Proof of Lemma 2.2 subject to Proposition 2.5.

Let $\varepsilon \in (0,1/2)$ , set $\eta := 1 - \varepsilon /2$ , and define $\mathcal {R} = \{r_1,\ldots ,r_R\}$ as

(2.9) $$ \begin{align} R &:= 2 + \lfloor(\log\log\xi\sqrt{d} - \log^{*4}\xi\sqrt{d}-\log\tfrac{4}{\varepsilon})/\log(1/\eta)\rfloor, \end{align} $$

(2.10) $$ \begin{align} r_i &:= (\xi\sqrt{d})^{\eta^{R-i}}, \quad\text{for} \quad i\in[R]. \end{align} $$

We show that this choice of $\mathcal {R}$ is $\varepsilon $ -well-spaced (Def. 2.4). Let $1-a\in [0,1)$ be the fractional part of the expression inside the $\lfloor \cdot \rfloor $ in (2.9). Then using that $\lfloor x\rfloor =x-1+a$ ,

(2.11) $$ \begin{align} r_1 = (\xi\sqrt{d})^{\eta^{R-1}} = (\xi\sqrt{d})^{\eta^{a}4\log^{*3}\xi\sqrt{d}/(\varepsilon\log\xi\sqrt{d})} = (\log\log\xi\sqrt{d})^{(1-\varepsilon/2)^{a}4/\varepsilon}. \end{align} $$

Since $a\in (0,1]$ and $\varepsilon <1/2$ , this implies that $r_1$ is a strictly larger power of $\log \log \xi \sqrt {d}$ than $1$ while R is of order $\log \log \xi \sqrt {d}$ . Since $\xi $ is large relative to $\varepsilon $ , Def. 2.4 (2.5) holds for $\varepsilon =\delta $ . For all $i \in [R]$ , since $\eta =1-\varepsilon /2$ :

$$\begin{align*}r_i/r_{i-1} = (\xi\sqrt{d})^{\eta^{R-i}(1-\eta)} \ge (\xi\sqrt{d})^{\eta^{R-1}\varepsilon/2} = r_1^{\varepsilon/2} = (\log\log\xi\sqrt{d})^{2(1-\varepsilon/2)^{a}}. \end{align*}$$

Since $a\le 1$ , $2(1-\varepsilon /2)^{a}>1$ for all $\varepsilon <1/2$ , so Def. 2.4 (2.6) holds even in $d=1$ , and $\mathcal {R}$ is $\varepsilon $ -well-spaced as claimed. Moreover, since $t \le \log \log \xi $ by hypothesis, by (2.11) we have $t \le (r_1/4\sqrt {d})^d$ . By Proposition 2.5, with probability at least $1-t\varepsilon $ , conditioned on $x_1,\ldots ,x_t\in \mathcal {V}$ , Q contains a strong $(\varepsilon ,\mathcal {R})$ -net $\mathcal {N}$ in Def. 2.3. We now show that a strong net is also a weak net (with the same $\varepsilon )$ . Fix an arbitrary $r \in [(\log \log \xi \sqrt {d})^{4/\varepsilon }, \xi \sqrt {d}]$ and $w \in [w_0,r^{d/(\tau -1)-\varepsilon }]$ as in Def. 2.1 (this interval is nonempty since $\tau \in (2,3), \varepsilon <1/2$ ), and consider a vertex $v \in \mathcal {N}$ . Since $a>0$ in (2.11), we have $r_1 \le r \le r_R$ . Let $r_j$ be such that $r\in [r_j,r_{j+1})$ ; thus $r^{\eta } \le r_j \le r$ by (2.10). Since $\xi $ is large relative to $\varepsilon $ , we have $w \le f_{R,\varepsilon }(r_j)$ (using (2.3)) for $f_{R,\varepsilon }$ ); thus by the definition of a strong $(\varepsilon , \mathcal {R})$ -net in Def. 2.3, $|\mathcal {N} \cap (B_{r_j}(v)\times [w/2, 2w])|\ge \ell (w)w^{-(\tau -1)}r_j^d/(2d)^{d+\tau +5}$ . Since $\xi $ is large relative to $\varepsilon $ and $r \ge r_j \ge r^{\eta }$ , the required inequality in Def. 2.1 follows since

$$\begin{align*}|\mathcal{N} \cap (B_{r}(v)\times[w/2, 2w])| \ge \ell(w)w^{-(\tau-1)}r^{d\eta}/(2d)^{d+\tau+5} \ge \ell(w)w^{-(\tau-1)}r^{d(1-\varepsilon)}. \end{align*}$$

Proof. We prove that given a $\delta $ -well-spaced $r_1< \ldots <r_R $ , there exist side lengths $r_1',\ldots ,r_R'$ that satisfy (P1) – (P3), that is, that $r_{i+1}'/r_i'$ is an integer, (2.12) holds, and $r_R'=\xi $ . We proceed by induction on i, starting from $i=R$ and decreasing i. We take $r_{R}' := r_{R}/\sqrt {d}=\xi $ , then (2.12) is satisfied immediately and (P2)–(P3) are vacuous. Suppose we have found $r_i',\ldots ,r_{R}'$ satisfying (P1)–(P3) for some $2 \le i \le R$ . Let

(2.14) $$ \begin{align} r_{i-1}' = \frac{r_i'}{\lceil \sqrt{d} r_{i}'/r_{i-1} \rceil}. \end{align} $$

This choice of $r_{i-1}'$ divides $r_i'$ , hence (P2) can be satisfied, and $r_{i-1}' \le r_i'/(\sqrt {d}r_i'/r_{i-1}) = r_{i-1}/\sqrt {d}$ . Moreover,

(2.15) $$ \begin{align} r_{i-1}' \ge \frac{r_i'}{1 + \sqrt{d} r_i'/r_{i-1}} = \frac{r_{i-1}}{r_{i-1}/r_i'+\sqrt{d}}. \end{align} $$

Since (2.12) holds for i (by the inductive assumption), $r_i' \ge r_i/2\sqrt {d}$ . Since $\mathcal {R}$ is also well-spaced, $r_{i-1} \le r_i/2$ by (2.6); hence $r_{i-1}/r_i' \le \sqrt {d}$ . so, by (2.15), $r_{i-1}' \ge r_{i-1}/2\sqrt {d}$ , and so (2.12) holds also for $i-1$ and the induction is advanced.

Given these $r_1',\ldots ,r_R'$ , we find an $\mathcal {R}$ -partition of Q by taking $\mathcal {P}_R = \{Q\}$ and iteratively forming each layer $\mathcal {P}_{i-1}$ by taking the unique partition of each box in $\mathcal {P}_i$ into $(r_i')^d/(r_{i-1}')^d$ sub-boxes of side length $r_{i-1}'$ . We first define each partition box to be of the form $\prod _{j=1}^d[a_j, b_j)$ , this allocates each point except d of the $d-1$ -dim faces of $\partial Q$ uniquely. Finally, we allocate the points $x\in \partial Q$ in $\mathcal {P}_i$ to the box in $\mathcal {P}_i$ that contains x in its closure, this box is unique except on $d-2$ dimensional faces. Here we again use half-open $d-1$ -dim boxes to determine the $d-2$ dim boundaries, and so on until only the corner-points are left which we allocate arbitrarily (but consistently across different i).

Proof. We start by showing (2.20). We write $I(w) = [w_-, w_+)$ . The definition of a base- $2$ -cover (Definition 2.8) ensures that $w_-,w_+ \in [w/2,2w]$ and $w_+/w_-=2$ . Thus by the lower bound (2.1) on $w_0$ , for all $w\in [w_0, f(r_R)]$ ,

$$ \begin{align*} \mathbb{P}(W \in I(w)) &= \ell(w_-)w_-^{-(\tau-1)} - \ell(w_+)w_+^{-(\tau-1)}\\ &\ge \ell(w)\left(\frac{99}{100}w_-^{-(\tau-1)} - \frac{101}{100}w_+^{-(\tau-1)}\right)\\ &= \ell(w)w_-^{-(\tau-1)}\left(\frac{99}{100} - \frac{101}{100}\cdot 2^{-(\tau-1)}\right). \end{align*} $$

Since $\tau> 2$ and $w_- \le w$ , it follows that $\mu _i(I(w)) \ge (r_i')^d\ell (w) w^{-(\tau -1)}/2^{\tau +1}$ . The required lower bound on $\mu _i(I(w))$ then follows by the lower bound (P1) in Definition 2.6 on $r_i'$ . By a very similar argument to the lower bound, (2.1) also implies that

$$ \begin{align*} \mathbb{P}(W \in I(w)) \le \frac{101}{100}\ell(w)w_-^{-(\tau-1)} \le 2^\tau\ell(w)w^{-(\tau-1)}, \end{align*} $$

so the required upper bound on $\mu _i(I(w))$ likewise follows by the upper bound (P1) on $r_i'$ . It remains to show (2.21). First we show that $f(r_R) \ge f(r_{R-1}) \ge \ldots \ge f(r_1) \ge w_0$ . Indeed, let $j \in [R]$ . By the definition of f in (2.3),

(2.22) $$ \begin{align} \frac{f(r_j)}{f(r_{j-1})} = \left(\frac{r_j}{r_{j-1}}\right)^{d/(\tau-1)} \cdot \left(\frac{1 \wedge \inf\big\{\ell(x)\colon x \in [w_0,r_j^{d/(\tau-1)}]\big\}}{1 \wedge \inf\big\{\ell(x)\colon x \in [w_0,r_{j-1}^{d/(\tau-1)}]\big\}}\right)^{1/(\tau-1)}. \end{align} $$

To bound the second factor, we will first observe that since $r_{j-1} \le r_j$ , we have $\inf \{\ell (x)\colon x \in [w_0,r_j^{d/(\tau -1)}]\} \le \inf \{\ell (x)\colon x \in [w_0,r_{j-1}^{d/(\tau -1)}]\}$ . By considering the two possible values of $1 \wedge \inf \{\ell (x)\colon x \in [w_0,r_j^{d/(\tau -1)}]\}$ separately, it follows that we can drop the minimum with $1$ in the ratio:

(2.23) $$ \begin{align} \left(\frac{1 \wedge \inf\big\{\ell(x)\colon x \in [w_0,r_j^{d/(\tau-1)}]\big\}}{1 \wedge \inf\big\{\ell(x)\colon x \in [w_0,r_{j-1}^{d/(\tau-1)}]\big\}}\right)^{1/(\tau-1)} \ge \left(\frac{\inf\big\{\ell(x)\colon x \in [w_0,r_j^{d/(\tau-1)}]\big\}}{\inf\big\{\ell(x)\colon x \in [w_0,r_{j-1}^{d/(\tau-1)}]\big\}}\right)^{1/(\tau-1)}. \end{align} $$

We now bound $\ell (x)$ on $[w_0, r_{j-1}^{d/(\tau -1)}]$ (in the denominator) by repeatedly applying (2.1). We write $\lg (\cdot )=\log _2(\cdot )$ . Then, we have $r_j^{d/(\tau -1)} = 2^{\lg ((r_j/r_{j-1} )^{d/(\tau -1)})} r_{j-1}^{d/(\tau -1)}$ , so we iterate the bound in (2.1) roughly $\lg ((r_j/r_{j-1} )^{d/(\tau -1)})$ many times to obtain that for all $x \in [r_{j-1}^{d/(\tau -1)}, r_j^{d/(\tau -1)}]$ we have

$$\begin{align*}\ell(x) \ge \left(\frac{99}{100}\right)^{1 + \lg ((r_j/r_{j-1} )^{d/(\tau-1)})} \ell(r_{j-1}). \end{align*}$$

Returning to (2.23), the ratio of the two infima in the smaller interval $[w_0,r_{j-1}^{d/(\tau -1)}]$ is one. And since $r_j \ge 2r_{j-1}$ by (R2) of Definition 2.4 (using $\delta <1/16 < 1$ ), it follows that

$$ \begin{align*} &\left(\frac{\inf\big\{\ell(x)\colon x \in [w_0,r_j^{d/(\tau-1)}]\big\}}{\inf\big\{\ell(x)\colon x \in [w_0,r_{j-1}^{d/(\tau-1)}]\big\}}\right)^{1/(\tau-1)} \ge \left(\frac{99}{100}\right)^{1 + \lg ((r_j/r_{j-1} )^{d/(\tau-1)}) } \\ &\qquad\qquad \qquad\ge \left(\frac{1}{2}\right)^{\frac{1}{\tau-1}\lg ((r_j/r_{j-1} )^{d/(\tau-1)}) } = \left(\frac{r_{j-1}}{r_j}\right)^{d/(\tau-1)^2}. \end{align*} $$

Combining this with (2.22), (2.23), and the fact that $\tau> 2$ , we obtain

$$\begin{align*}\frac{f(r_j)}{f(r_{j-1})} \ge \left(\frac{r_j}{r_{j-1}}\right)^{\frac{d}{\tau-1}(1-1/(\tau-1))} \ge 1. \end{align*}$$

Hence $f(r_R) \ge f(r_{R-1}) \ge \ldots \ge f(r_1)$ , as claimed. It is now relatively easy to prove the desired lower bound. From the definition of f in (2.3), we have $f(r_i) \le r_i^{d/(\tau -1)}$ , and (2.5) in the definition of well-spacedness ensures that $f(r_i)\ge w_0$ , so

$$ \begin{align*} \ell(f(r_i))f(r_i)^{-(\tau-1)} & = \frac{\ell(f(r_i))}{1 \wedge \inf\{\ell(x)\colon x \in [w_0,r_i^{d/(\tau-1)}]\}} \cdot \frac{(2d)^{2\tau+d+8}\log(16R/\delta)}{r_i^d} \\ &\ge \frac{(2d)^{2\tau+d+8}\log(16R/\delta)}{r_i^d}. \end{align*} $$

Multiplying by $r_i^d$ finishes the statement of (2.21).

Proof. Suppose that Q is R-good. The side length of Q equals $r_R'$ by (P3) in Def. 2.6, and $\delta \in (0,1/16)$ in the setting of Proposition 2.5, hence we may apply (B2) in Def. 2.10 for $i=R$ to get

$$\begin{align*}|\widetilde{\mathcal{G}}_R| \ge \left(\frac{1}{2} - \frac{2(R-1)\delta}{R}\right)\mathrm{Vol}(Q) \ge \left(\frac{1}{2} - 2\delta\right)\mathrm{Vol}(Q)> \mathrm{Vol}(Q)/4, \end{align*}$$

hence the cardinality assumption in Definition 2.3 is satisfied for $\widetilde {\mathcal {G}}_R$ . Recall that $B^i(v)$ is the box in $\mathcal {P}_i$ containing v. To show that $\widetilde {\mathcal {G}}_R$ satisfies Definitions 2.3, we first show that for all $v\in \mathcal {G}_R$ ,

(2.24) $$ \begin{align} \mathcal{G}_i \cap B^i(v)= \mathcal{G}_R \cap B^i(v). \end{align} $$

To show this, we need to show that every i-good vertex $u \in B^i(v)$ is also R-good, that is, that $B^j(u)$ is good for all $j \in [R]$ . By the definition of i-goodness (Def. 2.10), $B^j(u)$ is good for all $j \le i-1$ , and $B^R(u) = Q$ is good by hypothesis. Consider now a $j \in [i+1, R-1]$ . By Def. 2.6, the partition $\mathcal {P}_i$ is a refinement of the partition $\mathcal {P}_j$ , so $B^j(u) = B^j(v)$ . Since v is R-good, it follows that $B^j(u)$ is good. So, $B^j(u)$ is good for all $j \in [R]$ , so u is R-good, showing (2.24).

With $I(w)\subseteq [w/2, 2w]$ being the interval containing w in the base- $2$ -cover of $[w_0, f(r_i)]$ in Def.2.8, we now argue that

$$ \begin{align*} | \widetilde {\mathcal{G}}_R \cap (B_{r_i}(v) \times [w/2,2w])| \ge |\widetilde {\mathcal{G}}_R \cap (B^i(v)\times I(w))|=|\widetilde {\mathcal{G}}_i \cap (B^i(v)\times I(w))|. \end{align*} $$

Indeed, $B^{i}(v)\subseteq B_{r_i}(v)$ by (2.13) and $I(w)\subseteq [w/2, 2w]$ by Def. 2.8, and since all i-good vertices in $B^i(v)$ are also R-good by (2.24), the statement follows. Now we apply, on the right-hand side $|\widetilde {\mathcal {G}_i} \cap (B^i(v) \times I(w))|$ above, the lower bound (2.19) from Def. 2.10 (B3), then (2.20) to obtain

$$ \begin{align*} \big| \widetilde{\mathcal{G}}_R\cap (B^{i}(v)\times [w/2,2w])\big| &\ {\buildrel ({\scriptstyle 2.19}) \over \ge}\ \frac18\left(1-\frac{2i\delta}{R}\right) \mu_i(I(w)) \\&\ {\buildrel ({\scriptstyle 2.20}) \over \ge} \frac{1}{8} \left(1-\frac{2i\delta}{R}\right)r_i^d\ell(w)w^{-(\tau-1)}/(2d)^{\tau+d+1}. \end{align*} $$

Observing that $\delta <1/16$ and $i\le R$ ensures that the prefactor on the right-hand side of the last row is at least $1/8\cdot 1/2=1/2^4$ , establishing (2.4) for all $w\le f(r_i)$ , as required.

Proof. Recall $I_j$ from Def. 2.8, and let $B\in \mathcal {P}_i$ . Let

(2.27) $$ \begin{align} a(I_j) := (1-2i\delta/R)\mu_i(I_j)/8, \qquad b(I_j) := 8\mu_i(I_j) \end{align} $$

be the required lower and upper bounds in (2.19). Let $\Xi _j(B)=|\widetilde {\mathcal {G}_i}\cap (B\times I_j)|$ ; thus (B3) holds for $B\times I_j$ iff $\Xi _j(B) \in [a(I_j),b(I_j)]$ . Since B satisfies (B1)–(B2) on X, by a union bound,

(2.28) $$ \begin{align} \mathbb{P}\big(B\text{ is good}\mid \mathcal{X}_{i-1}(B) = X\big) \ge 1 - \sum_{I_j: j\le j_\star(i)} \mathbb{P}\big(\Xi_j(B) \notin [a(I_j),b(I_j)] \mid \mathcal{X}_{i-1}(B) = X \big). \end{align} $$

We proceed by bounding each term above. By the definition of $\mathcal {F}_{i-1}(B)$ in (2.25), we already exposed $\Xi _j(B)$ when $i> 1$ and $j\le j_*(i-1)$ ; the latter is equivalent to $\min (I_j) \le f(r_{i-1})$ .

Case 1: $\boldsymbol {i \!>\! 1}$ and $\boldsymbol {j\!\le \! j_\star (i\!-\!1)}$ . We first show that $\Xi _j \ge a(I_j)$ holds deterministically on $\{\mathcal {X}_{i-1}(B)=X\}$ . The goodness of each sub-box $B'\in \mathcal {P}_{i-1}$ of $B\in \mathcal {P}_i$ is revealed by $\mathcal {F}_{i-1}(B)$ . If $B'\in \mathcal {P}_{i-1}$ is a good box, all vertices in $\mathcal {G}_{i-1} \cap B'$ are also i-good by Definition 2.10. So

(2.29) $$ \begin{align} \Xi_j(B) = |\widetilde{\mathcal{G}_i}\cap (B\times I_j)|\ge \sum_{B'\in \mathcal{P}_{i-1}\colon B' \text{ good}} |\widetilde{\mathcal{G}}_{i-1}\cap (B'\times I_j)|. \end{align} $$

Since $j\le j_\star (i-1)$ , $\min (I_j) \le f(r_{i-1})$ , so we may apply (2.19) to the good subboxes:

(2.30) $$ \begin{align} |\widetilde{\mathcal{G}}_{i-1}\cap (B'\times I_j)|\ge \left(1-\frac{2(i-1)\delta}{R}\right)\frac{\mu_{i-1}(I_j)}{8} = \left(1-\frac{2(i-1)\delta}{R}\right)\frac{\mu_i(I_j)}{8}\left(\frac{r_{i-1}'}{r_i'}\right)^d, \end{align} $$

where $\mu _i(I_j)/\mu _{i-1}(I_j)=(r_i'/r_{i-1}')^d$ follows from (2.17). By (B1) holding on X, B contains at least $(1-2\delta /R)(r_i'/r_{i-1}')^d$ good sub-boxes in $\mathcal {P}_{i-1}$ . Combining that with (2.29)–(2.30) and (2.27) yields

$$ \begin{align*} \begin{aligned} \Xi_j(B)&\ge \left(1-\frac{2\delta}{R}\right) \cdot \left(1-\frac{2(i-1)\delta}{R}\right) \frac{\mu_i(I_j)}{8} \ge \left(1-\frac{2i\delta}{R}\right) \frac{\mu_i(I_j)}{8} = a(I_j). \end{aligned} \end{align*} $$

We show $\Xi _j(B)=|\widetilde {\mathcal {G}_i}\cap (B\times I_j)| \le b(I_j)$ also holds a.s. B contains $(r_i'/r_{i-1}')^d$ sub-boxes in $\mathcal {P}_{i-1}$ by Def. 2.6 (P2). If a sub-box is bad, it contains no i-good vertices. If it is good, (B3) holds and it contains at most $8\mu _i(I_j)(r_{i-1}'/r_i')^d i$ -good vertices with weights in $B\!\times \! I_j$ . We obtain $\Xi _j(B) \le 8\mu _i(I_j) = b(I_j)$ in (2.27). So overall we have shown that

(2.31) $$ \begin{align} \text{if } i>1 \text{ and } j \le j_\star(i-1):\quad \mathbb{P}\big(\Xi_j(B) \notin [a(I_j),b(I_j)] \mid \mathcal{F}_{i-1}(B) = F\big) = 0. \end{align} $$

Case 2: $\boldsymbol {i> 1}$ and $\boldsymbol {j> j_\star (i-1)}$ . Define the set

(2.32) $$ \begin{align} \mathcal{S}=\widetilde{\mathcal{G}_i}\cap \left(B\times \big([1,w_0) \cup \bigcup\nolimits_{j> j_\star(i-1)} I_j\big)\right), \end{align} $$

the i-good vertices in B with weights not revealed by $\mathcal {F}_{i-1}(B)$ . $\mathcal {F}_{i-1}(B)$ does reveal the positions of vertices in B, and all weighted vertices not in $\mathcal {S}$ ; thus $\mathcal {F}_{i-1}(B)$ reveals $|\mathcal {S}|$ , and the position of vertices in $\mathcal {S}$ . By Def. 1.3, vertex weights are independent of vertex positions and of each other. Further, for a vertex $v \in \widetilde {\mathcal {V}}$ , conditioning on $\mathcal {F}_{i-1}(B)$ is equivalent to either exposing its weight (if $w_v\in \cup _{j\le j_\star (i-1)}I_j$ ) or conditioning on $w_v \notin \cup _{j\le j_\star (i-1)} I_j$ (otherwise). So for each vertex in $\mathcal {S}$ , its weight distribution is the conditional distribution of W given that $W\notin \cup _{j\le j_\star (i-1)}I_j$ . Since $|\mathcal {S}|$ is determined by $\mathcal {F}_{i-1}(B)$ , for all $j\ge j_\star (i-1)$ , $\Xi _j(B)$ given $\mathcal {F}_{i-1}(B)$ is binomially distributed with parameters

(2.33) $$ \begin{align} \Xi_j(B) \mid \mathcal{F}_{i-1}(B)\ \ {\buildrel d \over =} \ \ \mathrm{Bin}\left(|\mathcal{S}|, \mathbb P(W\in I_j)/\mathbb P(W\notin \cup_{j\le j_\star(i-1)} I_j) \right). \end{align} $$

We next bound the conditional expectation of $\Xi _j(B)\mid \mathcal {F}_{i-1}(B)$ , starting with the upper bound. Using the lower bound (2.1) on $w_0$ , the success probability of the binomial in (2.33) is

(2.34) $$ \begin{align} \frac{\mathbb{P}(W\in I_j)}{\mathbb P(W\notin \cup_{j\le j_\star(i-1)} I_j)} \le \frac{\mathbb{P}(W\in I_j)}{\mathbb{P}(W\in [1,w_0))} = \frac{\mathbb{P}(W\in I_j)}{1-\ell(w_0)w_0^{\tau-1}}\le 2\mathbb{P}(W\in I_j). \end{align} $$

Since Def. 2.10 (B2) holds for B on $\mathcal {X}_{i-1}(B)=X$ , by (2.18) there are at most $2(r_i')^d i$ -good vertices in B, so $|\mathcal {S}| \le 2(r_i')^d$ . Recalling the definition of $\mu _i(I_j)$ from (2.17), and (2.27) we thus obtain

(2.35) $$ \begin{align} \mathbb{E}\big[\Xi_j(B) \mid \mathcal{X}_{i-1}(B) = X\big] \le 4(r_i')^d\mathbb{P}(W\in I_j) = 4\mu_i(I_j) = b(I_j)/2. \end{align} $$

We next prove the corresponding lower bound. We start by giving a lower bound on $|\mathcal {S}|$ . Clearly by (2.32), $|\mathcal {S}|=|\mathcal {G}_i\cap B|-|\cup _{j\le j_\star (i-1)} \widetilde {\mathcal {G}_i}\cap (B\times I_j)|$ , both terms revealed by X: the total number of i-good vertices in B minus the ones with revealed weight. We can bound $|\mathcal {G}_i\cap B|$ from below using (2.18) in Def. 2.10 (B2). By Def. 2.6 (P2), B contains $(r_i'/r_{i-1}')^d$ sub-boxes in $\mathcal {P}_{i-1}$ . Since there are no i-good vertices in bad boxes $B'\in \mathcal {P}_{i-1}$ , we can bound $|\cup _{j\le j_\star (i-1)} \widetilde {\mathcal {G}_i}\cap (B\times I_j)|$ from above by applying (2.19) to each good sub-box $B'\in \mathcal {P}_{i-1}$ of B, yielding

(2.36) $$ \begin{align} |\mathcal{S}|\ge \left(\frac{1}{2}-\frac{2(i-1)\delta}{R}\right)(r_i')^d - \left(\frac{r_i'}{r_{i-1}'}\right)^d\cdot\sum_{j \le j_\star(i-1)} 8\mu_{i-1}(I_j). \end{align} $$

Using that $I_j=[2^{j-1}w_0, w_0)$ for all $j\ge 1$ , Lemma 2.11 with $w=2^{j-1}w_0$ yields

(2.37) $$ \begin{align} \begin{aligned} \big(r_i'/r_{i-1}'\big)^d\sum_{j \le j_\star(i-1)} 8\mu_{i-1}(I_j)&\le 2^{\tau+3}(r_i')^d\sum_{j=1}^{j_\star(i-1)} \ell(2^{j-1}w_0)(2^{j-1}w_0)^{-(\tau-1)}\\ &< 2^{\tau+3}(r_i')^d\sum_{j = 0}^\infty \ell(2^jw_0)(2^jw_0)^{-(\tau-1)}, \end{aligned} \end{align} $$

where we switched indices in the last row. By the lower bound (2.1) on $w_0$ and since $\tau>2$ , for all $j \ge 0$ we have $\ell (2^{j+1}w_0)(2^{j+1}w_0)^{-(\tau -1)} < \tfrac {2}{3}\ell (2^jw_0)(2^jw_0)^{-(\tau -1)}$ , so the sum on the right-hand side is bounded above term-wise by a geometric series. It follows from (2.36) that

$$\begin{align*}|\mathcal{S}| \ge \left(\frac{1}{2} - \frac{2(i-1)\delta}{R} - 2^{\tau+5}\ell(w_0)w_0^{-(\tau-1)}\right)(r_i')^d \ge (r_i')^d/4, \end{align*}$$

where we used in the last step that $i \le R$ and $\delta < 1/16$ , and the lower bound (2.1) on $w_0$ . The success probability of the binomial in (2.33) is at least $\mathbb {P}(W\in I_j)$ . Since $a(I_j)= (1-2i\delta /R)\mu _i(I_j)/8$ in (2.27),

(2.38) $$ \begin{align} \mathbb{E}\big[\Xi_j(B) \mid \mathcal{X}_{i-1}(B) = X\big] \ge (r_i')^d\mathbb{P}(W\in I_j)/4 = \mu_i(I_j)/4 \ge 2a(I_j). \end{align} $$

Combining (2.35) with (2.38) yields that $\mathbb E[\Xi _j(B)\mid \mathcal {F}_{i-1}(B)=F] \in [2a(I_j), b(I_j)/2]$ , which allows us to bound $\mathbb {P}(\Xi _j(B) \notin [a(I_j),b(I_j)])$ with standard Chernoff bounds. By (2.35) and Theorem A.1 applied with $\varepsilon =1/2$ , we have shown that uniformly for all realisations F satisfying (B1)-(B2)

(2.39) $$ \begin{align} \begin{aligned} \text{for } i> 1, j > j_\star(i-1):\quad \mathbb{P}\big(\Xi_j(B) \notin [a(I_j), b(I_j)] \mid \mathcal{F}_{i-1}(B) = F\big) &\le 2\exp(-a(I_j)/6) \\ & \le 2\exp(-\mu_i(I_j)/96). \end{aligned} \end{align} $$

Case 3: $\boldsymbol {i\!=\!1}$ . When $i=1$ , we set $j_\star (i-1):=0$ naturally, since in $\mathcal {F}_0(B)$ we revealed the total number of vertices in $B\in \mathcal {P}_1$ , which are all $1$ -good by Def. 2.10. Conditioned on $\mathcal {X}_0(B)=X$ , (2.18) in (B2) is satisfied and $(r_1')^d/4 \le |\mathcal {S}| \le 2(r_1')^d$ directly. The rest of our previous calculations from Case 2 with $j>j_\star (0)=0$ all carry through for estimating the left-hand side of (2.19) in (B3). We obtain that the bound in (2.39) holds also for $i=1$ and all $j\ge j_\star (0)=0$ .

Combining the cases: By (2.31), (2.39), and Case 3, for all i and $j \le j_\star (i)$ the bound in (2.39) holds. Combining that with (2.28), for all $B\in \mathcal {P}_i$ and $ \mathcal {X}_{i-1}(B)=X$ satisfying (B1),(B2),

$$ \begin{align*} \begin{aligned} p_i:=&\ \mathbb{P}\big(B\in \mathcal{P}_i \text{ is } i\text{-good}\mid \mathcal{X}_{i-1}(B) = X\big) \ge 1 - 2\sum_{j \le j_\star(i)} \exp(-\mu_i(I_j)/96)\\ &\ge 1 - 2\sum_{j \le j_\star(i)} \exp\left({-}(2d)^{-(\tau+d+8)}r_i^d\ell(2^{j-1}w_0 )(2^{j-1}w_0)^{-(\tau-1)}\right), \end{aligned} \end{align*} $$

where the second line follows from $\mu _i(I_j)=\mu _i(I(2^{j-1}w_0))$ in Lemma 2.11. Here, the term in the sum with $j=j_\star (i)$ is maximal. We use then an reindexing $t=j_\star (i)-j, t\ge 0$ and argue that the sum can be dominated by a geometric series in t: By the lower bound (2.1) on $w_0$ and the fact that $\tau>2$ , for all $j\ge 1$ we have $ \ell (2^{j-1}w_0)(2^{j-1}w_0)^{-(\tau -1)} \ge \tfrac {3}{2}\ell (2^{j}w_0)(2^{j}w_0)^{-(\tau -1)}$ . Using the same geometric-sum argument as below (2.37) except now from the reversed viewpoint, writing $z:=w_02^{j_\star (i)-1}$ , the lower endpoint of $I_{j_\star (i)}$ , we have

(2.40) $$ \begin{align} p_i &\ge 1 - 2\sum_{t=0}^\infty \exp\left({-}\frac{1}{(2d)^{\tau+d+8}}r_i^d\left(\frac{3}{2}\right)^t\ell(z)z^{-(\tau-1)}\right). \end{align} $$

Since z is the lower endpoint of $I_{j_\star (i)}=I(f(r_i))$ , we have $z \le f(r_i) \le 2z$ by (2.16). Hence by (2.1)

$$\begin{align*}\ell(z)z^{-(\tau-1)} \ge 2^{-\tau}\ell(f(r_i))f(r_i)^{-(\tau-1)}. \end{align*}$$

Using this bound in (2.40) and combining it with (2.21) from Lemma 2.11, we obtain that

$$ \begin{align*} p_i \ge 1 - 2\sum_{t=0}^\infty \exp\left({-}\left(\frac{3}{2}\right)^t\log(16R/\delta)\right) \ge 1 - \frac{\delta}{8R} - 2\sum_{t=1}^\infty (\delta/16R)^t \ge 1 - \frac{\delta}{2R}, \end{align*} $$

where we used that $(3/2)^t \ge t$ for all $t \ge 1$ , and $\delta < 1/16$ . Independence across boxes in the same $\mathcal {P}_i$ is immediate, since whether B satisfies (B3) conditioned on $\mathcal {F}_{i-1}(B)=F$ satisfying (B1), (B2) only depends on vertices in B.

Proof. We prove the statement by induction on i, the base case being $i=1$ . Consider a box $B \in \mathcal {P}_1$ . By Def. 2.10, (B1) holds vacuously. We next consider (B2). Every vertex in B is $1$ -good, and so $|\mathcal {G}_1\cap B|=|\mathcal {V} \cap B|$ . In SFP, $\mathcal {V}$ is deterministic and $|\mathcal {G}_1\cap B|\in [(r_1')^d/2, 2(r_1')^d]$ holds with certainty by Def. 2.6 (P1). In (I)GIRG, $\mathcal {G}_1\cap B$ is a Poisson point process and the Palm theory (see, e.g., [Reference Last and Penrose61]) gives that under the conditioning $x_1, \dots , x_t\in \mathcal {V}$ , $|\mathcal {G}_1\cap B|-t$ is a Poisson variable with mean $(r_1')^d$ , so by a standard Chernoff bound (Theorem A.1 with $\varepsilon =1/2$ ), and using $r_i' \in [r_i/(2\sqrt {d}), r_i/\sqrt {d}]$ in Def. 2.6 (P1), and the bound (2.5) on $r_1$ in Def. 2.4,

$$ \begin{align*} p_1':=\mathbb{P}\left( |\mathcal{G}_1\cap B| \in[t+\tfrac12(r_1')^d, t+\tfrac32(r_1')^d]\mid x_1, \dots, x_t\in \mathcal{V}\right) \ge 1 - 2\mathrm{e}^{-(r_1/24d)^d} \ge 1-\delta/(2R), \end{align*} $$

so the lower bound in (2.18) in Def. 2.10 (B2) is satisfied. Moreover, since $t \le (r_1/4\sqrt {d})^d$ , by Def. 2.6 (P1), $\tfrac 32(r_1')^d + t \le 2(r_1')^d$ , the upper bound in Def. 2.10 (B2) also holds, so independently for all boxes $B\in \mathcal {P}_1$ , and regardless on where $x_1, \ldots , x_t$ fall,

$$ \begin{align*} \mathbb{P}\big(B \in \mathcal{P}_1 \text{ satisfies } (B2)\mid x_1,\ldots,x_t \in \mathcal{V}\big) \ge p_1'\ge 1 - \delta/(2R). \end{align*} $$

Lemma 2.14 ensures that (B3) holds uniformly with probability at least $1-\delta /(2R)$ conditioned on any realisation where (B1),(B2) holds. A union bound proves (2.41) for $B\in \mathcal {P}_1$ .

Now we advance the induction. Suppose that (2.41) holds for each $B\in \cup _{j\le i-1}\mathcal {P}_i$ , and let $B \in \mathcal {P}_i$ . B contains $(r_i'/r_{i-1}')^d$ sub-boxes in $\mathcal {P}_{i-1}$ (by Def. 2.6), and by induction these sub-boxes of B are good independently (regardless of the positions and weights of $x_1,\ldots ,x_t \in \mathcal {V}$ ), so the number of bad sub-boxes of B is binomial with mean at most $(r_i'/r_{i-1}')^d\delta /R$ . Let

(2.42) $$ \begin{align} \mathcal{A}_{i}(B):=\Big\{\big|\{B'\in \mathcal{P}_{i-1}\colon B'\subseteq B,\ B' \text{ not } (i-1)\text{-good}\}\big| < 2(r_i'/r_{i-1}')^d\delta/R\Big\}. \end{align} $$

Then, $\mathcal {A}_{i}(B)$ implies Def. 2.10 (B1). A Chernoff bound (Theorem A.1 with $\varepsilon =1$ ) yields that

$$\begin{align*}\mathbb{P}\left(\mathcal{A}_{i}(B)^\complement \mid x_1,\ldots,x_t\in\mathcal{V}\right) \le \exp\left({-}\frac{\delta}{3R}\cdot \left(\frac{r_i'}{r_{i-1}'}\right)^d \right). \end{align*}$$

By Def. 2.6 (P1), $(r_i'/r_{i-1}')^d \ge 2^{-d} (r_i/r_{i-1})^d$ , so by Def. 2.4 (2.6), regardless of the positions of $x_1,\ldots ,x_t\in \mathcal {V}$ , and independently across different boxes in $\mathcal {P}_i$ :

(2.43) $$ \begin{align} \mathbb{P}\left( \mathcal{A}_{i}(B)^{\complement}\mid x_1,\ldots,x_t\in\mathcal{V}\right) &\le \exp\left({-}\frac{\delta}{3R} \cdot 3^d R \cdot \frac{\log (2R/\delta)}{\delta}\right) \le \frac{\delta}{2R}. \end{align} $$

We now show that $\mathcal {A}_{i}(B)$ implies (B2) as well, by inductively applying (2.18) to the good sub-boxes of B. Consider an $(i-1)$ -good vertex v in a good sub-box $B'\in \mathcal {P}_{i-1}$ of B. Since v is $(i-1)$ -good, $B^1(v),\ldots ,B^{i-2}(v)$ must all be good; since $B^{i-1}(v)=B'$ is also good, it follows that v is in fact i-good. Thus, for all good $B'\in \mathcal {P}_{i-1}: \mathcal {G}_{i-1}\cap B'=\mathcal {G}_{i}\cap B'$ holds. Since $B'$ is $(i\!-\!1)$ -good, it satisfies (2.18), and we obtain:

$$\begin{align*}|\mathcal{G}_{i}\cap B| \ge \sum_{\text{good } B'\subset B} |\mathcal{G}_{i-1}\cap B'|\ge \big|\{B' \in \mathcal{P}_{i-1}\colon B'\subset B,\ B'\text{ good}\}\big|\left(\frac{1}{2} - \frac{2(i-2)\delta}{R}\right)(r_{i-1}')^d. \end{align*}$$

On $\mathcal {A}_i(B)$ in (2.42), there are $(1-2\delta /R)(r_i'/r_{i-1}')^d$ good sub-boxes of B, so a.s. on $\mathcal {A}_i(B)$

$$ \begin{align*} |\mathcal{G}_{i}\cap B| &\ge \left(1-\frac{2\delta}{R}\right)\left(\frac{r_i'}{r_{i-1}'}\right)^d \left(\frac{1}{2} - \frac{2(i-2)\delta}{R}\right)(r_{i-1}')^d \ge \left(\frac{1}{2}-\frac{2(i-1)\delta}{R}\right)(r_i')^d. \end{align*} $$

Vertices in bad sub-boxes cannot be i-good by Definition 2.10. So (2.18) similarly implies that on $\mathcal {A}_{i}(B)$ there are at most $2(r_i')^d i$ -good vertices in B, and thus $\mathcal {A}_i(B)$ in (2.42) implies both (B1)–(B2) for B; it follows from (2.43) that

(2.44) $$ \begin{align} \mathbb{P}\big((B1) \text{ and } (B2) \text{ hold for }B \mid x_1,\ldots,x_t \in \mathcal{V}\big) \ge 1 - \delta/(2R). \end{align} $$

By Lemma 2.14, B is good (i.e., (B3) also holds) with probability at least $1-\delta /(2R)$ for all $\mathcal {F}_{i-1}(B)=F$ with (B1)–(B2) holding for B; these events and $x_1,\ldots ,x_t \in \mathcal {V}$ are all determined by $\mathcal {F}_{i-1}(B)$ . So, a union bound on (2.26) and (2.44) yields that independently across boxes in $\mathcal {P}_i$ , and regardless of the vertices $x_1,\ldots ,x_t\in \mathcal {V}$ , (2.41) holds. This advances the induction and finishes the proof.