Structure of sets with nearly maximal Favard length

Let $E \subset B(1) \subset \mathbb R^{2}$ be an $\mathcal{H}^{1}$ measurable set with $\mathcal{H}^{1}(E)<\infty$, and let $L \subset \mathbb R^{2}$ be a line segment with $\mathcal{H}^{1}(L) = \mathcal{H}^{1}(E)$. It is not hard to see that $\mathrm{Fav}(E) \leq \mathrm{Fav}(L)$. We prove that in the case of near equality, that is, $$ \mathrm{Fav}(E) \geq \mathrm{Fav}(L) - \delta, $$ the set $E$ can be covered by an $\epsilon$-Lipschitz graph, up to a set of length $\epsilon$. The dependence between $\epsilon$ and $\delta$ is polynomial: in fact, the conclusions hold with $\epsilon = C\delta^{1/70}$ for an absolute constant $C>0$.


INTRODUCTION
Let E Ă R 2 be H 1 measurable with H 1 pEq ă 8.We recall the definition of Favard length: Here π θ : R 2 Ñ R is the orthogonal projection π θ pxq " x ¨pcos θ, sin θq.The definition of FavpEq can be posed without the assumption H 1 pEq ă 8, but this hypothesis will be crucial for most of the statements below, and it will be assumed unless otherwise stated.A fundamental result in geometric measure theory is the Besicovitch projection theorem [2] which relates Favard length and rectifiability: FavpEq ą 0 if and only if H 1 pE XΓq ą 0 for some Lipschitz graph Γ Ă R 2 -in other words, E is not purely 1-unrectifiable.
The proof of the Besicovitch projection theorem is famous for being difficult to quantify, partly because of its reliance on the Lebesgue differentiation theorem: it is hard to decipher from the argument just how large the intersection E X Γ is, and what the Lipschitz constant of Γ is.In fact, it is non-trivial to even find the right question: for example, if E Ă Bp1q, H 1 pEq " 1, and FavpEq ě δ for some small but fixed constant δ ą 0, then it is not true that H 1 pE X Γq ě for some ´1-Lipschitz graph Γ Ă R 2 , where " pδq ą 0. We construct a relevant counterexample in Section 6.
In Theorem 1.1, we show that similar counterexamples are no longer possible if the assumption "FavpEq ě δ" is upgraded to "FavpEq ě 2 H 1 pEq ´δ" for a sufficiently small constant δ ą 0. The number "2" comes from the fact that Favpr0, 1s ˆt0uq " 2 and that r0, 1s ˆt1u has the maximal Favard length among sets of length unity (see (2.4)).
Theorem 1.1.For every ą 0 there exists δ ą 0 such that the following holds.Let E Ă Bp1q be an H 1 measurable set with H 1 pEq ă 8, and assume that where L Ă R 2 is a line segment with H 1 pLq " H 1 pEq.Then, there exists an -Lipschitz graph Γ Ă R 2 such that H 1 pE X Γq ě H 1 pEq ´ .One can take δ " 70 {C for an absolute constant C ą 1.
Thus, if FavpEq is nearly maximal, the Besicovitch projection theorem can be quantified in a very strong way, whereas the example constructed in Section 6 shows that any similar conclusion fails completely if we make the weaker assumption FavpEq ě δ.However, it remains plausible that the assumption FavpEq ě δ is sufficient to guarantee a quantitative version of Besicovitch's theorem under the additional assumption that E is 1-Ahlfors regular, or satisfies other "multi-scale 1-dimensionality" hypotheses.For recent partial results, and more discussion on this question, see [8,17,21,24].The problem is closely related to Vitushkin's conjecture [25] on the connection between analytic capacity and Favard length, see [6,9].
We briefly mention another closely related topic: if E Ă R 2 is self-similar and purely 1-unrectifiable, then FavpEq " 0 by the Besicovitch projection theorem.It is an interesting and very popular question to attempt quantifying the (sharp) rate of decay at which FavpE n q Ñ 0, where E n is the "n th iteration" of the self-similar set.For recent developments, see [1,3,4,5,7,16,14,15,20,22].
1.1.Outline of the paper.A quick outline of the article is as follows: in Section 2 we introduce Crofton's formula and prove that line segments maximise Favard length.In Section 3 we show how to prove Theorem 1.1 using two main propositions, Proposition 3.3 and Proposition 3.10.The former allows us to cover a set with almost maximal Favard length by a bounded number of Lipschitz graphs with small constant.The latter says that, in fact, there can only be one such graph.These two propositions are then proven in Section 4 and Section 5, respectively.Section 6 contains the counterexample mentioned above Theorem 1.1.Finally, in Appendix A we give an exact formula for the measure of lines spanned by two rectifiable curves -this is used in Section 5 but it might be of independent interest.

Acknowledgements.
The paper was written while the authors were visiting the Hausdorff Research Institute for Mathematics in Bonn during the research trimester Interactions between Geometric measure theory, Singular integrals, and PDE.We would like to thank the institute and its staff for creating this opportunity for collaboration.

MEASURE THEORETIC PRELIMINARIES
2.1.Notation.For x P R d and r ą 0, the notation Bpx, rq stands for a closed ball of radius r centred at x.For A Ă R d , we denote the cardinality of A by #A, and we write Aprq :" tx P R d : distpx, Aq ď ru, where "dist" is Euclidean distance.For f, g ě 0, we write f g if there exists an absolute constant C ą 0 such that f ď Cg.The notation f g means the same as g f , and f " g is shorthand for f g f .If the constant C ą 0 is allowed to depend on some parameter "p", we signify this by writing f p g.

2.2.
Integralgeometry and Crofton's formula.One of the main tools is Crofton's formula for rectifiable sets, which states the following.
The equation (2.1) is false without the rectifiability assumption, but the inequality "ě" remains valid in this case.This formula (and the inequality) is a special case of a more general relation between Hausdorff measure and integralgeometric measure for n-rectifiable sets in R d , see Federer's paper [11,Theorem 9.7], or [12,Theorem 3.2.26].We next rephrase the formula (2.1) in slightly more abstract terms.We define the following measure η on the family A :" Ap2, 1q of all affine lines in R 2 : ηpLq " With this notation, the Crofton formula (2.1) can be rewritten as where LpEq :" t P A : E X ‰ Hu.

Lemma 2.3 (The line segment maximizes Favard length)
. and If E is rectifiable, then equality holds in (2.5).
If we replace E with the line segment L, then equality holds in both inequalities above.Thus, FavpLq " 2H 1 pLq " 2H 1 pEq, which combined with (2.6) (for E) proves (2.5).Next, (2.4) follows from the fact that the right-hand side of (2.5) is nonnegative.Finally, if E is rectifiable, then the second inequality in (2.6) becomes an equality, which implies that equality holds in (2.5).

Coarea formula.
We then record another tool in the proof of Theorem 1.1.It is closely related to Crofton's formula, but only considers the intersections with lines in a fixed direction.The price to pay is that the tangent of the rectifiable set enters the formula.It is a generalization of the following standard fact: Lemma 2.7 (Coarea formula).Let α ą 0. Let E Ă R 2 be a countable union of α-Lipschitz graphs over the x-axis.Then, for all H 1 measurable subsets A Ă E. (Recall that π 0 : R 2 Ñ R is the projection onto the x-axis.) Proof.This follows from the coarea formula for rectifiable sets.(See, e.g., [12,Theorem 3.2.22]or [13, Theorem 5.4.9].)

PROOF OF THEOREM 1.1 IN TWO MAIN STEPS
In this section we prove our main result using Proposition 3.3 and Proposition 3.10 introduced below.The former says that we can cover all of E, save for a tiny exceptional set, by a union of boundedly many Lipschitz graphs with small constant.The latter says that, in fact, there can be only one Lipschitz graph with small constant covering most of E, otherwise we run into contradiction with the assumption of almost maximal Favard length.

3.1.
Step 1.First reductions.Let E Ă R 2 be a Borel set with H 1 pEq ă 8.We start with the following simple lemma.Lemma 3.1.It suffices to prove Theorem 1.1 under the additional assumption that E is a finite union of disjoint C 1 curves.
Proof.We may assume that E Ă Bp1q is rectifiable, because by the Besicovitch projection theorem, the rectifiable part of E continues to satisfy all the assumptions of Theorem 1.1 (with the same constant δ ą 0).By this assumption, H 1 almost all of E can be covered by a countable union of C 1 -curves.Decomposing the curves further, we may assume that they are disjoint, and for any given η ą 0 we may write where H 1 pSq ď η, and H 1 pE X γ j q ě p1 ´ηqH 1 pγ j q.Now, the set Ē :" Ť M 1 j"1 γ j satisfies H 1 p Ēq ď p1 ´ηq ´1H 1 pEq and Favp Ēq ě FavpEq ´η, and is additionally a finite union of disjoint C 1 -curves.If Theorem 1.1 is already known under this additional assumption, we may now infer that H 1 p Ē z Γq ď , where Γ is an -Lipschitz graph.But then also H 1 pE z Γq ď H 1 pE z Ēq`H 1 p Ē z Γq ď η ` , and Theorem 1.1 follows for E by choosing the parameters , η appropriately.

3.2.
Step 2. Minigraphs and how to merge them.By Lemma 3.1, in the sequel we may assume that E is a finite union of disjoint C 1 -curves γ 1 , . . ., γ M 1 .We further chop up each curve γ j into connected pieces whose tangent varies by less than α, where α is a small constant depending on fixed later on (see (3.5)).At this point, we have managed to write E as a finite union of disjoint α-Lipschitz graphs γ 1 , . . ., γ M 1  1 , where M 1 ď M 1 1 ă `8.Each of these graphs will be called a minigraph, and their collection is denoted E. The main task in Theorem 1.1 is to combine the minigraphs into bigger graphs.
To begin with, each of the minigraphs is an α-Lipschitz graph over some line of the form The vector v k :" pcospkπ{M 2 q, sinpkπ{M 2 qq will be called the direction of the minigraph (if there are several suitable vectors for one minigraph, fix any one of them; we will only need to know that each minigraph is an α-Lipschitz graph over the line spanned by its direction).Statements about the (relative) angles of minigraphs should always be interpreted as statements about the relative angles of the direction vectors v k .For k P t0, . . ., M 2 u fixed, we write E k Ă E for the subset of minigraphs with direction v k .We suggest that the reader visualise the minigraphs as line segments I with =pI, spanpv k qq ď α.It seems likely that Theorem 1.1 could be reduced to the case where E is a finite union of line segments, but employing the minigraphs seems to spare us some unnecessary steps.
We write E k :" YE k .Thus It turns out that, except for a small error, each set E k is covered by a single Lipschitz graph with constant " α over spanpv k q.Indeed, note that Lemma 2.3 and (1.2) together imply ş LpEq #pE X q ´1 dηp q ď δ.Then we have the following proposition, whose proof will be carried out in Section 4.

Proposition 3.3.
There exist absolute constants C 0 , α 0 P p0, 1q and C lip ą 1 such that the following holds.Let δ, P p0, 1q and α P p0, α 0 q be such that δ ď C 0 α 3 2 .Let E Ă Bp1q be a set with H 1 pEq ă 8 of the form where E is a finite collection of disjoint α-Lipschitz graphs over a fixed line L Ă R 2 .Assume further that E satisfies ż LpEq #pE X q ´1 dηp q ď δ. (3.4) Then, there exists a C lip α-Lipschitz graph Γ over L, such that H 1 pE z Γq ď .

3.3.
Step 3.There can only be one graph.In Proposition 3.3 we managed to pack a majority of each set E j (as defined in (3.2)) to a Lipschitz graph of constant " α, up to errors which tend to zero as δ Ñ 0 in the main assumption (1.2).However, at this point there might be up to " α ´1 distinct Lipschitz graphs, and to prove Theorem 1.1, we would (roughly speaking) like to reduce their number to one.That this should be possible is not hard to believe: if E consists of several distinct Lipschitz graphs of substantial measure, which nevertheless cannot be fit into a single Lipschitz graph, then FavpEq cannot possibly be maximal.
We turn to the details.We recall the "given" constant ą 0 from the statement of Theorem 1.1, and we set for a sufficiently large absolute constant C thm ą 1.We define also α :" for some universal C alp ą 1.The universal constant C thm will depend on C alp , whereas C alp depends only on C lip and another constant C sep , which is introduced below.We record that Recall, once more, the decompositions E " E 0 Y . . .Y E M 2 and E " E 0 Y . . .Y E M 2 from the previous subsection: this decomposition depends on the parameter α fixed above.In addition to the decomposition E " E 0 Y . . .Y E M 2 , we will also need another, coarser, decomposition of E in this section.Write κ :" 1  10 , fix M 3 " α ´κ, and decompose E " F 0 Y . . .Y F M 3 in such a way that ‚ each F k is a union of finitely many consecutive families E j , and ‚ F k contains those minigraphs whose direction makes an angle ď α κ with w k " pcospkπ{M 3 q, sinpkπ{M 3 qq, for 0 ď k ď M 3 .We write At this point, we consider two distinct cases.Let C sep be a large constant depending only on the absolute constant C lip appearing in Proposition 3.3 (the letters "sep" stand for "separation").Thus, the constant C sep is also absolute, and we may (and will) assume that C alp is large relative to C sep .
Case 1.Given the constant ą 0 from Theorem 1.1, the first case is that we can find consecutive sets F k , F k`1 , . . ., F k`Csep with the property In this case we note that F :" F k Y. ..YF k`Csep is a union of minigraphs whose directions are within C sep α κ of the fixed vector w k .In particular, F can be expressed as a union of finitely many disjoint α 0 -Lipschitz graphs over the line spanpw k q, with α 0 " C sep α κ .This will place us in a positions to use Proposition 3.3 (with E replaced by F and α replaced by α 0 ).Of course also ż LpF q #pF X q ´1 dηp q ď ż LpEq #pE X q ´1 dηp q ď δ, so the analogue of the assumption (3.4) is valid for F in place of E. We also note that so if C thm is sufficiently large relative to C alp , then the hypothesis in Proposition 3.3 on the relation between δ, α 0 , and is satisfied (the constant C sep is large, so it can be safely ignored here).Consequently, there exists a Lipschitz graph and consequently H 1 pE z Γq ď 2 .By choosing C alp sufficiently large relative to C sep and C lip , we may ensure that Γ is an -Lipschitz graph, as desired.
Case 2. We then move to consider the other option, where E cannot be exhausted, up to measure , by a constant number of consecutive sets F k , F k`1 , . . ., F k`Csep .Since (3.7) fails for every k, we may find an index pair k, l P t0, . . ., M 3 u with |k ´l| ě C sep such that H 1 pF k q ě α 2κ and H 1 pF l q ě α 2κ .
(3.8)This follows immediately from the pigeonhole principle, recalling that the cardinality of the pieces F k is α ´κ, and also that α κ is much smaller than by (3.5).
Remark 3.9.Recall that the "separation" constant C sep above has been chosen to be large relative to the constant C lip in Proposition 3.3: morally, if Γ 1 , Γ 2 are two C lip α κ -Lipschitz graphs over lines L 1 , L 2 with =pL 1 , L 2 q ě C sep α κ , we need to know that Γ 1 and Γ 2 are still "transversal" (their tangents form angles ě 1 2 C sep α κ with each other).The next key proposition will imply that Case 2 cannot happen: Proposition 3.10.Suppose that C sep ą 0 is sufficiently large, and suppose that there are k, l P t0, . . ., M 3 u with |k ´l| ě C sep such that As we recorded in (3.6), we have α 7 " C thm C ´70 alp ¨δ.Thus, if C thm is chosen sufficiently large relative to C alp and the implicit absolute constants in (3.11), then (3.11) would lead to the contradiction δ ě ż LpEq #pE X q ´1 dηp q ą δ.
(For the first inequality, recall (2.5) and our main assumption (1.2).) Thus, with the choices of constants specified in this section, Case 2 cannot occur.This concludes the proof of Theorem 1.1.
In the next two sections we prove the two key results used above, Propositions 3.3 and 3.10.

PROOF OF PROPOSITION 3.3
Let E Ă R 2 be as in the proposition.With no loss of generality, we may assume that L is the x-axis, so the minigraphs in E are roughly horizontal.We introduce further notation.We write C β :" tpx, yq P R 2 : |y| ě β|x|u, β ą 0. Thus, the smaller the β, the wider the cone.We also write With this notation, if a set Γ Ă R 2 satisfies Γ X C β pxq " txu for all x P Γ, then Γ is (a subset of) a β-Lipschitz graph.Thus, in view of Proposition 3.3, it would be desirable to show that E X C C lip α pxq " txu for all x P E. In reality, we will prove a similar statement about a subset of E (of nearly full length).It is worth noting that a toy version of these statements is already present in our hypotheses: each minigraph γ P E is an α-Lipschitz graph over the x-axis.
Define the maximal conical density Lemma 4.1 says that points of high conical density are negligible, whereas Lemma 4.18 says that points of low conical density can be mostly contained in a Lipschitz graph.
If is a line, we let pwq denote the tube that is the w-neighborhood of .For a tube T " pwq, we denote wpT q " w.
To prove Lemma 4.1, we use the Besicovitch alternative: Then for all x P E and H ě 1, at least one of the following two alternatives holds: (A1) There exists a set (A2) There exists a set J x Ă Jpβq of measure H 1 pJ x q H ´1 and the following property: for every θ P J x , there is a tube T " T x,θ " x,θ pwpT qq centred around x,θ such that This alternative is part of Besicovitch's original argument [2] for the Besicovitch projection theorem.For a more recent presentation, see [10, p. 86-87].We include the details for completeness.
Proof of Lemma 4.3.Let E, x, β, H be as in the statement of the lemma.Let ε :" Θ E,β pxq, so that there exists an r ą 0 such that H 1 pC β px, rq X Eq ě εr.We set also J :" Jpβq.
If the alternative (A1) fails, then Since evidently x P C β px, rq X E X x,θ , this implies that most of the lines x,θ do not intersect the set C β px, rq X E outside x.Consequently, C β px, rq X E is contained in a union of narrow cones C 1 , C 2 , . . .which are centred around certain lines x,θ j with θ j P J, and whose opening angles β 1 , β 2 , . . .satisfy ř β j ď 2H ´1.We may arrange that the cones have the form C j :" CpI j q :" Yt x,θ : θ P I j u, where I j Ă J is a dyadic interval, |I j | " β j , and θ j P J is the midpoint of I j .We may also assume that the dyadic intervals I j are disjoint, so the sets C j z txu are disjoint.
To use these cones to arrive at alternative (A2), recall that H 1 pC β px, rq XEq ě εr, where ε " Θ E,β pxq.Now, we throw away cones which are not heavy: we call a cone heavy if it satisfies H 1 pC j X Bpx, rq X Eq ě 1 4 ¨εH|I j | ¨r.
(4.4) The total length of C β px, rq X E contained in the non-heavy cones is bounded from above by εHr 4 so at least half of the length in C β px, rq X E is contained in the union of the heavy cones.
In the sequel, we assume that all the cones C j are heavy.
Next, we would like to prove that ř This would be easy if the heavy cones also satisfied an upper bound roughly matching the lower bound in (4.4).If we knew this, then we could estimate This desired upper bound in (4.4) need not be true to begin with, but can be easily arranged.Fix a heavy cone CpI j q, and perform the following stopping time argument: the dyadic interval I j is successively replaced by its parent " Îj " until either the upper bound holds, or then Îj " J.This procedure gives rise to a new collection of cones Cp Îj q which are evidently still heavy, and whose union covers the union of the initial heavy cones.
Since the intervals Îj are dyadic, we may arrange that the new heavy cones are disjoint outside txu without violating the previous two properties.At this point, either Îj " J for some index j, in which case (4.5) is trivially true (using |J| " 1), or then the upper bound (4.6) holds for all the heavy cones.In this case the lower bound (4.5) holds by the very calculation shown in (4.5).
We are now fully equipped to establish alternative (A2).Consider a line x,θ contained in the union of the heavy cones.According to (4.5), the set of angles θ P J of such lines has length H ´1. This set of angles is the set J x Ă J whose existence is claimed in (A2).It remains to associate the tube T x,θ to each line x,θ with θ P J x .Let CpI j q " C j Ą x,θ be the (unique) heavy cone containing x,θ .The opening angle of C j is β j " |I j | P p0, |J|s, and it follows by elementary geometry that C j X Bpx, rq Ă x,θ p2β j rq ": T x,θ .
Finally, H 1 pE X T x,θ q ě H 1 pC j X Bpx, rq X Eq εHβ j ¨r " εH ¨wpT q, as claimed in alternative (A2).
Proof of Lemma 4.1.The main geometric observation is the following: every minigraph in E is an α ´1-Lipschitz graph over every line L θ :" spanpcos θ, sin θq " K 0,θ with θ P Jpα 1 q (recall that α 1 " C lip α{2).This is simply because the minigraphs in E are α-Lipschitz graphs over the x-axis, but for all θ P Jpα 1 q, the lines L θ form an angle α with the y-axis.Thus, E is a union of finitely many α ´1-Lipschitz graphs over L θ , for every θ P Jpα 1 q.This places us in a position to use the area formula (2.8): for every θ P Jpα 1 q and every H 1 measurable subset E 1 Ă E we have ż Fix H ě 1. (We will eventually choose H " 1{pαεq; see (4.16) below.)By Lemma 4.3 (with β " α 1 ), we can write R " R 1 Y R 2 , where alternative (A1) holds on R 1 and (A2) holds on R 2 .To prove (4.2), it suffices to show We first consider R 1 .Recall the sets I x Ă Jpα 1 q defined in (A1).Since E is a union of finitely many compact Lipschitz graphs, there are no measurability issues, and we may freely use Fubini's theorem: For θ P Jpα 1 q fixed, abbreviate R 1 θ :" tx P R 1 : θ P I x u.Write also We may now deduce from (4.7) applied to E 1 :" E 1 θ , and (4.10), that ż and finally ě αH ´1H 1 pR 1 q.
By (3.4) the left hand side is bounded from above by δ, so Recalling that we promised to choose H " 1{pαεq in the end, the bound above implies (4.8) for R 1 .
Next, we tackle R 2 .This time we define R 1 θ :" tx P R 2 : θ P J x u Ă E, and we deduce exactly as in (4.9) that Fix θ P Jpα 1 q with R 1 θ ‰ H.For each x P R 1 θ , by definition, there exists a tube T " T x,θ centred around x,θ with the property H 1 pE X T q εH ¨wpT q. (4.13) The tubes tT x,θ : x P R 1 θ u may overlap, but they are all parallel.It follows from an application of the Besicovitch covering theorem (to the projections I x,θ :" π θ pT x,θ q Ă R) that there exists a countable sub-collection T θ Ă tT x,θ : x P R 1 θ u, with the properties Fix T P T θ , and let BadpEXT q Ă EXT consist of those points x P EXT with #p x,θ XEq " 1.We apply the coarea formula (2.8) to the set A :" BadpE X T q Ă E. Recalling that for every θ P Jpα 1 q the set E is a union of finitely many α ´1-Lipschitz graphs over L θ (see remark above (4.7))we get that Now, for a suitable choice H " 1{pαεq, a combination of (4.13) and (4.15) shows that At this point, we simplify notation by setting By the definition of the sets BadpE X T q, if x P E θ , then #pE X x,θ q ě 2, and therefore Recalling once again from (3.4) that the left hand side above is ď δ, we deduce that which is (4.8) for R 2 .The proof of Lemma 4.1 is complete.
Next, repeating the classical "two cones" argument of Besicovitch, we show that we can pack most of points of low conical density into a single Lipschitz graph.Lemma 4.18 (Most low conical density points fit into a Lipschitz graph).Let E Ă Bp1q Ă R 2 and let ε P p0, 1q, β P p0, 1  2 q.Then, there exists a 2β-Lipschitz graph Γ Ă R 2 over the x-axis such that H 1 ptx P E : Θ E,β pxq ď εu z Γq ε{β.
Proof.Let G " tx P E : Θ E,β pxq ď εu.Our task is to find a subset Γ Ă G with H 1 pG z Γq ε{β and the property C 2β pxq X Γ " txu for all x P Γ.Then Γ extends to a 2β-Lipschitz graph, as desired.
Let B be the set of points x P G with the "bad" property that there exists a point y P G X C 2β pxq with y ‰ x.The goal is to show that H 1 pBq ε{β.For each x P B, let rpxq " supt|x ´y| : y P G X C 2β pxqu, so See Figure 1 for an illustration.Let T x be the tube around the vertical line passing through x with wpT x q :" 1 10 βrpxq.Then Ă Bpx, rpxqq.(4.21) Choose a point ypxq P GXC 2β pxq such that |x´ypxq| ě 9 10 rpxq.A slightly more delicate geometric fact is that This is an exercise in elementary geometry, see Figure 1 (or the proof in [18, Lemma 15.14] for a more formal argument): the disc Bpx, 1 2 βrpxqq, and in particular the intersection T x X Bpx, 1  2 βrpxqq, is contained in the cone C β pypxqq, whereas the rest of T x is contained in C β pxq, as already noted in (4.20).Consequently, using (4.21), the trivial in- clusion Bpx, rpxqq Ă Bpypxq, 2rpxqq, and x, ypxq P G, we have H 1 pB X T x q ď H 1 pC β pypxq, 2rpxqq X Eq `H1 pC β px, rpxqq X Eq ď 2εrpxq `εrpxq ď 30pε{βq ¨wpT x q.
We have now shown that every point x P B is contained on the central line of a vertical tube T x satisfying the estimate above.By the Besicovitch covering theorem, as in the proof of Lemma 4.1, we may then find a countable, boundedly overlapping sub-family T of these tubes which still cover B. All the tubes intersect Bp1q Ą B, so ř T PT wpT q 1.It follows that This completes the proof of Lemma 4.18.
We are then ready to prove Proposition 3.3: Proof of Proposition 3.3.Fix ą 0 as in the statement of the proposition, and set α 1 " C lip α{2.Define 1 :" α {C for a suitable absolute constant C ą 0. By Lemma 4.1 applied to ε " 1 , we know that the set R Ă E of bad points x P E with The set G :" E z R satisfies the hypotheses of Lemma 4.18 (with β " α 1 " C lip α{2 and ε " 1 ), so there exists a C lip α-Lipschitz graph Γ Ă R over the x-axis such that H 1 pG z Γq 1 {α " {C.If the constant C ą 0 was chosen large enough, we see that This concludes the proof of Proposition 3.3.

PROOF OF PROPOSITION 3.10
In this section we prove Proposition 3.10.Recall that we are assuming to be in "Case 2"; that is, E cannot be exhausted, up to measure , by a a constant number of consecutive sets F k , F k`1 , . . ., F k`Csep (recall this notation from Subsection 3.3).More precisely, this meant that (5.1) fails for every k; thus we found an index pair k, l P t0, . . ., M 3 u with |k ´l| ě C sep such that H 1 pF k q ě α 2κ and H 1 pF l q ě α 2κ .
(5.2) Recall that all the minigraphs in F k make an angle ď α κ with L k :" spanpw k q " spanpcospkπ{M 3 q, sinpkπ{M 3 qq and similarly all the minigraphs in F l make an angle ď α κ with L l " spanpw l q.
A configuration where positively many lines hit E twice.
The existence of F k and F l will imply a configuration such as the one depicted in Figure 2. A more precise definition is given in the lemma below.(1) Affine lines k and l with =p k , L k q ď α κ and =p l , L l q ď α κ .
Once the objects in Lemma 5.3 are found, it follows from a relatively simple geometric argument, presented below, that positively many lines intersect E twice (the lines in question are depicted in red colour in Figure 2): Lemma 5.5.There exists a set of lines LpG k , G l q of measure ηpLpG k , G l qq (5.6) Proposition 3.10 follows immediately by Lemma 5.5.We will next derive Lemma 5.5 from Lemma 5.3.(See Remark 5.10 and Appendix A for an alternative proof of Lemma 5.5.) Proof.The key geometric observation is the following: if Ă R 2 is any line with then must make an angle α 1{2 with both k and l , see Figure 2: indeed, if for example =p , l q !α 1{2 and X G l ‰ H, then X Bp1q Ă T l , and hence X G k " H by (5.4).It follows that both k , l are Cα ´1{2 -graphs over K , for any line connecting G k and G l .But since γ k , γ l were by definition C lip α-Lipschitz graphs over k , l , it follows that also γ k , γ l are Cα ´1{2 -Lipschitz graphs over K (assuming that α ą 0 is small enough).
With this notation, we claim that x P G l .
(5.7) Indeed, if tBpθ j , r j qu jPN is an arbitrary cover of Θpx, G k q, then the tubes x,θ j pCr j q cover G k , where C ą 0 is an absolute constant.This is because G k is covered by the cones C j :" Ť t x,θ : θ P Bpθ j , r j qu by definition, and each intersection G k X C j Ă Bp1q X C j is further covered by a tube of the form x,θ j pCr j q.Now recall that γ k Ą G k is an α ´1{2 -Lipschitz graph over each line K x,θ j : this gives which implies (5.7).We now infer from (5.7) and Fubini's theorem that ż π 0 To proceed, write G l pθq :" tx P G l : θ P Θpx, G k qu.We claim that H 1 pG l pθqq ‰ 0 ùñ H 1 pπ θ pG l pθqqq α 1{2 H 1 pG l pθqq, θ P r0, πq. (5.9) This will complete the proof of the corollary, because (5.8) then implies and the left hand side above is a lower bound for ηpLpG k , G l qq.
Remark 5.10.In fact, we have an exact expression for ηpLpG k , G l qq: In (5.11), τ k pxq denotes the unit tangent vector to γ k at x P γ k , and τ l pxq is defined similarly.For distinct x, x 1 P R 2 , θpx, x 1 q denotes the angle θ such that π θ pxq " π θ px 1 q.Now we show how (5.11) implies Lemma 5.5.By the key geometric observation in the first paragraph of the proof of Lemma 5.5 and the fact that G k , G l Ă Bp1q, the integrand in (5.11) We state and prove a more general form of (5.11) in Appendix A.
The remainder of this section is devoted to constructing the objects listed in Lemma 5.3.This is based on the assumption (3.8), that is, H 1 pF k q ě α 2κ and H 1 pF l q ě α 2κ .Recall also that F k , F l were the unions of the minigraphs in F k and F l .The minigraphs in F k make an angle ď α κ with L k , while the minigraphs in F l make an angle ď α κ with L l .Furthermore, =pL k , L l q ě C sep α κ , so the minigraphs from F k and F l point in quantitatively different directions.We also recall that F k (respectively F l q can be expressed as a union of certain consecutive families E i : (5.12) Some of these families may be empty, but not all, according to (5.2).Of course since there were no more than α ´1 of the families E j altogether.
5.1.Sketch of the proof.We now explain the proof strategy with a picture.In Figure 3, we have depicted the sets F k and F l , which are roughly speaking α κ -Lipschitz graphs over the lines L k , L l by Proposition 3.3 (details will follow).Both F k and F l are, moreover, tiled by α ´1 of the sets E j .Most of sets E j are (individually) contained on α-Lipschitz graphs γ j , by another application of Proposition 3.3.The red sets shown in Figure 3 illustrate sets of the form where B j is some ball of radius α with the property that H 1 pG j q " α H 1 pE j q.Each G j is contained in a tube T j of width α 1{2 (or even a tube of width α, which was also required The problem is that if we pick G k Ă F k and G l Ă F l arbitrarily, the tube T k associated with G k might intersect G l , or vice versa, violating (5.4).To satisfy (5.4), we need to pick G k , G l in such a way that the G k -tube avoids G l and the G l -tube avoids G k .To achieve this, we roughly choose 3 well-separated sets G l 1 , G l 2 , G l 3 Ă F l , and 2 further well-separated sets G k 1 , G k 2 Ă F k .Then, we use the "transversality" of the graphs F k , F l to deduce the following: each G k i -tube can intersect at most one of the sets G l j , and vice versa.At this point, we may deduce from the pigeonhole principle that there must exists a pair pG k i , G l j q such that the G k i -tube does not intersect G l j , and the G l j -tube does not intersect G k i .Indeed, there are six pairs pG k i , G l j q, but only five tubes.This will complete the proof.
5.2.Proof.We turn to the details.First, we apply Proposition 3.3 to the sets F k , F l , each of which can be written as a finite union of α κ -Lipschitz minigraphs over the lines L k , L l , respectively.It follows from the choice of constants δ " 70 {C thm and α " p {C alp q 10 made in Section 3.3 that δ !α 5κ , assuming that C thm is chosen sufficiently small compared to the absolute constant C alp .Writing α 5κ " pα κ q 3 α 2κ , this means that the main hypothesis of Proposition 3.3 is valid with constants "α κ " and " 1 2 α 2κ " in place of "α" and " ".It follows that there exist C lip α κ -Lipschitz graphs Γ k , Γ l over L k , L l , respectively, which cover most of F k and F l in the sense and each E j is a finite union of α-Lipschitz minigraphs E j over a certain line (which makes an angle ď α κ with L k ).Applying Proposition 3.3 again, for each E j with either j P ts, . . ., s `mu or j P tt, . . ., t `mu, we find Lipschitz graphs γ j with constant ď C lip α and the property For this application of Proposition 3.3 to be legitimate, we need δ !α 3 pα 2 q 2 " α 7 , which also follows from our choice of constants recalled above, taking C thm " C 70 alp .We write E 1 j :" E j Xγ j .With these choices, a major part of F 1 k is covered by the union of the graphs α.
Since H 1 pF 1 k q H 1 pF k q ě α 2κ , and κ " 1 10 , we infer that at least half of F 1 k is covered by the (subsets of) α-Lipschitz graphs E 1 j with s ď j ď s `m.The same conclusion mutatis mutandis holds for F 1 l and the sets E 1 j with t ď j ď t `m.We finally redefine This should cause no confusion, since the original sets F k , F l will no longer be used.We list all the properties of F k , F l we will need in the sequel: covered by the union of α ´1 Lipschitz graphs γ s , . . ., γ s`m with constant ď C lip α over certain lines s`j making an angle ď α κ with L k , ‚ F l is covered by the union of α ´1 Lipschitz graphs γ t , . . ., γ t`m with constant ď C lip α over certain lines t`j making an angle ď α κ with L l .We have now defined carefully the objects F k and F l in Figure 3.In defining the objects E k and E l in the same picture, there is the technical problem that the "initial" sets E j need not be localised, as the picture suggests.This will be easily fixed by intersecting the initial sets E j with balls.First, using that H 1 pF k q α 2κ , we choose two special points x 1 , x 2 P Fk with the properties |x 1 ´x2 | α 2κ and H 1 pF k X Bpx j , αqq ě α 2 for j P t1, 2u. ( This can be arranged, because the set of points x P F k with H 1 pF k X Bpx, αqq ď α 2 has total length at most α !H 1 pF k q.Thus, the admissible points for the second condition in (5.14) have total length ě 1 2 H 1 pF k q α 2κ .Then, to finish the selection, it remains to pick two of these points with separation α 2κ : this is possible because F k lies on a Lipschitz graph with constant ď 1, so in particular H 1 pF k X Bpx, rqq r for all r ą 0.
Next, we move attention from F k to F l .This time we pick 3 special points y 1 , y 2 , y 3 P F l with properties similar to those in (5.14): |y i ´yj | α 2κ for i ‰ j and H 1 pF l X Bpy j , αqq ě α 2 for j P t1, 2, 3u. (5.15) The details of the selection are the same as we have seen above.Next, recall that both F k and F l can be written as a finite union of (subsets of) C lip α-Lipschitz graphs: the covering graphs for F k were denoted γ s , . . ., γ s`m and the covering graphs for F l were denoted γ t , . . ., γ t`m , where m α ´1.Since H 1 pF k X Bpx 1 , αqq ě α 2 , at least one of the graphs γ s , . . ., γ s`m must have large intersection with F k X Bpx 1 , αq.We denote this graph by γ k 1 ; then we have (5.16) We find similarly a graph γ k 2 P tγ s , . . ., γ s`m u such that H 1 pF k X γ k 2 X Bpx 2 , αqq α 3 .Then, we also repeat the argument for the three balls Bpy j , αq: we find three graphs γ l 1 , γ l 2 , γ l 3 P tγ t , . . ., γ t`m u with the property H 1 pF l X Bpy j , αq X γ l j q α 3 , 1 ď j ď 3. (5.17) The sets , and G l j :" F l X γ l j X Bpy j , αq, j " 1, 2, 3 (5.18)are the ones we informally discussed below Figure 3. Next, we associate the lines and tubes (required by Lemma 5.3) to the sets G k i , G l j .We associate to each graph γ k i or γ l j an affine line k i or l j with the following properties: ‚ γ k i is a C lip α-Lipschitz graph over k i for i P t1, 2u, ‚ γ l j is a C lip α-Lipschitz graph over l j for j P t1, 2, 3u, ‚ The lines are chosen so that G k j Ă k i pCαq for i P t1, 2u and G l j Ă l j pCαq for j P t1, 2, 3u, where C " C lip .We now define pT k i q 1 :" k i pCαq and T k i :" k i pα 1{2 q for i P t1, 2u, and similarly pT l j q 1 :" l j pCαq and T l j :" l j pα 1{2 q for j P t1, 2, 3u.Thus, G k i Ă pT k i q 1 Ă T k i and G l j Ă pT l j q 1 Ă T l j .Since moreover H 1 pG k i q α 3 and H 1 pG l j q α 3 by (5.16)-(5.17),any pair pG k i , G l j q (with associated lines and tubes) would now satisfy all the requirements of Lemma 5.3, except perhaps the inclusions (5.4).
We will now use the pigeonhole principle to show that at least one of the pairs pG k i , G l j q also satisfies the inclusions (5.4).The main geometric observation is the following: The first inequality holds for i P t1, 2u, the second for j P t1, 2, 3u.The proof of (5.19) Transversality of T k i and Γ l .The angle between k j and L l is Cα κ . is contained in Figure 4. Recall that T k i is an α 1{2 -tube around a certain line k i with =p k i , L k q ď α κ .On the other hand, =pL k , L l q ě C sep α κ , so also =p k i , L l q ě pC sep ´1qα κ .Finally, Γ l is a C lip α κ -Lipschitz graph over L l , so every tangent of Γ l makes an angle C sep α κ with k i , since we chose C sep much larger than C lip in Section 3.3.Thus Γ l is an α ´κ-Lipschitz graph over p k j q K .It follows that . Now that we have proved (5.19), recall from (5.15) the three balls Bpy j , αq, all of which were centred at y j P F l Ă Γ l , and whose centres y j had pairwise separation α 2κ .Since κ " 1  10 , we have α 1{2´κ !α 2κ for α ą 0 small enough (or in other words assuming that the constant C alp ą 0 is chosen large enough), and therefore (5.19) implies that #tj P t1, 2, 3u : T k i X Bpy j , αq ‰ Hu ď 1, i P t1, 2u. (5.20) By a similar argument, #ti P t1, 2u : T l j X Bpx i , αq ‰ Hu ď 1, j P t1, 2, 3u. (5.21) We finally claim, as a consequence of (5.20)-(5.21)and the pigeonhole principle, that there exists a pair of balls pBpx i 0 , αq, Bpy j 0 , αqq, for some i 0 P t1, 2u and j 0 P t1, 2, 3u with the property T k i 0 X Bpy j 0 , αq " H and T l j 0 X Bpx i 0 , αq " H. Ă Bpx i 0 , αq z T l j 0 and G l j 0 (5.18) Ă Bpy j 0 , αq z T k i 0 , which (combined with (5.18)) completes the proof of the inclusions (5.4), and Lemma 5.3.
To prove (5.22), consider the bi-partite graph with 5 vertices tv 1 , v 2 u Y tw 1 , w 2 , w 3 u and the following edge set.
‚ For i P t1, 2u and j P t1, 2, 3u, the edge pv i , w j q is included if T k i X Bpy j , αq ‰ H. ‚ For j P t1, 2, 3u and i P t1, 2u, the edge pw j , v i q is included if T l j X Bpx i , αq ‰ H. Now, (5.20)-(5.21)can be restated as follows: for v i fixed, there can be at most one edge pv i , w j q, and for w i fixed, there can be at most one edge pw i , v j q.Thus, the edge set contains at most 5 edges.On the other hand, the product set tv 1 , v 2 u ˆtw 1 , w 2 , w 3 u contains 6 elements, so there must be a pair tv i , w j u so that neither pv i , w j q nor pw j , v i q lies in the edge set.This is equivalent to (5.22).This completes the proof of Lemma 5.3.

THE GRID EXAMPLE
In this section we provide an example showing that Theorem 1.1 is optimal in the sense that the assumption FavpEq ě FavpLq ´δ cannot be relaxed to FavpEq ě δ.Proposition 6.1.There exists an absolute constant δ ą 0 and a sequence of compact rectifiable sets E n Ă r0, 1s 2 Ă R 2 such that: (1) H 1 pE n q " 1, (2) FavpE n q ě δ, (3) for any α P r2n ´2, 1q and any curve Γ with H 1 pΓ X E n q ě α we have H 1 pΓq αn.In particular, property (3) implies that if M ě 1, then for any M -Lipschitz graph Γ we have H 1 pΓ X E n q M n ´1.
We begin the construction.Fix an integer n ě 2, and let rns :" t1, . . ., nu.For any j " pk, lq P rns 2 set Note that B j Ă r0, 1s 2 and if i, j P rns 2 , i ‰ j, then Define S j " BB j , and observe that H 1 pS j q " n ´2.
We define the set E n as E n :" Since H 1 pS j q " n ´2, we have H 1 pE n q " 1.This verifies property (1) for E n .It is also clear that E n is compact and rectifiable.Now we check property (3).We will use the following result.
Proof.Suppose that α P r2n ´2, 1q and let Γ be a curve with H 1 pΓ X E n q ě α.Since each circle S j comprising E n has length n ´2, we get that Γ intersects at least αn 2 different circles.Let J 0 Ă rns 2 be the set of indices such that for j P J 0 we have Γ X S j ‰ ∅, so that N :" #J 0 ě αn 2 .(6.6) To estimate H 1 pΓq, we are going to use (6.6) together with the fact that the circles S j are centered on a well-separated grid (6.2), (6.3).We provide the details below.
Let γ be the parametrisation of the curve Γ given by Lemma 6.4.Without loss of generality, we may assume that the curve Γ begins and ends on E n , i.e., γp0q, γp1q P Γ X E n .For all j P J 0 we choose a point y j P Γ X S j , and let t j P r0, 1s be such that γpt j q " y j (γ might be non-injective, in which case t j is non-unique, but in this case we pick t j arbitrarily among the admissible options).The only constraint we make on our choice of ty j u jPJ 0 is so that γp0q, γp1q P ty j u jPJ 0 .For convenience, we relabel the points t j in "ascending order": for all i P t1, . . ., N u we set t i :" t j for some j P J 0 , in such a way that t 1 ă t 2 ă ¨¨¨ă t N .We relabel in a similar way y j and S j .
Recalling that the circles S j are centered on a grid (6.2), it follows from the separation property (6.3) that for any i P t1, . . ., N u Summing over i P t1, . . ., N ´1u we get N ´1 2n ď Lippγq ¨pt N ´t1 q ď 32 H 1 pΓq ¨pt N ´t1 q.
Since we we assumed γp0q, γp1q P ty j u jPJ 0 , we get that t N " 1 and t 1 " 0. Thus, 32 H 1 pΓq ě N ´1 2n This completes the proof of the lemma.
It remains to prove the property (2), that is, FavpE n q ě δ.Let so that E n " BG n .Note that FavpE n q " FavpG n q.We define an auxiliary measure Recall that the 1-energy of µ is defined as Lemma 6.7.We have I 1 pµq 1.
We move on to estimating A 2 .Let Q j denote the square centered at x j with sidelength 1{pn `1q.Note that B j Ă Q j , and the squares Q j , j P rns 2 are pairwise disjoint.If x P B i and y P B j , with i ‰ j, then |x ´y| " distpB i , B j q " |x ´z| for any z P Q j .It follows that for a fixed x P B i ż dµpyq " distpB i , B j q ´1 µpB j q " distpB i , B j q ´1 L 2 pQ j q " ż Q j µpB i q " 1.
It follows that I 1 pµq 1.
This concludes the proof of Proposition 6.1.

APPENDIX A. LINES SPANNED BY RECTIFIABLE CURVES
We state and prove a generalization of (5.11), which was mentioned in Remark 5.10.

FIGURE 1 .
FIGURE 1. Containing the tube T x in the union of the cones C β pxq and C β pypxqq.The dotted cone illustrates C 2β pxq Q ypxq.

Lemma 5 . 3 .
If (5.2) holds, then there exists an absolute constant C " C lip (the constant from Proposition 3.3) such that the following objects exist: FIGURE 3. Finding the graphs and tubes claimed by Lemma 5.3.
To estimate A 1 we note that for any i P rns 2 and any fixed x P B i Bpx,2 ´k qzBpx,2 ´k´1 q 2 k dL 2 pzq 1.