3 Proof of Metric Space Carleson

This section is currently under construction, to incorporate the proof of Theorem 1.0.3

Let Borel sets $F$, $G$ in $X$ be given. Let a Borel function $f:X\to \mathbb{C}$ with $f\le {\mathbf{1}}_F$ be given. We have that

because every point of $X$ has finite distance from $o$.

We first show how Lemma 3.0.1 implies Theorem 1.0.2. As $R$ tends to $\mathrm{\infty }$, the integrand of the left-hand side of 3.0.2 grows monotonically toward the integrand of the left-hand side of 1.0.19 for all $x$. By Lebesgue’s monotone convergence theorem, the left-hand side of 3.0.2 converges to the left-hand side of 1.0.19. This verifies Theorem 1.0.2.

It remains to prove Lemma 3.0.1. Fix an integer $R>0$. By replacing $G$ with $G\cap B(o,R)$ if necessary, it suffices to show 3.0.2 under the assumption that $G$ is contained in $B(o,R)$. We make this assumption. For every $x\in G$, the domain of integration in 3.0.3 is contained in $B(o,2R)$. By replacing $F$ with $F\cap B(o,2R)$ if necessary, and correspondingly restricting $f$ to $B(o,2R)$, it suffices to show 3.0.2 under the assumption that $F$ is contained in $B(o,2R)$. We make this assumption.

Using the definition 2.0.5 of $K_s$ and the partition of unity 2.0.4, we express 3.0.3 as the sum of

and

where $s_1$ is the smallest integer such that $D^{s_1-2}{R}_2>\frac{1}{4D}$ and $s_2$ is the largest integer such that $D^{s_2+2}{R}_1<\frac{1}{2}$. We restrict the summation index $s$ by excluding summands with $s<s_1-2$ or $s>s_2+2$ because for these summands, the function $K_s$ vanishes on the domain of integration. We also omit the restriction in the integral for the summands in 3.0.4 because in these summands, the support of $K_s$ is contained in the set described by this restriction.

We apply the triangle inequality and estimate the versions of 3.0.2 separately with $T_{R_1,R_2,\vartheta }$ replaced by 3.0.4 and by each summand of 3.0.5. To handle the case 3.0.4, we employ the following lemma. Here, we utilize the fact that if $\frac{1}{R}\le R_1\le R_2\le R$, then $s_1$ and $s_2$ as in 3.0.4 are in an interval $[-S,S]$ for some sufficiently large $S$ depending on $R$.

To reduce Lemma 3.0.1 to Lemma 3.0.2, we need estimates for the summands in 3.0.5. Using Lemma 2.1.3, we obtain for arbitrary $s$ the inequality

where $M{\mathbf{1}}_F$ is as defined in Proposition 2.0.6. Now, the left-hand side of 3.0.2, with $T_{R_1,R_2,\vartheta }$ replaced by a summand of 3.0.5, can be estimated using Hölder’s inequality and Proposition 2.0.6 by

Applying the triangle inequality to estimate the left-hand side of 3.0.2 by contributions from the summands in 3.0.4 and 3.0.5, using Lemma 3.0.2 to control the first term, and the above to estimate the contribution from the four summands in 3.0.5, combined with $a\ge 4$ and $q<2$, completes the reduction of Lemma 3.0.1 to Lemma 3.0.2.

It remains to prove Lemma 3.0.2. Fix $S>0$.

We reduce Lemma 3.0.2 to Lemma 3.0.3. By the Lebesgue monotone convergence theorem, applied to an increasing sequence of finite sets $\tilde{\mathrm{\Theta }}$, inequality 3.0.9 continues to hold for countable $\tilde{\mathrm{\Theta }}$.

Let $ϵ=\frac{1}{2S+1}$. Pick some ${\vartheta }_0\in \mathrm{\Theta }$. For $k\ge 0$, let the set ${\tilde{\mathrm{\Theta }}}_k$ be a subset of $B_{B(o,2R)}({\vartheta }_0,k)$ of maximal size, such that for all $\vartheta ,\theta \in {\tilde{\mathrm{\Theta }}}_k$, it holds that $d_{B(o,2R)}(\vartheta ,\theta )\ge ϵ$. Such a set exists, since by Lemma 2.1.1 there exists an upper bound for the size of such subsets in $B_{B(o,2R)}({\vartheta }_0,k)$. Define

Then the set $\tilde{\mathrm{\Theta }}$ is at most countable, and it has the property that for any $\theta \in \mathrm{\Theta }$, there exists $\vartheta \in \tilde{\mathrm{\Theta }}$ with

For every $\vartheta \in \mathrm{\Theta }$, we have

Moreover, there is a $\tilde{\vartheta }\in \tilde{\mathrm{\Theta }}$ with $d_{B(o,2R)}(\vartheta ,\tilde{\vartheta })\le ϵ$. Hence,

Using Lemma 2.1.3, we can estimate the above expression by

We estimate the left-hand-side of 3.0.6 by the sum of left-hand-side of 3.0.9 and

which, as we have just shown, is estimated by

By Hölder’s inequality and Proposition 2.0.6 (more precisely, 2.0.46 with $p=q$), the above is no greater than

Combining this with Lemma 3.0.3 and the fact that

proves Lemma 3.0.2.

It remains to prove Lemma 3.0.3. Fix a finite set $\tilde{\mathrm{\Theta }}$.

We reduce Lemma 3.0.3 to Lemma 3.0.4. For each $x$, let ${\sigma }_1(x)$ be the minimal element $s^{\prime }\in [-S,S]$ such that

Similarly, let ${\sigma }_2(x)$ be the minimal element $s^{″}\in [-S,S]$ such that

Finally, choose a total order of the finite set $\tilde{\mathrm{\Theta }}$ and let $Q(x)$ be the minimal element $\vartheta$ with respect to this order such that

With these choices, and noting that

we conclude that the left-hand side of 3.0.9 and 3.0.11 are equal. Thus, Lemma 3.0.3 follows from Lemma 3.0.4.

It remains to prove Lemma 3.0.4. Fix ${\sigma }_1$, ${\sigma }_2$, and $Q$ as in the lemma. Applying Proposition 2.0.1 recursively, we obtain a sequence of sets $G_n$ with $G_0=G$ and, for each $n\ge 0$, $\mu (G_n)\le 2^{-n}\mu (G)$ and

Adding the first $n$ of these inequalities, we obtain by bounding a geometric series

As the integrand is non-negative and non-decreasing in $n$, we obtain by the monotone convergence theorem

This completes the proof of Lemma 3.0.4 and thus Theorem 1.0.2.