A d -dimensional Analyst's Travelling Salesman Theorem for general sets in R n
AAN ANALYST’S TRAVELLING SALESMAN THEOREM FORGENERAL SETS IN R n MATTHEW HYDE
Abstract.
In his 1990 paper, Jones proved the following: given E ⊆ R ,there exists a curve Γ such that E ⊆ Γ and H (Γ) ∼ diam E + (cid:88) Q β E (3 Q ) (cid:96) ( Q ) . Here, β E ( Q ) measures how far E deviates from a straight line inside Q . Thiswas extended by Okikiolu to subsets of R n and by Schul to subsets of a Hilbertspace.In 2018, Azzam and Schul introduced a variant of the Jones β -number.With this, they, and separately Villa, proved similar results for lower regularsubsets of R n . In particular, Villa proved that, given E ⊆ R n which is lowercontent regular, there exists a ‘nice’ d -dimensional surface F such that E ⊆ F and H d ( F ) ∼ diam( E ) d + (cid:88) Q β E (3 Q ) (cid:96) ( Q ) d . In this context, a set F is ‘nice’ if it satisfies a certain topological non degen-eracy condition, first introduced in a 2004 paper of David.In this paper we drop the lower regularity condition and prove an analogousresult for general d -dimensional subsets of R n . To do this, we introduce a new d -dimensional variant of the Jones β -number that is defined for any set in R n . Contents
1. Introduction 21.1. Background 21.2. Main Results 61.3. Acknowledgment 82. Preliminaries 82.1. Notation 82.2. Chirst-David cubes 82.3. Theorem of David and Toro 92.4. Hausdorff-type content 112.5. Preliminaries with β -numbers 163. Proof of Theorem 1.13 294. Proof of Theorem 1.14 37 Mathematics Subject Classification.
Key words and phrases.
Rectifiability, Travelling salesman theorem, beta numbers, Hausdorffcontent.M. Hyde was supported by The Maxwell Institute Graduate School in Analysis and its Ap-plications, a Centre for Doctoral Training funded by the UK Engineering and Physical SciencesResearch Council (grant EP/L016508/01), the Scottish Funding Council, Heriot-Watt Universityand the University of Edinburgh. a r X i v : . [ m a t h . C A ] J un MATTHEW HYDE F and its properties 374.2. Proof of (1.5) 44References 521. Introduction
Background.
The 1-dimensional Analyst’s Travelling Salesman Theorem wasfirst proven by Peter Jones [Jon90] for subsets of C , with the motivation of studyingthe boundedness of a certain class of singular integral operators. Roughly speaking,he proved that if E is flat enough at most scales and locations, then there is acurve Γ of finite length (with quantitative control on its length) such that E ⊆ Γ . Conversely, he proved that if E is a curve of finite length, then there is quantitativecontrol over how often E can be non-flat. The Jones β -number is the quantity heintroduced to measure flatness.Define for E, B ⊆ R n , β dE, ∞ ( B ) = 1 r B inf L sup { dist( y, L ) : y ∈ E ∩ B } where L ranges over d -planes in R n . Thus β dE, ∞ ( B ) r B is the width of the smallesttube containing E ∩ B. Jones’ result was then extended by Okikiolu [Oki92] to subsets of R n and furtherby Schul [Sch07b] to subsets of Hilbert spaces. We state the version for Euclideanspaces below. Theorem 1.1 (Jones: R [Jon90]; Okikiolu: R n [Oki92]; Schul [Sch07b] ) . Let n ≥ . There is a C = C ( n ) such that the following holds. Let E ⊆ R n . Then thereis a connected set Γ ⊇ E such that H (Γ) (cid:46) n diam E + (cid:88) Q ∈ ∆ Q ∩ E (cid:54) = ∅ β E, ∞ (3 Q ) (cid:96) ( Q ) , (1.1) where ∆ denotes the collection of all dyadic cubes in R n . Conversely, if Γ is con-nected and H (Γ) < ∞ , then diam Γ + (cid:88) Q ∈ ∆ Q ∩ Γ (cid:54) = ∅ β , ∞ (3 Q ) (cid:96) ( Q ) (cid:46) n H (Γ) . As a corollary of Theorem 1.1, if E ⊆ R n and the right hand side of (1.1) isfinite, then there exists a curve Γ ⊇ E such that H d (Γ) ∼ diam E + (cid:88) Q ∈ ∆ Q ∩ E (cid:54) = ∅ β E, ∞ (3 Q ) (cid:96) ( Q ) . Pajot [Paj96] proved an analogous result to the first half of Theorem 1.1 for 2-dimensional sets in R n . In this, he gave a sufficient condition in terms of the Jones β -number for when a set E ⊆ R n can be contained in a surface f ( R ) for somesmooth f : R → R n . It is natural to ask whether a d -dimensional analogue of theabove Jones’ theorem is true. That is: SP FOR GENERAL SETS 3 (1) Given a set E, can we find a ‘nice’ set F containing E such that H d ( F ) (cid:46) diam( E ) d + (cid:88) Q ∈ ∆ Q ∩ Γ (cid:54) = ∅ β dE, ∞ (3 Q ) (cid:96) ( Q ) d ?(2) Given that E is a ‘nice’ set, can we say thatdiam( E ) d + (cid:88) Q ∈ ∆ Q ∩ Γ (cid:54) = ∅ β dE, ∞ (3 Q ) (cid:96) ( Q ) d (cid:46) H d ( E )?The most natural candidate for a ‘nice’ set is a Lipschitz graph. However, in hisPhD thesis, Fang [Fan90] constructed a 3-dimensional Lipschitz graph whose β , ∞ sum was infinite. Thus, with the β -numbers as defined by Jones, a d -dimensionalanalogue of the second half of Theorem 1.1 was proven to be false. This issue wasresolved by David and Semmes [DS91] who introduced a new β -number and proveda Travelling Salesman Theorem for Ahlfors d -regular sets in R n . A set E ⊆ R n issaid to be Ahlfors d - regular if there is A > r d /A ≤ H d ( E ∩ B ( x, r )) ≤ Ar d for all x ∈ E, r ∈ (0 , diam E ) . They defined their β -number as follows. For E ⊆ R n and B a ball, setˆ β d,pE ( B ) = inf L (cid:32) r dB ˆ B (cid:18) dist( y, L ) r B (cid:19) p d H d | E ( y ) (cid:33) p , where L ranges over all d -planes in R n . Thus, ˆ β d,pE ( B ) measures the L p -averagedeviation of E from a plane. Let p ( d ) := (cid:40) dd − , d < ∞ , d ≤ . Their result goes as follows:
Theorem 1.2 (David, Semmes [DS91]) . Let E ⊆ R n be Ahlfors d -regular. Thefollowing are equivalent: (1) The set E has big pieces of Lipschitz images, meaning, there are constant L, c > such that for all x ∈ E and r ∈ (0 , diam E ) , there is an L -Lipschitzmap f : R d → R n satisfying H d ( f ( R d ) ∩ B ( x, r )) ≥ cr d . (2) For ≤ p < p ( d ) , ˆ β d,pE ( x, r ) dxdrr is a Carleson measure on E × (0 , ∞ ) . Recall that σ is a Carleson measure on E ∩ (0 , ∞ ) if σ ( B ( x, r ) × (0 , r )) (cid:46) r d . Remark 1.3.
An Ahlfors d -regular sets satisfying condition (1) of the above theo-rem is said to be uniformly rectifiable (UR). This is one of many characterizationsof UR sets.More recently Azzam and Schul [AS18] proved a Traveling Salesman Theoremfor d -dimensional sets in R n , under the weakened assumption of ( c, d )-lower contentregularity. A set E ⊆ R n is said to be ( c, d )- lower content regular in a ball B if forall x ∈ E ∩ B and r ∈ (0 , r B ) , H d ∞ ( E ∩ B ( x, r )) ≥ cr d . Here, H ∞ denotes that Hausdorff content. See [Mat99] for more details. Notice,under this relaxed condition, the measure H d | E may not be locally finite. As aresult, Azzam and Schul were required to introduce a new β -number. The β -number MATTHEW HYDE they defined is analogous to that of David and Semmes but they instead ‘integrate’with respect to Hausdorff content. For E ⊆ R n and a ball B , they definedˇ β d,pE ( B ) = inf L (cid:32) r dB ˆ H d ∞ ( { x ∈ E ∩ B : dist( x, L ) > tr B } ) t p − dt (cid:33) p , where L ranges over all d -planes in R n .To state their results, we need some additional notation. For closed sets E, F ⊆ R n and a set B , we define d B ( E, F ) = 2diam( B ) max (cid:40) sup y ∈ E ∩ B dist( y, F ) , sup y ∈ F ∩ B dist( y, E ) (cid:41) . In the case where B = B ( x, r ) , we may write d x,r ( E, F ) to denote the abovequantity.For C > ε > , letBWGL( C , ε ) = { Q ∈ D : d C B Q ( E, P ) ≥ ε for all d -planes P } . Remark 1.4.
Above, BWGL stands for bi-lateral weak geometric lemma . Davidand Semmes [DS93] gave another characterization of UR sets in terms of BWGL.They showed that an Ahflors d -regular set is UR if and only if for every C ≥ , thereexists ε > C , ε ) satisfies a Carleson condition with constantdepending on ε. We now state the result from [AS18]. In fact, we state the reformulation pre-sented in [AV19].
Theorem 1.5 (Azzam, Schul [AS18]) . Let ≤ d < n and E ⊆ R n be a closed set.Suppose that E is ( c, d ) -lower content regular and let D denote the Christ-Davidcubes for E . Let C > . Then there is ε > small enough so that the followingholds. Let ≤ p < p ( d ) . For R ∈ D , let BWGL( R ) = BWGL( R, ε, C ) = (cid:88) Q ∈ BWGL( ε,C ) Q ⊆ R (cid:96) ( Q ) d and ˇ β E,A,p ( R ) := (cid:96) ( R ) d + (cid:88) Q ⊆ R ˇ β d,pE ( AB Q ) (cid:96) ( Q ) d . Then, for R ∈ D , H d ( R ) + BWGL( R, ε, C ) ∼ A,n,c,p,C ,ε ˇ β E,A,p ( R ) . (1.2)We should mention the work of Edelen, Naber and Valtorta [ENV16], who de-scribe how well the size of a Radon measure µ can be bounded from above by thecorresponding ˆ β d,pµ -number (these are defined analogously to ˆ β d,pE with the integraltaking over µ instead of H | dE ). We state a corollary of their results for Hausdorffmeasure and compare this to Theorem 1.5. Theorem 1.6.
Let E ⊆ R n . Set µ = H d | E and assume ˆ β µ, ( x, r ) drr ≤ M for µ -a.e x ∈ B . SP FOR GENERAL SETS 5
Then E is rectifiable and for every x ∈ B and < r ≤ , we have H d ( E ∩ B ( x, r )) (cid:46) n (1 + M ) r d . As mentioned, this is just one corollary of much more general theorem for generalRadon measures (see [ENV16, Theorem 1.3]). Both Theorem 1.5 and Theorem 1.6do not require E to be Ahlfors d -regular. Instead, Theorem 1.5 requires that E must be lower content regular, whereas Theorem 1.6 requires the existence of alocally finite measure. Furthermore, Theorem 1.5 also provides lower bounds forHausdorff measure.Azzam and Villa further generalise Theorem 1.5 in [AV19]. Here, they introducethe notion of a quantitative property which is a way of splitting the surface cubes ofa set E into “good” and “bad” parts. They prove estimates of the form of Theorem1.5, where BWGL( R ) is instead replaced with other quantitative properties which‘guarantee uniform rectifiability’. Here, a quantitative property is said to guaranteeuniform rectifiability if whenever the bad set of cubes is small (quantified by aCarleson packing condition) then E is uniform rectifiability. The BWGL conditionis an example of a quantitative property which guarantees uniform rectifiability. Wedirect the reader to [AV19] for a more precise description and more example of thesequantitative properties. We state one of their results for the bilateral approximationuniformly by planes (BAUP) condition, which we explain below (this will be usedlater to prove the second main result of the paper). Theorem 1.7 (Azzam, Villa [AV19]) . Let E ⊆ R n be a ( c, d ) -lower content regularset with Christ-David cubes D . Let
BAUP( C , ε ) = { Q ∈ D : d C B Q ( E, U ) ≥ ε, U is a union of d -plane } . For R ∈ D , define BAUP(
R, C , ε ) = (cid:88) Q ⊆ RQ ∈ BAUP( C ,ε ) (cid:96) ( Q ) d . and ˇ β E ( R ) = (cid:96) ( R ) d + (cid:88) Q ⊆ R ˇ β d, E (3 B Q ) (cid:96) ( Q ) d . Then, for all R ∈ D , C > , and ε > small enough depending on C and c , H d ( R ) + BAUP( R, C , ε ) ∼ ˇ β E ( R ) . (1.3)Notice, Theorem 1.5 is more concerned with establishing quantitative estimatesof the form seen in Theorem 1.1, rather than studying what types of surfaces couldmimic the role of finite length curves, as in the 1-dimensional case. Very recently,Villa [Vil19] proved a Travelling Salesman Theorem for lower content regular setsmore closely resembling that of Jones’ original theorem. The ‘nice’ sets he used werea certain class of topological non-degenerate surfaces, first introduced by David[Dav04]. Definition 1.8.
Let 0 < α < . Consider a one parameter family of Lipschitz maps { ϕ t } ≤ t ≤ , defined on R n . We say { ϕ } ≤ t ≤ is an allowed Lipschitz deformation with parameters α , or an α -ALD, if it satisfies the following condition:(1) ϕ t ⊆ B ( x, r ) for each t ∈ [0 , y ∈ R n , t (cid:55)→ ϕ t ( y ) is a continuous function on [0 , MATTHEW HYDE (3) ϕ ( y ) = y and ϕ t ( y ) = y for t ∈ [0 ,
1] whenever y ∈ R n \ B ( x, r );(4) dist( ϕ t ( y ) , E ) ≤ α r for t ∈ [0 ,
1] and y ∈ E ∩ B ( x, r ) , where 0 < α < . Definition 1.9.
Fix parameters r , α , δ and η . We say E ⊆ R n is a topologicallystable d -surface with parameters r , α , δ and η , if for all α -ALD { ϕ } , and forall x ∈ E and 0 < r < r , we have H d ( B ( x, (1 − η ) r ) ∩ ϕ ( E )) ≥ δ r d . Amongst other things, Villa proved the following.
Theorem 1.10.
Let E ⊆ R n be a ( c, d ) -lower content regular set and let Q ∈ D . Given two parameters < ε, κ < , there exists a set Σ = Σ( ε, κ, Q ) such that (1) Q ⊆ Σ . (2) Σ is topologinally stable d -surface with parameters r = diam( Q ) / , <η > / , and α and δ sufficiently small with respect to ε and κ. (3) We have the estimate H d (Σ) ∼ c ,n,d,ε diam( Q ) d + (cid:88) Q ∈ D Q ⊆ Q ˇ β d,pE ( C B Q ) (cid:96) ( Q ) d Before stating our main results, we mention that Travelling Salesman type prob-lems have been considered in a variety of other setting outside of R n . For suchresults in the Heisenberg group see [FFP + Main Results.
We prove a d -dimensional analogue of Theorem 1.1 for generalsets in R n . In particular, we do not assume E to be Ahlfors regular or lower contentregular and we do not assume the existence of a locally finite measure on E (as wasthe case in Theorem 1.2, Theorem 1.5 and Theorem 1.6, respectively). We shouldemphasize that while Theorem 1.5 and Theorem 1.6 concentrate on proving boundsfor measures, our result differs in the sense that we construct a nice surface whichcontains our set, and the measure of this surface is controlled by our β -numbers.Observe that if E does not satisfy any lower regularity condition, it may be that H d ∞ ( E ) = 0. Thus, ˇ β d,pE may trivially return a zero value even if there is someinherent non-flatness, for example if E is a dense collection of point in some purelyunrectifiable set. We shall introduce a new β -numbers, β d,pE , to deal with this. Wefirst define a variant of the Hausdorff content, where we ‘force’ sets to have somelower regularity with respect to this content, and define β d,pE (analogously to Azzamand Schul) by integrating with respect to this new content. Definition 1.11.
Let E ⊆ R n , B a ball and 0 < c ≤ c < ∞ be constants to befixed later. We say a collection of balls B which covers E ∩ B is good if r B (cid:48) ≤ r B for every B (cid:48) ∈ B and for all x ∈ E ∩ B and 0 < r < r B , we have (cid:88) B (cid:48) ∈ B B (cid:48) ∩ B ( x,r ) ∩ E ∩ B (cid:54) = ∅ r dB (cid:48) ≥ c r d and (cid:88) B (cid:48) ∈ B B (cid:48) ∩ B ( x,r ) ∩ E ∩ B (cid:54) = ∅ r B (cid:48) ≤ r r dB (cid:48) ≤ c r d . SP FOR GENERAL SETS 7
Then, for A ⊆ E ∩ B, define H d,EB, ∞ ( A ) = inf (cid:88) B (cid:48) ∈ B B (cid:48) ∩ A (cid:54) = ∅ r dB (cid:48) : B is good for E ∩ B Definition 1.12.
Let 1 ≤ p < ∞ , E ⊆ R n , B a ball centered on E and L a d -plane.Define β d,pE ( B, L ) p = 1 r dB ˆ (cid:18) dist( x, L ) r B (cid:19) p d H d,EB, ∞ = 1 r dB ˆ H d,EB, ∞ ( { x ∈ E ∩ B : dist( x, L ) > tr B } ) t p − dt, and β d,pE ( B ) = inf { β d,pE ( B, L ) : L is a d -plane } . In Section 2 we study the above definitions. We are more explicit about theconstant c , c appearing in Definition 1.11 and we shall prove some basic propertiesof H d,EB, ∞ and β d,pE . Our main results read as follows:
Theorem 1.13.
Let ≤ d < n , C > and ≤ p < p ( d ) . There exists a constant c > such that the following holds. Suppose E ⊆ F ⊆ R n , where F is ( c, d ) -lowercontent regular for some c ≥ c . Let D E and D F be the Christ-David cubes for E and F respectively. Let Q E ∈ D E and let Q F be the cube in D F with the samecenter and side length as Q E . Then (1.4) diam( Q E ) d + (cid:88) Q ∈ D E Q ⊆ Q E β d,pE ( C Q ) (cid:96) ( Q ) d (cid:46) C ,c,n,p diam( Q F ) d + (cid:88) Q ∈ D F Q ⊆ Q F β d,pF ( C Q ) (cid:96) ( Q ) d . Theorem 1.14.
Let ≤ d < n , C > and ≤ p < p ( d ) . Let E ⊆ R n , D E denotethe Christ-David cubes for E and let Q E ∈ D E be such that diam( Q E ) ≥ λ(cid:96) ( Q E ) for some < λ ≤ . Then there exists a ( c , d ) -lower content regular set F (with c as in the previous theorem) such that the following holds. Let D F denote theChrist-David cubes for F and let Q F denote the cube in D F with the same centerand side length as Q E . Then (1.5) diam( Q F ) d + (cid:88) Q ∈ D F Q ⊆ Q F β d,pF ( C Q ) (cid:96) ( Q ) d (cid:46) C ,c,n,p,λ diam( Q E ) d + (cid:88) Q ∈ D E Q ⊆ Q E β d,pE ( C Q ) (cid:96) ( Q ) d . Remark 1.15.
The condition that diam( Q E ) ≥ λ(cid:96) ( Q ) is not too stringent. Itsimply ensures that the top cube is suitably sized for the set we are considering. MATTHEW HYDE
As an immediate corollary of Theorem 1.13 and Theorem 1.14, along with The-orem 1.10, we obtain a Travelling Salesman Theorem for general sets in R n resem-bling that of Jones original theorem. Corollary 1.16.
Let ≤ d ≤ n , ≤ p ≤ p ( d ) , and C > . Suppose E ⊆ R n and Q ∈ D . Then there exists r , α , δ and η and a topologically stable d -surface Σ , with parameters r , α , δ and η such that E ⊆ Σ and H d (Σ) ∼ diam( Q ) d + (cid:88) Q ∈ D Q ⊆ Q β d,pE ( C B Q ) (cid:96) ( Q ) d . Acknowledgment.
I would like to thank Jonas Azzam, my supervisor, forhis invaluable support, guidance and patience throughout this project.2.
Preliminaries
Notation.
If there exists
C > a ≤ Cb, then we shall write a (cid:46) b. If the constant C depends of a parameter t , we shall write a (cid:46) t b. We shall write a ∼ b if a (cid:46) b and b (cid:46) a, similarly we define a ∼ t b. For
A, B ⊆ R n , letdist( A, B ) = inf {| x − y | : x ∈ A, y ∈ B } and diam( A ) = sup {| x − y | : x, y ∈ A } . We recall the d -dimensional Hausdorff measure and content. For A ⊆ R n , d ≥ < δ ≤ ∞ define H dδ ( A ) = inf (cid:88) i diam( A i ) d : A ⊆ (cid:91) i A i and diam A i ≤ δ . (2.1)The d -dimensional Hausdorff content of A is defined to be H d ∞ ( A ) and the d -dimensional Hausdorff measure of A is defined to be H d ( A ) = lim δ → H dδ ( A ) . Chirst-David cubes.
For a set E ⊆ R n we shall need a version of “dyadiccubes”. These were first introduced by David [Dav88] and generalised in [Chr90]and [HM12]. Lemma 2.1.
Let X be a doubling metric space and X k be a sequence of maximal ρ k -separated nets, where ρ = 1 / and let c = 1 / . Then, for each n ∈ Z ,there is a collection D k of cubes such that the following hold. (1) For each k ∈ Z , X = (cid:83) Q ∈ D k Q. (2) If Q , Q ∈ D = (cid:83) k D k and Q ∩ Q (cid:54) = ∅ , then Q ⊆ Q or Q ⊆ Q . (3) For Q ∈ D , let k ( Q ) be the unique integer so that Q ∈ D k and set (cid:96) ( Q ) =5 ρ k . Then there is x Q ∈ X k such that B ( x Q , c (cid:96) ( Q )) ⊆ Q ⊆ B ( x Q , (cid:96) ( Q )) . Given a collection of cubes D and Q ∈ D , define D ( Q ) = { R ∈ D : R ⊆ Q } . SP FOR GENERAL SETS 9
Let Child k ( Q ) denote the k th generational descendants of Q (where we often writeChild( Q ) to mean Child ( Q )) and Q ( k ) denote the k th generational ancestor. Weshall denote the descendants up to the k th by Des k ( Q ) , that is,Des k ( Q ) = k (cid:91) i =0 Child i ( Q ) . (2.2)Finally define a distance function, d C , to a collection of cubes C ⊆ D by setting d C ( x ) = inf { (cid:96) ( R ) + dist( x, R ) : R ∈ C } , and for Q ∈ D , set d C ( Q ) = inf { d C ( x ) : x ∈ Q } . The following lemma is standard and can be found in, for example, [AS18].
Lemma 2.2.
Let C ⊆ D and Q, Q (cid:48) ∈ D . Then d C ( Q ) ≤ (cid:96) ( Q ) + dist( Q, Q (cid:48) ) + 2 (cid:96) ( Q (cid:48) ) + d C ( Q (cid:48) ) . (2.3)2.3. Theorem of David and Toro.
The surface F from Theorem 1.14 will be aunion of surfaces constructed using the following Reifenberg parametrization theo-rem of David and Toro [DT12]. Theorem 2.3 ([DT12, Sections 1 - 9]) . Let P a plane. Let k ∈ N and set r k =10 − k . Let { x j,k } j ∈ J k be an r k -separated net. To each x j,k , associate a ball B jk = B ( x jk , r k ) and a plane P jk containing x jk . Assume { x j } j ∈ J ⊂ P , and x ik ∈ V k − , where V λk := (cid:83) j ∈ J k λB jk . Define ε k ( x ) = sup { d x il , r l ( P jk , P il ) : j ∈ J k , | l − k | ≤ , i ∈ J k , x ∈ B jk ∩ B il } . Then, there is ε > such that if ε ∈ (0 , ε ) and ε k ( x jk ) < ε, for all k ≥ and j ∈ J k , then there is a bijection f : R n → R n such that: (1) We have E ∞ := ∞ (cid:92) K =1 ∞ (cid:91) k = K { x j,k } j ∈ J k ⊆ Σ := f ( R n ) . (2) f ( x ) = x when dist( z, P ) > . (3) For x, y ∈ R n , | x − y | τ ≤ | f ( x ) − f ( y ) | ≤ | x − y | − τ . (4) | f ( x ) − x | (cid:46) ε for x ∈ R n . (5) For x ∈ P , f ( x ) = lim k σ k ◦ · · · ◦ σ , where σ k ( y ) = ψ k ( y ) + (cid:88) j ∈ J k θ j,k ( y )[ π j,k ( y ) − y ] . Here, { x j,k } j ∈ L k is a maximal r k -separated set in R n \ V k ,B j,k = B ( x j,k , r k / for j ∈ L k , { θ j,k } j ∈ J k ∪ L k is a partition of unity such that B j,k ≤ θ j,k ≤ B j,k forall k and j ∈ L k ∪ J k , and ψ k = (cid:80) j ∈ L k θ j,k . (6) For k ≥ , (2.4) σ k ( y ) and Dσ k ( y ) = I for y ∈ R n \ V k . (7) Let Σ = P and Σ k = σ k (Σ k − ) . There is a function A j,k : P j,k ∩ B j,k → P ⊥ j,k of class C such that (cid:12)(cid:12) A j,k ( x j,k ) (cid:12)(cid:12) (cid:46) εr k , (cid:12)(cid:12) DA j,k (cid:12)(cid:12) (cid:46) ε on P j,k ∩ B j,k , and if Γ j,k is its graphover P j,k , then Σ k ∩ D ( x j,k , P j,k , r k ) = Γ k ∩ D ( x j,k , P j,k , r k ) where D ( x j,k , P j,k , r k ) = { z + w : z ∈ P ∩ B ( x, r ) , w ∈ P ⊥ ∩ B (0 , r ) } . In particular, (2.5) d x jk , r k (Σ k , P jk ) (cid:46) ε. (8) For k ≥ and y ∈ Σ k , there is an affine d -plane P through y and a Cε -Lipschitz and C -function A : P → P ⊥ so that if Γ is the graph of A over P , then Σ k ∩ B ( y, r k ) = Γ ∩ B ( y, r k ) . (9) Have
Σ = f ( P ) is Cε -Reifenberg flat in the sense that for all z ∈ Σ , and t ∈ (0 , , there is P = P ( z, t ) so that d z,t (Σ , P ) (cid:46) ε. (10) For all y ∈ Σ k , (cid:12)(cid:12) σ k ( y ) − y (cid:12)(cid:12) (cid:46) ε k ( y ) r k (2.6) and moreover, dist( y, Σ) (cid:46) εr k , for y ∈ Σ k (2.7)(11) For k ≥ , y ∈ Σ j ∩ V k , choose i ∈ J k such that y ∈ B ik . Then (cid:12)(cid:12) σ k ( y ) − π ik ( y ) (cid:12)(cid:12) (cid:46) ε k ( y ) r k . (12) For x ∈ Σ and r > , H d ∞ (Σ ∩ B ( x, r )) ≥ (1 − Cε ) ω d r d (2.8) where ω d is the volume of the unit ball in R d . Remark 2.4.
We conjecture the main results hold for subsets of an infinite di-mensional Hilbert space. With this in mind, we have tried to make as much of ourwork here dimension free. We shall indicate the places where the estimates dependon the ambient dimension. As such, many of the volume arguments rely on thefollowing result. Put simply, it states that a collection of disjoint balls lying closeenough to a d -dimensional plane will satisfy a d -dimensional packing condition. SP FOR GENERAL SETS 11
Lemma 2.5 ([ENV18, Lemma 3.1]) . Let V be an affine d -dimensional plane ina Banach space X , and { B ( x i , r i ) } i ∈ I be a family of pairwise disjoint balls with r i ≤ R, B ( x i , r i ) ∈ B ( x, R ) , for some x ∈ R n and dist( x i , V ) < r i / . Then, thereis a constant κ = κ ( d ) such that (2.9) (cid:88) i ∈ I r di ≤ κR d . Lemma 2.6.
Let E ⊆ R n , Q ∈ D and M ≥ . If < ε ≤ c ρ M and β dE, ∞ ( M B Q ) ≤ ε, then Q has at most K = K ( M, d ) children, i.e. independent of n. Proof.
The balls { c B R } R ∈ Child( Q ) are pairwise disjoint (recall c from Lemma 2.1),contained in M B Q , and have radius less than or equal to r MB Q . Since β dE, ∞ ( M B Q ) ≤ ε, there exists a d -plane P Q such thatdist( y, P Q ) ≤ εM (cid:96) ( Q )for all y ∈ M B Q . In particular, for any R ∈ Child( Q ) , we havedist( x R , P Q ) ≤ εM (cid:96) ( Q ) ≤ r c B R / . By Lemma 2.5, { R : R ∈ Child( Q ) } c ρ d (cid:96) ( Q ) d = (cid:88) R ∈ Child( Q ) ( c (cid:96) ( R )) d (2.9) ≤ κ ( M (cid:96) ( Q )) d , from which the lemma follows by dividing through by c ρ d (cid:96) ( Q ) d . (cid:3) As a simple corollary of the above lemma, we also get a bound on the number ofdescendants up to a specified generation. The constant here ends up also dependingon the generation.
Lemma 2.7.
Let E ⊆ R n , Q ∈ D , M ≥ and k ≥ . If < ε < c ρ M and β dE, ∞ ( M B Q ) ≤ ε for all R ∈ Des k ( Q ) , then (cid:88) R ∈ Des k ( Q ) (cid:96) ( R ) d (cid:46) d,M,k (cid:96) ( Q ) d . (2.10)2.4. Hausdorff-type content.
In this section we study the Hausdorff content H d,EB, ∞ that we defined in the introduction. For the convenience of the reader, westate the definition again. Remark 2.8.
Let us be explicit about the constant c appearing in Theorem 1.13and Theorem 1.14. We fix now c := ω d { − d − , ρ/ } where ρ is the constant appearing in Theorem 2.1 and ω d is the volume of the unitball in R d . We fix another constant c := 18 d κ, with κ as in Lemma 2.5. We comment on this choice of constants in Remark 2.22.Basically, we have chosen c sufficiently small and c sufficiently large. Definition 2.9.
Let E ⊆ R n , B a ball. We say a collection of balls B which covers E ∩ B is good if r B (cid:48) ≤ r B for every B (cid:48) ∈ B and for all x ∈ E ∩ B and 0 < r < r B , we have (cid:88) B (cid:48) ∈ B B (cid:48) ∩ B ( x,r ) ∩ E ∩ B (cid:54) = ∅ r dB (cid:48) ≥ c r d (2.11)and (cid:88) B (cid:48) ∈ B B (cid:48) ∩ B ( x,r ) ∩ E ∩ B (cid:54) = ∅ r B (cid:48) ≤ r r dB (cid:48) ≤ c r d . (2.12)Then, for A ⊆ E ∩ B, define H d,EB, ∞ ( A ) = inf (cid:88) B (cid:48) ∈ B B (cid:48) ∩ A (cid:54) = ∅ r dB (cid:48) : B is good for E ∩ B See Figure 1 for an example of a good cover.
Remark 2.10.
For the usual Hausdorff content, H d ∞ , all coverings of a set arepermissible (see (2.1)). In defining our new content, we restrict the permissiblecoverings to ensure all sets will have a lower regularity property with respect tothis content (this is the role of (2.11)). In addition, we require an upper regularitycondition, (2.12). This is to ensure any cover we choose is sensible. In particular, itstops us constructing lower regular covers by just repeatedly adding the same ballover and over again. Remark 2.11. If E ⊆ R n and B is a ball, then B = { B } is a good cover for E ∩ B. In particular, every set has a good cover.
Remark 2.12.
The constant 100 is not important, it can be chosen to be any largenumber.
Remark 2.13.
It is easy to see that for any A ⊆ E ∩ B, we have H d ∞ ( A ) ≤ H d,EB, ∞ ( A ) . We shall now prove some basic properties of H d,EB, ∞ . Before doing so we need somepreliminary lemmas. The first is [Mat99, Lemma 2.5] and the second is modificationof [Mat99, Lemma 2.6], whose proof is essentially the same.
Lemma 2.14.
Suppose a, b ∈ R , < | a | ≤ | a − b | and < | b | ≤ | a − b | . Then theangle between the vectors a and b is at least ◦ , that is, | a/ | a | − b/ | b || ≥ . Lemma 2.15.
There is N ( n ) ∈ N with the following property. Let B be a ball andsuppose there are k disjoint balls B , . . . , B k such that r B i ≥ r B and B ∩ B i (cid:54) = ∅ for all i = 1 , . . . , k. Then k ≤ N ( n ) . SP FOR GENERAL SETS 13
Figure 1.
Example of a good cover for E ∩ B . Notice we needto cover the star segment with a large ball to ensure the cover islower regular. We could equally have added many smaller balls aslong as the upper regularity condition is not violated. Proof.
We may assume B is centered at the origin. If one of the B i is centeredat the origin then k = 1 so assume this is not the case. Let B i = B ( x i , r i ) . Since B i ∩ B j (cid:54) = ∅ , we have | x i − x j | > r i + r j > r i + r B , and so 0 < | x i | ≤ r B + r i ≤ | x i − x j | for i (cid:54) = j. Applying Lemma 2.14 with a = x i and b = x j for i (cid:54) = j in the two dimensionalplane containing 0 , x i , x j , we obtain | x i / | x i | − x j / | x j || ≥ i (cid:54) = j. Since the unit sphere S n − is compact there are at most N ( n ) such points. (cid:3) Lemma 2.16.
Let E ⊆ R n , and B be a ball. Then, (1) H d,EB, ∞ ( E ∩ B ( x, r )) ≥ c r d for all x ∈ E ∩ B, < r ≤ r B . (2) If A ⊆ A ⊆ E ∩ B , then H d,EB, ∞ ( A ) ≤ H d,EB, ∞ ( A ) . (3) If B (cid:48) ⊆ B and A ⊆ E ∩ B (cid:48) , then H d,EB (cid:48) , ∞ ( A ) ≤ H d,EB, ∞ ( A ) . (4) Suppose E ∩ B = E ∪ E . Then H d,EB, ∞ ( E ∩ B ) (cid:46) H d,EB, ∞ ( E ) + H d,EB, ∞ ( E ) . Proof.
Property (1) is an immediate consequence of Definition 2.9 since any goodcover B of E ∩ B satisfies (2.11). Property (2) is also clear from Definition 2.9. If B (cid:48) ⊆ B then any good cover for B is also a good cover for B (cid:48) , and (3) follows.To prove (4), let ε > B i , i = 1 , E ∩ B suchthat (cid:88) B (cid:48) ∈ B i B (cid:48) ∩ E i (cid:54) = ∅ r dB ≤ H d,EB, ∞ ( E i ) + ε/ . (2.13)Let B (cid:48) be the collection of balls B (cid:48) ∈ B such that B (cid:48) ∩ E = ∅ . We partition B (cid:48) into two further collection. Define B (cid:48) , = { B (cid:48) ∈ B (cid:48) : there is B (cid:48)(cid:48) ∈ B with B (cid:48) ∩ B (cid:48)(cid:48) ∩ E ∩ B (cid:54) = ∅ and r B (cid:48)(cid:48) ≥ r B (cid:48) } and B (cid:48) , = { B (cid:48) ∈ B (cid:48) : r B (cid:48)(cid:48) < r B (cid:48) for all B (cid:48)(cid:48) ∈ B such that B (cid:48) ∩ B (cid:48)(cid:48) ∩ E ∩ B (cid:54) = ∅} . For B (cid:48) ∈ B (cid:48) , let ˜ B be the ball in B such that ˜ B ∩ B (cid:48) ∩ E ∩ B (cid:54) = ∅ and r ˜ B ≥ r B (cid:48) . Since B (cid:48) ∩ E = ∅ and E ∩ B = E ∪ E , it must be that B (cid:48) ∩ E ∩ B ⊆ E , hence˜ B ∩ E (cid:54) = ∅ . If there is more than one such ball we can choose ˜ B arbitrarily. Then (cid:88) B (cid:48) ∈ B (cid:48) , r dB (cid:48) = (cid:88) B (cid:48)(cid:48) ∈ B B (cid:48)(cid:48) ∩ E (cid:54) = ∅ (cid:88) B (cid:48) ∈ B (cid:48) , ˜ B = B (cid:48)(cid:48) r dB (cid:48) ≤ (cid:88) B (cid:48)(cid:48) ∈ B B (cid:48)(cid:48) ∩ E (cid:54) = ∅ (cid:88) B (cid:48) ∈ B B (cid:48) ∩ B (cid:48)(cid:48) ∩ E ∩ B (cid:54) = ∅ r B (cid:48) ≤ r B (cid:48)(cid:48) r dB (cid:48) (2.12) ≤ c (cid:88) B (cid:48)(cid:48) ∈ B B (cid:48)(cid:48) ∩ E (cid:54) = ∅ r dB (cid:48)(cid:48) . We turn our attention to B (cid:48) , . Let B be the largest ball in B (cid:48) , . Then, given B , . . . , B k , define B k +1 to be the largest ball B (cid:48) ∈ B (cid:48) , such that E ∩ B ∩ B (cid:48) ∩ k (cid:91) i =1 B i = ∅ . Let { B i } ∞ i =1 be the resulting disjoint collection of balls. Any ball B (cid:48) ∈ B (cid:48) , suchthat E ∩ B ∩ B (cid:48) (cid:54) = ∅ is contained in 201 B (recall from Definition 2.9 that each ball B (cid:48) has radius at most 100 r B ). So by compactness, for any R > B (cid:48) ∈ B (cid:48) , such that r B (cid:48) ≥ R and E ∩ B ∩ B (cid:48) (cid:54) = ∅ .Hence, for any B (cid:48) ∈ B (cid:48) , there exists B i such that B (cid:48) ∩ B i ∩ E ∩ B (cid:54) = ∅ . Moreover,for any B i such that B (cid:48) ∩ B i ∩ E ∩ B (cid:54) = ∅ , we have r B (cid:48) ≤ r B i . Thus, (cid:88) B (cid:48) ∈ B (cid:48) , r dB (cid:48) ≤ ∞ (cid:88) i =1 (cid:88) B (cid:48) ∈ B (cid:48) , B (cid:48) ∩ B i ∩ E ∩ B (cid:54) = ∅ r dB (cid:48) (2.12) ≤ c ∞ (cid:88) i =1 r dB i . By Lemma 2.15, for any B (cid:48) ∈ B , we have { B i : B (cid:48) ∩ B i ∩ E ∩ B (cid:54) = ∅} (cid:46) n . (2.14) SP FOR GENERAL SETS 15
As before, since B i ∩ E = ∅ , if B (cid:48) ∈ B satisfies B (cid:48) ∩ B i ∩ E ∩ B (cid:54) = ∅ for some i, then B (cid:48) ∩ E (cid:54) = ∅ . Hence (cid:88) B (cid:48) ∈ B (cid:48) , r dB ≤ c ∞ (cid:88) i =1 r dB i (2.11) ≤ c c ∞ (cid:88) i =1 (cid:88) B (cid:48) ∈ B B (cid:48) ∩ E (cid:54) = ∅ B (cid:48) ∩ B i ∩ E ∩ B (cid:54) = ∅ r dB (cid:48) ≤ c c (cid:88) B (cid:48) ∈ B B (cid:48) ∩ E (cid:54) = ∅ (cid:88) iB (cid:48) ∩ B i ∩ E ∩ B (cid:54) = ∅ r dB (cid:48) (2.14) (cid:46) c c (cid:88) B (cid:48) ∈ B B (cid:48) ∩ E (cid:54) = ∅ r dB (cid:48) We concluding by noting that, since B is a good cover, we have H d,EB, ∞ ( E ∩ B ) ≤ (cid:88) B (cid:48) ∈ B B (cid:48) ∩ E ∩ B (cid:54) = ∅ r dB (cid:48) = (cid:88) B (cid:48) ∈ B B (cid:48) ∩ E (cid:54) = ∅ r dB (cid:48) + (cid:88) B (cid:48) ∈ B (cid:48) r dB (cid:48) (cid:46) (cid:88) B (cid:48) ∈ B B (cid:48) ∩ E (cid:54) = ∅ r dB (cid:48) + (cid:88) B (cid:48) ∈ B B (cid:48) ∩ E (cid:54) = ∅ r dB (cid:48) (2.13) ≤ H d,EB, ∞ ( E ) + H d,EB, ∞ ( E ) + ε. Since ε was arbitrary, (4) follows. (cid:3) For a function f : R n → [0 , ∞ ) , define integration with respect to H d,EB, ∞ via theChoquet integral: ˆ f d H d,EB, ∞ := ˆ ∞ H d,EB, ∞ ( { x ∈ E ∩ B : f ( x ) > t } ) dt. We state some basic properties of the above Choquet integral, see [Wan11] formore details. Note, in [Wan11] there are additional upper and lower continuityassumption, but these are not required for the following lemma.
Lemma 2.17.
Let f, g : R n → [0 , ∞ ) such that f ≤ g and α > . Then (1) ´ f d H d,EB, ∞ ≤ ´ g d H d,EB, ∞ ;(2) ´ ( f + α ) d H d,EB, ∞ = ´ f d H d,EB, ∞ + α H d,EB, ∞ ( E ∩ B );(3) ´ αf d H d,EB, ∞ = α ´ f d H d,EB, ∞ . The Choquet integral also satisfies a Jensen-type inequality. The proof is basedon the proof of the usual Jensen’s inequality for general measures, see [Rud06].
Lemma 2.18.
Suppose
E, B ⊆ R n , φ : R → R is convex and f : R n → R isbounded. Then, φ H d,EB, ∞ ( E ∩ B ) ˆ f d H d,EB, ∞ ≤ H d,EB, ∞ ( E ∩ B ) ˆ φ ◦ f d H d,EB, ∞ . (2.15) Proof.
Since f is bounded and H d,EB, ∞ ( E ∩ B ) (cid:38) r dB , we can set t = 1 H d,EB, ∞ ( E ∩ B ) ˆ f d H d,EB, ∞ < ∞ . Since φ is convex, if −∞ < s < t < u < ∞ , then φ ( t ) − φ ( s ) t − s ≤ φ ( u ) − φ ( t ) u − t . (2.16)Let γ be the supremum of the left hand side of (2.16) taken over all s ∈ ( −∞ , t ) . It is clear then that φ ( t ) ≤ φ ( s ) + γ · ( t − s )for all s ∈ R . By rearranging the above inequality we have that for any x ∈ R n ,γf ( x ) + φ ( t ) ≤ φ ( f ( x )) + γt. Integrating both with respect to x, and using Lemma 2.17, we have γ ˆ f d H d,EB, ∞ + H d,EB, ∞ ( E ∩ B ) φ ( t ) ≤ ˆ φ ◦ f d H d,EB, ∞ + γ H d,EB, ∞ ( E ∩ B ) t, thus, H d,EB, ∞ ( E ∩ B ) φ ( t ) ≤ ˆ φ ◦ f d H d,EB, ∞ + γ (cid:18) H d,EB, ∞ ( E ∩ B ) t − ˆ f d H d,EB, ∞ (cid:19) . By definition of t , the second term on the right hand side of the above inequalityis zero, from which the lemma follows. (cid:3) Integration with respect to H d ∞ can be defined similarly and satisfies identicalproperties. In the special case of integration with respect to H d ∞ we also have thefollowing: Lemma 2.19 ([AS18, Lemma 2.1]) . Let < p < ∞ . Let f i be a countable collectionof Borel functions in R n . If the sets supp f i = { f i > } have bounded overlap,meaning there exists a C < ∞ such that (cid:88) supp f i ≤ C, then ˆ (cid:16)(cid:88) f i (cid:17) p d H d ∞ ≤ C p (cid:88) ˆ f pi d H d ∞ . Preliminaries with β -numbers. In this section we prove some basic prop-erties of β d,pE . Again, we restate the definition for the readers convenience.
Definition 2.20.
Let 1 ≤ p < ∞ , E ⊆ R n , B a ball centered on E and L a d -plane.Define β d,pE ( B, L ) p = 1 r dB ˆ (cid:18) dist( x, L ) r B (cid:19) p d H d,EB, ∞ = 1 r dB ˆ H d,EB, ∞ ( { x ∈ E ∩ B : dist( x, L ) > tr B } ) t p − dt, and β d,pE ( B ) = inf { β d,pE ( B, L ) : L is a d -plane } . Remark 2.21.
It is easy to show that H d ∞ ≤ H d,EB, ∞ which implies ˇ β d,pE ≤ β d,pE . If c ≥ c and E is a ( c, d )-lower regular set, we get the reverse inequality up to aconstant, see Corollary 2.26. SP FOR GENERAL SETS 17
Remark 2.22.
It is possible to define a β -number β d,p,cE where the additionalparameter c replaces the fixed lower regularity constant c from (2.11). We couldprove a version of Theorem 1.14 for β d,p,c , that is, if E ⊆ F and the sum of these β d,p,c coefficients for E is finite, then we can find a ( c , d )-lower content regular set F such that E ⊆ F (notice that the regularity constant for F is independent of c ).Thus to prove Corollary 1.16 we could just as well have used β d,p,c instead of β d,p for any c ≤ c . We have fixed our lower regularity parameter for clarity.
Lemma 2.23.
Let ≤ p < ∞ , E ⊆ R n and B a ball centred on E . Then β d, E ( B ) (cid:46) β d,pE ( B )(2.17) Proof.
This is a direct consequence of Lemma 2.18 and Definition 2.20. The in-equality obviusly holds for p = 1 , so assume p > . Then the map x (cid:55)→ x p is convex.Let P be the d -plane such that β d,pE ( B ) = β d,pE ( B, P ) . Since dist( x, P ) ≤ r B forall x ∈ E ∩ B and H d,EB, ∞ ( E ∩ B ) ∼ r dB , we have, by Lemma 2.18, β d, E ( B ) ≤ r dB ˆ (cid:18) dist( x, P ) r B (cid:19) d H d,EB, ∞ (2.15) (cid:46) (cid:32) r dB ˆ (cid:18) dist( x, P ) r B (cid:19) p d H d,EB, ∞ (cid:33) p = β d,pE ( B ) . (cid:3) Lemma 2.24.
Let ≤ p < ∞ and E ⊆ R n . Then for all balls B (cid:48) ⊆ B centered on E, β d,pE ( B (cid:48) ) ≤ (cid:18) r B r B (cid:48) (cid:19) dp β d,pE ( B ) . Proof.
Let P be the d -plane such that β d,pE ( B ) = β d,pE ( B, P ) . By Lemma 2.16 (2),(3)and a change of variables, we have β d,pE ( B (cid:48) ) p ≤ r dB (cid:48) ˆ H d,EB (cid:48) , ∞ ( { x ∈ E ∩ B (cid:48) : dist( x, P ) > tr B (cid:48) } ) t p − dt (2) , (3) ≤ r dB r dB (cid:48) r dB ˆ H d,EB, ∞ ( { x ∈ E ∩ B : dist( x, P ) > tr B (cid:48) } ) t p − dt ≤ (cid:18) r B r B (cid:48) (cid:19) d + p r dB ˆ H d,EB, ∞ ( { x ∈ E ∩ B : dist( x, P ) > tr B } ) t p − dt = β d,pE ( B ) p . (cid:3) By restricting the covers of our sets as we have done, we lose monotonicityof β d,p · ( B, L ) with respect to set inclusion. This is because we are imposing thecondition that all sets have large measure, even singleton points. We illustrate thispoint with an example. Let ε > B be the unit ball and L a d -plane throughthe origin. Consider a set F consisting of the d -plane L with the segment throughthe origin replaced by two sides of an equilateral triangle with height ε and the set E ⊆ F which is just the singleton point at the tip of the equilateral triangle (see Figure 2. β d,p · is not monotone.Figure 2). Since F is lower 1-regular and we have comparability for ˇ β d,p and β d,p on lower regular sets (see Corollary 2.26), it is easy to show β , F ( B , L ) (cid:46) ˇ β , F (2 B , L ) ∼ (cid:15) . On the other hand, by Lemma 2.16 (1), we must have β , E ( B , L ) = ˆ H ,E B , ∞ ( { x ∈ E ∩ B : dist( x, L ) ≥ t } ) dt = ˆ ε H ,E B , ∞ ( E ∩ B ) dt (1) (cid:38) (cid:15), which for (cid:15) small enough implies β , F ( B , L ) ≤ β , E ( B , L ) . We do however have the following result at least in the case where the larger setis lower regular. Roughly speaking it states that if E ⊆ F and F is lower regular,then we can control the β -number of E by the β -number of F with some error termdependent on the average distance of F from E. Lemma 2.25.
Let F ⊆ R n be ( c, d ) -lower content regular for some c ≥ c and let E ⊆ F. Let B be a ball and L a d -dimensional plane. Then β d,pE ( B, L ) (cid:46) c,d,p β d,pF (2 B, L ) + (cid:32) r dB ˆ F ∩ B (cid:18) dist( x, E ) r B (cid:19) p d H d ∞ ( x ) (cid:33) p . SP FOR GENERAL SETS 19
Proof.
We actually prove a stronger version of the above statement, with β d,pF re-placed by ˇ β d,pF . Note, by re-scaling and translating, we may assume B = B . For t > E t = { x ∈ E ∩ B : dist( x, L ) > t } . To prove the lemma, it suffices to show H d,E B , ∞ ( E t ) (cid:46) H d ∞ (cid:0) { x ∈ F ∩ B : dist( x, L ) > t/ } (cid:1) + H d ∞ (cid:0) { x ∈ F ∩ B : dist( x, E ) > t/ } (cid:1) (2.18)since if the above were true, β d,pE ( B , L ) p = ˆ H d,E B , ∞ ( E t ) t p − dt (2.18) (cid:46) ˆ H d ∞ ( { x ∈ F ∩ B : dist( x, L ) > t/ } ) t p − dt + ˆ H d ∞ ( { x ∈ F ∩ B : dist( x, L ) > t/ } ) t p − dt (cid:46) ˆ H d ∞ ( { x ∈ F ∩ B : dist( x, L ) > t } ) t p − dt + ˆ H d ∞ ( { x ∈ F ∩ B : dist( x, L ) > t } ) t p − dt ∼ ˇ β d,pF (2 B , L ) p + ˆ F ∩ B dist( x, E ) p d H d ∞ ( x ) . For the rest of the proof we focus on (2.18). Fix t > . We must first constructa suitable good cover for E ∩ B . For x ∈ F ∩ B , let δ ( x ) = max { dist( x, L ) , dist( x, E ) } + t/ X = { x i } i ∈ I be a maximal net in F ∩ B such that | x i − x j | ≥ { δ ( x i ) , δ ( x j ) } (2.20)for all i (cid:54) = j. For each i ∈ I let B (cid:48) i = B ( x i , δ ( x i )) and B (cid:48) = { B (cid:48) i } i ∈ I . By (2.19), δ ( x ) > x ∈ F ∩ B , so the balls B (cid:48) i are non-degenerate. Furthermore, define B = { B (cid:48) i } i ∈ I = { B i } i ∈ I . Claim: B is a good cover for E ∩ B . Since E ⊆ F, it follows that B covers E by maximaility. We are left to verify (2.11)and (2.12). Let x ∈ E ∩ B and 0 < r < . We look first at (2.11).Assume B (cid:48) i ∩ B ( x, r/ ∩ F (cid:54) = ∅ for some i ∈ I. If 4 δ ( x i ) ≥ r/ B ( x, r/ ⊆ B (cid:48) i , hence 3 B (cid:48) i ∩ B ( x, r/ ∩ E (cid:54) = ∅ . If 4 δ ( x i ) < r/
3, then B (cid:48) i ⊆ B ( x, r ) andsince dist( x i , E ) ≤ δ ( x i ) there exists some y ∈ E ∩ B ( x, r ) ∩ B (cid:48) i , in particular, B (cid:48) i ∩ B ( x, r ) ∩ E (cid:54) = ∅ . In either case, we conclude E ∩ B ( x, r ) ∩ B (cid:48) i (cid:54) = ∅ . Then, since F is ( c, d )-lower content regular in B , we have (cid:88) B ∈ B B ∩ E ∩ B ( x,r ) (cid:54) = ∅ (cid:18) r B (cid:19) d ≥ (cid:88) B (cid:48) ∈ B (cid:48) B (cid:48) ∩ F ∩ B ( x,r/ (cid:54) = ∅ r dB (cid:48) ≥ c (cid:18) r (cid:19) d , hence B satisfies the lower bound (2.11). Now for the upper bound (2.12). If B ∈ B satisfies B ∩ B ( x, r ) ∩ E (cid:54) = ∅ and r B ≤ r , then B ⊆ B ( x, r ). Furthermore, since the balls { B } B ∈ B are disjoint andsatisfy dist( x B , L ) ≤ r B / , we have by Lemma 2.5 (cid:88) B ∈ B B ∩ E ∩ B ( x,r ) (cid:54) = ∅ r B ≤ r r dB ≤ d (cid:88) B ∈ B B ⊆ B ( x, r ) r d B (2.9) ≤ d κr d Since c ≥ c and, recalling that c = 18 d κ, it follows that B satisfies (2.12), thus, B is a good cover for E ∩ B which proves the claim.We partition the balls in B as follows, let B E = { B i ∈ B : r i = 12dist( x i , E ) + t/ } , B L = { B i ∈ B : r i = 12dist( x i , L ) + t/ } . If dist( x i , E ) = dist( x i , L ) then we put B i in B E or B L arbitrarily. Then, since B is good for E ∩ B , we have H d,E B , ∞ ( E t ) ≤ (cid:88) B ∈ B B ∩ E t (cid:54) = ∅ r dB = (cid:88) B ∈ B E B ∩ E t (cid:54) = ∅ r dB + (cid:88) B ∈ B L B ∩ E t (cid:54) = ∅ r dB . (2.21)If we can show that (cid:88) B ∈ B E B ∩ E t (cid:54) = ∅ r dB (cid:46) H d ∞ (cid:0) { x ∈ F ∩ B : dist( x, E ) > t/ } (cid:1) (2.22)and (cid:88) B ∈ B L B ∩ E t (cid:54) = ∅ r dB (cid:46) H d ∞ (cid:0) { x ∈ F ∩ B : dist( x, L ) > t/ } (cid:1) , (2.23)then (2.18) follows from the above two inequalities and (2.21). We first prove (2.22).The proof of (2.23) is similar and we shall comment on the necessary changes afterwe are done with (2.22).Let A := { x ∈ F ∩ B : dist( x, E ) > t/ } and let B A be a cover of A such that each ball B ∈ B A is centered on A , has r B ≤ r B and H d ∞ ( A ) ∼ (cid:88) B ∈ B A r dB . (2.24)Let B i ∈ B E satisfy E t ∩ B i (cid:54) = ∅ and y i ∈ E t ∩ B i . Recall that since B i ∈ B E wehave dist( x i , L ) ≤ dist( x i , E ) and B i = B ( x i ,
12 dist( x i , E ) + t/ t < dist( y i , L ) ≤ | y i − x i | + dist( x i , E ) ≤
12 dist( x i , E ) + t/
10 + dist( x i , E ) ≤
13 dist( x i , E ) + t/ . Rearranging, we find that dist( x i , E ) > t/ . (2.25) SP FOR GENERAL SETS 21
Figure 3.
Examples of balls in C (left) and C (right).This implies that F ∩ B i ⊆ A, (2.26)since for any y ∈ F ∩ B i we havedist( y, E ) ≥ dist( x i , E ) − r B i = 12 dist( x i , E ) − t (2.25) ≥ t . By (2.26), since B A covers A , there is B ∈ B A such that B i ∩ B (cid:54) = ∅ . Wepartition B E further by setting C = { B i : there exists B ∈ B A such that B i ∩ B (cid:54) = ∅ and r B ≥ r B i / } , C = B E \ C . See Figure 3. We first control the sum over balls in C . Assume B ∈ B A is suchthat there exists some B i ∈ C such that B i ∩ B (cid:54) = ∅ and r B ≥ r B i / . For y ∈ B i , we have | y − x B | ≤ | y − x i | + | x i − x B | ≤ r B i + r B + r B i / ≤ r B , from which we conclude B i ⊆ B. So, if C B = { B i ∈ C : B i ∩ B (cid:54) = ∅ and r B ≥ r B i / } , the balls { B i } B i ∈ C B are pairwise disjoint, contained in 26 B and satisfydist( x i , L ) ≤ r B i / B i = B ( x i , δ ( x i ))). By Lemma 2.5 and because B is a good cover,this gives (cid:88) B i ∈ C B r dB i (cid:46) (cid:88) B i ∈ C B r d B i (2.9) (cid:46) r dB . Thus, (cid:88) B i ∈ C r dB i ≤ (cid:88) B ∈ B A (cid:88) B i ∈ C B r dB i (cid:46) (cid:88) B ∈ B A r dB (2.24) (cid:46) H d ∞ ( A ) . (2.27)Now, the sum over balls in C . If B i ∈ C and B ∈ B A is such that B i ∩ B (cid:54) = ∅ , then r B < r B i / . Furthermore B ∩ B j = ∅ for all B j ∈ C , j (cid:54) = i, since otherwise | x i − x j | ≤ | x i − x B | + | x B − x j | ≤ r B i /
24 + 2 r B + r B j / ≤ (cid:18)
124 + 112 + 124 (cid:19) max { r B i , r B j } ≤
16 max { r B i , r B j }≤ { δ ( x i ) , δ ( x j ) } , contradicting (2.20). Thus,(2.28) { B i ∈ C : B ∩ B i (cid:54) = ∅} ≤ . Since B A forms a cover for A and F ∩ B i ⊆ A by (2.26), it follows that { B ∈ B A : B ∩ B i (cid:54) = ∅} forms a cover for F ∩ B i . Using then that F is ( c, d )-lowercontent regular, we have (cid:88) B i ∈ C r dB i (cid:46) (cid:88) B i ∈ C H d ∞ ( F ∩ B i ) ≤ (cid:88) B i ∈ C (cid:88) B ∈ B A B ∩ B i (cid:54) = ∅ r dB = (cid:88) B ∈ B A (cid:88) B i ∈ C B ∩ B i (cid:54) = ∅ r dB (2.28) ≤ (cid:88) B ∈ B A r dB (2.24) (cid:46) H d ∞ ( A ) . (2.29)Combining (2.27) and (2.29) completes the proof of (2.22). The proof of (2.23)follows exactly the same reasoning: For each B i ∈ B L we havedist( x i , L ) > t/ , the proof of which is the same as (2.25). Analogously to (2.26), this implies F ∩ B i ⊆ A (cid:48) , where A (cid:48) = { x ∈ F ∩ B : dist( x, L ) > t/ } . The rest of the proof is identical. This completes the proof of (2.18) which in turncompletes the proof of the lemma. (cid:3)
SP FOR GENERAL SETS 23
As a corollary of the above proof of Lemma 2.25 and Remark 2.21, we have thefollowing:
Corollary 2.26.
Suppose c ≥ c and E ⊆ R n is ( c, d ) -lower content regular. Forany ball B and d -plane L, we have ˇ β d,pE ( B, L ) ≤ β d,pE ( B, L ) (cid:46) ˇ β d,pE (2 B, L ) . Lemma 2.27.
Assume E ⊆ R n and there is B centered on E so that for all B (cid:48) ⊆ B centered on E we have H d ∞ ( E ∩ B (cid:48) ) ≥ cr dB (cid:48) . Then β dE, ∞ (cid:18) B (cid:19) (cid:46) β d, E ( B ) d +1 . (2.30) Proof.
Azzam and Schul prove the same inequality for ˇ β d,pE (see [AS18, Lemma2.12]). Then, by Corollary 2.26, β dE, ∞ (cid:18) B (cid:19) (cid:46) ˇ β d, E ( B ) d +1 (cid:46) β d, E ( B ) d +1 . (cid:3) Remark 2.28.
By (2.30) and Lemma 2.6, if ε > M ) and β d,pE ( M B Q ) ≤ ε for some Q ∈ D , then Q has at most K children where K depends only on M and d and not the ambient dimension n. The following is analogous to Lemma 2.21 in [AS18]. It says the β -number of alower regular set can be controlled by the β -number of a nearby set, with an errordepending on the average distance between the two. The proof is very similar tothe proof of Lemma 2.25. Lemma 2.29.
Let ≤ p < ∞ . Suppose
E, F ⊆ R n , B is a ball centered on E and B is a ball of same radius but centered on F such that B ⊆ B . Suppose for allballs B ⊆ B centered on E we have H d ∞ ( B ∩ E ) ≥ cr Bd for some c > . Then ˇ β d,pE ( B , P ) (cid:46) c,p,d β d,pF (2 B , P )(2.31) + (cid:32) r dB ˆ E ∩ B (cid:18) dist( y, F ) r B (cid:19) p d H d ∞ ( y ) (cid:33) p . Proof.
By scaling, we can assume that B = B . For t >
0, set E t = { x ∈ E ∩ B : dist( x, L ) > t } . To prove (2.31), it suffices to show H d ∞ ( E t ) (cid:46) H d,F B , ∞ ( { x ∈ F ∩ B : dist( x, L ) > tr B / } )(2.32) + H d ∞ ( { x ∈ E ∩ B : dist( x, F ) > tr B / } ) , which gives β d,pE ( B , L ) p = ˆ H d ∞ ( E t ) t p − dt (2.32) (cid:46) ˆ H d,F B , ∞ ( { x ∈ F ∩ B : dist( x, L ) > t/ } ) t p − dt + ˆ H d ∞ ( { x ∈ E ∩ B : dist( x, F ) > t/ } ) t p − dt (cid:46) ˆ H d,F B , ∞ ( { x ∈ F ∩ B : dist( x, L ) > t } ) t p − dt + ˆ H d ∞ ( { x ∈ E ∩ B : dist( x, F ) > t } ) t p − dt ∼ β d,pF (2 B , L ) p + ˆ E ∩ B dist( x, F ) p d H d ∞ ( x ) , So, let us prove (2.32). We first need to construct a suitable cover for E t . For x ∈ E t , let δ ( x ) = max { dist( x, L ) , x, F ) } and set X t to be a maximally separated net in E t such that, for x, y ∈ X t , we have | x − y | ≥ { δ ( x ) , δ ( y ) } . (2.33)Enumerate X t = { x i } i ∈ I . For x i ∈ X t denote B i = B ( x i , δ ( x i )) . Notice, these ballsare non-degenerate since x i ∈ X t ⊆ E t . By maximality we know { B i } covers E t so, H d ∞ ( E t ) (cid:46) (cid:88) i ∈ I ( r B i ) d . (2.34)We partition I = I ∪ I , where I = { i ∈ I : δ ( x i ) = dist( x i , L ) } and I = { i ∈ I : δ ( x i ) = 16dist( x i , F ) } . If dist( x i , L ) = 16dist( x i , F ), we put i in I or I arbitrarily. By (2.34), it followsthat H d ∞ ( E t ) (cid:46) (cid:88) i ∈ I ( r B i ) d + (cid:88) i ∈ I ( r B i ) d . We will show that (cid:88) i ∈ I ( r B i ) d (cid:46) H d,E B , ∞ ( { x ∈ F ∩ B : dist( x, L ) > t/ } )(2.35)and (cid:88) i ∈ I ( r B i ) d (cid:46) H d ∞ ( { x ∈ E ∩ B : dist( x, F ) > t/ } )(2.36)from which (2.32) follows. The rest of the proof is dedicated to proving (2.35) and(2.36). Let us begin with (2.35). Let F t = { x ∈ F ∩ B : dist( x, L ) > t/ } SP FOR GENERAL SETS 25 and B be a good cover for F ∩ B such that H d,F B , ∞ ( F t ) ∼ (cid:88) B ∈ B B ∩ F t (cid:54) = ∅ r dB . (2.37)If i ∈ I then F ∩ B i (cid:54) = ∅ and F ∩ B i ⊆ F t (2.38)since dist( x i , L ) > t (by virtue of that fact that x i ∈ E t ) and B i = B ( x i , dist( x i , L ) / ⊇ B ( x i , x i , F )) . Since B forms a cover of F t there exists at least one B ∈ B such that B ∩ B i (cid:54) = ∅ . We further partition I . Let I , = { i ∈ I : there exists B ∈ B such that B i ∩ B (cid:54) = ∅ and r B ≥ r B i } ,I , = I \ I , . We first control the sum over I , . If B ∈ B is such that there is B i satisfying B i ∩ B (cid:54) = ∅ and r B ≥ r B i (which by definition implies i ∈ I , ), then 2 B i ⊆ B. By (2.33) we know the { B i } are disjoint and satisfy dist( x i , L ) ≤ r B i / . ByLemma 2.5, we have (cid:88) i ∈ I , B i ∩ B (cid:54) = ∅ r Bi ≤ r B r d B i (cid:46) (cid:88) i ∈ I , B i ⊆ B r d B i (2.9) (cid:46) r dB . Thus, (cid:88) i ∈ I , r d B i ≤ (cid:88) B ∈ B B ∩ F t (cid:54) = ∅ (cid:88) i ∈ I , B i ∩ B (cid:54) = ∅ r Bi ≤ r B r d B i (cid:46) (cid:88) B ∈ B B ∩ F t (cid:54) = ∅ r dB . (2.39)We now turn our attention to I , . For i ∈ I , , let x (cid:48) i be the point in F closest to x i and set B (cid:48) i = B ( x (cid:48) i , dist( x i , L ) / . Note that B (cid:48) i ⊆ B i , since for y ∈ B (cid:48) i we have | y − x i | ≤
14 dist( x i , L ) + | x − x i | ( i ∈ I ) ≤ (cid:18)
14 + 116 (cid:19) dist( x i , L ) ≤
12 dist( x i , L ) . Since F ∩ B i ⊆ F t by (2.38), and B forms a cover for F t , the balls { B ∈ B : B ∩ F ∩ B i (cid:54) = ∅} form a cover for F ∩ B i . Furthermore, if B ∩ B i (cid:54) = ∅ then B ∩ B j = ∅ for all i (cid:54) = j , that is { i ∈ I , : B i ∩ B (cid:54) = ∅} ≤ , (2.40)since otherwise | x i − x j | < { δ ( x i ) , δ ( x j ) } , contradicting (2.33). By Lemma2.16 (1), we know H d,F B , ∞ ( F ∩ B (cid:48) i ) (cid:38) r dB (cid:48) i (cid:38) r dB i , and since B is a good cover for F ∩ B , we have(2.41) (cid:88) i ∈ I , ( r B i ) d (cid:46) (cid:88) i ∈ I , H d,F B , ∞ ( F ∩ B (cid:48) i ) ≤ (cid:88) i ∈ I , H d,F B , ∞ ( F ∩ B i ) ≤ (cid:88) i ∈ I , (cid:88) B ∈ B B ∩ B i ∩ F (cid:54) = ∅ r dB = (cid:88) B ∈ B B ∩ F t (cid:54) = ∅ (cid:88) i ∈ I , B ∩ B i ∩ F (cid:54) = ∅ r dB (2.40) ≤ (cid:88) B ∈ B B ∩ F t (cid:54) = ∅ r dB . Combining (2.39) and (2.41), we conclude (cid:88) i ∈ I r d B i (cid:46) (cid:88) B ∈ B B ∩ F t (cid:54) = ∅ r dB (2.37) (cid:46) H d,F B , ∞ ( F t )which is (2.35).We turn our attention to proving (2.36), the proof of which follows much thesame as that for (2.35). Let E (cid:48) t = { x ∈ E ∩ B : dist( x, F ) > t/ } and B (cid:48) be a collection of balls covering E (cid:48) t such that each B ∈ B (cid:48) is centered on E (cid:48) t , has r B ≤ r B and H d ∞ ( E (cid:48) t ) ∼ (cid:88) B ∈ B (cid:48) r dB . (2.42)As before, we partition I . If i ∈ I , since x i ∈ E t , we have dist( x i , L ) > t anddist( x i , F ) ≥ dist( x i , L ) / ≥ t/ , hence E ∩ B i = E ∩ B ( x i , dist( x i , F ) / ⊆ E (cid:48) t . (2.43)Thus for each B i , since B (cid:48) forms a cover for E (cid:48) t , there exists B ∈ B (cid:48) such that B ∩ B i (cid:54) = ∅ . We partition I by letting I , = { i ∈ I : there exists B ∈ B (cid:48) such that B i ∩ B (cid:54) = ∅ and r B ≥ r B i } ,I , = I \ I , . If B ∈ B (cid:48) and B ∩ B i (cid:54) = ∅ with r B ≥ r B i then 2 B i ⊆ B. Furthermore, by (2.33),we know the { B i } are disjoint and satisfy dist( x i , L ) < dist( x i , F ) /
16 = r B i / , so by Lemma 2.5, we have (cid:88) i ∈ I , B i ∩ B (cid:54) = ∅ r Bi ≤ r B r d B i (cid:46) (cid:88) i ∈ I , B i ⊆ B r d B i (2.9) (cid:46) r dB . Thus, (cid:88) i ∈ I , r d B i ≤ (cid:88) B ∈ B (cid:48) (cid:88) i ∈ I , B i ∩ B (cid:54) = ∅ r Bi ≤ r B r d B i ≤ (cid:88) B ∈ B (cid:48) r dB . (2.44) SP FOR GENERAL SETS 27
We now deal with I , . Since by (2.43), E ∩ B i ⊆ E (cid:48) t , the balls { B ∈ B (cid:48) : B ∩ B i ∩ E (cid:54) = ∅} form a cover for E ∩ B i . As before, if B ∩ B i (cid:54) = ∅ then B ∩ B j = ∅ for all i (cid:54) = j by (2.33). By lower regularity of E , we know H d ∞ ( E ∩ B i ) (cid:38) r dB i , fromwhich we conclude (cid:88) i ∈ I , r d B i (cid:46) (cid:88) i ∈ I , H d ∞ ( E ∩ B i ) ≤ (cid:88) i ∈ I , (cid:88) B ∈ B (cid:48) B ∩ B i ∩ F (cid:54) = ∅ r dB = (cid:88) B ∈ B (cid:48) (cid:88) i ∈ I , B ∩ B i ∩ F (cid:54) = ∅ r dB (cid:46) (cid:88) B ∈ B (cid:48) r dB . (2.45)The proof of (2.36) (and hence the proof of the lemma) is completed since (cid:88) i ∈ I r d B i = (cid:88) i ∈ I , r d B i + (cid:88) i ∈ I , r d B i (2.44)(2.45) (cid:46) (cid:88) B ∈ B (cid:48) r dB (2.42) (cid:46) H d ∞ ( E (cid:48) t ) . (cid:3) In Section 3 and Section 4, we want to apply the construction of David and Toro(Theorem 2.3). We wish to do this by controlling the angles between pairs of planes,by their corresponding β -numbers. The following series of lemmas, culminating inLemma 2.34, will allow us to do so. We first introduce some more notation.For two planes P, P (cid:48) containing the origin, we define ∠ ( P, P (cid:48) ) = d B (0 , ( P, P (cid:48) ) . If P, P (cid:48) are general affine planes with x ∈ P and y ∈ P (cid:48) , we define ∠ ( P, P (cid:48) ) = ∠ ( P − x, P (cid:48) − y ) . For planes P , P and P , it is not difficult to show that ∠ ( P , P ) ≤ ∠ ( P , P ) + ∠ ( P , P ) . Lemma 2.30 ([AT15, Lemma 6.4]) . Suppose P and P are d -planes in R n and X = { x , . . . , x d } are points so that (1) η ∈ (0 , , where η = η ( X ) = min { dist( x i , span( X \ { x i } ) } / diam( X )(2) dist( x i , P j ) < ε diam( X ) for i = 0 , . . . , d and j = 1 , , where ε < ηd − / . Then dist( y, P ) ≤ ε (cid:18) dη dist( y, X ) + diam( X ) (cid:19) for all y ∈ P . In order to control angles between d -planes, we need to know that E is sufficientlyspread out in at least d directions. This is quantified below. Definition 2.31.
Let 0 < α < . We say a ball B has ( d + 1 , α )-separated pointsif there exist points X = { x , . . . , x d } in E ∩ B such that, for each i = 1 , . . . , d, wehave dist( x i +1 , span { x , . . . , x d } ) ≥ αr B . (2.46) Lemma 2.32.
Suppose E ⊆ R n and there is B (cid:48) and B both centered on E with B (cid:48) ⊆ B. Suppose further that there exists < α < such that B (cid:48) has ( d + 1 , α ) -separated points. Let P and P (cid:48) be two d -planes. Then d B (cid:48) ( P, P (cid:48) ) (cid:46) α d +2 (cid:34)(cid:18) r B r B (cid:48) (cid:19) d +1 β d, E (2 B, P ) + β d, E (2 B (cid:48) , P (cid:48) ) (cid:35) . Proof.
Since B (cid:48) has ( d + 1 , α )-separated points, we can find X = { x , . . . , x d } satisfying (2.46). This implies that α < η ( X ) ≤ . Let B i = B ( x i , αr B (cid:48) /
4) and for t > E t,i = { x ∈ E ∩ B i : dist( x, P ) > tr B (cid:48) or dist( x, P (cid:48) ) > tr B (cid:48) } . Let
T > E t,i = E ∩ B i for all t ≤ T. We shall bound T . By Lemma2.16 (1), H d,EB i , ∞ ( E ∩ B i ) ≥ c r dB i = c α d d r dB (cid:48) . Using this, along with Lemma 2.16 (4), we get T ≤ H d,EB i , ∞ ( E ∩ B i ) − ˆ T H d,EB i , ∞ ( E t,i ) dt (cid:46) α d r dB (cid:48) ˆ T H d,EB i , ∞ ( E t,i ) dt (4) (cid:46) α d r dB (cid:48) ˆ T H d,EB i , ∞ { x ∈ E ∩ B i : dist( x, P ) > tr B (cid:48) } dt + 1 α d r dB (cid:48) ˆ T H d,EB i , ∞ { x ∈ E ∩ B i : dist( x, P (cid:48) ) > tr B (cid:48) } dt (cid:46) α d r dB (cid:48) ˆ T H d,E B (cid:48) , ∞ { x ∈ E ∩ B (cid:48) : dist( x, P ) > tr B (cid:48) } dt + r d +1 B r d +1 B (cid:48) α d r dB (cid:48) ˆ T H d,E B, ∞ { x ∈ E ∩ B : dist( x, P (cid:48) ) > tr B } dt (cid:46) α d (cid:34)(cid:18) r B r B (cid:48) (cid:19) d +1 β d, E (2 B, P ) + β d, E (2 B (cid:48) , P (cid:48) ) (cid:35) =: λα . Note, we define λ like this for convenience in the forthcoming estimates. Thus, thereis a constant C such that T ≤ Cλα . This implies for each i = 0 , , . . . , d , thereexists some y i ∈ ( E ∩ B i ) \ E λα ,i . Let Y = { y , . . . , y d } . Since | x i − x j | ≥ αr B (cid:48) for all i (cid:54) = j, and y i ∈ B i , it follows thatdiam( Y ) ≥ αr B (cid:48) / . (2.47)Thus, dist( y i , P j ) ≤ Cλα r B (cid:48) = 2 Cλα r B (cid:48) diam( Y ) diam( Y ) (2.47) ≤ Cλα diam( Y ) . Because d B (cid:48) ( P, P (cid:48) ) ≤ , if λ ≥ Cd then the lemma follows. Assume instead that λ < Cd . By (2.47) we can show that α/ ≤ η ( Y ) ≤ , (2.48)which gives 4 Cλα ≤ αd − (2.48) ≤ η ( Y ) d − / , SP FOR GENERAL SETS 29 so, taking ε = 4 Cλα in Lemma 2.30, we get d B (cid:48) ( P, P (cid:48) ) ≤ (cid:15) (cid:18) dη ( Y ) + 1 (cid:19) (2.47) ≤ Cλα (cid:18) dα + 1 (cid:19) ≤ Cdλ, which proves the lemma. (cid:3)
Remark 2.33.
The following lemma is essentially Lemma 2.18 from [AS18] andthe proof is the same. The main difference is that since E is not necessarily lowerregular, we need to assume that E has ( d + 1 , α )-separated points in each cube.The final constant then also ends up depending on α. Lemma 2.34.
Let
M > , α > and E a Borel set. Let D be the cubes for E from Lemma 2.1 and Q ∈ D . Let P Q satisfy β d, E ( M B Q ) = β d, E ( M B Q , P Q ) . Let Q, R ∈ D , Q, R ⊆ Q and suppose for all cubes T ⊆ Q such that T contains either Q or R that β d, E ( M B T ) < ε and T has ( α, d + 1) -separated points. Then for Λ > ,if dist( Q, R ) ≤ Λ max { (cid:96) ( Q ) , (cid:96) ( R ) } ≤ Λ min { (cid:96) ( Q ) , (cid:96) ( R ) } , then ∠ ( P Q , P R ) (cid:46) M, Λ εα d +2 . Proof of Theorem 1.13
Let X Ek ⊆ X Fk be sequences of maximally ρ k -separated nets in E and F respec-tively. Let D E and D F be the cubes from Theorem 2.1 with respect to X Ek and X Fk . Let Q E ∈ D E and let Q F ∈ D F be the cube with the same center and sidelength as Q E . To simplify notation we will write D = D F and Q = Q F . We firstreduce to proof of (1.4) to the proof of (3.1) below.
Lemma 3.1. If (cid:88) Q ∈ D Q ⊆ Q β d,pE ( C B Q ) (cid:96) ( Q ) d (cid:46) H d ( Q ) + (cid:88) Q ∈ D Q ⊆ Q ˇ β d, F ( AB Q ) (cid:96) ( Q ) d (3.1) for some A ≥ C , then (1.4) holds.Proof. Assume (3.1) holds. First, it is easy to see thatdiam( Q E ) d + (cid:88) Q ∈ D E Q ⊆ Q E β d,pE ( C B Q ) (cid:96) ( Q ) d (cid:46) diam( Q ) d + (cid:88) Q ∈ D Q ⊆ Q β d,pE ( C B Q ) (cid:96) ( Q ) d . By (3.1), Lemma 2.23 and Theorem 1.5 (using the fact that F is lower contentregular), we have (cid:88) Q ∈ D Q ⊆ Q β d,pE ( C B Q ) (cid:96) ( Q ) d (3.1) (cid:46) H d ( Q ) + (cid:88) Q ∈ D Q ⊆ Q ˇ β d, F ( AB Q ) (cid:96) ( Q ) d (2.17) (cid:46) H d ( Q ) + (cid:88) Q ∈ D Q ⊆ Q ˇ β d,pF ( AB Q ) (cid:96) ( Q ) d (1.2) (cid:46) diam( Q ) d + (cid:88) Q ∈ D Q ⊆ Q ˇ β d,pF ( AB Q ) (cid:96) ( Q ) d . Now, let K = K ( C , A ) be the smallest integer such that (1 + Aρ K ) ≤ C . Let Q ∈ D k for some k ≥ K and y ∈ AB Q . Then | y − x Q ( K ) | ≤ A(cid:96) ( Q ) + (cid:96) ( Q ( K ) ) = ( Aρ K + 1) (cid:96) ( Q ( K ) ) ≤ C (cid:96) ( Q ( K ) ) . Hence AB Q ⊆ C B Q ( K ) . Each cube Q ∈ D has at most C = C ( n ) children (noticethe number of descendants is dependent on the the ambient dimension since we donot necessarily know β d, F is small for an arbitrary cube in D ). It follows that each Q has at most KC descendants up to the K th generation. In particular, this is alsotrue for Q . By this and Lemma 2.24, we have (cid:88) Q ∈ D Q ⊆ Q ˇ β d,pF ( AB Q ) (cid:96) ( Q ) d (cid:46) K − (cid:88) k =0 (cid:88) Q ∈ D k Q ⊆ Q ˇ β d,pF ( AB Q ) (cid:96) ( Q ) d + ∞ (cid:88) k = K (cid:88) Q ∈ D k Q ⊆ Q ˇ β d,pF ( AB Q ) (cid:96) ( Q ) d (cid:46) diam( Q ) d + (cid:88) Q ∈ D ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d . Combing each of the above sets of inequalities gives (1.5). (cid:3)
The rest of this section is devoted to proving (3.1) holds for A ≥ C . Definition 3.2.
A collection of cubes S ⊆ D is called a stopping time region if thefollowing hold.(1) There is a cube Q ( S ) ∈ S such that Q ( S ) contains all cubes in S .(2) If Q ∈ S and Q ⊆ R ⊆ Q ( S ) , then R ∈ S .(3) If Q ∈ S , then all siblings of Q are also in S .We let: • Q ( S ) denote the maximal cube in S . • min( S ) denote the cubes in S which have a child not contained in S . • S ( Q ) denote the unique stopping time regions S such that Q ∈ S. We split D ( Q ) into a collection of stopping time regions S where in eachstopping time region, F is well-approximated by E and there is good control ona certain Jones type function. Observe that if Q ∈ D and C B Q ∩ E = ∅ then β d,pE ( C B Q ) = 0 , and so we will not restart our stopping times on these cubes.Let M > ε > Q ∈ D ( Q ) such that E ∩ C B Q (cid:54) = ∅ , we define a stopping timeregion S Q as follows. Begin by adding Q to S Q and inductively, on scales, addcubes R to S Q if each of the following holds,(1) R (1) ∈ S Q ,(2) for every sibling R (cid:48) of R , if x ∈ R (cid:48) then dist( x, E ) ≤ ε(cid:96) ( R (cid:48) ) . (3) every sibling R (cid:48) of R satisfies (cid:88) R (cid:48) ⊆ T ⊆ Q ˇ β d, F ( M B T ) < ε . SP FOR GENERAL SETS 31
Remark 3.3.
If ˇ β d, F ( M B Q ) ≥ ε or if there exists x ∈ Q such that dist( x, E ) >ε(cid:96) ( Q ) then S Q = { Q } . Remark 3.4.
Firstly, ε will be chosen sufficiently small so that each cube R con-tained in some stopping time region has at most K children, where K depends onlyon M and d. See Remark 2.28 for why this is possible.We partition { Q ∈ D ( Q ) : E ∩ C B Q (cid:54) = ∅} as follows. First, add S Q to S . Then, if S has been added to S and if Q ∈ Child( R ) for some R ∈ min( S ) suchthat E ∩ C B Q (cid:54) = ∅ , also add S Q to S . Let S be the collection of stopping timeregions obtained by repeating this process indefinitely. Note that (cid:88) Q ∈ D ( Q ) β d,pE ( C B Q ) (cid:96) ( Q ) d = (cid:88) S ∈ S (cid:88) Q ∈ S β d,pE ( C B Q ) (cid:96) ( Q ) d . For each S ∈ S which is not a singleton (i.e. S (cid:54) = { Q } ) we plan to find abi-Lipschitz surface which well approximates F inside S . With some additionalconstraints, the surfaces produced by Theorem 2.3 will be bi-Lipschitz. Theorem 3.5 ([DT12, Theorem 2.5]) . With the same notation and assumptionsas Theorem 2.3, assume additionally that there exists
K < ∞ such that (cid:88) k ≥ ε (cid:48) k ( f k ( z )) ≤ K for z ∈ Σ with ε (cid:48) k ( x ) = sup { d x il , r l ( P jk , P il ) : j ∈ J k , | l − k | ≤ , i ∈ J k , x ∈ B jk ∩ B il } . Then f = lim f N = lim N σ ◦ · · · ◦ σ N : Σ → Σ is C ( K ) -bi-Lipschitz. Lemma 3.6.
There exists ε > small enough so that for each S ∈ S , which isnot a singleton, there is a surface Σ S such that dist( y, Σ S ) (cid:46) ε d +1 (cid:96) ( R )(3.2) for each y ∈ F ∩ M B R where R ∈ S . Also, for each ball B , centered on Σ S andcontained in M B Q ( S ) , we have ω d r dB ≤ H d (Σ S ∩ B ) (cid:46) r dB (3.3) Proof.
For k ≥ s ( k ) be such that 5 ρ s ( k ) ≤ r k ≤ ρ s ( k ) − . For each Q ∈ S , let L Q be the d -plane through x Q such thatˇ β d, F ( M B Q , L Q ) ≤ β d, F ( M B Q ) . For each k , let C (cid:48) k be a maximal r k -separated net for C k = { x Q : Q ∈ D s ( k ) ∩ S } . By Lemma 2.34, C (cid:48) k , with planes { L Q } Q ∈ C (cid:48) k satisfy the assumptions of Theorem2.3 with ε k ( x ) (cid:46) ˇ β d, F ( M B Q ) < ε (3.4)for x ∈ Q ∈ C k . Let Σ (cid:48) S be the resulting surface from Theorem 2.3. By (2.8), wecan choose ε small enough so that for all balls B centered on Σ (cid:48) S we have H d ∞ (Σ (cid:48) S ∩ B ) ≥ ω d r dB . (3.5) We verify that the additional assumption of Theorem 3.5 is satisfied, thus Σ (cid:48) S is in fact a bi-Lipschitz surface. Let x = f ( z ) ∈ Σ (cid:48) S and set x k = f k ( z ). By thetriangle inequality, we have | x − x k | ≤ (cid:88) j ≥ k | x j +1 − x j | ≤ (cid:88) j ≥ k | σ j ( x j ) − x j | (2.6) (cid:46) (cid:88) j ≥ k ε j ( x ) r j (3.4) (cid:46) (cid:88) j ≥ k εr j (cid:46) εr k . Thus, for ε small enough, x k ∈ B ( x, r k ) . Let Q ∈ D s ( k ) ∩ S such that x ∈ Q. Suppose (cid:96) ∈ { k, k − } and x Q (cid:48) ∈ C (cid:48) (cid:96) is such that x k ∈ B ( x Q (cid:48) , r (cid:96) ) (the existence of x Q (cid:48) is guaranteed by maximality). It follows that | x Q (cid:48) − x Q | ≤ | x Q (cid:48) − x k | + | x k − x | + | x − x Q | ≤ r (cid:96) + 2 r k + r k ≤ r k For M large enough, this gives100 B ( x Q (cid:48) , r (cid:96) ) ⊆ B ( x Q , r k + 100 r (cid:96) ) ⊆ B ( x Q , r k ) ⊆ B ( x Q , M (cid:96) ( Q )) . Then by Lemma 2.32, ε (cid:48) k ( x k ) (cid:46) ˇ β d, F ( M B Q ). Since, by our stopping time condi-tion, we have control of the sum of ˇ β d, F ( M B Q ), we have verified the additionalassumption. We define Σ S = Σ (cid:48) S ∩ M B Q ( S ) . Thus, for all balls B centered on Σ S such that B ⊆ M B Q ( S ) , left-most inequalityin (3.3) follows by (3.5) and the right-most inequality follows since Σ (cid:48) S is the bi-Lipschitz image of R d . We check (3.2). Let R ∈ D s ( k ) ∩ S and y ∈ M B R . Let R (cid:48) ∈ D s ( k ) ∩ S Q be a cubesuch that x R (cid:48) ∈ C (cid:48) k and | x R − x R (cid:48) | ≤ r k . By the triangle inequality | y − x R (cid:48) | ≤ | y − x R | + | x R − x R (cid:48) | ≤ M (cid:96) ( R ) + r k ≤ (cid:18) M ρ − (cid:19) (cid:96) ( R ) . Choosing M ≥ ρ − gives M B R ⊆ M B R (cid:48) . Since ˇ β d, F ( M B R (cid:48) , L R (cid:48) ) < ε, by (2.30), β dE, ∞ ( M B R (cid:48) ) (cid:46) ε d +1 , hencedist( y, L R (cid:48) ) (cid:46) ε d +1 r k (cid:46) ε d +1 (cid:96) ( R ) . By (2.5), there exists z ∈ Σ k such that (cid:12)(cid:12) π L R (cid:48) ( y ) − z (cid:12)(cid:12) (cid:46) εr k . Furthermore, by (2.7),dist( z, Σ S ) (cid:46) εr k . Combing the previous estimates we see that (3.2) holds. (cid:3) Lemma 3.7.
For M ≥ C , (cid:88) S ∈ S (cid:88) Q ∈ S β d,pE ( C B Q ) (cid:96) ( Q ) d (cid:46) (cid:88) S ∈ S (cid:88) Q ∈ S ˇ β d,pF ( M B Q ) (cid:96) ( Q ) d + (cid:88) S ∈ S (cid:96) ( Q ( S )) d . To prove Lemma 3.7 we will need apply the smoothing procedure of David andSemmes (see for example [DS91, Chapter 8]). Let us introduce these smoothedcubes and prove a general fact about them. Let τ = ρ ) (3.6)and, for each S ∈ S , let Stop( S ) be the collection of maximal cubes in D such that (cid:96) ( Q ) < τ d S ( Q ) , that is,Stop( S ) = { Q ∈ D : Q is maximal so that (cid:96) ( Q ) < τ d S ( Q ) } . SP FOR GENERAL SETS 33
Lemma 3.8.
Let S ∈ S and suppose R ∈ Stop( S ) is such that R ∩ C B Q ( S ) (cid:54) = ∅ . Then there exists Q = Q ( R ) ∈ S such that τ dist( R, Q ) ≤ ρ (cid:96) ( R )(3.7) and τ ρ (cid:96) ( Q ) ≤ (cid:96) ( R ) ≤ C τ (cid:96) ( Q ) . (3.8) Proof.
Let Q (cid:48) ∈ S be a cube such that2 d S ( R ) ≥ (cid:96) ( Q (cid:48) ) + dist( Q (cid:48) , R ) . (3.9)By maximality,1 ρ (cid:96) ( R ) = (cid:96) ( R (1) ) > τ d S ( R (1) ) (2.3) ≥ τ (cid:16) d S ( R ) − (cid:96) ( R ) − (cid:96) ( R (1) ) (cid:17) = τ d S ( R ) − ρ − ) τ (cid:96) ( R ) . By our choice of τ (see (3.6)), this gives τ d S ( R ) ≤ ρ (cid:96) ( R ) . Then, τ (cid:0) (cid:96) ( Q (cid:48) ) + dist( Q (cid:48) , R ) (cid:1) (3.9) ≤ τ d S ( R ) ≤ ρ (cid:96) ( R ) . From here, we see that (3.7) and the left hand inequality in (3.8) are true. If (cid:96) ( R ) ≤ C τ (cid:96) ( Q (cid:48) ) , we can set Q = Q (cid:48) and the lemma follows. Otherwise, since R ∩ C Q ( S ) (cid:54) = ∅ , we have (cid:96) ( R ) < τ ( (cid:96) ( Q ( S )) + dist( R, Q ( S ))) ≤ C τ (cid:96) ( Q ( S )) . We can then choose Q to be the smallest ancestor of Q (cid:48) contained in Q ( S ) such that(3.8) holds. The existence of such a Q is guaranteed by the above inequality. (cid:3) Proof of Lemma 3.7.
First, by Lemma 2.25, we have (cid:88) S ∈ S (cid:88) Q ∈ S β d,pE ( C B Q ) (cid:96) ( Q ) d (cid:46) (cid:88) S ∈ S (cid:88) Q ∈ S ˇ β d,pF (2 C B Q ) (cid:96) ( Q ) d + (cid:88) S ∈ S (cid:88) Q ∈ S (cid:32) (cid:96) ( Q ) d ˆ F ∩ C B Q (cid:18) dist( x, E ) (cid:96) ( Q ) (cid:19) p d H d ∞ ( x ) (cid:33) p (cid:96) ( Q ) d . Since M ≥ C , we have the desired bound on the first term. To deal with thesecond term, let I pS := (cid:88) Q ∈ S (cid:32) (cid:96) ( Q ) d ˆ F ∩ C B Q (cid:18) dist( x, E ) (cid:96) ( Q ) (cid:19) p d H d ∞ ( x ) (cid:33) p (cid:96) ( Q ) d . If we can show that I pS (cid:46) (cid:96) ( Q ( S )) d (3.10)for each S ∈ S then the lemma follows. The rest of the proof is devoted to proving(3.10). Note, we may assume that p ≥ , since for p < , we have I pS (cid:46) I S byJensen’s inequality (Lemma 2.18).Let S ∈ S . In the case that S = { Q } is a singleton, since C B Q ∩ E (cid:54) = ∅ , wehave dist( x, E ) (cid:46) (cid:96) ( Q ) for each x ∈ F ∩ C B Q . Then, I pS reduces to (cid:96) ( Q ) and(3.10) follows. Assume then that S is not a singleton. First, by Lemma 2.19, we can write I pS (cid:46) (cid:88) Q ∈ S (cid:96) ( Q ) d (cid:88) R ∈ Stop( S ) R ∩ C B Q (cid:54) = ∅ ˆ R (cid:18) dist( x, E ) (cid:96) ( Q ) (cid:19) p d H d ∞ ( x ) p (cid:96) ( Q ) d . Let R ∈ Stop( S ) be such that R ∩ C B Q (cid:54) = ∅ for some Q ∈ S . By Lemma 3.8, wecan find a cube Q (cid:48) ∈ S such thatdist( Q (cid:48) , R ) (cid:46) (cid:96) ( R ) and (cid:96) ( R ) ∼ (cid:96) ( Q (cid:48) ) . (3.11)By our stopping time condition (2), since Q (cid:48) ∈ S, we havedist( y (cid:48) , E ) ≤ ε(cid:96) ( Q (cid:48) )(3.12)for all y (cid:48) ∈ Q (cid:48) . Let y Q (cid:48) ∈ Q (cid:48) and y R ∈ R be points such thatdist( R, Q (cid:48) ) = | y Q (cid:48) − y R | . (3.13)Then, for any y ∈ R , we havedist( y, E ) ≤ | y − y Q (cid:48) | + dist( y Q (cid:48) , E ) (3.12) ≤ | y − y Q (cid:48) | + | y Q (cid:48) − y R | + ε(cid:96) ( Q (cid:48) ) (3.13) (cid:46) (cid:96) ( R ) + dist( R, Q (cid:48) ) + ε(cid:96) ( Q (cid:48) ) (3.11) (cid:46) (cid:96) ( R ) . Using this, along with the that fact that p ≤
1, we get I pS (cid:46) (cid:88) Q ∈ S (cid:88) R ∈ Stop( S ) R ∩ C B Q (cid:54) = ∅ (cid:96) ( R ) d + p (cid:96) ( Q ) d + p p (cid:96) ( Q ) d (cid:46) (cid:88) Q ∈ S (cid:88) R ∈ Stop( S ) R ∩ C B Q (cid:54) = ∅ (cid:96) ( R ) d p +2 (cid:96) ( Q ) d ( p − . Claim:
For each k ∈ N , we have { Q ∈ D k ∩ S : R ∩ C B Q (cid:54) = ∅} (cid:46) . (3.14) Proof of Claim.
Let Q (cid:48) be the cube from Lemma 3.8 for R . By (3.11), we canchoose M large enough so that R ⊆ M B Q (cid:48) (taking M ≥ C /ρ is sufficient). Inparticular, x R ∈ M B Q (cid:48) . Then, by Lemma 3.6,dist( x R , Σ S ) (3.2) (cid:46) ε d +1 (cid:96) ( Q (cid:48) ) (3.8) (cid:46) ε d +1 τ (cid:96) ( R ) . (3.15)Recall the definition of τ from (3.6). For any Q ∈ D k ∩ S such that R ∩ C B Q (cid:54) = ∅ , we have (cid:96) ( R ) < τ d S ( R ) ≤ τ ( (cid:96) ( Q ) + dist( Q, R )) (cid:46) τ (cid:96) ( Q ) . (3.16)Let x (cid:48) R be the point in Σ S closest to x R . By (3.15) and (3.16) there exists
A > Q ∈ D k ∩ S is such that R ∩ C B Q (cid:54) = ∅ then Q ⊆ B ( x (cid:48) R , A(cid:96) ( Q )) := B. Since Σ S is ( Cε, d )-Reifenberg flat we can find a plane P through x (cid:48) R such that d B ( P, Σ S ) (cid:46) ε. (3.17) SP FOR GENERAL SETS 35
Let x (cid:48) Q be the point in Σ S which is closest to x Q , thendist( x Q , P ) ≤ | x Q − x (cid:48) Q | + dist( x (cid:48) Q , P ) (3.2)(3.17) (cid:46) ( ε d +1 + ε ) (cid:96) ( Q ) . So, for ε small enough, we have dist( x Q , P ) ≤ c (cid:96) ( Q ) / . Then (3.14) follows fromLemma 2.5. (cid:3)
Returning to I pS , since by assumption, p < d/ ( d − , or equivalently, d (2 /p −
1) + 2 > , we can swap the order of integration, apply (3.14), and sum over ageometric series and obtain I pS (cid:46) (cid:88) R ∈ Stop( S ) R ∩ C B Q ( S ) (cid:54) = ∅ (cid:88) Q ∈ SR ∩ C B Q (cid:54) = ∅ (cid:96) ( R ) d p +2 (cid:96) ( Q ) d ( p − (cid:46) (cid:88) R ∈ Stop( S ) R ∩ C B Q ( S ) (cid:54) = ∅ (cid:96) ( R ) d . By (3.15), for ε (cid:28) τ d +1 , each c B R carves out a large proportion of Σ S i.e. H d ( c B R ∩ Σ S ) (cid:38) (cid:96) ( R ) d . Since the c B R are disjoint and Σ S is bi-Lipschitz, wehave I pS (cid:46) (cid:88) R ∈ Stop R ∩ C B Q ( S ) (cid:54) = ∅ H d ( c B R ∩ Σ S ) ≤ H d ( B Q ( S ) ∩ Σ S ) (cid:46) (cid:96) ( Q ( S )) d , completing the proof of (3.10) and hence the proof of the lemma. (cid:3) Lemma 3.9. (cid:88) S ∈ S (cid:96) ( Q ( S )) d (cid:46) H d ( Q ) + (cid:88) Q ∈ D ( Q ) ˇ β d, F ( M B Q ) (cid:96) ( Q ) d . Lemma 3.9 along with Lemma 3.7 finishes the proof of (3.1) (and hence finishesthe proof of Theorem 1.13). We prove Lemma 3.9 via the following two lemma.Let min S be the collection of minimal cubes from S , i.e.min S = (cid:91) S ∈ S { Q ∈ min( S ) } . Lemma 3.10. (cid:88) Q ∈ min S (cid:96) ( Q ) d (cid:46) H d ( Q ) + (cid:88) Q ∈ D ( Q ) ˇ β d, F ( M B Q ) (cid:96) ( Q ) d . Proof of Lemma 3.10.
We split min S into two sub families, F and F where F is the collection of cubes R in min S such that R has a child R (cid:48) with (cid:88) Q ∈ S ( R ) Q ⊇ R (cid:48) ˇ β d, F ( M B Q ) ≥ ε (3.18)and F = min S \ F (recall the definition of S ( R ) from Definition 3.2). We dealwith F first. Let R ∈ F and let R (cid:48) be its child satisfying (3.18). Note that ifˇ β d, F ( M B R (cid:48) ) ≤ ε / , then (cid:88) Q ∈ S ( R ) Q ⊇ R ˇ β d, F ( M B Q ) ≥ ε / . If, instead, ˇ β d, F ( M B R (cid:48) ) > ε / , we have (cid:88) Q ∈ S ( R ) Q ⊇ R ˇ β d, F ( M B Q ) ≥ ˇ β d, F ( M B R ) (cid:38) ˇ β d, F ( M B R (cid:48) ) ≥ ε / . In either case it follows that (cid:88) Q ∈ S ( R ) Q ⊇ R ˇ β d, F ( M B Q ) (cid:38) ε . Thus, (cid:88) R ∈ F (cid:96) ( R ) d (cid:46) ε (cid:88) R ∈ F (cid:96) ( R ) d (cid:88) Q ∈ S ( R ) Q ⊇ R ˇ β d, F ( M B Q ) (cid:46) (cid:88) S ∈ S (cid:88) Q ∈ S ˇ β d, F ( M B Q ) (cid:88) R ∈ S ∩ F R ⊆ Q (cid:96) ( R ) d . (3.19)In the case that S is not a singleton, by Lemma 3.6, dist( x R , Σ S ( R ) ) (cid:46) ε d +1 (cid:96) ( R ) . Thus, for ε >
0, small enough c B R carves out a large proportion of Σ S , hence H d ( c B R ∩ Σ S ( R ) ) (cid:38) (cid:96) ( R ) d . The balls c B R are disjoint and contained in B Q , so (cid:88) R ∈ S ∩ F R ⊆ Q (cid:96) ( R ) d (cid:46) (cid:88) R ∈ S ∩ F R ⊆ Q H d ( c B R ∩ Σ S ( R ) ) ≤ H d ( B Q ∩ Σ S ( R ) ) (cid:46) (cid:96) ( Q ) d . (3.20)If S is a singleton, Σ S was not defined but we have S = { Q } = { R } , so the aboveestimate holds trivially. In either case, combining (3.19) and (3.20) gives (cid:88) R ∈ F (cid:96) ( R ) d (cid:46) (cid:88) S ∈ S (cid:88) Q ∈ S ˇ β d, F ( M B Q ) (cid:96) ( Q ) d . (3.21)Let us consider F . For R ∈ F , we know there exists a child R (cid:48) of R and a point x (cid:48) ∈ R (cid:48) such that dist( x (cid:48) , E ) ≥ ε(cid:96) ( R (cid:48) ) . We let C R be the maximal cube containing x (cid:48) such that (cid:96) ( C R ) ≤ ερC (cid:96) ( R ) . This implies that ερ C (cid:96) ( R ) ≤ (cid:96) ( C R ) ≤ ερC (cid:96) ( R ) . (3.22)In this way, we have dist( y, E ) > y ∈ C B C R since ε(cid:96) ( R ) ≤ dist( x (cid:48) , E ) ≤ | x (cid:48) − y | + dist( y, E ) ≤ C (cid:96) ( C R ) + dist( y, E ) ≤ ερ(cid:96) ( R ) + dist( y, E ) . In particular, 2 C B C R ∩ E = ∅ . (3.23)Now, let x ∈ F, N ∈ N and suppose { R i } Ni =1 is a finite collection of distinct cubesin F such that x ∈ N (cid:92) i =1 C R i . SP FOR GENERAL SETS 37
We assume without loss of generality that (cid:96) ( R i ) > (cid:96) ( R j ) for all i < j. It followsthat (cid:96) ( R j ) ≥ ερ C (cid:96) ( R )(3.24)for all j = 2 , . . . , N. Otherwise, by (3.22) we have R N ⊆ C R which by (3.23) givesthat 2 C B R N ∩ E = ∅ . Thus, we arrive at a contradiction since E ∩ C B Q (cid:54) = ∅ forany Q ∈ F by our stopping time condition (2). From (3.24), it follows that N (cid:46) ε,C . (3.25)Now, since F is lower content regular, we get (cid:88) R ∈ F (cid:96) ( Q ) d (3.22) (cid:46) (cid:88) R ∈ F H d ( C R ) (3.25) (cid:46) H d ( Q ) . (3.26)Combining (3.21) and (3.26) completes the proof of the lemma. (cid:3) Proof of Theorem 1.14
Constructing F and its properties. Let X Ek be a sequence of maximal ρ k -separated nets in E and let D E be the cubes from Theorem 2.1 with respect tothese maximal nets. By scaling and translation, we can assume there is Q ∈ D E which contains 0 . Let M ≥ ε, α > Remark 4.1.
The constants M and ε may be different from the previous section.We shall choose M sufficiently large, ε and α shall be chosen sufficiently small.We wish to construct F using the Reifenberg parametrisation theorem of Davidand Toro. Since E is not lower regular, we need the condition that we have ( d +1 , α )-separated points to apply the construction. This will be added to our stopping timeconditions.For Q ∈ D E ( Q ), we define a stopping time region S Q as follows. First, add Q to S Q . Then, inductively on scales, we add a cube R to S Q if R (1) ∈ S Q and eachsibling R (cid:48) of R satisfies:(1) (cid:88) Q ⊇ T ⊇ R (cid:48) β d, E ( M B Q ) ≤ ε . (2) M B R (cid:48) has ( d + 1 , α ) separated points (in the sense of Definition 2.31).We partition D E ( Q ) into a collection of stopping time regions S : Begin by adding S Q to S . Then, if S has been added to S , add S Q to S if Q ∈ Child( R ) for some R ∈ min( S ) . We continue in this way to generate S . Letmin S = (cid:91) S ∈ S { Q ∈ min( S ) } denote the collection of all minimal cubes in S . Remark 4.2.
The following is essentially Lemma 3.6. The proof follows the sameif we take (cid:15) ≤ α d +4 . Lemma 4.3.
There exists ε > small enough so that for each S ∈ S , which isnot a singleton, there is a surface Σ S which satisfies the following. Firstly, if Σ (cid:48) S denotes the bi-Lipschitz surface from Theorem 3.5, then Σ S = Σ (cid:48) S ∩ M B Q ( S ) . (4.1) Second, for each R ∈ S and y ∈ F ∩ M B R , dist( y, Σ S ) (cid:46) ε d +1 (cid:96) ( R )(4.2) Finally, for each ball B centered on Σ S and contained in M B Q ( S ) , ω d r dB ≤ H d (Σ S ∩ B ) (cid:46) r dB . (4.3) Remark 4.4. If S = { Q } is a singleton, we define Σ S = P Q ∩ M B Q , where P Q issome d -plane through x Q such that β d,pE ( M B Q , P Q ) ≤ β d,pE ( M B Q ) . Let B α ⊆ D E ( Q ) be the set of cubes Q which have a sibling Q (cid:48) which fails thestopping time condition (2) i.e. M B Q (cid:48) does not have ( d + 1 , α )-separated points.Consider those points in Q for which we stopped a finite number of times or wenever stopped. That is, we consider G := x ∈ Q : (cid:88) β d, E ( M B Q ) χ Q ( x ) < ∞ and (cid:88) QQ ∈ B α χ Q ( x ) < ∞ . Observe that G ⊆ (cid:83) S ∈ S Σ S . We define E (cid:48) = Q \ G and set F := E (cid:48) ∪ (cid:91) S ∈ S Σ S . Lemma 4.5. F is ( c , d ) -lower content regular.Proof. Fix x ∈ F and r > . Assume first of all that x ∈ Σ S for some S ∈ S . Let Q = Q ( S ) . For any k ≥ , recalling that Q ( k ) is the k th generational ancestor of Q ,we have | x − x Q ( k ) | ≤ | x − x Q | + | x Q − x Q ( k ) | ≤ M (cid:96) ( Q ) + (cid:96) ( Q ( k ) ) ≤ M (cid:96) ( Q ( k ) )Assume first that r ≥ M (cid:96) ( Q ) and let k = k ( r ) ≥ M (cid:96) ( Q ( k ) ) ≤ r ≤ M (cid:96) ( Q ( k +1) ) . Then,
M B Q ( k ) ⊆ B ( x, M (cid:96) ( Q ( k ) ) + M (cid:96) ( Q ( k ) )) ⊆ B ( x, r ) . The lower regularity follow since H d ∞ ( F ∩ B ( x, r )) ≥ H d ∞ (Σ S ( Q ( k ) ) ) (4.3) ≥ ω d M (cid:96) ( Q ( k ) )) d ≥ ω d ρ r d ≥ c (2 r ) d . Assume now that r < M (cid:96) ( Q ( S )) . If B ( x, r ) ⊆ M B Q ( S ) we can trivially apply thelower regularity estimates for Σ S , so it suffices to consider the case when B ( x, r ) (cid:54)⊆ M B Q ( S ) . We split this into two further sub-cases: either B ( x, r ) ∩ B Q ( S ) = ∅ or B ( x, r ) ∩ B Q ( S ) (cid:54) = ∅ . In the first sub-case, since by (2.4), we have Σ S \ B Q ( S ) = P Q ( S ) \ B Q ( S ) , theportion of Σ S contained in B ( x, r ) is just a d -plane through x. Since x is contained SP FOR GENERAL SETS 39 in M B Q ( S ) , we can find y ∈ B ( x, r ) ∩ Σ S such that B ( y, r/ ⊆ M B Q ( S ) ∩ B ( x, r )and apply the lower regularity estimates for Σ S inside B ( y, r/
4) to obtain H d ∞ ( F ∩ B ( x, r )) ≥ H d ∞ (Σ S ∩ B ( y, r/ (4.3) ≥ ω d d +1 r d ≥ c (2 r ) d . See for example the left image in Figure 4.In the second sub-case, since B ( x, r ) (cid:54)⊆ M B Q ( S ) but B ( x, r ) ∩ B Q ( S ) (cid:54) = ∅ , itmust be that r is comparable with M (cid:96) ( Q ) . For M sufficiently large ( M ≥
100 issufficient), we can certainly find y ∈ B ( x, r ) ∩ Σ S such that B ( y, r/ ⊆ M B Q ( S ) ∩ B ( x, r ) and apply the lower regularity estimates for Σ S inside B ( y, r/
10) to get H d ∞ ( F ∩ B ( x, r )) ≥ H d ∞ (Σ S ∩ B ( y, r/ (4.3) ≥ ω d d +1 r d ≥ c (2 r ) d . See for example the right image in Figure 4.
Figure 4.
Sub-cases 1 and 2Suppose now that x ∈ E (cid:48) . Since E (cid:48) is the collection of points where we stoppedan infinite number of times, we may find a sequence of stopping time regions S i such that x ∈ Q ( S i ) and (cid:96) ( Q ( S i )) ↓ . We denote by S the stopping time region forwhich x ∈ S and (cid:96) ( Q ( S )) ≤ r/ . Then, by (4.2),dist( x, Σ S ) (cid:46) ε d +1 (cid:96) ( Q ( S )) ≤ ε d +1 r . Let x (cid:48) be the point in Σ S closest to x. For ε small enough, B ( x (cid:48) , r/ ⊆ B ( x, r ),which gives H d ( F ∩ B ( x, r )) ≥ H d ( F ∩ B ( x (cid:48) , r/ ≥ c r d , where the final inequality follows by the first case we considered. (cid:3) To finish the proof of Theorem 1.14, it remains to show (1.5). If the right handside of (1.5) is infinite then this proves Theorem 1.14. Hence from now on, we shallassume the right hand side of (1.5) is finite. We have the following.
Lemma 4.6.
Assume that (cid:88) Q ∈ D E ( Q ) β d, E ( M B Q ) (cid:96) ( Q ) d < ∞ . (4.4) Then H d ( E (cid:48) ) = 0 , from which it follows that F is rectifiable and H d ( F ) (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E ( Q ) β d, E ( M B Q ) (cid:96) ( Q ) d . (4.5)Before proving Lemma 4.6, we shall need two preliminary results. Lemma 4.7.
Let S ∈ S and Q ∈ S. For ε > small enough, we have (cid:88) R ∈ min SR ⊆ Q (cid:96) ( R ) d (cid:46) (cid:96) ( Q ) d . Proof.
Let R ∈ min S satisfy R ⊆ Q. By (4.2), the ball c B R carves out a largeproportion of Σ S , in particular, H d (Σ S ∩ c B R ) (cid:38) (cid:96) ( R ) d . Furthermore, the balls { c B R } are disjoint and contained in B Q . Using this, we have (cid:88) R ∈ min SR ⊆ Q (cid:96) ( R ) d (cid:46) (cid:88) R ∈ min SR ⊆ Q H d (Σ S ∩ c B R ) ≤ H d (Σ S ∩ B Q ) (cid:46) (cid:96) ( Q ) d . (4.6) (cid:3) Lemma 4.8. (cid:88) R ∈ min S (cid:96) ( R ) d (cid:46) (cid:96) ( Q ) d + (cid:88) S ∈ S (cid:88) Q ∈ S β d, E ( M B Q ) (cid:96) ( Q ) d . Proof.
We split min S into two sub families, Type I and Type II , where Type I is thecollection of cubes R ∈ min S such that each child R (cid:48) of R has ( d + 1 , α )-separatedpoints, but there is a child R (cid:48)(cid:48) such that (cid:88) Q ∈ S ( R ) Q ⊇ R (cid:48)(cid:48) β d, E ( M B Q ) ≥ ε and Type II = min S \ Type I . Controlling cubes in Type I is done in a similar way tohow we controlled the cubes in F in the proof of Lemma 3.10, the only differencebeing that we construct the surfaces Σ S by Lemma 4.3 instead of Lemma 3.6. Thisis because E is not necessary lower regular but we know each cube in a stoppingtime region (which is not a singleton) has ( d + 1 , α )-separated points. This gives (cid:88) Q ∈ Type I (cid:96) ( Q ) d (cid:46) (cid:88) Q ∈ D E ( Q ) β d, E ( M B Q ) (cid:96) ( Q ) d . We now consider cubes in Type II . We will in fact show that (cid:88) Q ∈ Type II (cid:96) ( Q ) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ Type I (cid:96) ( Q ) d (4.7)from which the lemma follows immediately. We state and prove three preliminaryclaims before we proceed with the proof of (4.7). SP FOR GENERAL SETS 41
Claim 1:
For all δ > α > Q ∈ Type II and k ∗ = (cid:22) log(2 αM ρ/c )log ρ (cid:23) , then (cid:88) R ∈ Child k ∗ ( Q ) (cid:96) ( R ) d ≤ δ(cid:96) ( Q ) d . (4.8)Let δ > Q ∈ Type II . By definition, there exists Q (cid:48) ∈ Child( Q )such that β d − E, ∞ ( M B Q (cid:48) ) ≤ α. If L is the ( d − β d − E, ∞ ( M B Q (cid:48) ) = β d − E, ∞ ( M B Q (cid:48) , L ) , then each R ⊆ Q satisfiesdist( x R , L ) ≤ αM (cid:96) ( Q (cid:48) ) = αM ρ(cid:96) ( Q ) . By our choice of k ∗ , if R ⊆ Q is such that (cid:96) ( R ) = ρ k ∗ (cid:96) ( Q ) then dist( x R , L ) ≤ c (cid:96) ( R ) . Thus, by Lemma 2.5, (cid:88) R ∈ Child k ∗ ( Q ) (cid:96) ( R ) d − ≤ C(cid:96) ( Q ) d − . Multiplying both sides by (cid:96) ( R ) , we obtain (cid:88) R ∈ Child k ∗ ( Q ) (cid:96) ( R ) d ≤ C(cid:96) ( Q ) d − (cid:96) ( R ) ≤ Cρ k ∗ (cid:96) ( Q ) d . By taking α > Cρ k ∗ < δ. This proves the claim.Let Q ∈ Type II. Define Type I ( Q ) to be the maximal collection of cubes inType I contained in Q . Let Tree( Q ) be the collection of cubes R ∈ D E such that R ⊆ Q and R is not properly contained in any cube from Type I ( Q ) andType II ( Q ) = Tree( Q ) ∩ Type II . We also define sequences of subsets of Tree( Q ), T k and M k as follows: Let T = { Q } and M = min S Q . Then, supposing that T k and M k have been defined for some integer k ≥ , we let T k +1 = { T ∈ Tree( Q ) : T ∈ Child k ∗ ( R ) for some R ∈ M k } and M k +1 = { T ∈ Type II ( Q ) : T is max so that T ⊆ R for some R ∈ T k +1 } Recalling the definition of Des k ( R ) from (2.2), we have the following: Claim 2:
Type II ( Q ) ⊆ { Q } ∪ ∞ (cid:91) k =0 (cid:91) R ∈ M k Des k ∗ − ( R ) . (4.9) Let R ∈ Type II ( Q ) . The claim is clearly true for R = Q so let us assume R (cid:54) = Q. Let k R be the largest integer k ≥ T ∈ M k with R ⊆ T. The existence of such a k is guaranteed since there exists ˜ T ∈ M such that R ⊆ ˜ T and each cube T ∈ M k satisfies (cid:96) ( T ) ≤ ρ kk ∗ (cid:96) ( Q ) . Let T R ∈ M k R be such that R ⊆ T R . Assume R ∈ Child k ( T R ) for some k ≥ k ∗ . If this is the case then we can finda cube R (cid:48) ∈ Child k ∗ ( T R ) such that R ⊆ R (cid:48) , recall by definition that R (cid:48) ∈ T k R +1 . By maximality this implies that there exists some R (cid:48)(cid:48) ∈ M k R +1 such that R ⊆ R (cid:48)(cid:48) ,which contradicts the definition of k R . It follows that R ∈ Des k ∗ − ( T R ) whichcompletes the proof of the claim. Claim 3:
There exists α > k ≥ , (cid:88) R ∈ M k (cid:96) ( R ) d (cid:46) (cid:18) (cid:19) k (cid:96) ( Q ) d . (4.10)The results holds for k = 0 by Lemma 4.7. Assume k ≥ k − ≥
0. We will show that it holds for k . Let R ∈ M k and S ( R ) bethe stopping time region S ∈ S such that R ∈ S. By maximality, then there exists T ∈ T k ∩ S ( R ) such that R ⊆ T. By Lemma 4.7, Claim 1 and the definitions of T k and M k , we have (cid:88) R ∈ M k (cid:96) ( R ) d ≤ (cid:88) T ∈ T k (cid:88) R ∈ M k R ⊆ T (cid:96) ( R ) d (4.6) ≤ C (cid:88) T ∈ T k (cid:96) ( T ) d (4.8) ≤ Cδ (cid:88) T ∈ M k − (cid:96) ( T ) d (cid:46) Cδ (cid:18) (cid:19) k − (cid:96) ( Q ) d . Choosing δ (and hence α ) small enough so that Cδ < / II ( Q ) . Suppose
R, R (cid:48) ∈ Tree( Q ) are such that R (cid:48) ⊂ R. Any cube T ∈ D such that R (cid:48) ⊂ T ⊆ R is either the child of a Type II cube or is contained insome stopping that is not a singleton. In either case β d − E, ∞ ( M B T ) ≤ α or β d, E ( M B T ) ≤ ε. So, by Lemma 2.7 (for α and ε small enough), (cid:88) T ∈ Des k ∗− ( R ) (cid:96) ( T ) d (cid:46) α,d,M (cid:96) ( R ) d . Combing all the above, we get (cid:88) R ∈ Type II ( Q ) (cid:96) ( R ) d (4.9) ≤ (cid:96) ( Q ) d + ∞ (cid:88) k =0 (cid:88) R ∈ M k (cid:88) T ∈ Des k ∗− ( R ) (cid:96) ( T ) d (2.10) (cid:46) (cid:96) ( Q ) d + ∞ (cid:88) k =0 (cid:88) R ∈ M k (cid:96) ( R ) d (4.10) ≤ (cid:96) ( Q ) d + ∞ (cid:88) k =0 (cid:18) (cid:19) k (cid:96) ( Q ) d (cid:46) (cid:96) ( Q ) d . SP FOR GENERAL SETS 43
Now, for each Q ∈ Type II , let R ( Q ) be the smallest cube in Type I which contains Q . By construction, if T ∈ Tree( Q ) for some Q ∈ Type II then R ( T ) = R ( Q ) . Giventhis, if R ∈ Type I , (cid:88) Q ∈ Type II R ( Q )= R (cid:96) ( Q ) d = (cid:88) Q ∈ Child( R ) ∩ Type II (cid:88) T ∈ Type II ( Q ) (cid:96) ( T ) d (cid:46) (cid:88) Q ∈ Child( R ) ∩ Type II (cid:96) ( Q ) d (cid:46) (cid:96) ( R ) d . It may be that there is a cube Q in Type II which is not contained in any cube fromType I . If this is the case, then Q ∈ Type II and Q ∈ Type II ( Q ) . In any case, weget (cid:88) Q ∈ Type II (cid:96) ( Q ) d ≤ (cid:88) Q ∈ Type II ( Q ) (cid:96) ( Q ) d + (cid:88) R ∈ Type I (cid:88) Q ∈ Type II R ( Q )= R (cid:96) ( Q ) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ Type I (cid:96) ( Q ) d . (cid:3) Proof of Lemma 4.6.
Let x ∈ E (cid:48) . By definition, for each r > Q ∈ min S such that (cid:96) ( Q ) < r and x ∈ Q. By induction, we can construct a sequenceof distinct covers C k ⊆ min S of E (cid:48) such that (cid:96) ( Q ) < k for each Q ∈ C k . With thefinite assumption (4.4), it must be thatlim k →∞ (cid:88) Q ∈ C k (cid:96) ( Q ) d = 0 . (4.11)Assume towards a contradiction that there exists δ > K ∈ N such that (cid:80) Q ∈ C k (cid:96) ( Q ) d > δ for all k ≥ K. By this and Lemma 4.8 it follows that (cid:96) ( Q ) d + (cid:88) S ∈ S (cid:88) Q ∈ S β d, E ( M B Q ) (cid:96) ( Q ) d (cid:38) (cid:88) Q ∈ min S (cid:96) ( Q ) d ≥ ∞ (cid:88) k =1 (cid:88) Q ∈ C k (cid:96) ( Q ) d = ∞ which contradicts (4.4) and proves (4.11). The fact that H d ( E (cid:48) ) = 0 follows from(4.11) since H d ( E (cid:48) ) = lim k →∞ H d k ( E (cid:48) ) ≤ lim k →∞ (cid:88) Q ∈ C k (cid:96) ( Q ) d = 0 . Furthermore H d ( F ) ≤ H d (cid:91) S ∈ S Σ S ≤ (cid:88) S ∈ S (cid:96) ( Q ( S )) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ min S (cid:88) R ∈ Child( Q ) (cid:96) ( R ) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ min S (cid:96) ( Q ) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E β d, E ( M B Q ) (cid:96) ( Q ) d , (4.12) which proves (4.5). The inequality from the second the third lines follows becauseany Q ∈ min S has at most K = K ( M, d ) children by Lemma 2.6. (cid:3)
Proof of (1.5).
Recall the definition of Q F from the statement of Theorem1.14. Just like at the beginning of the proof of Theorem 1.13, we can reduce proving(1.5) to proving diam( Q F ) d + (cid:88) Q ∈ D F Q ⊆ Q F ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d (4.13) (cid:46) diam( Q ) d + (cid:88) Q ∈ D E Q ⊆ Q β d, E ( AB Q ) (cid:96) ( Q ) d , that is, we can replace the constant C by some larger constant A and set p = 1on the right-hand side. The rest of this section is devoted to proving (4.13). In thestatement of Theorem 1.14 we assume diam( Q ) ≥ λ(cid:96) ( Q ) . We have the followingbound on the first term:diam( Q F ) ≤ (cid:96) ( Q F ) = (cid:96) ( Q ) (cid:46) λ diam( Q ) . So, in order to prove (4.13), it suffices to bound the second term. Let { S i } be anenumeration of the stopping time regions in S which are not singletons. First, weobserve that (cid:88) Q ∈ D E ( Q ) ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d = (cid:88) i (cid:88) Q ∈ S i ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d + (cid:88) S ∈ S S = { Q } ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d . If S ∈ S is such that S = { Q } then Q ∈ min S . Then, since ˇ β d,pF ( · ) (cid:46) , thesecond term on the right hand side of the above equation is at most some constantmultiple of (cid:88) Q ∈ min S (cid:96) ( Q ) d , which we bound by Lemma 4.8. Thus, to prove (4.13) it suffices to bound the firstterm. Using the β -error estimate (Lemma 2.29), we obtain (cid:88) i (cid:88) Q ∈ S i β d,pF ( C B Q ) (cid:96) ( Q ) d (2.31) (cid:46) (cid:88) i (cid:88) Q ∈ S i β d,pE (2 C B Q ) (cid:96) ( Q ) d + (cid:88) i (cid:88) Q ∈ S i (cid:32) (cid:96) ( Q ) d ˆ F ∩ C B Q (cid:18) dist( x, E ) (cid:96) ( Q ) (cid:19) p d H d ∞ ( x ) (cid:33) p (cid:96) ( Q ) d . We have a trivial bound for the first term so we now focus on bounding the secondterm, the proof of which is very similar to the proof of Lemma 3.7. First, let D (cid:48) = (cid:91) i S i . SP FOR GENERAL SETS 45
We split D (cid:48) into two families. Let δ > G = { Q ∈ D (cid:48) : dist( x, E ) ≤ δ(cid:96) ( Q ) for all x ∈ F ∩ M B Q } , B = D (cid:48) \ G . To each cubes Q ∈ B we shall assign a patch C Q of F . By definition, if Q ∈ B then there exists a point y Q ∈ F ∩ M B Q such that dist( y Q , E ) > δ(cid:96) ( Q ) . We define C Q = B ( y Q , δ(cid:96) ( Q ) / . We claim that the balls { C Q } Q ∈ B have bounded overlap in F . This is the contentof the following lemma. Lemma 4.9.
The collection of balls { C Q } Q ∈ B have bounded overlap in F .Proof. Let x ∈ F . We will show B x = { Q ∈ B : x ∈ C Q } (cid:46) . (4.14)First, note that C Q ⊆ M B Q for all Q ∈ B . Let
Q, Q (cid:48) ∈ B . If (cid:96) ( Q (cid:48) ) ≤ δ M (cid:96) ( Q ) and y ∈ C Q we have | y − x Q (cid:48) | ≥ | y Q − x Q (cid:48) | − | y − y Q | ≥ δ(cid:96) ( Q ) − δ(cid:96) ( Q )2 ≥ M (cid:96) ( Q (cid:48) ) , in particular, C Q ∩ M B Q (cid:48) = ∅ . Since C Q (cid:48) ⊆ M B Q (cid:48) , we must have that C Q ∩ C Q (cid:48) = ∅ . Reversing the role of Q and Q (cid:48) above, we conclude that if C Q ∩ C Q (cid:48) (cid:54) = ∅ then (cid:96) ( Q ) ∼ M,δ (cid:96) ( Q (cid:48) ) . In particular, (4.14) follows if we can show that for each k , { Q ∈ B ∩ D k : x ∈ C Q } (cid:46) , (4.15)with constant independent of k . Fix k and let Q, Q (cid:48) ∈ B ∩ D k such that C Q ∩ C Q (cid:48) (cid:54) = ∅ . Then M B Q ∩ M B Q (cid:48) (cid:54) = ∅ which implies | x Q − x Q (cid:48) | ≤ M (cid:96) ( Q ) . Since, β d,pE ( M B Q ) ≤ ε, we have dist( x Q (cid:48) , P Q ) (cid:46) ε d +1 (cid:96) ( Q ) = ε d +1 (cid:96) ( Q (cid:48) ) . For ε small enough, (4.15) followsfrom Lemma 2.5. (cid:3) Lemma 4.10.
We have (cid:88) i (cid:88) Q ∈ S i (cid:32) (cid:96) ( Q ) d ˆ F ∩ C B Q (cid:18) dist( x, E ) (cid:96) ( Q ) (cid:19) p d H d ∞ ( x ) (cid:33) p (cid:96) ( Q ) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E β d, E ( M B Q ) (cid:96) ( Q ) d . Proof.
By Jensen’s inequality (Lemma 2.18), we may assume p ≥
2. Let D (cid:48) , G and B be as above Lemma 4.9. First, since F is lower regular and the C Q have boundedoverlap, we get (cid:88) Q ∈ B (cid:32) (cid:96) ( Q ) d ˆ F ∩ C B Q (cid:18) dist( x, E ) (cid:96) ( Q ) (cid:19) p d H d ∞ ( x ) (cid:33) p (cid:96) ( Q ) d ≤ (cid:88) Q ∈ B (cid:96) ( Q ) d (cid:46) (cid:88) Q ∈ B H d ( C Q ) (4.14) (cid:46) H d ( F ) (4.5) (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E β d, E ( M B Q ) (cid:96) ( Q ) d . Consider a single S = S i . Let S G denote S ∩ G . Let D F denote the Christ-Davidcubes for F and let S F G be the smoothed out cubes in F with respect to S G . Define τ = 1 / ρ ) (as in (3.6)) then let S F G = { Q ∈ D F : Q is maximal with (cid:96) ( Q ) < τ d S G ( Q ) } . Let R ∈ S F G . Claim 1.
For each y ∈ R, dist( y, E ) (cid:46) (cid:96) ( R ) . By a direct analogue of Lemma 3.8 (whose proof is exactly the same), thereexists Q ∈ S G such that τ dist( Q, R ) (cid:46) (cid:96) ( R ) and τ (cid:96) ( Q ) ∼ (cid:96) ( R ) . Let y Q be the pointin Q closest to y. Then,dist( y, E ) ≤ dist( y, y Q ) + dist( y Q , E ) (cid:46) (cid:96) ( R ) τ + δ(cid:96) ( Q ) (cid:46) (cid:96) ( R ) . Claim 2.
We have dist( x R , Σ S ) (cid:46) ε d +1 τ (cid:96) ( R ) . Let Q ∈ S G be as in the proof of Claim 1. We can chose M large enough(depending on τ ) so that R ⊆ M B Q . Then, by Lemma 4.3 and the fact that (cid:96) ( R ) ∼ τ (cid:96) ( Q ) , dist( x R , Σ S ) (4.2) (cid:46) ε d +1 (cid:96) ( Q ) (cid:46) ε d +1 τ (cid:96) ( R ) . (4.16) Claim 3.
For each k ∈ N , { Q ∈ D Ek ∩ S G : R ∩ C B Q (cid:54) = ∅} (cid:46) . If Q ∈ D Ek ∩ S G is such that R ∩ C B Q (cid:54) = ∅ , then (cid:96) ( R ) ≤ τ ( (cid:96) ( Q ) + dist( R, Q )) (cid:46) (cid:96) ( Q ) . Let x (cid:48) R be the point in Σ S closest to x R . By the above and (4.16) there exists aconstant
A > Q ∈ D k ∩ S is such that R ∩ C B Q (cid:54) = ∅ then Q ⊆ B ( x (cid:48) R , A(cid:96) ( Q )) := B. Since dist( x Q , Σ S ) (cid:46) ε d +1 (cid:96) ( Q ) , the balls c B Q carve out a large proportion of Σ S . Then, by (4.3), (cid:88) Q ∈ S G R ∩ C B Q (cid:54) = ∅ (cid:96) ( Q ) d (cid:46) (cid:88) Q ∈ S G R ∩ C B Q (cid:54) = ∅ H d (Σ S ∩ c B Q ) ≤ H d (Σ S ∩ B ) (cid:46) (cid:96) ( Q ) d , which proves the claim. SP FOR GENERAL SETS 47
Since we have assumed p >
2, we apply Jensen’s inequality (Lemma 2.18) andClaim 1 to get I := (cid:88) Q ∈ S G (cid:32) (cid:96) ( Q ) d ˆ F ∩ C B Q (cid:18) dist( x, E ) (cid:96) ( Q ) (cid:19) p d H d ∞ ( x ) (cid:33) p (cid:96) ( Q ) d (cid:46) (cid:88) Q ∈ S G (cid:88) R ∈ S F G R ∩ C B Q (cid:54) = ∅ (cid:96) ( R ) d + p (cid:96) ( Q ) d + p p (cid:96) ( Q ) d (cid:46) (cid:88) Q ∈ S G (cid:88) R ∈ S F G R ∩ C B Q (cid:54) = ∅ (cid:96) ( R ) d p +2 (cid:96) ( Q ) d ( p − By Claim 3, we swap the order of integration and sum over a geometric series toget, I (cid:46) (cid:88) R ∈ S F G R ∩ C B Q ( S ) (cid:54) = ∅ (cid:88) Q ∈ S G R ∩ C B Q (cid:54) = ∅ (cid:96) ( R ) d p +2 (cid:96) ( Q ) d ( p − (cid:46) (cid:88) R ∈ S F G R ∩ C B Q ( Si ) (cid:54) = ∅ (cid:96) ( R ) d . By (4.16), for ε small enough, the ball c B R carves out a large proportion of Σ S for each R ∈ S F G , i.e. H d ( c B R ∩ Σ S ) (cid:38) (cid:96) ( R ) d . By (4.3), using the fact the c B R are disjoint, we have I (cid:46) (cid:88) R ∈ S F G R ∩ C B Q ( S ) (cid:54) = ∅ H d ( c B R ∩ Σ S ) ≤ H d ( B Q ( S ) ∩ Σ S ) (cid:46) (cid:96) ( Q ( S )) d . Hence, (cid:88) i (cid:88) Q ∈ S i (cid:32) (cid:96) ( Q ) d ˆ F ∩ C B Q (cid:18) dist( x, E ) (cid:96) ( Q ) (cid:19) p d H d ∞ ( x ) (cid:33) p (cid:96) ( Q ) d (cid:46) (cid:88) S ∈ S (cid:96) ( Q ( S )) d (4.12) (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E β d, E ( M B Q ) (cid:96) ( Q ) d (cid:3) We have proved so far that(4.17) diam( Q F ) d + (cid:88) Q ∈ D E ( Q ) ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d (cid:46) diam( Q ) d + (cid:88) Q ∈ D E ( Q ) β d, E ( M B Q ) (cid:96) ( Q ) d . In order to finish the proof of Theorem 1.14, we wish to prove the same inequalitybut with the sum on the left hand side of (4.17) taken over all cubes in D F ( Q F ) . We do this by partitioning the cubes in F into those which lie close to E andthose which do not. We control the sum over F -cubes which lie close to E by thecorresponding sum over E -cubes and shall control the sum over the F -cubes faraway from E by Theorem 1.7. We define a Whitney decomposition of F \ E. For k ≥ , we let B Ek = (cid:91) Q ∈ D Ek M B Q . Lemma 4.11.
Define
Top F = { Q ∈ D F ( Q F ) : Q is max such that ( C + M ) (cid:96) ( Q ) < dist( x Q , E ) } . Let k ∈ N and Q ∈ Top F ∩ D Fk . Then C B Q ∩ B El = ∅ for all l ≥ k (4.18) and C B Q ⊆ B El for all ≤ l ≤ k − Proof.
Let Q ∈ Top F ∩ D Fk . By definition, (4.18) is immediate so let us prove(4.19). By maximality, we havedist( x Q , E ) ≤ | x Q − x Q (1) | + dist( x Q (1) , E ) ≤ (1 + C + M ) ρ − (cid:96) ( Q ) . (4.20)Let z Q be the point in E closest to x Q and let Q (cid:48) be the cube in B Ek − such that z Q ∈ Q (cid:48) . Then, for y ∈ C B Q , we have | y − x Q (cid:48) | ≤ | y − x Q | + | x Q − z Q | + | z Q − x Q (cid:48) | (4.20) ≤ (cid:96) ( Q ) + (1 + C + M ) ρ − (cid:96) ( Q ) + (cid:96) ( Q (cid:48) ) ≤ M ρ − (cid:96) ( Q ) = 3 M (cid:96) ( Q (cid:48) )This implies y ∈ M B Q (cid:48) . Since y is arbitrary point in C B Q , we have C B Q ⊆ M B Q (cid:48) ⊆ B Ek − . Clearly, this also implies that C B Q ⊆ B El for all 0 ≤ l ≤ k − . (cid:3) Lemma 4.12.
The collection of balls { B Q } Q ∈ Top F have bounded overlap with con-stant dependent on n .Proof. Let x ∈ R n and let Q x = { Q ∈ Top F : x ∈ B Q } . We first show that each cube in Q x has comparable size. Let Q, Q (cid:48) ∈ Q x and assumewithout loss of generality that (cid:96) ( Q ) ≤ (cid:96) ( Q (cid:48) ) . Since Q (cid:48) ∈ Top F and B Q ∩ B Q (cid:48) (cid:54) = ∅ we have (cid:96) ( Q (cid:48) ) ≤ ( C + M ) − dist( x Q (cid:48) , E ) ≤ ( C + M ) − ( | x Q − x Q (cid:48) | + dist( x Q , E )) ≤ ( C + M ) − (2 (cid:96) ( Q (cid:48) ) + dist( x Q , E )) . Taking M large enough so that 2( C + M ) − ≤ , we can rearrange the aboveequation to give (cid:96) ( Q (cid:48) ) ≤ C + M dist( x Q , E ) (4.20) (cid:46) (cid:96) ( Q )which proves that (cid:96) ( Q ) ∼ (cid:96) ( Q (cid:48) ) . (4.21) SP FOR GENERAL SETS 49
By a standard volume argument, for each k ∈ N we have Q x ∩ D k ) (cid:46) n , which when combined with (4.21) finishes the proof of the lemma. (cid:3) Lemma 4.13. (cid:88) Q ∈ Top F (cid:88) R ⊆ Q ˇ β d,pF ( C B R ) (cid:96) ( R ) d (cid:46) (cid:96) ( Q E ) d + (cid:88) Q ∈ D E β d,pE ( M B Q ) (cid:96) ( Q ) d . Proof.
Let Q ∈ Top F ∩ D Fk and let S Q ⊆ S be the collection of stopping timeregions such that C B Q ∩ Σ Q ( S ) (cid:54) = ∅ . Since C B Q ∩ B El = ∅ for all l ≥ k, by (4.18),it must be that Q ( S ) ∈ D El for some 0 ≤ l ≤ k − . By (4.19) it follows that C B Q ⊆ M B Q ( S ) for each S ∈ S Q . We wish to use that fact that in each of these balls, F is wellapproximated by a union of planes. At the minute this is not quite the case. Itcould be that C B Q intersects M B Q ( S ) at the boundary for some S ∈ S Q (see forexample Figure 5). As such we must extend each of the surfaces Σ S . The resultingunion of these extended surfaces still has comparable measure to F . Figure 5. F is not necessarily well approximated by a union of planes.Recall from Lemma 4.3 that Σ (cid:48) S is the unbounded bi-Lipschitz surface fromTheorem 3.5. For each S ∈ S , let ˜Σ S be the surface obtained by restricting Σ (cid:48) S to6 M B Q ( S ) , i.e. ˜Σ S = Σ (cid:48) S ∩ M B Q ( S ) . Compare this to how we define Σ S in (4.1). This ensures that each ˜Σ S is ( Cε, d )-Reifenberg flat in 3
M B Q ( S ) . Clearly we have F ∩ C B Q ⊆ E (cid:48) ∪ (cid:91) S ∈ S Q ˜Σ S ∩ C B Q . Define F Q = E (cid:48) ∪ (cid:91) S ∈ S Q ˜Σ S ∪ (cid:91) S ∈ S \ S Q Σ S . We can show that F Q is lower regular in exactly the same way as we did for F (weshall omit the details). Let D F Q denote the cubes for F Q from Theorem 2.1. Inthis way, for each R ∈ D F there exists a corresponding cube ˜ R ∈ D F Q such that x R = x ˜ R and (cid:96) ( R ) = (cid:96) ( ˜ R ) . It is clear then, that C B R = C B ˜ R . Now, let R ⊆ Q and let ˜ R ∈ D F Q be the cube described above. If S ∈ S Q and˜Σ S ∩ C B ˜ R (cid:54) = ∅ , let x S ∈ ˜Σ S ∩ C B ˜ R be a point of intersection. Since C B ˜ R ⊆ B ( x S , C (cid:96) ( ˜ R )) ⊆ M B Q ( S ) and ˜Σ S is ( Cε, d )-Reifenberg flat in 3
M B Q ( S ) , we can find a plane L S through x S such that d C B ˜ R ( ˜Σ S , L S ) ≤ Cε.
Let U ˜ R := (cid:91) S ∈ S Q C B ˜ R ∩ ˜Σ S (cid:54) = ∅ L S . Since by construction we have F Q ∩ C B ˜ R = (cid:91) S ∈ S Q C B ˜ R ∩ ˜Σ S (cid:54) = ∅ ˜Σ S , it follows that d C B ˜ R ( F Q , U ˜ R ) ≤ Cε, i.e. ˜ R (cid:54)∈ BAUP( C , Cε ). See the below Figure 6. Since Q ∈ Top F and R ⊆ Q Figure 6.
An illustration of the above argument for R = Q .Compare the extended surface shown above to the original sur-face shown in Figure 5.were arbitrary, we have BAUP( ˜ Q, C , Cε ) = 0 for all Q ∈ Top F . Using that fact
SP FOR GENERAL SETS 51 that F ⊆ F Q , the correspondence between cubes in D F and D F Q and Theorem 1.7,we get (cid:88) R ⊆ Q ˇ β d,pF ( C B R ) (cid:96) ( R ) d ≤ (cid:88) ˜ R ⊆ ˜ Q ˇ β d,pF Q ( C B ˜ R ) (cid:96) ( ˜ R ) d (1.3) (cid:46) H d ( ˜ Q ) . Since by Lemma 4.12 the collection of balls { B Q } Q ∈ Top F have bounded overlap,the same is true for the cubes { ˜ Q } Q ∈ Top F . If we define˜ F = E (cid:48) ∪ (cid:91) S ∈ S ˜Σ S , then ˜ Q ⊆ ˜ F F for all Q ∈ Top F which gives (cid:88) Q ∈ Top F (cid:88) R ⊆ Q ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d (cid:46) (cid:88) Q ∈ Top F H d ( ˜ Q ) (cid:46) H d ( ˜ F ) (cid:46) (cid:88) S ∈ S (cid:96) ( Q ( S )) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E ( Q ) β d, E ( M B Q ) (cid:96) ( Q ) d , where the last inequality follows from (4.12). (cid:3) Lemma 4.14.
Let Up F be the collection of cubes which are not properly containedin any cube from Top F . Then (cid:88) Q ∈ Up F ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E ( Q ) β d, E (2 M B Q ) (cid:96) ( Q ) d . Proof.
Let Q ∈ Up F ∩ D Fk . By construction, we have dist( x Q , E ) < ( C + M ) (cid:96) ( Q )so there exists a cube Q (cid:48) ∈ D Ek such that C B Q ∩ M B Q (cid:48) (cid:54) = ∅ . In particular C B Q ⊆ M B Q (cid:48) , which by Lemma 2.24 implies β d,pF ( C B Q ) (cid:46) β d,pF ( M B Q (cid:48) ) . (4.22)For Q ∈ D E , a standard volume argument gives { R ∈ Up F : R (cid:48) = Q } (cid:46) n . (4.23)Then, (cid:88) Q ∈ Up F ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d ≤ (cid:88) Q ∈ D E ( Q ) (cid:88) R ∈ Up F R (cid:48) = Q ˇ β d,pF ( C B R ) (cid:96) ( R ) d (4.22)(4.23) (cid:46) n (cid:88) Q ∈ D E ( Q ) ˇ β d,pF (2 M B Q ) (cid:96) ( Q ) d (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E ( Q ) ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d (4.17) (cid:46) (cid:96) ( Q ) d + (cid:88) Q ∈ D E ( Q ) β d, E (2 M B Q ) (cid:96) ( Q ) d . The third inequality follows for the following reason: For Q small enough (depend-ing on C and M ) we can find a larger cube Q (cid:48) such that M B Q ⊆ C B Q (cid:48) . We usethis along with the fact that any cube has a bounded number of descendants upto the K th generation, say, with constant dependent on n and K . The sum of thelarger cubes is absorbed into the first term since again we can control the numberof these cubes. This is what we did in the proof of Lemma 3.1, see there for moredetails. (cid:3) The proof of Theorem 1.14 is finished by noting that (cid:88) Q ∈ D F ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d = (cid:88) Q ∈ Up F ˇ β d,pF ( C B Q ) (cid:96) ( Q ) d + (cid:88) Q ∈ Top F (cid:88) R ⊆ Q ˇ β d,pF ( C B R ) (cid:96) ( R ) d and applying Lemma 4.13 and Lemma 4.2. References [AS18] Jonas Azzam and Raanan Schul. An analyst’s traveling salesman theorem for sets ofdimension larger than one.
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Matthew Hyde, School of Mathematics, University of Edinburgh, JCMB, Kings Build-ings, Mayfield Road, Edinburgh, EH9 3JZ, Scotland.
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