Proof. Observe that since $0\in Q_0$ , and by (2·7) $Q_0\subset B(p_{Q_0},\Lambda\ell(Q_0))$ , we have $Q_0\subset B(0,2\Lambda\ell(Q_0)).$ Hence,

\begin{align*} T_{\kappa} &\subset \left\{ (z,t)\in B(0,2\Lambda\ell(Q_0))\ |\ |z|\le \kappa\, 2^{-N} \ell(Q_0) \right\}\\[5pt] &\subset \left\{ (z,t)\in {\mathbb{H}}^n\ |\ |z|\le \kappa\, 2^{-N} \ell(Q_0),\ 16|t|^2\le (2\Lambda \ell(Q_0))^4 \right\}\;=\!:\; \widetilde{T}_{\kappa}. \end{align*}

By (2·3),

\begin{equation*} \mathcal{H}^{D}(\widetilde{T}_{\kappa}) = C\mathcal{L}^{2n+1}(\widetilde{T}_{\kappa}) \approx (\kappa 2^{-N} \ell(Q_0))^{2n}(2\Lambda \ell(Q_0))^2 \approx_{\kappa} 2^{-2nN}\ell(Q_0)^{D}. \end{equation*}

It follows that $\mathcal{H}^{D}(T_{\kappa})\lesssim_{\kappa}2^{-2nN}\ell(Q_0)^{D}.$ On the other hand, recall that for any cube Q with sidelength $\ell(Q)=2^{-N}\ell(Q_0)$ we have $\mathcal{H}^{D}(Q)\approx 2^{-ND}\ell(Q_0)^{D}$ . Since all such cubes are pairwise disjoint, we get

\begin{equation*} \#\left\{Q\in\mathcal{D}\ |\ \ell(Q)=2^{-N}\ell(Q_0),\ Q\subset T_{\kappa} \right\} \lesssim \frac{\mathcal{H}^{D}(T_{\kappa})}{2^{-ND}\ell(Q_0)^{D}}\lesssim_{\kappa} \frac{2^{-2nN}\ell(Q_0)^{D}}{2^{-N(2n+2)}\ell(Q_0)^{D}}=2^{2N}. \end{equation*}

Proof. We argue by contradiction. To this end, let us assume that there exists a cube $Q_2 \subset Q_0 \setminus T$ with $\ell(Q_2)=2^{-AN}\ell(Q_0)$ such that (3·5) does not hold - that is

(3·6) \begin{equation} \mu(Q_2) \geq \frac{\mu(Q_0)}{\Theta_0 \,2^{AND}}. \end{equation}

Let $0< \delta < \text{dist}(Q_1, Q_2)$ , let $p\in Q_2$ be arbitrary, and consider

\begin{equation*} R_{\mu,\delta}(\unicode{x1d7d9}_{Q_1})(p) = \int_{Q_1} K(q^{-1} \cdot p) \, d\mu(q). \end{equation*}

By triangle inequality,

(3·7) \begin{equation} |R_{\mu,\delta}(\unicode{x1d7d9}_{Q_1})(p)| \geq \left| \int_{Q_1} K(p) \, d\mu(q) \right| - \left| \int_{Q_1} K(q^{-1}\cdot p) - K(p) \, d\mu(q) \right|. \end{equation}

We estimate the first term as follows. Note that, since $p \in Q_2$ and $Q_2$ lies outside T, then, writing $p=(z, t)$ and using (2·4), we have

\begin{align*} |K(p)|^2 \approx \frac{|z|^2}{(|z|^4 + 16t^2)^{n+1}} \gtrsim \frac{|z|^2}{\ell(Q_0)^{4(n+1)}} \geq 2^{-2N}\ell(Q_0)^{-4n-2} = 2^{-2N}\ell(Q_0)^{-2D+2}. \end{align*}

And thus we also have

(3·8) \begin{equation} \left| \int_{Q_1} K(p)\, d\mu(q) \right| = \left| K(p) \right|\mu(Q_1) \gtrsim 2^{-N} \, \frac{\mu(Q_1)}{\ell(Q_0)^{D-1}}. \end{equation}

For the second term in (3·7) we use the continuity of the kernel K (2·6) and the fact that $d(p,q) \approx \|p\|_{{\mathbb{H}}}\ge 2^{-N}\ell(Q_0)$ (because $p\in Q_2\subset Q_0\setminus T$ ):

(3·9) \begin{align} |K(q^{-1} \cdot p) - K(p)| \lesssim \frac{\|q\|}{\min(\|p\|_{{\mathbb{H}}}, d(p,q))^{D}}\lesssim \frac{2^{-AN}\ell(Q_0)}{(2^{-N}\ell(Q_0))^{D}} = \frac{2^{-AN+DN}}{\ell(Q_0)^{D-1}}. \end{align}

Taking $A\ge 2D$ we get

\begin{equation*} \left| \int_{Q_1} K(q^{-1} \cdot p) - K(p) \, d\mu(q) \right|\lesssim 2^{-AN/2}\frac{\mu(Q_1)}{\ell(Q_0)^{D-1}}. \end{equation*}

Together with (3·8) and (3·7), assuming $N_0$ bigger than some absolute constant (recall that $N\ge N_0$ ), this gives

\begin{equation*} |R_{\mu,\delta}(\unicode{x1d7d9}_{Q_1})(p)|\gtrsim 2^{-N} \, \frac{\mu(Q_1)}{\ell(Q_0)^{D-1}} \end{equation*}

for all $p\in Q_2$ .

Now, we use the estimate above and the $L^2(\mu)$ boundedness of $R_{\mu}$ to get

\begin{equation*} 2^{-N}\frac{\mu(Q_1)}{\ell(Q_0)^{D-1}}\mu(Q_2)^{\frac{1}{2}} \lesssim \left( \int |R_{\mu,\delta}(\unicode{x1d7d9}_{Q_1})(p)|^2 \, d\mu(p) \right)^{\frac{1}{2}} \leq C_1 \mu(Q_1)^{\frac{1}{2}}. \end{equation*}

Our assumptions on $Q_1$ (3·4) and $Q_2$ (3·6) yield

\begin{align*} C_1 & \gtrsim 2^{-N}\frac{\mu(Q_1)^{\frac{1}{2}}\mu(Q_2)^{\frac{1}{2}}}{\ell(Q_0)^{D-1}}\gtrsim 2^{-N}\frac{\mu(Q_0)}{2^{AND}\ell(Q_0)^{D-1}}\Theta_0^{-1/2} = 2^{-AND-N}\Theta_0^{1/2} \notag\\[5pt] & \overset{\scriptsize{({3\!\cdot\!3})}}{\approx} 2^{-AND-N}\, 2^{A^2 N/2}. \end{align*}

Taking $A\ge 5D$ we can bound the last term from below in the following way:

\begin{equation*} 2^{-AND-N+A^2 N/2}\ge 2^{A^2N/4}\overset{\scriptsize{({3\!\cdot\!3})}}{\gtrsim} M^{1/4}. \end{equation*}

Putting together the estimates above gives $C_1\gtrsim M^{1/4}$ , which is a contradiction for $M=M(C_1,n)$ big enough.

We immediately get the following corollary.

Proof. Observe that if $Q\in\mathcal{D}$ satisfies $\ell(Q) = \ell(Q_1) = 2^{-AN}\ell(Q_0)$ and $Q\not\subset T_2$ , then we have $Q\cap T=\varnothing$ (assuming A large enough with respect to $\Lambda$ ). It follows that Q satisfies the assumptions of Lemma 3·3, and so

\begin{equation*} \mu(Q)\le2^{-AND}\Theta_0^{-1}\mu(Q_0). \end{equation*}

Summing over all such Q and using (2·8) yields

\begin{equation*} \mu(Q_0\setminus T_2) \le \Theta_0^{-1}\mu(Q_0). \end{equation*}

Proof. We will prove (3·11) by contradiction. Suppose that

(3·12) \begin{equation} \sum_{Q\in{\mathsf{HD}}(Q_0)}\mu(Q) < (1-\Theta_0^{-s})\mu(Q_0). \end{equation}

Set

\begin{equation*} {\mathsf{LD}}(Q_0) = \left\{Q\in\mathcal{D}\ |\ Q\subset T_{2\Lambda},\ \ell(Q)=2^{-N}\ell(Q_0),\ \Theta(Q)\le 2\Theta_0 \right\}. \end{equation*}

It is easy to see that the cubes from ${\mathsf{HD}}(Q_0)\cup {\mathsf{LD}}(Q_0)$ cover $T_{2}$ . If we assume $\Theta_0\ge M>100$ , and $s<1/2$ , then $\Theta_0^{-s}/2\ge \Theta_0^{-1}$ , and so by (3·10) and (3·12) we get

(3·13) \begin{equation} \sum_{Q\in{\mathsf{LD}}(Q_0)}\mu(Q) \ge \frac{\Theta_0^{-s}}{2}\mu(Q_0). \end{equation}

On the other hand, recall from Lemma 3·2 that there are at most $C2^{2N}$ cubes of sidelength $2^{-N}\ell(Q_0)$ contained in $T_{2\Lambda}$ , where $C=C(\Lambda,n)$ . Moreover, for any $Q\in {\mathsf{LD}}(Q_0)$ we have

\begin{equation*} \mu(Q)\le 2 \Theta_0 \ell(Q)^{D-1}= 2\, \mu(Q_0)\frac{\ell(Q)^{D-1}}{\ell(Q_0)^{D-1}} =2^{-N(D-1)+1}\mu(Q_0). \end{equation*}

In consequence,

\begin{equation*} \sum_{Q\in{\mathsf{LD}}(Q_0)}\mu(Q) \le C 2^{2N} 2^{-N(D-1)+1}\mu(Q_0). \end{equation*}

This contradicts (3·13) because

\begin{equation*} C\, 2^{-ND+3N+1} = 2\, C\, (2^{-A^2N})^{(-D+3)A^{-2}} \overset{\scriptsize{({3\!\cdot\!3})}}{\le}\widetilde{C}(n)\Theta_0^{(-D+3)A^{-2}}\le \frac{\Theta_0^{-s}}{2}, \end{equation*}

choosing $s=s(A,n)$ small enough.

Proof. We will prove that for $Q\in{\mathsf{HD}}(Q_0)$ we have $Q\cap T_2\neq\varnothing.$ Then, since $\ell(Q)=2^{-N}\ell(Q_0)$ , it follows easily from (2·7) that indeed $Q\subset T_{\Lambda+2}(Q_0)\subset T_{2\Lambda}(Q_0)$ .

We argue by contradiction. Suppose that $Q\in {\mathsf{HD}}(Q_0)$ and $Q\cap T_2=\varnothing$ . Consider the cubes $\{P_i\}_{i\in I}$ with $\ell(P_i)=2^{-AN}\ell(Q_0) = 2^{-(A-1)N}\ell(Q)$ and $P_i\subset Q$ . Then, $Q=\bigcup_i P_i,$ for all $i\in I$ we have $P_i\cap T_2=\varnothing,$ and $\# I\approx 2^{(A-1)ND}$ by (2·8).

We use Lemma 3·3 to conclude that for all $i\in I$

\begin{equation*} \mu(P_i) \le \frac{\mu(Q_0)}{\Theta_0 \, 2^{AND}}. \end{equation*}

Summing over $i\in I$ yields

\begin{equation*} \mu(Q) = \sum_{i\in I}\mu(P_i)\le \# I\cdot \frac{\mu(Q_0)}{\Theta_0 \, 2^{AND}} \approx 2^{(A-1)ND} \frac{\mu(Q_0)}{\Theta_0 \, 2^{AND}} = \frac{\mu(Q_0)}{\Theta_0 \, 2^{ND}}, \end{equation*}

so that

\begin{equation*} \Theta(Q) = \frac{\mu(Q)}{(2^{-N}\ell(Q_0))^{D-1}} \lesssim \frac{\mu(Q_0)}{\Theta_0 \, 2^{ND}}\cdot \frac{1}{2^{-N(D-1)}\ell(Q_0)^{D-1}} = \frac{\Theta_0}{\Theta_0 \, 2^{N}}=2^{-N}\le 1. \end{equation*}

But this contradicts the assumption $Q\in {\mathsf{HD}}(Q_0)$ :

\begin{equation*} \Theta(Q)\ge 2\Theta_0\ge 2M>1, \end{equation*}

and so the proof of (3·14) is finished.

Concerning (3·15), note that by (3·14) and Lemma 3·2 we have

(3·16) \begin{equation} \# {\mathsf{HD}}(Q_0) \lesssim 2^{2N}. \end{equation}

Hence,

\begin{align*} \sum_{Q \in {\mathsf{HD}}(Q_0)} \ell(Q)^2 = \ell(Q_0)^2\, 2^{-2N} \sum_{Q \in {\mathsf{HD}}(Q_0)} 1 \lesssim \ell(Q_0)^2. \end{align*}

Proof. Observe that for $Q\in{\mathsf{HD}}_j$ we have

(4·1) \begin{equation} \Theta(Q)\ge 2^{j}\Theta(Q_0)\ge 2^{j}M.\end{equation}

In particular, $\Theta(Q)\ge M$ and so we may apply Lemma 3·1 to Q. It follows that for any $j \geq 0$ we have

\begin{align*} \mu(Z_{j+1})=\sum_{Q \in {\mathsf{HD}}_{j+1}} \mu(Q) = \sum_{Q \in {\mathsf{HD}}_{j}} \sum_{P \in {\mathsf{HD}}(Q)} \mu(P) \stackrel{\scriptsize{({3\!\cdot\!1})}}{\geq} \sum_{Q \in {\mathsf{HD}}_{j}} (1-\Theta(Q)^{-s})\mu(Q)\\[5pt] \overset{\scriptsize{({4\!\cdot\!1})}}{\ge} \sum_{Q \in {\mathsf{HD}}_{j}} (1-2^{-js}M^{-s})\mu(Q) = (1-2^{-js}M^{-s}) \mu(Z_{j}).\end{align*}

Using this estimate $(j+1)$ times we arrive at

(4·2) \begin{equation} \mu(Z_{j+1}) \geq \prod_{i=0}^{j}(1-2^{-is}M^{-s}) \mu(Q_0).\end{equation}

Since $Z_j$ form a sequence of decreasing sets, we get by the continuity of measure

\begin{equation*} \mu(Z) = \lim_{j\to\infty} \mu(Z_j)\ge \prod_{i=0}^{\infty}(1-2^{-is}M^{-s}) \mu(Q_0) = C(s,M)\mu(Q_0),\end{equation*}

where C(s, M) is positive and finite because $\sum_{i=0}^{\infty} 2^{-is}<\infty$ .

Proof. Recall that $N(Q) = \left\lfloor A^{-2} \log\!(\Theta(Q)) \right\rfloor$ . It follows from (4·1) that for $Q\in{\mathsf{HD}}_j$ we have $N(Q)\ge C_3jA^{-2}$ for some absolute constant $C_3>0$ . Thus, for $Q\in{\mathsf{HD}}_j$ and $P\in{\mathsf{HD}}(Q)$

\begin{equation*} \ell(P)=2^{-N(Q)}\ell(Q)\le 2^{-C_3jA^{-2}}\ell(Q).\end{equation*}

Using this observation j times we get that for $P\in{\mathsf{HD}}_{j+1}$

\begin{equation*} \ell(P)\le 2^{-C_4 j(j+1)A^{-2}}\ell(Q_0),\end{equation*}

where $C_4=C_3/2.$ Hence, the cubes from ${\mathsf{HD}}_j$ form coverings of Z with decreasing diameters, well suited for estimating the Hausdorff measure of Z.

Let $0<\varepsilon<1,\ 0<\delta<1$ be small. Let $j\ge 0$ be so big that for $Q\in{\mathsf{HD}}_{j}$ we have $diam(Q)\le\Lambda\ell(Q)\le\delta$ . Then,

(4·3) \begin{equation} \mathcal{H}_{\delta}^{2+\epsilon}(Z) \le \Lambda^{2+\varepsilon}\sum_{Q\in{\mathsf{HD}}_j}\ell(Q)^{2+\varepsilon}\le \Lambda^{2+\varepsilon}(2^{-C_4j(j-1)A^{-2}}\ell(Q_0))^{\varepsilon}\sum_{Q\in{\mathsf{HD}}_j}\ell(Q)^{2}.\end{equation}

It follows by (3·2) that

\begin{equation*} \sum_{Q\in{\mathsf{HD}}_j}\ell(Q)^{2} = \sum_{P\in{\mathsf{HD}}_{j-1}}\sum_{Q\in{\mathsf{HD}}(P)}\ell(Q)^{2}\le C_p\sum_{P\in{\mathsf{HD}}_{j-1}}\ell(P)^{2}.\end{equation*}

Using the estimate above j times, and putting it together with (4·3) we arrive at

\begin{equation*} \mathcal{H}_{\delta}^{2+\epsilon}(Z) \le \Lambda^{2+\varepsilon} (C_p)^j\, \,2^{-\varepsilon C_4 j(j-1)A^{-2}}\, \ell(Q_0)^{2+\varepsilon}.\end{equation*}

The right hand side above converges to 0 as $j\to\infty$ (just note that the exponent at $C_p$ is linear in j while the exponent at 2 is quadratic in j). Hence, $\mathcal{H}_{\delta}^{2+\epsilon}(Z)=0.$ Letting $\delta\to 0$ we get $\mathcal{H}^{2+\epsilon}(Z)=0.$ Since this is true for arbitrarily small $\epsilon>0$ , it follows that

\begin{equation*} \textstyle{\text{dim}_H(Z)} = \inf\{t\ge 0\ :\ \mathcal{H}^t(Z)=0\} \le 2.\end{equation*}

Proof of Theorem 1·1. We have found a set $Z \subset \mathbb{H}^{n}$ of dimension smaller than or equal to 2 (Claim ) but which nevertheless has positive $\mu$ -measure (Claim ). This contradicts the assumptions of Theorem 1·1. Thus, there exists $C_2=C_2(n,C_1)$ such that $\mu(B(x,r)) \leq C_2 r^{2n+1}$ for all $x\in{\mathbb{H}}^n$ and $r>0$ .