair. Lemma 10.6
now implies $A^{\prime},B^{\prime}$ is a tame pair, so we may apply Lemma
10.13 to conclude that $A,B$ is a tame pair. ∎

The equation $\mu_{*}(A_{0}+B_{0})=\mu(A_{0})+\mu(B_{0})$ in conclusion (II)
of Proposition 10.2 allows one to apply Proposition 10.2 to $A_{0},B_{0}$ and
obtain Corollary 10.15. The next lemma is required to perform the iteration.
We use the term “type” as in Definition 10.5.

######  Lemma 10.14.

Suppose $\mu_{*}(A+B)=\mu(A)+\mu(B)$ and conclusion (II) holds in Proposition
10.2 with some subgroup $\mathbf{K}\leq\mathbf{G}$ and quasi-periodic
decompositions $A=A_{1}\cup A_{0}$, $B=B_{1}\cup B_{0}$. Suppose further that
$A_{0}$ and $B_{0}$ are $\mu$-nonnull, and that $A_{0},B_{0}$ has type (II)
with respect to a subgroup $\mathbf{K}^{\prime}\leq\mathbf{G}$, with quasi-
periodic decompositions $A_{0}=A_{1}^{\prime}\cup A_{0}^{\prime}$,
$B_{0}=B_{1}^{\prime}\cup B_{0}^{\prime}$. Then $A,B,\mathbf{K}^{\prime}$ has
type (II), with quasi-periodic decompositions $A_{1}^{\prime\prime}\cup
A_{0}^{\prime\prime}$, where $A_{0}^{\prime\prime}=A_{0}^{\prime}$, and
$A_{1}^{\prime\prime}:=A\setminus A_{0}^{\prime}$, and similarly for $B$.

###### Proof.

Since $A_{0}$ and $B_{0}$, and therefore $A_{0}^{\prime}$ and
$B_{0}^{\prime}$, are each contained in cosets of $\mathbf{K}$, and
$A_{1}^{\prime}\sim_{\mu}A_{1}^{\prime}+\mathbf{K}^{\prime}$,
$B_{1}^{\prime}\sim_{\mu}B_{1}^{\prime}+\mathbf{K}^{\prime}$, we have
$\mathbf{K}^{\prime}\subseteq\mathbf{K}$. It follows that
$A_{1}\sim_{\mu}A_{1}+\mathbf{K}^{\prime}$,
$B_{1}\sim_{\mu}B_{1}+\mathbf{K}^{\prime}$. Consequently,
$A_{1}^{\prime\prime}$ and $B_{1}^{\prime\prime}$, as defined in the statement
of the lemma, satisfy the same similarities. To see that
$A+B\nsim_{\mu}A+B+\mathbf{K}^{\prime}$, note that our hypothesis implies
$A_{0}+B_{0}\nsim_{\mu}A_{0}+B_{0}+\mathbf{K}^{\prime}$. Remark 10.3 implies
$A+B$ can be written as $C\cup(A_{0}+B_{0})$, where $C=(A_{1}+B)\cup(A+B_{1})$
is disjoint from $A_{0}+B_{0}+\mathbf{K}$, so the last non-similarity implies
$A+B\nsim_{\mu}A+B+\mathbf{K}^{\prime}$. ∎

We now iterate Proposition 10.2 and obtain a more detailed conclusion with the
same hypothesis.

######  Corollary 10.15.

If $A,B\subseteq\mathbf{G}$ are $\mu$-measurable sets such that $\mu(A)>0$,
$\mu(B)>0$, and $\mu_{*}(A+B)=\mu(A)+\mu(B)$, then at least one of the
following is true.

  1. (a)

$A+B\sim_{\mu}A+B+\mathbf{K}$ for some $\mu$-measurable finite index internal
subgroup $\mathbf{K}\leq\mathbf{G}$.

  2. (b)

$A,B$ has type (III).

  3. (c)

For all $\eta>0$, there is a $\mu$-measurable finite index internal subgroup
$\mathbf{K}\leq\mathbf{G}$ with $\mu(\mathbf{K})<\eta$ such that
$A,B,\mathbf{K}$ has type (II).

  4. (d)

$A,B$ has type (II) with quasi-periodic decompositions $A=A_{1}\cup A_{0}$,
$B=B_{1}\cup B_{0}$ where at least one of $\mu(A_{0})=0$ or $\mu(B_{0})=0$.

  5. (e)

$A,B$ has type (II) with quasi-periodic decompositions $A=A_{1}\cup A_{0}$,
$B=B_{1}\cup B_{0}$ and $A_{0},B_{0}$ has type (III) with $\mu(A_{0})>0$ and
$\mu(B_{0})>0$.

###### Proof.

Assuming $A$ and $B$ are as in the hypothesis, we apply Proposition 10.2 and
consider the possible conclusions. In conclusions (I) or (III) of the
proposition, we have conclusion (a) or (b), respectively, of the present
corollary. We therefore assume $A,B$ satisfies (II) in Proposition 10.2. Let
$\eta_{0}$ be the infimum of all the numbers $\mu(\mathbf{K})$ such that $A,B$
and $\mathbf{K}$ satisfy (II) in Proposition 10.2. If $\eta_{0}=0$, then
conclusion (c) holds here. If $\eta_{0}>0$ we choose a subgroup
$\mathbf{K}\leq\mathbf{G}$ with $\mu(\mathbf{K})<2\eta_{0}$ such that
$A,B,\mathbf{K}$ has type (II) and fix the corresponding quasi-periodic
decompositions $A=A_{1}\cup A_{0}$ and $B=B_{1}\cup B_{0}$, where
$\mu_{*}(A_{0}+B_{0})=\mu(A_{0})+\mu(B_{0})$. If one of $A_{0}$ or $B_{0}$ has
measure $0$, then $A,B$ satisfies conclusion (d) here. Otherwise, we apply
Proposition 10.2 to $A_{0},B_{0}$. If this pair satisfies conclusion (I) or
(III) in the proposition, then $A,B$ satisfies conclusion (a) or (e),
respectively, in the present corollary. If $A_{0},B_{0}$ has type (II), then
we will derive a contradiction: in this case Lemma 10.14 implies
$A,B,\mathbf{K}^{\prime}$ has type (II), with $\mathbf{K}^{\prime}$ a proper
subgroup of $\mathbf{K}$. Thus
$\mu(\mathbf{K}^{\prime})\leq\frac{1}{2}\mu(\mathbf{K})<\eta_{0}$,
contradicting the definition of $\eta_{0}$. ∎

##  11 Proof of Theorem 1.15

Suppose, to get a contradiction, that Theorem 1.15 fails for a given
$\varepsilon>0$. Then for all $n\in\mathbb{N}$, there exists a compact abelian
group $G_{n}$ with Haar probability measure $m_{n}$, inner Haar measure
$m_{n*}$, and sets $A_{n},B_{n}\subseteq G_{n}$ such that
$m_{n}(A_{n}),m_{n}(B_{n})>\varepsilon$,

$m_{n*}(A_{n}+B_{n})\leq m_{n}(A_{n})+m_{n}(B_{n})+\tfrac{1}{n},$ (11.1)

while the pair $A_{n},B_{n}$ satisfies none of the conclusions (I)-(III) of
Theorem 1.15 with a subgroup $K=K_{n}\leq G_{n}$ having index at most $n$. We
will derive a contradiction by showing that for some $n$ (in fact for
infinitely many $n$) the pair $A_{n},B_{n}$ satisfies the conclusion of
Theorem 1.15 with a subgroup $K=K_{n}$ having index less than $n$.

Let $\mathcal{U}$ be a nonprincipal ultrafilter on $\mathbb{N}$. Form the
ultraproduct $\mathbf{G}:=\prod_{n\to\mathcal{U}}G_{n}$ and the corresponding
Loeb measure $\mu$ and consider the internal sets
$\mathbf{A}:=\prod_{n\to\mathcal{U}}A_{n}$,
$\mathbf{B}:=\prod_{n\to\mathcal{U}}B_{n}$ with
$\mu(\mathbf{A})=\lim_{n\to\mathcal{U}}m_{n}(A_{n})\geq\varepsilon$,
$\mu(\mathbf{B})=\lim_{n\to\mathcal{U}}m_{n}(B_{n})\geq\varepsilon$, and
$\mu_{*}(\mathbf{A}+\mathbf{B})=\lim_{n\to\mathcal{U}}m_{n*}(A_{n}+B_{n})$ by
Lemma A.6. Together with Inequality (11.1) this implies