4 Proof of Finitary Carleson

To prove Proposition 2.0.1, we already fixed in Chapter 2 measurable functions $σ_{1}, σ_{2}, Q$ and Borel sets $F, G$ . We have also defined $S$ to be the smallest integer such that the ranges of $σ_{1}$ and $σ_{2}$ are contained in $[- S, S]$ and $F$ and $G$ are contained in the ball $B (o, \frac{1}{4} D^{S})$ .

The proof of the next lemma is done in Section 4.1, following the construction of dyadic cubes in [ .

Lemma 4.0.1 grid existence

✓

There exists a grid structure $(D, c, s)$ .

The next lemma, which we prove in Section 4.2, should be compared with the construction in [ .

Lemma 4.0.2 tile structure

✓

For a given grid structure $(D, c, s)$ , there exists a tile structure $(P, I, Ω, Q, c, s)$ .

Choose a grid structure $(D, c, s)$ with Lemma 4.0.1 and a tile structure for this grid structure $(P, I, Ω, Q, c, s)$ with Lemma 4.0.2. Applying Proposition 2.0.2, we obtain a Borel set $G^{'}$ in $X$ with $2 μ (G^{'}) \leq μ (G)$ such that for all Borel functions $f : X \to C$ with $| f | \leq 1_{F}$ we have 2.0.22.

Lemma 4.0.3 tile sum operator

✓

We have for all $x \in G ∖ G^{'}$

\sum_{p \in P} T_{p} f (x) = \sum_{s = σ_{1} (x)}^{σ_{2} (x)} \int K_{s} (x, y) f (y) e (Q (x) (y) - Q (x) (x)) d μ (y) .

4.0.1

Proof ▶

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

\sum_{p \in P : s (p) = s} T_{p} f (x) = 0

4.0.2

if $s \notin [σ_{1} (x), σ_{2} (x)]$ and

\sum_{p \in P : s (p) = s} T_{p} f (x) = \int K_{s} (x, y) f (y) e (Q (x) (y) - Q (x) (x)) d μ (y) .

4.0.3

if $s \in [σ_{1} (x), σ_{2} (x)]$ . If $s \notin [σ_{1} (x), σ_{2} (x)]$ , then by definition of $E (p)$ we have $x \notin E (p)$ for any $p$ with $s (p) = s$ and thus $T_{p} f (x) = 0$ . This proves 4.0.2.

Now assume $s \in [σ_{1} (x), σ_{2} (x)]$ . By 2.0.7, 2.0.9, 2.0.10, the fact that $c (I_{0}) = o$ and $G \subset B (o, \frac{1}{4} D^{S})$ , there is at least one $I \in D$ with $s (I) = s$ and $x \in I$ . By 2.0.8, this $I$ is unique. By 2.0.13, there is precisely one $p \in P (I)$ such that $Q (x) \in Ω (p)$ . Hence there is precisely one $p \in P$ with $s (p) = s$ such that $x \in E (p)$ . For this $p$ , the value $T_{p} (x)$ by its definition in 2.0.21 equals the right-hand side of 4.0.3. This proves the lemma.

We use this to prove Proposition 2.0.1.

Proof of Proposition 2.0.1 ▶

We now estimate with Lemma 4.0.3 and Proposition 2.0.2

\int_{G ∖ G^{'}} | \sum_{s = σ_{1} (x)}^{σ_{2} (x)} \int K_{s} (x, y) f (y) e (Q (x) (y)) d μ (y) | d μ (x)

4.0.4

= \int_{G ∖ G^{'}} | \sum_{s = σ_{1} (x)}^{σ_{2} (x)} \int K_{s} (x, y) f (y) e (Q (x) (y) - Q (x) (x)) d μ (y) | d μ (x)

4.0.5

= \int_{G ∖ G^{'}} | \sum_{p \in P} T_{p} f (x) | d μ (x) \leq \frac{2^{434 a^{3}}}{(q - 1)^{5}} μ (G)^{1 - \frac{1}{q}} μ (F)^{\frac{1}{q}} .

4.0.6

This proves 2.0.6 for the chosen set $G^{'}$ and arbitrary $f$ and thus completes the proof of Proposition 2.0.1.

4.1 Proof of Grid Existence Lemma

We begin with the construction of the centers of the dyadic cubes.

Lemma 4.1.1 counting balls

✓

Let $- S \leq k \leq S$ . Consider $Y \subset X$ such that for any $y \in Y$ , we have

y \in B (o, 4 D^{S} - D^{k}),

4.1.1

furthermore, for any $y^{'} \in Y$ with $y \neq y^{'}$ , we have

B (y, D^{k}) \cap B (y^{'}, D^{k}) = \emptyset .

4.1.2

Then the cardinality of $Y$ is bounded by

| Y | \leq 2^{3 a + 200 S a^{3}} .

4.1.3

Proof ▶

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

μ (B (y, 2^{j} D^{k})) \leq 2^{a j} μ (B (y, D^{k})) .

4.1.4

Since $X$ is the union of the balls $B (y, 2^{j} D^{k})$ and $μ$ 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 $j^{'} = \ln_{2} (8 D^{2 S}) = 3 + 2 S \cdot 100 a^{2}$ by 2.0.1, using $- S \leq k \leq S$ , $y \in B (o, 4 D^{S})$ , and the triangle inequality, we have

B (o, 4 D^{S}) \subset B (y, 8 D^{S}) \subset B (y, 2^{j^{'}} D^{k}) .

4.1.5

Using the disjointedness of the balls in 4.1.2, 4.1.1, and the triangle inequality for $ρ$ , we obtain

| Y | μ (B (o, 4 D^{S})) \leq 2^{j^{'} a} \sum_{y \in Y} μ (B (y, D^{k}))

4.1.6

\leq 2^{j^{'} a} μ (⋃_{y \in Y} B (y, D^{k})) \leq 2^{j^{'} a} μ (o, 4 D^{S}) .

4.1.7

As $μ (o, 4 D^{S})$ is not zero, the lemma follows.

For each $- S \leq k \leq S$ , let $Y_{k}$ be a set of maximal cardinality in $X$ such that $Y = Y_{k}$ satisfies the properties 4.1.1 and 4.1.2 and such that $o \in Y_{k}$ . By the upper bound of Lemma 4.1.1, such a set exists.

For each $- S \leq k \leq S$ , choose an enumeration of the points in the finite set $Y_{k}$ and thus a total order $<$ on $Y_{k}$ .

Lemma 4.1.2 cover big ball

✓

For each $- S \leq k \leq S$ , the ball $B (o, 4 D^{S} - D^{k})$ is contained in the union of the balls $B (y, 2 D^{k})$ with $y \in Y_{k}$ .

Proof ▶

Let $x$ be any point of $B (o, 4 D^{S} - D^{k})$ . 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, $x \in B (y, 2 D^{k})$ .

Define the set

C := {(y, k) : - S \leq k \leq S, y \in Y_{k}}

4.1.8

We totally order the set $C$ lexicographically by setting $(y, k) < (y^{'}, k^{'})$ if $k < k^{'}$ or both $k = k^{'}$ and $y < y^{'}$ . In what follows, we define recursively in the sense of this order a function

(I_{1}, I_{2}, I_{3}) : C \to P (X) \times P (X) \times P (X) .

4.1.9

Assume the sets $I_{j} (y^{'}, k^{'})$ have already been defined for $j = 1, 2, 3$ if $k^{'} < k$ and if $k = k^{'}$ and $y^{'} < y$ .

If $k = - S$ , define for $j \in {1, 2}$ the set $I_{j} (y, k)$ to be $B (y, j D^{- S})$ . If $- S < k$ , define for $j \in {1, 2}$ and $y \in Y_{k}$ the set $I_{j} (y, k)$ to be

⋃ {I_{3} (y^{'}, k - 1) : y^{'} \in Y_{k - 1} \cap B (y, j D^{k})} .

4.1.10

Define for $- S \leq k \leq S$ and $y \in Y_{k}$

I_{3} (y, k) := I_{1} (y, k) \cup [I_{2} (y, k) ∖ [X_{k} \cup ⋃ {I_{3} (y^{'}, k) : y^{'} \in Y_{k}, y^{'} < y})]]

4.1.11

with

X_{k} := ⋃ {I_{1} (y^{'}, k) : y^{'} \in Y_{k}} .

4.1.12

Lemma 4.1.3 basic grid structure

✓

For each $- S \leq k \leq S$ and $1 \leq j \leq 3$ the following holds.

If $j \neq 2$ and for some $x \in X$ and $y_{1}, y_{2} \in Y_{k}$ we have

x \in I_{j} (y_{1}, k) \cap I_{j} (y_{2}, k),

4.1.13

then $y_{1} = y_{2}$ .

If $j \neq 1$ , then

B (o, 4 D^{S} - 2 D^{k}) \subset ⋃_{y \in Y_{k}} I_{j} (y, k) .

4.1.14

We have for each $y \in Y_{k}$ ,

B (y, \frac{1}{2} D^{k}) \subset I_{3} (y, k) \subset B (y, 4 D^{k}) .

4.1.15

Proof ▶

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

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 $k > - S$ , 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 $z < y_{1}$ or $z < y_{2}$ . Hence, neither $y_{1} < y_{2}$ nor $y_{2} < y_{1}$ , 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 $k > - S$ . Let $x$ be a point of $B (o, 4 D^{S} - 2 D^{k})$ . 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 $x \in B (y^{'}, 4 D^{k - 1})$ . As $D > 4$ , by applying the triangle inequality with the points, $o$ , $x$ , and $y^{'}$ , we obtain that $y^{'} \in B (o, 4 D^{S} - D^{k})$ . By Lemma 4.1.2, $y^{'}$ is in $B (y, 2 D^{k})$ 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 $x \in B (o, 4 D^{S} - 2 D^{k})$ . 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 $x \notin X_{k}$ . 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 $x \in B (y, \frac{1}{2} D^{k})$ . 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 $k > - S$ . By the inductive statement 4.1.14 and $D > 4$ , 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 $x \in B (y^{'}, 4 D^{k - 1})$ . 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 $y^{'} \in Y_{k - 1} \cap B (y, 2 D^{k})$ with $x \in I_{3} (y^{'}, k - 1)$ . By induction, $x \in B (y^{'}, 4 D^{k - 1})$ . By the triangle inequality applied to the points $x, y^{'}, y$ and $D > 4$ , we conclude $x \in B (y, 4 D^{k})$ . This shows the second inclusion in 4.1.15 and completes the proof of the lemma.

Lemma 4.1.4 cover by cubes

✓

Let $- S \leq l \leq k \leq S$ and $y \in Y_{k}$ . We have

I_{3} (y, k) \subset ⋃_{y^{'} \in Y_{l}} I_{3} (y^{'}, l) .

4.1.16

Proof ▶

Let $- S \leq l \leq k \leq S$ 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 $k > l$ 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.

Lemma 4.1.5 dyadic property

✓

Let $- S \leq l \leq k \leq S$ and $y \in Y_{k}$ and $y^{'} \in Y_{l}$ with $I_{3} (y^{'}, l) \cap I_{3} (y, k) \neq \emptyset$ . Then

I_{3} (y^{'}, l) \subset I_{3} (y, k) .

4.1.17

Proof ▶

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 $l < k$ . 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

I_{3} (y^{″}, k - 1) \subset I_{3} (y, k) .

4.1.18

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 $u < y$ in $Y_{k}$ , we have $x \notin I_{3} (u, k)$ . Let $u < y$ . By transitivity of the order in $Y_{k}$ , we conclude $x \notin I_{2} (u, k)$ . 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 $u < y$ 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.

For $- S \leq k^{'} \leq k \leq S$ and $y^{'} \in Y_{k^{'}}$ , $y \in Y_{k}$ write $(y^{'}, k^{'} | y, k)$ if $I_{3} (y^{'}, k^{'}) \subset I_{3} (y, k)$ and

inf_{x \in X ∖ I_{3} (y, k)} ρ (y^{'}, x) < 6 D^{k^{'}} .

4.1.19

Lemma 4.1.6 transitive boundary

✓

Assume $- S \leq k^{″} < k^{'} < k \leq S$ and $y^{″} \in Y_{k^{″}}$ , $y^{'} \in Y_{k^{'}}$ , $y \in Y_{k}$ . Assume there is $x \in X$ such that

x \in I_{3} (y^{″}, k^{″}) \cap I_{3} (y^{'}, k^{'}) \cap I_{3} (y, k) .

4.1.20

If $(y^{″}, k^{″} | y, k)$ , the also $(y^{″}, k^{″} | y^{'}, k^{'})$ and $(y^{'}, k^{'} | y, k)$

Proof ▶

As $x \in I_{3} (y^{″}, k^{″}) \cap I_{3} (y^{'}, k^{'})$ and $k^{″} < k^{'}$ , 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 $x^{'} \in X ∖ I_{3} (y, k)$ such that

ρ (y^{″}, x^{'}) < 6 D^{k^{″}},

4.1.21

which exists as $(y^{″}, k^{″} | y, k)$ . As $x^{'} \in X ∖ I_{3} (y^{'}, k^{'})$ as well, we conclude $(y^{″}, k^{″} | y^{'}, k^{'})$ . By the triangle inequality, we have

ρ (y^{'}, x^{'}) \leq ρ (y^{'}, x) + ρ (x, y^{″}) + ρ (y^{″}, x^{'})

4.1.22

Using the choice of $x$ and 4.1.15 as well as 4.1.21, we estimate this by

< 4 D^{k^{'}} + 4 D^{k^{″}} + 6 D^{k^{″}} \leq 6 D^{k^{'}},

4.1.23

where we have used $D > 5$ and $k^{″} < k^{'}$ . We conclude $(y^{'}, k^{'} | y, k)$ .

Lemma 4.1.7 small boundary

✓

Let $K = 2^{4 a + 1}$ . For each $- S + K \leq k \leq S$ and $y \in Y_{k}$ we have

\sum_{z \in Y_{k - K} : (z, k - K | y, k)} μ (I_{3} (z, k - K)) \leq \frac{1}{2} μ (I_{3} (y, k)) .

4.1.24

Proof ▶

Let $K$ be as in the lemma. Let $- S + K \leq k \leq S$ and $y \in Y_{k}$ .

Pick $k^{'}$ so that $k - K \leq k^{'} \leq k$ . 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

I_{3} (y^{″}, k - K) \subset I_{3} (y^{'}, k^{'}) \subset I_{3} (y, k) .

4.1.25

Using Lemma 4.1.6, $(y^{'}, k^{'} | y, k)$ .

We conclude using the disjointedness property of Lemma 4.1.3 that

\sum_{y^{″} : (y^{″}, k - K | y, k)} μ (I_{3} (y^{″}, k - K)) \leq \sum_{y^{'} : (y^{'}, k^{'} | y, k)} μ (I_{3} (y^{'}, k^{'})) .

4.1.26

Adding over $k - K < k^{'} \leq k$ , and using

μ (I_{3} (y^{'}, k^{'})) \leq 2^{4 a} μ (B (y^{'}, \frac{1}{4} D^{k^{'}}))

from the doubling property 1.0.5 and 4.1.15 gives

K \sum_{y^{″} : (y^{″}, k - K | y, k)} μ (I_{3} (y^{″}, k - K))

4.1.27

\leq 2^{4 a} \sum_{k - K < k^{'} \leq k} [\sum_{y^{'} : (y^{'}, k^{'} | y, k)} μ (B (y^{'}, \frac{1}{4} D^{k^{'}}))]

4.1.28

Each ball $B (y^{'}, \frac{1}{4} D^{k^{'}})$ 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,

K \sum_{y^{″} : (y^{″}, k - K | y, k)} μ (I_{3} (y^{″}, k - K)) \leq 2^{4 a} μ (I_{3} (y, k))

4.1.29

which by $K = 2^{4 a + 1}$ 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 $B (u, \frac{1}{4} D^{l})$ and $B (u^{'}, \frac{1}{4} D^{l^{'}})$ 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 $l^{'} < l$ . Towards a contradiction, assume that

B (u, \frac{1}{4} D^{l}) \cap B (u^{'}, \frac{1}{4} D^{l^{'}}) \neq \emptyset

4.1.30

As $(u^{'}, l^{'} | y, k)$ , there is a point $x$ in $X ∖ I_{3} (y, k)$ with $ρ (x, u^{'}) < 6 D^{l^{'}}$ . Using $D > 25$ , we conclude from the triangle inequality and 4.1.30 that $x \in B (u, \frac{1}{2} D^{l})$ . However, $B (u, \frac{1}{2} D^{l}) \subset I_{3} (u, l)$ , and $I_{3} (u, l) \subset I_{3} (y, k)$ , a contradiction to $x \notin I_{3} (y, k)$ . This proves the lemma.

Lemma 4.1.8 smaller boundary

✓

Let $K = 2^{4 a + 1}$ and let $n \geq 0$ be an integer. Then for each $- S + n K \leq k \leq S$ we have

\sum_{y^{'} \in Y_{k - n K} : (y^{'}, k - n K | y, k)} μ (I_{3} (y^{'}, k - n K)) \leq 2^{- n} μ (I_{3} (y, k)) .

4.1.31

Proof ▶

We prove this by induction on $n$ . If $n = 0$ , both sides of 4.1.31 are equal to $μ (I_{3} (y, k))$ by 4.1.13. If $n = 1$ , this follows from Lemma 4.1.7.

Assume $n > 1$ and 4.1.31 has been proven for $n - 1$ . We write 4.1.31

\sum_{y^{″} \in Y_{k - n K} : (y^{″}, k - n K | y, k)} μ (I_{3} (y^{″}, k - n K))

4.1.32

= \sum_{y^{'} \in Y_{k - K} : (y^{'}, k - K | y, k)} [\sum_{y^{″} \in Y_{k - n K} : (y^{″}, k - n K | y^{'}, k - K)} μ (I_{3} (y^{″}, k - n K))]

4.1.33

Applying the induction hypothesis, this is bounded by

= \sum_{y^{'} \in Y_{k - K} : (y^{'}, k - K | y, k)} 2^{1 - n} μ (I_{3} (y^{'}, k - K))

4.1.34

Applying 4.1.24 gives 4.1.31, and proves the lemma.

Lemma 4.1.9 boundary measure

✓

For each $- S \leq k \leq S$ and $y \in Y_{k}$ and $0 < t < 1$ with $t D^{k} \geq D^{- S}$ we have

μ ({x \in I_{3} (y, k) : ρ (x, X ∖ I_{3} (y, k)) \leq t D^{k}}) \leq 2 t^{κ} μ (I_{3} (y, k)) .

4.1.35

Proof ▶

Let $x \in I_{3} (y, k)$ with $ρ (x, X ∖ I_{3} (y, k)) \leq t D^{k}$ . Let $K = 2^{4 a + 1}$ as in Lemma 4.1.8. Let $n$ be the largest integer such that $D^{n K} \leq \frac{1}{t}$ , so that $t D^{k} \leq D^{k - n K}$ and

D^{n K} > \frac{1}{t D^{K}} .

4.1.36

Let $k^{'} = k - n K$ , by the assumption $t D^{k} \geq D^{- S}$ , we have $k^{'} \geq - S$ . 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

ρ (y^{'}, X ∖ I_{3} (y, k)) \leq ρ (x, y^{'}) + ρ (x, X ∖ I_{3} (y, k)) \leq 4 D^{k^{'}} + t D^{k} .

By the assumption on $n$ and the definition of $k^{'}$ , this is

\leq 4 D^{k^{'}} + D^{k - n K} < 6 D^{k^{'}} .

Together with 4.1.17 thus $(y^{'}, k^{'} | y, k)$ . We have shown that

{x \in I_{3} (y, k) : ρ (x, X ∖ I_{3} (y, k)) \leq t D^{k}}

\subset ⋃_{y^{'} \in Y_{k - n K} : (y^{'}, k - n K | y, k)} I_{3} (y^{'}, k - n K) .

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

μ ({x \in I_{3} (y, k) : ρ (x, X ∖ I_{3} (y, k)) \leq t D^{k}}) \leq 2^{- n} μ (I_{3} (y, k)) .

By 4.1.36 and the definition 2.0.1 of $D$ , this is bounded by

2 t^{1 / (100 a^{2} K)} μ (I_{3} (y, k)),

which completes the proof by the definition 2.0.2 of $κ$ .

Let $\tilde{D}$ be the set of all $I_{3} (y, k)$ with $k \in [- S, S]$ and $y \in Y_{k}$ . Define

s (I_{3} (y, k)) := k

4.1.37

c (I_{3} (y, k)) := y .

4.1.38

We define $D$ to be the set of all $I \in \tilde{D}$ such that $I \subset I_{3} (o, S)$ .

Proof of Lemma 4.0.1 ▶

We first show that $(\tilde{D}, c, s)$ 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 $(\tilde{D}, c, s)$ and $x$ .

We first show 2.0.7 for $(\tilde{D}, c, s)$ 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 $- S \leq k < l$ ; in particular $- S < l$ . 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 $(\tilde{D}, c, s)$ . Assume to get a contradiction that there are non-disjoint $I, J \in \tilde{D}$ with $s (I) \leq s (J)$ and $I ⊄ J$ . 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 $I ⊄ J$ . 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 $s (J) > k$ . Choose $y \in I \cap J$ . By property 2.0.7, there is $K \in \tilde{D}$ 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 $(D, c, s)$ by restriction. 2.0.9 is true for $(D, c, s)$ by definition, and 2.0.7 follows from 2.0.7 and 2.0.8 for $(\tilde{D}, c, s)$ .

4.2 Proof of Tile Structure Lemma

Choose a grid structure $(D, c, s)$ with Lemma 4.0.1 Let $I \in D$ . Suppose that

Z \subset Q (X)

4.2.1

is such that for any $ϑ, θ \in Z$ with $ϑ \neq θ$ we have

B_{I^{\circ}} (ϑ, 0.3) \cap B_{I^{\circ}} (θ, 0.3) \cap Q (X) = \emptyset .

4.2.2

Since $Q (X)$ is finite, there exists a set $Z$ satisfying both 4.2.1 and 4.2.2 of maximal cardinality among all such sets. We pick for each $I \in D$ such a set $Z (I)$ .

Lemma 4.2.1 frequency ball cover

✓

For each $I \in D$ , we have

Q (X) \subset ⋃_{z \in Z (I)} B_{I^{\circ}} (z, 0.7) .

4.2.3

Proof ▶

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

d_{I^{\circ}} (z, θ) \leq d_{I^{\circ}} (z, z_{1}) + d_{I^{\circ}} (θ, z_{1}) < 0.3 + 0.3 = 0.6,

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

We define

P = {(I, z) : I \in D, z \in Z (I)},

I ((I, z)) = I and Q ((I, z)) = z .

We further set

s (p) = s (I (p)), c (p) = c (I (p)) .

Then 2.0.18, 2.0.19 hold by definition.

It remains to construct the map $Ω$ , and verify properties 2.0.13, 2.0.14 and 2.0.15. We first construct an auxiliary map $Ω_{1}$ . For each $I \in D$ , we pick an enumeration of the finite set $Z (I)$

Z (I) = {z_{1}, \dots, z_{M}} .

We define $Ω_{1} : P \mapsto P (Θ)$ as below. Set

Ω_{1} ((I, z_{1})) = B_{I^{\circ}} (z_{1}, 0.7) ∖ ⋃_{z \in Z (I) ∖ {z_{1}}} B_{I^{\circ}} (z, 0.3)

and then define iteratively

Ω_{1} ((I, z_{k})) = B_{I^{\circ}} (z_{k}, 0.7) ∖ ⋃_{z \in Z (I) ∖ {z_{k}}} B_{I^{\circ}} (z, 0.3) ∖ ⋃_{i = 1}^{k - 1} Ω_{1} ((I, z_{i})) .

4.2.4

Lemma 4.2.2 disjoint frequency cubes

✓

For each $I \in D$ , and $p_{1}, p_{2} \in P (I)$ , if

Ω_{1} (p_{1}) \cap Ω_{1} (p_{2}) \neq \emptyset,

then $p_{1} = p_{2}$ .

Proof ▶

By the definition of the map $I$ , we have

P (I) = {(I, z) : z \in Z (I)} .

By 4.2.4, the set $Ω_{1} ((I, z_{k}))$ is disjoint from each $Ω_{1} ((I, z_{i}))$ with $i < k$ . Thus the sets $Ω_{1} (p)$ , $p \in P (I)$ are pairwise disjoint.

Lemma 4.2.3 frequency cube cover

✓

For each $I \in D$ , it holds that

⋃_{z \in Z (I)} B_{I^{\circ}} (z, 0.7) \subset ⋃_{p \in P (I)} Ω_{1} (p) .

4.2.5

For every $p \in P$ , it holds that

B_{p} (Q (p), 0.3) \subset Ω_{1} (p) \subset B_{p} (Q (p), 0.7) .

4.2.6

Proof ▶

For 4.2.6 let $p = (I, z)$ . The second inclusion in 4.2.6 then follows from 4.2.4 and the equality $B_{p} (Q (p), 0.7) = B_{I^{\circ}} (z, 0.7)$ , which is true by definition. For the first inclusion in 4.2.6 let $ϑ \in B_{p} (Q (p), 0.3)$ . 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 $ϑ \notin Ω_{1} ((I, z_{i}))$ for all $i < k$ . Thus, again from 4.2.4, we have $ϑ \in Ω_{1} ((I, z_{k}))$ .

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

z \in Ω_{1} ((I, z)) \subset ⋃_{p \in P (I)} Ω_{1} (p)

by the first inclusion in 4.2.6.

Now suppose that there exists no $z \in Z (I)$ with $ϑ \in B_{I^{\circ}} (z, 0.3)$ . Let $k$ be minimal such that $ϑ \in B_{I^{\circ}} (z_{k}, 0.7)$ . Since $Ω_{1} ((I, z_{i})) \subset B_{I^{\circ}} (z_{i}, 0.7)$ for each $i$ by 4.2.4, we have that $ϑ \notin Ω_{1} ((I, z_{i}))$ for all $i < k$ . Hence $ϑ \in Ω_{1} ((I, z_{k}))$ , again by 4.2.4.

Now we are ready to define the function $Ω$ . We define for all $p \in P (I_{0})$

Ω (p) = Ω_{1} (p) .

4.2.7

For all other cubes $I \in D$ , $I \neq I_{0}$ , there exists, by 2.0.8 and 2.0.9, $J \in D$ with $I \subset J$ and $I \neq J$ . On such $I$ we define $Ω$ by recursion. We can pick an inclusion minimal $J \in D$ among the finitely many cubes such that $I \subset J$ and $I \neq J$ . This $J$ is unique: Suppose that $J^{'}$ is another inclusion minimal cube with $I \subset J^{'}$ and $I \neq J^{'}$ . Without loss of generality, we have that $s (J) \leq s (J^{'})$ . By 2.0.8, it follows that $J \subset J^{'}$ . Since $J^{'}$ is minimal with respect to inclusion, it follows that $J = J^{'}$ . Then we define

Ω (p) = ⋃_{z \in Z (J) \cap Ω_{1} (p)} Ω ((J, z)) \cup B_{p} (Q (p), 0.2) .

4.2.8

We now verify that $(P, I, Ω, Q, c, s)$ forms a tile structure.

Proof of Lemma 4.0.2 ▶

First, we prove 2.0.15. If $I = I_{0}$ , then 2.0.15 holds for all $p \in P (I)$ by 4.2.7 and 4.2.6. Now suppose that $I$ is not maximal in $D$ with respect to set inclusion. Then we may assume by induction that for all $J \in D$ with $I \subset J$ and all $p^{'} \in P (J)$ , 2.0.15 holds. Let $J$ be the unique minimal cube in $D$ with $I ⊊ J$ .

Suppose that $ϑ \in Ω (p)$ . If $ϑ \in B_{p} (Q (p), 0.2)$ , then since

B_{p} (Q (p), 0.2) \subset B_{p} (Q (p), 1),

we conclude that $ϑ \in B_{p} (Q (p), 0.7)$ . If not, by 4.2.8, there exists $z \in Z (J) \cap Ω_{1} (p)$ with $ϑ \in Ω (J, z)$ . Using the triangle inequality and 4.2.6, we obtain

d_{I^{\circ}} (Q (p), ϑ) \leq d_{I^{\circ}} (Q (p), z) + d_{I^{\circ}} (z, ϑ) \leq 0.7 + d_{I^{\circ}} (z, ϑ) .

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

\leq 0.7 + 2^{- 95 a} d_{J^{\circ}} (z, ϑ) \leq 0.7 + 2^{- 95 a} \cdot 1 < 1 .

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 $I \in D$ .

If $I = I_{0}$ , then disjointedness of the sets $Ω (p)$ for $p \in P (I)$ 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 $J \in D$ with $I \subset J$ and $I \neq J$ . In this case we use induction: It suffices to show 2.0.13 under the assumption that it holds for all cubes $J \in D$ 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 $Ω (p), p \in P (I)$ we pick two tiles $p, p^{'} \in P (I)$ and $ϑ \in Ω (p) \cap Ω (p^{'})$ . Then we are by 4.2.8 in one of the following four cases.

1. There exist $z \in Z (J) \cap Ω_{1} (p)$ such that $ϑ \in Ω (J, z)$ , and there exists $z^{'} \in Z (J) \cap Ω_{1} (p^{'})$ such that $ϑ \in Ω (J, z^{'})$ . 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 $p = p^{'}$ .

2. There exists $z \in Z (J) \cap Ω_{1} (p)$ such that $ϑ \in Ω (J, z)$ , and $ϑ \in B_{p^{'}} (Q (p^{'}), 0.2)$ . Using the triangle inequality, Lemma 2.1.2 and 2.0.15, we obtain

d_{p^{'}} (Q (p^{'}), z) \leq d_{p^{'}} (Q (p^{'}), ϑ) + d_{p^{'}} (z, ϑ) \leq 0.2 + 2^{- 95 a} \cdot 1 < 0.3 .

Thus $z \in Ω_{1} (p^{'})$ by 4.2.6. By Lemma 4.2.2, it follows that $p = p^{'}$ .

3. There exists $z^{'} \in Z (J) \cap Ω_{1} (p^{'})$ such that $ϑ \in Ω (J, z^{'})$ , and $ϑ \in B_{p} (Q (p), 0.2)$ . This case is the same as case 2., after swapping $p$ and $p^{'}$ .

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

To show the inclusion in 2.0.13, let $ϑ \in Q (X)$ . By the induction hypothesis, there exists $p \in P (J)$ such that $ϑ \in Ω (p)$ . By definition of the set $P$ , we have $p = (J, z)$ for some $z \in Z (J)$ . Thus, by 4.2.3, there exists $z^{'} \in Z (I)$ with $z \in B_{I^{\circ}} (z^{'}, 0.7)$ . Then by Lemma 4.2.3 there exists $p^{'} \in P (I)$ with $z \in Z (J) \cap Ω_{1} (p^{'})$ . Consequently, by 4.2.8, $ϑ \in Ω (p^{'})$ . This completes the proof of 2.0.13.

Finally, we show 2.0.14. Let $p, q \in P$ with $I (p) \subset I (q)$ and $Ω (p) \cap Ω (q) \neq \emptyset$ . If we have $s (p) \geq s (q)$ , then it follows from 2.0.8 that $I = J$ , thus $p, q \in P (I)$ . By 2.0.13 we have then either $Ω (p) \cap Ω (q) = \emptyset$ or $Ω (p) = Ω (q)$ . By the assumption in 2.0.14 we have $Ω (p) \cap Ω (q) \neq \emptyset$ , so we must have $Ω (p) = Ω (q)$ and in particular $Ω (q) \subset Ω (p)$ .

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

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