Recall that our goal is to prove that the family of all non-BAUP cellsP∈Dis Carleson. In view of the result of the previous section, it will suffice to show that we can chooseA, α>0 such that for every fixed linear hyperplaneH and for every integerN, the corresponding familyF=F(A, α, H, N) of all non-BAUP cellsP∈D containing an (H, A, α)-flat cellQ at mostN levels down from P is Carleson. The result of this section can be stated as follows.
Lemma 8. If F is not Carleson, then for every positive integer K and every η >0, there exist a cell P∈F and K+1 alternating pairs of finite layers Pk,Qk⊂D (k=
0, ..., K)such that
• P0={P}.
• Pk⊂FP for all k=0, ..., K.
• All layers Qk consist of (H, A, α)-flat cells only.
• Each individual layer (either Pk or Qk)consists of pairwise disjoint cells.
• If Q∈Qk,then there exists P0∈Pk such that Q⊂P0 (k=0, ..., K).
• If P0∈Pk+1,then there exists Q∈Qk such that P0⊂Q (k=0, ..., K−1).
• P
Q∈QKµ(Q)>(1−η)µ(P).
In other words, each layer tiles P up to a set of negligible measure and they have the usual Cantor-type hierarchy (due to this hierarchy, it suffices to look only at the very bottom layer to evaluate the efficiency of the tiling for all of them). The construction in this section is rather universal and does not depend on the meaning of the words
“non-BAUP” in any way. All that we need to know here is that some cells are BAUP and some are not. Note that we do not exclude here the possibility that the same cell is used in several different layers. This will never really happen because the non-BAUPness
condition is, in fact, just a particular quantitative negation of the flatness condition, so, when finally choosing our parameters, we will ensure that no non-BAUP cell can be an (H, A, α)-flat cell as well, thus guaranteeing that we always go down when moving from each layer to the next. Also our construction will be done in such a way that no two different P layers can contain the same cell. However, the disjointness of layers is not a part of the formal statement we have just made and the results of this and the next sections remain perfectly valid even if all layers we construct here consist of the single starting cellP, which, in that case, must be simultaneously non-BAUP and (H, A, α)-flat.
Proof. SupposeF is not Carleson. By Lemma7, for everyη0>0 and every positive integer M, we can find a cellP∈F andM+1 non-Carleson layers L0, ...,LM⊂FP that have the desired Cantor-type hierarchy and satisfyP
P0∈LMµ(P0)>(1−η0)µ(P) (see§13 for details).
We shall start with describing the main step of the construction, which will allow us to go from each layerPkto the next layerPk+1creating the intermediate layerQk on the way. Letmbe much smaller thanM, so that there are as many available non-Carleson layers down frommas we may possibly need. Fix a large integerS >0.
LetL0m⊂Lm. We shall call a cellP00∈Lm+sN (s=1, ..., S) exceptional if it is con-tained in some cellP0∈L0mbut there is no (H, A, α)-flat cellQ∈Dsuch thatP00⊂Q⊂P0. We claim that for eachs=1, ..., S, the sum ofµ-measures of all exceptional cells inLm+sN
does not exceed (1−c16−N d)sµ(P).
The proof goes by induction ons. To prove the bases=1, just recall that every cell P0∈L0m⊂Lmcontains some (H, A, α)-flat cellQ(P0)∈D at mostN levels down fromP0. Since every cellP00∈Lm+N that is contained inP0∈L0m must be at leastN levels down from P0 (the non-Carleson layers constructed in §13 cannot have repeating cells), we conclude that every cell P00∈Lm+N contained in P0 is either contained in Q(P0) or disjoint to Q(P0). In the first case P00 is, certainly, not exceptional, so the sum of the measures of all exceptional cells inLm+N that are contained inP0 is at most
µ(P0)−µ(Q(P0))6(1−c16−N d)µ(P0),
whence the total sum of measures of all exceptional cells inLm+N is at most (1−c16−N d) X
P0∈L0m
µ(P0)6(1−c16−N d)µ(P).
To make the induction step, assume that we already know that the claim holds for some s. Note that every exceptional cell P00∈Lm+(s+1)N is contained in some cell Pe00∈Lm+sN. We claim that Pe00 must be exceptional as well. Indeed, let P0 be the cell
inL0mcontainingP00. ThenPe00∩P06=∅, which, due to the hierarchy of the non-Carleson layers, is possible only ifPe00⊂P0. If there had been any (H, A, α)-flat cell Qsatisfying Pe00⊂Q⊂P0, we would also haveP00⊂Q⊂P0, so the cellP00would not be exceptional. Now it remains to note thatP00 must also be disjoint toQ(Pe00) and to repeat the argument above to conclude that the sum of measures of all exceptional cells inLm+(s+1)N is at most (1−c16−N d) times the sum of measures of all exceptional cells inLm+sN. It remains to apply the induction assumption and to combine two factors into one.
Now let L0m+SN⊂Lm+SN be the set of all cells in Lm+SN that are contained in some cell fromL0m but are not exceptional. Then, for every cell P00∈L0m+SN and the corresponding cellP0∈L0m containing P00, there exists an (H, A, α)-flat cell Q∈D such thatP00⊂Q⊂P0. LetQ be the set of all cellsQthat can arise in this way and letQ∗be the set of all maximal cells inQ(i.e., cells that are not contained in any larger cell from Q). Then the cellsQ∈Q∗ are pairwise disjoint and form an intermediate layer between L0m and L0m+SN in the sense that every Q∈Q∗ is contained in some cell P0∈L0m and everyP00∈L0m+SN is contained in some cellQ∈Q∗.
Moreover, X
P00∈L0m+SN
µ(P00)> X
P00∈Lm+SN
− X
P00∈Lm+SN
P006⊂P0for anyP0∈L0m
− X
P00∈Lm+SN
P00is exceptional
>(1−η0)µ(P)−
µ(P)− X
P0∈L0m
µ(P0)
−(1−c16−N d)Sµ(P)
= X
P0∈L0m
µ(P0)−[η0+(1−c16−N d)S]µ(P).
Now assume that M >(K+1)SN. Then we can start with L00=L0={P} and apply this construction inductively withm=0, SN,2SN, ..., KSN. The resulting layersL0kSN (k=0, ..., K) will satisfy all properties ofPk and the intermediate layersQ∗(one of those will arise during each step) will satisfy all properties ofQk except, perhaps, the measure estimate.
However, sinceL00coversP completely and during each step the total measure loss is bounded by [η0+(1−c16−N d)S]µ(P), we have
X
Q∈QK
µ(Q)> X
P0∈L0(K+1)SN
µ(P0)>µ(P)−(K+1)[η0+(1−c16−N d)S]µ(P) and it remains to note that for any fixedK, we can always make
(K+1)[η0+(1−c16−N d)S] less thanη if we chooseη0 small enough andS large enough.