Fix \(x\in G\setminus G'\). Sorting the tiles \({\mathfrak p}\) on the left-hand-side of 4.0.1 by the value \({\mathrm{s}}({\mathfrak p})\in [-S,S]\), it suffices to prove for every \(-S\le s\le S\) that

if \(s\not\in [\sigma _1(x), \sigma _2(x)]\) and

if \(s\in [\sigma _1(x),\sigma _2(x)]\). If \(s\not\in [\sigma _1(x), \sigma _2(x)]\), then by definition of \(E({\mathfrak p})\) we have \(x\not\in E({\mathfrak p})\) for any \({\mathfrak p}\) with \({\mathrm{s}}({\mathfrak p})=s\) and thus \(T_{{\mathfrak p}} f(x)=0\). This proves 4.0.2.

Now assume \(s\in [\sigma _1(x),\sigma _2(x)]\). By 2.0.7, 2.0.9, 2.0.10, the fact that \(c(I_0) = o\) and \(G\subset B(o,\frac14 D^S)\), there is at least one \(I\in \mathcal{D}\) with \(s(I)=s\) and \(x\in I\). By 2.0.8, this \(I\) is unique. By 2.0.13, there is precisely one \({\mathfrak p}\in {\mathfrak P}(I)\) such that \({Q}(x)\in {\Omega }({\mathfrak p})\). Hence there is precisely one \({\mathfrak p}\in {\mathfrak P}\) with \({\mathrm{s}}({\mathfrak p})=s\) such that \(x\in E({\mathfrak p})\). For this \({\mathfrak p}\), the value \(T_{{\mathfrak p}}(x)\) by its definition in 2.0.21 equals the right-hand side of 4.0.3. This proves the lemma.

We now estimate with Lemma 4.0.3 and Proposition 2.0.2

This proves 2.0.6 for the chosen set \(G'\) and arbitrary \(f\) and thus completes the proof of Proposition 2.0.1.

Let \(k\) and \(Y\) be given. By applying the doubling property 1.0.5 inductively, we have for each integer \(j\ge 0\)

Since \(X\) is the union of the balls \(B(y,2^{j}D^k)\) and \(\mu \) is not zero, at least one of the balls \(B(y,2^{j}D^k)\) has positive measure, thus \(B(y,D^k)\) has positive measure.

Applying 4.1.4 for \(j' = \ln _2(8D^{2S}) = 3 + 2S \cdot 100a^2\) by 2.0.1, using \(-S\le k\le S\), \(y\in B(o,4D^S)\), and the triangle inequality, we have

Using the disjointedness of the balls in 4.1.2, 4.1.1, and the triangle inequality for \(\rho \), we obtain

As \(\mu (o,4D^S)\) is not zero, the lemma follows.

Let \(x\) be any point of \(B(o, 4D^S-D^k)\). By maximality of \(|Y_k|\), the ball \(B(x, D^k)\) intersects one of the balls \(B(y, D^k)\) with \(y\in Y_k\). By the triangle inequality, \(x\in B(y,2D^k)\).

We prove these statements simultaneously by induction on the ordered set of pairs \((y,k)\). Let \(-S\le k\le S\).

We first consider 4.1.13 for \(j=1\). If \(k=-S\), disjointedness of the sets \(I_1(y,-S)\) follows by definition of \(I_1\) and \(Y_k\). If \(k{\gt}-S\), assume \(x\) is in \(I_1(y_m,k)\) for \(m=1,2\). Then, for \(m=1,2\), there is \(z_m\in Y_{k-1}\cap B(y_m,D^k)\) with \(x\in I_3(z_m,k-1)\). Using 4.1.13 inductively for \(j=3\), we conclude \(z_1=z_2\). This implies that the balls \(B(y_1, D^k)\) and \(B(y_2, D^k)\) intersect. By construction of \(Y_k\), this implies \(y_1=y_2\). This proves 4.1.13 for \(j=1\).

We next consider 4.1.13 for \(j=3\). Assume \(x\) is in \(I_3(y_m,k)\) for \(m=1,2\) and \(y_m\in Y_k\). If \(x\) is in \(X_k\), then by definition 4.1.11, \(x\in I_1(y_m,k)\) for \(m=1,2\). As we have already shown 4.1.13 for \(j=1\), we conclude \(y_1=y_2\). This completes the proof in case \(x\in X_k\), and we may assume \(x\) is not in \(X_k\). By definition 4.1.11, \(x\) is not in \(I_3(z,k)\) for any \(z\) with \(z{\lt}y_1\) or \(z{\lt}y_2\). Hence, neither \(y_1{\lt}y_2\) nor \(y_2{\lt}y_1\), and by totality of the order of \(Y_k\), we have \(y_1=y_2\). This completes the proof of 4.1.13 for \(j=3\).

We show 4.1.14 for \(j=2\). In case \(k=-S\), this follows from Lemma 4.1.2. Assume \(k{\gt}-S\). Let \(x\) be a point of \(B(o, 4D^S-2D^k)\). By induction, there is \(y'\in Y_{k-1}\) such that \(x\in I_3(y',k-1)\). Using the inductive statement 4.1.15, we obtain \(x\in B(y',4D^{k-1})\). As \(D{\gt}4\), by applying the triangle inequality with the points, \(o\), \(x\), and \(y'\), we obtain that \(y'\in B(o, 4D^S-D^k)\). By Lemma 4.1.2, \(y'\) is in \(B(y,2D^k)\) for some \(y\in Y_k\). It follows that \(x\in I_2(y,k)\). This proves 4.1.14 for \(j=2\).

We show 4.1.14 for \(j=3\). Let \(x\in B(o, 4D^S-2D^k)\). In case \(x\in X_k\), then by definition of \(X_k\) we have \(x\in I_1(y,k)\) for some \(y\in Y_k\) and thus \(x\in I_3(y,k)\). We may thus assume \(x\not\in X_k\). As we have already seen 4.1.14 for \(j=2\), there is \(y\in Y_k\) such that \(x\in I_2(y,k)\). We may assume this \(y\) is minimal with respect to the order in \(Y_k\). Then \(x\in I_3(y,k)\). This proves 4.1.14 for \(j=3\).

Next, we show the first inclusion in 4.1.15. Let \(x\in B(y,\frac12 D^k)\). As \(I_1(y,k)\subset I_3(y,k)\), it suffices to show \(x\in I_1(y,k)\). If \(k=-S\), this follows immediately from the assumption on \(x\) and the definition of \(I_1\). Assume \(k{\gt}-S\). By the inductive statement 4.1.14 and \(D{\gt}4\), there is a \(y'\in Y_{k-1}\) such that \(x\in I_3(y',k-1)\). By the inductive statement 4.1.15, we conclude \(x\in B(y',4D^{k-1})\). By the triangle inequality with points \(x\), \(y\), \(y'\), and \(D\geq 8\), we have \(y'\in B(y,D^k)\). It follows by definition 4.1.10 that \(I_3(y',k-1)\subset I_1(y,k)\), and thus \(x\in I_3(y,k)\). This proves the first inclusion in 4.1.15.

We show the second inclusion in 4.1.15. Let \(x\in I_3(y,k)\). As \(I_1(y,k)\subset I_2(y,k)\) directly from the definition 4.1.10, it follows by definition 4.1.11 that \(x\in I_2(y,k)\). By definition 4.1.10, there is \(y'\in Y_{k-1}\cap B(y,2D^k)\) with \(x\in I_3(y',k-1)\). By induction, \(x\in B(y', 4D^{k-1})\). By the triangle inequality applied to the points \(x,y',y\) and \(D{\gt}4\), we conclude \(x\in B(y,4D^k)\). This shows the second inclusion in 4.1.15 and completes the proof of the lemma.

Let \(-S\le l\le k\le S\) and \(y\in Y_k\). If \(l=k\), the inclusion 4.1.16 is true from the definition of set union. We may then assume inductively that \(k{\gt}l\) and the statement of the lemma is true if \(k\) is replaced by \(k-1\). Let \(x\in I_3(y,k)\). By definition 4.1.11, \(x\in I_j(y,k)\) for some \(j\in \{ 1,2\} \). By 4.1.10, \(x\in I_3(w,k-1)\) for some \(w\in Y_{k-1}\). We conclude 4.1.16 by induction.

Let \(l,k,y,y'\) be as in the lemma. Pick \(x\in I_3(y',l)\cap I_3(y,k)\). Assume first \(l=k\). By 4.1.13 of Lemma 4.1.3, we conclude \(y'=y\), and thus 4.1.17. Now assume \(l{\lt}k\). By induction, we may assume that the statement of the lemma is proven for \(k-1\) in place of \(k\).

By Lemma 4.1.4, there is a \(y''\in Y_{k-1}\) such that \(x\in I_3(y'',k-1)\). By induction, we have \(I_3(y',l)\subset I_3(y'',k-1)\). It remains to prove

We make a case distinction and assume first \(x\in X_k\). By Definition 4.1.11, we have \(x\in I_1(y,k)\). By Definition 4.1.10, there is a \(v\in Y_{k-1}\cap B(y,D^k)\) with \(x\in I_3(v,k-1)\). By 4.1.13 of Lemma 4.1.3, we have \(v=y''\). By Definition 4.1.10, we then have \(I_3(y'',k-1)\subset I_1(y,k)\). Then 4.1.18 follows by Definition 4.1.11 in the given case.

Assume now the case \(x\notin X_k\). By 4.1.11, we have \(x\in I_2(y,k)\). Moreover, for any \(u{\lt}y\) in \(Y_k\), we have \(x\not\in I_3(u,k)\). Let \(u{\lt}y\). By transitivity of the order in \(Y_k\), we conclude \(x\not\in I_2(u,k)\). By 4.1.10 and the disjointedness property of Lemma 4.1.3, we have \(I_3(y'',k-1)\cap I_2(u,k)= \emptyset \). Similarly, \(I_3(y'',k-1)\cap I_1(u,k)= \emptyset \). Hence \(I_3(y'',k-1)\cap I_3(u,k)=\emptyset \). As \(u{\lt}y\) was arbitrary, we conclude with 4.1.11 the claim in the given case. This completes the proof of 4.1.18, and thus also 4.1.17.

As \(x\in I_3(y'',k'')\cap I_3(y',k')\) and \(k''{\lt} k'\), we have by Lemma 4.1.5 that \(I_3(y'',k'')\subset I_3(y',k')\). Similarly, \(I_3(y',k')\subset I_3(y,k)\). Pick \(x'\in X\setminus I_3(y,k)\) such that

which exists as \((y'',k''|y,k)\). As \(x'\in X\setminus I_3(y',k')\) as well, we conclude \((y'',k''| y',k')\). By the triangle inequality, we have

Using the choice of \(x\) and 4.1.15 as well as 4.1.21, we estimate this by

where we have used \(D{\gt}5\) and \(k''{\lt}k'\). We conclude \((y',k'|y,k)\).

Let \(K\) be as in the lemma. Let \(-S+K\le k\le S\) and \(y\in Y_k\).

Pick \(k'\) so that \(k-K\le k'\le k\). For each \(y''\in Y_{k-K}\) with \((y'',k-K| y,k)\), by Lemma 4.1.4 and Lemma 4.1.5, there is a unique \(y'\in Y_{k'}\) such that

Using Lemma 4.1.6, \((y',k'|y,k)\).

We conclude using the disjointedness property of Lemma 4.1.3 that

Adding over \(k-K{\lt}k'\le k\), and using

from the doubling property 1.0.5 and 4.1.15 gives

Each ball \(B(y', \frac14 D^{k'})\) occurring in 4.1.28 is contained in \(I_3(y',k')\) by 4.1.15 and in turn contained in \(I_3(y,k)\) by 4.1.25. Assume for the moment all these balls are pairwise disjoint. Then by additivity of the measure,

which by \(K=2^{4a+1}\) implies 4.1.24.

It thus remains to prove that the balls occurring in 4.1.28 are pairwise disjoint. Let \((u,l)\) and \((u',l')\) be two parameter pairs occurring in the sum of 4.1.28 and let \( B(u, \frac14 D^l)\) and \(B({u’}, \frac14 D^{l'})\) be the corresponding balls. If \(l=l'\), then the balls are equal or disjoint by 4.1.15 and 4.1.13 of Lemma 4.1.3. Assume then without loss of generality that \(l'{\lt}l\). Towards a contradiction, assume that

As \((u',l'|y,k)\), there is a point \(x\) in \(X\setminus I_3(y,k)\) with \(\rho (x,u'){\lt}6D^{l'}\). Using \(D{\gt}25\), we conclude from the triangle inequality and 4.1.30 that \(x\in B(u,\frac12D^l)\). However, \(B(u,\frac12 D^l)\subset I_3(u,l)\), and \(I_3(u,l)\subset I_3(y,k)\), a contradiction to \(x\not\in I_3(y,k)\). This proves the lemma.

We prove this by induction on \(n\). If \(n=0\), both sides of 4.1.31 are equal to \(\mu (I_3(y,k))\) by 4.1.13. If \(n=1\), this follows from Lemma 4.1.7.

Assume \(n{\gt}1\) and 4.1.31 has been proven for \(n-1\). We write 4.1.31

Applying the induction hypothesis, this is bounded by

Applying 4.1.24 gives 4.1.31, and proves the lemma.

Let \(x\in I_3(y,k)\) with \(\rho (x, X \setminus I_3(y,k)) \leq t D^{k}\). Let \(K = 2^{4a+1}\) as in Lemma 4.1.8. Let \(n\) be the largest integer such that \(D^{nK} \le \frac{1}{t}\), so that \(tD^k \le D^{k-nK}\) and

Let \(k' = k - nK\), by the assumption \(tD^k \ge D^{-S}\), we have \(k' \ge -S\). By 4.1.16, there exists \(y' \in Y_{k'}\) with \(x \in I_3(y',k')\). By the squeezing property 4.1.15 and the assumption on \(x\), we have

By the assumption on \(n\) and the definition of \(k'\), this is

Together with 4.1.17 thus \((y',k'|y,k)\). We have shown that

Using monotonicity and additivity of the measure and Lemma 4.1.8, we obtain

By 4.1.36 and the definition 2.0.1 of \(D\), this is bounded by

which completes the proof by the definition 2.0.2 of \(\kappa \).

We first show that \((\tilde{\mathcal{D}},c,s)\) satisfies properties 2.0.7, 2.0.8, 2.0.10 and 2.0.11. Property 2.0.10 follows from 4.1.15, while 2.0.11 follows from Lemma 4.1.9.

Let \(x\in B(o, D^S)\). We show properties 2.0.7 and 2.0.8 for \((\tilde{\mathcal{D}},c,s)\) and \(x\).

We first show 2.0.7 for \((\tilde{\mathcal{D}},c,s)\) by contradiction. Then there is an \(I\) violating the conclusion of 2.0.7. Pick such \(I=I_3(y,l)\) such that \(l\) is minimal. By assumption, we have \(-S\le k{\lt}l\); in particular \(-S{\lt}l\). By definition, \(I_3(y, l)\) is contained in \(I_1(y, l)\cup I_2(y, l)\), which is contained in the union of \(I_3(y',l-1)\) with \(y'\in Y_{l-1}\). By minimality of \(l\), each such \(I_3(y',l-1)\) is contained in the union of all \(I_3(z,k)\) with \(z\in Y_k\). This proves 2.0.7.

We now show 2.0.8 for \((\tilde{\mathcal{D}},c,s)\). Assume to get a contradiction that there are non-disjoint \(I, J\in \tilde{\mathcal{D}}\) with \(s(I)\le s(J)\) and \(I \not\subset J\). We may assume the existence of such \(I\) and \(J\) with minimal \(s(J)-s(I)\). Let \(k=s(I)\). Assume first \(s(J)=k\). Let \(I=I_3(y_1,k)\) and \(J=I_3(y_2,k)\) with \(y_1,y_2\in Y_k\). If \(y_1=y_2\), then \(I=J\), a contradiction to \(I\not\subset J\). If \(y_1\neq y_2\), then \(I\cap J=\emptyset \) by 4.1.13, a contradiction to the non-disjointedness of \(I,J\). Assume now \(s(J){\gt}k\). Choose \(y\in I\cap J\). By property 2.0.7, there is \(K\in \tilde{\mathcal{D}}\) with \(s(K)=s(J)-1\) and \(y\in K\). By construction of \(J\), and pairwise disjointedness of all \(I_3(w,s(J)-1)\) that we have already seen, we have \(K\subset J\). By minimality of \(s(J)\), we have \(I\subset K\). This proves \(I\subset J\) and thus 2.0.8.

Now note that properties 2.0.8, 2.0.10 and 2.0.11 immediately carry over to \((\mathcal{D},c, s)\) by restriction. 2.0.9 is true for \((\mathcal{D}, c, s)\) by definition, and 2.0.7 follows from 2.0.7 and 2.0.8 for \((\tilde{\mathcal{D}}, c, s)\).

Let \({\theta }\in \bigcup _{{\vartheta }\in {Q}(X)} B_{I^\circ }({\vartheta }, 1)\). By maximality of \(\mathcal{Z}(I)\), there must be a point \(z \in \mathcal{Z}(I)\) such that \(B_{I^\circ }(z, 0.3) \cap B_{I^\circ }({\theta }, 0.3) \ne \emptyset \). Else, \(\mathcal{Z}(I) \cup \{ {\theta }\} \) would be a set of larger cardinality than \(\mathcal{Z}(I)\) satisfying 4.2.1 and 4.2.2. Fix such \(z\), and fix a point \(z_1 \in B_{I^\circ }(z, 0.3) \cap B_{I^\circ }({\theta }, 0.3)\). By the triangle inequality, we deduce that

and hence \({\theta }\in B_{I^\circ }(z, 0.7)\).

By the definition of the map \({\mathcal{I}}\), we have

By 4.2.4, the set \(\Omega _1((I, z_k))\) is disjoint from each \(\Omega _1((I, z_i))\) with \(i {\lt} k\). Thus the sets \(\Omega _1({\mathfrak p})\), \({\mathfrak p}\in {\mathfrak P}(I)\) are pairwise disjoint.

For 4.2.6 let \({\mathfrak p}= (I, z)\). The second inclusion in 4.2.6 then follows from 4.2.4 and the equality \(B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), 0.7) = B_{I^\circ }(z, 0.7)\), which is true by definition. For the first inclusion in 4.2.6 let \({\vartheta }\in B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}),0.3)\). Let \(k\) be such that \(z = z_k\) in the enumeration we chose above. It follows immediately from 4.2.4 and 4.2.2 that \({\vartheta }\notin \Omega _1((I, z_i))\) for all \(i {\lt} k\). Thus, again from 4.2.4, we have \({\vartheta }\in \Omega _1((I,z_k))\).

To show 4.2.5 let \({\vartheta }\in \bigcup _{z \in \mathcal{Z}(I)} B_{I^\circ }(z,0.7)\). If there exists \(z \in \mathcal{Z}(I)\) with \({\vartheta }\in B_{I^\circ }(z,0.3)\), then

by the first inclusion in 4.2.6.

Now suppose that there exists no \(z \in \mathcal{Z}(I)\) with \({\vartheta }\in B_{I^\circ }(z, 0.3)\). Let \(k\) be minimal such that \({\vartheta }\in B_{I^\circ }(z_k, 0.7)\). Since \(\Omega _1((I, z_i)) \subset B_{I^\circ }(z_i, 0.7)\) for each \(i\) by 4.2.4, we have that \({\vartheta }\notin \Omega _1((I, z_i))\) for all \(i {\lt} k\). Hence \({\vartheta }\in \Omega _1((I, z_k))\), again by 4.2.4.

First, we prove 2.0.15. If \(I =I_0\), then 2.0.15 holds for all \({\mathfrak p}\in {\mathfrak P}(I)\) by 4.2.7 and 4.2.6. Now suppose that \(I\) is not maximal in \(\mathcal{D}\) with respect to set inclusion. Then we may assume by induction that for all \(J \in \mathcal{D}\) with \(I \subset J\) and all \({\mathfrak p}' \in {\mathfrak P}(J)\), 2.0.15 holds. Let \(J\) be the unique minimal cube in \(\mathcal{D}\) with \(I \subsetneq J\).

Suppose that \({\vartheta }\in \Omega ({\mathfrak p})\). If \({\vartheta }\in B_{{\mathfrak p}}(\mathcal{Q}({\mathfrak p}), 0.2)\), then since

we conclude that \({\vartheta }\in B_{{\mathfrak p}}(\mathcal{Q}({\mathfrak p}), 0.7)\). If not, by 4.2.8, there exists \(z \in \mathcal{Z}(J) \cap \Omega _1({\mathfrak p})\) with \({\vartheta }\in \Omega (J,z)\). Using the triangle inequality and 4.2.6, we obtain

By Lemma 2.1.2 and the induction hypothesis, this is estimated by

This shows the second inclusion in 2.0.15. The first inclusion is immediate from 4.2.8.

Next, we show 2.0.13. Let \(I \in \mathcal{D}\).

If \(I = I_0\), then disjointedness of the sets \({\Omega }({\mathfrak p})\) for \({\mathfrak p}\in {\mathfrak P}(I)\) follows from the definition 4.2.7 and Lemma 4.2.2. To obtain the inclusion in 2.0.13 one combines the inclusions 4.2.3 and 4.2.5 of Lemma 4.2.3 with 4.2.7.

Now we turn to the case where there exists \(J \in \mathcal{D}\) with \(I \subset J\) and \(I\ne J\). In this case we use induction: It suffices to show 2.0.13 under the assumption that it holds for all cubes \(J \in \mathcal{D}\) with \(I \subset J\). As shown before definition 4.2.8, we may choose the unique inclusion minimal such \(J\). To show disjointedness of the sets \({\Omega }({\mathfrak p}), {\mathfrak p}\in {\mathfrak P}(I)\) we pick two tiles \({\mathfrak p}, {\mathfrak p}' \in {\mathfrak P}(I)\) and \({\vartheta }\in {\Omega }({\mathfrak p}) \cap {\Omega }({\mathfrak p}')\). Then we are by 4.2.8 in one of the following four cases.

1. There exist \(z \in \mathcal{Z}(J) \cap \Omega _1({\mathfrak p})\) such that \({\vartheta }\in \Omega (J, z)\), and there exists \(z' \in \mathcal{Z}(J) \cap \Omega _1({\mathfrak p}')\) such that \({\vartheta }\in \Omega (J, z')\). By the induction hypothesis, that 2.0.13 holds for \(J\), we must have \(z = z'\). By Lemma 4.2.2, we must then have \({\mathfrak p}= {\mathfrak p}'\).

2. There exists \(z \in \mathcal{Z}(J) \cap \Omega _1({\mathfrak p})\) such that \({\vartheta }\in \Omega (J,z)\), and \({\vartheta }\in B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'), 0.2)\). Using the triangle inequality, Lemma 2.1.2 and 2.0.15, we obtain

Thus \(z \in \Omega _1({\mathfrak p}')\) by 4.2.6. By Lemma 4.2.2, it follows that \({\mathfrak p}= {\mathfrak p}'\).

3. There exists \(z' \in \mathcal{Z}(J) \cap \Omega _1({\mathfrak p}')\) such that \({\vartheta }\in \Omega (J,z')\), and \({\vartheta }\in B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), 0.2)\). This case is the same as case 2., after swapping \({\mathfrak p}\) and \({\mathfrak p}'\).

4. We have \({\vartheta }\in B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), 0.2) \cap B_{{\mathfrak p}'}({\mathcal{Q}}({\mathfrak p}'), 0.2)\). In this case it follows that \({\mathfrak p}= {\mathfrak p}'\) since the sets \(B_{{\mathfrak p}}({\mathcal{Q}}({\mathfrak p}), 0.2)\) are pairwise disjoint by the inclusion 4.2.6 and Lemma 4.2.2.

To show the inclusion in 2.0.13, let \({\vartheta }\in {Q}(X)\). By the induction hypothesis, there exists \({\mathfrak p}\in {\mathfrak P}(J)\) such that \({\vartheta }\in \Omega ({\mathfrak p})\). By definition of the set \({\mathfrak P}\), we have \({\mathfrak p}= (J, z)\) for some \(z \in \mathcal{Z}(J)\). Thus, by 4.2.3, there exists \(z' \in \mathcal{Z}(I)\) with \(z \in B_{I^\circ }(z', 0.7)\). Then by Lemma 4.2.3 there exists \({\mathfrak p}' \in {\mathfrak P}(I)\) with \(z \in \mathcal{Z}(J) \cap \Omega _1({\mathfrak p}')\). Consequently, by 4.2.8, \({\vartheta }\in {\Omega }({\mathfrak p}')\). This completes the proof of 2.0.13.

Finally, we show 2.0.14. Let \({\mathfrak p}, {\mathfrak q}\in {\mathfrak P}\) with \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak q})\) and \({\Omega }({\mathfrak p}) \cap {\Omega }({\mathfrak q}) \ne \emptyset \). If we have \({\mathrm{s}}({\mathfrak p}) \ge {\mathrm{s}}({\mathfrak q})\), then it follows from 2.0.8 that \(I = J\), thus \({\mathfrak p}, {\mathfrak q}\in {\mathfrak P}(I)\). By 2.0.13 we have then either \({\Omega }({\mathfrak p}) \cap {\Omega }({\mathfrak q}) = \emptyset \) or \({\Omega }({\mathfrak p}) = {\Omega }({\mathfrak q})\). By the assumption in 2.0.14 we have \({\Omega }({\mathfrak p}) \cap {\Omega }({\mathfrak q}) \ne \emptyset \), so we must have \({\Omega }({\mathfrak p}) = {\Omega }({\mathfrak q})\) and in particular \({\Omega }({\mathfrak q}) \subset {\Omega }({\mathfrak p})\).

So it remains to show 2.0.14 under the additional assumption that \({\mathrm{s}}({\mathfrak q}) {\gt} {\mathrm{s}}({\mathfrak p})\). In this case, we argue by induction on \({\mathrm{s}}({\mathfrak q})-{\mathrm{s}}({\mathfrak p})\). By 2.0.7, there exists a cube \(J \in \mathcal{D}\) with \(s(J) = {\mathrm{s}}({\mathfrak q}) - 1\) and \(J \cap {\mathcal{I}}({\mathfrak p}) \ne \emptyset \). We pick one such \(J\). By 2.0.8, we have \({\mathcal{I}}({\mathfrak p}) \subset J \subset {\mathcal{I}}({\mathfrak q})\).

Thus, by 4.2.3, there exists \(z' \in \mathcal{Z}(J)\) with \({\mathcal{Q}}({\mathfrak q}) \in B_{J^\circ }(z', 0.7)\). Then by Lemma 4.2.3 there exists \({\mathfrak q}' \in {\mathfrak P}(J)\) with \({\mathcal{Q}}({\mathfrak q}) \in \Omega _1({\mathfrak q}')\). By 4.2.8, it follows that \(\Omega ({\mathfrak q}) \subset \Omega ({\mathfrak q}')\). Note that then \({\mathcal{I}}({\mathfrak p}) \subset {\mathcal{I}}({\mathfrak q}')\) and \({\Omega }({\mathfrak p}) \cap {\Omega }({\mathfrak q}') \ne \emptyset \) and \({\mathrm{s}}({\mathfrak q}') - {\mathrm{s}}({\mathfrak p}) = {\mathrm{s}}({\mathfrak q}) - {\mathrm{s}}({\mathfrak p}) - 1\). Thus, we have by the induction hypothesis that \(\Omega ({\mathfrak q}') \subset \Omega ({\mathfrak p})\). This completes the proof.