Proposition 2.0.2 follows by applying the triangle inequality to 2.0.22 according to the splitting in Lemma 5.1.2 and Lemma 5.1.3 and using both Lemmas as well as the bound on the set \(G'\) given by Lemma 5.1.1.

Let

For each \({\mathfrak p}\in {\mathfrak P}_{F,G}\) pick a \(r({\mathfrak p}){\gt}4D^{{\mathrm{s}}({\mathfrak p})}\) with

This ball exists by definition of \({\mathfrak P}_{F,G}\) and \(\operatorname{\operatorname {dens}}_2\). By applying Proposition 2.0.6 to the collection of balls

and the function \(u = \mathbf{1}_F\), we obtain

We conclude with 2.0.10 and \(r({\mathfrak p}){\gt}4D^{{\mathrm{s}}({\mathfrak p})}\)

The union of dyadic cubes in \(\mathcal{C}(G,k)\) is contained the union of elements of the set \(\mathcal{M}(k)\) of all dyadic cubes \(J\) with \({\mu (G \cap J)} {\gt} 2^{-k-1}{\mu (J)}\). The union of elements in the set \(\mathcal{M}(k)\) is contained in the union of elements in the set \(\mathcal{M}^*(k)\) of maximal elements in \(\mathcal{M}(k)\) with respect to set inclusion. Hence

Using the definition of \(\mathcal{M}(k)\) and then the pairwise disjointedness of elements in \(\mathcal{M}^*(k)\), we estimate 5.2.2 by

This proves the lemma.

Let \({\mathfrak p},{\mathfrak p}'\) be as in the lemma. By definition of \(E_1\), we have \(E_1({\mathfrak p})\subset {\mathcal{I}}({\mathfrak p})\) and analogously for \({\mathfrak p}'\), we conclude from 5.2.4 that \({\mathcal{I}}({\mathfrak p})\cap {\mathcal{I}}({\mathfrak p}')\neq \emptyset \). Let without loss of generality \({\mathcal{I}}({\mathfrak p})\) be maximal in \(\{ {\mathcal{I}}({\mathfrak p}),{\mathcal{I}}({\mathfrak p}')\} \), then \({\mathcal{I}}({\mathfrak p}')\subset {\mathcal{I}}({\mathfrak p})\). By 5.2.4, we conclude by definition of \(E_1\) that \({\Omega }({\mathfrak p})\cap {\Omega }({\mathfrak p}')\neq \emptyset \). By 2.0.14 we conclude \({\Omega }({\mathfrak p})\subset {\Omega }({\mathfrak p}')\). It follows that \({\mathfrak p}'\le {\mathfrak p}\). By maximality 5.1.5 of \({\mathfrak p}'\), we have \({\mathfrak p}'={\mathfrak p}\). This proves the lemma.

Fix \(k,n,\lambda ,x\) as in the lemma such that \(x\in A(\lambda ,k,n)\). Let \(\mathcal{M}\) be the set of dyadic cubes \({\mathcal{I}}({\mathfrak p})\) with \({\mathfrak p}\) in \(\mathfrak {M}(k,n)\) and \(x\in {\mathcal{I}}({\mathfrak p})\). By definition of \(A(\lambda ,k,n)\), the cardinality of \(\mathcal{M}\) is at least \(1+\lambda 2^{n+1}\). Let \(I\) be a cube of smallest scale in \(\mathcal{M}\). Then \(I\) is contained in all cubes of \(\mathcal{M}\). It follows that \(I\subset A(\lambda ,k,n)\).

Fix \(k,n\) as in the lemma and suppress notation to write \(A(\lambda )\) for \(A(\lambda ,k,n)\). We prove the lemma by induction on \(\lambda \). For \(\lambda =0\), we use that \(A(\lambda )\) by definition of \(\mathfrak {M}(k,n)\) is contained in the union of elements in \( \mathcal{C}(G,k)\). Lemma 5.2.2 then completes the base of the induction.

Now assume that the statement of Lemma 5.2.5 is proven for some integer \(\lambda \ge 0\). The set \(A(\lambda +1)\) is contained in the set \(A(\lambda )\). Let \(\mathcal{M}\) be the set of dyadic cubes which are a subset of \(A(\lambda )\). By Lemma 5.2.4, the union of \(\mathcal{M}\) is \(A(\lambda )\). Let \(\mathcal{M}^*\) be the set of maximal dyadic cubes in \(\mathcal{M}\).

Let \(x\in A(\lambda +1)\) and \(L\in \mathcal{M}^*\) such that \(x\in L\). Then by the dyadic property 2.0.8

We now show

The left-hand side of 5.2.6 is strictly greater than \((\lambda +1)2^{n+1}\). If \(L\) is the top cube the second sum \(Q_2\) on the right-hand side of 5.2.6 is zero and 5.2.7 follows immediately. Otherwise consider the inclusion-minimal cube \(\hat{L}\) with \(L\subsetneq \hat{L}\); all tiles \({\mathfrak p}\) over which \(Q_2\) is summed over satisfy \(\hat{L}\subset {\mathfrak p}\), so \(Q_2\) is constant for all \(x\in \hat{L}\). By maximality of \(L\), \(Q_2\) is at most \(\lambda 2^{n+1}\) somewhere on \(\hat{L}\), thus on all of \(\hat{L}\) and consequently also at \(x\). Rearranging the inequality yields 5.2.7.

By Lemma 5.2.3, we have

Multiplying by \(2^n\) and applying 5.1.4, we obtain

We then have with 5.2.7 and 5.2.9

Hence

Using the induction hypothesis, this proves 5.2.5 for \(\lambda +1\) and completes the proof of the lemma.

We use Lemma 5.2.5 and sum twice a geometric series to obtain

This proves the lemma.

We write the left-hand side of 5.2.16

Using Lemma 5.2.5 and then summing a geometric series, we estimate this by

This proves the lemma.

Let \(x\in X\). For each \({\mathfrak u}\in {\mathfrak U}_1(k,n,j)\) with \(x\in {\mathcal{I}}({\mathfrak u})\), as \({\mathfrak u}\in {\mathfrak C}_1(k,n,j)\), there are at least \(2^{j}\) elements \(\mathfrak {m}\in \mathfrak {M}(k,n)\) with \(100{\mathfrak u}\lesssim \mathfrak {m}\) and in particular \(x\in {\mathcal{I}}(\mathfrak {m})\). Hence

Conversely, for each \(\mathfrak {m}\in \mathfrak {M}(k,n)\) with \(x\in {\mathcal{I}}(\mathfrak {m})\), let \({\mathfrak U}(\mathfrak {m})\) be the set of \({\mathfrak u}\in {\mathfrak U}_1(k,n,j)\) with \(x\in {\mathcal{I}}({\mathfrak u})\) and \(100{\mathfrak u}\lesssim \mathfrak {m}\). Summing 5.2.20 over \({\mathfrak u}\) and counting the pairs \(({\mathfrak u},\mathfrak {m})\) with \(100{\mathfrak u}\lesssim \mathfrak {m}\) differently gives

We estimate the number of elements in \({\mathfrak U}(\mathfrak {m})\). Let \({\mathfrak u}\in {\mathfrak U}(\mathfrak {m})\). Then by definition of \({\mathfrak U}(\mathfrak {m})\)

If \({\mathfrak u}'\) is a further element in \({\mathfrak U}(\mathfrak {m})\) with \({\mathfrak u}\neq {\mathfrak u}'\), then

By the last display and definition of \({\mathfrak U}_1(k,n,j)\), none of \({\mathcal{I}}({\mathfrak u})\), \({\mathcal{I}}({\mathfrak u}')\) is strictly contained in the other. As both contain \(x\), we have \({\mathcal{I}}({\mathfrak u})={\mathcal{I}}({\mathfrak u}')\). We then have \(d_{{\mathfrak u}}=d_{{\mathfrak u}'}\).

By 2.0.15, the balls \(B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}),0.2)\) and \(B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}'),0.2)\) are contained respectively in \({\Omega }({\mathfrak u})\) and \({\Omega }({\mathfrak u}')\) and thus are disjoint by 2.0.13. By 5.2.22 and the triangle inequality, both balls are contained in \(B_{{\mathfrak u}}({\mathcal{Q}}(\mathfrak {m}), 100.2)\).

By 1.0.11 applied nine times, there is a collection of at most \(2^{9a}\) balls of radius \(0.2\) with respect to the metric \(d_{{\mathfrak u}}\) which cover the ball \(B_{{\mathfrak u}}({\mathcal{Q}}(\mathfrak {m}),100.2)\). Let \(B'\) be a ball in this cover. As the center of \(B'\) can be in at most one of the disjoint balls \(B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}),0.2)\) and \(B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}'),0.2)\), the ball \(B'\) can contain at most one of the points \({\mathcal{Q}}({\mathfrak u})\), \({\mathcal{Q}}({\mathfrak u}')\).

Hence the image of \({\mathfrak U}(\mathfrak {m})\) under \({\mathcal{Q}}\) has at most \(2^{9a}\) elements; since \({\mathcal{Q}}\) is injective on \({\mathfrak U}(\mathfrak {m})\), the same is true of \({\mathfrak U}(\mathfrak {m})\). Inserting this into 5.2.21 proves the lemma.

Let \({\mathfrak u}\in {\mathfrak U}_1(k,n,l)\). Let \(I \in \mathcal{L}({\mathfrak u})\). Then we have \(s(I) = {\mathrm{s}}({\mathfrak u}) - Z(n+1) - 1\) and \(I \subset {\mathcal{I}}({\mathfrak u})\) and \(B(c(I), 8D^{s(I)}) \not\subset {\mathcal{I}}({\mathfrak u})\). By 2.0.10, the set \(I\) is contained in \(B(c(I), 4D^{s(I)})\). By the triangle inequality, the set \(I\) is contained in

By the small boundary property 2.0.11, noting that

we have

Using \(\kappa {\lt}1\) and \(D \ge 12\), this proves the lemma.

As each \({\mathfrak p}\in {\mathfrak L}_4(k,n,j)\) is contained in \(\cup \mathcal{L}({\mathfrak u})\) for some \({\mathfrak u}\in {\mathfrak U}_1(k,n,l)\), we have

Using Lemma 5.2.9 and then Lemma 5.2.8, we estimate this further by

Using Lemma 5.2.7, we estimate this by

Now we estimate \(G_3\) defined in 5.1.28 by

Summing geometric series, using that \(D^{\kappa Z}\ge 8\) by 2.0.1, 2.0.2 and 2.0.3, we estimate this by

Using \(D = 2^{100a^2}\) and \(a \ge 4\) and \(\kappa Z \ge 2\) by 2.0.1 and 2.0.2 proves the lemma.

Adding up the bounds in Lemmas 5.2.1, 5.2.6, and 5.2.10 proves Lemma 5.1.1.

This follows immediately from the definition 2.0.24 of \(\lesssim \) and the two inclusions \(B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), n) \subset B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), n')\) and \(B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'), m') \subset B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'), m)\).

The assumption 5.3.1 together with the definition 2.0.24 of \(\lesssim \) implies that \({\mathcal{I}}({\mathfrak p}) \subsetneq {\mathcal{I}}({\mathfrak p}')\). Let \({\vartheta }\in B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'), m)\). Then we have by the triangle inequality

The first summand is bounded by \(n\) since

using 2.0.24. For the second summand we use Lemma 2.1.2 to show that the sum is estimated by

Thus \(B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'),m) \subset B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}),n + 2^{-95a}m)\). Combined with \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak p}')\), this yields 5.3.2.

5.3.4 and 5.3.5 are easy consequences of Lemma 5.3.1, Lemma 5.3.2 and the fact that \(a \ge 4\). For 5.3.3, if \({\mathcal{I}}({\mathfrak q}) = {\mathcal{I}}({\mathfrak q}')\) then we get \({\mathfrak q}= {\mathfrak q}'\) by 2.0.13 and 2.0.23. If \({\mathcal{I}}({\mathfrak q}) \ne {\mathcal{I}}({\mathfrak q}')\), then from 2.0.23, 2.0.24 and 2.0.15 it follows that \({\mathfrak q}\lesssim 0.2{\mathfrak q}'\), and 5.3.3 follows from an easy calculation using Lemma 5.3.2.

Suppose that \({\mathfrak p}\le {\mathfrak p}' \le {\mathfrak p}''\) and \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak P}(k)\). By 5.1.3 we have \({\mathcal{I}}({\mathfrak p}), {\mathcal{I}}({\mathfrak p}'') \in \mathcal{C}(G,k)\), so there exists by 5.1.1 some \(J \in \mathcal{D}\) with

and \(\mu (G \cap J) {\gt} 2^{-k-1} \mu (J)\). Thus 5.1.1 holds for \({\mathcal{I}}({\mathfrak p}')\). On the other hand, by 5.1.2, there exists no \(J \in \mathcal{D}\) with \({\mathcal{I}}({\mathfrak p}) \subset J\) and \(\mu (G \cap J) {\gt} 2^{-k} \mu (J)\). Since \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak p}')\), this implies that 5.1.2 holds for \({\mathcal{I}}({\mathfrak p}')\). Hence \({\mathcal{I}}({\mathfrak p}') \in \mathcal{C}(G,k)\), and therefore by 5.1.3 \({\mathfrak p}' \in {\mathfrak P}(k)\).

Let \({\mathfrak p}\le {\mathfrak p}' \le {\mathfrak p}''\) with \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak C}(k,n)\). Then, in particular, \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak P}(k)\), so, by Lemma 5.3.4, \({\mathfrak p}' \in {\mathfrak P}(k)\). Next, we show that if \({\mathfrak q}\le {\mathfrak q}' \in {\mathfrak P}(k)\) then \(\operatorname{\operatorname {dens}}'_k(\{ {\mathfrak q}\} ) \ge \operatorname{\operatorname {dens}}_k'(\{ {\mathfrak q}'\} )\). If \({\mathfrak p}\in {\mathfrak P}(k)\) and \(\lambda \ge 2\) with \(\lambda {\mathfrak q}' \lesssim \lambda {\mathfrak p}\), then it follows from \({\mathfrak q}\le {\mathfrak q}'\), 5.3.3 of Lemma 5.3.3 and transitivity of \(\lesssim \) that \(\lambda {\mathfrak q}\lesssim \lambda {\mathfrak p}\). Thus the supremum in the definition 5.1.6 of \(\operatorname{\operatorname {dens}}_k'(\{ {\mathfrak q}\} )\) is over a superset of the set the supremum in the definition of \(\operatorname{\operatorname {dens}}_k'(\{ {\mathfrak q}'\} )\) is taken over, which shows \(\operatorname{\operatorname {dens}}'_k(\{ {\mathfrak q}\} ) \ge \operatorname{\operatorname {dens}}_k'(\{ {\mathfrak q}'\} )\). From \({\mathfrak p}' \le {\mathfrak p}''\), \({\mathfrak p}'' \in {\mathfrak C}(k,n)\) and 5.1.7 it then follows that

Similarly, it follows from \({\mathfrak p}\le {\mathfrak p}'\), \({\mathfrak p}\in {\mathfrak C}(k,n)\) and 5.1.7 that

Thus \({\mathfrak p}' \in {\mathfrak C}(k,n)\).

Let \({\mathfrak p}\le {\mathfrak p}'\le {\mathfrak p}''\) with \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak C}_1(k,n,j)\). By Lemma 5.3.5 and the inclusion \({\mathfrak C}_1(k,n,j) \subset {\mathfrak C}(k,n)\), which holds by definition 5.1.9, we have \({\mathfrak p}' \in {\mathfrak C}(k,n)\). By 5.3.3 and transitivity of \(\lesssim \) we have that \({\mathfrak q}\le {\mathfrak q}'\) and \(100 {\mathfrak q}' \lesssim \mathfrak {m}\) imply \(100 {\mathfrak q}\lesssim \mathfrak {m}\). So, by 5.1.8, \(\mathfrak {B}({\mathfrak p}'') \subset \mathfrak {B}({\mathfrak p}') \subset \mathfrak {B}({\mathfrak p})\). Consequently, by 5.1.9

thus \({\mathfrak p}' \in {\mathfrak C}_1(k,n,j)\).

Let \({\mathfrak p}\le {\mathfrak p}' \le {\mathfrak p}''\) with \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak C}_2(k,n,j)\). By 5.1.13, we have

Combined with Lemma 5.3.6, it follows that \({\mathfrak p}' \in {\mathfrak C}_1(k,n,j)\). If \({\mathfrak p}={\mathfrak p}'\) the statement is trivially true, otherwise suppose that \({\mathfrak p}' \notin {\mathfrak C}_2(k,n,j)\). By 5.1.13, this implies that there exists \(0 \le l' \le Z(n+1)\) with \({\mathfrak p}' \in {\mathfrak L}_1(k,n,j,l')\). By the definition 5.1.11 of \({\mathfrak L}_1(k,n,j,l')\), this implies that \({\mathfrak p}'\) is minimal with respect to \(\le \) in \({\mathfrak C}_1(k,n,j) \setminus \bigcup _{l {\lt} l'} {\mathfrak L}_1(k,n,j,l)\). Since \({\mathfrak p}\le {\mathfrak p}'\) and \({\mathfrak p}\in {\mathfrak C}_2(k,n,j)\), \({\mathfrak p}={\mathfrak p}'\), a contradiction.

Let \({\mathfrak p}\le {\mathfrak p}' \le {\mathfrak p}''\) with \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak C}_3(k,n,j)\). By 5.1.16 and Lemma 5.3.7 it follows that \({\mathfrak p}' \in {\mathfrak C}_2(k,n,j)\). By 5.1.16 and 5.1.15, there exists \({\mathfrak u}\in {\mathfrak U}_1(k,n,j)\) with \(2{\mathfrak p}'' \lesssim {\mathfrak u}\) and \({\mathcal{I}}({\mathfrak p}'') \subsetneq {\mathcal{I}}({\mathfrak u})\). From \({\mathfrak p}' \le {\mathfrak p}''\), 5.3.3 and transitivity of \(\lesssim \) we then have \(2{\mathfrak p}' \lesssim {\mathfrak u}\) and \({\mathcal{I}}({\mathfrak p}') \subsetneq {\mathcal{I}}({\mathfrak u})\), so \({\mathfrak p}' \in {\mathfrak C}_3(k,n,j)\).

The proof is entirely analogous to Lemma 5.3.7, substituting \({\mathfrak C}_4\) for \({\mathfrak C}_2\), \({\mathfrak C}_3\) for \({\mathfrak C}_1\) and \({\mathfrak p}'\le {\mathfrak p}''\) for \({\mathfrak p}\le {\mathfrak p}'\).

Let \({\mathfrak p}\le {\mathfrak p}' \le {\mathfrak p}''\) with \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak C}_5(k,n,j)\). Then \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak C}_4(k,n,j)\) by 5.1.23, and thus by Lemma 5.3.9 \({\mathfrak p}' \in {\mathfrak C}_4(k,n,j)\). It suffices to show that if \({\mathfrak p}' \in {\mathfrak L}_4(k,n,j)\) then \({\mathfrak p}\in {\mathfrak L}_4(k,n,j)\) by contraposition; this is true by 5.1.22 and \({\mathfrak p}\le {\mathfrak p}'\).

It suffices to show that for all \({\mathfrak p}'\in {\mathfrak P}'\) and \(\lambda \ge 2\) and \({\mathfrak p}\in {\mathfrak P}({\mathfrak P}')\) with \(\lambda {\mathfrak p}' \lesssim \lambda {\mathfrak p}\) we have

Let such \({\mathfrak p}'\), \(\lambda \), \({\mathfrak p}\) be given. It suffices to show that \({\mathfrak p}\in {\mathfrak P}(k)\), that is, it satisfies 5.1.1 and 5.1.2.

We show 5.1.1. As \({\mathfrak p}\in {\mathfrak P}({\mathfrak P}')\), there exists \({\mathfrak p}''\in {\mathfrak P}'\) with \({\mathcal{I}}({\mathfrak p}')\subset {\mathcal{I}}({\mathfrak p}'')\). By assumption on \({\mathfrak P}'\), we have \({\mathfrak p}''\in {\mathfrak P}(k)\) and there exists \(J\in \mathcal{D}\) with \({\mathcal{I}}({\mathfrak p}'')\subset J\) and

Then also \({\mathcal{I}}({\mathfrak p}')\subset J\), which proves 5.1.1 for \({\mathfrak p}\).

We show 5.1.2. Assume to get a contradiction that there exists \(J\in \mathcal{D}\) with \({\mathcal{I}}({\mathfrak p})\subset J\) and

As \(\lambda {\mathfrak p}'\lesssim \lambda {\mathfrak p}\), we have \({\mathcal{I}}({\mathfrak p}')\subset {\mathcal{I}}({\mathfrak p})\), and therefore \({\mathcal{I}}({\mathfrak p}')\subset J\). This contradicts \({\mathfrak p}'\in {\mathfrak P}'\subset {\mathfrak P}(k)\). This proves 5.1.2 for \({\mathfrak p}\).

We have by Lemma 5.3.11 that \(\operatorname{\operatorname {dens}}_1(\mathfrak {A}) \le \operatorname{\operatorname {dens}}_k'(\mathfrak {A})\). Since \(\mathfrak {A} \subset {\mathfrak C}(k,n)\), it follows from monotonicity of suprema and the definition 5.1.6 that \( \operatorname{\operatorname {dens}}_k'(\mathfrak {A}) \le \operatorname{\operatorname {dens}}_k'({\mathfrak C}(k,n))\, . \) By 5.1.6 and 5.1.7, we have

Let \({\mathfrak u}, {\mathfrak u}' \in {\mathfrak U}_2(k,n,j)\) with \({\mathfrak u}\sim {\mathfrak u}'\). If \({\mathfrak u}= {\mathfrak u}'\) then the conclusion of the Lemma clearly holds. Else, there exists \({\mathfrak p}\in {\mathfrak C}_1(k,n,j)\) such that \({\mathcal{I}}({\mathfrak p}) \ne {\mathcal{I}}({\mathfrak u})\) and \(2 {\mathfrak p}\lesssim {\mathfrak u}\) and \(10 {\mathfrak p}\lesssim {\mathfrak u}'\). Using Lemma 5.3.1 and 5.3.4 of Lemma 5.3.3, we deduce that

Now suppose that \(B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}), 100) \cap B_{{\mathfrak u}'}({\mathcal{Q}}({\mathfrak u}'), 100) = \emptyset \). Then we have \(\mathfrak {B}({\mathfrak u}) \cap \mathfrak {B}({\mathfrak u}') = \emptyset \), by the definition 5.1.8 of \(\mathfrak {B}\) and the definition 2.0.24 of \(\lesssim \), but also \(\mathfrak {B}({\mathfrak u}) \subset \mathfrak {B}({\mathfrak p})\) and \(\mathfrak {B}({\mathfrak u}') \subset \mathfrak {B}({\mathfrak p})\), by 5.1.8, 2.0.24 and 5.4.3. Hence,

which contradicts \({\mathfrak p}\in {\mathfrak C}_1(k,n,j)\). Therefore we must have

It follows from \(2{\mathfrak p}\lesssim {\mathfrak u}\) and \(10{\mathfrak p}\lesssim {\mathfrak u}'\) that \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u})\) and \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u}')\). By 2.0.8, it follows that \({\mathcal{I}}({\mathfrak u})\) and \({\mathcal{I}}({\mathfrak u}')\) are nested. Combining this with the conclusion of the last paragraph and definition 5.1.14 of \({\mathfrak U}_1(k,n,j)\), we obtain that \({\mathcal{I}}({\mathfrak u}) = {\mathcal{I}}({\mathfrak u}')\).

Reflexivity holds by definition. For transitivity, suppose that

and \({\mathfrak u}\sim {\mathfrak u}'\), \({\mathfrak u}' \sim {\mathfrak u}''\). By Lemma 5.4.1, it follows that \({\mathcal{I}}({\mathfrak u}) ={\mathcal{I}}({\mathfrak u}') = {\mathcal{I}}({\mathfrak u}'')\), that there exists

and that there exists

If \({\mathfrak u}= {\mathfrak u}'\), then \({\mathfrak u}\sim {\mathfrak u}''\) holds by assumption. Else, there exists by the definition of \(\sim \) some \({\mathfrak p}\in \mathfrak {T}_1({\mathfrak u})\) with \(10{\mathfrak p}\lesssim {\mathfrak u}'\). Then we have \(2{\mathfrak p}\lesssim {\mathfrak u}\) and \({\mathfrak p}\ne {\mathfrak u}\) by definition of \(\mathfrak {T}_1({\mathfrak u})\), so \(4 {\mathfrak p}\lesssim 500 {\mathfrak u}\) by 5.3.5. For \(q \in B_{{\mathfrak u}''}({\mathcal{Q}}({\mathfrak u}''), 1)\) it follows by the triangle inequality that

Using 2.0.17 and the fact that \({\mathcal{I}}({\mathfrak u}) = {\mathcal{I}}({\mathfrak u}') = {\mathcal{I}}({\mathfrak u}'')\) this equals

Since \(4{\mathfrak p}\lesssim 500 {\mathfrak u}\), it follows that \(d_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), q) {\lt} 4 {\lt} 10\). We have shown that \(B_{{\mathfrak u}''}({\mathcal{Q}}({\mathfrak u}''), 1) \subset B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), 10)\), combining this with \({\mathcal{I}}({\mathfrak u}'') = {\mathcal{I}}({\mathfrak u})\) gives \({\mathfrak u}\sim {\mathfrak u}''\).

For symmetry suppose that \({\mathfrak u}\sim {\mathfrak u}'\). By Lemma 5.4.1, it follows that \({\mathcal{I}}({\mathfrak u}) = {\mathcal{I}}({\mathfrak u}')\) and that there exists \({\vartheta }\in B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}), 100) \cap B_{{\mathfrak u}'}({\mathcal{Q}}({\mathfrak u}'), 100)\). Again, for \({\mathfrak u}= {\mathfrak u}'\) symmetry is obvious, so suppose that \({\mathfrak u}\ne {\mathfrak u}'\). There exists \({\mathfrak p}\in \mathfrak {T}_1({\mathfrak u}')\), which then satisfies \(2{\mathfrak p}\lesssim {\mathfrak u}'\) and \({\mathcal{I}}({\mathfrak p}) \neq {\mathcal{I}}({\mathfrak u}')\). By Lemma 5.3.1 and 5.3.5, it follows that

If \(q \in B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}),1)\) then we have from the triangle inequality and the fact that \({\mathcal{I}}({\mathfrak u}) = {\mathcal{I}}({\mathfrak u}')\):

Combining this with 5.4.4 and 2.0.24, we get

Since \(2{\mathfrak p}\lesssim {\mathfrak u}'\), we have \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u}') = {\mathcal{I}}({\mathfrak u})\). Thus, \(10{\mathfrak p}\lesssim {\mathfrak u}\) which completes the proof of \({\mathfrak u}' \sim {\mathfrak u}\).

Let \({\mathfrak p}\in {\mathfrak C}_6(k,n,j)\). By 5.1.19 and 5.1.23, we have \({\mathfrak p}\in {\mathfrak C}_3(k,n,j)\). By 5.1.15 and 5.1.16, there exists \({\mathfrak u}\in {\mathfrak U}_1(k,n,j)\) with \(2{\mathfrak p}\lesssim {\mathfrak u}\) and \({\mathcal{I}}({\mathfrak p}) \ne {\mathcal{I}}({\mathfrak u})\), that is, with \({\mathfrak p}\in \mathfrak {T}_1({\mathfrak u})\). Then \(\mathfrak {T}_1({\mathfrak u})\) is clearly nonempty, so \({\mathfrak u}\in {\mathfrak U}_2(k,n,j)\). By the definition of \({\mathfrak U}_3(k,n,j)\), there exists \({\mathfrak u}' \in {\mathfrak U}_3(k,n,j)\) with \({\mathfrak u}\sim {\mathfrak u}'\). By 5.4.5, we have \({\mathfrak p}\in \mathfrak {T}_2({\mathfrak u}')\).

Let \({\mathfrak p}\in \mathfrak {T}_2({\mathfrak u})\). By 5.4.5, there exists \({\mathfrak u}' \sim {\mathfrak u}\) with \({\mathfrak p}\in \mathfrak {T}_1({\mathfrak u}')\). Then we have \(2{\mathfrak p}\lesssim {\mathfrak u}'\) and \({\mathcal{I}}({\mathfrak p}) \ne {\mathcal{I}}({\mathfrak u}')\), so by 5.3.5 \(4{\mathfrak p}\lesssim 500{\mathfrak u}'\). Further, by Lemma 5.4.1, we have that \({\mathcal{I}}({\mathfrak u}') = {\mathcal{I}}({\mathfrak u})\) and there exists \({\vartheta }\in B_{{\mathfrak u}'}({\mathcal{Q}}({\mathfrak u}'),100) \cap B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}),100)\). Let \({\theta }\in B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}), 1)\). Using the triangle inequality and the fact that \({\mathcal{I}}({\mathfrak u}') ={\mathcal{I}}({\mathfrak u})\), we obtain

Combining this with \(4{\mathfrak p}\lesssim 500{\mathfrak u}'\), we obtain

Together with \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u}') = {\mathcal{I}}({\mathfrak u})\), this gives \(4{\mathfrak p}\lesssim {\mathfrak u}\), which is 2.0.32.

Let \({\mathfrak p}, {\mathfrak p}'' \in \mathfrak {T}_2({\mathfrak u})\) and \({\mathfrak p}' \in {\mathfrak P}\) with \({\mathfrak p}\le {\mathfrak p}' \le {\mathfrak p}''\). By 5.4.5 we have \({\mathfrak p}, {\mathfrak p}'' \in {\mathfrak C}_6(k,n,j) \subset {\mathfrak C}_5(k,n,j)\). By Lemma 5.3.10, we have \({\mathfrak p}' \in {\mathfrak C}_5(k,n,j)\). Since \({\mathfrak p}\in {\mathfrak C}_6(k,n,j)\) we have \({\mathcal{I}}({\mathfrak p}) \not\subset G'\), so \({\mathcal{I}}({\mathfrak p}') \not\subset G'\) and therefore also \({\mathfrak p}' \in {\mathfrak C}_6(k,n,j)\).

By 5.4.5 there exists \({\mathfrak u}' \in {\mathfrak U}_2(k,n,j)\) with \({\mathfrak p}'' \in \mathfrak {T}_1({\mathfrak u}')\) and hence \(2{\mathfrak p}'' \lesssim {\mathfrak u}'\) and \({\mathcal{I}}({\mathfrak p}'') \ne {\mathcal{I}}({\mathfrak u}')\). Together this implies \({\mathcal{I}}({\mathfrak p}'') \subsetneq {\mathcal{I}}({\mathfrak u}')\). With the inclusion \({\mathcal{I}}({\mathfrak p}') \subset {\mathcal{I}}({\mathfrak p}'')\) from \({\mathfrak p}' \le {\mathfrak p}''\), it follows that \({\mathcal{I}}({\mathfrak p}') \subsetneq {\mathcal{I}}({\mathfrak u}')\) and hence \({\mathcal{I}}({\mathfrak p}') \ne {\mathcal{I}}({\mathfrak u}')\). By 5.3.3 and transitivity of \(\lesssim \) we further have \(2{\mathfrak p}' \lesssim {\mathfrak u}'\), so \({\mathfrak p}' \in \mathfrak {T}_1({\mathfrak u}')\). It follows that \({\mathfrak p}' \in \mathfrak {T}_2({\mathfrak u})\), which shows 2.0.33.

By the definition 5.1.13 of \({\mathfrak C}_2(k,n,j)\), there exists a tile \({\mathfrak p}' \in {\mathfrak C}_1(k,n,j)\) with \({\mathfrak p}' \le {\mathfrak p}\) and \({\mathrm{s}}({\mathfrak p}') \le {\mathrm{s}}({\mathfrak p})- Z(n+1)\). By Lemma 2.1.2 we have

By 5.3.3 we have \(2{\mathfrak p}' \lesssim 2{\mathfrak p}\), so by transitivity of \(\lesssim \) there exists \(\mathfrak {v} \sim {\mathfrak u}\) with \(2{\mathfrak p}' \lesssim \mathfrak {v}\) and \({\mathcal{I}}({\mathfrak p}') \ne {\mathcal{I}}(\mathfrak {v})\). Since \({\mathfrak u}, {\mathfrak u}'\) are not equivalent under \(\sim \), we have \(\mathfrak {v} \not\sim {\mathfrak u}'\), thus \(10{\mathfrak p}' \not\lesssim {\mathfrak u}'\). This implies that there exists \(q \in B_{{\mathfrak u}'}({\mathcal{Q}}({\mathfrak u}'), 1) \setminus B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'), 10)\).

From \({\mathfrak p}' \le {\mathfrak p}\), \({\mathcal{I}}({\mathfrak p}') \subset {\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak u}')\) and Lemma 2.1.2 it then follows that

The lemma follows by combining the two displays with the fact that \(95 a \ge 1\).

Let \({\mathfrak p}\in \mathfrak {T}_2({\mathfrak u})\). Then \({\mathfrak p}\in {\mathfrak C}_4(k,n,j)\), hence there exists a chain

of distinct tiles in \({\mathfrak C}_3(n,k,j)\). We pick such a chain and set \({\mathfrak q}= {\mathfrak p}_0\). Then we have from distinctness of the tiles in the chain that \({\mathrm{s}}({\mathfrak p}) \le {\mathrm{s}}({\mathfrak q}) - Z(n+1)\). By 5.1.16 there exists \({\mathfrak u}'' \in {\mathfrak U}_1(k,n,j)\) with \(2{\mathfrak q}\lesssim {\mathfrak u}''\) and \({\mathrm{s}}({\mathfrak q}) {\lt} {\mathrm{s}}({\mathfrak u}'')\). Then we have in particular by Lemma 5.3.1 that \(10 {\mathfrak p}\lesssim {\mathfrak u}''\). Let \({\mathfrak u}' \sim {\mathfrak u}\) be such that \({\mathfrak p}\in \mathfrak {T}_1({\mathfrak u}')\). By the definition of \(\sim \), it follows that \({\mathfrak u}' \sim {\mathfrak u}''\). By transitivity of \(\sim \), we have \({\mathfrak u}\sim {\mathfrak u}''\). By Lemma 5.4.1, we have \(\sc ({\mathfrak u}'') = \sc ({\mathfrak u})\), hence \({\mathrm{s}}({\mathfrak q}) {\lt} {\mathrm{s}}({\mathfrak u})\) and \({\mathrm{s}}({\mathfrak p}) \le {\mathrm{s}}({\mathfrak q}) - Z(n+1) \le {\mathrm{s}}({\mathfrak u}) - Z(n+1) - 1\).

Thus, there exists some cube \(I \in \mathcal{D}\) with \(s(I) = {\mathrm{s}}({\mathfrak u}) - Z(n+1) - 1\) and \(I \subset {\mathcal{I}}({\mathfrak u})\) and \({\mathcal{I}}({\mathfrak p}) \subset I\). Since \({\mathfrak p}\in {\mathfrak C}_5(k,n,j)\), we have that \(I \notin \mathcal{L}({\mathfrak u})\), so \(B(c(I), 8D^{s(I)}) \subset {\mathcal{I}}({\mathfrak u})\). By the triangle inequality, 2.0.1 and \(a \ge 4\), the same then holds for the subcube \({\mathcal{I}}({\mathfrak p}) \subset I\).

Suppose that a point \(x\) is contained in more than \((4n + 12)2^n\) cubes \({\mathcal{I}}({\mathfrak u})\) with \({\mathfrak u}\in {\mathfrak U}_3(k,n,j)\). Since \({\mathfrak U}_3(k,n,j) \subset {\mathfrak C}_1(k,n,j)\) for each such \({\mathfrak u}\), there exists \(\mathfrak {m} \in \mathfrak {M}(k,n)\) such that \(100{\mathfrak u}\lesssim \mathfrak {m}\). We fix such an \(\mathfrak {m}({\mathfrak u}) := \mathfrak {m}\) for each \({\mathfrak u}\), and claim that the map \({\mathfrak u}\mapsto \mathfrak {m}({\mathfrak u})\) is injective. Indeed, assume for \({\mathfrak u}\neq {\mathfrak u}'\) there is \(\mathfrak {m} \in \mathfrak {M}(k,n)\) such that \(100{\mathfrak u}\lesssim \mathfrak {m}\) and \(100{\mathfrak u}' \lesssim \mathfrak {m}\). By 2.0.8, either \({\mathcal{I}}({\mathfrak u}) \subset {\mathcal{I}}({\mathfrak u}')\) or \({\mathcal{I}}({\mathfrak u}') \subset {\mathcal{I}}({\mathfrak u})\). By 5.1.14, \(B_{{\mathfrak u}}({\mathcal{Q}}({\mathfrak u}),100) \cap B_{{\mathfrak u}'}({\mathcal{Q}}({\mathfrak u}'), 100) = \emptyset \). This contradicts \(\Omega (\mathfrak {m})\) being contained in both sets by 2.0.15. Thus \(x\) is contained in more than \((4n + 12)2^n\) cubes \({\mathcal{I}}(\mathfrak {m})\), \(\mathfrak {m} \in \mathfrak {M}(k,n)\). Consequently, we have by 5.1.26 that \(x \in A(2n + 6, k,n) \subset G_2\). Let \({\mathcal{I}}({\mathfrak u})\) be an inclusion minimal cube among the \({\mathcal{I}}({\mathfrak u}'), {\mathfrak u}' \in {\mathfrak U}_3(k,n,j)\) with \(x \in {\mathcal{I}}({\mathfrak u})\). By the dyadic property 2.0.8, we have \({\mathcal{I}}({\mathfrak u}) \subset {\mathcal{I}}({\mathfrak u}')\) for all cubes \({\mathcal{I}}({\mathfrak u}')\) containing \(x\). Thus

Thus \(\mathfrak {T}_1({\mathfrak u}) \cap {\mathfrak C}_6(k,n,j) = \emptyset \). This contradicts \({\mathfrak u}\in {\mathfrak U}_2(k,n,j)\).

We first fix \(k,n, j\). By 2.0.21 and 2.0.20, we have that \(\mathbf{1}_{{\mathcal{I}}({\mathfrak p})} T_{{\mathfrak p}}f(x) = T_{{\mathfrak p}}f(x)\) and hence \(\mathbf{1}_{G \setminus G'} T_{{\mathfrak p}}f(x)= 0\) for all \({\mathfrak p}\in {\mathfrak C}_5(k,n,j) \setminus {\mathfrak C}_6(k,n,j)\). Thus it suffices to estimate the contribution of the sets \({\mathfrak C}_6(k,n,j)\). By Lemma 5.4.8, we can decompose \({\mathfrak U}_3(k,n,j)\) as a disjoint union of at most \(4n + 13\) collections \({\mathfrak U}_4(k,n,j,l)\), \(1 \le l \le 4n+13\), each satisfying

By Lemmas 5.4.4, 5.4.5, 5.4.6, 5.4.7 and 5.3.12, the pairs

are \(n\)-forests for each \(k,n,j,l\), and by Lemma 5.4.3, we have

Since \({\mathcal{I}}({\mathfrak p}) \not\subset G_1\) for all \({\mathfrak p}\in {\mathfrak C}_6(k,n,j)\), we have \({\mathfrak C}_6(k,n,j) \cap {\mathfrak P}_{F,G} = \emptyset \) and hence

Using the triangle inequality according to the splitting by \(k,n,j\) and \(l\) in 5.1.31 and applying Proposition 2.0.4 to each term, we obtain the estimate

for the left hand side of 5.1.31. Since \(|f| \le \mathbf{1}_F\), we have \(\| f\| _2 \le \mu (F)^{1/2}\), and we have \(\| \mathbf{1}_{G\setminus G'}\| _2 \le \mu (G)^{1/2}\). Combining this with \(a \ge 4\), we estimate by

Interchanging the order of summation, the sum equals

which completes the proof of the lemma.

Let \({\mathfrak p}\in {\mathfrak P}_2 \cap {\mathfrak P}_{G \setminus G'}\). Clearly, for every cube \(J = {\mathcal{I}}({\mathfrak p})\) with \({\mathfrak p}\in {\mathfrak P}_{G \setminus G'}\) there exists some \(k \ge 0\) such that 5.1.1 holds, and for no cube \(J \in \mathcal{D}\) and no \(k {\lt} 0\) does 5.1.2 hold. Thus \({\mathfrak p}\in {\mathfrak P}(k)\) for some \(k \ge 0\).

Next, since \(E_2(\lambda , {\mathfrak p}') \subset {\mathcal{I}}({\mathfrak p}')\cap G\) for every \(\lambda \ge 2\) and every tile \({\mathfrak p}' \in {\mathfrak P}(k)\) with \(\lambda {\mathfrak p}\lesssim \lambda {\mathfrak p}'\), it follows from 5.1.2 that \(\mu (E_2(\lambda , {\mathfrak p}')) \le 2^{-k} \mu ({\mathcal{I}}({\mathfrak p}'))\) for every such \({\mathfrak p}'\), so \(\operatorname{\operatorname {dens}}_k'(\{ {\mathfrak p}\} ) \le 2^{-k}\). Combining this with \(a \ge 0\), it follows from 5.1.7 that there exists \(n\ge k\) with \({\mathfrak p}\in {\mathfrak C}(k,n)\).

Since \({\mathfrak p}\in {\mathfrak P}_{G \setminus G'}\), we have in particular \({\mathcal{I}}({\mathfrak p}) \not\subset A(2n + 6, k, n)\), so there exist at most \(1 + (4n + 12)2^n {\lt} 2^{2n+4}\) tiles \(\mathfrak {m} \in \mathfrak {M}(k,n)\) with \({\mathfrak p}\le \mathfrak {m}\). It follows that \({\mathfrak p}\in {\mathfrak L}_0(k,n)\) or \({\mathfrak p}\in {\mathfrak C}_1(k,n,j)\) for some \(1 \le j \le 2n + 3\). In the former case we are done, in the latter case the equality to be shown follows from the definitions of the collections \({\mathfrak C}_i\) and \({\mathfrak L}_i\).

It suffices to show that \({\mathfrak L}_0(k,n)\) contains no chain of length \(n + 1\). Suppose that we had such a chain \({\mathfrak p}_0 \le {\mathfrak p}_1 \le \dotsb \le {\mathfrak p}_{n}\) with \({\mathfrak p}_i \ne {\mathfrak p}_{i+1}\) for \(i =0, \dotsc , n-1\). By 5.1.7, we have that \(\operatorname{\operatorname {dens}}_k'(\{ {\mathfrak p}_n\} ) {\gt} 2^{-n}\). Thus, by 5.1.6, there exists \({\mathfrak p}' \in {\mathfrak P}(k)\) and \(\lambda \ge 2\) with \(\lambda {\mathfrak p}_n \le \lambda {\mathfrak p}'\) and

Let \(\mathfrak {O}\) be the set of all \({\mathfrak p}'' \in {\mathfrak P}(k)\) such that we have \( {\mathcal{I}}({\mathfrak p}'') = {\mathcal{I}}({\mathfrak p}')\) and \(B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'), \lambda ) \cap \Omega ({\mathfrak p}'') \neq \emptyset \). We now show that

The balls \(B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}''), 0.2)\), \({\mathfrak p}'' \in \mathfrak {O}\) are disjoint by 2.0.15, and by the triangle inequality contained in \(B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'), \lambda +1.2)\). By assumption 1.0.11 on \(\Theta \), this ball can be covered with

many \(d_{{\mathfrak p}'}\)-balls of radius \(0.2\). Here we have used that for \(\lambda \ge 2\)

By the triangle inequality, each such ball contains at most one \({\mathcal{Q}}({\mathfrak p}'')\), and each \({\mathcal{Q}}({\mathfrak p}'')\) is contained in one of the balls. Thus we get 5.5.7.

By 2.0.26 and 2.0.27 we have \(E_2(\lambda , {\mathfrak p}') \subset \bigcup _{{\mathfrak p}'' \in \mathfrak {O}} E_1({\mathfrak p}'')\), thus

Hence there exists a tile \({\mathfrak p}'' \in \mathfrak {O}\) with

By the definition 5.1.5 of \(\mathfrak {M}(k,n)\), there exists a tile \(\mathfrak {m} \in \mathfrak {M}(k,n)\) with \({\mathfrak p}' \leq \mathfrak {m}\). From 5.5.6, the inclusion \(E_2(\lambda , {\mathfrak p}') \subset {\mathcal{I}}({\mathfrak p}')\) and \(a\ge 1\) we obtain

From the triangle inequality, Lemma 2.1.2 and \(a \ge 1\), we now obtain for all \({\vartheta }\in B_{\mathfrak {m}}({\mathcal{Q}}(\mathfrak {m}), 1)\) that

Thus, by 2.0.23, \(100{\mathfrak p}_0 \lesssim \mathfrak {m}\), a contradiction to \({\mathfrak p}_0 \notin {\mathfrak C}(k,n)\).

Suppose that there are \({\mathfrak p}_0, {\mathfrak p}_1 \in {\mathfrak L}_2(k,n,j)\) with \({\mathfrak p}_0 \ne {\mathfrak p}_1\) and \({\mathfrak p}_0 \le {\mathfrak p}_1\). By Lemma 5.3.1 and Lemma 5.3.2, it follows that \(2{\mathfrak p}_0 \lesssim 200{\mathfrak p}_1\). Since \({\mathfrak L}_2(k,n,j)\) is finite, there exists a maximal \(l \ge 1\) such that there exists a chain \(2{\mathfrak p}_0 \lesssim 200 {\mathfrak p}_1 \lesssim \dotsb \lesssim 200 {\mathfrak p}_l\) with all \({\mathfrak p}_i\) in \({\mathfrak C}_1(k,n,j)\) and \({\mathfrak p}_i \ne {\mathfrak p}_{i+1}\) for \(i = 0, \dotsc , l-1\). If we have \({\mathfrak p}_l \in {\mathfrak U}_1(k,n,j)\), then it follows from \(2{\mathfrak p}_0 \lesssim 200 {\mathfrak p}_l \lesssim {\mathfrak p}_l\) and 5.1.15 that \({\mathfrak p}_0 \not\in {\mathfrak L}_2(k,n,j)\), a contradiction. Thus, by the definition 5.1.14 of \({\mathfrak U}_1(k,n,j)\), there exists \({\mathfrak p}_{l+1} \in {\mathfrak C}_1(k,n,j)\) with \({\mathcal{I}}({\mathfrak p}_l) \subsetneq {\mathcal{I}}({\mathfrak p}_{l+1}) \) and \({\vartheta }\in B_{{\mathfrak p}_l}({\mathcal{Q}}({\mathfrak p}_l), 100) \cap B_{{\mathfrak p}_{l+1}}({\mathcal{Q}}({\mathfrak p}_{l+1}), 100)\). Using the triangle inequality and Lemma 2.1.2, one deduces that \(200 {\mathfrak p}_l \lesssim 200{\mathfrak p}_{l+1}\). This contradicts maximality of \(l\).

By its definition 5.1.11, each set \({\mathfrak L}_1(k,n,j,l)\) is a set of minimal elements in some set of tiles with respect to \(\le \). If there were distinct \({\mathfrak p}, {\mathfrak q}\in {\mathfrak L}_1(k,n,j,l)\) with \({\mathfrak p}\le {\mathfrak q}\), then \({\mathfrak q}\) would not be minimal. Hence such \({\mathfrak p}, {\mathfrak q}\) do not exist. Similarly, by 5.1.17, each set \({\mathfrak L}_3(k,n,j,l)\) is a set of maximal elements in some set of tiles with respect to \(\le \). If there were distinct \({\mathfrak p}, {\mathfrak q}\in {\mathfrak L}_3(k,n,j,l)\) with \({\mathfrak p}\le {\mathfrak q}\), then \({\mathfrak p}\) would not be maximal.

If \({\mathfrak p}\not\in {\mathfrak P}_{G \setminus G'}\), then \(\mu ({\mathcal{I}}({\mathfrak p}) \cap (G \setminus G')) = 0\). By 2.0.21 and 2.0.26, it follows that \(\mathbf{1}_{G \setminus G'} T_{{\mathfrak p}}f(x) = 0\). We thus have

Let \({\mathfrak L}(k,n)\) denote any of the terms \({\mathfrak L}_i(k,n,j,l) \cap {\mathfrak P}_{G \setminus G'}\) on the right hand side of 5.5.1, where the indices \(j, l\) may be void. Then \({\mathfrak L}(k,n)\) is an antichain, by Lemmas 5.5.2,5.5.3, 5.5.4. Further, we have

by Lemma 5.3.12, and we have

since

Applying now the triangle inequality according to the decomposition in Lemma 5.5.1, and then applying Proposition 2.0.3 to each term, we obtain the estimate

Because \(|f| \le \mathbf{1}_F\), we have \(\| f\| _2 \le \mu (F)^{1/2}\), and we have \(\| \mathbf{1}_{G\setminus G'}\| _2 \le \mu (G)^{1/2}\). Using this and 2.0.3, we bound

The last sum equals, by changing the order of summation,

This completes the proof.