Proof of Theorem 1.2.2. Theorem 1.2.2 follows by applying the triangle inequality to disclesssim according to the splitting in Lemma 1.5.1.2 and Lemma 1.5.1.3 and using both lemmas as well as the bound on the set G' given by Lemma 1.5.1.1.

Proof. Let K = 2^{2a+5}\frac{\mu(F)}{\mu(G)}. For each \fp\in \fP_{F,G} pick a r(\fp)>4D^{\ps(\fp)} with {\mu(F\cap B(\pc(\fp),r(\fp)))}\ge K{\mu(B(\pc(\fp),r(\fp)))}. This ball exists by definition of \fP_{F,G} and dens_2. By applying Theorem 1.2.6 to the collection of balls \mathcal{B} = \{B(\pc(\fp),r(\fp)) \ : \ \fp \in \fP_{F,G}\} and the function u = \mathbf{1}_F, we obtain \mu(\bigcup \mathcal{B}) \le 2^{2a} K^{-1} \mu(F). We conclude with eq-vol-sp-cube and r(\fp)>4D^{\ps(\fp)} \mu(G_1)= \mu(\bigcup_{\fp\in \fP_{F,G}} \scI(\fp)) \le \mu(\bigcup \mathcal{B})\le 2^{2a} K^{-1} \mu (F) = 2^{-5}\mu(G).

Proof. The union of dyadic cubes in \mathcal{C}(G,k) is contained in the union of elements of the set \mathcal{M}(k) of all dyadic cubes J with {\mu(G \cap J)} > 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 \mu (\bigcup \mathcal{C}(G,k))\le \mu (\bigcup \mathcal{M}^*(k))\le \sum_{J\in \mathcal{M}^*(k)}\mu(J). Using the definition of \mathcal{M}(k) and then the pairwise disjointedness of elements in \mathcal{M}^*(k), we estimate this by \le 2^{k+1}\sum_{J\in \mathcal{M}^*(k)}\mu(J\cap G) \le 2^{k+1}\mu(G). This proves the lemma.

Proof. Let \fp,\fp' be as in the lemma. By definition of E_1, we have E_1(\fp)\subset \scI(\fp) and analogously for \fp', we conclude from eintersect that \scI(\fp)\cap \scI(\fp')\neq \emptyset. Let without loss of generality \scI(\fp) be maximal in \{\scI(\fp),\scI(\fp')\}, then \scI(\fp')\subset \scI(\fp). By eintersect, we conclude by definition of E_1 that \fc(\fp)\cap \fc(\fp')\neq \emptyset. By eq-freq-dyadic we conclude \fc(\fp)\subset \fc(\fp'). It follows that \fp'\le \fp. By maximality mnkmax of \fp', we have \fp'=\fp. This proves the lemma.

Proof. 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 \scI(\fp) with \fp in \mathfrak{M}(k,n) and x\in \scI(\fp). 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).

Proof. 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 1.5.2.2 then completes the base of the induction. Now assume that the statement of Lemma 1.5.2.5 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 1.5.2.4, 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 dyadicproperty \sum_{\fp \in {\mathfrak{M}}(k,n)} \mathbf{1}_{\scI(\fp)}(x) = \sum_{\fp \in {\mathfrak{M}}(k,n):\scI(\fp) \subset L} \mathbf{1}_{\scI(\fp)}(x) + \sum_{\fp \in {\mathfrak{M}}(k,n):L \subsetneq \scI(\fp)} \mathbf{1}_{\scI(\fp)}(x). We now show \sum_{\fp \in {\mathfrak{M}}(k,n):\scI(\fp) \subset L} \mathbf{1}_{\scI(\fp)}(x)\ge 2^{n+1}. The left-hand side of the previous display is strictly greater than (\lambda+1)2^{n+1}. If L is the top cube the second sum is zero and the claim follows immediately. Otherwise consider the inclusion-minimal cube \hat{L} with L\subsetneq\hat{L}; all tiles in the second sum satisfy \hat{L}\subset\fp, so the second sum is constant for all x\in\hat{L}. By maximality of L, the second sum 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 the claim. By Lemma 1.5.2.3, we have \sum_{\fp \in {\mathfrak{M}}(k,n):\scI(\fp) \subset L} \mu({E_1}(\fp)) \leq \mu(L). Multiplying by 2^n and applying ebardense, we obtain \sum_{\fp \in {\mathfrak{M}}(k,n):\scI(\fp) \subset L} \mu(\scI(\fp)) \leq 2^n \mu(L). We then have with the previous two estimates 2^{n+1}\mu(A(\lambda+1)\cap L) = \int_{A(\lambda+1)\cap L} 2^{n+1} d\mu \le \int \sum_{\fp \in {\mathfrak{M}}(k,n):\scI(\fp) \subset L} \mathbf{1}_{\scI(\fp)} d\mu \le 2^n \mu(L). Hence 2\mu(A(\lambda+1))=2\sum_{L\in \mathcal{M}^*} \mu(A(\lambda+1)\cap L)\le \sum_{L\in \mathcal{M}^*}\mu( L)= \mu(A(\lambda)). Using the induction hypothesis, this proves alambdameasure for \lambda+1 and completes the proof of the lemma.

Proof. We use Lemma 1.5.2.5 and sum twice a geometric series to obtain \sum_{0\le k}\sum_{k\le n} \mu(A(2n+6,k,n))\le \sum_{0\le k}\sum_{k\le n} 2^{k-5-2n}\mu(G) \le \sum_{0\le k} 2^{-k-3}\mu(G)\le 2^{-2}\mu(G). This proves the lemma.

Proof. We write the left-hand side of eq-musum \int \sum_{\mathfrak{m} \in \mathfrak{M}(k,n)} \mathbf{1}_{\scI(\mathfrak{m})}(x) \, d\mu(x) \le 2^{n+1} \sum_{\lambda=0}^{|\mathfrak{M}|}\mu(A(\lambda, k,n)). Using Lemma 1.5.2.5 and then summing a geometric series, we estimate this by \le 2^{n+1}\sum_{\lambda=0}^{|\mathfrak{M}|} 2^{k+1-\lambda}\mu(G) \le 2^{n+1}2^{k+2}\mu(G). This proves the lemma.

Proof. Let x\in X. For each \fu\in \fU_1(k,n,j) with x\in \scI(\fu), as \fu \in \fC_1(k,n,j), there are at least 2^{j} elements \mathfrak{m}\in \mathfrak{M}(k,n) with 100\fu \lesssim \mathfrak{m} and in particular x\in \scI(\mathfrak{m}). Hence \mathbf{1}_{\scI(\fu)}(x) \le 2^{-j}\sum_{\mathfrak{m} \in \mathfrak{M}(k,n): 100\fu\lesssim \mathfrak{m}} \mathbf{1}_{\scI(\mathfrak{m})}(x). Conversely, for each \mathfrak{m}\in \mathfrak{M}(k,n) with x\in \scI(\mathfrak{m}), let \fU(\mathfrak{m}) be the set of \fu\in \fU_1(k,n,j) with x\in \scI(\fu) and 100\fu \lesssim \mathfrak{m}. Summing the previous bound over \fu and counting the pairs (\fu,\mathfrak{m}) with 100\fu \lesssim \mathfrak{m} differently gives \sum_{\fu\in \fU_1(k,n,j)} \mathbf{1}_{\scI(\fu)}(x) \le 2^{-j}\sum_{\mathfrak{m} \in \mathfrak{M}(k,n)} \sum_{\fu \in \fU(\mathfrak{m})} \mathbf{1}_{\scI(\mathfrak{m})}(x). We estimate the number of elements in \fU(\mathfrak{m}). Let \fu \in \fU(\mathfrak{m}). Then by definition of \fU(\mathfrak{m}) d_{\fu}(\fcc(\fu),\fcc(\mathfrak{m}))\le 100. If \fu' is a further element in \fU(\mathfrak{m}) with \fu\neq \fu', then \fcc(\mathfrak{m}) \in B_{\fu}(\fcc(\fu),100)\cap B_{\fu'}(\fcc(\fu'),100). By the last display and definition of \fU_1(k,n,j), none of \scI(\fu), \scI(\fu') is strictly contained in the other. As both contain x, we have \scI(\fu)=\scI(\fu'). We then have d_{\fu}=d_{\fu'}. By eq-freq-comp-ball, the balls B_{\fu}(\fcc(\fu),0.2) and B_{\fu}(\fcc(\fu'),0.2) are contained respectively in \fc(\fu) and \fc(\fu') and thus are disjoint by eq-dis-freq-cover. By dby2 and the triangle inequality, both balls are contained in B_{\fu}(\fcc(\mathfrak{m}), 100.2). By thirddb applied nine times, there is a collection of at most 2^{9a} balls of radius 0.2 with respect to the metric d_{\fu} which cover the ball B_{\fu}(\fcc(\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_{\fu}(\fcc(\fu),0.2) and B_{\fu}(\fcc(\fu'),0.2), the ball B' can contain at most one of the points \fcc(\fu), \fcc(\fu'). Hence the image of \fU(\mathfrak{m}) under \fcc has at most 2^{9a} elements; since \fcc is injective on \fU(\mathfrak{m}), the same is true of \fU(\mathfrak{m}). Inserting this into usumbymsum proves the lemma.

Proof. Let \fu\in \fU_1(k,n,l). Let I \in \mathcal{L}(\fu). Then we have s(I) = \ps(\fu) - Z(n+1) - 1 and I \subset \scI(\fu) and B(c(I), 8D^{s(I)}) \not \subset \scI(\fu). By eq-vol-sp-cube, the set I is contained in B(c(I), 4D^{s(I)}). By the triangle inequality, the set I is contained in X(\fu):=\{x \in \scI(\fu) \, : \, \rho(x, X \setminus \scI(\fu)) \leq 12 D^{\ps(\fu) - Z(n+1)-1}\}. By the small boundary property eq-small-boundary, noting that 12D^{\ps(\fu) - Z(n+1) - 1} = 12D^{s(I)} > D^{-S}, we have \mu(X(\fu)) \le 2\cdot(12 D^{-Z(n+1)-1})^\kappa \mu(\scI(\mathfrak{u})). Using \kappa<1 and D \ge 12, this proves the lemma.

Proof. As each \fp\in \fL_4(k,n,j) is contained in \cup\mathcal{L}(\fu) for some \fu\in \fU_1(k,n,l), we have \mu(\bigcup_{\fp \in \fL_4 (k,n,j)}\scI(\fp)) \le \sum_{\fu\in \fU_1(k,n,j)} \mu(\bigcup_{I \in \mathcal{L} (\fu)}I). Using Lemma 1.5.2.9 and then Lemma 1.5.2.8, we estimate this further by \le \sum_{\fu\in \fU_1(k,n,j)} D^{1-\kappa Z(n+1)} \mu(\scI(\mathfrak{u})) \le 2^{100a^2+9a+1-j} \sum_{\mathfrak{m}\in \mathfrak{M}(k,n)} D^{-\kappa Z(n+1)} \mu(\scI(\mathfrak{m})). Using Lemma 1.5.2.7, we estimate this by \le 2^{100a^2 + 9a + 1-j} D^{-\kappa Z(n+1)} 2^{n+k+3}\mu(G). Now we estimate G_3 defined in defineg3 by \mu(G_3)\le \sum_{k\ge 0}\, \sum_{n \geq k}\, \sum_{0\le j\le 2n+3} \mu(\bigcup_{\fp \in \fL_4 (k,n,j)} \scI(\fp)) \le \sum_{k\ge 0}\, \sum_{n \geq k}\, \sum_{0\le j\le 2n+3} 2^{100a^2 + 9a + 3 + n + k -j} D^{-\kappa Z(n+1)}\mu(G). Summing geometric series, using that D^{\kappa Z}\ge 8 by defineD, definekappa and defineZ, we estimate this by \le \sum_{k\ge 0}\, \sum_{n \geq k}\, 2^{100a^2 + 9a + 4 + n + k} D^{-\kappa Z(n+1)}\mu(G) = \sum_{k\ge 0} 2^{100a^2 + 9a + 4 + 2k} D^{-\kappa Z(k+1)} \sum_{n \geq k}\, 2^{n - k} D^{-\kappa Z(n-k)}\mu(G) \le \sum_{k\ge 0} 2^{100a^2 + 9a + 5 + 2k} D^{-\kappa Z(k+1)}\mu(G) \le 2^{100a^2 + 9a + 6} D^{-\kappa Z}\mu(G). Using D = 2^{100a^2} and a \ge 4 and \kappa Z \ge 2 by defineD and definekappa proves the lemma.

Adding up the bounds in Lemmas Lemma 1.5.2.1, Lemma 1.5.2.6, and Lemma 1.5.2.10 proves Lemma 1.5.1.1.

Proof. This follows immediately from the definition wiggleorder of \lesssim and the two inclusions B_{\fp}(\fcc(\fp), n) \subset B_{\fp}(\fcc(\fp), n') and B_{\fp'}(\fcc(\fp'), m') \subset B_{\fp'}(\fcc(\fp'), m).

Proof. The assumption eq-wiggle1 together with the definition wiggleorder of \lesssim implies that \scI(\fp) \subsetneq \scI(\fp'). Let \mfa \in B_{\fp'}(\fcc(\fp'), m). Then we have by the triangle inequality d_{\fp}(\fcc(\fp), \mfa) \le d_{\fp}(\fcc(\fp), \fcc(\fp')) + d_{\fp}(\fcc(\fp'), \mfa). The first summand is bounded by n since \fcc(\fp') \in B_{\fp'}(\fcc(\fp'), k) \subset B_{\fp}(\fcc(\fp), n), using wiggleorder. For the second summand we use Lemma 1.2.1.2 to show that the sum is estimated by n + 2^{-95a} d_{\fp'}(\fcc(\fp'), \mfa) < n + 2^{-95a} m. Thus B_{\fp'}(\fcc(\fp'),m) \subset B_{\fp}(\fcc(\fp),n + 2^{-95a}m). Combined with \scI(\fp) \subset \scI(\fp'), this yields eq-wiggle2.