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# Rough approximate subgroups

Arturo Rodríguez Fanlo, Frank O Wagner

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Arturo Rodríguez Fanlo, Frank O Wagner. Rough approximate subgroups. 2023. [https://hal.science/hal-03952372](https://hal.science/hal-03952372)

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Preprint submitted on 23 Jan 2023

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# Rough approximate subgroups

Arturo Rodríguez Fanlo, Frank O Wagner

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Arturo Rodríguez Fanlo, Frank O Wagner. Rough approximate subgroups. Notre Dame Journal of Formal Logic, In press. hal-03952372v2

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Submitted on 8 Dec 2024

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entific research documents, whether they are pub-
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abroad, or from public or private research centers.

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destinée au dépôt et à la diffusion de documents
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Distributed under a Creative Commons Attribution 4.0 International License
# ROUGH APPROXIMATE SUBGROUPS

ARTURO RODRÍGUEZ FANLO AND FRANK O. WAGNER

**ABSTRACT.** Given a rough definably amenable rough approximate subgroup $A$ of a group in some first-order structure, there is a type-definable subgroup $H$ normalised by $A$ and contained in $A^4$ of bounded index in $\langle A \rangle$.

## INTRODUCTION

In [3, Theorem 4.2], Hrushovski found a fundamental connexion between approximate subgroups and Lie groups. This result, now known as the Lie Model Theorem, was the starting point for the complete classification of finite approximate subgroups by Breuillard, Green and Tao [1].

The proof of [3, Theorem 4.2] first shows that, for any definable approximate subgroup $A$, under a certain definable amenability condition, there is a type-definable normal subgroup $S$ of bounded index of $\langle A \rangle$ contained in $A^4$. For this first fundamental step, Hrushovski proves a general Stabilizer Theorem [3, Theorem 3.5], working with an $S_1$ ideal of definable sets which, in addition, is invariant under translations and automorphisms. Then, in [3, Theorem 4.2], this general Stabilizer Theorem is applied for the $S_1$ ideal of sets of measure zero. However, in order for this ideal to become invariant under automorphisms, one needs to work with an expansion of the initial language.

In a joint work with Massicot, the second named author provided in [5] an alternative proof of Hrushovski's Lie Model Theorem avoiding any expansion of the language. While Hrushovski's original proof only uses the ideal of measure zero subsets, the argument in [5] fundamentally exploits the additivity of the measure.

In [6], the first named author adapted [3, Theorem 3.5] and [3, Theorem 4.2] for hyperdefinable sets, providing in particular a generalisation of the Lie Model Theorem for rough approximate subgroups. This result was later used in [4] for the particular case of metric groups. The argument in [6] uses again a general version of

<footnote>Date: 23rd January 2023.</footnote>

<footnote n="2010">2010 Mathematics Subject Classification. 03C98, 11P70, 20A15.</footnote>

<footnote>Key words and phrases. rough approximate subgroup, rough definable amenability, Lie model.</footnote>

<footnote>First author supported by the Israel Academy of Sciences and Humanities & Council for Higher Education Excellence Fellowship Program for International Postdoctoral Researchers.</footnote>

<footnote>Second author supported by ANR-DFG AAPG2019 GeoMod.</footnote>
# ROUGH APPROXIMATE SUBGROUPS

ARTURO RODRÍGUEZ FANLO AND FRANK O. WAGNER

**ABSTRACT.** Given a rough definably amenable rough approximate subgroup $A$ of a group in some first-order structure there is a type-definable subgroup $N$ normalised by $A$ and contained in $A^4$ of bounded index in the subgroup $\langle A \rangle$ generated by $A$.

## INTRODUCTION

In [Hru12, Theorem 4.2], Hrushovski found a fundamental connexion between approximate subgroups and Lie groups. This result, now known as the Lie Model Theorem, was the starting point for the complete classification of finite approximate subgroups by Breuillard, Green and Tao [BGT12].

The proof of [Hru12, Theorem 4.2] first shows that, for any definable approximate subgroup $A$, under a certain definable amenability condition, there is a type-definable normal subgroup $S$ of bounded index of $\langle A \rangle$ contained in $A^4$. For this first fundamental step, Hrushovski proves a general Stabilizer Theorem [Hru12, Theorem 3.5], working with an $S_1$ ideal of definable sets which, in addition, is invariant under translations and automorphisms. Then, in [Hru12, Theorem 4.2], this general Stabilizer Theorem is applied for the $S_1$ ideal of sets of measure zero. However, in order for this ideal to become invariant under automorphisms, one needs to work with an expansion of the initial language.

Based on the work of Sanders and Croot-Sisask, an alternative combinatorial proof of the Lie Model Theorem, avoiding the Stabiliser Theorem, was presented in [BGT12]. This proof, however, still required working with an expansion of the initial language. In joint work with Massicot, the second named author provided in [MW15] an improvement of the combinatorial proof that avoids the need for an expansion of the language. While Hrushovski's original proof only uses the ideal of subsets of measure zero, the combinatorial arguments in [BGT12] and [MW15] fundamentally exploit the additivity of the measure.

<footnote>Date: 8th December 2024.</footnote>

<footnote n="2010">2010 Mathematics Subject Classification. 03C98, 11P70, 20A15.</footnote>

<footnote>Key words and phrases. rough approximate subgroup, rough definable amenability, Lie model.</footnote>

<footnote>The first author is supported by the Israel Academy of Sciences and Humanities & Council for Higher Education Excellence Fellowship Program for International Postdoctoral Researchers.</footnote>

<footnote>The second author is supported by ANR-DFG AAPG2019 GeoMod.</footnote>
the Stabilizer Theorem (now for hyperdefinable sets), and so the applications in [4]
require again to work with an expansion of the initial language.

The aim of this paper is to adapt the alternative technique developed in [5] for
the case studied in [4]. This is not a straightforward task — while the ideal used for
the original Lie Model Theorem is directly given by a measure, the ideal in [4] is not
associated with a measure at all. Instead, the ideal in [4] is built as the inf-limit of
the zero ideals of a sequence of subadditive set functions. Hence, in [4], we lack the
additivity property that is fundamentally used in [5]. Fortunately, a careful analysis
of the properties of these subadditive set functions shows that, although they are not
completely additive, they still keep some rough additivity that suffices to conclude
the proof.

Our main result is the following:

**Theorem 1.** Assume $A$ is $T$-rough definably amenable with respect to $(\mu_i)_{i \in \mathbb{N}}$ and suppose that for every $m \in \mathbb{N}$ there is $i_m \in \mathbb{N}$ such that $\mu_i(A^m) < \infty$ for all $i > i_m$. Then, there is a type-definable normal subgroup $N$ of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4T_\omega$.

As a corollary, we get the following improved version of [4, Theorem 20]:

**Corollary 2.** Let $(G_m, A_m, r_{i,m})_{i \le m \in \mathbb{N}}$ be a sequence such that, for some $\ell \in \mathbb{N}$ and $K = (K_i : i \in \mathbb{N})$ sequence of natural numbers,

(1) $G_m$ is a metric group,

(2) $A_m$ is an $(\ell, r_{0,m})$-Lipschitz symmetric subset,

(3) $(r_{i,m})_{i \le m}$ is a sequence of positive reals with $\max\{l, 2\} \cdot r_{i,m} \le r_{i-1,m}$ and

$$
N_{r_{i,m}}(A_m^4) \le K_i \cdot N_{r_{i,m}}(A_m) < \infty.
$$

Let $(G, A, r)$ be a non-principal ultraproduct in the group language with a predicate
for $A$ and predicates $T_i$ for each ball of radius $r_i$. Let $T_\omega = \bigcap_{i \in \mathbb{N}} T_i$. Then:

(1) There is a type-definable normal subgroup $N$ of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4T_\omega$.

(2) If $K$ is constant, $A^2$ is a $T_\omega$-rough $K$-approximate subgroup. In particular, $AT_\omega$ has a connected Lie model $\pi : H \to L$, i.e. a surjective group homomorphism to a connected Lie group $L$ defined on a subgroup $H \subseteq \langle AT_\omega \rangle$ with kernel $T_\omega \subseteq \ker(\pi) \subseteq A^8T_\omega$ such that

• $H \cap A^4T_\omega$ and $A^2T_\omega$ are commensurable,

• $H \cap A^8T_\omega$ generates $H$,

• $\pi(H \cap A^4)$ is a compact neighbourhood of the identity in $L$, and

• $\pi$ is continuous and proper from the logic topology using countably many parameters.
In [RF23], the first named author adapted [Hru12, Theorem 3.5] and [Hru12, Theorem 4.2] for hyperdefinable sets, providing in particular a generalisation of the Lie Model Theorem for rough approximate subgroups. This result was later used in [HRF22] for the particular case of metric groups. The argument in [RF23] uses again a general version of the Stabilizer Theorem (now for hyperdefinable sets), and so the applications in [HRF22] require again to work with an expansion of the initial language.

The aim of this paper is to adapt the alternative technique developed in [MW15] for the case studied in [HRF22]. This is not a straightforward task — while the ideal used for the original Lie Model Theorem is directly given by a measure, the ideal in [HRF22] is not associated with a measure at all. Instead, the ideal in [HRF22] is built as the inf-limit of the zero ideals of a sequence of subadditive set functions. Hence, in [HRF22], we lack the additivity property that is fundamentally used in [MW15]. Fortunately, a careful analysis of the properties of these subadditive set functions shows that, although they are not completely additive, they still keep some rough additivity that suffices to conclude the proof.

Our main result is the following (see Definitions 5, 13 and 19 for more details):

**Theorem 1.** Assume $A$ is $T$-rough definably amenable with respect to the rough measures $(\mu_i)_{i \in \mathbb{N}}$ and suppose that for every $m \in \mathbb{N}$ there is $i_m \in \mathbb{N}$ such that $\mu_i(A^m) < \infty$ for all $i > i_m$. Then, there is a type-definable normal subgroup $N$ of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4T_\omega$.

In particular, we obtain the following rigorous version of the statement in the abstract:

**Corollary 2.** *Assume $A$ is $T$-rough definably amenable and a $T_i$-rough approximate subgroup for each $i \in \mathbb{N}$ (possibly with different doubling constants) such that $\mu_i(\langle AT_\omega \rangle) < \infty$ for all $i$. Then there is a type-definable normal subgroup $N$ of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4T_\omega$.*

Theorem 1 provides an alternative to the Stabilizer Theorem [RF23, Theorem 3.1]. In particular, we can use Theorem 1 to replace [RF23, Theorem 3.1] in the proofs of [RF23, Theorem 3.2 and 3.3]. For example, we obtain the following theorem, which corresponds to [RF23, Theorem 3.2].

**Theorem 3 (Rough Lie Model).** *Assume $A$ is $T$-rough definably amenable and $A^n$ is a $T_\omega$-rough approximate subgroup. Then $\langle AT_\omega \rangle$ has a connected Lie model $\pi: H \to L$, i.e. a surjective group homomorphism to a connected Lie group $L$ defined on a subgroup $H \subseteq \langle AT_\omega \rangle$ with kernel $T_\omega \subseteq \ker(\pi) \subseteq A^{2n+4}T_\omega$ such that*

* $H \cap A^{2n}T_\omega$ is type-definable and commensurable to $A^nT_\omega$,

* $\pi(H \cap A^{2n})$ is a compact neighbourhood of the identity in $L$,
**Notations:** Through this paper, fix a many sorted first order language L, a κ-saturated L-structure M with κ ≥ |L|, a definable group G and a definable symmetric subset A of G. We always use product notation and, unless otherwise stated, we consider the group acting on itself on the left. A subset X of G is called symmetric if 1 ∈ X = X⁻¹. For subsets X and Y of G, we write XY := {xy : x ∈ X, y ∈ Y} for the set of pairwise products and call it the Y-thickening of X. For an integer n > 0, abbreviate Xⁿ := XXⁿ⁻¹ and Xⁿ⁻¹ := (Xⁿ⁻¹)ⁿ, and put X⁰ := {1}. Write Y⁸ := {x⁻¹yx : x ∈ X, y ∈ Y}, and say that X normalises Y if Y⁸ ⊆ Y.

# 1. ROUGH MEASURES

Through this section, let $T \subseteq G$ be a symmetric subset and $\mathcal{A}$ some subalgebra of the boolean algebra of definable subsets of $G$ containing $A$ and closed under $T$-thickenings.

**Definition 3.** A *T-rough measure* on $\mathcal{A}$ is an increasing subadditive set function $\mu : \mathcal{A} \to \mathbb{R}_{\ge 0} \cup \{\infty\}$ which is additive modulo $T$, i.e. for $X, Y \in \mathcal{A}$ with $X \cap YT = \emptyset$ we have $\mu(\bar{X} \cup Y) = \mu(X) + \mu(Y)$. For a type-definable set $X$ we put $\mu(X) := \inf\{\mu(\bar{X}) : \bar{X} \supseteq X \text{ definable}\}$. We say that $\mu$ is *normalised* at $A$ if $\mu(A) = 1$.

**Lemma 4.** Let $\mu$ be a T-rough measure on $\mathcal{A}$. Then, for any $S_0, \dots, S_{n-1} \in \mathcal{A}$, we have

$$ \mu(\bigcup_{i<n} S_i) \geq \sum_{i<n} \mu(S_i) - \sum_{i<j<n} \min\{\mu(S_i \cap S_j T), \mu(S_i T \cap S_j)\}. $$

*Proof.* For $n < 2$, there is nothing to prove. For $n = 2$, by subadditivity,

$$ \mu(S_0) \leq \mu(S_0 \setminus S_1 T) + \mu(S_0 \cap S_1 T). $$

Hence, by additivity modulo $T$, we get

$$ \begin{aligned} \mu(S_0 \cup S_1) &\geq \mu((S_0 \setminus S_1 T) \cup S_1) = \mu(S_0 \setminus S_1 T) + \mu(S_1) \\ &\geq \mu(S_0) + \mu(S_1) - \mu(S_0 \cap S_1 T). \end{aligned} $$

By symmetry, $\mu(S_0 \cup S_1) \geq \mu(S_0) + \mu(S_1) - \mu(S_0 T \cap S_1)$, and the result follows.

Suppose the assertion is true for $n$, and consider $S_0, \dots, S_n \in \mathcal{A}$. Put

$$ I = \{i < n : \mu(S_i \cap S_n T) \leq \mu(S_i T \cap S_n)\}, $$

and set $X = \bigcup_{i \in I} S_i$ and $X' = \bigcup_{i \in n \setminus I} S_i$. Suppose $x \in X' \cap (S_n \setminus X'T)T$; say $x = yt$ with $y \in S_n \setminus X'T$ and $t \in T$. Then, $y = xt^{-1} \in X'T$, a contradiction. Therefore,
* H is generated by $H \cap A^{2n+4}T_\omega$ and

* $\pi$ is continuous and proper from the B-logic topology (i.e. the topology whose closed sets are the B-type-definable sets) for some countable set of parameters B.

The proof of Theorem 3 is exactly the same as the one of [RF23, Theorem 3.2], but replacing the use of [RF23, Theorem 3.1] by Theorem 1 in the first line in the proof of [RF23, Proposition 3.20].

Finally, as a corollary, we get the following improvement of [HRF22, Theorem 20]:

**Corollary 4.** Let $(G_m, A_m, r_{i,m})_{i \le m \in \mathbb{N}}$ be a sequence such that, for some $\ell \in \mathbb{N}$ and $K = (K_i : i \in \mathbb{N})$ sequence of natural numbers,

(i) $G_m$ is a metric group (i.e. a group with a metric invariant under left translations),

(ii) $A_m$ is a symmetric subset and $(\ell, r_{0,m})$-Lipschitz (i.e. every right translation by an element of $A_m$ is $\ell$-Lipschitz in the ball of radius $r_{0,m}$ at the identity),

(iii) $(r_{i,m})_{i \le m}$ is a sequence of positive reals with $\max\{l, 2\} \cdot r_{i,m} \le r_{i-1,m}$ and

$$N_{r_{i,m}/2}(A_m^4) \le K_i \cdot N_{r_{i,m}}(A_m) < \infty;$$

where $N_r(X)$ denotes the maximum size of an $r$-separated subset of $X$.

Let $(G, A, r)$ be a non-principal ultraproduct in the group language with a predicate for $A$ and predicates $T_i$ for each ball of radius $r_i$. Let $T_\omega = \bigcap_{i \in \mathbb{N}} T_i$. Then:

(1) There is a type-definable normal subgroup $N$ of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4T_\omega$.

(2) If $K$ is constant, $A^2$ is a $T_\omega$-rough $K$-approximate subgroup. In particular, $AT_\omega$ has a connected Lie model $\pi: H \to L$, i.e. a surjective group homomorphism to a connected Lie group $L$ defined on a subgroup $H \subseteq \langle AT_\omega \rangle$ with kernel $T_\omega \subseteq \ker(\pi) \subseteq A^8T_\omega$ such that

* $H \cap A^4T_\omega$ and $A^2T_\omega$ are commensurable,

* $H \cap A^8T_\omega$ generates $H$,

* $\pi(H \cap A^4)$ is a compact neighbourhood of the identity in $L$, and

* $\pi$ is continuous and proper from the B-logic topology (i.e. the topology whose closed sets are the B-type-definable sets) for some countable set of parameters B.

Corollary 4 improves [HRF22, Theorem 20] in two ways.

First, condition (iii) is weaker (and more natural) than the corresponding condition in [HRF22, Theorem 20], which was

$$N_{r_{i,m}/9}(A_m^9) \le K_i \cdot N_{r_{i,m}}(A_m) < \infty.$$

Second, Corollary 4 works in the group language with a predicate for $A$ and predicates for $T_i$, while [HRF22, Theorem 20] requires a further expansion of this language.
$$
\begin{align*}
(X \cup X') \cap (S_n \setminus X'T)T \subseteq X \cap S_nT. \text{ Thus,} \\
\mu (\bigcup_{i \le n} S_i) &\ge \mu(X \cup X' \cup (S_n \setminus X'T)) \\
&\ge \mu(X \cup X') + \mu(S_n \setminus X'T) - \mu((X \cup X') \cap (S_n \setminus X'T)T) \\
&\ge \mu(X \cup X') + \mu(S_n) - \mu(S_n \cap X'T) - \mu(X \cap S_nT) \\
&\ge \sum_{i \le n} \mu(S_i) - \sum_{i<j<n} \min\{\mu(S_i \cap S_jT), \mu(S_iT \cap S_j)\} \\
&\quad - \sum_{i \in n \setminus I} \mu(S_iT \cap S_n) - \sum_{i \in I} \mu(S_i \cap S_nT) \\
&= \sum_{i \le n} \mu(S_i) - \sum_{i<j \le n} \min\{\mu(S_i \cap S_jT), \mu(S_iT \cap S_j)\}.
\end{align*}
$$

□

**Remark 5.** The above lemma also holds for type-definable sets, i.e. bounded intersections of definable sets in $\mathcal{A}$, by taking limits.

**Definition 6.** Let $X$ be a definable subset of $G$. A definable symmetric subset $Y$ of $G$ is $t$-thick in $X$ if, for any $t+1$ elements $g_0, \dots, g_t$ of $X$, there are $0 \le i \ne j \le t$ such that $g_i^{-1}g_j \in Y$. A type-definable symmetric subset of $G$ is thick in $X$ if every definable symmetric superset is $t$-thick in $X$ for some $t \in \mathbb{N}$.

Note that, since $Y$ is symmetric, $g_i^{-1}g_j \in Y$ if and only if $g_j^{-1}g_i \in Y$.

**Lemma 7.** If $Y$ is $t$-thick in $X$, then $t$ left translates of $Y$ cover $X$. Conversely, if $t$ left translates of $Y$ cover $X$, then $Y^{-1}Y$ is $t$-thick in $X$.

*Proof.* Take $Z \subseteq X$ maximal such that $z_1^{-1}z_2 \notin Y$ for $z_1, z_2 \in Z$ with $z_1 \ne z_2$. Since $Y$ is $t$-thick, $Z$ exists and $|Z| \le t$. Since $Z$ is maximal, for any $g \in X$, there is $z \in Z$ such that $z^{-1}g \in Y$; in other words, $X \subseteq ZY$.

Conversely, if $X \subseteq ZY$ with $|Z| \le t$, by the pigeonhole principle, for any $g_0, \dots, g_t$ in $X$, there are $0 \le i \ne j \le t$ and $z \in Z$ such that $g_i, g_j \in zY$, so $g_i^{-1}g_j \in Y^{-1}Y$. Since $g_0, \dots, g_t$ are arbitrary, we conclude that $Y^{-1}Y$ is $t$-thick □

**Lemma 8.** Let $X$ be a definable subset of $G$. An arbitrary intersection of type-definable subsets of $G$ which are thick in $X$ is again thick in $X$.

*Proof.* Let us see first that a finite intersection of definable thick subsets in $X$ is thick in $X$. If $Y \subseteq G$ is definable and $t$-thick in $X$ and $Y' \subseteq G$ is definable and $t'$-thick in $X$, then the intersection $Y \cap Y'$ is $R(2, t+1, t'+1)$-thick in $X$, where $R(\cdot, \cdot, \cdot)$ is the Ramsey number with respect to a 3-colouring. Indeed, $Y \cap Y'$ is obviously definable and symmetric. Now, given any $Z$ subset of $X$ of size $R(2, t+1, t'+1)$, colour the edges of the complete graph on $Z$ by
**Notations.** Through this paper, fix a many sorted first-order language L, a κ-saturated L-structure M with κ > |L| + ℵ₀, a definable group G and a definable symmetric subset A of G. We always use product notation and, unless otherwise stated, we consider the group acting on itself on the left. A subset X of G is called symmetric if 1 ∈ X = X⁻<fnref n="1" />. For subsets X and Y of G, we write XY := {xy : x ∈ X, y ∈ Y} for the set of pairwise products and call it the Y-thickening of X. For an integer n > 0, abbreviate Xⁿ := XXⁿ⁻<fnref n="1" /> and Xⁿ⁻<fnref n="1" /> := (X⁻<fnref n="1" />)ⁿ, and put X⁰ := {1}. Write Y⁸ := {x⁻¹yx : x ∈ X, y ∈ Y}, and say that X normalises Y if Y⁸ ⊆ Y.

# 1. ROUGH MEASURES

Through this section, let $T \subseteq G$ be a symmetric subset and $\mathcal{A}$ some subalgebra of the Boolean algebra of definable subsets of $G$ containing $A$ and closed under $T$-thickenings. A set in $\mathcal{A}$ will be called $\mathcal{A}$-definable, and an intersection of $\mathcal{A}$-definable sets $\mathcal{A}$-type-definable.

**Definition 5.** A *T-rough measure*<fnref n="1" /> on $\mathcal{A}$ is an increasing subadditive set function $\mu: \mathcal{A} \to \mathbb{R}_{\ge 0} \cup \{\infty\}$ which is additive modulo $T$, i.e. for $X, Y \in \mathcal{A}$ with $X \cap YT = \emptyset$ we have $\mu(\bar{X} \cup Y) = \mu(X) + \mu(Y)$. For an $\mathcal{A}$-type-definable set $X$ we put $\mu(X) := \inf\{\mu(\bar{X}) : X \subseteq \bar{X} \in \mathcal{A}\}$. We say that $\mu$ is *normalised* at some $A \in \mathcal{A}$ if $\mu(A) = 1$.

**Lemma 6.** Let $\mu$ be a T-rough measure on $\mathcal{A}$. Then, for any $S_0, \dots, S_{n-1} \in \mathcal{A}$, we have

$$ \mu(\bigcup_{i<n} S_i) \geq \sum_{i<n} \mu(S_i) - \sum_{i<j<n} \min\{\mu(S_i \cap S_j T), \mu(S_i T \cap S_j)\}. $$

*Proof.* For $n < 2$, there is nothing to prove. For $n = 2$, by subadditivity,

$$ \mu(S_0) \leq \mu(S_0 \setminus S_1 T) + \mu(S_0 \cap S_1 T). $$

Hence, by additivity modulo $T$, we get

$$ \begin{aligned} \mu(S_0 \cup S_1) &\geq \mu((S_0 \setminus S_1 T) \cup S_1) = \mu(S_0 \setminus S_1 T) + \mu(S_1) \\ &\geq \mu(S_0) + \mu(S_1) - \mu(S_0 \cap S_1 T). \end{aligned} $$

By symmetry, $\mu(S_0 \cup S_1) \geq \mu(S_0) + \mu(S_1) - \mu(S_0 T \cap S_1)$, and the result follows.

Suppose the assertion is true for $n$, and consider $S_0, \dots, S_n \in \mathcal{A}$. Put

$$ I = \{i < n : \mu(S_i \cap S_n T) \leq \mu(S_i T \cap S_n)\}, $$

and set $X = \bigcup_{i \in I} S_i$ and $X' = \bigcup_{i \in n \setminus I} S_i$. Suppose $x \in X' \cap (S_n \setminus X'T)T$; say $x = yt$ with $y \in S_n \setminus X'T$ and $t \in T$. Then, $y = xt^{-1} \in X'T$, a contradiction. Therefore,

<footnote n="1">¹This notion is similar to the concept of pre-mean considered in [HKP22].</footnote>
• {$z, z'$} is yellow if $z^{-1}z' \in Y \cap Y'$,

• {$z, z'$} is blue if $z^{-1}z' \notin Y$,

• {$z, z'$} is red if $z^{-1}z' \in Y \setminus Y'$.

Since $Y$ is $t$-thick, $Z$ cannot contain a complete blue subgraph of size $t+1$ and, as $Y'$ is $t'$-thick, $Z$ cannot contain a complete red subgraph of size $t'+1$ either. Thus, it must contain a yellow edge, i.e. there are $z, z' \in Z$ with $z \neq z'$ such that $z^{-1}z' \in Y \cap Y'$. As $Z$ is arbitrary, we conclude that $Y \cap Y'$ is $R(2, t+1, t'+1)$-thick in $X$. By induction, from the two case, we conclude that any finite intersection of definable thick subsets in $X$ is again thick in $X$.

Now, let $\bigcap_I Y_i$ be an intersection of type-definable subsets of $G$ thick in $X$ and $\bar{Y}$ be a definable symmetric superset of $\bigcap_I Y_i$. By compactness, there are a finite $I_0 \subseteq I$ and definable symmetric supersets $\bar{Y}_i$ of $Y_i$ for each $i \in I_0$ such that $\bigcap_{I_0} \bar{Y}_i \subseteq \bar{Y}$. As $\bigcap_{I_0} \bar{Y}_i$ is thick in $X$, so is $\bar{Y}$; as $\bar{Y}$ is arbitrary, $\bigcap_I Y_i$ is thick in $X$. $\square$

**Lemma 9.** Let $\mu$ be a $T$-rough measure on $\mathcal{A}$ normalised at $A$. Suppose that $A^{2m} \in \mathcal{A}$ and $\mu(A^{2m}) < \infty$. Let $t > 0$ be an integer and $B \in \mathcal{A}$ with $B \subseteq A^m$ and $\mu(B) \ge 2\mu(A^{2m})/t$. Put

$$S(B) = \left\{ g \in A^{2m} : \min\{\mu(B \cap gBT), \mu(gB \cap BT)\} \ge \frac{2\mu(A^{2m})}{t^2} \right\}.$$

Then, $S(B)$ is $t$-thick in $A^m$.

Note that $S(B)$ is symmetric, since $A$ is symmetric and $\mu$ is left invariant. Also, note that any $g \in G$ with $B \cap gBT$ or $gB \cap BT$ non-empty must be in $BTB^{-1}$, so, in particular, $S(B) \subseteq BTB^{-1}$.

*Proof.* Suppose, aiming for a contradiction, that the conclusion is false, and consider a counter-example $(g_i : i \le t)$. Then,

$$\min\{\mu(g_iB \cap g_jBT), \mu(g_iBT \cap g_jB)\} < \frac{2\mu(A^{2m})}{t^2},$$

for all $0 \le i < j \le t$. By Lemma 4,

$$\begin{align*}
\mu(A^{2m}) &\ge \mu\left(\bigcup_{i \le t} g_i B\right) \\
&\ge \sum_{i \le t} \mu(g_i B) - \sum_{0 \le i < j \le t} \min\{\mu(g_i B \cap g_j BT), \mu(g_i BT \cap g_j B)\} \\
&> (t+1) \cdot \frac{2\mu(A^{2m})}{t} - \frac{(t+1)t}{2} \cdot \frac{2\mu(A^{2m})}{t^2} \\
&= \mu(A^{2m}) \cdot \frac{t+1}{t} > \mu(A^{2m}),
\end{align*}$$

a contradiction. $\square$
$(X \cup X') \cap (S_n \setminus X'T)T \subseteq X \cap S_nT$. Thus,

$$
\begin{align*}
\mu(\bigcup_{i \le n} S_i) &\ge \mu(X \cup X' \cup (S_n \setminus X'T)) \\
&\ge \mu(X \cup X') + \mu(S_n \setminus X'T) - \mu((X \cup X') \cap (S_n \setminus X'T)T) \\
&\ge \mu(X \cup X') + \mu(S_n) - \mu(S_n \cap X'T) - \mu(X \cap S_nT) \\
&\ge \sum_{i \le n} \mu(S_i) - \sum_{i<j<n} \min\{\mu(S_i \cap S_jT), \mu(S_iT \cap S_j)\} \\
&\quad - \sum_{i \in n \setminus I} \mu(S_iT \cap S_n) - \sum_{i \in I} \mu(S_i \cap S_nT) \\
&= \sum_{i \le n} \mu(S_i) - \sum_{i<j \le n} \min\{\mu(S_i \cap S_jT), \mu(S_iT \cap S_j)\};
\end{align*}
$$

where the first inequality follows from monotonicity, the second from the case $n = 2$, the third from subadditivity and monotonicity, the fourth from induction hypothesis and subadditivity, and the last equality from the choice of $I$. $\square$

**Remark 7.** The above lemma also holds for type-definable sets, i.e. bounded intersections of definable sets in $\mathcal{A}$, by taking limits.

**Definition 8.** Let $X$ be a subset of $G$. A symmetric subset $Y$ of $G$ is $t$-thick$^2$ in $X$ if, for any $t+1$ elements $g_0, \dots, g_t$ of $X$, there are $0 \le i \ne j \le t$ such that $g_i^{-1}g_j \in Y$.

A type-definable symmetric subset of $G$ is thick in $X$ if every definable symmetric superset is $t$-thick in $X$ for some $t \in \mathbb{N}$.

**Remark 9.** Note that, since $Y$ is symmetric, $g_i^{-1}g_j \in Y$ if and only if $g_j^{-1}g_i \in Y$.

It is easy to see that a type-definable subgroup $H$ of a group $G$ is thick if and only if it has bounded index.

The following lemmas are adaptations to our relative notion of thickness of [Gis11, Lemma 3.2(4)] and [Gis12, Lemma 1.2(1)], respectively.

**Lemma 10.** If $Y$ is $t$-thick in $X$, then $t$ left translates of $Y$ cover $X$. Conversely, if $t$ left translates of $Y$ cover $X$, then $Y^{-1}Y$ is $t$-thick in $X$.

*Proof.* Take $Z \subseteq X$ maximal such that $z_1^{-1}z_2 \notin Y$ for $z_1, z_2 \in Z$ with $z_1 \ne z_2$. Since $Y$ is $t$-thick, $Z$ exists and $|Z| \le t$. Since $Z$ is maximal, for any $g \in X$, there is $z \in Z$ such that $z^{-1}g \in Y$; in other words, $X \subseteq ZY$.

<footnote>$^2$This notion of thickness is inspired by the classical notion of thick formulas [Zie02, CLPZ01]. The notion of thickness in the context of groups was first considered in [Gis11]. Its connection with approximate subgroups was noted in [HKP22]. The novelty of our definition here is that we consider thickness relative to some subset.</footnote>
## 2. ROUGH DEFINABLE AMENABILITY

Throughout this section, let $T = (T_i : i \in \mathbb{N})$ be a sequence of definable symmetric subsets of $G$ such that $T_{i+1}^2 \subseteq T_i$ and $T_{i+1}^A \subseteq T_i$ for all $i \in \mathbb{N}$. Write $T_\omega = \bigcap_{i \in \mathbb{N}} T_i$; note that $T_\omega$ is a type-definable subgroup of $G$ normalised by $A$.

**Definition 10.** We say that $A$ is *T-rough definably amenable* if for all $i \in \mathbb{N}$ there is a $T_i$-rough measure $\mu_i$ on the boolean algebra of definable subsets of $G$ invariant under left translation and normalised at $A$.

**Remark 11.** We do not really need $\mu_i$ to be defined on all definable subsets of $G$, but just on a boolean subalgebra $\mathcal{A}$ which contains $A$ and is closed under left translations, $T_i$-thickenings for $i \in \mathbb{N}$, and setwise products (i.e. $XY \in \mathcal{A}$ for any $X, Y \in \mathcal{A}$).

**Lemma 12.** Let $P$ be a partially ordered set and $f_0, \dots, f_{r-1} : P \to \mathbb{R}_{>0}$ be increasing functions. Suppose there is a $a > 0$ such that $f_k(x) \ge a$ for all $x \in P$ and each $k < r$. Then, for any $\varepsilon_0, \dots, \varepsilon_{r-1} > 0$, there is $x \in P$ such that $f_k(y) > (1 - \varepsilon_k) \cdot f_k(x)$ for any $y \in P$ with $y < x$ and each $k < r$.

*Proof.* Aiming for a contradiction, suppose otherwise. Then, we can recursively construct a decreasing sequence $(x_n)_{n \in \mathbb{N}}$ in $P$ such that for any $n \in \mathbb{N}$ there is $i < r$ with $f_i(x_{n+1}) \le (1 - \varepsilon_i) \cdot f_i(x_n)$. Since $r \in \mathbb{N}$ is finite, by the pigeonhole principle, there is $i < r$ and $N \subseteq \mathbb{N}$ infinite such that $f_i(x_{n+1}) \le (1 - \varepsilon_i) \cdot f_i(x_n)$ for any $n \in N$. Since $f_i$ is monotonous, we get that $f_i(x_m) \le (1 - \varepsilon_i) \cdot f_i(x_n)$ for any $n \in N$ and $m > n$. Let $n_0 := \min N$ and $b := f_i(x_{n_0})$. For any $n \in N$, put $\alpha := |\{m \in N : m < n\}|$. Then,

$$a \le f_i(x_n) \le (1 - \varepsilon_i)^\alpha \cdot f_i(x_{n_0}) = (1 - \varepsilon_i)^\alpha \cdot b,$$

getting a contradiction when $\alpha > \ln(a/b)/\ln(1 - \varepsilon_i)$. $\square$

<figure><img src="image_3.png" /></figure>

**Proposition 13.** Let $A \subseteq G$ be T-rough definably amenable, and $m$ and $r$ integers with $\mu_i(A^{2m(r+1)}) < \infty$ for each $i \in \mathbb{N}$. Suppose $B \subseteq A^m$ is type-definable and thick in $A$. Then, there is a type-definable subset $S \subseteq B^2T_\omega \cap A^{2m}$ thick in $A^m$ such that $S^r \subseteq B^4T_\omega$.

*Proof.* For all $i, t \in \mathbb{N}$, define recursively on $j \in \mathbb{N}$ a family $\mathcal{T}_{i,j}^t$ of definable subsets of $A^m$. For $j = 0$, let $X \in \mathcal{T}_{i,0}^t$ if $X \ne \emptyset$. Given $\mathcal{T}_{i,j}^t$, let $X \in \mathcal{T}_{i,j+1}^t$ if

$$S_{i,j}^t(X) := \{g \in A^{2m} : X \cap gXT_i, X \cap g^{-1}XT_i \in \mathcal{T}_{i,j}^{t^2}\}$$

is $t$-thick in $A^m$. Note that $S_{i,j}^t(X) \subseteq XT_iX^{-1}$ is symmetric.

By induction on $n$, we easily see that these families are definable, i.e. $\{b : X(b) \in \mathcal{T}_{i,j}^t\}$ is definable for any definable family $X(b)$ of subsets of $A^m$. Also, we easily see
Conversely, if $X \subseteq ZY$ with $|Z| \le t$, by the pigeonhole principle, for any $g_0, \dots, g_t$ in $X$, there are $0 \le i \ne j \le t$ and $z \in Z$ such that $g_i, g_j \in zY$, so $g_i^{-1}g_j \in Y^{-1}Y$. Since $g_0, \dots, g_t$ are arbitrary, we conclude that $Y^{-1}Y$ is $t$-thick $\square$

**Lemma 11.** An arbitrary intersection of type-definable subsets of $G$ which are thick in $X$ is again thick in $X$.

*Proof.* Let us see first that if $Y \subseteq G$ is $t$-thick in $X$ and $Y' \subseteq G$ is $t'$-thick in $X$, then $Y \cap Y'$ is $(R(t+1, t'+1) - 1)$-thick in $X$, where $R(\cdot, \cdot)$ is the Ramsey number with respect to a 2-colouring. Indeed, $Y \cap Y'$ is obviously symmetric. Now, given any subset $Z$ of $X$ of size $R(t+1, t'+1)$, colour the edges of the complete graph on $Z$ by

* $\{z, z'\}$ is black if $z^{-1}z' \notin Y$,

* $\{z, z'\}$ is white otherwise.

Since $Y$ is $t$-thick, $Z$ cannot contain a complete black subgraph of size $t+1$. By Ramsey's Theorem, it must contain a complete white subgraph of size $t'+1$, i.e. there is a subset $Z_0$ of $Z$ of size $t'+1$ such that $z^{-1}z' \in Y$ for any two elements $z, z' \in Z_0$. As $Y'$ is $t'$-thick, there are $z, z' \in Z_0$ with $z \ne z'$ such that $z^{-1}z' \in Y'$, so $z^{-1}z' \in Y \cap Y'$. As $Z$ is arbitrary, we conclude that $Y \cap Y'$ is $(R(t+1, t'+1) - 1)$-thick in $X$.

By induction, it follows that if $Y_i$ is $t_i$-thick in $X$ for $i < n$ then $\bigcap_{i<n} Y_i$ is $t$-thick in $X$ for some $t$.

Now, let $\bigcap_I Y_i$ be an intersection of type-definable subsets of $G$ thick in $X$ and $\bar{Y}$ be a definable symmetric superset of $\bigcap_I Y_i$. By compactness, there are a finite $I_0 \subseteq I$ and definable symmetric supersets $\bar{Y}_i$ of $Y_i$ for each $i \in I_0$ such that $\bigcap_{I_0} \bar{Y}_i \subseteq \bar{Y}$. By the above $\bigcap_{I_0} \bar{Y}_i$ is $t$-thick in $X$ for some $t$, and so is $\bar{Y}$; as $\bar{Y}$ is arbitrary, $\bigcap_I Y_i$ is thick in $X$. $\square$

**Lemma 12.** Let $\mu$ be a $T$-rough measure on $\mathcal{A}$ normalised at $A$. Suppose that $A^{2m} \in \mathcal{A}$ and $\mu(A^{2m}) < \infty$. Let $t > 0$ be an integer and $B \in \mathcal{A}$ with $B \subseteq A^m$ and $\mu(B) \ge 2\mu(A^{2m})/t$. Put

$$S(B) = \left\{ g \in A^{2m} : \min\{\mu(B \cap gBT), \mu(gB \cap BT)\} \ge \frac{2\mu(A^{2m})}{t^2} \right\}.$$

Then, $S(B)$ is $t$-thick in $A^m$.

Note that $S(B)$ is symmetric, since $A$ is symmetric and $\mu$ is left invariant. Also, note that any $g \in G$ with $B \cap gBT$ or $gB \cap BT$ non-empty must be in $BTB^{-1}$, so, in particular, $S(B) \subseteq BTB^{-1}$.
by induction that $Y \in \mathcal{T}_{i,j}^t$ implies $X \in \mathcal{T}_{i,\ell}^s$ and $S_{i,j}^t(Y) \subseteq S_{i,\ell}^s(X)$ whenever $j \ge \ell$, $t \le s$ and $Y \subseteq X$.

Next, we claim that, if $X \subseteq A^m$ is definable with $\mu_i(X) \ge 2\mu_i(A^{2m})/t$, then $X \in \mathcal{T}_{i,j}^t$ for all $j \in \mathbb{N}$. This is trivial for $j = 0$. Assume it holds for $j$; by Lemma 9, we have that

$$S_i(X) := \{g \in A^{2m} : \min\{\mu_i(X \cap gXT_i), \mu_i(X \cap g^{-1}XT_i)\} \ge 2\mu_i(A^{2m})/t^2\}$$

is $t$-thick in $A^m$ and, by induction hypothesis, $S_i(X) \subseteq S_{i,j}^{t^2}(X)$, concluding that $X \in \mathcal{T}_{i,j+1}^t$.

Pick a definable symmetric superset $\bar{B}$ of $B$ contained in $A^m$ and fix $i \in \mathbb{N}$. By Lemma 7, finitely many left translates of $\bar{B}$ cover $A$, so we have $\mu_i(\bar{B}) > 0$. Let $K_i = \mu_i(A^{2m(r+1)})$ and $\varepsilon_i = \mu_i(\bar{B})/2rK_i$. Finally, set $t_{i,n} := \lceil 2K_i/\mu_i(\bar{B}) \rceil^{2n}$, so $\bar{B} \in \mathcal{T}_{i,j}^{t_{i,n}}$ for all $j \in \mathbb{N}$.

We define recursively sequences $(X_{i,n} : n \in \mathbb{N})$ and $(S_{i,n} : n \in \mathbb{N})$ such that $(X_{i,n} : n \in \mathbb{N})$ is decreasing, $S_{i,n}$ is thick in $A^m$, $S_{i,n} \subseteq X_{i,n}T_iX_{i,n} \cap A^{2m}$ is symmetric and $X_{i,n} \in \mathcal{T}_{i,j}^{t_{i,n}}$ for all $j \in \mathbb{N}$.

For $n = 0$, set $X_{i,0} = \bar{B}$. Given $X_{i,n}$, set $S_{i,n} = \bigcap_{j \in \mathbb{N}} S_{i,j}^{t_{i,n}}(X_{i,n})$; since $X_{i,n} \in \mathcal{T}_{i,j}^{t_{i,n}}$ for all $j \in \mathbb{N}$, it follows that $S_{i,n}$ is thick in $A^m$ by Lemma 8. Also, note that $S_{i,j}(X_{i,n}) \subseteq X_{i,n}T_iX_{i,n}^{-1} \cap A^{2m}$ is symmetric. Given $X_{i,n}$ and $S_{i,n}$, we take $X_{i,n+1} = X_{i,n} \cap gX_{i,n}T_i$, where $g \in S_{i,n}$ satisfies for some $k < r$ that

$$\mu_i(X_{i,n+1}T_i^k \bar{B}T_i^k \cap A^{2m(r+1)}) \le (1 - \varepsilon_i)\mu_i(X_{i,n}T_i^k \bar{B}T_i^k \cap A^{2m(r+1)}),$$

or, if there is no such $g \in S_{i,n}$, we take $X_{i,n+1} = X_{i,n}$. Since $X_{i,n} \in \mathcal{T}_{i,j}^{t_{i,n}}$ and $g \in S_{i,j}^{t_{i,n}}(X_{i,n})$ for all $j \in \mathbb{N}$, we conclude that $X_{i,n+1} \in \mathcal{T}_{i,j}^{t_{i,n+1}}$ for all $j \in \mathbb{N}$.

Since $1 \in T_i$ and $X_{i,n} \subseteq \bar{B} \subseteq A^m$ for every $n \in \mathbb{N}$, we have by monotonicity and left invariance that

$$0 < \mu_i(\bar{B}) \le \mu_i(X_{i,n+1}T_i^k \bar{B}T_i^k \cap A^{2m(r+1)}) \le \mu_i(A^{2m(r+1)}) < \infty.$$

Applying Lemma 12 to the functions $f_k : X \mapsto \mu_i(XT_i^k \bar{B}T_i^k \cap A^{2m(r+1)})$, $k < r$, on the set $\{X_{i,n} : n \in \mathbb{N}\}$ linearly ordered by inclusion, we conclude that there is $n_i \in \mathbb{N}$ such that

$$\mu_i(X_{i,n}T_i^k \bar{B}T_i^k \cap A^{2m(r+1)}) > (1 - \varepsilon_i) \cdot \mu_i(X_{i,n_i}T_i^k \bar{B}T_i^k \cap A^{2m(r+1)})$$

for all $n > n_i$ and each $k < r$. By construction of the sequence $(X_{i,n})_{n \in \mathbb{N}}$, we see that $X_{i,n_i} = X_{i,n_{i+1}} =: X_i$, and

$$\mu_i((X_i \cap gX_iT_i)T_i^k \bar{B}T_i^k \cap A^{2m(r+1)}) > (1 - \varepsilon_i) \cdot \mu_i(X_iT_i^k \bar{B}T_i^k \cap A^{2m(r+1)})$$
*Proof.* Suppose, aiming for a contradiction, that the conclusion is false, and consider a counter-example $(g_i : i \le t)$. Then, $g_i \in A^m$ for each $i \le t$ and

$$ \min\{\mu(g_i B \cap g_j BT), \mu(g_i BT \cap g_j B)\} < \frac{2\mu(A^{2m})}{t^2}, $$

for all $0 \le i < j \le t$. Hence, we get the contradiction

$$ \begin{aligned} \mu(A^{2m}) &\ge \mu(\bigcup_{i \le t} g_i B) \\ &\ge \sum_{i \le t} \mu(g_i B) - \sum_{0 \le i < j \le t} \min\{\mu(g_i B \cap g_j BT), \mu(g_i BT \cap g_j B)\} \\ &> (t+1) \cdot \frac{2\mu(A^{2m})}{t} - \frac{(t+1)t}{2} \cdot \frac{2\mu(A^{2m})}{t^2} \\ &= \mu(A^{2m}) \cdot \frac{t+1}{t} > \mu(A^{2m}), \end{aligned} $$

where the first inequality follows from monotonicity as $g_i \in A^m$ for each $i$, the second from Lemma 6 and the third by the choice of $(g_i : i \le t)$. $\square$

## 2. ROUGH DEFINABLE AMENABILITY

Throughout this section, let $T = (T_i : i \in \mathbb{N})$ be a sequence of definable symmetric subsets of $G$ such that $T_{i+1}^2 \subseteq T_i$ and $T_{i+1}^A \subseteq T_i$ for all $i \in \mathbb{N}$. Write $T_\omega = \bigcap_{i \in \mathbb{N}} T_i$; note that $T_\omega$ is a type-definable subgroup of $G$ normalised by $A$.

**Definition 13.** We say that $A$ is *T-rough definably amenable* if for all $i \in \mathbb{N}$ there is a $T_i$-rough measure $\mu_i$ on the Boolean algebra of definable subsets of $G$ invariant under left translations and normalised at $A$.

**Lemma 14.** Let $P$ be a partially ordered set and $f_0, \dots, f_{r-1} : P \to \mathbb{R}_{>0}$ be increasing functions. Suppose there is $a > 0$ such that $f_k(x) \ge a$ for all $x \in P$ and each $k < r$. Then, for any $\varepsilon_0, \dots, \varepsilon_{r-1} > 0$, there is $x \in P$ such that $f_k(y) > (1 - \varepsilon_k) \cdot f_k(x)$ for any $y \in P$ with $y < x$ and each $k < r$.

*Proof.* Aiming for a contradiction, suppose otherwise. Then, we can recursively construct a decreasing sequence $(x_n)_{n \in \mathbb{N}}$ in $P$ such that for any $n \in \mathbb{N}$ there is $i < r$ with $f_i(x_{n+1}) \le (1 - \varepsilon_i) \cdot f_i(x_n)$. Since $r \in \mathbb{N}$ is finite, by the pigeonhole principle, there is $i < r$ and $N \subseteq \mathbb{N}$ infinite such that $f_i(x_{n+1}) \le (1 - \varepsilon_i) \cdot f_i(x_n)$ for any $n \in N$. Since $f_i$ is monotonous, we get that $f_i(x_m) \le (1 - \varepsilon_i) \cdot f_i(x_n)$ for any $n \in N$ and $m > n$. Let $n_0 := \min N$ and $b := f_i(x_{n_0})$. For any $n \in N$, put $\alpha := |\{m \in N : m < n\}|$. Then,

$$ a \le f_i(x_n) \le (1 - \varepsilon_i)^\alpha \cdot f_i(x_{n_0}) = (1 - \varepsilon_i)^\alpha \cdot b, $$

getting a contradiction when $\alpha > \ln(a/b)/\ln(1 - \varepsilon_i)$. $\square$
for all $g \in S_i := S_{i,n_i}$ and each $k < r$. Thus

$$ \mu_i(X_i T_i^k \bar{B} T_i^k \cap g X_i T_i^{k+1} \bar{B} T_i^k \cap A^{2m(r+1)}) \ge \mu_i((X_i \cap g X_i T_i) T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) \\ > (1 - \varepsilon_i) \cdot \mu_i(X_i T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) $$

for all $g \in S_i$ and each $k < r$. Hence,

$$ \mu_i((X_i T_i^k \bar{B} T_i^k \setminus g X_i T_i^{k+1} \bar{B} T_i^{k+1}) \cap A^{2m(r+1)}) \\ \le \mu_i(X_i T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) - \mu_i(X_i T_i^k \bar{B} T_i^k \cap g X_i T_i^{k+1} \bar{B} T_i^k \cap A^{2m(r+1)}) \\ < (1 - (1 - \varepsilon_i)) \cdot \mu_i(X_i T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) \\ \le \varepsilon_i K_i. $$

Take $g_0, \dots, g_{r-1} \in S_i$ arbitrary and write $g_{<k} := g_0 \cdots g_{k-1}$ for each $k < r$. It follows that

$$ \mu_i((X_i \bar{B} \setminus g_{<r} X_i T_i^r \bar{B} T_i^r) \cap A^{2m}) \\ \le \mu_i\left(\bigcup_{k<r} g_{<k}(X_i T_i^k \bar{B} T_i^k \setminus g_k X_i T_i^{k+1} \bar{B} T_i^{k+1}) \cap A^{2m}\right) \\ \le \sum_{k<r} \mu_i(g_{<k}(X_i T_i^k \bar{B} T_i^k \setminus g_k X_i T_i^{k+1} \bar{B} T_i^{k+1}) \cap A^{2m}) \\ \le \sum_{k<r} \mu_i((X_i T_i^k \bar{B} T_i^k \setminus g_k X_i T_i^{k+1} \bar{B} T_i^{k+1}) \cap A^{2m(r+1)}) \\ < r \varepsilon_i K_i < \mu_i(\bar{B}) \le \mu_i(X_i \bar{B} \cap A^{2m}). $$

In particular, $X_i \bar{B} \cap g_{<r} X_i T_i^r \bar{B} T_i^r \ne \emptyset$, so $g_{<r} \in X_i \bar{B} T_i^r \bar{B} T_i^r X_i \subseteq \bar{B}^2(T_i^r \bar{B})^2$.

Now, put $S(\bar{B}) := \bigcap_{i \in \mathbb{N}} S_i$ and $S = \bigcap_{\bar{B} \in \mathcal{B}} S(\bar{B})$ where $\mathcal{B} = \{\bar{B} : B \subseteq \bar{B} \subseteq A^m \text{ definable}\}$. Then, for any $g_0, \dots, g_{r-1} \in S$, we conclude that

$$ g_{<r} \in \bigcap_{\bar{B} \in \mathcal{B}} \bar{B}^2(T_i^r \bar{B})^2. $$

By saturation, $\bigcap_{\bar{B} \in \mathcal{B}} \bigcap_{i \in \mathbb{N}} \bar{B}^2(T_i^r \bar{B})^2 = B^2(T_\omega^r B)^2 = B^4 T_\omega$ — recall that $T_\omega$ is a subgroup normalised by $A$, whence also normalised by $B$. Hence, $S^r \subseteq B^4 T_\omega \subseteq A^{4m} T_\omega$. Also note that $S \subseteq B^2 T_\omega \cap A^{2m}$. $\square$

We can now prove Theorem 1:

**Theorem 14.** Assume $A$ is $T$-rough definably amenable and suppose that for every $m \in \mathbb{N}$ there is $i_m \in \mathbb{N}$ such that $\mu_i(A^m) < \infty$ for all $i > i_m$. Then, there is a type-definable normal subgroup $N$ of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4 T_\omega$.
**Proposition 15.** Let $A \subseteq G$ be $T$-rough definably amenable, and $m$ and $r$ integers with $\mu_i(A^{2m(r+1)}) < \infty$ for each $i \in \mathbb{N}$. Suppose $B \subseteq A^m$ is type-definable and thick in $A$. Then, there is a type-definable subset $S \subseteq B^2T_\omega \cap A^{2m}$ thick in $A^m$ such that $S^r \subseteq B^4T_\omega$.

*Proof.* For all $i, t \in \mathbb{N}$, define recursively on $j \in \mathbb{N}$ a family $\mathcal{T}_{i,j}^t$ of definable subsets of $A^m$. For $j = 0$, let $X \in \mathcal{T}_{i,0}^t$ if $X \neq \emptyset$. Given $\mathcal{T}_{i,j}^t$, let $X \in \mathcal{T}_{i,j+1}^t$ if

$$S_{i,j}^t(X) := \{g \in A^{2m} : X \cap gXT_i, X \cap g^{-1}XT_i \in \mathcal{T}_{i,j}^{t2}\}$$

is $t$-thick in $A^m$. Note that $S_{i,j}^t(X) \subseteq XT_iX^{-1}$ is symmetric.

By induction on $n$, we easily see that these families are definable, i.e. $\{b : X(b) \in \mathcal{T}_{i,j}^t\}$ is definable for any definable family $X(b)$ of subsets of $A^m$. Also, we easily see by induction that $Y \in \mathcal{T}_{i,j}^t$ implies $X \in \mathcal{T}_{i,l}^s$ and $S_{i,j}^t(Y) \subseteq S_{i,l}^s(X)$ whenever $j \ge l$, $t \le s$ and $Y \subseteq X$.

Next, we claim that for any $t > 0$, if $X \subseteq A^m$ is definable with $\mu_i(X) \ge 2\mu_i(A^{2m})/t$, then $X \in \mathcal{T}_{i,j}^t$ for all $j \in \mathbb{N}$. This is trivial for $j = 0$. Assume it holds for $j$; by Lemma 12 the set

$$S_i(X) := \{g \in A^{2m} : \min\{\mu_i(X \cap gXT_i), \mu_i(X \cap g^{-1}XT_i)\} \ge 2\mu_i(A^{2m})/t^2\}$$

is $t$-thick in $A^m$. By the inductive hypothesis for $t^2$ we have $X \cap gXT_i, X \cap g^{-1}XT_i \in \mathcal{T}_{i,j}^{t2}$ for any $g \in S_i(X)$, whence $S_i(X) \subseteq S_{i,j}^t(X)$. Thus $S_{i,j}^t(X)$ is $t$-thick in $A^m$, implying that $X \in \mathcal{T}_{i,j+1}^t$.

Let $\mathcal{B}$ be the set of definable symmetric supersets of $\mathcal{B}$ contained in $A^m$. Fix $\bar{B} \in \mathcal{B}$ and $i \in \mathbb{N}$. By Lemma 10, finitely many left translates of $\bar{B}$ cover $A$, so we have $\mu_i(\bar{B}) > 0$. Let $K_i = \mu_i(A^{2m(r+1)})$ and $\varepsilon_i = \mu_i(\bar{B})/2rK_i$. Finally, set $t_{i,n} := \lceil 2K_i/\mu_i(\bar{B}) \rceil^{2n}$, so $\bar{B} \in \mathcal{T}_{i,j}^{t_{i,n}}$ for all $j \in \mathbb{N}$.

We recursively define sequences $(X_{i,n} : n \in \mathbb{N})$ and $(S_{i,n} : n \in \mathbb{N})$ such that $(X_{i,n} : n \in \mathbb{N})$ is decreasing and $X_{i,n} \in \mathcal{T}_{i,j}^{t_{i,n}}$ for all $j \in \mathbb{N}$, and $S_{i,n} \subseteq X_{i,n}T_iX_{i,n} \cap A^{2m}$ is symmetric and thick in $A^m$.

For $n = 0$, set $X_{i,0} = \bar{B}$. Given $X_{i,n}$, set $S_{i,n} = \bigcap_{j \in \mathbb{N}} S_{i,j}^{t_{i,n}}(X_{i,n})$; since $X_{i,n} \in \mathcal{T}_{i,j}^{t_{i,n}}$ for all $j \in \mathbb{N}$, it follows that $S_{i,n}$ is thick in $A^m$ by Lemma 11. Also, note that $S_{i,j}(X_{i,n}) \subseteq X_{i,n}T_iX_{i,n}^{-1} \cap A^{2m}$ is symmetric. Given $X_{i,n}$ and $S_{i,n}$, we take $X_{i,n+1} = X_{i,n} \cap gX_{i,n}T_i$ where $g \in S_{i,n}$ satisfies, for some $k < r$, that

$$\mu_i(X_{i,n+1}T_i^k \bar{B}T_i^k \cap A^{2m(r+1)}) \le (1 - \varepsilon_i)\mu_i(X_{i,n}T_i^k \bar{B}T_i^k \cap A^{2m(r+1)}),$$

or, if there is no such $g \in S_{i,n}$, we take $X_{i,n+1} = X_{i,n}$. Since $X_{i,n} \in \mathcal{T}_{i,j}^{t_{i,n}}$ and $g \in S_{i,j}^{t_{i,n}}(X_{i,n})$ for all $j \in \mathbb{N}$, we conclude that $X_{i,n+1} \in \mathcal{T}_{i,j}^{t_{i,n+1}}$ for all $j \in \mathbb{N}$.
**Proof.** We iterate Proposition 13. Put $A_0 = A$ and, given $A_n \subseteq A^{2^n}$, we apply Proposition 13 with $2^n$ in place of $m$, 8 in place of $r$, the subsequences $(\mu_i : i > i_{9.2^{n+1}})$ and $(T_i : i > i_{9.2^{n+1}})$ in place of $(\mu_i)_{i \in \mathbb{N}}$ and $(T_i)_{i \in \mathbb{N}}$ respectively, $A_n$ in place of $B$ and set $A_{n+1}$ in place of $S$. Note that $A_{n+1} \subseteq A_n^2 \subseteq A^{2^{n+1}}$.

Clearly, $H = \bigcap_{n \in \mathbb{N}} A_n^4 T_\omega$ is a type-definable subgroup contained in $A^4 T_\omega$. Furthermore, $H$ is thick in $A^m$ for each $m \in \mathbb{N}$. Thus, the index of $H$ in $\langle AT_\omega \rangle$ is bounded by Lemma 7. Now, as $H$ has bounded index in $\langle AT_\omega \rangle$, the intersection of all $\langle AT_\omega \rangle$-conjugates of $H$ is a bounded intersection, whence a type-definable normal subgroup $N$ in $\langle AT_\omega \rangle$ of bounded index. $\square$

**Definition 15.** Let $K > 0$ be an integer and $Z \subseteq G$ a symmetric subset. We say that $A \subseteq G$ is a Z-rough K-approximate subgroup if there is a finite set $E$ of size $K$ such that $A^2 \subseteq EAZ$.

**Remark 16.** The notion of rough approximate subgroups has been studied by several authors before (e.g. [7], [2]). Typically, rough approximate subgroups are only defined in the context of metric groups where $Z$ is a ball at the identity of positive radius. Our definition is more flexible as it does not requires a metric. Nevertheless, note that, when $Z$ is symmetric, both definitions coincide as one can (a posteriori) pick the discrete metric given by $d(x, y) = \min\{n : x^{-1}y \in Z^n\}$.

**Lemma 17.** Assume $A$ is T-rough definably amenable, and suppose $\mu_{i+3}(A^4) \le K_i$ for some $i \in \mathbb{N}$. Then, $A^2$ is a $T_i$-rough $K_i^2$-approximate subgroup.

**Proof.** This is just a rough version of Ruzsa's Covering Lemma. Suppose $(g_k : k \le K_i)$ are in $A^3$ with $g_k A \cap g_{k'} AT_{i+3} = \emptyset$ for $k \ne k'$. Hence,

$$K_i \ge \mu_{i+3}(A^4) \ge (K_i + 1)\mu_{i+3}(A) = K_i + 1,$$

a contradiction. Thus, there is $E \subseteq A^3$ of size at most $K_i$ such that, for any $g \in A^3$, there is $e \in E$ with $gA \cap eAT_{i+3} \ne \emptyset$, whence $A^3 \subseteq EAT_{i+3}A \subseteq EA^2T_{i+2}$.

It follows that

$$(A^2)^2 = A^3 \cdot A \subseteq EA^2T_{i+2} \cdot A = EA^3T_{i+1} \subseteq E^2A^2T_{i+2}T_{i+1} \subseteq E^2A^2T_i,$$

which yields that $A^2$ is a $T_i$-rough $K_i^2$-approximate subgroup. $\square$

**Corollary 18.** Assume $A \subseteq G$ is T-rough definably amenable, and suppose that $\mu_i(A^4 T_\omega) < \infty$ for all $i \in \mathbb{N}$. Then, $\mu_i(A^m) < \infty$ for all $i, m \in \mathbb{N}$. In particular, by Theorem 14, there is a type-definable normal subgroup of bounded index of $\langle AT_\omega \rangle$ contained in $A^4 T_\omega$.

**Proof.** By Lemma 17, $A^2$ is a $T_j$-rough approximate subgroup for all $j \in \mathbb{N}$. Thus, for any $m, j \in \mathbb{N}$, there is $E_{m,j}$ finite such that $A^m \subseteq E_{m,j}A^2T_j$. Since $\mu_i(A^2T_\omega) < \infty$, there is $j_0 \in \mathbb{N}$ such that $\mu_i(A^2T_{j_0}) < \infty$, so $\mu_i(A^m) \le |E_{m,j_0}| \cdot \mu_i(A^2T_{j_0}) < \infty$. $\square$
Since $1 \in T_i$ and $X_{i,n} \subseteq \bar{B} \subseteq A^m$ for every $n \in \mathbb{N}$, by monotonicity and left invariance we have that

$$0 < \mu_i(\bar{B}) \le \mu_i(X_{i,n+1} T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) \le \mu_i(A^{2m(r+1)}) < \infty.$$

Applying Lemma 14 to the functions $f_k: X \mapsto \mu_i(XT_i^k \bar{B}T_i^k \cap A^{2m(r+1)})$ for $k < r$, on the set $\{X_{i,n} : n \in \mathbb{N}\}$ linearly ordered by inclusion, we conclude that there is $n_i \in \mathbb{N}$ such that

$$\mu_i(X_{i,n} T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) > (1 - \varepsilon_i) \cdot \mu_i(X_{i,n_i} T_i^k \bar{B} T_i^k \cap A^{2m(r+1)})$$

for all $n > n_i$ and each $k < r$. By construction of the sequence $(X_{i,n})_{n \in \mathbb{N}}$, we see that $X_{i,n_i} = X_{i,n_i+1} =: X_i$, and

$$\mu_i((X_i \cap gX_i T_i) T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) > (1 - \varepsilon_i) \cdot \mu_i(X_i T_i^k \bar{B} T_i^k \cap A^{2m(r+1)})$$

for all $g \in S_i := S_{i,n_i}$ and each $k < r$. Thus for all $g \in S_i$ and each $k < r$ we have

$$\mu_i(X_i T_i^k \bar{B} T_i^k \cap gX_i T_i^{k+1} \bar{B} T_i^k \cap A^{2m(r+1)}) \ge \mu_i((X_i \cap gX_i T_i) T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) \\ > (1 - \varepsilon_i) \cdot \mu_i(X_i T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}),$$

where the first inequality follows from monotonicity using the general inequality $(V \cap W)Z \subseteq VZ \cap WZ$. Hence,

$$\mu_i((X_i T_i^k \bar{B} T_i^k \setminus gX_i T_i^{k+1} \bar{B} T_i^{k+1}) \cap A^{2m(r+1)}) \\ \le \mu_i(X_i T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) - \mu_i(X_i T_i^k \bar{B} T_i^k \cap gX_i T_i^{k+1} \bar{B} T_i^{k+1} \cap A^{2m(r+1)}) \\ < (1 - (1 - \varepsilon_i)) \cdot \mu_i(X_i T_i^k \bar{B} T_i^k \cap A^{2m(r+1)}) \\ \le \varepsilon_i K_i,$$

where the first inequality follows from $T_i$-rough additivity and monotonicity.

Take $g_0, \dots, g_{r-1} \in S_i$ arbitrary and write $g_{<k} := g_0 \cdots g_{k-1}$ for each $k < r$. It follows that

$$\mu_i((X_i \bar{B} \setminus g_{<r} X_i T_i^r \bar{B} T_i^r) \cap A^{2m}) \\ \le \mu_i \left( \bigcup_{k<r} g_{<k} (X_i T_i^k \bar{B} T_i^k \setminus g_k X_i T_i^{k+1} \bar{B} T_i^{k+1}) \cap A^{2m} \right) \\ \le \sum_{k<r} \mu_i(g_{<k} (X_i T_i^k \bar{B} T_i^k \setminus g_k X_i T_i^{k+1} \bar{B} T_i^{k+1}) \cap A^{2m}) \\ \le \sum_{k<r} \mu_i((X_i T_i^k \bar{B} T_i^k \setminus g_k X_i T_i^{k+1} \bar{B} T_i^{k+1}) \cap A^{2m(r+1)}) \\ < r \varepsilon_i K_i < \mu_i(\bar{B}) \le \mu_i(X_i \bar{B} \cap A^{2m}),$$

where the first inequality follows from monotonicity using the general telescopic inequality $Z_0 \setminus Z_n \subseteq \bigcup_{i=0}^{n-1} Z_i \setminus Z_{i+1}$, the second follows from subadditivity, and the
### 3. APPLICATIONS TO METRIC GROUPS

In this section we explain how to apply Theorem 14 to get Corollary 2 (i.e. [4, Theorem 20]). Recall that a *metric group* is a group $G$ together with a metric $d$ invariant under left translations. We denote by $\bar{D}_r$ the closed ball of radius $r$ at the identity. For $\ell \in \mathbb{N}_{>0}$ and $r \in \mathbb{R}_{>0}$, a subset $A \subseteq G$ is $(\ell, r)$-Lipschitz if every right translation by an element of $A$ is $\ell$-Lipschitz in $\bar{D}_r$, i.e. $d(xa, ya) \le \ell \cdot d(x, y)$ for any $a \in A$ and $x, y \in \bar{D}_r$. For a subset $X \subseteq G$, we define $N_r(X) := \max\{|Z| : Z \subseteq X \text{ and } d(z, z') > r \text{ for all } z, z' \in Z \text{ with } z \ne z'\}$.

**Remark 19.** Note that $N_r$ is a $\bar{D}_r$-rough measure invariant under left translations. Indeed, as the metric is invariant under left translations, so is $N_r$. If $Z \subseteq X \cup Y$ is $r$-separated, $|Z| \le |Z \cap X| + |Z \cap Y|$ where $Z \cap X \subseteq X$ and $Z \cap Y \subseteq Y$ are $r$-separated, so $N_r$ is subadditive. Finally, if $X \cap Y \bar{D}_r = \emptyset$, then $X$ and $Y$ are $r$-separated. As the union of two $r$-separated subsets which are $r$-separated is $r$-separated, $N_r(X \cup Y) = N_r(X) + N_r(Y)$, so $N_r$ is $\bar{D}_r$-rough additive.

Fix natural numbers $\ell$ and $(K_i : i \in \mathbb{N})$. Let $(G_m, A_m, r_{i,m})_{i \le m \in \mathbb{N}}$ be such that

(1) $G_m$ is a metric group,

(2) $A_m$ is an $(\ell, r_{0,m})$-Lipschitz symmetric subset,

(3) $r_m = (r_{i,m})_{i \le m}$ is a finite sequence of positive real numbers with $\max\{\ell, 2\} \cdot r_{i,m} \le r_{i-1,m}$ and

$$N_{r_{i,m}}(A_m^4) \le K_i \cdot N_{r_{i,m}}(A_m) < \infty.$$

Consider the language of groups enriched with a predicate for $A_m$ and predicates for $\bar{D}_{r_{i,m}}$ for each $i \in \mathbb{N}$ — where $r_{i,m} = r_{m,m}$ for $i \ge m$; we consider $(G_m, \bar{X}_m, r_{i,m})_{i \le m \in \mathbb{N}}$ as a sequence of structures in this language. Pick a non-principal ultrafilter $\mathcal{u}$ of $\mathbb{N}$ and let $(G, A, r)$ be the corresponding ultraproduct — recall that this is $\aleph_1$-saturated $[8, \text{Exercise 5.2.3}]$. Set $T_i := \bar{D}_{r_i}$ and $T_\omega := \bigcap T_i$.

**Lemma 20.** A is $T$-rough definably amenable with respect to $\mu_i(Y) := \frac{N_{r_i}(Y)}{N_{r_i}(A)}$ and $\mu_i(A^4T_\omega) < \infty$ for any $i \in \mathbb{N}_{>0}$.

*Proof.* As $A$ is $(\ell, r_0)$-Lipschitz and $\ell r_{i+1} \le r_i$, we get $T_{i+1}^A \subseteq T_i$. Since $2r_{i+1} \le r_i$, we have $T_{i+1}T_{i+1} \subseteq T_i$. As $N_{r_i}$ is a $T_i$-rough measure on the boolean algebra of definable subsets invariant under left translations, we conclude that $A$ is $T$-rough definably amenable with respect to the normalisation $\mu_i$. Finally, note that $\mu_i(A^4T_\omega) \le \mu_i(A^4T_{i+2}) \le \mu_{i-1}(A^4) \le K_{i-1}$ for any $i \in \mathbb{N}_{>0}$. $\square$

By Corollary 18, there is a type-definable normal subgroup $N$ of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4T_\omega$; in other words, we get Corollary 2(1).

When $(K_i : i \in \mathbb{N})$ is a constant $K$, $A^2$ is a $T_\omega$-rough $K$-approximate subgroup by Lemma 17 and saturation. Thus, by [6, Theorem 3.2], we get Corollary 2(2).
third follows by left invariance and monotonicity as $g_k \in A^{2m}$ for each $k$ — note that for this reason we use $A^{2m(r+1)}$ rather than $A^{2m}$.

In particular, $X_i \bar{B} \cap g_{<r} X_i T_i^r \bar{B} T_i^r \neq \emptyset$, so $g_{<r} \in X_i \bar{B} T_i^r \bar{B} T_i^r X_i \subseteq \bar{B}^2 (T_i^r \bar{B})^2$.

Now, put $S(\bar{B}) := \bigcap_{i \in \mathbb{N}} S_i$ and $S = \bigcap_{\bar{B} \in \mathcal{B}} S(\bar{B})$. Then, for any $g_0, \dots, g_{r-1} \in S$, we conclude that

$$g_{<r} \in \bigcap_{\bar{B} \in \mathcal{B}} \bigcap_{i \in \mathbb{N}} \bar{B}^2 (T_i^r \bar{B})^2.$$

By saturation, $\bigcap_{\bar{B} \in \mathcal{B}} \bigcap_{i \in \mathbb{N}} \bar{B}^2 (T_i^r \bar{B})^2 = B^2 (T_\omega^r B)^2 = B^4 T_\omega$ — recall that $T_\omega$ is a subgroup normalised by $A$, whence also normalised by $B$. Hence, $S^r \subseteq B^4 T_\omega \subseteq A^{4m} T_\omega$. Similarly, $S \subseteq \bigcap_{\bar{B} \in \mathcal{B}} \bigcap_{i \in \mathbb{N}} \bar{B} T_i \bar{B} \cap A^{2m} = B^2 T_\omega \cap A^{2m}$. $\square$

**Remark 16.** In Proposition 15 we do not really need $\mu_i$ to be defined on all definable subsets of $G$, but just on a Boolean subalgebra $\mathcal{A}$ which contains $A$ and is closed under left translations, $T_i$-thickenings for $i \in \mathbb{N}$ and setwise products (i.e. $XY \in \mathcal{A}$ for any $X, Y \in \mathcal{A}$). In that case, Proposition 15 can be applied under the extra hypothesis that $B$ is an intersection of definable subsets from $\mathcal{A}$. Note however that the set $S$ obtained need not be of this form.

**Remark 17.** In Proposition 15, $S(\bar{B})$ is obtained as a countable intersection of definable subsets. Consequently, if $B$ is an intersection of $\lambda$ many definable subsets, $S$ is an intersection of $\lambda + \aleph_0$ many definable subsets. In particular, if $B$ is countably type-definable (i.e. a countable intersection of definable sets), so is $S$.

We can now prove Theorem 1:

**Theorem 18.** Assume $A$ is $T$-rough definably amenable and suppose that for every $m \in \mathbb{N}$ there is $i_m \in \mathbb{N}$ such that $\mu_i(A^m) < \infty$ for all $i > i_m$. Then there is a countably type-definable normal subgroup $N$ of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4 T_\omega$.

*Proof.* We iterate Proposition 15. Put $A_0 = A$ and, given $A_n \subseteq A^{2^n}$, we apply Proposition 15 with $2^n$ in place of $m$, 8 in place of $r$, the subsequences $(\mu_i : i > i_{9.2^{n+1}})$ and $(T_i : i > i_{9.2^{n+1}})$ in place of $(\mu_i)_{i \in \mathbb{N}}$ and $(T_i)_{i \in \mathbb{N}}$ respectively, $A_n$ in place of $B$ and set $A_{n+1}$ in place of $S$. Note that $A_{n+1} \subseteq A^{2m} = A^{2^{n+1}}$.

As all $A_n^4 T_\omega$ are symmetric and $(A_{n+1}^4 T_\omega)^2 = A_n^8 T_\omega \subseteq A_n^4 T_\omega$, we have that $H = \bigcap_{n \in \mathbb{N}} A_n^4 T_\omega$ is a type-definable subgroup contained in $A^4 T_\omega$. Furthermore, $H$ is thick in $A^m$ for each $m \in \mathbb{N}$.

We now argue by working in some $(2^{\aleph_0})^+$-saturated elementary extension $\hat{G}$; write $\hat{H}$, $\hat{A}$ and $\hat{T}_\omega$ for the corresponding interpretations in the elementary extension. $H$ is countably type-definable by Remark 17; as it is thick in $A^m$, the index of $\hat{H}$ in $\langle \hat{A} \hat{T}_\omega \rangle$ is bounded by $2^{\aleph_0}$ by Lemma 10. Since $\hat{H}$ has bounded index in $\langle \hat{A} \hat{T}_\omega \rangle$, the
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[2] Gowers, W. T. & Long, J. (2020). “Partial associativity and rough approximate groups”, *Geo-metric and Functional Analysis* **30**(6): 1583–1647. DOI:10.1007/s00039-020-00553-1.

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[7] Tao, T. (2014, March 19). *Metric entropy analogues of sum set theory*. Retrieved on 6th November, 2021, from https://terrytao.wordpress.com/2014/03/19/metric-entropy-analogues-of-sum-set-theory/

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HEBREW UNIVERSITY OF JERUSALEM; EINSTEIN INSTITUTE OF MATHEMATICS; JERUSALEM,
9190401, ISRAEL

UNIVERSITÉ LYON 1; CNRS; INSTITUT CAMILLE JORDAN UMR 5208, 21 AVENUE CLAUDE BERNARD, 69622 VILLEURBANNE CEDEX, FRANCE

Email address: arturo.rodriguez@mail.huji.ac.il

Email address: wagner@math.univ-lyon1.fr
intersection $\hat{N}$ of all $\langle \hat{A}\hat{T}_{\omega} \rangle$-conjugates of $\hat{H}$ is a bounded intersection, whence a type-definable normal subgroup in $\langle \hat{A}\hat{T}_{\omega} \rangle$ of bounded index, whence thick. Note that it is again countably type-definable over the same parameters, as if $\hat{H} = \bigcap_{i<\omega} \hat{X}_i$ we have $\hat{N} = \bigcap_{i,m<\omega} \{x : \forall y \in \hat{A}^m x^y \in \hat{X}_i\}$. So the index of $\hat{N}$ in $\langle \hat{A}\hat{T}_{\omega} \rangle$ is also bounded by $2^{\aleph_0}$. In particular, we can consider the interpretation $N$ of $\hat{N}$ in the original structure, getting a countably type-definable normal subgroup of $\langle AT_{\omega} \rangle$ of bounded index contained in $A_n^4T_{\omega}$. $\square$

**Definition 19.** Let $K > 0$ be an integer and $Z \subseteq G$ a symmetric subset. We say that $A \subseteq G$ is a Z-rough K-approximate subgroup if there is a finite set $E$ of size $K$ such that $A^2 \subseteq EAZ$.

**Remark 20.** The notion of rough approximate subgroups has been studied by several authors before (e.g. [Tao14], [GL20]). Typically, rough approximate subgroups are only defined in the context of metric groups where $Z$ is a ball at the identity of positive radius. Our definition is more flexible as it does not require a metric. Nevertheless, note that when $Z$ is symmetric, both definitions coincide as one can (a posteriori) pick the discrete metric given by $d(x, y) = \min\{n : x^{-1}y \in Z^n\}$.

Hence, Corollary 2 follows immediately from Theorem 1 and the definition of rough approximate subgroups:

**Corollary 21.** Assume $A$ is T-rough definably amenable and a $T_i$-rough approximate subgroup for each $i \in \mathbb{N}$ (possibly with different doubling constants $K_i$) such that $\mu_i(AT_{\omega}) < \infty$ for all $i$. Then there is a countably type-definable normal subgroup $N$ of $\langle AT_{\omega} \rangle$ of bounded index such that $T_{\omega} \subseteq N \subseteq A^4T_{\omega}$.

*Proof.* As $A$ is a $T_i$-rough approximate subgroup for every $i$, we conclude by induction that for any $m, j \in \mathbb{N}$, there is $E_{m,j}$ finite such that $A^m \subseteq E_{m,j}AT_j$. Since $\mu_i(AT_{\omega}) < \infty$, there is $j$ such that $\mu_i(AT_j) < \infty$. Hence, $\mu_i(A^m) \le |E_{m,j}|\mu_i(AT_j) < \infty$ by subadditivity. By Theorem 1, we conclude. $\square$

Using a rough version of Ruzsa's Covering argument, we obtain the following related result (Corollary 23).

**Lemma 22.** Assume $A$ is T-rough definably amenable, and suppose $\mu_{i+3}(A^4) \le K_i$ for some $i \in \mathbb{N}$. Then $A^2$ is a $T_i$-rough $K_i^2$-approximate subgroup.

*Proof.* This is just a rough version of Ruzsa's Covering Lemma. Suppose $(g_k : k \le K_i)$ are in $A^3$ with $g_kA \cap g_{k'}AT_{i+3} = \emptyset$ for $k \ne k'$. Hence,

$$K_i \ge \mu_{i+3}(A^4) \ge (K_i + 1)\mu_{i+3}(A) = K_i + 1,$$

a contradiction. Thus, there is $E \subseteq A^3$ of size at most $K_i$ such that, for any $g \in A^3$, there is $e \in E$ with $gA \cap eAT_{i+3} \ne \emptyset$, whence $A^3 \subseteq EAT_{i+3}A \subseteq EA^2T_{i+2}$.
It follows that

$$(A^2)^2 = A^3 \cdot A \subseteq EA^2T_{i+2} \cdot A \subseteq EA^3T_{i+1} \subseteq E^2A^2T_{i+2}T_{i+1} \subseteq E^2A^2T_i,$$

which yields that $A^2$ is a $T_i$-rough $K_i^2$-approximate subgroup. $\square$

<figure><img src="image_3.png" /></figure>

**Corollary 23.** Assume $A \subseteq G$ is $T$-rough definably amenable, and suppose that $\mu_i(A^4T_\omega) < \infty$ for all $i \in \mathbb{N}$. Then $\mu_i(A^m) < \infty$ for all $i, m \in \mathbb{N}$. In particular, by Theorem 1, there is a countably type-definable normal subgroup of bounded index of $\langle AT_\omega \rangle$ contained in $A^4T_\omega$.

*Proof.* By Lemma 22, $A^2$ is a $T_j$-rough approximate subgroup for all $j \in \mathbb{N}$. Thus, for any $m, j \in \mathbb{N}$, there is $E_{m,j}$ finite such that $A^m \subseteq E_{m,j}A^2T_j$. Since $\mu_i(A^2T_\omega) < \infty$, there is $j_0 \in \mathbb{N}$ such that $\mu_i(A^2T_{j_0}) < \infty$, so $\mu_i(A^m) \le |E_{m,j_0}| \cdot \mu_i(A^2T_{j_0}) < \infty$. $\square$

<figure><img src="image_4.png" /></figure>

### 3. APPLICATIONS TO METRIC GROUPS

In this section we explain how to apply Theorem 1 to get Corollary 4 (i.e. [HRF22, Theorem 20]). Recall that a *metric group* is a group $G$ together with a metric $d$ invariant under left translations. We denote by $\overline{D}_r$ the closed ball of radius $r$ at the identity. For $\ell \in \mathbb{N}_{>0}$ and $r \in \mathbb{R}_{>0}$, a subset $A \subseteq G$ is $(\ell, r)$-Lipschitz if every right translation by an element of $A$ is $\ell$-Lipschitz in $\overline{D}_r$, i.e. $d(xa, ya) \le \ell \cdot d(x, y)$ for any $a \in A$ and $x, y \in \overline{D}_r$. For a subset $X \subseteq G$, we define $N_r(X) := \max\{|Z| : Z \subseteq X \text{ and } d(z, z') > r \text{ for all } z, z' \in Z \text{ with } z \ne z'\}$.

**Remark 24.** Note that $N_r$ is a $\overline{D}_r$-rough measure invariant under left translations. Indeed, as the metric is invariant under left translations, so is $N_r$. If $Z \subseteq X \cup Y$ is $r$-separated, $|Z| \le |Z \cap X| + |Z \cap Y|$ where $Z \cap X \subseteq X$ and $Z \cap Y \subseteq Y$ are $r$-separated, so $N_r$ is subadditive. Finally, if $X \cap Y \overline{D}_r = \emptyset$, then $X$ and $Y$ are $r$-separated. As the union of two $r$-separated subsets which are $r$-separated is $r$-separated, $N_r(X \cup Y) = N_r(X) + N_r(Y)$, so $N_r$ is $\overline{D}_r$-rough additive.

Fix natural numbers $\ell$ and $(K_i : i \in \mathbb{N})$. Let $(G_m, A_m, r_{i,m})_{i \le m \in \mathbb{N}}$ be such that

(i) $G_m$ is a metric group,

(ii) $A_m$ is an $(\ell, r_{0,m})$-Lipschitz symmetric subset,

(iii) $r_m = (r_{i,m})_{i \le m}$ is a finite sequence of positive real numbers with $\max\{\ell, 2\} \cdot r_{i,m} \le r_{i-1,m}$ and

$$N_{r_{i,m}/2}(A_m^4) \le K_i \cdot N_{r_{i,m}}(A_m) < \infty.$$

Consider the language of groups enriched with a predicate for $A_m$ and predicates for $\overline{D}_{r_{i,m}}$ for each $i \in \mathbb{N}$ — where $r_{i,m} = r_{m,m}$ for $i \ge m$; we consider $(G_m, A_m, r_{i,m})_{i \le m \in \mathbb{N}}$ as a sequence of structures in this language. Pick a non-principal ultrafilter $\mathfrak{u}$ of $\mathbb{N}$ and let $(G, A, r)$ be the corresponding ultraproduct — recall that this is $\aleph_1$-saturated [TZ12, Exercise 5.2.3]. Set $T_i := \overline{D}_{r_i}$ and $T_\omega := \bigcap T_i$.
**Lemma 25.** A is T-rough definably amenable with respect to $\mu_i(Y) := \frac{N_{r_i}(Y)}{N_{r_i}(A)}$ and $\mu_i(A^4T_\omega) < \infty$ for any $i \in \mathbb{N}_{>0}$.

*Proof.* As A is $(\ell, r_0)$-Lipschitz and $\ell r_{i+1} \le r_i$, we get $T_{i+1}^A \subseteq T_i$. Since $2r_{i+1} \le r_i$, we have $T_{i+1}T_{i+1} \subseteq T_i$. As $N_{r_i}$ is a $T_i$-rough measure on the Boolean algebra of definable subsets invariant under left translations, we conclude that A is T-rough definably amenable with respect to the normalisation $\mu_i$.

We now claim that $N_{r_i}(A^4T_{i+2}) \le N_{r_i/2}(A^4)$. So pick any $r_i$-separated subset $Z \subseteq A^4T_{i+2}$ and consider $Z' = ZT_{i+2} \cap A^4$. Then for any $z_1, z_2 \in Z$ with $z_1 \ne z_2$ we have $z_1 = a_1t_1$ and $z_2 = a_2t_2$ for some $a_1, a_2 \in Z'$ and $t_1, t_2 \in T_{i+2}$. By left invariance, $d(z_1, a_1) = d(t_1, 1) \le r_{i+2}$ and $d(z_2, a_2) = d(t_2, 1) \le r_{i+2}$. Thus, by the triangular inequality and the doubling condition of the sequence $(r_i)_{i \in \mathbb{N}}$ we obtain

$$r_i \le d(z_1, z_2) \le d(a_1, a_2) + d(a_1, z_1) + d(a_2, z_2) \le d(a_1, a_2) + 2r_{i+2} \le d(a_1, a_2) + r_{i+1}.$$

Therefore, $d(a_1, a_2) \ge r_i - r_{i+1} \ge r_i/2$. As $z_1, z_2$ are arbitrary, we conclude that $Z'$ is $r_i/2$-separated and $|Z'| \ge |Z|$. Hence, as Z was arbitrary, $N_{r_i}(A^4T_{i+2}) \le N_{r_i/2}(A^4)$. But then for any $i \in \mathbb{N}_{>0}$ we have

$$\mu_i(A^4T_\omega) \le \mu_i(A^4T_{i+2}) = \frac{N_{r_i}(A^4T_{i+2})}{N_{r_i}(A)} \le \frac{N_{r_i/2}(A^4)}{N_{r_i}(A)} \le K_i < \infty. \quad \square$$

By Corollary 23, there is a countably type-definable normal subgroup N of $\langle AT_\omega \rangle$ of bounded index such that $T_\omega \subseteq N \subseteq A^4T_\omega$. Since N is thick in $A^mT_\omega$ for all m, all cosets of N in $\langle AT_\omega \rangle$ are already realized in $\langle AT_\omega \rangle$ by $\aleph_1$-saturation. Thus $\pi : \langle AT_\omega \rangle \to \langle AT_\omega \rangle/N$ is the required Lie model for Corollary 4(1).

When $(K_i : i \in \mathbb{N})$ is a constant K, $A^2$ is a $T_\omega$-rough $K^2$-approximate subgroup by Lemma 22 and saturation. Thus, by Theorem 3, we also obtain Corollary 4(2).

**Acknowledgments.** We would like to thank the referee for the thorough report.

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(A. Rodríguez Fanlo) HEBREW UNIVERSITY OF JERUSALEM; EINSTEIN INSTITUTE OF MATHEMATICS; JERUSALEM, 9190401, ISRAEL

Email address, A. Rodríguez Fanlo: arturo.rodriguez@mail.huji.ac.il

(F. Wagner) UNIVERSITÉ LYON 1; CNRS; INSTITUT CAMILLE JORDAN UMR 5208, 21 AVENUE CLAUDE BERNARD, 69622 VILLEURBANNE CEDEX, FRANCE

Email address, F. Wagner: wagner@math.univ-lyon1.fr