Proof. By [Reference Widmer21, Theorem 5.4] the set $S$ is measurable, and moreover,

(3.1) $$\begin{eqnarray}\biggl|\#(S\cap \unicode[STIX]{x1D6EC})-\frac{\operatorname{Vol}S}{\det \unicode[STIX]{x1D6EC}}\biggr|\leqslant D^{3D^{2}/2}M\max _{1\leqslant j<D}\biggl\{1,\frac{L^{j}}{\unicode[STIX]{x1D706}_{1}\cdots \unicode[STIX]{x1D706}_{j}}\biggr\}.\end{eqnarray}$$

First suppose $L\geqslant \unicode[STIX]{x1D706}_{1}$ . Then the lemma follows immediately from (3.1). Next we assume $L<\unicode[STIX]{x1D706}_{1}$ . We distinguish two subcases. First suppose $S\cap \unicode[STIX]{x1D6EC}\neq \emptyset$ . Then

$$\begin{eqnarray}\max _{1\leqslant j<D}\biggl\{1,\frac{L^{j}}{\unicode[STIX]{x1D706}_{1}\cdots \unicode[STIX]{x1D706}_{j}}\biggr\}=1=1^{\ast }(S\cap \unicode[STIX]{x1D6EC})\leqslant \biggl(\frac{L}{\unicode[STIX]{x1D706}_{1}}\biggr)^{D-1}+1^{\ast }(S\cap \unicode[STIX]{x1D6EC}).\end{eqnarray}$$

Now suppose $S\cap \unicode[STIX]{x1D6EC}=\emptyset$ . As $L<\unicode[STIX]{x1D706}_{1}$ we get, using Minkowski’s second theorem,

$$\begin{eqnarray}\biggl|\#(S\cap \unicode[STIX]{x1D6EC})-\frac{\operatorname{Vol}S}{\det \unicode[STIX]{x1D6EC}}\biggr|=\frac{\operatorname{Vol}S}{\det \unicode[STIX]{x1D6EC}}\leqslant \frac{(2L)^{D}}{\unicode[STIX]{x1D706}_{1}\cdots \unicode[STIX]{x1D706}_{D}}\leqslant 2^{D}\biggl(\frac{L}{\unicode[STIX]{x1D706}_{1}}\biggr)^{D-1}.\end{eqnarray}$$

This proves the lemma. ◻

Proof. We have

$$\begin{eqnarray}\unicode[STIX]{x1D719}(Z_{\mathbf{Q}})\subset \unicode[STIX]{x1D719}(B_{\mathbf{y}_{1}}(Q_{1})\times \cdots \times B_{\mathbf{y}_{n}}(Q_{n}))=B_{\unicode[STIX]{x1D703}_{1}\mathbf{y}_{1}}(\overline{Q})\times \cdots \times B_{\unicode[STIX]{x1D703}_{n}\mathbf{y}_{n}}(\overline{Q}),\end{eqnarray}$$

and hence, $\unicode[STIX]{x1D719}(Z_{\mathbf{Q}})\subset B_{\unicode[STIX]{x1D719}\text{}\underline{\mathbf{y}}}(n^{1/2}\overline{Q})$ . As $Z_{\mathbf{Q}}\in {\mathcal{F}}_{\unicode[STIX]{x1D705},M}$ the claim follows.◻

Proof. Since

$$\begin{eqnarray}\#(Z_{\mathbf{Q}}\cap \unicode[STIX]{x1D6EC})=\#(\unicode[STIX]{x1D719}Z_{\mathbf{Q}}\cap \unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC})\quad \text{and}\quad \operatorname{Vol}Z_{\mathbf{Q}}/\det \unicode[STIX]{x1D6EC}=\operatorname{Vol}\unicode[STIX]{x1D719}Z_{\mathbf{Q}}/\det \unicode[STIX]{x1D719}\unicode[STIX]{x1D6EC}\end{eqnarray}$$

this follows immediately from Lemmas 3.1 and 4.1. ◻

Proof. By (4.1) we have $\unicode[STIX]{x1D703}_{i}\geqslant 1$ (for all $i\notin I$ ). Moreover, if $\text{}\underline{\mathbf{x}}\in \unicode[STIX]{x1D6EC}\cap C_{I}$ then $\mathbf{x}_{i}=\mathbf{0}$ (for all $i\in I$ ), and thus

$$\begin{eqnarray}|\unicode[STIX]{x1D719}(\text{}\underline{\mathbf{x}})|^{2}=\mathop{\sum }_{\substack{ 1\leqslant i\leqslant n \\ i\notin I}}|\unicode[STIX]{x1D703}_{i}\mathbf{x}_{i}|^{2}\geqslant \mathop{\sum }_{\substack{ 1\leqslant i\leqslant n \\ i\notin I}}|\mathbf{x}_{i}|^{2}=|\text{}\underline{\mathbf{x}}|^{2}.\end{eqnarray}$$

Hence, if $\text{}\underline{\mathbf{x}}\in \unicode[STIX]{x1D6EC}\cap C_{I}$ and $\text{}\underline{\mathbf{x}}\neq 0$ then $|\unicode[STIX]{x1D719}(\text{}\underline{\mathbf{x}})|\geqslant \unicode[STIX]{x1D706}_{1}(\unicode[STIX]{x1D6EC}\cap C_{I})$ .

Now suppose that $\text{}\underline{\mathbf{x}}\in \unicode[STIX]{x1D6EC}\backslash C_{I}$ . If $\text{}\underline{\mathbf{z}}$ is an arbitrary point in $\mathbf{R}^{N}$ then, by the weighted arithmetic geometric mean inequality, we have

$$\begin{eqnarray}|\text{}\underline{\mathbf{z}}|^{2}=\mathop{\sum }_{i=1}^{n}|\mathbf{z}_{i}|^{2}\geqslant \frac{1}{\max _{i}\unicode[STIX]{x1D6FD}_{i}}\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D6FD}_{i}|\mathbf{z}_{i}|^{2}\geqslant \frac{t}{\max _{i}\unicode[STIX]{x1D6FD}_{i}}\biggl(\mathop{\prod }_{i=1}^{n}|\mathbf{z}_{i}|^{2\unicode[STIX]{x1D6FD}_{i}}\biggr)^{1/t}\geqslant \operatorname{Nm}_{\boldsymbol{\unicode[STIX]{x1D6FD}}}(\text{}\underline{\mathbf{z}})^{2/t},\end{eqnarray}$$

and thus

(4.3) $$\begin{eqnarray}|\text{}\underline{\mathbf{z}}|\geqslant \operatorname{Nm}_{\boldsymbol{\unicode[STIX]{x1D6FD}}}(\text{}\underline{\mathbf{z}})^{1/t}.\end{eqnarray}$$

Using (4.3) and (4.2) we conclude that

$$\begin{eqnarray}|\unicode[STIX]{x1D719}(\text{}\underline{\mathbf{x}})|\geqslant \operatorname{Nm}_{\boldsymbol{\unicode[STIX]{x1D6FD}}}(\unicode[STIX]{x1D719}\text{}\underline{\mathbf{x}})^{1/t}=\operatorname{Nm}_{\boldsymbol{\unicode[STIX]{x1D6FD}}}(\text{}\underline{\mathbf{x}})^{1/t}.\end{eqnarray}$$

First suppose that $|\text{}\underline{\mathbf{x}}|<B$ . Then we have by the definition of $\unicode[STIX]{x1D708}(\cdot ,\cdot )$

$$\begin{eqnarray}\operatorname{Nm}_{\boldsymbol{\unicode[STIX]{x1D6FD}}}(\text{}\underline{\mathbf{x}})^{1/t}\geqslant \unicode[STIX]{x1D708}(\unicode[STIX]{x1D6EC},B),\end{eqnarray}$$

and hence $|\unicode[STIX]{x1D719}(\text{}\underline{\mathbf{x}})|\geqslant \unicode[STIX]{x1D708}(\unicode[STIX]{x1D6EC},B)$ . Now suppose $|\text{}\underline{\mathbf{x}}|\geqslant B$ . Then we have

$$\begin{eqnarray}\displaystyle |\unicode[STIX]{x1D719}(\text{}\underline{\mathbf{x}})| & = & \displaystyle \unicode[STIX]{x1D703}_{\min }|(\unicode[STIX]{x1D703}_{1}\mathbf{x}_{1}/\unicode[STIX]{x1D703}_{\min },\ldots ,\unicode[STIX]{x1D703}_{n}\mathbf{x}_{n}/\unicode[STIX]{x1D703}_{\min })|\geqslant \unicode[STIX]{x1D703}_{\min }|(\mathbf{x}_{1},\ldots ,\mathbf{x}_{n})|\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D703}_{\min }|\text{}\underline{\mathbf{x}}|\geqslant \unicode[STIX]{x1D703}_{\min }B.\nonumber\end{eqnarray}$$

This proves the lemma. ◻

Proof. First assume $A=I_{D}$ so that $\unicode[STIX]{x1D6EC}=\mathbf{Z}^{D}$ . Suppose $v=(a_{1},\ldots ,a_{D})\in \mathbf{Z}^{D}$ is non-zero, $kv\in B_{P}(R)$ and $P=(x_{1},\ldots ,x_{D})$ . Then $ka_{i}$ lies in $[x_{i}-R,x_{i}+R]$ for $1\leqslant i\leqslant D$ . As $v\neq \mathbf{0}$ there exists an $i$ with $a_{i}\neq 0$ . We conclude that $k$ is a divisor of some non-zero integer in $[x_{i}-R,x_{i}+R]$ . There are at most $2R+1$ integers in this interval, each of which of modulus at most $R+|P|$ . Hence the number of possibilities for $k$ is ${\leqslant}T(R+|P|)(2R+1)$ . This proves the lemma for $A=I_{D}$ . Next note that

$$\begin{eqnarray}\#(B_{P}(R)\backslash \{\mathbf{0}\}\cap k\unicode[STIX]{x1D6EC})=\#(A^{-1}B_{P}(R)\backslash \{\mathbf{0}\}\cap k\mathbf{Z}^{D}).\end{eqnarray}$$

Hence, the general case follows from the case $A=I_{D}$ upon noticing $A^{-1}B_{P}(R)\subset B_{A^{-1}(P)}(R\Vert A^{-1}\Vert )$ , and $|A^{-1}(P)|\leqslant \Vert A^{-1}\Vert |P|$ .◻

Proof. Any lattice $\unicode[STIX]{x1D6EC}$ in $\mathbf{R}^{D}$ has a basis $v_{1},\ldots ,v_{D}$ with

$$\begin{eqnarray}\frac{|v_{1}|\cdots |v_{D}|}{|\text{det}[v_{1}\ldots v_{D}]|}\leqslant D^{2D},\end{eqnarray}$$

see, e.g., [Reference Widmer21, Lemma 4.4]. Let $A$ be the matrix that sends the canonical basis $e_{1},\ldots ,e_{D}$ to $v_{1},\ldots ,v_{D}$ . Now suppose $A^{-1}$ sends $e_{i}$ to $(\unicode[STIX]{x1D71A}_{1},\ldots ,\unicode[STIX]{x1D71A}_{D})$ then by Cramer’s rule

$$\begin{eqnarray}|\unicode[STIX]{x1D71A}_{j}|=\biggl|\frac{\det [v_{1}\ldots e_{i}\ldots v_{D}]}{\det [v_{1}\ldots v_{j}\ldots v_{D}]}\biggr|\leqslant \frac{|\text{det}[v_{1}\ldots e_{i}\ldots v_{D}]|}{|v_{1}|\cdots |v_{j}|\cdots |v_{D}|}D^{2D}.\end{eqnarray}$$

Now we apply Hadamard’s inequality to obtain

$$\begin{eqnarray}\frac{|\text{det}[v_{1}\ldots e_{i}\ldots v_{D}]|}{|v_{1}|\cdots |v_{j}|\cdots |v_{D}|}\leqslant \frac{|v_{1}|\cdots |e_{j}|\cdots |v_{D}|}{|v_{1}|\cdots |v_{i}|\cdots |v_{D}|}=\frac{1}{|v_{i}|}\leqslant \frac{1}{\unicode[STIX]{x1D706}_{1}}.\end{eqnarray}$$

Next we use that for a $D\times D$ matrix $[a_{ij}]$ with real entries we have $\Vert [a_{ij}]\Vert \leqslant D\max _{ij}|a_{ij}|$ , and this proves the lemma.◻

Proof. Note that $\sum _{k=1}^{\infty }1^{\ast }(B_{P}(R)\backslash \{\mathbf{0}\}\,\cap \,k\unicode[STIX]{x1D6EC})=\#\{k\in \mathbf{N};B_{P}(R)\backslash \{\mathbf{0}\}\,\cap \,k\unicode[STIX]{x1D6EC}\neq \emptyset \}$ . Hence, the lemma follows immediately from Lemmas 5.1 and 5.2.◻

Proof. Lemma 4.1 implies that $\unicode[STIX]{x2202}Z_{\mathbf{Q}}^{\ast }\in \operatorname{Lip}(N,M+1,L)$ with $L=2n^{1/2}\unicode[STIX]{x1D705}\overline{Q}$ . As noted in the proof of the latter lemma we have $\unicode[STIX]{x1D719}(Z_{\mathbf{Q}}^{\ast })\subset B_{\unicode[STIX]{x1D719}\text{}\underline{\mathbf{y}}}(R)\backslash \{\text{}\underline{\mathbf{0}}\}$ . We conclude as in Lemma 4.2.◻

Proof. Let $v_{1},\ldots ,v_{n-1}$ be linearly independent lattice points in $\unicode[STIX]{x1D706}_{n-1}(\unicode[STIX]{x1D6EC},B_{\text{}\underline{\mathbf{x}}_{j}})B_{\text{}\underline{\mathbf{x}}_{j}}$ . Then the lattice points $\sum _{l=1}^{n-1}m_{l}v_{l}$ with $-N_{j}/(c_{j}n)\leqslant m_{l}\leqslant N_{j}/(c_{j}n)$ are all distinct and lie all in $Z_{\mathbf{Q}_{j}}$ . Since $2[N_{j}/(c_{j}n)]+1\geqslant N_{j}/(c_{j}n)$ and $\operatorname{Vol}Z_{\mathbf{Q}_{j}}=(2\overline{Q})^{n}\leqslant 2^{n}akN_{j}^{n-1}$ the claimed inequality follows at once. Recall that $\unicode[STIX]{x1D6EC}$ is not admissible. Therefore, $\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})$ and $\overline{Q}/Q_{\max }=\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})^{1/n}/\text{max}_{i}\{|x_{ji}|\}$ both tend to $0$ , and $N_{j}$ tends to infinity.◻

Proof. This follows immediately from Lemma 7.1. ◻

Proof. We have $Q_{\max }\leqslant N_{j}|\text{}\underline{\mathbf{x}}_{j}|$ , and so ignoring the first few members of the sequence $\text{}\underline{\mathbf{x}}_{j}$ , we can assume that

$$\begin{eqnarray}\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6EC},Q_{\max })\geqslant \unicode[STIX]{x1D708}(N_{j}|\text{}\underline{\mathbf{x}}_{j}|)=\unicode[STIX]{x1D708}(a|\text{}\underline{\mathbf{x}}_{j}|/\unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)^{n})\geqslant \unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|/\unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)^{n})\geqslant \unicode[STIX]{x1D6FE}\unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|).\end{eqnarray}$$

Hence,

$$\begin{eqnarray}\displaystyle \inf _{0<B\leqslant Q_{\max }}\biggl(\frac{\overline{Q}}{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6EC},B)}+\frac{Q_{\max }}{B}\biggr) & {\leqslant} & \displaystyle \biggl(\frac{\overline{Q}}{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6EC},Q_{\max })}+1\biggr)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl(\frac{\overline{Q}}{\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)}+1\biggr)\leqslant (2k^{1/n}/\unicode[STIX]{x1D6FE})N_{j}\nonumber\end{eqnarray}$$

for all $j$ large enough. This, in conjunction with (7.3), shows that (7.4) holds true.◻

Proof. Let $\unicode[STIX]{x1D6FC}$ be an irrational real number, and consider the lattice $\unicode[STIX]{x1D6EC}$ given by the vectors $(p-q\unicode[STIX]{x1D6FC},q)$ with $p,q\in \mathbf{Z}$ . Then $\unicode[STIX]{x1D6EC}$ is unimodular and weakly admissible for $({\mathcal{S}},C)$ . To choose an appropriate $\unicode[STIX]{x1D6FC}$ we consider its continued fraction expansion $\unicode[STIX]{x1D6FC}=[a_{0},a_{1},a_{2},\ldots ]$ . Using the recurrence relation $q_{j+1}=a_{j+1}q_{j}+q_{j-1}$ for the denominator $q_{j}$ of the $j$ th convergent $p_{j}/q_{j}$ (in lowest terms) we can define $\unicode[STIX]{x1D6FC}$ by setting $a_{0}=a_{1}=1$ (so that $q_{0}=q_{1}=1$ ) and $a_{j+1}=[\log q_{j}]+1$ . Next we note that $a_{j+1}=[\log (a_{j}q_{j-1}+q_{j-2})]+1\leqslant \log ((a_{j}+1)q_{j-1})+1\leqslant \log (a_{j}+1)+a_{j}+1\leqslant 3a_{j}$ . Similarly we find $a_{j}+\log a_{j}-1\leqslant a_{j+1}$ , and hence,

$$\begin{eqnarray}a_{j}+\log a_{j}-1\leqslant a_{j+1}\leqslant 3a_{j}.\end{eqnarray}$$

Put $\text{}\underline{\mathbf{x}}_{j}=(p_{j}-q_{j}\unicode[STIX]{x1D6FC},q_{j})\in \unicode[STIX]{x1D6EC}\backslash C$ so that $|\text{}\underline{\mathbf{x}}_{j}|>|\text{}\underline{\mathbf{x}}_{j-1}|$ , at least for $j$ large enough. From the theory of continued fractions we know that for $\text{}\underline{\mathbf{x}}\in \unicode[STIX]{x1D6EC}\backslash C$ the inequality $\operatorname{Nm}(\text{}\underline{\mathbf{x}})<1/2$ implies that $\text{}\underline{\mathbf{x}}=c\text{}\underline{\mathbf{x}}_{j}$ for some non-zero integer $c$ and $j\in \mathbf{N}$ . We conclude that for all sufficiently large $\unicode[STIX]{x1D71A}$ we have $\unicode[STIX]{x1D708}(\unicode[STIX]{x1D71A})^{2}=\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})$ for some  $j$ . Also by the theory of continued fractions we know that

$$\begin{eqnarray}1/(a_{j+1}+2)<\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})<1/a_{j+1}.\end{eqnarray}$$

Since, for $j$ sufficiently large, $a_{j}>a_{j-1}+2$ , we conclude $\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j-1})<\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j-2})$ and thus

$$\begin{eqnarray}\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j-1})=\unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)^{2}\end{eqnarray}$$

for $j$ large enough; so we can take $k=1$ . We also easily find that $|\text{}\underline{\mathbf{x}}_{j}|/\unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)^{2}\leqslant |\text{}\underline{\mathbf{x}}_{j+1}|$ for $j$ large enough. It is now straightforward to verify (7.5). Moreover, for $j$ large enough, (2.3) holds true, and so $Z_{\mathbf{Q}_{j}}$ is an eligible set. Since $n=2$ we automatically have (7.2) with $c_{\unicode[STIX]{x1D6EC}}=1$ . Hence we can apply Proposition 7.1. Finally, we note that $\operatorname{Vol}Z_{\mathbf{Q}_{j}}=4N_{j}^{2}\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})=(2a)^{2}\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j-1})^{-2}\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})\geqslant 2^{-6}a_{j}^{2}/(a_{j+1}+2)$ which tends to infinity, and moreover, that the boxes $Z_{\mathbf{Q}_{j}}$ are increasingly distorted by Lemma 7.1. This completes the proof.◻

Proof. Let $\unicode[STIX]{x1D6FC}=[a_{0},a_{1},a_{2},\ldots ]$ be a badly approximable real number, so that the partial quotients $a_{i}$ are bounded. We set $a_{M}=\max a_{i}$ , and we consider the lattice

(7.7) $$\begin{eqnarray}\unicode[STIX]{x1D6EC}=\{(p_{1}-q\unicode[STIX]{x1D6FC},p_{2}-q\unicode[STIX]{x1D6FC},q);p_{1},p_{2},q\in \mathbf{Z}\}.\end{eqnarray}$$

The lattice $\unicode[STIX]{x1D6EC}$ is unimodular and weakly admissible for $({\mathcal{S}},C)$ . In this proof we write $h\ll g$ to mean $h\leqslant cg$ for a constant $c=c(a_{M})$ depending only on $a_{M}$ . First we note that

$$\begin{eqnarray}\operatorname{Nm}(\text{}\underline{\mathbf{x}})\gg |\text{}\underline{\mathbf{x}}|^{-1}\end{eqnarray}$$

for every $\text{}\underline{\mathbf{x}}\in \unicode[STIX]{x1D6EC}\backslash C$ . Hence,

(7.8) $$\begin{eqnarray}\unicode[STIX]{x1D708}(\unicode[STIX]{x1D71A})\gg \unicode[STIX]{x1D71A}^{-1/3}.\end{eqnarray}$$

Now suppose $p_{j}/q_{j}$ is the $j$ th convergent of $\unicode[STIX]{x1D6FC}$ , and put $\text{}\underline{\mathbf{x}}_{j}=(p_{j}-q_{j}\unicode[STIX]{x1D6FC},p_{j}-q_{j}\unicode[STIX]{x1D6FC},q_{j})\in \unicode[STIX]{x1D6EC}\backslash C$ . Then, for $j$ large enough, (2.3) holds true, and so $Z_{\mathbf{Q}_{j}}$ is an eligible set. Since

$$\begin{eqnarray}\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})\ll |\text{}\underline{\mathbf{x}}_{j}|^{-1},\end{eqnarray}$$

we also conclude that there exists $k=k(a_{M})\geqslant 1$ such that

$$\begin{eqnarray}\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})\leqslant k\unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)^{3}.\end{eqnarray}$$

Since $q_{j+1}=a_{j+1}q_{j}+q_{j-1}$ we get $q_{j+1}\ll q_{j}$ and, as is well-known, $|p_{j+1}-q_{j+1}\unicode[STIX]{x1D6FC}|<|p_{j}-q_{j}\unicode[STIX]{x1D6FC}|$ . Furthermore, $(p_{j},q_{j})$ and $(p_{j+1},q_{j+1})$ are linearly independent, and thus $\text{}\underline{\mathbf{x}}_{j}$ and $\text{}\underline{\mathbf{x}}_{j+1}$ are linearly independent. Hence, we conclude

$$\begin{eqnarray}c_{j}:=\unicode[STIX]{x1D706}_{2}(\unicode[STIX]{x1D6EC},B_{\text{}\underline{\mathbf{x}}_{j}})\ll 1,\end{eqnarray}$$

and thus, by Lemma 7.2, we get ${\mathcal{E}}_{\unicode[STIX]{x1D6EC}}(Z_{\mathbf{Q}_{j}})\gg N_{j}^{2}$ . Moreover, for $j$ sufficiently large, we have

(7.9) $$\begin{eqnarray}|\text{}\underline{\mathbf{x}}_{j-1}|<|\text{}\underline{\mathbf{x}}_{j}|\ll |\text{}\underline{\mathbf{x}}_{j-1}|,\end{eqnarray}$$

and thus

(7.10) $$\begin{eqnarray}\unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)\leqslant \operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j-1})^{1/3}\ll |\text{}\underline{\mathbf{x}}_{j-1}|^{-1/3}\ll |\text{}\underline{\mathbf{x}}_{j}|^{-1/3}.\end{eqnarray}$$

Combining (7.8), (7.9), and (7.10) implies that

$$\begin{eqnarray}\unicode[STIX]{x1D71A}^{-1/3}\ll \unicode[STIX]{x1D708}(\unicode[STIX]{x1D71A})\ll \unicode[STIX]{x1D71A}^{-1/3}.\end{eqnarray}$$

Therefore, we have

$$\begin{eqnarray}N_{j}\ll \unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)^{-3}\ll |\text{}\underline{\mathbf{x}}_{j}|\ll q_{j}\leqslant |\text{}\underline{\mathbf{x}}_{j}|\ll \unicode[STIX]{x1D708}(|\text{}\underline{\mathbf{x}}_{j}|)^{-3}\ll N_{j}.\end{eqnarray}$$

Thus, $N_{j}^{2}\ll Q_{\max }=N_{j}q_{j}\ll N_{j}^{2}$ , and due to (7.1), $\overline{Q}\ll N_{j}^{2/3}$ . Hence, with $B=N_{j}$ we have

$$\begin{eqnarray}\frac{\overline{Q}}{\unicode[STIX]{x1D708}(B)}\ll \frac{Q_{\max }}{B},\end{eqnarray}$$

and thus for all $j$ large enough

$$\begin{eqnarray}\inf _{0<B\leqslant Q_{\max }}\biggl(\frac{\overline{Q}}{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6EC},B)}+\frac{Q_{\max }}{B}\biggr)^{2}\ll \biggl(\frac{Q_{\max }}{B}\biggr)^{2}\ll N_{j}^{2}\ll {\mathcal{E}}_{\unicode[STIX]{x1D6EC}}(Z_{\mathbf{Q}_{j}}).\end{eqnarray}$$

Hence, we have shown that (7.4) holds true. Finally, we observe that $\operatorname{Vol}Z_{\mathbf{Q}_{j}}=8N_{j}^{3}\operatorname{Nm}(\text{}\underline{\mathbf{x}}_{j})\gg N_{j}^{2}$ which, due to Lemma 7.1, completes the proof.◻

Proof. First note that if $\#Z_{T}\leqslant 1$ then $\unicode[STIX]{x2202}Z_{T}$ lies in $\operatorname{Lip}(D,1,0)$ . Hence, it suffices to prove the claim for those $T$ with $\#Z_{T}\geqslant 2$ . By replacing $Z$ with the definable set $\{(T,x)\in Z;(\exists x,y\in Z_{T})(x\neq y)\}$ we can assume that $\#Z_{T}\geqslant 2$ for all $T\in \unicode[STIX]{x1D70B}(Z)$ , where $\unicode[STIX]{x1D70B}$ is the projection onto the first $d$ coordinates. We use the existence of definable Skolem functions. By [Reference van den Dries20, Ch. 6, Proposition (1.2)] there exists a definable map $f:\unicode[STIX]{x1D70B}(Z)\rightarrow \mathbf{R}^{D}$ whose graph $\unicode[STIX]{x1D6E4}(f)\subset Z$ . The proof of said (1.2) Proposition actually shows that there is an algorithmic way to construct the Skolem function $f$ . We will use the fact that this choice of $f$ is determined by $Z$ and $\unicode[STIX]{x1D70B}$ , and hence can be seen as part of the data of $Z$ .

Now we consider the set $Z^{\prime }=\{(T,y);(T,x)\in Z,y=x-f(T)\}$ . This set is again definable, and each non-empty fiber contains the origin, i.e., $0\in Z_{T}^{\prime }$ for all $T\in \unicode[STIX]{x1D70B}(Z)$ . Next we scale the fibers and translate by the point $y_{0}=(-1/2)(1,\ldots ,1)\in \mathbf{R}^{D}$ to get a new definable set whose fibers all lie in $(0,1)^{D}$ . We put $Z^{\prime \prime }=\{(T,z);(T,y)\in Z^{\prime },z=(3\cdot \operatorname{diam}(Z_{T}^{\prime }))^{-1}y-y_{0}\}$ (recall that $\operatorname{diam}(Z_{T}^{\prime })=\operatorname{diam}(Z_{T})>0$ since $Z_{T}$ has at least two points). We note that the graph of the function $T\rightarrow \operatorname{diam}(Z_{T})$ from $\unicode[STIX]{x1D70B}(Z)$ to $\mathbf{R}$ is given by

$$\begin{eqnarray}\{(T,t)\in \unicode[STIX]{x1D70B}(Z)\times \mathbf{R};\unicode[STIX]{x1D719}(T,t)\wedge \neg ((\exists u\in \mathbf{R})(\unicode[STIX]{x1D719}(T,u)\wedge u<t))\},\end{eqnarray}$$

where $\unicode[STIX]{x1D719}(T,t)$ stands for $(\forall x,y\in Z_{T})(|x-y|\leqslant t)$ . This shows that the aforementioned map is definable and hence, so is $Z^{\prime \prime }$ . Also we have $Z_{T}^{\prime \prime }\subset (0,1)^{D}$ for all $T$ . By [Reference Barroero and Widmer3, Lemma 3.15] the set $Z^{\prime \prime \prime }=\{(T,w);w\in \unicode[STIX]{x2202}Z_{T}^{\prime \prime }\}$ is also definable. The fibers of a definable set are again definable (cf. [Reference Barroero and Widmer3, Lemma 3.1]), and hence by [Reference van den Dries20, Ch. 4, Corollary (1.10)] we have $\dim (\unicode[STIX]{x2202}Z_{T}^{\prime \prime })\leqslant D-1$ . From Pila and Wilkie’s reparameterization lemma for definable families [Reference Pila and Wilkie11, Corollary 5.2] we concludeFootnote ¹² that $\unicode[STIX]{x2202}Z_{T}^{\prime \prime }$ lies in $\operatorname{Lip}(D,M_{Z^{\prime \prime \prime }},\unicode[STIX]{x1D705}_{Z^{\prime \prime \prime }})$ for all $T\in \mathbf{R}^{d}$ with certain constants $\unicode[STIX]{x1D705}_{Z^{\prime \prime \prime }}$ and $M_{Z^{\prime \prime \prime }}$ . Rescaling and retranslating gives $\unicode[STIX]{x2202}Z_{T}\in \operatorname{Lip}(D,M_{Z^{\prime \prime \prime }},\unicode[STIX]{x1D705}_{Z^{\prime \prime \prime }}\cdot \operatorname{diam}(Z_{T}))$ . Finally, we note that $Z^{\prime \prime \prime }$ depends only on $Z$ and $f$ which itself can be seen as part of the data of $Z$ , so that the constants $\unicode[STIX]{x1D705}_{Z^{\prime \prime \prime }}$ and $M_{Z^{\prime \prime \prime }}$ may be chosen to depend only on $Z$ . This completes the proof of the lemma.◻