Carleson density selection

Write $d(t)={\int }_{E\cap \{N\in {\omega }_t\}}{\chi }_{I_t}$. Consider all tiles t dominating some s in S and satisfying $d(t)>\delta \mathrm{/}2$. They form a finite nonempty set: $d(t)\le m(E)\mathrm{/}\mathrm{\mid }{I}_t\mathrm{\mid }$ implies $\mathrm{\mid }{I}_t\mathrm{\mid }<2m(E)\mathrm{/}\delta$, while domination implies $\mathrm{\mid }{I}_t\mathrm{\mid }\ge {\min }_{s\in S}\mathrm{\mid }{I}_s\mathrm{\mid }>0$. Only finitely many dyadic scales lie in this range; at each scale there is one spatial ancestor of each fixed s and finitely many frequency subintervals of its frequency interval. Nonemptiness follows from the definition of positive supremum delta. Let Tops be the maximal members of this finite set. Assign each s lying below a member of Tops to one such top using a fixed finite ordering. This gives disjoint trees. Every s left over has density at most $\delta \mathrm{/}2$: otherwise a witness t with $d(t)>\delta \mathrm{/}2$ would dominate s and extend to a maximal candidate.

For a top t let $I=I_t$, $\omega ={\omega }_t$, and let $2^{k}I$ be the interval with the same center and length $2^k\mathrm{\mid }I\mathrm{\mid }$. There is an integer $k\ge 0$ with $m(E\cap \{N\in \omega \}\cap 2^{k}I)\ge c\delta 2^{2k}\mathrm{\mid }I\mathrm{\mid }$, where c is an absolute positive constant fixed below. Indeed, partition the line into I and the shells $2^{k}I\setminus 2^{k-1}I$ for k>=1. On the kth shell ${\chi }_I\le A\cdot 2^{-20k}\mathrm{/}\mathrm{\mid }I\mathrm{\mid }$ for $A=2^{40}$, and the same bound with k=0 dominates its value on I. If every proposed mass bound failed, then $d(t)\le Ac\delta {\sum }_{k\ge 0}{2}^{-18k}$. Taking $c=(8A{)}^{-1}$ makes this less than $\delta \mathrm{/}2$, contradicting the choice of t. Assign the least successful k to t, and call its assigned family ${\mathcal{T}}_k$.

Fix k. Greedily choose from ${\mathcal{T}}_k$ a tile with greatest spatial length, retain it, and remove every remaining tile whose enlarged rectangle $2^k{I}_s\times {\omega }_s$ intersects its enlarged rectangle. Continue until none remain; this is a finite procedure. Retained enlarged rectangles are disjoint, so their sets $E\cap \{N\in {\omega }_t\}\cap 2^k{I}_t$ are disjoint. Step 2.1 gives ${\sum }_{t\text{ retained}}\mathrm{\mid }{I}_t\mathrm{\mid }\le (c\delta 2^{2k}{)}^{-1}m(E)$.

A tile s removed by a retained t has $\mathrm{\mid }{I}_s\mathrm{\mid }\le \mathrm{\mid }{I}_t\mathrm{\mid }$ and overlapping frequency intervals; dyadic nesting and reciprocal lengths imply ${\omega }_t\subseteq {\omega }_s$. All tiles removed by this same t therefore have mutually overlapping, hence nested, frequency intervals. Since Tops is an antichain, their original spatial intervals must be pairwise disjoint: otherwise dyadic nesting of both coordinates would make two tops comparable. Also overlap of the enlarged spatial intervals implies each original $I_s$ lies in the interval centered at $c(I_t)$ of radius $(3\mathrm{/}2)2^k\mathrm{\mid }{I}_t\mathrm{\mid }$. Thus the sum of their lengths is at most $3\cdot 2^k\mathrm{\mid }{I}_t\mathrm{\mid }$. Including t itself in its removed family, and using step 3.1, yields ${\sum }_{s\in {\mathcal{T}}_k}\mathrm{\mid }{I}_s\mathrm{\mid }\le (3\mathrm{/}c)2^{-k}{\delta }^{-1}m(E)$. Summing over k gives ${\sum }_{t\in \mathrm{T}\mathrm{o}\mathrm{p}\mathrm{s}}\mathrm{\mid }{I}_t\mathrm{\mid }\le (6\mathrm{/}c){\delta }^{-1}m(E)$. Together with step 1.1 this proves the partition and count assertion. Empty assigned trees may be deleted, only reducing the count. AC is inherited; the selections here are finite greedy choices and least integers.