Proof. Let $F"$ be a subset of maximal cardinality satisfying the assumption. Then, by maximality, for every $y \in Y$ , there are $i \in I$ and $f\in F"$ such that $fX_i \cap yX_i \neq \emptyset $ , that is, $y \in fX_iX_i^{-1}$ . Since $F"$ has size at most C, this concludes the proof.

Proof. Let R be a connected solvable group of dimension d. For any compact subset ${A \subset R}$ , we have $\mu (A^2) \geq 2^{d/2}\mu (A)$ , see [Reference An, Jing, Tran and ZhangAJTZ21, §3] and references within. Hence, if A is a K-approximate subgroup, then $d = O(\log K)$ . By [Reference MachadoMac25, Theorem 1.6], the internal space of minimal dimension is a connected solvable Lie group. The result follows now from standard results about approximate subgroups [Reference MachadoMac25, §2.2].

Proof. To prove (3.1), notice that if $Ax \cap B \neq \emptyset $ , then $x \in A^{-1}B$ . So,

$$ \begin{align*} \int_B \sum_{x \in X} \mathbf{1}_{Ax}(g)\,d\mu_G(g) & = \sum_{x \in X}\int_B \mathbf{1}_{Ax}(g)\,d\mu_G(g) \\ & = \sum_{x \in X \cap A^{-1}B}\int_B \mathbf{1}_{Ax}(g)\,d\mu_G(g)\\ & \leq \mu_G(A)|X \cap A^{-1}B|. \end{align*} $$

Let us now prove (3.2). If $x \in B$ , then $Ax \in AB$ . Therefore,

$$ \begin{align*} \mu_G(A)|X \cap B| &\leq \int_{AB}\sum_{x \in X \cap B} \mathbf{1}_{Ax}(g)\,d\mu_G(g) \\ &\leq \int_{AB}\sum_{x \in X} \mathbf{1}_{Ax}(g)\,d\mu_G(g).\\[-45pt] \end{align*} $$

Proof. By adapting an argument from [Reference HrushovskiHru20, Lemma A.3], there is $\mathcal {F}$ Borel such that $Y^{-1}Y\mathcal {F} = G$ and the multiplication map $Y \times \mathcal {F} \rightarrow G$ is one-to-one. Indeed, let V be a compact neighbourhood of the identity such that $VV^{-1} \cap Y^{-1}Y =\{e\}$ and let $(g_n)_{n \geq 0}$ be a sequence of elements of G such that $G=\bigcup _{n \geq 0} Vg_n$ . Define inductively $B_n:=Vg_n \setminus \bigcup _{m<n} Y^{-1}Y B_m$ and $B:=\bigcup _{n\geq 0}B_n$ . Then, $BB^{-1} \cap Y^{-1}Y =\{e\}$ and $Y^{-1}YB=G$ . We can in fact choose such an $\mathcal {F}$ with null boundary. Therefore, the multiplication map $X \times \mathcal {F}\rightarrow G$ is $|F|$ -to-one. So for every $g \in G$ , $|gX \cap V^{-1}| \leq |F|$ , where V denotes the interior of $\mathcal {F}$ . We obtain $|Y \cap V^{-1}| \leq |F|$ for every $Y \in \Omega _X$ since $V^{-1}$ is open. Hence,

$$ \begin{align*}|F| \geq \int_{\Omega_X}|Y \cap V^{-1}|\, d\nu(Y) = \mathcal{P}_X^*\nu(V^{-1}).\end{align*} $$

However, $\mathcal {P}_X^*\nu $ dominates a Haar measure. Thus, V has finite Haar measure, and so has $\mathcal {F}$ .

Proof. Take V a compact neighbourhood of the identity such that $X^{-1}X \cap V^{-1}V =\{e\}$ . Then, we can reformulate (3.2) of Lemma 3.1 as follows:

(4.1) $$ \begin{align} \mu_G(V) |X^{-1} \cap F_n| \leq \int_{VF_n}\mathcal{P}_X \mathbf{1}_{V}(gX)\,d\mu_G(g). \end{align} $$

Consider $\phi \in C^0_c(G)$ with non-negative values and $\phi \geq \mathbf {1}_{V}$ . Let $\mu _n$ denote the measure with density $\mu _G(VF_n)^{-1}\mathbf {1}_{VF_n}$ against the Haar measure and write

$$ \begin{align*}\liminf \frac{|X \cap F_n|}{\mu_G(F_n)} = \alpha.\end{align*} $$

Since $F_n$ is a Følner sequence, $\mu _G(VF_n) \sim \mu _G(F_n)$ and $(VF_n)_{\geq 0}$ is a Følner sequence. Then, (4.1) implies that

$$ \begin{align*} \liminf\int_{\Omega_X} \mathcal{P}_X\phi(Y)\,d(\mu_n * \delta_X)(Y) &\geq \liminf\int_{VF_n} \frac{\mathcal{P}_X \mathbf{1}_{V}(g)}{\mu_G(VF_n)}\,d\mu_G(g)\\ &\geq \alpha\mu_G(V)> 0, \end{align*} $$

where $\delta _X$ denotes the Dirac mass at X. Therefore, any weak-* limit $\nu $ of $\mu _n * \delta _X$ satisfies

(4.2) $$ \begin{align} \int_{\Omega_X} \mathcal{P}_X\phi(Y)\,d\nu(Y) \geq \alpha\mu_G(V). \end{align} $$

Since $\mathcal {P}_X\phi (Y)$ has compact support contained in $\Omega _X \setminus \{\emptyset \}$ , $\nu $ is not supported on $\{\emptyset \}$ . However, $\nu $ is a G-invariant Borel probability measure since $(VF_n)_{n \geq 0}$ is a Følner sequence. Note moreover that (4.2) is valid for any such $\phi $ , so

$$ \begin{align*}\mathcal{P}_X^*\nu(V) = \int_{\Omega_X} \mathcal{P}_X\mathbf{1}_V(Y)\,d\nu(Y) \geq \alpha \mu_G(V).\end{align*} $$

Thus, a proper G-invariant ergodic Borel probability measure $\nu '$ such that $\nu '(U^V) \geq \alpha \mu _G(V)$ must appear in the ergodic decomposition of $\nu $ . The result now follows from the fact that $\mathcal {P}_X^*\nu '$ is a Haar measure.

Proof. Let $m: A_0^{-1}A_1 \times \cdots \times A_{k}^{-1}A_{k+1} \rightarrow G$ denote the multiplication map. Take an element $a \in A_0^{-1}S_1\cdots S_{k}A_{k+1}$ and choose $\alpha _0 \in A_0, \alpha _{k+1} \in A_{k+1}, s_1\in S_1, \ldots , s_k \in S_k$ such that $a=\alpha _0^{-1}s_1\cdots s_k\alpha _{k+1}$ . Now, for every $l \in \{1, \ldots , k\}$ , take $\alpha _l \in A_l \cap s_lA_l$ . We have $\alpha _l^{-1}s_l\alpha _{l+1} \in A_l^{-1}A_{l+1}$ . Moreover,

$$ \begin{align*}a = m(\alpha_0^{-1}s_1\alpha_1,\ldots,\alpha_k^{-1}s_k\alpha_{k+1}).\end{align*} $$

So a belongs to the range of the map m and $m^{-1}(a)$ has size at least $|A_1 \cap s_1A_1| \cdots |A_k \cap s_kA_k|$ . Therefore,

$$ \begin{align*}|m^{-1}(a)| \geq (4K)^{-k}|A_1|\cdots |A_k|.\end{align*} $$

However,

$$ \begin{align*}|A_0^{-1}A_1 \times \cdots \times A_{k}^{-1}A_{k+1}| \leq K^{k+1}|A_1| \cdots |A_{k+1}|.\end{align*} $$

So, there are at most $4^kK^{2k+1}|A_{k+1}|$ such elements a.

Proof of Proposition 4.2

We will rephrase condition (2) as follows. Choose $\alpha> 0$ such that $\limsup |X \cap F_n|/\mu _G(F_n)> \alpha $ . Upon considering a subsequence of $(F_n)_{n \geq 0}$ , we may assume that

$$ \begin{align*}|X \cap F_nW|> \alpha \mu_G(F_nW)\end{align*} $$

for every $n \geq 0$ .

We will be able to rephrase condition (1) in a similar fashion. Fix $\beta _0> 1$ . First of all, upon considering a further subsequence, we may assume that the sequence $(F_n)_{n \geq 0}$ satisfies the Shulman assumption [Reference ShulmanShu88]. In other words,

$$ \begin{align*} \text{ for every } m> n \geq 0,\quad \mu_G(F_n^{-1}F_m) \leq \beta_0\mu_G(F_m).\end{align*} $$

Note that $\beta _0> 0$ may be chosen as close to $1$ as one wishes.

Let W be as in condition (1). Since $(F_n)_{n \geq 0}$ is bi-Følner, we may furthermore assume that

$$ \begin{align*} \text{ for every } m> n \geq 0,\quad \mu_G(W^{-1}F_n^{-1}F_mW^2) \leq \beta_0\mu_G(F_m)\end{align*} $$

for the same $\beta _0$ as above.

We have

$$ \begin{align*} \mu_G(W)|X^{-1}X \cap W^{-1}F_n^{-1}F_mW|& = \mu_G( (X^{-1}X \cap W^{-1}F_n^{-1}F_mW)W ) \\ & \leq \mu_G(W^{-1}F_n^{-1}F_mW^2) \\ & \leq \beta_0\mu_G(F_m). \end{align*} $$

Define now the subsets $\tilde {F}_n:=F_nW$ and $A_n:= X \cap \tilde {F}_n$ for every $n \geq 0$ . The above discussion implies

(4.3) $$ \begin{align} &\text{ for every } n \geq 0, \quad\quad\quad\kern0.4pt |A_n| \geq \alpha \mu_G(F_n), \end{align} $$

(4.4) $$ \begin{align} &\text{ for every } m> n \geq 0, \quad |A_n^{-1}A_m| \leq \beta \mu_G(F_m), \end{align} $$

where $\beta = {\beta _0}/{\mu _G(W)}$ .

Choose a non-principal ultrafilter U on $\mathbb {N}$ . Let $P_{W}(G)$ be the set of subsets Y of G such that the multiplication map $Y \times W \rightarrow G$ is one-to-one. For $Y \in P_{W}(G)$ , define

$$ \begin{align*}M(Y) = \lim_{n \rightarrow U} \frac{\mu_G(YW \cap \tilde{F}_nW)}{\mu_G(\tilde{F}_n)}.\end{align*} $$

Since $(F_n)_{n \geq 0}$ is Følner, it easily seen that M is a left-invariant finitely additive measure defined on $P_{W}(G)$ . According to (4.3), we have $M(X) \geq \alpha \mu _G(W)$ . By (4.4), we have $M(X^{-1}X)\leq \beta \mu _G(W)$ . Define

$$ \begin{align*} S:=\bigg\{ g \in G : M(gX \cap X) \geq \frac{\alpha}{2\beta}M(X)\bigg\}. \end{align*} $$

Proof. Let $m \geq 0$ and suppose that there is no subset F of size at most m such that ${X^{-1}\subset FS}$ . Therefore, we can choose by induction $f_1, \ldots , f_m \in X^{-1}$ such that for every $1 \leq i \leq m$ , we have

$$ \begin{align*} f_i \notin \bigcup_{j=1}^{i-1}f_jS.\end{align*} $$

Equivalently, for every $1 \leq i< j \leq m$ ,

$$ \begin{align*} M(f_iX \cap f_jX) < \frac{\alpha}{2\beta}M(X).\end{align*} $$

Hence,

$$ \begin{align*} M(X^{-1}X) & \geq M\bigg(\bigcup_{i=1}^m f_i X\bigg) \\ & \geq \sum_{i=1}^m M(f_iX) - \sum_{1\leq i < j \leq m }M(f_iX \cap f_jX) \\ &> mM(X) - \frac{m(m-1)}{2}\frac{\alpha}{2\beta}M(X) \\ & = m\bigg(1 - \frac{(m-1)\alpha}{4\beta}\bigg)M(X). \end{align*} $$

This yields

$$ \begin{align*} \beta> \alpha m\bigg(1-\frac{(m-1)\alpha}{4\beta}\bigg).\end{align*} $$

Additionally, this inequality fails for $m =\lfloor {\alpha }/{2\beta }\rfloor $ . So, for $m =\lfloor {\alpha }/{2\beta }\rfloor $ , $X^{-1}\subset FS$ for some $F \subset X^{-1}$ of cardinality at most m.

We will now show that $S^n$ is uniformly discrete for every $n \geq 0$ . To do so, we will prove that $S^n \cap W$ is finite for every $n \geq 0$ . Notice first that $S\subset \bigcup _{n \geq 0} S_n$ , where

$$ \begin{align*}S_n:=\bigg\{g \in G : |A_n \cap gA_n| \geq \frac{\alpha}{4\beta}|A_n|\bigg\}\end{align*} $$

for every $n \geq 0$ . More precisely, each element of S is contained in infinitely many $S_n$ terms. Indeed, for every $Y \in P_W(G)$ , we have

$$ \begin{align*} |Y \cap \tilde{F}_n| \leq\mu_G(YW \cap F_nW^2).\end{align*} $$

Additionally,

$$ \begin{align*} \mu_G(YW \cap F_nW^2) & \leq |Y \cap \tilde{F}_n| + \mu_G(YW \cap (F_nW^2 \setminus F_n)) \\ & = |Y \cap \tilde{F}_n| + o(\mu_G(F_n)). \end{align*} $$

However, if $g \in S$ , for some $A \in U$ and all $n \in A$ , we have

$$ \begin{align*}\mu_G( (X \cap gX)W \cap \tilde{F}_nW ) \geq \frac{\alpha}{3\beta} \mu_G( XW \cap \tilde{F}_nW),\end{align*} $$

which yields

$$ \begin{align*} |X \cap gX \cap \tilde{F}_n| + o(\mu_G(F_n)) \geq \frac{\alpha}{3\beta} |X \cap \tilde{F}_n|.\end{align*} $$

Since $|X \cap \tilde {F}_n| \geq \alpha \mu _G(\tilde {F}_n)$ , if n is chosen sufficiently large with respect to U, then

$$ \begin{align*} |X \cap gX \cap \tilde{F}_n| \geq \frac{\alpha}{4\beta} |X \cap \tilde{F}_n|.\end{align*} $$

So our claim is proved up to considering a subsequence.

Choose any finite subset $F' \subset S^n \cap W$ such that the subsets $fX$ are pairwise disjoint when f runs through $F'$ . Then, there are $k_n>\cdots > k_0 \geq 0$ such that $F' \subset S_{k_1}\cdots S_{k_n}$ . Choose an integer $k_{n+1}$ such that $k_n < k_{n+1}$ . We thus have by Lemma 4.3 (note that we had to specify $k_0$ to invoke Lemma 4.3, although it is not needed from now on),

$$ \begin{align*}|F'||A_{k_{n+1}}| = |F'A_{k_{n+1}}| \leq |S_{k_1}\cdots S_{k_{n}}A_{k_{n+1}}| \leq 4^n\frac{\beta^{2n+1}}{\alpha^{2n+1}}|A_{k_{n+1}}|.\end{align*} $$

Therefore, $|F'|\leq 4^n({\beta ^{2n+1}}/{\alpha ^{2n+1}})$ . So take $F'$ as above of maximal size. By Ruzsa’s covering lemma, we have $S^n\cap W \subset F'XX^{-1}$ . Since $F' \subset W$ , we finally have

$$ \begin{align*}S^n\cap W \subset F'(XX^{-1} \cap W^2)=F'.\end{align*} $$

Let us finally prove that $S^2$ is an approximate lattice. Notice first that

$$ \begin{align*} \limsup \frac{|S \cap F_nF|}{\mu_G(F_n)} & \geq \limsup \frac{|SF^{-1} \cap F_n|}{|F|\mu_G(F_n)} \\ & \geq \limsup \frac{|X \cap F_n|}{|F|\mu_G(F_n)} \\ & \geq \limsup \frac{|X \cap F_n|}{|F|\mu_G(F_n)}> 0. \end{align*} $$

Since $(F_n)_{n \geq 0}$ is bi-Følner, $(F_nF)_{ n \geq 0}$ is a Følner sequence and $\mu _G(F_nF) \sim \mu _G(F_n)$ . Hence, according to Proposition 4.1, there is a proper G-invariant Borel probability measure $\nu $ on $\Omega _S$ . By Lemma 3.2, that $S^5$ is uniformly discrete implies that $S^2$ is an approximate subgroup and, hence, an approximate lattice. The bounds in the statement of Proposition 4.2 are a consequence of the bounds on the size of $F'$ and those found in Lemma 3.2.

Proof. Since $\mu _G$ is bi-invariant, for every $h \in G$ ,

$$ \begin{align*}\int_{\mathcal{F}}\sum_{x \in X} \mathbf{1}_V(hgx)\, d\mu_G(g) & = \sum_{x \in X} \int_{\mathcal{F}}\mathbf{1}_V(hgx)\,d\mu_G(g) \\ & = \sum_{x \in X} \int_{\mathcal{F}x}\mathbf{1}_V(hg)\,d\mu_G(g) \\ & \geq \int_{\mathcal{F}X}\mathbf{1}_V(hg)\,d\mu_G(g) = \mu_G(V). \end{align*} $$

If now $K \subset \mathcal {F}$ , we have

$$ \begin{align*} \int_{K}\sum_{x \in X} \mathbf{1}_V(hgx)\, d\mu_G(g) & \geq \mu_G(V) - \int_{\mathcal{F}\setminus K}\sum_{x \in X} \mathbf{1}_V(hgx)\, d\mu_G(g). \end{align*} $$

However,

$$ \begin{align*}\sum_{x \in X}\mathbf{1}_V(hgx) = | X \cap g^{-1}h^{-1}V |.\end{align*} $$

Additionally, if $\mathbf {1}_V(hgx)> 0$ , then $ x \in g^{-1}h^{-1}V$ . Hence,

$$ \begin{align*}\sum_{x \in X}\mathbf{1}_V(hgx) \leq |X^{-1}X\cap V^{-1}V| = 1.\end{align*} $$

So,

$$ \begin{align*} \int_{K}\sum_{x \in X} \mathbf{1}_V(hgx)\, d\mu_G(g) \geq \mu_G(V) - \mu_G(\mathcal{F}\setminus K). \end{align*} $$

Consider K compact such that $\mu _G(\mathcal {F}\setminus K) \leq \epsilon \mu _G(V)$ . Then,

$$ \begin{align*}\int_{B^{-1}} \int_{K}\sum_{x \in X} \mathbf{1}_V(hgx)\, d\mu_G(g)\, d\mu_G(h)\geq (1-\epsilon)\mu_G(B)\mu_G(V).\end{align*} $$

However, Lemma 3.1 implies

$$ \begin{align*} \int_{B^{-1}}\sum_{x \in X}\mathbf{1}_{V}(hgx)\, d\mu_G(h) \leq |gX \cap BV|\mu_G(V).\end{align*} $$

Therefore,

$$ \begin{align*}\int_K \mu_G(V) |gX \cap BV |\, d\mu_G(g) \geq (1-\epsilon)\mu_G(B)\mu_G(V).\end{align*} $$

Now, if $\sup _{g \in K} |gX \cap BV | = + \infty $ , then the result is certainly satisfied. Otherwise, since $g \mapsto |gX \cap BV |$ takes integer values, we have

$$ \begin{align*}\sup_{g \in K} |gX \cap BV | = \max_{g \in K} |gX \cap BV |.\end{align*} $$

Then,

$$ \begin{align*} \Big(\sup_{g \in K} |gX \cap BV |\Big) \mu_G(K) \geq \int_K |gX \cap BV |\, d\mu_G(g) \geq (1-\epsilon)\mu_G(B),\end{align*} $$

which concludes the proof.

Proof of Proposition 4.6

Let $(K_n)_{n \geq 0}$ be an exhaustion of G by compact subsets and let V be a compact neighbourhood of the identity such that $XX^{-1} \cap V^{-1}V =\{e\}$ . Fix $n \geq 0$ . The set $X^{-1}$ satisfies the hypotheses of Lemma 4.7. Let $K_n'$ be given by Lemma 4.7 applied to $X^{-1}$ and $\epsilon =1/n$ .

Since G is amenable, there is a symmetric compact subset $F_n$ such that

$$ \begin{align*}\mu_G(K_nK_n'F_nVK_n) \leq \bigg(1+\frac{1}{n}\bigg)\mu_G(F_n).\end{align*} $$

Define now $F_n':=K_n^{\prime -1}F_nV$ for every $n \geq 0$ . Then, $(F_n')_{\geq 0}$ is a bi-Følner sequence, so is $(F_n^{\prime -1})_{n \geq 0}$ , and

$$ \begin{align*}\kern-15pt |X \cap F_n^{\prime-1}| = |X^{-1} \cap F_n'| \geq \bigg(1 - \frac{1}{n}\bigg)\frac{\mu_G(F_n)}{\mu_G(\mathcal{F})} \geq \frac{(1 - {1}/{n})}{1 + {1}/{n}}\frac{\mu_G(F_n^{\prime-1})}{\mu_G(\mathcal{F})}.\\[-43pt] \end{align*} $$

Proof of Theorem 1.1

By Proposition 4.6 for some bi-Følner sequence $(F_n)_{n \geq 0}$ ,

$$ \begin{align*} \liminf \frac{|X \cap F_n |}{\mu_G(F_n)} \geq \frac{1}{\mu_G(\mathcal{F})}.\end{align*} $$

So, Theorem 1.1 is a consequence of Proposition 4.2.

Proof. Let $(F_n)_{n \geq 0}$ be a Følner sequence and let $\mu _n$ be the probability measure with density $\mu _G(F_n)^{-1}\mathbf {1}_{F_n}$ against the Haar measure. By the mean ergodic theorem, we know that $\phi _n: Y \rightarrow \int _G |gY \cap V|\,d\mu _n(g)$ converges in the $L^2$ -norm to its average $\int _{\Omega _X} \mathcal {P}_X\mathbf {1}_V(Y)\,d\nu (Y)$ . However, $\mathcal {P}_X\mathbf {1}_V = \mathbf {1}_{U^V}$ . So $\phi _n$ converges in the $L^2$ -norm to $\nu (U^V)$ . Upon considering a subsequence, we may assume that the convergence is $\nu $ -almost everywhere. Therefore, for $\nu $ -almost all Y, we have $\int _G |gY \cap V|\, d\mu _n(g) \rightarrow \nu (U^V)$ . According to Lemma 3.1, we find

$$ \begin{align*}\liminf \frac{|Y \cap VF_n|}{\mu_G(VF_n)} \geq \frac{\nu(U^V)}{\mu_G(V)}=1.\\[-44pt]\end{align*} $$

Proof of Theorem 1.2

According to Proposition 4.6, there is a bi-Følner sequence $(F_n)_{n \geq 0}$ such that

$$ \begin{align*}\liminf \frac{|Y \cap VF_n|}{\mu_G(VF_n)} \geq 1\end{align*} $$

for $\nu $ -almost every $Y \in \Omega _X$ , where $\mu _G:=\mathcal {P}_X^*\nu $ . Fix one such $Y_0 \in \Omega _X$ . By Proposition 4.2 applied to $Y_0$ , there is an approximate subgroup S contained in $Y_0Y_0^{-1}$ and a finite subset F such that $Y_0 \subset SF$ . So, S is an approximate lattice according to Lemma 3.3.

Proof. Let $\nu _0$ denote a G-invariant probability measure on S. Take $F \subset (Y')^{-1}$ finite such that for $\nu _0$ -almost every $s \in S$ , the subsets $\{\Phi (s)f: f \in F\}$ are pairwise disjoint. Let ${V \subset G}$ be a neighbourhood of the identity such that $YX^{-1}XY^{-1} \cap V^{-1}V =\{e\}$ . Therefore, for every $g \in G$ and $X' \in \Omega _X$ , we have $|X'Y^{-1} \cap gV| \leq 1$ for every $X' \in \Omega _X$ . Since F is finite and $G \cdot Y$ is dense in $\Omega _Y$ , there is $h \in G$ such that $F \subset Y^{-1}h$ . Now,

$$ \begin{align*}\int_S \sum_{x \in \Phi(s)F}\mathbf{1}_{Vh}(x)\,d\nu_0(s)& = \int_S \sum_{x \in \Phi(s)Fh^{-1}}\mathbf{1}_{V}(x)\,d\nu_0(s) \\ & \leq \int_S \sum_{x \in \Phi(s)Y^{-1}}\mathbf{1}_{V}(x)\,d\nu_0(s) \\ & \leq 1. \end{align*} $$

However,

$$ \begin{align*} \int_S \sum_{x \in \Phi(s)F}\mathbf{1}_{Vh}(x)\,d\nu_0(s) & = \sum_{f \in F} \int_S \sum_{x \in \Phi(s)f}\mathbf{1}_{Vh}(x)\,d\nu_0(s) \\ & = \sum_{f \in F} \int_S \sum_{x \in \Phi(s)}\mathbf{1}_{Vhf^{-1}}(x)\,d\nu_0(s) \\ & = \sum_{f \in F} \mathcal{P}_X^*\Phi_*\nu_0(Vhf^{-1}) = |F|\mathcal{P}_X^*\Phi_*\nu_0(V), \end{align*} $$

where the interversion in the first line was because of the disjointness property of F and we use in the last line that $\mathcal {P}_X^*\Phi _*\nu _0$ is a Haar measure of a unimodular locally compact group. Thus, $|F| \leq \mathcal {P}_X^*\Phi _*\nu _0(V)^{-1}$ . Hence,

$$ \begin{align*}(Y')^{-1} \subset \bigcup_{s \in S} \Phi^{-1}(s)\Phi(s)F'\end{align*} $$

for some $F' \subset (Y')^{-1}$ of size at most $\mathcal {P}_X^*\Phi _*\nu _0(V)^{-1}$ by the covering lemma (Lemma 2.3).

Suppose now that $\nu _0$ is ergodic. For every $g \in G$ , the set

$$ \begin{align*}S_g:=\{s \in S : g \in \Phi^{-1}(s)\Phi(s)\}\end{align*} $$

is G-invariant and Borel. Indeed, since $\Phi (s)^{-1}\Phi (s)$ is a uniformly discrete subset, we have $g \in \Phi ^{-1}(s)\Phi (s)$ if and only if

$$ \begin{align*}\Phi(s) \in \bigcap_{n \geq 0}\bigcup_{d \in D} U^{dV_n} \cap U^{dV_ng},\end{align*} $$

where $(V_n)_{n \geq 0}$ is a neighbourhood basis at e and D is a dense subset of G. Because $\nu $ is ergodic, we thus have that $S_g$ is either null or co-null. Now, $\bigcup _{s \in S} \Phi ^{-1}(s)\Phi (s)$ is contained in $X^{-1}X$ , which is countable. Define

$$ \begin{align*}\tilde{S}=\bigcap_{g : \nu(S_g)=1}S_g \setminus \bigg( \bigcup_{g : \nu(S_g)=0}S_g\bigg).\end{align*} $$

Then, for every $s \in \tilde {S}$ , we have $\bigcup _{s' \in \tilde {S}} \Phi ^{-1}(s')\Phi (s') = \Phi ^{-1}(s)\Phi (s)$ . Hence,

$$ \begin{align*}Y' \subset \bigcup_{s' \in \tilde{S}} \Phi^{-1}(s')\Phi(s')F' = \Phi^{-1}(s)\Phi(s)F'\end{align*} $$

for any $s \in \tilde {S}$ .

Proof. We have the equivalences,

$$ \begin{align*}g_0 \in T_h(gY,B) \Leftrightarrow g_0gY \in B \Leftrightarrow g_0g \in T_h(Y,B).\end{align*} $$

So, $T_h(gY,B) = T_h(Y,B)g^{-1}$ . This proves part (1). Take $g_1, g_2 \in T_h(Y,B)$ . Set $Y_1:=g_1Y$ and $Y_2:=g_2Y$ . Then, $Y_1,Y_2 \in B$ and $g_1g_2^{-1}Y_2 = Y_1$ . So taking $Y'=Y_2$ works and part (2) is proved. Furthermore, since $e \in Y_1$ , there is $y \in Y_2$ such that $g_1g_2^{-1}y=e$ , that is, $g_1g_2^{-1} \in Y_2^{-1}$ . However, $Y_2 \in B$ so the first inclusion of part (3) is proved. If, now, we take any $Z \in B$ , then $Z \in \mathcal {T}_e$ . In other words, $e \in Z$ . By [Reference Björklund and HartnickBH18], we have $Z \subset X^{-1}X$ . This proves part (3). Finally, remark that

$$ \begin{align*}\{Y \in \Omega_X : T(Y,B) \cap W \neq \emptyset \} = \{Y \in \Omega_X: WY \cap B \neq \emptyset\} = W^{-1}B.\end{align*} $$

And part (4) is proved.

Proof. By Lemma 5.1, it suffices to show that

$$ \begin{align*} \Omega_X & \longrightarrow \mathcal{C}(G) \\ Y & \longmapsto T_h(Y,B)^{-1} \end{align*} $$

is a well-defined G-equivariant Borel map, takes values in $\Omega _{X'}$ for $X'$ such that $X'Z_0$ is uniformly discrete and takes non-empty values almost always. We will see this as a consequence of Lemma 5.6. Part (1) of Lemma 5.6 implies G-equivariance and part (2) implies measurability. Since $\nu $ is ergodic, there is $Y'$ such that for almost every $Y \in \Omega _X$ , $T_h(Y,B)^{-1} \in \Omega _{X'}$ with $X'=T_h(Y',B)$ . In particular, $X^{\prime -1}X' \subset X^{-1}X$ so $X'Z_0^{-1}$ is uniformly discrete. It remains to prove that $T_h(Y,B)$ takes non-empty values for $\nu $ -almost every Y. However, $T_h(Y,B)$ is non-empty as soon as $Y \in G\cdot B$ . Since B cannot be null, $G\cdot B$ is a G-invariant Borel subset of $\Omega _X$ of full measure and, hence, $\nu (G \cdot B)=1$ as $\nu $ is ergodic.

Proof. We explain the proof under assumption (ii), the proof of part (1) will then follow with minor changes. By a Fubini argument, for $\nu $ -almost every $Y \in \Omega _X$ and almost every $g \in G$ , we have

(5.1) $$ \begin{align} \tau \circ \alpha_s(g,Y) = \phi(gY)^{-1}\pi(g)\phi(Y). \end{align} $$

According to [Reference ZimmerZim84, Lemma B.8], we can therefore find a G-invariant Borel subset $\Omega \subset \Omega _X$ of measure $1$ and a Borel map $s':\Omega \rightarrow G$ such that for every $Y \in \Omega $ and almost every $g \in G$ , (5.1) holds at $(g,s'(Y)Y)$ —in other words,

(5.2) $$ \begin{align} \tau \circ \alpha_s(g,s'(Y)Y) = \phi(gs'(Y)Y)^{-1}\pi(g)\phi(s'(Y)Y). \end{align} $$

Note moreover that $s'(Y)=e$ for almost every $e \in \Omega $ . By a further Fubini argument, the G-invariant subset $\{Y \in \Omega : s'(gY)=e \text { for almost every } g \in G\}$ is Borel and co-null in $\Omega $ . So, we may assume that it is equal to $\Omega $ . For every $Y \in \Omega $ , set $s"(Y)=s'(Y)^{-1}s(s'(Y)Y)$ . Since s is a Borel section, we get

$$ \begin{align*}s"(Y) =s'(Y)^{-1}s(s'(Y)Y) \in s'(Y)^{-1}s'(Y)Y = Y.\end{align*} $$

So $s"$ is a Borel section as well. Fix $Y \in \Omega $ . For almost every $g \in G$ , we have $s'(gY)=e$ . Hence, (5.2) holds at $(s'(gY)gs'(Y)^{-1},s'(Y)Y)$ for almost every $g\in G$ . So, for every $Y \in \Omega $ and almost every $g \in G$ , (5.2) becomes

(5.3) $$ \begin{align} \tau\circ \alpha_{s"}(g,Y) =\tau\circ \alpha_{s}(s'(gY)gs'(Y)^{-1},Y) = \psi(gY)^{-1}\pi(g)\psi(Y), \end{align} $$

where $\psi (Y) := \pi (s'(Y))^{-1}\phi (s'(Y)Y)$ . Therefore, for every $Y \in \Omega $ and $h \in G$ , we have

$$ \begin{align*} \pi(h)\psi(Y) & = \pi(g)^{-1}\psi(ghY)\tau\circ \alpha_{s"}(gh,Y) \ \qquad\qquad\qquad\text{for almost every } g \in G \\ & = \pi(g)^{-1}\psi(ghY)\tau\circ \alpha_{s"}(g,hY)\tau\circ \alpha_{s"}(h,Y)\quad\text{for almost every } g \in G \\ & = \psi(hY)\tau\circ \alpha_{s"}(h,Y), \end{align*} $$

where we have used (5.3) in the first line, the cocycle identity to deduce the second line and (5.3) again to get to the last line. Notice now that for every $Y \in \Omega $ , we have $s(s'(Y)Y)^{-1}s'(Y)s(Y) \in Y^{-1}Y \subset X^{-1}X$ . So, we can unfold the definitions of $s"$ and $\alpha _{s"}$ to find that, for every $g \in G$ and every $Y \in \Omega $ , we have

$$ \begin{align*}\tau\circ \alpha_{s}(g,Y) = \chi(gY)^{-1}\pi(g)\chi(Y),\end{align*} $$

where

$$ \begin{align*}\chi(Y):=\pi(s'(Y))^{-1}\phi(Y)\tau(s(s'(Y)Y)^{-1}s'(Y)s(Y)).\end{align*} $$

Finally, since for $Y \in \Omega \cap \mathcal {T}_e$ and $g \in T_{h}(Y,\Omega \cap \mathcal {T}_e)$ we have $s(Y)=s(gY)=e$ , we get Lemma 5.9.

Proof. Let $\mu _G$ denote the Haar measure on G. By a Fubini-type argument, there is $\Omega _0 \subset \Omega _{X}$ of full measure such that for every $Y \in \Omega _0$ and almost every $ g \in G$ , we have $ \alpha _s(g,Y)\phi (Y) = \phi (gY)$ . By [Reference ZimmerZim84, Lemma B.8], we can find $\Omega \subset \Omega _X G$ -invariant and of measure $1$ and a Borel map $s':\Omega \rightarrow G$ such that for every $Y \in \Omega $ and almost every $g \in G$ , we have

$$ \begin{align*} \alpha_s(g,s'(Y)Y)\phi(s'(Y)Y) = \phi(gs'(Y)Y).\end{align*} $$

Set $s"(Y):=s'(Y)^{-1}s(s'(Y)Y)$ for every $Y \in \Omega $ . Then, $s"$ is a Borel section of $\Omega $ (Definition 5.8) that satisfies $s"(Y)=e$ for every $Y \in \Omega \cap \mathcal {T}_e$ . Furthermore, one can ensure—proceeding as in the proof of Lemma 5.9—that for every $Y \in \Omega $ and almost every $g \in G$ , we have

$$ \begin{align*}\alpha_{s"}(g,Y) \phi(s'(Y)Y)= \phi(s'(gY)gY).\end{align*} $$

Let $\psi : \Omega \rightarrow B$ be defined by $\psi (Y)= \phi (s'(Y)Y)$ . For every $Y \in \Omega $ , we have

$$ \begin{align*}\alpha_{s"}(g,Y)^{-1}\psi(gY)=\psi(Y)\end{align*} $$

for almost every $g \in G$ . Moreover, we have for every $Y \in \Omega $ and $h \in G$ ,

$$ \begin{align*}\psi(hY) & = \alpha_{s"}(g,hY)^{-1}\psi(ghY) \quad \qquad\qquad \text{for almost every } g \in G\\ & = \alpha_{s"}(h,Y)\alpha_{s"}(gh,Y)^{-1}\psi(ghY) \quad \text{for almost every } g \in G \\ & = \alpha_{s"}(h,Y)\psi(Y), \end{align*} $$

where we went from the first line to the second using the cocycle identity. We conclude as in the proof of Lemma 5.9.

Proof. Fix a Borel section $s: \Omega _X \rightarrow G$ . Since $\dim _k(\mathbb {H}) \leq \dim (G)$ and G is a property (T) simple algebraic group, we can apply [Reference MachadoMac23, Proposition 14]. So, $\tau \circ \alpha _s$ satisfies the conclusions of Zimmer’s cocycle superrigidity. Therefore, we may apply Lemma 5.9.

If we have a measurable map $\psi : \Omega \cap \mathcal {T}_e \rightarrow \mathbb {H}(k)$ and a compact subgroup $K\subset \mathbb {H}(k)$ such that $\psi (gY)\tau (g)\psi (Y)^{-1} \subset K$ for every $Y \in \Omega \cap \mathcal {T}_e$ and $g \in T_h(Y, \Omega \cap \mathcal {T}_e)$ (part (1) of Lemma 5.9), we will prove that condition (1) of Theorem 6.1 holds. There is $B \subset \mathcal {T}_e$ non-null such that $K':=\overline {\psi (B)}$ is compact (Lemma 5.4). So, if we have $g \in G$ , $Z \in B$ and $gZ \in B$ , then $\tau (g) \in K^{\prime -1}KK'$ . Therefore, for every $ Y \in \Omega _X$ , we have

$$ \begin{align*}\tau(T(Y,B)T(Y,B)^{-1}) \subset K^{\prime-1}KK'.\end{align*} $$

However, X is covered by finitely many translates of $\bigcup _Y T(Y,B)T(Y,B)^{-1}$ (Lemma 5.7), so $\tau (X)$ is relatively compact.

Suppose that we are given a measurable map $\psi : \Omega \cap \mathcal {T}_e \rightarrow \mathbb {H}(k)$ and a continuous group homomorphism $\pi : G \rightarrow \mathbb {H}(k)$ such that for every $Y \in \Omega \cap \mathcal {T}_e$ and $g \in T_h(Y, \Omega \cap \mathcal {T}_e)$ , we have $\tau (g) = \psi (gY) ^{-1}\pi (g) \psi (Y)$ (corresponding to part (2) of Lemma 5.9). We will prove that condition (2) of Theorem 6.1 must hold. We will follow the strategy of [Reference MachadoMac23]. By Lemma 5.4, there is $h \in \mathbb {H}(k)$ such that for every compact neighbourhood V of h, there is $B_V \subset \mathcal {T}_e$ non-null such that $\psi (B_V) \subset V$ . Upon modifying $\psi $ , we may assume without loss of generality that $h=e$ . Now, for any $Y_0 \in \mathcal {T}_e$ and any $Y \in \Omega _X$ , we have that $YY_0^{-1}$ is uniformly discrete. Therefore, for any V neighbourhood of e, we know by Lemma 5.7 that there is a finite set $F=F(Y_0,V)$ such that

$$ \begin{align*}Y_0 \subset FT_h(Y,B_V)T_h(Y,B_V)^{-1}.\end{align*} $$

Take $y \in Y_0$ , then there is $f \in F$ such that $f^{-1}y \in T_h(Y,B_V)T_h(Y,B_V)^{-1}$ . By Lemma 5.6, there is $Y_1 \in B_V$ such that $f^{-1}yY_1 \in B_V$ as well. In particular, this yields

$$ \begin{align*}\tau(y) = \tau(f) \tau(f^{-1}y) =\tau(f)\psi(f^{-1}yY_1)\pi(f^{-1}y)\psi(Y_1) \in \tau(F)V\pi(F)^{-1}\pi(y)V.\end{align*} $$

Write $K_V = \tau (F)V\pi (F)^{-1}$ , then we have the equality

$$ \begin{align*}\tau(y) \in K_V \pi(y) V\end{align*} $$

for every $y \in Y_0$ .

We have shown in [Reference MachadoMac23, §3.4] that if $\gamma \in G$ satisfies that the projection of $\pi (Y_0 \cap Y_0\gamma ^{-1})$ to $K \setminus H$ has dense closure in the (right-)visual boundary $\partial (K \setminus H)$ —where K denotes a maximal compact subgroup—then $\tau (\gamma )=\pi (\gamma )$ . Since $\pi $ is surjective according to our assumption on dimensions, it in fact suffices to show: (*) the projection of $Y_0 \cap Y_0\gamma ^{-1}$ to $K \setminus G$ has dense closure in the (right-)visual boundary $\partial (K \setminus G)$ —where K denotes a maximal compact subgroup of G this time. We will show that this condition is fulfilled for many $\gamma \in G$ by invoking pointwise ergodic theorems.

Note first that the map

$$ \begin{align*} I_{\gamma}: \Omega_X & \longrightarrow \Omega_X^{\mathrm{ext}} \\ Y & \longmapsto Y \cap Y\gamma^{-1} \end{align*} $$

is a well-defined G-equivariant map. Let us show that it is Borel when $\gamma \in X$ . As a consequence of the uniform discreteness of $XX^{-1}$ , there is a neighbourhood of the identity $W \subset G$ such that for every $Y \in \Omega _X$ and every $g \in G$ , we have $|YX^{-1} \cap gW| \leq 1$ . So,

$$ \begin{align*} |Y \cap gW| + |Y\gamma^{-1} \cap gW| - |Y \cap Y\gamma^{-1} \cap gW| &= |Y \cup Y\gamma^{-1} \cap gW| \\ & \leq |YX^{-1} \cap gW| \\ & \leq 1. \end{align*} $$

Therefore, $ Y \cap Y\gamma ^{-1} \cap gW$ is non-empty if and only if both $Y \cap gW$ and $Y\gamma ^{-1} \cap gW$ are non-empty. In other words,

$$ \begin{align*}\{Y \in \Omega_X : Y \cap Y\gamma^{-1} \cap gW \neq \emptyset \} = U^{gW} \cap U^{gW\gamma}.\end{align*} $$

So, $I_{\gamma }$ is Borel (see e.g. [Reference MachadoMac23, §2.1.2]). As a consequence, the push-forward $\nu _{\gamma }$ of $\nu $ is a G-invariant ergodic measure. It is either proper or concentrated on $\{\emptyset \}$ .

Fix $D \subset G$ countable such that the sets of $\xi \in \partial (K \setminus G)$ such that $ x_0 \cdot d^n \rightarrow \xi $ is dense. Here, $x_0$ denotes the class of e in $(K \setminus G)$ . Take $\gamma \in G$ and suppose that $\nu _{\gamma }$ is proper. Then, the semi-group generated by $d^{-1}$ acts ergodically on $(\Omega _X,\nu )$ according to the Howe–Moore property (e.g. [Reference ZimmerZim84, 2.2.20]. So, for every $W \subset G$ open and relatively compact, and almost every $Y \in \Omega _X$ , we have that

$$ \begin{align*}\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=0}^n |d^{-k}(Y \cap Y\gamma^{-1}) \cap W | = \nu(U^W).\end{align*} $$

In particular, there is a sequence of elements $(y_n)$ of $Y \cap Y\gamma ^{-1}$ such that $d(y_n,e) \rightarrow \infty $ as n goes to $\infty $ and $y_n \in d^{i_n}W$ for some $i_n \geq 0$ . By considering the above for W running over a countable basis of neighbourhoods of the identity, we find that $\gamma $ satisfies condition (*) for almost every $Y \in \Omega _X$ .

Let us now show that there are many $\gamma \in G$ such that $\nu _{\gamma }$ is proper. Define

$$ \begin{align*}S := \{ x \in X^{-1}X : \nu_x \text{ is proper}\}\end{align*} $$

and fix W a relatively compact open neighbourhood of the identity. If $X^{-1}$ is not covered by finitely many right-translates of S, then we can find a sequence $(x_n)_{n \geq 0}$ of elements of $X^{-1}$ such that for any two $i \neq j$ , $\nu _{x_ix_j^{-1}}$ is concentrated on $\{\emptyset \}$ . In particular,

$$ \begin{align*}\int_{\Omega_X} |Yx_j \cap Yx_i \cap W|\,d\nu(Y) = \int_{\Omega_X} |Y \cap Yx_ix_j^{-1} \cap Wx_j^{-1}|\,d\nu(Y) = 0\end{align*} $$

for every $i,j \geq 0$ . However, for every $N \geq 0$ and $Y \in \Omega _X$ , we have

(6.1) $$ \begin{align} |YX^{-1} \cap W | \geq \bigg|\kern-1.5pt\bigcup_{i=0}^N Yx_i \cap W\bigg| \geq \sum_{i=0}^N |Yx_i \cap W| - \sum_{0\leq i < j \leq N}|Yx_j \cap Yx_i \cap W|. \end{align} $$

Write $\mu _G$ the pull-back of $\nu $ via the periodization map. We know that $\mu _G$ is a Haar measure. In addition, since $XX^{-1}$ is uniformly discrete, $|YX^{-1} \cap W | \leq C$ for some constant C. Integrating (6.1) against $\nu $ , we find

$$ \begin{align*} C \geq \sum_{i=0}^N \mu_G(Wx_i^{-1}) = (N+1)\mu_G(W)\end{align*} $$

for every $N \geq 0$ . This is a contradiction. So, there is $F \subset X^{-1}$ finite such that $X^{-1} \subset SF$ . Since $\tau _{|S} = \pi _{|S}$ , we conclude that S is the subset for which we are looking.

Proof. Following Zimmer’s approach [Reference ZimmerZim84] (see also [Reference MachadoMac23]), we can find a non-trivial $\phi : \Omega _X \rightarrow G$ such that for every $g \in G$ and $\nu $ -almost every $Y \in \Omega _X$ , we have

$$ \begin{align*} \pi (\alpha_s(g,Y)) (\phi(Y)) = \phi(gY).\end{align*} $$

In particular, $||\phi (Y)|| =||\phi (gY)||$ . So $||\phi (Y)||$ is constant for almost every Y by ergodicity of $\nu $ . We may therefore assume that $\phi $ takes values in the set of unit vectors of $\mathcal {H}$ . By Lemma 5.11, there are thus $\Omega \subset \Omega _X$ that is G-invariant and of full measure, and $\psi : \Omega \cap \mathcal {T}_e \rightarrow \mathcal {H}$ measurable such that

$$ \begin{align*} \pi(g)\cdot \psi(Y) = \psi(gY)\end{align*} $$

for every $Y \in \Omega \cap \mathcal {T}_e$ and $g\in T_h(Y, \Omega \cap \mathcal {T}_e)$ .

Take $\xi \in \mathcal {H}$ as provided by Lemma 5.4. Then, for every $\epsilon> 0$ , the subset

$$ \begin{align*}\mathcal{T}_{e,\epsilon}:=\psi^{-1}(B(\xi, \epsilon))\end{align*} $$

is non-null. According to Lemma 5.7,

$$ \begin{align*}X \subset FT_h(Y, \mathcal{T}_{e,\epsilon})T_h(Y, \mathcal{T}_{e,\epsilon})^{-1}\end{align*} $$

for some finite subset F and some $Y \in \Omega $ . For every $g \in T_h(Y, \mathcal {T}_{e,\epsilon })T_h(Y, \mathcal {T}_{e,\epsilon })^{-1}$ , we can find $Y' \in \mathcal {T}_{e,\epsilon }$ such that $gY' \in \mathcal {T}_{e,\epsilon }$ . So,

$$ \begin{align*}\psi(gY') = \pi(g)(\psi(Y')).\end{align*} $$

Since $\psi (Y'), \psi (gY') \in B(\xi , \epsilon )$ , we have $\pi (g)(\xi ) \in B(\xi , 2\epsilon )$ . To sum up, we have $\pi (X)(\xi ) \subset \pi (F)(B(\xi , 2\epsilon ))$ . Since there is such an F for every $\epsilon $ and $\mathcal {H}$ is complete, $\pi (X)(\xi )$ is relatively compact.

Proof of Proposition 6.2

The proof of [Reference MachadoMac23, Proposition 13] holds almost verbatim. We reproduce it here for the sake of completeness. Let $\mathcal {H}$ denote $\bigoplus _{\Delta \subset \langle X \rangle \text { f.g. }} L^2(\langle X \rangle /\Delta )$ . The sum of quasi-regular representations $(\pi , \mathcal {H})$ almost has invariant vectors (e.g. [Reference Bekka, de la Harpe and ValetteBdlHV08]). So, we can find $\phi \in \mathcal {H}$ such that $X \cdot \phi $ has compact closure in the norm topology. Upon projecting $\phi $ to a well-chosen factor of $\mathcal {H}$ , we may assume that ${\phi \in L^2(\langle X \rangle /\Delta )}$ for some finitely generated subgroup $\Delta $ of $\langle X \rangle $ . Now, define

$$ \begin{align*}X(\delta, \phi):=\{x \in X^{-1}X : ||x\cdot \phi - \phi|| \leq \delta\}.\end{align*} $$

Then, $(X(\delta , \phi ))_{\delta> 0}$ is a family of subsets such that X is covered by finitely many translates of $X(\delta , \phi )$ and $X(\delta , \phi )$ is contained in $X^{-1}X$ (see [Reference MachadoMac23]). Now, let $p: \langle X\rangle \rightarrow \langle X\rangle /\Delta $ denote the natural projection. Take $\gamma \in \langle X \rangle $ such that $\phi (p(\gamma ))=\alpha> 0$ . So, for every $x \in X(\alpha /2,\phi )$ , we have

meaning $x^{-1}\gamma \in \phi ^{-1}([\alpha /2; +\infty ))$ . Since $\phi ^{-1}([\alpha /2; +\infty ))$ is finite, we can find a finite set F of representatives of $\phi ^{-1}([\alpha /2; +\infty ))$ in $\langle X \rangle $ . Then, $x^{-1} \gamma \Delta \cap F \Delta \neq \emptyset $ and $X(\alpha /2,\phi )$ is contained in $F\Delta \gamma ^{-1}$ . However, there is a finite subset $F' \subset \langle \Lambda \rangle $ such that $X \subset F'X(\alpha /2, \phi ) \subset F'F\Delta \gamma ^{-1}$ . Since

$$ \begin{align*}X \subset F' \cup F \cup \Delta \cup \{\gamma^{-1}\} \subset \langle X \rangle,\end{align*} $$

$F' \cup F \cup \Delta \cup \{\gamma ^{-1}\}$ generates $\langle X \rangle $ . However, $\Delta $ is finitely generated. So, $\langle X \rangle $ is finitely generated.

Proof. Suppose as we may that G is a closed subgroup of $\operatorname {\mathrm {GL}}_n(k)$ for some n. According to Proposition 6.2, the subgroup $\langle X \rangle $ is finitely generated. So the field K generated by entries of elements of $\langle X \rangle $ is finitely generated. Let $S \subset XX^{-1}$ be the subset provided by Theorem 6.1. We will show that $S^m$ is uniformly discrete for every $m \geq 0$ . Let W be a compact neighbourhood of the identity and take an integer $m \geq 0$ . If $S^{2m} \cap W$ is infinite, then the set E of entries of elements in $S^{2m} \cap W$ is infinite as well. By a result of Breuillard and Gelander [Reference Breuillard and GelanderBG03], there is a local field l and a field embedding $\sigma : K \rightarrow l$ such that $\sigma (E)$ is unbounded. Denote by $\sigma _0$ the natural group homomorphism $\operatorname {\mathrm {GL}}_m(K) \rightarrow \operatorname {\mathrm {GL}}_m(l)$ given by applying $\sigma $ entry-wise. Then, $\sigma _0(S^{2m} \cap W)$ is unbounded. As

$$ \begin{align*}\sigma_0(S^{2m} \cap W) \subset \sigma_0((XX^{-1})^{2m}),\end{align*} $$

the subset $\sigma _0(X)$ must be unbounded. Moreover, the Zariski-closure of $\sigma _0(\langle X \rangle )$ is semi-simple and has dimension at most $\dim G$ . By Theorem 6.1, there is a continuous group homomorphism $\pi : G \rightarrow \operatorname {\mathrm {GL}}_m(l)$ such that $(\sigma _0)_{|\langle S \rangle } = \pi _{|\langle S \rangle }$ . However,

$$ \begin{align*}\sigma_0(S^{2m} \cap W ) = \pi(S^{2m} \cap W ) \subset \pi(W).\end{align*} $$

So, $\sigma _0(S^{2m} \cap W )$ is bounded. This is a contradiction.

Therefore, $S^{2m} \cap W$ is finite for every $m \geq 0$ . So, $S^m$ is uniformly discrete for every $m \geq 0$ . However, there is F finite such that $X \subset FS$ . By Lemma 3.3, there is thus a Borel subset $\mathcal {F}$ of finite Haar measure such that $S^2 \mathcal {F} = G$ . Since $S^5$ is uniformly discrete, we find that $S^2$ is an approximate subgroup as a consequence of Lemma 3.2. Hence, $S^2$ is an approximate lattice.