Remark 2·2. Note that the condition that each $\psi_{i}(q) \ll {1}/{q}$ is necessary, since the p-adic distance between any two rational integers can be made arbitrarily small. This is in stark contrast to the real case where $\psi(q)< {1}/{2}$ is sufficient to ensure rectangles in the same ‘layer’ $\mathcal{A}_{q}(\Psi)$ are non-intersecting.

Remark 2·5. We note that the condition on the summation of the exponent vector $\boldsymbol{\tau}$ is present due to the fact that if $\sum_{i=1}^{n}\tau_{i} \leq n+1$ , then, by the p-adic version of Dirichlet’s Theorem, we have that $\mathfrak{W}_{n}(\boldsymbol{\tau})=\mathbb{Z}_{p}^{n}$ .

Remark 2·6. The condition that each $\tau_{i}>1$ may seem unnecessarily restrictive, however, the following reasoning shows why this must be the case. Similarly to Remark 2·2 the key reasoning behind the condition is that $\mathbb{Z}$ is dense in $\mathbb{Z}_{p}$ so in any coordinate axis where $\tau_{i}<1$ all points along the axis can be approximated, regardless of the choice of $a_{0}$ in our approximation sets. If, for example, we considered the approximation set $\mathfrak{W}_{2}((1-\varepsilon, \tau_{2}))$ for $\varepsilon>0$ and $\tau_{2}>2$ then the above argument gives us that $\mathfrak{W}_{2}((1-\varepsilon,\tau_{2}))=\mathbb{Z}_{p} \times \mathfrak{W}_{1}(\tau_{2})$ . Using well–known bounds on the Hausdorff dimension of product spaces (see e.g [ Reference Tricot66 ]) we have that

\begin{equation*} \dim \mathfrak{W}_{1}(\tau_{2}) + \dim \mathbb{Z}_{p} \leq \dim \mathfrak{W}_{2}((1-\varepsilon, \tau_{2})) \leq \dim \mathfrak{W}_{1}(\tau_{2}) +\dim_{B}\mathbb{Z}_{p}, \end{equation*}

where $\dim_{B}$ is the box-counting dimension, we have that

(2·4) \begin{equation} \dim \mathfrak{W}_{2}((1-\varepsilon, \tau_{2}))=\frac{2}{\tau_{2}}+1. \end{equation}

However, if Theorem 2·4 was applicable we would have that

\begin{equation*} \dim \mathfrak{W}_{2}((1-\varepsilon, \tau_{2}))=\min \!\left\{ \frac{3+(\tau_{2}-(1-\varepsilon))}{\tau_{2}}, \frac{3}{1-\varepsilon} \right\}=\frac{2}{\tau_{2}}+\frac{\tau_{2}+\varepsilon}{\tau_{2}}\,, \end{equation*}

contrary to (2·4).

Proof. By the condition that each function $\psi_{i}$ has corresponding positive limit (2·5), for any $\epsilon>0$ we have that

\begin{equation*} q^{-(v_{i}+\epsilon)} \leq \psi_i(q) \leq q^{-(v_{i}-\epsilon)} \quad (1 \leq i \leq n) \end{equation*}

for all sufficiently large $q \in \mathbb{N}$ . Let $\epsilon =(\epsilon, \dots, \epsilon) \in \mathbb{R}^{n}_{+}$ . Then, we have that

\begin{equation*} \mathfrak{W}_{n}(\mathbf{v}+ {\boldsymbol{\epsilon}} ) \subseteq \mathfrak{W}_{n}(\Psi) \subseteq \mathfrak{W}_{n}(\mathbf{v}- {\boldsymbol{\epsilon}}). \end{equation*}

By letting $\epsilon \to 0$ , and applying Theorem 2·4 we get the required result.

Remark 2·9. Note that if the right–hand side of (2·6) was simply $o\!\left(\max_{1 \leq j \leq d} |y_{j}-x_{j}|_{p}\right)$ then f would be simply differentiable at $\boldsymbol{{x}}$ . The above definition imposes a stronger condition than differentiability in the sense that the error term in (2·6) is quadratic. It is readily verified that any $C^2$ function, as defined in [ Reference Schikhof58 ] (see also [ Reference Kleinbock and Tomanov51 ] for a brief survey of p-adic $C^k$ functions), is DQE at every point. The converse may not be true. Mahler’s normal functions are $C^{\infty}$ and so they are DQE.

Remark 2·13. Note that the dimension results of Theorems 2·10–2·11 are contained within Theorem 2·12. However, due to the method of proof we are not able to obtain the Hausdorff measure result in Theorem 2·12. Also note that the statements remain true if the assumptions imposed on $\mathbf{f}$ are imposed on a sufficiently small ball $B\subset\mathcal{U}$ instead of $\mathcal{U}$ .

Remark 2·14. The assumption that the approximations over the independent variables ( $\tau_1$ in Theorem 2·11 and $\tau_{1}, \dots , \tau_{d}$ in Theorem 2·12) are larger than the approximations over the dependent variables is merely technical. Observe that this condition is not needed amongst the approximations over each respective variable, since we may permute the variables to obtain the desired ordering. However, the other requirements placed on $\boldsymbol{\tau}$ are necessary to allow the result to hold for as general set of manifolds as possible. In particular, the conditions that $\sum_{i=1}^{m} \tau_{d+i}<m+1$ and $\tau_{i}>1$ for $1 \leq i \leq n$ ensure that even if the manifold is a hyperplane passing through badly approximable points we will still have an infinite number of rational approximations. If these conditions do not hold a counterexample can be readily obtained on modifying the example of Remark 3 in [ Reference Beresnevich, Lee, Vaughan and Velani12 ]. It is also easy to see that the lower bound $\tau_{1} \geq n+1-\tilde\tau$ is necessary for otherwise (2·9) would be false. The upper bound $\tilde\tau<n$ on $\tilde\tau$ can likely be improved, however this will require imposing additional conditions on the curves such as non-degeneracy (meaning $1,x,f_1(x),\dots,f_{n-1}(x)$ are linearly independent over $\mathbb{Z}_p$ in the case these are analytic/Mahler’s normal functions), and will require a different approach such as that of [ Reference Beresnevich7 ]. We plan to address the problem for non-degenerate curves separately in a subsequent publication.

Remark 2·15. We expect that the lower bound of Theorem 2·10-2·12 is sharp and each dimension result should indeed be equality at least for non-degenerate curves. Obtaining the upper bounds represents a challenging open problem even in dimension 2. We would like to stress that there is currently no equivalent to Huxley’s estimate [ Reference Huxley46 ] in the p-adic setting, let alone the sharper Vaughan–Velani result [ Reference Vaughan and Velani67 ].

Proof. The proof of this lemma is standard and uses Dirichlet’s pigeonhole principle. Nevertheless, as it is used in the proofs of our main results, for completeness we provide its details here. To begin with, note that there are $T^{n+1}$ different rational integer vectors $ \boldsymbol{{x}}=(x_{0}, \dots , x_{n})$ satisfying (3·2), subject to the condition that $ x_i \geq 0 $ for each i. Let $\varepsilon\in(0,1)$ and $T_{\varepsilon}=T-\varepsilon$ . For each $i=1, \dots, n$ let $\delta_{i}$ be the unique integer such that

(3·4) \begin{equation}p^{\delta_{i}-1} \leq p^{-\sigma_i}T_{\varepsilon}^{\tau_{i}} < p^{\delta_{i}}\,.\end{equation}

Assuming $H_{\boldsymbol{\sigma}}$ , which can be found explicitly, is sufficiently large we ensure that $\delta_i\ge0$ for each i. Clearly, for each $\boldsymbol{{x}}\in\mathbb{Z}^n$ we have that $L(\boldsymbol{{x}})\,:\!=\,(L_1(\boldsymbol{{x}}),\dots,L_n(\boldsymbol{{x}}))\in\mathbb{Z}_p^n$ . Split $\mathbb{Z}_p^n$ into the subsets $S(\boldsymbol{{a}})$ given by

\begin{equation*}S(\boldsymbol{{a}})=\prod_{i=1}^n \left\{x_i\in\mathbb{Z}_p\,:\,|x_i-a_i|_p\le p^{-\delta_i}\right\}\end{equation*}

for each $\boldsymbol{{a}}=(a_1,\dots,a_n)\in\mathbb{Z}^n$ with $0\le a_i< p^{\delta_i}$ . It is readily seen that the sets $S(\boldsymbol{{a}})$ are disjoint and cover the whole of $\mathbb{Z}_p^n$ . Furthermore, using the facts that $\sum_i\tau_i=n+1$ , $\sum_i\sigma_i=n$ and (3·4), we find that the number of sets $S(\boldsymbol{{a}})$ is

\begin{equation*}p^{\sum_i\delta_i}\le T_{\varepsilon}^{\sum_i\tau_i}=T_{\varepsilon}^{n+1}<T^{n+1}\,.\end{equation*}

Hence, by the pigeonhole principle, at least one of the sets $S(\boldsymbol{{a}})$ contains $L(\boldsymbol{{x}}_i)$ for at least two distinct integer points $\boldsymbol{{x}}_1$ and $\boldsymbol{{x}}_2$ as specified above. Let $\boldsymbol{{x}}=\boldsymbol{{x}}_1-\boldsymbol{{x}}_2$ . Clearly, (3·2) is satisfied and $\boldsymbol{{x}}$ is non-zero. Furthermore, for each $i=1,\dots,n$ we have that

(3·5) \begin{equation} |L_i(\boldsymbol{{x}})|=|L_{i}(\boldsymbol{{x}}_{1}-\boldsymbol{{x}}_{2})|_{p} = |L_{i}(\boldsymbol{{x}}_{1})-L_i(\boldsymbol{{x}}_{2})|_{p} \leq p^{-\delta_{i}} \stackrel{(3.4)}{<} p^{\sigma_i}T_{\varepsilon}^{-\tau_{i}}\,. \end{equation}

Since there are only finitely many integer vectors $ \boldsymbol{{x}}=(x_{0}, \dots , x_{n})$ satisfying (3·2), there is a non-zero $\boldsymbol{{x}}$ subject to (3·2) satisfying (3·5) for every $\varepsilon\in(0,1)$ . Letting $\varepsilon\to0$ verifies (3·3) and completes the proof.

Proof. Let $\mathcal{U}_j\,:\!=\,\bigcup_{i\ge j}U_i$ and $\mathcal{D}_j\,:\!=\,\mathcal{B}\setminus\mathcal{U}_j$ . Then, $\mathcal{D}\,:\!=\,\mathcal{B}\setminus\mathcal{U}=\bigcup_{j}\mathcal{D}_j$ and we need to prove that $\mathcal{D}$ has $\mu$ -measure zero. Assume the contrary. Then, since every set $\mathcal{D}_j$ is $\mu$ -measurable and $\mathcal{D}_j\subseteq \mathcal{D}_{j+1}$ for all j, by the continuity of $\mu$ , there is an $\ell\in\mathbb{N}$ such that $\mu(D_\ell)>0$ . Since $\mu$ is Borel regular $\mu(\mathcal{D_\ell})=\inf\{\mu(A)\,:\,\mathcal{D_\ell}\subset A,\;A\text{ is open}\}$ . Since X is separable and ultrametric, every open set A can be written as a disjoint countable union of balls. Hence, for any $\varepsilon>0$ there exists a countable collection of disjoint balls $(A_i)$ such that

(4·2) \begin{equation}\mathcal{D}_\ell\subset \bigcup_i A_i\quad\text{and}\quad \sum_i\mu(A_i)-\varepsilon\le \mu(\mathcal{D_\ell})\le \sum_i\mu(A_i)\,.\end{equation}

Let

\begin{equation*}\lambda\,:\!=\,\sup\!\left\{\frac{\mu(A_i\cap\mathcal{D}_\ell)}{\mu(A_i)}\,:\,i\in\mathbb{N},\;\mu(A_i)>0\right\}\,.\end{equation*}

Note that, since $\mu(\mathcal{D}_\ell)>0$ , the above set is non-empty and therefore $\lambda\in[0,1]$ . Then, by (4·2), we have that

\begin{equation*}\mu(\mathcal{D}_\ell)=\sum_i\mu(A_i\cap \mathcal{D}_\ell)\le \lambda\sum_i\mu(A_i)\le \lambda(\mu(\mathcal{D}_\ell)+\varepsilon)\,.\end{equation*}

Therefore,

\begin{equation*}\lambda\ge\frac{\mu(\mathcal{D}_\ell)}{\mu(\mathcal{D}_\ell)+\varepsilon}\,.\end{equation*}

Since $\mu(\mathcal{D}_\ell)>0$ , on taking $\varepsilon>0$ small enough, we can ensure that $\lambda>1-c$ . Then, by the definition of $\lambda$ , there exists $i_0\in\mathbb{N}$ such that $\mu\!\left(A_{i_0}\right)>0$ and

(4·3) \begin{equation}\frac{\mu\!\left(A_{i_0}\cap\mathcal{D}_\ell\right)}{\mu\!\left(A_{i_0}\right)}>1-c\,.\end{equation}

Take $j\ge \ell$ sufficiently large so that for every $i\ge j$ the radius of $B_i$ is less than the radius of $A_{i_0}$ . Then, since X is ultrametric, for all $i\ge j$ if $B_i\cap A_{i_0}\neq\varnothing$ then $B_i\subset A_{i_0}$ . Since $\mathcal{D}_\ell\subset \mathcal{D}\subset\mathcal{B}\subset \bigcup_{i\ge j}B_i$ , we have that

(4·4) \begin{equation} A_{i_0}\cap\mathcal{D}_\ell\subset \bigcup_{i\ge j,\; B_i\cap A_{i_0}\neq\varnothing}B_i\cap\mathcal{D}_\ell\,.\end{equation}

Without loss of generality assume the $B_{i}$ over $i \geq j$ are disjoint, since if not we can take a disjoint sub-collection of $(B_{i})_{i\ge j}$ such that the union of balls in this subcollection is again $\bigcup_{i\ge j}B_i$ and so the sub-collection would satisfy (4·4). Such sub-collection is possible to choose since X is ultrametric. Therefore, by (4·4), we have that

(4·5) \begin{equation} \mu\!\left(A_{i_0}\cap\mathcal{D}_\ell\right)\le \sum_{i\ge j,\; B_i\cap A_{i_0}\neq\varnothing}\mu(B_i\cap\mathcal{D}_\ell)\,.\end{equation}

By construction $\mathcal{D}_{i}\cap U_{i}=\emptyset$ for every i. Thus, in view of (4·1) and the fact that $U_i\subset B_i$ we have that

\begin{equation*}\mu(B_{i})\ge\mu(U_{i}\cap B_i)+\mu(\mathcal{D}_{i}\cap B_{i})\ge c\mu(B_{i})+\mu(\mathcal{D}_{i}\cap B_{i})\end{equation*}

and so $\mu(\mathcal{D}_{i}\cap B_{i})\le(1-c) \ \mu(B_{i}) $ for all i. In particular, since $\mathcal{D}_{i}\subset \mathcal{D}_{i+1}$ for all i and $j\ge \ell $ we get that

\begin{equation*}\mu(\mathcal{D}_{\ell}\cap B_{i})\le \mu(\mathcal{D}_{i}\cap B_{i})\le(1-c) \ \mu(B_{i}) \quad\text{for all }i\ge j\,.\end{equation*}

Hence, by (4·5) and the assumption that the $B_i$ for $i\ge j$ are disjoint, we get that

\begin{equation*} \mu\!\left(A_{i_0}\cap\mathcal{D}_\ell\right)\le \sum_{i\ge j,\; B_i\cap A_{i_0}\neq\varnothing}(1-c)\mu(B_i) =(1-c)\mu\!\left(\bigcup_{i\ge j,\; B_i\cap A_{i_0}\neq\varnothing}B_i\right)\le(1-c)\mu\!\left(A_{i_0}\right)\,.\end{equation*}

This contradicts (4·3). The proof is thus complete.

Proof. We initially prove that

\begin{equation*}\mu \!\left( \limsup_{i \to \infty} B_{i}^{(1)} \times \prod_{j=2}^{n}B_{i}^{(j)} \right)= \mu \!\left( \limsup_{i \to \infty} U_{i}^{(1)} \times \prod_{j=2}^{n}B_{i}^{(j)} \right),\end{equation*}

and note that Lemma 4·2 follows inductively. For ease of notation let

\begin{equation*}\hat{\mu}=\prod_{j=2}^{n}\mu_{j}, \quad \hat{B}_{i}=\prod_{j=2}^{n}B_{i}^{(j)}, \quad \hat{X}=\prod_{j=2}^{n}X_{j}.\end{equation*}

For any $y \in \hat{X}$ let

\begin{equation*}I_{y}= \{ i\,:\, y \in \hat{B}_{i}\},\end{equation*}

and for any $F \subseteq X$ let $F_y$ denote the fiber of F at y, that is

\begin{equation*}F_y=\{ x\,:\, (x,y) \in F\} \subseteq X_{1}.\end{equation*}

Observe that

(4·8) \begin{equation}A\,:\!=\,\left( \limsup_{i \to \infty} B_{i}^{(1)} \times \hat{B}_{i} \right)_{y}=\underset{i \in I_y}{\limsup_{i \to \infty}}B^{(1)}_{i}\,=\!:\,D.\end{equation}

Indeed, if $x \in A$ then it implies there exists an infinite sequence $\{i_{k}\}$ such that

\begin{equation*}(x,y) \in B_{i_{k}}^{(1)} \times \hat{B}_{i_k} \quad \text{ for all } i_k.\end{equation*}

Hence $\{i_k\} \subseteq I_y$ and so $x \in D$ .

Conversely, if $x \in D$ then D is non-empty and so $I_y$ must be infinite. By the definition of $I_y$ and the fact that $x \in D$ we have that $x \in B^{(1)}_{i}$ for infinitely many $i \in I_y$ , and so $x \in A$ .

Similarly, we have that

(4·9) \begin{equation}\left( \limsup_{i \to \infty} U_{i}^{(1)} \times \hat{B}_{i} \right)_{y}=\underset{i \in I_y}{\limsup_{i \to \infty}}U^{(1)}_{i}.\end{equation}

Applying Fubini’s theorem we have that

\begin{align*}\mu \!\left( \limsup_{i \to \infty} B_{i}^{(1)} \times \hat{B}_{i} \right) & = \int_{\hat X} \mu_{1} \!\left( \left( \limsup_{i \to \infty} B_{i}^{(1)} \times \hat{B}_{i} \right)_{y} \right) d \hat{\mu}, \\& \!\!\stackrel{(4.8)}{=} \int_{\hat X} \mu_{1} \!\left( \underset{i \in I_y }{\limsup_{i \to \infty}} B_{i}^{(1)} \right) d \hat{\mu},\\& \!\!\!\!\!\!\!\overset{\text{Lemma 4.1}}{=} \int_{\hat X} \mu_{1} \!\left( \underset{i \in I_y }{\limsup_{i \to \infty}}\; U_{i}^{(1)} \right)_{y} d \hat{\mu}, \\& \!\!\stackrel{(4.9)}{=} \int_{\hat X} \mu_{1} \!\left( \left( \limsup_{i \to \infty} U_{i}^{(1)} \times \hat{B}_{i} \right)_{y} \right) d \hat{\mu}, \\&=\mu \!\left( \limsup_{i \to \infty} U_{i}^{(1)} \times \hat{B}_{i} \right)\,.\end{align*}

Note that in the above argument we have not made use of the fact $\hat B_i$ are products of balls; we only used the fact that these are measurable sets. Hence, the above argument can be repeated $n-1$ more times, for $\ell=2,\dots,n-1$ each time replacing $B^{(\ell)}_i$ by $U^{(\ell)}_i$ so that at step $\ell$ we get that

\begin{align*}\mu \!\left( \limsup_{i \to \infty} \prod_{j=1}^{\ell-1} U_{i}^{(j)} \times \prod_{j=\ell}^n B_{i}^{(j)} \right) & =\mu \!\left( \limsup_{i \to \infty}\prod_{j=1}^\ell U_{i}^{(j)} \times \prod_{j=1}^{\ell+1} B_{i}^{(j)} \right)\,.\end{align*}

Putting all these equations for $\ell=1,\dots,n$ together we get (4·7) as claimed.

Remark 4·5. Haynes proved this result for the more general setting of S-arithmetic approximation. We note that Lemma 4·4 can also be proven in the S-arithmetic setting, however we limit ourselves to the p-adic case to avoid introducing further notation and concepts which are not dealt with in this paper.

Proof. Firstly, note that the sets $\mathfrak{A}^{\prime}_{a_{0}}(\Psi)$ used to construct our $\limsup$ set have the property that if $p \mid a_{0}$ , then $\mathfrak{A}^{\prime}_{a_{0}} (\Psi)= \emptyset$ or $\mathbb{Z}_{p}^{n}$ , so assume $p \nmid a_{0}$ . Define the map $\pi\,:\, \mathbb{Z}_{p} \to \mathbb{Z}_{p}$ as follows. For a p-adic integer $x \in \mathbb{Z}_{p}$ with p-adic expansion

\begin{equation*}x=\sum_{i=0}^{\infty}u_{i}p^{i}, \quad u_{i} \in \{0, \dots , p-1\},\end{equation*}

define

\begin{equation*}\pi(x)= \begin{cases}\sum_{i=0}^{\infty}u_{i+1}p^{i}, \quad \text{ if } u_{0}=0, \\[5pt]1+ \sum_{i=0}^{\infty} u_{i+1}p^{i}, \quad \text{ otherwise}.\end{cases}\end{equation*}

Let $\pi_{n}\,:\,\mathbb{Z}_{p}^{n} \to \mathbb{Z}_{p}^{n}$ be the transformation $(x_{1}, \dots, x_{n}) \mapsto (\pi(x_{1}), \dots , \pi(x_{n}))$ . By using the fact that $p \nmid a_{0}$ , and that each $(a_{i},a_{0})=1$ , it can be shown that under such mapping

\begin{equation*}\pi_{n}\!\left( \mathfrak{W}^{\prime}_{n}(\Psi)\right) \subseteq \mathfrak{W}^{\prime}_{n}(p\Psi),\end{equation*}

where $p\Psi$ means each component of $\Psi$ has to be multiplied by p. This can be repeated inductively to show that $\pi_{n}^{K}\!\left( \mathfrak{W}^{\prime}_{n}(\Psi)\right) \subseteq \mathfrak{W}^{\prime}_{n}\!\left(p^{K}\Psi\right)$ for any $K \in \mathbb{N}$ . Assuming that $\mu_{p,n}\!\left(\mathfrak{W}^{\prime}_{n}(\Psi)\right)>0$ , then by a p-adic version of the Lebesgue density theorem (see e.g. Lemma 1 in [ Reference Sprindžuk63 , part II, chapter 1]) for any $\epsilon>0$ there exists integer vector $\boldsymbol{{x}}_{0} \in \mathbb{Z}^{n}$ and $N \in \mathbb{N}$ such that

\begin{equation*}\mu_{p,n}\!\left( \left\{ \boldsymbol{{x}} \in \mathfrak{W}^{\prime}_{n}(\Psi)\,:\, |\boldsymbol{{x}}-\boldsymbol{{x}}_{0}|_{p} \leq p^{-N} \right\} \right) \geq (1-\epsilon)p^{-N}.\end{equation*}

Further, we have that

\begin{equation*}\pi_{n}^{N}\!\left( \left\{ \boldsymbol{{x}} \in \mathfrak{W}^{\prime}_{n}(\Psi)\,:\, |\boldsymbol{{x}}-\boldsymbol{{x}}_{0}|_{p} \leq p^{-N} \right\} \right) \subseteq \mathfrak{W}^{\prime}_{n}\!\left(p^{N} \Psi\right),\end{equation*}

and so

\begin{align*}\mu_{p,n} \!\left( \mathfrak{W}^{\prime}_{n}\!\left(p^{N}\Psi\right) \right) & \geq \mu_{p,n} \left( \pi_{n}^{N}\left( \left\{ \boldsymbol{{x}} \in \mathfrak{W}^{\prime}_{n}(\Psi)\,:\, |\boldsymbol{{x}}-\boldsymbol{{x}}_{0}|_{p} \leq p^{-N} \right\} \right) \right) \\&\geq p^{N}(1-\epsilon) p^{-N} \\&=(1- \epsilon).\end{align*}

Since $\epsilon$ is arbitrary we have that $\mu_{p,n}\!\left(\bigcup_{N=1}^\infty\mathfrak{W}^{\prime}_{n}\left(p^{N}\Psi\right)\right)=1$ . Now observe that

\begin{equation*}\mathfrak{W}^{\prime}_{n}(\Psi)\subset \mathfrak{W}^{\prime}_{n}(p\Psi)\subset \mathfrak{W}^{\prime}_{n}\left(p^{2}\Psi\right)\subset\cdots\end{equation*}

and so, by Lemma 4·2 with $X=\mathbb{Z}_p^n$ , d given by the sup norm, and $\mu=\mu_{p,n}$ , we have that $\mu_{p,n}\!\left(\mathfrak{W}^{\prime}_{n}(\Psi)\right)=\mu_{p,n}\!\left(\mathfrak{W}^{\prime}_{n}\left(p^N\Psi\right)\right)$ for every $N\in\mathbb{N}$ . Hence,

\begin{equation*}\mu_{p,n}\!\left(\mathfrak{W}^{\prime}_{n}(\Psi)\right)=\lim_{N\to\infty}\mu_{p,n}\!\left(\mathfrak{W}^{\prime}_{n}\left(p^{N}\Psi\right)\right)=\mu_{p,n}\!\left(\bigcup_{N=1}^\infty\mathfrak{W}^{\prime}_{n}\left(p^{N}\Psi\right)\right)=1\,,\end{equation*}

thus finishing the proof.

Proof. Fix some ball $B=B(\boldsymbol{{y}},r)$ for some $\boldsymbol{{y}} \in \mathbb{Z}_{p}^{n}$ and $r \in \left\{p^{i}\,:\, i \in \mathbb{N}\cup\{0\} \right\}$ . We will assume that k is sufficiently large so that $M^{k}r\geq 1$ . In view of (6·1) and the fact that $\alpha_i>1$ for all i, by Lemma 3·3, we have that for any $\boldsymbol{{x}}=(x_{1}, \dots, x_{n}) \in B$ there exists $(a_{0},\dots,a_{n}) \in \mathbb{Z}^{n+1}$ , satisfying

(6·2) \begin{equation}|a_{i}| \;\le\; M^{k}\qquad(1 \leq i \leq n),\qquad 0<a_0< M^{k+1}\end{equation}

and

(6·3) \begin{equation} |a_{0}x_{i}-a_{i}|_{p}<p\!\left(M^{k+\frac{1}{n+1}}\right)^{-\alpha_{i}}\qquad(1\leq i \leq n).\end{equation}

Since $\alpha_{i}>1$ for each $1 \leq i \leq n$ , (6·3) combined with $0<a_{0} \leq M^{k+1}$ implies that $|a_{i}|_{p} \leq |a_{0}|_{p}$ for each $1 \leq i \leq n$ , provided that k is sufficiently large. Let $\lambda$ be the integer such that $|a_{0}|_{p}=p^{-\lambda}$ . Write $a^{\prime}_{0}=a_{0}p^{-\lambda}$ and $a^{\prime}_{i}=a_{i}p^{-\lambda}$ . Since $|a_i|_p\le |a_0|_p$ , we have that $a^{\prime}_{0},a^{\prime}_{i} \in \mathbb{Z}$ . Also, by definition,

(6·4) \begin{equation} 0<a^{\prime}_{0}\leq p^{-\lambda}M^{k+1}, \quad |a^{\prime}_{i}|\le p^{-\lambda}M^{k}\end{equation}

for each $1 \leq i \leq n$ and, by (6·3) and the fact that $\gcd\!(a^{\prime}_0,p)=1$ , that

(6·5) \begin{align}\left|x_{i}-\frac{a^{\prime}_{i}}{a^{\prime}_{0}}\right|_{p}&=\left|a^{\prime}_0x_{i}-{a^{\prime}_{i}}\right|_{p}=p^{\lambda}|a_{0}x_{i}-a_{i}|_{p}<p^{\lambda+1}\left(M^{k+\frac{1}{n+1}}\right)^{-\alpha_{i}} \end{align}

for $1 \leq i \leq n$ . We want to remove the $a^{\prime}_{0}$ values that are divisible by too high a power of p, that is $|a^{\prime}_{0}|_{p} < p^{-\lambda_{0}}$ for some fixed $\lambda_{0} \in \mathbb{N}$ to be chosen later. We consider the integer vectors $\left(a^{\prime}_{0}, \dots , a^{\prime}_{n}\right)$ satisfying (6·4) such that

\begin{equation*}\left( \frac{a^{\prime}_{1}}{a^{\prime}_{0}}, \dots , \frac{a^{\prime}_{n}}{a^{\prime}_{0}} \right) \in B(\boldsymbol{{y}},r)\,,\end{equation*}

where r is a poser of p. Considering the congruence equations for $a^{\prime}_{0}$ fixed we have that there are

\begin{equation*}\left(2p^{-\lambda}M^{k}r+1 \right)^{n}< \left(3 p^{-\lambda} M^{k}r \right)^{n}\end{equation*}

such points. Hence

\begin{align*}&\mu_{p,n} \!\left( B \cap \underset{\lambda \geq \lambda_{0}}{\bigcup} \underset{ 0< a^{\prime}_{0} \leq \frac{M^{k+1}}{p^{\lambda}}}{\bigcup_{|a^{\prime}_{i}|\leq \frac{M^{k}}{p^{\lambda}}}} \bigcup_{\left( \frac{a^{\prime}_{1}}{a^{\prime}_{0}}, \dots , \frac{a^{\prime}_{n}}{a^{\prime}_{0}} \right) \in B} \prod_{i=1}^{n} B\!\left( \frac{a^{\prime}_{i}}{a^{\prime}_{0}}, p^{\lambda+1}\left(M^{k+\frac{1}{n+1}}\right)^{-\alpha_{i}} \right) \right)\\& \qquad\qquad\qquad \leq \sum_{\lambda \geq \lambda_{0}} \frac{M^{k+1}}{p^{\lambda}} \left(3 \frac{M^{k}}{p^{\lambda}}r \right)^{n} p^{n \lambda + n} M^{-k(n+1)-1}, \\& \qquad\qquad\qquad = \sum_{\lambda \geq \lambda_{0}} \mu_{p,n}(B)3^{n}p^{n-\lambda}, \\& \qquad\qquad\qquad \leq 3^{n}\frac{p^{n+1-\lambda_{0}}}{p-1}\mu_{p,n}(B).\end{align*}

Taking $\lambda_{0}$ sufficiently large, e.g. $p^{\lambda_{0}} > 4({3^{n}p^{n+1}})/({p-1})$ , then we have that

(6·6) \begin{equation}\mu_{p,n} \!\left( B \cap \underset{\lambda \geq \lambda_{0}}{\bigcup} \underset{ 0< a^{\prime}_{0} \leq \frac{M^{k+1}}{p^{\lambda}}}{\bigcup_{|a_{i}|\leq \frac{M^{k}}{p^{\lambda}}}} \bigcup_{\left( \frac{a^{\prime}_{1}}{a^{\prime}_{0}}, \dots , \frac{a^{\prime}_{n}}{a^{\prime}_{0}} \right) \in B} \prod_{i=1}^{n} B\!\left( \frac{a^{\prime}_{i}}{a^{\prime}_{0}}, p^{\lambda_{0}+1}\left(M^{k+\frac{1}{n+1}}\right)^{-\alpha_{i}} \right) \right) \leq \tfrac{1}{4} \mu_{p,n}(B).\end{equation}

Using similar calculations to those of above we have that

(6·7) \begin{align}\mu_{p,n}\!\left( B \cap \underset{0 < a_{0} \leq M^{k}\,:\, |a_{0}|_{p} \geq p^{-\lambda_{0}}}{\bigcup_{|a_{i}| \leq M^{k}}} \prod_{i=1}^{n} B\!\left( \frac{a_{i}}{a_{0}} , p^{\lambda_{0}+1}\left(M^{k+\frac{1}{n+1}}\right)^{-\alpha_{i}} \right) \right) & \leq \frac{3^{n}p^{n\lambda_{0}+n}}{M} \mu_{p,n}(B)\nonumber\\& \le\tfrac14\mu_{p,n}(B)\end{align}

provided that $M>4 \, 3^{n} p^{n\lambda_{0}+n}$ . Combining (6·6) and (6·7) with the fact that (6·2) and (6·3) can be solved in integers $(a_0,\dots,a_n)$ for all $\boldsymbol{{x}}$ , we get that

\begin{equation*}\mu_{p,n} \!\left( B \cap \underset{M^{k} < a_{0} \leq M^{k+1}\,:\, |a_{0}|_{p} \geq p^{-\lambda_{0}}}{\bigcup_{|a_{i}| \leq M^{k} }}\prod_{i=1}^{n}B\!\left( \frac{a_{i}}{a_{0}}, p^{\lambda_{0}+1}\left(M^{k+\frac{1}{n+1}}\right)^{-\alpha_{i}} \right) \right) \geq \tfrac12\, \mu_{p,n}(B).\end{equation*}

Taking the constant

\begin{equation*}c_{1}=\max_{1 \leq i \leq n} \, p^{\frac{\lambda_{0}+1}{\alpha_{i}}}M^{1-\frac{1}{n+1}}\end{equation*}

completes the proof.

Proof. By Lemma 3·3 with $\boldsymbol{\sigma}=((n+m\lambda)/d,\dots,(n+m\lambda)/d,-\lambda,\dots,-\lambda)$ , $H_0=\dots=H_n=H$ and $T=H+1$ , for any integer $H \ge H_{\boldsymbol{\sigma}}^{1/(n+1)}$ the following system

(7·4) \begin{align}\left\{\begin{array}{l@{\quad}l@{\quad}l}|b_{0}x_{i}-b_{i}|_{p} &< p^{(n+m\lambda)/d}H^{-v_{i}} & (1 \leq i \leq d), \\[10pt]\displaystyle\left| b_{0}p^{\lambda}f_{j}(\boldsymbol{{x}})-\sum_{i=1}^{d}p^{\lambda}\dfrac{\partial f_{j}}{\partial x_{i}}(\boldsymbol{{x}}) \left( b_0x_{i}-b_{i} \right)-p^{\lambda}b_{d+j} \right|_{p} &< p^{-\lambda}H^{-\tau_{j}} & (1 \leq j \leq m), \\[15pt]\underset{0\leq i\leq n}{\max} |b_{i}| & \leq H\end{array}\right.\end{align}

has a non-zero integer solution $(b_{0},b_{1},\dots, b_{n}) \in \mathbb{Z}^{n+1}$ . Without loss of generality we can assume that $d=\gcd\!(b_{0},b_{1},\dots, b_{n})$ is a power of p as otherwise we can divide (7·4) through by any other prime powers in the factorisation of d without affecting (7·4). Let $C>0$ and $0<\varepsilon<1$ be the constants that satisfy Definition 2·8 for all $f_j$ simultaneously. In particular, we have that $B(\boldsymbol{{x}},\varepsilon) \subseteq \mathcal{U}$ . Let $v_{\min}\,:\!=\,\min_{1 \leq i \leq d}v_{i}$ and $\tau_{\max}\,:\!=\,\max_{1 \leq j \leq m} \tau_{j}$ . Let $H_{0}$ be defined as follows

\begin{equation*}H_{0}\,:\!=\,\max \!\left\{ \begin{array}{cc} C^{\frac{2}{2v_{\min} - \tau_{\max}}}, & (\alpha1) \\ C^{\frac{1}{v_{\min} - 1}}, & (\alpha2) \\ \left(\varepsilon^{-1}p^{(n+m\lambda)/d}\right)^{\frac{1}{v_{\min}}-1}, & (\beta) \\ p^{\frac{n+n\lambda}{d(v_{\min} -1)}}, & (\gamma)\\ H_{\boldsymbol{\sigma}}^{1/(n+1)} & (\delta) \end{array} \right\}\,.\end{equation*}

Note that $H_{0}$ is a well defined positive real number since $v_{\min}-1>0$ and $2v_{\min}-\tau_{\max}>0$ . The latter follows from the facts that each $\tau_{j}>1$ and $\sum_{j=1}^{m}\tau_{j}<m+1$ and so $\tau_{j}<2$ , and the condition that each $v_{i}>1$ . Note that $(\gamma)$ implies that $p^{(n+m\lambda)/d}H^{-v_{i}}<H^{-1}$ whenever $H> H_0$ . We will use this observation a few times in this proof.

We will now prove two statements concerning the integer solution $(b_{0},b_{1},\dots, b_{n})$ to (7·4). First we verify that $b_0\neq0$ . Suppose the contrary, that is $b_{0}=0$ . Then by the first inequality of (7·4) we have that $|b_{i}|_{p}<p^{(n+m\lambda)/d}H^{-v_{i}}<H^{-1}$ . As $|b_{i}| \leq H$ and $H>H_{0}$ , we have that $b_{i}=0$ for $1 \leq i \leq d$ . Considering the second set of inequalities of (7·4), for each $1 \leq j \leq m$ we have that $|b_{d+j}|_{p}<H^{-\tau_{j}}$ which also forces us to conclude that $b_{d+j}=0$ , since $\tau_{j}>1$ for each $1 \leq j \leq m$ . Thus $(b_{0}, b_{1}, \dots, b_{n})=\boldsymbol{0}$ , a contradiction. So we must have that $b_{0}\neq 0$ .

Now we show that ${b_{i}}/{b_{0}}$ is a p-adic integer for all $1 \leq i \leq d$ . Since $b_{0} \neq 0$ , we may rewrite the first inequality of (7·4) to get

\begin{equation*} |b_{0}|_{p}\left| x_{i}-\frac{b_{i}}{b_{0}} \right|_{p} <p^{(n+m\lambda)/d}H^{-v_{i}}, \quad 1 \leq i \leq d. \end{equation*}

Suppose that $\left|{b_{i}}/{b_{0}}\right|_{p}>1$ for some $1 \leq i \leq d$ , then $\left|{b_{i}}/{b_{0}}\right|_{p}>|x_{i}|_{p}$ since $\boldsymbol{{x}} \in \mathcal{U} \subseteq \mathbb{Z}_{p}^{d}$ so, by the strong triangle inequality, we have that

\begin{equation*} |b_{i}|_{p}=|b_{0}|_{p} \max \!\left\{ |x_{i}|_{p},\left|\frac{b_{i}}{b_{0}}\right|_{p} \right\}= |b_{0}|_{p}\left| x_{i}-\frac{b_{i}}{b_{0}} \right|_{p}<p^{(n+m\lambda)/d}H^{-v_{i}}<H^{-1} \end{equation*}

for $H>H_0$ . Such inequality fails unless $b_{i}=0$ , since $|b_{i}| \leq H$ . Thus, ${b_{i}}/{b_{0}} \in \mathbb{Z}_{p}$ for all $1 \leq i \leq d$ .

Now we are ready to construct $(a_0,\dots,a_n)$ with $(a_{0},p)=1$ . Let $k\ge0$ be the unique integer such that $p^{k}|b_{0}$ but $p^{k+1} \nmid b_{0}$ . Then, since ${b_{i}}/{b_{0}} \in \mathbb{Z}_{p}$ so we have that $p^{k}|b_{i}$ for all $1 \leq i \leq d$ . By (7·4), we get that

\begin{align*} |b_{d+j}|_p &\le \max\!\left\{\left| b_0f_{j}(\boldsymbol{{x}})-\sum_{i=1}^{d}\frac{\partial f_{j}}{\partial x_{i}}(\boldsymbol{{x}}) \left(b_0x_{i}-b_{i}\right)-b_{d+j}\right|_{p}, |b_0f_{j}(\boldsymbol{{x}})|_p,\left|\sum_{i=1}^{d}\frac{\partial f_{j}}{\partial x_{i}}(\boldsymbol{{x}}) \left(b_0x_{i}-b_{i}\right)\right|_p\right\}\\[2ex] & \le \max\!\left\{H^{-\tau_{j}},p^{-k},p^\lambda p^{(n+m\lambda)/d}H^{-v_{\min}} \right\}=p^{-k}, \end{align*}

since $\tau_j>1$ and $H>H_0$ . Therefore, $p^k|b_{d+j}$ and we have that ${b_{d+j}}/{b_{0}} \in \mathbb{Z}_{p}$ for each $1 \leq j \leq m$ . In particular we have that $d=\gcd\!(b_{0},b_{1},\dots, b_{n})=p^k$ . For $0 \leq i \leq n$ define the numbers $a_{i}=p^{-k}b_{i}$ , which, by what we have proven above, are all integers satisfying $\gcd\!(a_{0},a_{1},\dots, a_{n})=1$ and, by the choice of k, $(a_{0},p)=1$ . By the third inequality of (7·4), we have that $\max_{0 \leq i \leq n} |a_{i}| \leq p^{-k}H$ , which verifies the third inequality in (7·2). Further, using the first set of inequalities of (7·4), we get that

(7·5) \begin{equation} |a_{0}x-a_{i}|_{p} = |p^{-k}b_{0}x-p^{-k}b_{i}|_{p} =p^{k}|b_{0}x-b_{i}|_{p} <p^{(n+m\lambda)/d} p^kH^{-v_{i}} \end{equation}

for each $1 \leq i \leq d$ , since $v_i>1$ . This verifies the first set of inequalities in (7·2).

By (7·5) and the fact that $p^k\le H$ , we get that

\begin{equation*}\left( \frac{a_{1}}{a_{0}}, \dots , \frac{a_{d}}{a_{0}} \right)\in B\!\left( \boldsymbol{{x}}, p^{(n+m\lambda)/d}H^{-v_{\min}+1} \right) \subseteq B(\boldsymbol{{x}},\varepsilon)\subseteq \mathcal{U},\end{equation*}

where the last inclusion follows from condition $(\beta)$ on $H_{0}$ . Thus, $\boldsymbol{{y}}=\left( {a_{1}}/{a_{0}}, \dots , {a_{d}}/{a_{0}} \right) \in \mathcal{U}$ and, in particular, $f_{j}\left({a_{1}}/{a_{0}}, \dots , {a_{d}}/{a_{0}}\right)$ is well defined and (2·6) is applicable to $f=f_j$ for each $1 \leq j \leq m$ .

Using the fact that each $f_{j}$ is DQE at $\boldsymbol{{x}}$ we get that

(7·6) \begin{align} \left| f_{j} \left(\frac{a_{1}}{a_{0}}, \dots , \frac{a_{d}}{a_{0}} \right)-f_{j}(\boldsymbol{{x}})-\sum_{1 \leq i \leq d} \frac{\partial f_{j}}{\partial x_{i}}(\boldsymbol{{x}}) \left(\frac{a_{i}}{a_{0}}-x_{i} \right) \right|_{p} & < C\max_{1 \leq i \leq d} \left| \frac{a_{i}}{a_{0}}-x_{i} \right|_{p}^{2}\end{align}

\begin{align}& < (p^{-k}H)^{-\tau_{j}} \nonumber\end{align}

for each $1 \leq j \leq m$ , where the last inequality follows since

\begin{align*}C\max_{1 \leq i \leq d} \left| \frac{a_{i}}{a_{0}}-x_{i} \right|_{p}^{2} & \;\stackrel{(7.5)}{<}\; C p^{(2n+2m\lambda)/d} p^{2k}H^{-2v_{\min}}\\& = C p^{(2n+2m\lambda)/d}p^{-2k(v_{\min}-1)} (p^{-k}H)^{-2v_{\min}}\\[1ex]& \stackrel{(*)}{\le} (p^{-k}H)^{-\tau_{\max}} \le (p^{-k}H)^{-\tau_{j}}\,.\end{align*}

Here $(*)$ follows from condition $(\alpha1)$ on $H_{0}$ if $p^k\le H_0^{1/2}$ and it follows from condition $(\alpha2)$ on $H_{0}$ if $p^k>H_0^{1/2}$ , and we also use the facts that $v_{\min}>1$ and $2v_{\min}>\tau_{\max}$ .

For each $1 \leq j \leq m$ in the second row of inequalities of (7·4) we may divide through by $p^k=|b_{0}|^{-1}_{p}$ and $p^\lambda$ , and combine with (7·6) to obtain

\begin{equation*}\left| f_{j} \left( \frac{a_{1}}{a_{0}}, \dots , \frac{a_{d}}{a_{0}} \right)-\frac{a_{d+j}}{a_{0}} \right|_{p}< (p^{-k}H)^{-\tau_{j}}\end{equation*}

for each $1 \leq j \leq m$ . This verifies the second set of inequalities in (7·2), while the first set of inequalities in (7·2) follows from (7·5). The proof is thus complete.

Proof. First, observe that (7·7) is a consequence of (7·2) since $h=\max_{0 \leq i \leq n} |a_{i}| \leq p^{-k}H$ and $v_i>1$ for all i. So we only need to verify that there are infinitely many different solutions $(a_{0}, \dots, a_{n}) $ to (7·2) as H varies. Suppose the contrary. Then, since $\boldsymbol{{x}} \in \mathbb{Z}_{p}^{d} \backslash \mathbb{Q}^{d}$ , there is $1 \leq i \leq d$ such that $x_{i}-{a_{i}}/{a_{0}} \neq 0$ and so

(7·8) \begin{equation} \delta\,:\!=\,\min \!\left|x_{i}-\frac{a_{i}}{a_{0}}\right|_{p}>0,\end{equation}

where the minimum is taken amongst the solutions $(a_{0}, a_{1},\dots, a_{n})$ to (7·2) over all $H\ge H_0$ . On the other hand, by (7·2), we have $\delta< p^{(n+m\lambda)/d}p^{k}H^{-v_i}\le p^{(n+m\lambda)/d}H^{-v_i+1}\to0$ as $H\to\infty$ since $v_i>1$ , giving a contradiction for large H.

Proof. Define the set of integer points

(7·10) \begin{equation}S_{\boldsymbol{\tau}}=\left\{(a_{0},\dots, a_{n}) \in \mathbb{Z}^{n+1}:\, \begin{array}{l}\text{(7.3) holds and for all }1 \leq j \leq m\\[1ex]\left| f_{j}\left(\frac{a_{1}}{a_{0}}, \dots , \frac{a_{d}}{a_{0}} \right) -\frac{a_{d+j}}{a_{0}}\right|_{p}<h^{-\tau_{d+j}}\\[1.5ex]\text{where }\underset{0 \leq i \leq n}{\max} |a_{i}| = h\end{array}\right\},\end{equation}

and for each $\boldsymbol{{a}}\in S_{\boldsymbol{\tau}}$ and $\delta>0$ consider the hyperrectangles

(7·11) \begin{equation}B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\,\delta)=\left\{ \boldsymbol{{x}} \in \mathbb{Z}_{p}^{d}\,:\, \left|x_{i}-\frac{a_{i}}{a_{0}}\right|_{p}<\delta h^{-\tau_{i}} \qquad (1 \leq i \leq d ) \right\}.\end{equation}

By Corollary 7·2,

(7·12) \begin{equation}\bigcup_{\delta>0}\underset{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}\limsup B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\,\delta)=\mathcal{U}\setminus\mathbb{Q}_p^d\end{equation}

and therefore this union has full measure in $\mathcal{U}$ , since the sequence of sets in (7·12) is increasing as $\delta$ increases. These are Borel sets and therefore measurable. Hence, by the continuity of measure, we have that

(7·13) \begin{equation}\lim_{\delta\to+\infty}\mu_{p, n}\!\left(\underset{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}\limsup B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\,\delta)\right)=\mu_{p, n}\!\left(\bigcup_{\delta>0}\underset{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}\limsup B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\,\delta)\right)=\mu_{p, n}(\mathcal{U})\,.\end{equation}

By Lemma 4·2, every limsup set in (7·13) is of the same measure. Hence,

\begin{equation*}\mu_{p, n}\!\left(\underset{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}\limsup B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\,\delta)\right)=\mu_{p, n}(\mathcal{U})\end{equation*}

for every $\delta>0$ . This is exactly what we had to prove.

Proof. Suppose $\boldsymbol{{x}}\in \mathcal{U}^*\cap B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\,\delta)$ . Then

\begin{align*}\left|f_{j}(\boldsymbol{{x}})-f_{j}\left(\frac{a_{1}}{a_{0}}, \dots , \frac{a_{d}}{a_{0}} \right)\right|_{p} & <\max\!\left\{\max_{1\le i\le d}\left|\frac{\partial f_j(\boldsymbol{{x}})}{\partial x_i}\right|_p\max_{1 \leq i \leq d}\left|x_{i}-\frac{a_{i}}{a_{0}} \right|_{p},C\max_{1 \leq i \leq d}\left|x_{i}-\frac{a_{i}}{a_{0}} \right|_{p}^2 \right\}\\[2ex] & <\max\!\left\{\max_{1\le i\le d}\left|\frac{\partial f_j(\boldsymbol{{x}})}{\partial x_i}\right|_p \delta h^{-\tau_{\min}},C\delta^2 h^{-2\tau_{\min}} \right\} < h^{-\tau_{d+j}}\end{align*}

for any $1 \leq j \leq m$ and all sufficiently large h if (8·2) holds. In turn, if (8·3) holds, we use the fact that $\mathbf{f}$ is Lipschitz:

\begin{align*}\left|f_{j}(\boldsymbol{{x}})-f_{j}\left(\frac{a_{1}}{a_{0}}, \dots , \frac{a_{d}}{a_{0}} \right)\right|_{p} & < L\max_{1 \leq i \leq d}\left|x_{i}-\frac{a_{i}}{a_{0}} \right|_{p} < L\delta h^{-\tau_{\min}} \le h^{-\tau_{d+j}}\end{align*}

for any $1 \leq j \leq m$ and all sufficiently large h since $0<\delta\le L^{-1}$ . In either case, if $\boldsymbol{{a}}\in S_{\boldsymbol{\tau}}$ , then

\begin{equation*}\left|f_{j}(\boldsymbol{{x}})-\frac{a_{d+j}}{a_0}\right|_{p}\le \max\!\left\{\left|f_{j}(\boldsymbol{{x}})-f_{j}\left(\frac{a_{1}}{a_{0}}, \dots , \frac{a_{d}}{a_{0}} \right)\right|_{p},\left|\frac{a_{d+j}}{a_0}-f_{j}\left(\frac{a_{1}}{a_{0}}, \dots , \frac{a_{d}}{a_{0}} \right)\right|_{p}\right\} < h^{-\tau_{d+j}}\end{equation*}

provided that h is sufficiently large. Hence, assuming that $\boldsymbol{{x}}\in \mathcal{U}^*\cap \underset{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}\limsup B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\delta)$ we conclude that the system of inequalities

(8·4) \begin{align}\begin{cases}|a_{0}x_{i}-a_{i}|_{p}<\delta h^{-\tau_{i}}\le h^{-\tau_{i}}, \quad& (1 \leq i \leq d), \\|a_{0}f_{j}(\boldsymbol{{x}})-a_{d+j}|_{p}< h^{-\tau_{d+j}} & (1 \leq j \leq m), \\\max \{|a_{0}|,\dots,|a_{n}| \} =h,\end{cases}\end{align}

holds for infinitely many $\boldsymbol{{a}}\in\mathbb{Z}^{n+1}$ . Therefore, $\boldsymbol{{x}}\in \mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau}))$ and the proof is complete.

Proof of Theorems 2·10—2·11. First of all, note that (2·7) and (2·9) follow from Theorem 2·12. Thus we only need to verify the measure part of these theorems, that is (2·8) and (2·10). Consequently, we will assume that $\mathbf{f}$ is Lipschitz on $\mathcal{U}$ . Let $0<\delta\le \min\{1,L^{-1}\}$ , where L is the Lipschitz constant of $\mathbf{f}$ . With reference to the General mass transference principle (Theorem 3·5), take the function $g(x)=x^{d}$ as our dimension function. Note that g is doubling and that $\mathcal{H}^{g}\asymp \mu_{p,d}$ . For any ball $B=B(x,r)$ and dimension function $f(x)=x^{s}$ , define $B^{s}= B\!\left(x,g^{-1}(x^{s})\right)$ . Note that in Theorems 2·10 and 2·11 we have that $\tau_{1}=\tau_{2}= \dots=\tau_{d}$ . Therefore the sets $B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\,\delta)$ defined by (7·11) are balls. Let the vector $\mathbf{v}=(v_{1}, \dots , v_{d})$ be of the form $\mathbf{v}=(v, \dots , v)$ , where

\begin{equation*}v= \frac{n+1-\sum_{i=1}^{m}\tau_{d+i}}{d}.\end{equation*}

Note that this $\mathbf{v}$ satisfies the requirements of Theorem 7·1 and its corollaries. Let

\begin{equation*}s=\frac{n+1-\sum_{i=1}^{m}\tau_{d+i}}{\tau_{d}},\end{equation*}

then

\begin{equation*}B^{s}_{\boldsymbol{{a}}}(\tau_{d};\,\delta)= \left\{ \boldsymbol{{x}} \in \mathbb{Z}_{p}^{d} \,:\, \max_{1 \leq i \leq d}\left|x_{i}-\frac{a_{i}}{a_{0}}\right|_{p}<\delta^{s/d}h^{-v} \right\}\end{equation*}

and, by Corollary 7·3,

\begin{equation*} \mu_{p,d}\!\left(\underset{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}{\limsup}B^{s}_{\boldsymbol{{a}}}(\tau_{d};\,\delta) \right)=\mu_{p,d}(\mathcal{U}). \end{equation*}

Hence, for any ball $B \subset \mathcal{U}$ ,

\begin{equation*}\mathcal{H}^{g}\!\left( B \cap \limsup_{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}B^{s}_{\boldsymbol{{a}}}(\tau_{d};\,\delta) \right) = \mathcal{H}^{g}\!\left(B \right).\end{equation*}

By the Mass transference principle (Theorem 3·5), we have that for any ball $B \subseteq \mathcal{U}$ ,

(8·5) \begin{equation} \mathcal{H}^{s}\!\left( B \cap \limsup_{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}B^{g}_{\boldsymbol{{a}}}(\tau_{d};\,\delta) \right) = \mathcal{H}^{s}\!\left(B \right).\end{equation}

By Proposition 8·1 and the choice of $\delta$ , we have that (8·1) holds, where $\mathcal{U}^*=\mathcal{U}$ . Combining (8·5) and (8·1) gives the required Hausdorff measure results and completes the proof.

Proof of Theorem 2·12. First of all, without loss of generality we can assume throughout this proof that (8·2) holds. Otherwise we could consider $\boldsymbol{\tau}^{\prime}=(\tau_1+\varepsilon,\dots,\tau_d+\varepsilon,\tau_{d+1},\dots,\tau_{n})$ for a suitably small $\varepsilon>0$ and note that $\mathbf{F}^{-1}\!\left(\mathfrak{W}_n(\boldsymbol{\tau}^{\prime})\right)\subset \mathbf{F}^{-1}\!\left(\mathfrak{W}_n(\boldsymbol{\tau})\right)$ . Hence, the validity of (2·11) for $\boldsymbol{\tau}^{\prime}$ would give us the bound

\begin{equation*}\dim \!\left(\mathbf{F}^{-1}\left(\mathfrak{W}_n(\boldsymbol{\tau})\right)\right) \ge \dim \!\left(\mathbf{F}^{-1}\left(\mathfrak{W}_n(\boldsymbol{\tau}^{\prime})\right)\right) \geq \min_{1 \leq i \leq d}\left\{ \frac{n+1+\sum_{\tau_{j} < \tau_{i}}\big(\tau_{i}-\tau_{j}\big)}{\tau_{i}+\varepsilon}-m \right\}\end{equation*}

and on letting $\varepsilon\to0$ we would get the required result for $\boldsymbol{\tau}$ .

Now, since (8·2) holds, by Proposition 8·1 with $\delta=1$ , get that

(8·6) \begin{equation}\underset{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}\limsup B_{\boldsymbol{{a}}}(\boldsymbol{\tau};\,1)\subset \mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau})).\end{equation}

Corollary 7·3 provides us with a full measure statement, which will be the basis for applying the Mass transference principle from rectangles to rectangles without Ubiquity (Theorem 3·9). With reference to the notation used in Theorem 3·9 take

\begin{equation*}\begin{array}{ll}J=S_{\boldsymbol{\tau}}, & \rho(q)=q^{-1},\\[1ex]R_{\alpha}=\left\{ \left(\frac{a_{1}}{a_{0}}, \dots , \frac{a_{n}}{a_{0}}\right)\right\},\quad & \beta_{\alpha}=a_0 \qquad \text{for }\alpha=(a_0,\dots,a_n) \in S_{\boldsymbol{\tau}}\end{array}\end{equation*}

and so

(8·7) \begin{equation}\limsup_{\boldsymbol{{a}} \in S_{\boldsymbol{\tau}}}B_{\boldsymbol{{a}}}(\mathbf{v};\,1)= \limsup_{\alpha \in J} \Delta\!\left(R_{\alpha},\rho(\beta_{\alpha})^{-\mathbf{v}}\right).\end{equation}

By Corollary 7·3 and (8·7), we have that

(8·8) \begin{equation}\mu_{p,d}\!\left(\limsup_{\alpha \in J} \Delta\!\left(R_{\alpha},\rho(\beta_{\alpha})^{-\mathbf{v}}\right) \right)=\mu_{p,d}(\mathcal{U})\end{equation}

for any $\mathbf{v}=(v_{1}, \dots , v_{d}) \in \mathbb{R}^{d}_{+}$ satisfying

(8·9) \begin{equation} v_{i}>1, \quad \sum_{i=1}^{d}v_{i}=n+1-\sum_{j=1}^{m}\tau_{j}.\end{equation}

Without loss of generality we will assume that $\tau_{1} > \tau_{2} > \dots > \tau_{d}$ . Similarly to what proceeds the proof of Proposition 6·1 define each $v_{i}$ recursively, starting with $r=0$ , by

\begin{equation*}v_{d-r}=\min\!\left\{ \tau_{d-r}, \frac{n+1-\sum_{j=1}^{m}\tau_{d+j}-\sum_{i=d-r+1}^{d}v_{i}}{d-i} \right\}.\end{equation*}

Observe that this choice of $\mathbf{v}$ satisfies (8·9). Furthermore, there exists a $1 \leq b \leq d$ such that

\begin{equation*}v_{c}=\frac{n+1-\sum_{j=1}^{m}\tau_{d+j}-\sum_{i=d-b}^{d}v_{i}}{d-b}\end{equation*}

for all $1 \leq c \leq d-b$ . Define $t_1,\dots,t_d$ from the equations

\begin{equation*}\tau_{j}=v_{j}+t_{j}\end{equation*}

then note that $\boldsymbol{{t}}=(t_{1}, \dots , t_{d}) \in \mathbb{R}^{d}_{\geq0}$ and thus satisfies the conditions of Theorem 3·9. Thus, the set $W(\boldsymbol{{t}})$ , defined in Theorem 3·9, is exactly the right–hand side of (8·6). Hence, by (8·6), we get that

\begin{equation*}\dim \mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau}))\ge \dim W(\boldsymbol{{t}})\,.\end{equation*}

Also, in view of (8·8), Theorem 3·9 is applicable and so $\dim \mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau}))\ge s$ , where s is the same as in Theorem 3·9. The proof is now split into the following three cases.

1. (i) $A_{i} \in \{ v_{1}, \dots v_{d-b} \}$ : For these values of $A_{i}$ , which are defined in Theorem 3·9, we have that

   \begin{equation*}\begin{array}{ccc}K_{1}=\{1, \dots , d-b \}, & K_{2}=\{ d-b+1, \dots , d\}, & K_{3}=\emptyset.\end{array}\end{equation*}
   Applying Theorem 3·9 gives
   \begin{align*}\dim \mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau})) & \ge \dim W(\boldsymbol{{t}}) \\& \geq \min_{1 \leq i \leq d-b} \left\{ \frac{(d-b)v_{i}+(d-(d-b+1)+1)v_{i} -\sum_{j=d-b}^{n}t_{j}}{v_{i}} \right\}, \\&=\min_{1 \leq i \leq d-b} \left\{ d-\frac{\sum_{j=d-b+1}^{d}t_{j}}{v_{i}} \right\}. \\\end{align*}

   Since $t_{i}=0$ for $d-b+1 \leq i \leq d$ we have that $\dim \mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau}))\geq d$ , which is the maximal possible dimension for $\mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau}))$ .

2. (ii) $A_{i} \in \{ v_{d-b+1}, \dots , v_{d} \}$ : For such values of $A_{i}$ observe that

   \begin{equation*}\begin{array}{ccc}K_{1}=\{1, \dots , i \}, & K_{2}=\{i+1, \dots , d\}, & K_{3}=\emptyset.\end{array}\end{equation*}
   Then in this case we have that
   \begin{equation*}\dim \mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau}))\ge \dim W(\boldsymbol{{t}}) \geq \min_{d-b+1 \leq i \leq d}\left\{ \frac{iv_{i} + (d-i)v_{i} - \sum_{j=i+1}^{d} t_{j}}{v_{i}} \right\}.\end{equation*}
   Similarly to the previous case, since $t_{j}=0$ for $d-b+1 \leq j \leq d$ the r.h.s of the above equation is d.

3. (iii) $A_{i} \in \{\tau_{1}, \dots , \tau_{d}\}$ : Since $\tau_{i}=v_{i}$ for $d-b+1 \leq i \leq d$ , ii) covers such result. So we only need to consider the set of $A_{i} \in \{ \tau_{1}, \dots \tau_{d-b} \}$ . If $A_{i}$ is contained in such set, then

   \begin{equation*}\begin{array}{ccc}K_{1}=\emptyset, & K_{2}=\{ i, \dots , d\}, & K_{3}=\{ 1, \dots , i-1\}.\end{array}\end{equation*}
   Thus, by Theorem 3·9, we have that
   \begin{align*}& \dim \mathbf{F}^{-1}(\mathfrak{W}_n(\boldsymbol{\tau}))\geq \min_{1 \leq i \leq d} \left\{ \frac{(d-i+1)\tau_{i}+\sum_{j=1}^{i-1}v_{j}-\sum_{j=i}^{d}t_{j}}{\tau_{i}} \right\} \\&= \min_{1 \leq i \leq d} \left\{ \frac{(d-i+1)\tau_{i}+(i-1)\left( \frac{n+1-\sum_{j=1}^{m}\tau_{d+j}-\sum_{j=d-b+1}^{d}v_{j}}{d-b} \right)-\sum_{j=i}^{d-b}(\tau_{j}-v_{j})-\sum_{j=d-b+1}^{d}t_{j}}{\tau_{i}} \right\} \\& = \min_{1 \leq i \leq d}\left\{ \frac{(d-i+1)\tau_{i}+(d-b)\left( \frac{n+1-\sum_{j=1}^{m}\tau_{d+j}-\sum_{j=d-b+1}^{d}v_{j}}{d-b} \right)-\sum_{j=i}^{d-b}\tau_{j}-\sum_{j=d-b+1}^{d}t_{j}}{\tau_{i}} \right\} \\&=\min_{1 \leq i \leq d} \left\{ \frac{n+1+\sum_{j=i}^{d}\big(\tau_{i}-\tau_{j}\big)-\sum_{j=1}^{m}\tau_{d+j}}{\tau_{i}} \right\} \\&=\min_{1 \leq i \leq d}\left\{ \frac{n+1+\sum_{j=i}^{n}\big(\tau_{i}-\tau_{j}\big)}{\tau_{i}}-m \right\}.\end{align*}

Considering all cases we have that

\begin{equation*}\dim \mathbf{F}^{-1}(\mathfrak{W}(\boldsymbol{\tau}) )\geq \dim W(\boldsymbol{{t}}) \geq \min_{1 \leq i \leq d}\left\{ \frac{n+1+\sum_{j=i}^{n}\big(\tau_{i}-\tau_{j}\big)}{\tau_{i}}-m \right\}\end{equation*}

as required.