Mapping $n$ grid points onto a square forces an arbitrarily large Lipschitz constant

We prove that the regular $n\times n$ square grid of points in the integer lattice $\mathbb{Z}^{2}$ cannot be recovered from an arbitrary $n^{2}$-element subset of $\mathbb{Z}^{2}$ via a mapping with prescribed Lipschitz constant (independent of $n$). This answers negatively a question of Feige from 2002. Our resolution of Feige's question takes place largely in a continuous setting and is based on some new results for Lipschitz mappings falling into two broad areas of interest, which we study independently. Firstly the present work contains a detailed investigation of Lipschitz regular mappings on Euclidean spaces, with emphasis on their bilipschitz decomposability in a sense comparable to that of the well known result of Jones. Secondly, we build on work of Burago and Kleiner and McMullen on non-realisable densities. We verify the existence, and further prevalence, of strongly non-realisable densities inside spaces of continuous functions.


Introduction
The main objective of this paper is to answer a question of Feige originating in the 1990s, which asks whether every subset of the integer lattice Z 2 with cardinality n 2 for some n ∈ N can be mapped via a bijection f onto the regular n × n grid {1, . . . , n} 2 in such a way that the Lipschitz constant of f may be bounded above independently of n: Question 1.1 ('Feige's Question'). Is there a constant L > 0 such that for every n ∈ N and every set S ⊂ Z 2 with cardinality |S| = n 2 there is a bijection f : S → {1, . . . , n} 2 with Lipschitz constant Lip(f ) ≤ L?
Feige's question appears to have first come to wider attention at the workshop 'Discrete Metric Spaces and their Algorithmic Applications' held in Haifa in March 2002 [21] and appeared in the technical report [22,Question 2.12], which lists many open problems arising from this meeting and, in several cases, their subsequent solutions. However, to the best of the authors' knowledge, there has been almost no progress on Feige's question prior to the present work. The second author worked on Feige's question in his master thesis [19]; some ideas contained there were helpful in the development of the present work.
Feige's motivation to ask the question stemmed from his work on the so-called 'graph bandwidth' problem. In this problem the goal is to find a bijection l : V → {1, . . . , n} for a given n-vertex graph (V, E) with the 'bandwidth' as small as possible, that is, minimising the quantity max uv∈E |l(u) − l(v)|. It is known that finding the optimal solution to this problem is NP-hard [25]. In [12] Feige designated a randomised approximation algorithm that produces a solution with bandwidth larger than the optimum by a factor polylogarithmic in n. His algorithm can also be adapted to a generalisation of the bandwidth problem to two dimensions; however, in this case it does not produce a bijection between the set of vertices V and the grid {1, . . . , √ n} 2 , but rather an injection into a larger grid leaving some of the grid points unused [13]. Feige then asked whether one can map such a set bijectively onto the grid {1, . . . , √ n} 2 without increasing the bandwidth 'too much'. We will prove that the answer to Feige's question is negative in all dimensions d ≥ 2: Moreover, we show how almost any positive continuous function on the unit cube [0, 1] d can be used to construct a sequence of sets S n ∈ F n verifying Theorem 1.2. It is then natural to ask how fast C n grows. Whilst it is straightforward to verify that C n ≤ √ dn, the results of the present article do not permit finer estimates of C n , either from above or below. The question of the rate of growth of the sequence (C n ) will be an interesting topic for future research.
In order to answer Feige's question, we adapt a technique developed independently by Burago and Kleiner [4] and McMullen [24], which translates questions about Lipschitz mappings on discrete sets into a continuous setting. The papers [4] and [24] present a negative answer to the question of whether every two separated nets in the plane are bilipschitz equivalent, or put differently, whether every separated net M ⊆ Z 2 admits a bilipschitz bijection f : M → Z 2 . This 'discrete' question is shown to be equivalent to the following 'continuous' one: Does there exist for every measurable function ρ : [0, 1] 2 → (0, ∞) with 0 < inf ρ < sup ρ < ∞ a bilipschitz mapping f : [0, 1] 2 → R 2 with ρ = Jac(f ) almost everywhere.  [4] and McMullen [24] comes from Jiří Matoušek, who shared it with the second author while supervising his master thesis [19]. By encoding measurable functions ρ : [0, 1] d → (0, ∞) as sequences of discrete sets, we will show that a negative answer to Feige's question is implied by the existence of a special type of non-realisable density function. We require here a function non-realisable in a stronger sense than that provided by Burago and Kleiner and McMullen: Our density ρ must exclude solutions to a natural generalisation of equation (1.1), not only in the class of bilipschitz mappings, but in the wider class of Lipschitz regular mappings. Lipschitz regular mappings can be thought of as 'non-degenerate' Lipschitz mappings and were introduced by David in [7]; we give a more detailed introduction to this class in Section 2. Given a measurable function ρ : Then E is a σ-porous subset of C([0, 1] d , R).
In the above, C([0, 1] d , R) denotes the Banach space of continuous real-valued functions on the unit cube [0, 1] d . Further, the term σ-porous refers to a class of negligible subsets of complete metric spaces. This notion is discussed in greater detail at the beginning of Section 4.
The fact that Lipschitz regular mappings need not be bijective presents significant additional difficulties in constructing non-realisable density functions for this class. To produce bilipschitz non-realisable densities on [0, 1] d it suffices to take any non-empty open subset U ⊆ [0, 1] d and to define a 'badly behaving' function ρ there in the style of McMullen or Burago and Kleiner. This function can then be extended arbitrarily to the whole of [0, 1] d . However, for a Lipschitz regular mapping f : [0, 1] d → R d and T ⊆ f ([0, 1] d ) the pre-image f −1 (T ) may consist of several 'pieces' whose images overlap in T . Therefore any local 'bad behaviour' of ρ on one piece of f −1 (T ) may be compensated for by its values on other pieces, allowing f to satisfy equation (1.2). So excluding Lipschitz regular solutions of (1.2) crucially requires global control of the density ρ.
We overcome these problems, in part, by using a bilipschitz decomposition result for Lipschitz regular mappings, which derives from a result of Bonk  This result turns out to be very useful. Given a Lipschitz regular mapping f : [0, 1] d → R d , it allows us to carefully choose the set T ⊆ f ([0, 1] d ), referred to in the discussion above so that its pre-image f −1 (T ) decomposes precisely as a finite union of open sets on which f is bilipschitz; see Proposition 2. 19. We then extend the existing techniques for constructing bilipschitz non-realisable densities to fit this more complex situation, where there are multiple bilipschitz mappings in play instead of just one.
Structure of the paper.
In Section 2 we investigate Lipschitz regular mappings on Euclidean spaces, in particular proving Theorem 2.11. Besides the material necessary for the proof of Theorem 2.11, and subsequently for the proof of Proposition 2.19, which we use to answer Question 1.1, Section 2 also contains a discussion of optimality and limits of Theorem 2.11.
The groundwork for our verification of non-realisable densities is laid in Section 3, where we develop a construction of Burago and Kleiner in order to derive certain powerful properties of bilipschitz Jacobians; see Lemma 3.1. This section is the most technical part of the present paper; readers willing to accept Lemma 3.1 may safely skip this Section during the first pass-over, and perhaps, return to it only later. Both Sections 2 and 3 can be read independently of the rest of the paper. Section 4 is devoted to the proof of existence of non-realisable densities and Theorem 4.1. The only material from the previous sections needed in Section 4 is Proposition 2.19 and Lemma 3.1. Having gathered together all the necessary ingredients, we resolve Feige's question in Section 5.
In an effort to make the paper easier to read and to bring the more important ideas of the arguments to the forefront, we postpone the proofs of technical lemmas, or formal verifications of intuitive statements until an appendix. The appendix is divided into four subsections, each corresponding to a section of the paper. Where specific notation is introduced within a given section, we adopt the same notation in its appendix.

Notation.
We conclude this introduction with a summary of the notation and key definitions common to all sections of the paper: Sets and measures. For i ∈ N we write [i] for the set {1, 2, . . . , i}. Given a finite set F , we let |F | denote the cardinality of F . For an infinite set F we put |F | equal to ∞. Throughout the paper I will denote the unit interval I := [0, 1]. The closure, interior and boundary of a set A are written as A, int A and ∂A respectively. We adopt the convention of Mattila [23] and do not distinguish between outer measures and measures. The symbol L will refer to the Lebesgue measure and we write a.e. instead of 'almost everywhere' or 'almost every' with respect to L. We let − S ρ := 1 L(S) S ρ dL denote the average value of a measurable, real-valued function ρ on a measurable set S of positive Lebesgue measure. We will use the term density to refer to a non-negative, measurable real valued function. Thus, each density ρ : I d → [0, ∞) can be associated to a measure ρL on I d as defined earlier in this introduction.
Norms and balls. We write − 2 for the Euclidean norm and − ∞ for both the supremum norm and (briefly) the L ∞ norm. In addition, the symbol T op will represent the operator norm of a linear mapping T . An open ball with centre a and radius r will be denoted by B(a, r). Most of the time balls will be in Euclidean spaces, but we sometimes consider balls in general metric spaces including spaces of functions. It will be clear from the context which norm or metric is relevant. Occasionally we will extend this notation to denote neighbourhoods of sets: Given a set A ⊆ R d and r > 0 we let B(A, r) := a∈A B(a, r). To denote the closure of a ball or set neighbourhood we write B instead of B.
Mappings. For L ≥ 1, we call a mapping f : then we say that f is (b, L)-bilipschitz. Moreover, we use the term L-bilipschitz to refer to the special case where b = 1/L. Occasionally we will consider Lipschitz and bilipschitz mappings between more general metric spaces and generalise the above notions in the standard way. Given a mapping f , we write Df (x) for the (Fréchét) derivative of f at a point x and Jac(f ) := det(Df ) for the Jacobian of f whenever these notions make sense. When f is Lipschitz and the domain of f is open, then the classical theorem of Rademacher [23,Thm. 7.3] asserts that Df (x) exists a.e.

Lipschitz regular mappings.
Lipschitz regular mappings constitute an intermediate class between Lipschitz and bilipschitz mappings. While bilipschitz mappings are sometimes too rigid, Lipschitz mappings can be very degenerate; they can map many points onto a single one or map sets of positive measure onto sets of measure zero. Various classes of mappings lying somewhere in between Lipschitz and bilipschitz have been studied, for instance, in [1], [17], [20] We provide answers to these questions in the case that the dimension of the domain is equal to the dimension of the co-domain.
After the first pre-print of the present paper was published online the authors became aware of a result of Bonk and Kleiner [3,Theorem 3.4] from which the main result of the present section derives very quickly. Indeed an answer to the first part of the above question is implicitly present in [3]. For readers primarily interested in the resolution of Feige's question and comfortable with the result [3,Theorem 3.4], there is a quick short cut through the present section, beginning after the introduction of necessary notation and background on Lipschitz regular mappings; see the paragraph under the heading 'Short cut' on page 9.
Our argument for the proof of Theorem 2.11 appears to be new. Moreover, since [3, Theorem 3.4] is a statement for more general mappings then we consider here, many of the difficulties which occur in the proof in [3] are not present in our setting. We therefore present the full details of our argument here. We discuss the result [3, Theorem 3.4] of Bonk and Kleiner and its proof in further detail later on. The material of the present section remaining after Proposition 2.19, which appears under the heading 'Optimality of Theorem 2.11' on page 17, can be seen as a complement of Theorem 2.11 and Proposition 2.19 and is independent of Feige's question and of [3].
Notation and convention. We use the term (C, L)-regular mapping to denote a Lipschitz regular mapping f with Lip(f ) ≤ L and Reg(f ) ≤ C. Let R > 0, we call a set S in a metric space (M, dist) R-separated if for every two distinct points x, y ∈ S we have dist(x, y) > R. We write H d for the d-dimensional Hausdorff measure. A ball is always assumed to be open if not said otherwise.
Before we start the exposition of our results we list general properties of Lipschitz regular mappings that will be needed later.
The upper bound in the previous lemma comes from the Lipschitz property, while the lower bound can be derived easily using Lipschitz regularity and the definition of Hausdorff measure. As a corollary, Lipschitz regular mappings possess Luzin's properties (N ) and (N −1 ): A converse of Lemma 2.2 holds in the setting of Ahlfors regular spaces. We do not introduce the definition of Ahlfors regularity here 2 , since we will work only in the setting of Euclidean spaces, which are also Ahlfors regular.
for every x ∈ M and every r > 0, then f is Lipschitz regular.
To prove the previous lemma, it is sufficient to observe that the definition of Lipschitz regularity can be restated in the following way: Lemma 2.5 (Alternative definition of Lipschitz regularity). A Lipschitz mapping f : M → M is Lipschitz regular if and only if there is C > 0 such that for every x ∈ M and every r > 0 the set f −1 (B M (x, r)) does not contain any 2Cr-separated set with more than C elements.
We also add an easy observation, which, however, will prove useful later.
Observation 2.6. Let f : M → M be Lipschitz regular and y ∈ M . Then we have Proof. To the contrary, we assume there are pairwise distinct points x 1 , . . . , x Reg(f )+1 ∈ f −1 ({y}). Let us denote by r the minimum distance between x i and x j for 1 ≤ i < j ≤ Reg(f ) + 1. Then no ball in M of radius r 2 can contain more than one of the points x 1 , . . . , x Reg(f )+1 . Therefore, f −1 (B M (y, r 2 Reg(f ) )) cannot be covered with at most Reg(f ) balls of radius r 2 in M , a contradiction.
does not exist or does not have full rank}) occupies almost all of U . Notice that for every x ∈ N (f ) we have that Df (x) exists and is invertible 3 , and moreover, that f −1 ({f (x)}) ⊆ N (f ). We use the set N (f ) several times later on.
Occasionally, we will be given an open set U and a Lipschitz regular mapping f defined on U , the closure of U . Then by N (f ) we mean the set N (f | U ) ⊆ U . Note that it is then still true that f (N (f )) has full measure in f (U )-we will use this fact several times.
Topological degree. An important tool that we use in our work, besides differentiability, is the notion of topological degree. We briefly introduce it here; for a detailed treatment of this topic, we refer to [10,.
The degree function is uniquely determined by the following three properties [10, Theorem 1.1]: The degree function is defined explicitly in [10,Chapter 2]. We just point out that in the special case where g ∈ C 1 (U , R d ) and for every point is invertible, then the degree function is given by the expression (see [10,Definition 2.1]). In particular, we have that deg(g, U, y) = 0 whenever y ∈ R d \ g(U ). We will require some further properties of the degree which follow easily from the properties (d1), (d2) and (d3). All of the statements of the next Proposition are contained in [10, Theorem 3.1].
Proposition 2.7. Let U ⊆ R d be an open, bounded set, f ∈ C(U , R d ) and y ∈ R d \f (∂U ).
(i) If y and y belong to the same connected component of R d \f (∂U ) then deg(f, U, y) = deg(f, U, y ).
In the next Proposition, we extend the formula (2.1) to Lipschitz mappings. sign(Jac(f )(x)).
Since the proof of Proposition 2.8 is a rather technical exercise, we include it only in Appendix A.
Before we present our version of the proof of Theorem 2.11, let us state one additional auxiliary lemma. It says that whenever a continuous mapping in R d has derivative of full rank at a point, it preserves neighbourhoods of this point. We believe that such a statement may be a folklore; however, we did not find any reference. Lemma 2.9. Let a ∈ R d , r > 0 and f : B(a, r) → R d be a continuous mapping (Fréchet) differentiable at the point a with rank(Df (a)) = d. Then there is δ 0 > 0 such that for every δ ∈ (0, δ 0 ] we have Up to an affine transformation, the lemma can be restated in the following way: The easy proof of Lemma 2.9 using Lemma 2.10 can be found in Appendix A. Since our proof of Lemma 2.10 is very short, we provide it here. It relies on Proposition 2.7.
Bilipschitz decomposition of Lipschitz regular mappings. Our main goal in this section is to show that Lipschitz regular mappings in Euclidean spaces decompose into bilipschitz mappings in a nice way: Theorem 2.11. Let U ⊆ R d be open and f : U → R d be a Lipschitz regular mapping. Then there exist pairwise disjoint, open sets (A n ) ∞ n=1 in U such that ∞ n=1 A n is dense in U and f | An is bilipschitz for each n ∈ N with lower bilipschitz constant b = b(Reg(f )).
Before we prove Theorem 2.11, let us put it briefly into context. For a general Lipschitz mapping h : R d → R d it is known that one can obtain a different bilipschitz decomposition using Sard's theorem; see e.g. [11,Lemma 3.2.2]. One can start with sets defined for every k ∈ N and then cut these sets into pieces of diameter less than 1/k forming a decomposition (A n ) ∞ n=1 . Then Sard's theorem implies that L h R d \ n∈N A n = 0. When compared to the decomposition of Theorem 2.11, the difference is that the sets A n are not necessarily open, the lower bilipschitz constant of each h| An may depend on n and ∞ n=1 A n need not be a large subset of the domain in any sense. If the decomposition that was just described is applied to a Lipschitz regular mapping, the resulting sets A n occupy almost all of the domain, since the set N (f ) has a full measure in the domain. But the sets A n still need not be open. The fact that for Lipschitz regular mappings we are able to ensure the openness of bilipschitz pieces A n will be of crucial importance to us.
The first quantitative version of the decomposition using Sard's theorem was provided by David [6,Proposition 1] for general Lipschitz mappings f : David shows that for any ball B ⊂ R d , if L(f (B)) is large in measure, then B contains a set E large in measure such that f | E is bilipschitz. When applied to a Lipschitz regular mapping f , using the measure-preserving property expressed in Lemma 2.2, the condition that L(f (B)) is large in measure is satisfied automatically; for this version of David's result, see [9,Theorem 4.1].
A well-known result of Jones [18] provides another quantitative version of the decomposition for Lipschitz mappings I d → R m . In the decomposition of Jones as well as that of David the bilipschitz pieces may have empty interior.
Question. Can we hope for any control of the measure of the bilipschitz pieces in a bilipschitz decomposition of Lipschitz regular mappings if one requires the pieces being open? For example, can we hope for any control of the measure of the set ∞ n=1 A n given by the conclusion of Theorem 2.11?
The answer to the previous question is no: The decomposition from Theorem 2.11 cannot be strengthened in this way for a general Lipschitz regular mapping. A detailed discussion of these questions is contained in the subsection 'Optimality of Theorem 2.11' at the end of this section.
The proof of Theorem 2.11 can be divided into three main parts. The first one is to find, for any given open set in the domain, an open subset on which the given mapping is almost injective (this notion is formalised below). The second part is to show that a Lipschitz regular, almost injective map on an open set is injective on some open subset and the third part is to prove that a Lipschitz regular, injective map on an open set with a convex image is bilipschitz. In each of these steps we rely on the Lipschitz regularity of the mapping in question.
Let us remark that the first two steps described above, which comprise Lemma 2.14, 2.15 and 2.16 in the following, may be replaced by an application of [3,Theorem 3.4]. Bonk and Kleiner work in [3] with much more general mappings; instead of assuming that f : U ⊆ R d → R d is Lipschitz regular, they only require that f is continuous and that there is some constant C > 0 such that f −1 ({y}) ≤ C for all y ∈ R d . The latter condition is referred to as 'bounded multiplicity'. Moreover, the domain U may be replaced by any compact metric space X with the property that every non-empty open subset of X has topological dimension d.
The argument we present below is different to that of Bonk and Kleiner in [3]. However, a key aspect of both proofs appears to be finding points x in the domain such that f (x) is an interior point of the image f (O) for every neighbourhood O of x. The most difficult part of Bonk and Kleiner's argument is to show that such points exist. However, for Lipschitz regular mappings we can easily find many such points using almost everywhere differentiability of Lipschitz mappings, the regularity condition and Lemma 2.9. Indeed, note that all points in the set N (f ) have this property. Therefore, our argument below may be a more accessible approach to [3,Theorem 3.4] for the special case where the mappings considered are Lipschitz regular.
We start with the following definitions.
That is, n(h, x, B) denotes the number of preimages of h(x) in B.
Definition 2.13. We say that a mapping h : As advertised above, we begin by showing that a Lipschitz regular mapping is almost injective on some ball: We inductively construct open balls V k , k ∈ N, such that for every point x ∈ V k we have n(f, x, U \ V k ) ≥ k − 1; see Figure 1. For k > Reg(f ) this will contradict Observation 2.6.
We choose as V 1 any non-empty open ball in U . We describe how to construct V k+1 provided the set V k has already been defined.
By the assumption on B(V k ), there are two points x k , y k ∈ B(V k ) ∩ N (f ) such that f (x k ) = f (y k ). We find two disjoint open balls C k , D k ⊂ V k , the first one centred at x k , the other one at y k . By Lemma 2.9, there is an open ball We can find a non-empty open ball V k+1 in C k ∩ f −1 (G k ). By the construction, for every x ∈ V k+1 there is a point y ∈ D k such that f (x) = f (y). Moreover, by the assumption on V k , for every x ∈ V k+1 we have n(f, x, U \ V k ) ≥ k − 1. Combining these two facts, we obtain that n(f, x, U \ V k+1 ) ≥ k holds for every x ∈ V k+1 .
As the next step, we use the degree to show that whenever a Lipschitz regular mapping is almost injective on a bounded open set U , it is injective on an open subset of U . As a preliminary step, we investigate the behaviour of the degree of a Lipschitz regular, almost injective mapping. Proof. First, we observe that n(f, x, U ) = 1 for every x ∈ N (f )\f −1 (f (∂U )). We assume the contrary and find a pointx  Proof. Assume the contrary and find a point y ∈ f (U ) \ f (∂U ) which has more than one preimage. By Lemma 2.15 we have that deg(f, U, y) = ±1. Let which is a contradiction.
The argument presented above was inspired by an argument in [16], which takes place in a different setting, but is similar in spirit. Lemma 2.16 is sufficient for the proof of Theorem 2.11, but it is not the strongest statement one could hope for. It is possible to show the following, as one might expect: open set, f : U → R d be a Lipschitz regular, almost injective mapping. Then f | U is a homeomorphism onto its image.
We use Lemma 2.17 only in one complementary result at the end of this section (namely, in the proof of Lemma 2.20), which can be viewed independently of the rest of this paper. Therefore, a reader may safely skip its proof and possibly come back here later.
Proof. We show the lemma in the case that U = B(0, 1). It is then an easy exercise to prove the general case.
We write B t for the ball B(0, t), where t > 0, and B 0 for the set containing the single point 0. We divide the proof into two claims.
Proof. To the contrary, assume that which has a positive measure, is mapped by f inside f (B t ). This contradicts almost injectivity of f .
Proof. To the contrary, assume that there is For a fixed x there can be only finitely many such numbers t by Lipschitz regularity of f , as expressed in Observation 2.6. We choose t as the largest such number. Consequently, for every l ∈ (t, 1) we have that f (x) / ∈ f (∂B l ). Let us write x for a point in ∂B t such that f (x) = f (x ). Fix any l ∈ (t, 1). Then x ∈ B l \ f −1 (f (∂B l )). By Lemma 2.16 the mapping f is injective on the set B l \ f −1 (f (∂B l )). Therefore, we can choose r > 0 such that B(x , r) ⊂ B l and that f | B(x ,r) is injective. By Brouwer's invariance of domain [15,Thm. 2B.3], the mapping f | B(x ,r) is a homeomorphism onto its image. But this gives us the inclusions Since l ∈ (0, 1), this is a contradiction to Claim 2.17.1. The third step towards the proof of Theorem 2.11 is to show that a Lipschitz regular, injective mapping with a convex image is bilipschitz. We note that the same statement also appears in [3,Lemma 3.8]. For reader's convenience, we include its short proof here as well.
Proof. By Brouwer's invariance of domain [15,Thm. 2B.3] the mapping f is a homeomorphism onto its image.
For every two distinct points x, y ∈ U we consider the line segment f (x)f (y) ⊂ f (U ) connecting their images. Its preimage under f , we denote it by γ(x, y) , is a curve with endpoints x and y. By Lipschitz regularity, the curve γ(x, y) can be covered by at most Reg(f ) balls of radius Reg(f ) f (y) − f (x) 2 . Consequently, the distance between x and y cannot be larger than Finally, we have gathered all the ingredients needed for the proof of Theorem 2.11.
Proof of Theorem 2.11. We start with a countable basis (U n ) n∈N for the subspace topology on U . By Lemma 2.14 we get a collection of open balls (V n ) n∈N such that for every n ∈ N we have V n ⊆ U n and that f | Vn is almost injective. Since each V n is a ball, L(∂V n ) = 0. Therefore, f is also almost injective on the closed ball V n .
Next, we apply Lemma 2.16 to each set V n and find a non-empty open set W n ⊆ V n such that f | Wn is injective. Furthermore, using Brouwer's invariance of domain [15, Thm.

2B.3], we can find a non-empty open ball
Subsequently, we apply to each f | Gn Lemma 2.18 and infer that f | Gn is bilipschitz with Now we set A 1 := G 1 and inductively define A n := G n \ n−1 j=1 A j . It is clear that the collection (A n ) n∈N consists of pairwise disjoint open sets such that for every n ∈ N the mapping f | An is bilipschitz with lower bilipschitz constant b.
It remains to show that the set ∞ n=1 A n is dense in U , and hence, also in U . But this is immediate since for every element U n of the countable basis for the topology on U the set A n is either a non-empty subset of U n , or it is empty, in which case at least one of the sets A 1 , . . . , A n−1 intersects G n , which is a non-empty subset of U n .
Using Theorem 2.11 we can deduce that a Lipschitz regular mapping on an open set can be expressed, on some open subset of the image, as a sum of bilipschitz homeomorphisms. Such form of a decomposition is needed for the non-realisability results contained in the next section and, ultimately, for the resolution of Feige's question 1.1.
n=1 be the open sets from the conclusion of Theorem 2.11 applied to the mapping f . Let y ∈ f (U ) be such that the number Assuming that this claim is valid we may define the desired sets . Thus the proof can be completed by verifying the earlier claim, that is, by proving that However, this contradicts the fact that f is bilipschitz on A β i , and therefore also bilipschitz Optimality of Theorem 2.11.
The remainder of the current section is devoted to discussion of limits and optimality of Theorem 2.11. The content here is independent of the rest of the article, so the reader interested mainly in the resolution of Feige's question 1.1 can safely skip the rest of this section.
In the above, we have raised a question of optimality of Theorem 2.11 in terms of the measure of the bilipschitz pieces A n . Theorem 2.11 does not offer any control of their measure; below we will show that this is unavoidable. However, in a special case that a Lipschitz regular mapping f has Reg(f ) ≤ 2, we can provide a stronger bilipschitz decomposition; namely, the bilipschitz pieces A n , in addition to the conclusions of Theorem 2.11, can cover almost all of the domain.
Proof. From Observation 2.6 we know that every point y ∈ f (U ) has either one or two preimages. Since If deg(f, U, y) ≡ 1 (mod 2), then Proposition 2.8 implies that y has exactly one preimage. Using Proposition 2.7, part (i), we deduce that the same is true of all points y ∈ f (N (f )) ∩ B(y, r). Thus the mapping f : f −1 (B(y, r)) → B(y, r) is almost injective.
We may now apply Lemma 2.17 and then Lemma 2.18 to conclude that f | f −1 (B(y,r)) is bilipschitz with lower bilipschitz constant On the other hand, if deg(f, U, y) ≡ 0 (mod 2) then y must have two distinct preimages contains a non-empty open ball G containing the point y. Then every point in G has exactly one pre-image in each of the balls B 1 and B 2 . Hence f | f −1 (G)∩B i is injective for i = 1, 2 and, applying Lemma 2.18, we conclude that these mappings are also bilipschitz with lower bilipschitz constant 1 2 Reg(f ) 2 .
In the above we established that for every point y ∈ f (N (f )) \ f (∂U ) there is an open ball B containing y such that f −1 (B) decomposes precisely as the union of at most two sets on which f is bilipschitz with lower bilipschitz constant On the other hand, for every ε > 0 we provide an example of a regular mapping f : I d → R d with Reg(f ) = 3 and the following property. The set of points x such that there is an open neighbourhood of x on which f is injective has measure at most ε. Consequently, for Lipschitz regular mappings f with Reg(f ) ≥ 3 we cannot hope for any control of the measure of the bilipschitz pieces A n if we insist on A n being open.
dI d and a set X ⊂ I d with the following properties: (iii) For every x ∈ X and every δ > 0 the mapping f is not injective on the ball B(x, δ). Moreover, there are disjoint, non-empty, open balls U 1 , We give a proof for the case d = 1. The example for d ≥ 1 can easily be constructed from this: If f : I → R is the example for the case d = 1 with an appropriate choice of ε, then the function h : Given a point a ∈ (0, 1) and c > 0 we will denote by F a,c the interval [a, a + 3c]. Next, we define a 1-Lipschitz function g(a, c) : I → I that makes two folds on F a,c in a sense; see Figure 2. More precisely, we let The function g(a, c) and the interval F a,c together with their images.
We will now summarise various properties of the mapping g(a, c) which will be needed in the following construction. It is clear that g(a, c) is 1-Lipschitz and g(a, c) − id ∞ ≤ 2c. Moreover, g isometrically maps each of the three subintervals [a + (i − 1)c, a + ic] of F a,c for i ∈ [3] onto the same interval [a, a + c]. Denoting by J 1 , J 2 the two components of the set I \ F a,c we further point out that the sets g(a, c)(J 1 ), g(a, c)(J 2 ) and g(a, c)(F a,c ) are pairwise disjoint subsets of I, and that g restricted to each J i is a translation. Therefore, for any interval U ⊆ g(a, c)(I), the preimage g(a, c) −1 (U ) is an isometric copy of U whenever U does not intersect g(a, c)(F a,c ), and g(a, c) −1 (U ) may be covered by 3 intervals of length L(U ) whenever U intersects g(a, c)(F a,c ).
Let (D n ) ∞ n=1 be a strictly increasing sequence of numbers D n ∈ [1, 3) such that We will construct sequences (f n ) ∞ n=1 , (a n ) ∞ n=1 and (c n ) ∞ n=1 of functions f n : I → R, points a n ∈ I and numbers c n > 0 satisfying the following conditions for every n ∈ N.
(1 n ) f n is 1-Lipschitz and for any interval U ⊆ f n (I), f −1 n (U ) can be covered by 3 intervals of length D n L(U ).
isometrically onto the same interval.
(6 n ) The point a n is the midpoint of a component of n=1 is a Cauchy sequence in C(I, R). Let f := lim n→∞ f n and X = I \ ∞ n=1 F an,cn . We will show that f and X have all of the desired properties.
Conditions (i) and (ii) follow immediately from (2 n ) and (3 n ) respectively. To see that f is (3, 1)-regular, note first that f is 1-Lipschitz because f is the uniform limit of a sequence of 1-Lipschitz functions. Further, given any non- Finally, we verify (iii). Fix x ∈ X and δ > 0. The condition (6 n ) ensures that the set ∞ n=1 F an,cn is dense in I, whilst (3 n ) forces the sequence (c n ) ∞ n=1 to converge to zero. Therefore, we may choose i such that F a i ,c i ⊆ (x − δ, x + δ). Then for all n ≥ i, f n behaves according to (5 n ) on F a i ,c i . From this it follows easily that f also satisfies the condition of (5 n ) on F a i ,c i , which verifies (iii).
To complete the proof, we describe how to construct the sequences (f n ) ∞ n=1 , (a n ) ∞ n=1 and (c n ) ∞ n=1 satisfying (1 n )-(6 n ) for all n ∈ N. The construction is illustrated in Figure 3.
We start with f 1 := g(a 1 , c 1 ) on I. It is clear that the conditions (1 n ), -(6 n ) are satisfied for n = 1. Suppose that n ≥ 1 and that we have sequences . We now describe how to define f n+1 , a n+1 and c n+1 extending these sequences. Let a n+1 ∈ I be a point satisfying (6 n+1 ) and choose c n+1 ≤ ε2 −(n+2) sufficiently small so that (3 n+1 ) is satisfied and It is possible to choose c n+1 sufficiently small so that the last inequality is satisfied because the right hand side converges to a positive constant when we send c n+1 to zero. We define Then From (4 n ) and the fact that With this in mind, (4 n+1 ) and (5 n+1 ) can be derived easily from (4 n ), (5 n ) and the properties of the mapping g( · , · ) discussed at the start of the proof.
It only remains to verify (1 n+1 ). We fix an interval U ⊆ f n+1 (I) and verify that f −1 n+1 (U ) can be covered by 3 intervals of length D n+1 L(U ). We will use the condition (1 n ), the identity and the covering property of g(·, ·) −1 (U ) described at the start of the proof.
. Then, using (2.2) and (4 n ), we deduce that f −1 n+1 (U ) can be covered by 3 intervals of length at most L(U ) ≤ D n+1 L(U ). In the remaining case, we have that and, by (1 n ), the latter set can be covered by 3 intervals of length This completes the construction of the sequences (f n ) ∞ n=1 , (a n ) ∞ n=1 and (c n ) ∞ n=1 .
A natural question that can come to the reader's mind is where we can put Lipschitz regular mappings on the imaginary scale between bilipschitz and Lipschitz mappings? Are they closer to general Lipschitz mappings or rather to bilipschitz ones?
We can show that a typical 1-Lipschitz mapping, in the sense of the Baire Category Theorem, is not injective on any open subset of the domain, and hence, in the light of Theorem 2.11, a typical 1-Lipschitz mapping is not Lipschitz regular: Let D = B(u, r) ∈ B, g ∈ I(D) and η > 0. To verify that I(D) is nowhere dense we will find g ∈ V and s > 0 such that g − g ∞ < η and B(g , s) ∩ I(D) = ∅.
Choose ε < min {L(B(u, r/2), r/2, η} and let f : I d → I d and X ⊆ I d be given by Another question that a curious reader may ask is whether Lipschitz regular mappings can be characterised as Lipschitz mappings admitting a bilipschitz decomposition as in Theorem 2.11.
However, this turns out not to be the case. It is easy to construct an example with infinitely many overlapping images of bilipschitz pieces. But even more is true: It is possible to construct an injective 1-Lipschitz function on the unit interval that has a decomposition as in Theorem 2.11, but, at the same time, is not Lipschitz regular. An example f is given by the formula where g : I → I is any positive, bounded, measurable function which is constant and equal to one on a dense collection of open subintervals of I and not a.e. bounded away from zero.

Geometric properties of bilipschitz mappings.
Bilipschitz mappings of a Euclidean space R d transform volume according to the formula L(f (A)) = A |Jac(f )| dL. In this section we establish that bilipschitz mappings cannot transform volume too wildly. In some sense we show that sufficiently fine grids of cubes must witness 'continuity' of the volume transform. This in turn places rather restrictive conditions on the Jacobian of a bilipschitz mapping, which we will exploit in Section 4 in order to find non-realisable densities. Our work in this section is an interpretation of the construction of Burago and Kleiner [4], which we modify in various ways, leading to some extensions of the results in [4]. The main result of this section will be the following lemma:

For any
and e 1 -adjacent cubes S, S ∈ S i such that Statement 1 expresses that each collection of cubes S i+1 is much finer than the previous collection S i . The inequality of statement 2 can be interpreted geometrically as stating that the volume of the image of the cube S under h j is very close to the volume of the image of its neighbour S . Put differently, we may rewrite the inequality of 2 in the following form: It is possible to assemble Lemma 3.1 using predominantly arguments contained in the article [4] of Burago and Kleiner. However, Burago and Kleiner do not state any version of Lemma 3.1 explicitly and to prove Lemma 3.1 it is not sufficient to just take some continuous part of their argument. One needs to inspect their whole proof in detail and work considerably to put together all of the pieces correctly. Therefore, we present a complete proof of Lemma 3.1 in which we introduce some new elements. The proof of Lemma 3.1 requires some preparation and will be given at the end of this section.
We remark that in [5] Cortez and Navas came up with a variant of the Burago-Kleiner construction carried out completely in a discrete setting; they used it to produce a repetitive, non-rectifiable separated net in R d .
Lying behind all of the results of the present section is a simple property of Lipschitz mappings of an interval: If [0, c] ⊆ R is an interval and a Lipschitz mapping h : [0, c] → R n stretches the endpoints 0, c almost as much as its Lipschitz constant allows, then it is intuitively clear that the mapping h is close to affine. The next dichotomy can be thought of as a 'discretised' version of this statement: Statement 2 is a discrete formulation of the condition that the Lipschitz constant of h is not almost realised by the endpoints 0, c. Statement 1 expresses in a discrete way that h is close to affine; after partitioning the interval [0, c] into N subintervals of equal length this statement asserts that h looks like an affine mapping on nearly all pairs of adjacent subintervals.
We now formulate a multi-dimensional version of Lemma 3.2; see Figure 4. We consider thin cuboids in R d of the form [0, c] × [0, c/N ] d−1 and prove that when such a cuboid is sufficiently thin, that is when N is sufficiently large, then the one-dimensional statement for L-Lipschitz mappings f : [0, c] → R n given in Lemma 3.2 can, in a sense, be extended to L-bilipschitz mappings f :  For now, let us demonstrate how we intend to apply Lemma 3.3. First, we show that whenever statement 1 of Lemma 3.3 holds for a bilipschitz mapping h into R d , there are adjacent cubes S i and S i+1 whose images under h have almost the same measure. Eventually this will lead to conclusion 2 of Lemma 3.1.
Let the mappings f 1 : Then f 1 , f 2 are both L-bilipschitz mappings of the cube S i ∈ Q d c/N which satisfy f 1 − f 2 ∞ ≤ cε/N , due to (3.1). These conditions imply a bound on the difference in volume of the images f 1 (S i ) and f 2 (S i ), namely corresponding to 0 and ce 1 respectively. If, again, it is the case that statement 2 is valid for this mapping, then we find points a 2 , b 2 inside the new cuboid which g stretches by a factor (1 + ϕ) more than it stretches the pair a 1 and b 1 , and so a factor (1 + ϕ) 2 times more than it stretches a 0 and b 0 . The process is illustrated in Figure 5. We iterate this procedure as long as possible to obtain sequences (a i ) and (b i ) satisfying where the final bound is given by the lower bilipschitz inequality for g. It is clear now that the procedure described above cannot continue forever: Otherwise, for i sufficiently large, the inequality above contradicts the L-Lipschitz condition on g. Thus, Lemma 3.3 tells us that after at most r-iterations of the procedure, where r ∈ N is a number determined by d, L and ε, we must have that statement 1 is valid for the appropriate restriction of the mapping g. Figure 5: An illustration of the strategy to eliminate statement 2 from Lemma 3.3. The stretch factor of g on the points a 1 and b 1 is at least (1 + ϕ)-times larger then the stretch factor of g on the points 0 and ce 1 . If statement 2 applies in the next iteration, we find two points stretched by g with factor at least (1 + ϕ) 2 -times the stretch factor on 0 and ce 1 .
Let us now present the conclusions of the above sketch formally. We postpone the formal proof of this statement until the appendix.
such that statement 1 of Lemma 3.3 is valid for the mapping g p : [0, We are now ready to give a proof of Lemma 3.1:  Let us now verify that the above defined families S 1 , . . . , S r satisfy condition 1 in the statement of Lemma 3.1. It is immediate from Definition 3.6 that S r ⊆ S r−1 ⊆ . . . ⊆ S 1 .
Thus, given 1 ≤ i < r and S ∈ S i , we have that Note that any cube in the collections S i+1 has the form for some w ∈ c i+1 Z d and l ∈ [N ]. Since S ∈ Q c i /N and c i /N = M c i+1 , such a cube can only intersect S in a set of positive Lebesgue measure when w ∈ S. Therefore, the number of cubes in S i+1 that can intersect S ∈ Q c i /N in a set of positive Lebesgue measure is bounded above by

It follows that
where, in the above, we use c i+1 = c i /N M and L(S) = (c i /N ) d and prescribe that N is sufficiently large so that the inequality holds. Thus, statement 1 is satisfied. Turning now to statement 2, we consider a k-tuple (h 1 , . . . , h k ) of L-bilipschitz mappings h i : U → R d and define a mapping g : U → R kd co-ordinate-wise by We fix i ∈ Ω and impose the condition ε < 1 2L on ε. Then the conditions of Lemma 3.4 are satisfied for L = L √ k, ε, d, N , c = c p , h = h t,p for each t ∈ [k] and i. Hence, when we prescribe that ε ≤ ζ 2(L √ k) d+1 d . Set S = p j=1 z j + S i and S = p j=1 z j + S i+1 . It is clear upon reference to Definition 3.6 that S and S are e 1 -adjacent cubes belonging to the family S p . Moreover, we have h t (S) = h t,p (S i ) and h t (S ) = h t,p (S i+1 ) for all t ∈ [k]. Therefore S and S verify statement 2 of Lemma 3.1 for the k-tuple (h 1 , . . . , h k ). This completes the proof of Lemma 3.1.

Realisability in spaces of functions.
The objective of the present section is to prove that in some sense almost all continuous functions ρ ∈ C(I d , R) do not admit a Lipschitz regular mapping f : where we view C(I d , R) as a Banach space with the supremum norm − ∞ . More precisely, we prove the following result: Then E is a σ-porous subset of C(I d , R).

Remark 4.2.
To be able to work with functions ρ ∈ C(I d , R) attaining negative values as well, we extend the definition of the pushforward measure to such functions: where by ρ + , ρ − we mean the positive and the negative part of ρ. Technically speaking, the pushforward measure is no longer a measure, but a difference of two measures 4 . However, we will use it only in the form of (4.1), that is, when the result is again a measure. This is only a technical tool that helps us treat functions attaining negative values properly, but it does not bring in any additional difficulty to the present work. An alternative option would be to say that, by definition, no function with negative values satisfies (4.1), but the statement of Theorem 4.1 would be then seemingly weaker. [24] prove the existence of a positive function ρ : I 2 → R for which equation (4.1) has no bilipschitz solutions f : I 2 → R 2 . We point out that Theorem 4.1 extends this result in various ways. Firstly, Lipschitz regular mappings of I d into R d form a larger class than the class of bilipschitz mappings from I d to R d . Thus, Theorem 4.1 establishes the existence of a density ρ which admits no solutions f to equation (4.1) inside a larger class of mappings. Secondly, Theorem 4.1 asserts the existence of not only one such density ρ, but states that almost all continuous functions ρ ∈ C(I d , R) are not realisable in the sense of (4.1) for Lipschitz regular mappings f : I d → R d . That bilipschitz non-realisable functions contain a dense G δ subset of both the set of positive continuous functions and the set of positive, L ∞ -bounded, measurable functions on the unit square [0, 1] 2 , was recently proved by Viera in [26], but [26] is completely independent from the present work. Remark 4.3. We point out that there are positive, bilipschitz non-realisable densities in C(I d , R) which fail to be Lipschitz regular non-realisable, i.e. positive functions ρ ∈ C(I d , R) for which equation (4.1) admits Lipschitz regular but not bilipschitz solutions f : I d → R d . An example may be constructed as follows. We split the unit cube I d in half, distinguishing two pieces D 1 := [0, 1 2 ]×I d−1 and D 2 := [ 1 2 , 1]×I d−1 and write f : I d → D 1 for the mapping which 'folds D 2 onto D 1 '. More precisely, the mapping f is defined as the identity mapping on D 1 and as the reflection in the hyperplane 1 2 × R d−1 on D 2 . Let ψ ∈ C(D 1 , R) be a positive, bilipschitz non-realisable density with values in (0, 1). We impose the additional mild condition that ψ is constant with value 1 2 inside the hyperplane

Burago and Kleiner [4] and McMullen
The existence of such a density ψ follows easily from the d-dimensional analog of [4, Theorem 1.2]. Set ρ = ψ on D 1 . The bilipschitz non-realisability of ρ is now already assured, no matter how we define ρ on D 2 . To make ρ Lipschitz regular realisable, we define ρ on The function ρ : I d → R is continuous and positive, whilst the mapping f : I d → R d is Lipschitz regular and satisfies f (I d ) = D 1 . Moreover, for any measurable set S ⊆ D 1 we have where, for the penultimate equation, we use the change of variables formula and the fact that f restricted to the set D 2 is an affine isometry. This verifies the Lipschitz regular realisability of ρ.
Porous and σ-porous sets. We recall the definitions of porosity and σ-porosity according to [27, Definition 2.1], where they are referred to as 'lower porosity' and 'lower σ-porosity' respectively. (i) A set P ⊆ X is called porous at a point x ∈ X if there exist ε 0 > 0 and α ∈ (0, 1) such that for every ε ∈ (0, ε 0 ) there exists y ∈ X such that y − x ≤ ε and B(y, αε) ∩ P = ∅.
(ii) A set P ⊆ X is called porous if P is porous at x for every point x ∈ P .
(iii) A set E ⊆ X is called σ-porous if E may be expressed as a countable union of porous subsets of X.
The class of σ-porous subsets of a Banach space X is strictly contained in the class of subsets of X of the first category in the sense of the Baire Category Theorem. Further, the notions of porosity and σ-porosity extend to metric spaces in the natural way. For a survey on porous and σ-porous sets we refer the reader to [27]. Due to its relevance later in the paper, we point out that porosity of a set P ⊆ X is a weaker condition than requiring P to be porous at all points x ∈ X (not just at points x ∈ P ). For example, the set 1 n : n ∈ Z \ {0} is porous in R but is not porous at the point 0. Let us begin working towards the proof of Theorem 4.1.
Porous decompositon of E. In the present paragraph we describe how to partition the set E into a countable family of porous sets (E C,L,n ). We will need the lower bilipschitz constant b(·) given by the conclusion of Proposition 2.19. Let (O n ) ∞ n=1 be a countable basis for the topology of I d . For C, L, n ∈ N we let E C,L,n denote the set of all for a.e. y ∈ O n . (4.2) Note that the basis set O n 'generates' the diagram of bilipschitz homeomorphisms f i : Figure 7. However, the critical role of O n in the definition above is to prescribe the portion of the domain I d on which all functions ρ ∈ E C,L,n have the special form given by (4.2). Figure 7: The diagram of a bilipschtz decomposition for a density ρ ∈ E C,L,n .
To explain the origins of equation (4.2), we refer the reader back to Proposition 2.19. Consider a Lipschitz regular mapping f : I d → R d and the non-empty open set T ⊆ f (I d ) given by the conclusion of Proposition 2.19. Because the preimage f −1 (T ) decomposes precisely as a union of N sets on which f defines a bilipschitz homeomorphism to T , a pushforward (signed) measure of the form f ρL with ρ ∈ C(I d , R) can be expressed on T as a sum of integrals involving ρ and Jacobians of N bilipschitz homeomorphisms f 1 , . . . , f N . Thus, whenever f ρL = L| f (I d ) we obtain some equation relating ρ to finitely many bilipschitz homeomorphisms and their Jacobians. We will see that this equation has precisely the form of (4.2).
For C, L ∈ N, let E C,L denote the subset of C(I d , R) consisting of all functions ρ for which there exists a (C, L)-regular mapping f : Clearly we have E = C,L∈N E C,L . In the next lemma we prove that E C,L is covered by the sets (E C,L,n ).
To see that these choices witness that ρ ∈ E C,L,n , it only remains to verify equation (4.2).
We conclude that Recall that the sets Y i and V are all bilipschitz homeomorphic via the mappings f i : Y i → V . Therefore, we may make the substitution x = f 1 (y) in the above equation, after which a simple rearrangement and an application of a 'chain rule identity' for Jacobians yields (4.2).
If, for the time being, we treat the terms ρ(f −1 i • f 1 (y)) in (4.2) as constants, then, on the open set O n , functions ρ ∈ E C,L,n are linear combinations of at most C Jacobians of L/b(C)-bilipschitz mappings. The purpose of the next lemma is to provide, for given constants k and L, a function ψ ∈ C(I d , R) which is small in supremum norm, but far away from being a linear combination of k L-bilipschitz Jacobians. Lemma 4.6. Let ε, ζ ∈ (0, 1), L ≥ 1, k ∈ N and U ⊆ I d be an open set. Then there exists a function ψ ∈ C(I d , R) such that ψ ∞ ≤ ε, supp(ψ) ⊆ U and for every k- To prove Lemma 4.6 it suffices to consider the families of tiled cubes S 1 , . . . , S r given by the conclusion of Lemma 3.1 applied to d, k, U , L, ζ and some very small η ∈ (0, 1), and to define ψ as a 'chessboard function' whose average value on e 1 -adjacent cubes makes jumps of size at least ε. From the conclusion 1 of Lemma 3.1 we may essentially regard the cubes from two different families S i , S j as pairwise disjoint; choosing η sufficiently small ensures that the values of ψ on S j have negligible impact on the average values of ψ on cubes in S i for i < j. We postpone the formal description of this construction until Appendix C, since it is a standard argument without any deep ideas. For now, let us proceed to the key proof of the present section, namely the verification of porosity of the sets (E C,L,n ). We actually prove that the sets (E C,L,n ) possess a stronger property: Lemma 4.7. For every C, L, n ∈ N, E C,L,n is a porous subset of C(I d , R). In fact, the set E C,L,n is porous at every point φ ∈ C(I d , R).
Before we begin the proof, we will outline the strategy. For given C, L, n ∈ N, φ ∈ C(I d , R) and ε ∈ (0, 1), our task is to find a function φ ∈ C(I d , R) so that φ − φ ∞ ≤ ε and B( φ, αε) ∩ E C,L,n = ∅ for some α = α(C, L, n, φ). We will exploit the uniform continuity of φ: By prescribing at the start a sufficiently small open set U ⊆ O n we may treat φ as constant (relative to ε) on U and indeed on any L/b(C)-bilipschitz image of U . Thus, when using the condition (4.2) for functions ρ ∈ E C,L,n we will always be able to treat the terms ρ(f −1 i • f 1 (y)) as constant. In other words, on U we will have that all functions in E C,L,n are linear combinations of at most C L/b(C)-bilipschitz Jacobians. We set φ = φ+ψ where ψ is given by the conclusion of Lemma 4.6 for ζ = αε, L = L/b(C) and an appropriate choice of k. If B( φ, αε) ∩ E C,L,n is non-empty, then, up until addition by the 'constant' φ, ψ is approximately a linear combination of at most C L/b(C)-bilipschitz Jacobians on U . This will be incompatible with the conclusion of Lemma 4.6.
and (4.4) holds for ψ. Using (4.2), we may now write for a.e. y ∈ O n . To complete the proof we will show that the average value of the final expression over the cube S is too close to its average value over S , that is, closer than the condition (4.4) on ψ allows.
Let i ∈ {2, 3, . . . , N }. Then we have that whenever y, z ∈ S ∪ S ⊆ U . Therefore, in light of (4.5) and the fact that for all y ∈ S ∪ S . Thus, we have (4.8) For the final inequality above we used ρ diam(S ∪ S ) ≤ diam(U ) < δ we deduce that the average value of the right hand side of (4.6) over the cube S differs from its average value over S by at most However, with the setting ζ = ε 2(4 + C( φ ∞ + 1 + 6(L/b(C)) d )) this number is strictly less than ε, contrary to (4.4). Thus, we conclude that B φ, ε 2(4 + C( φ ∞ + 1 + 6(L/b(C)) d )) ∩ E C,L,n = ∅, which demonstrates the porosity of E C,L,n at φ.
It is now a simple task to combine the previous Lemmas for a proof of Theorem 4.1.
Proof of Theorem 4.1. From Lemma 4.5 we have whilst Lemma 4.7 asserts that each of the sets in the union on the right hand side is porous.
Readers interested in the resolution of Feige's question may proceed immediately to Section 5. In the remainder of the present section we discuss an independent topic of interest.

Realisability in L ∞ spaces.
Until now we have only studied realisability in spaces of continuous functions. However, functions ρ admitting a bilipschitz or Lipschitz regular solution f : need not be continuous. Therefore, it is natural to study the set of realisable functions in the less restrictive setting of L ∞ (I d ), the space of all Lebesgue measurable, real-valued functions ρ defined on I d , which are bounded with respect to the L ∞ -norm ρ ∞ := inf C > 0 : |ρ(x)| ≤ C for a.e. x ∈ I d .
We will prove that the set of all bilipschitz realisable functions in L ∞ (I d ) is a σ-porous set. For bilipschitz mappings f , (4.9) is equivalent to the equation |Jac(f )| = ρ a.e. (4.10) The question of whether Lipschitz regular realisable densities are also σ-porous, or in some sense negligible, in L ∞ spaces remains open.
Then G is a σ-porous subset of L ∞ (I d ). In fact G may be decomposed as a countable union of sets (G L ) ∞ L=1 so that each G L is porous at every point ρ ∈ L ∞ (I d ).
Remark. For 1 ≤ p < ∞, the question of whether the set of bilipschitz realisable densities is small in L p (I d ) is not interesting because the set of all a.e. bounded functions is already σ-porous in this space.
The proof of Theorem 4.8 will require the following lemma, for which we recall the notation of Section 3. We postpone the proof of the lemma until Appendix C, since it is based on a slightly more delicate version of the construction employed in the proof of Lemma 4.6. Lemma 4.9. Let λ > 0, S ⊆ Q d λ be a finite collection of tiled cubes in I d , ρ ∈ L ∞ (I d ) and ε > 0. Then there exists a function ψ = ψ(S, ρ, ε) ∈ L ∞ (I d ) such that ψ − ρ ∞ ≤ ε and − S ψ − − S ψ ≥ ε whenever S, S ∈ S are e 1 -adjacent cubes.
Proof of Theorem 4.8. We decompose G as G = ∞ L=1 G L where G L := ρ ∈ L ∞ (I d ) : (4.10) admits an L-bilipschitz solution f : Fix L ≥ 1, ρ ∈ L ∞ (I d ) and ε > 0. We will find ρ ∈ L ∞ (I d ) with ρ − ρ ∞ ≤ ε and B( ρ, ε/16) ∩ G L = ∅. This will verify the porosity of the set G L and complete the proof of the theorem. Let U ⊆ I d be an arbitrary, non-empty, open set, ζ = ε/2 and let η ∈ (0, 1) be a parameter to be determined later in the proof. Let r ∈ N and the tiled families S 1 , . . . , S r of cubes contained in U be given by the conclusion of Lemma 3.1 applied to d, k = 1, L, η and ζ. We define a sequence of functions where ψ(S i , ρ, ε) is given by the conclusion of Lemma 4.9. Now let ρ ∈ L ∞ (I d ) be defined by It is clear that ρ − ρ ∞ ≤ ε. Let φ ∈ B( ρ, ε/16). Then, given i ∈ [r] and e 1 -adjacent cubes S, S ∈ S i we let T := S ∩ r j=i+1 S j and T := S ∩ r j=i+1 S j . From Lemma 3.1, part 1 we have that max {L(T ), L(T )} ≤ ηL(S). We deduce when we set η = 1 16 . Together with Lemma 3.1, part 2 and the setting ζ = ε/2, this implies that equation (4.10) with ρ = φ has no L-bilipschitz solutions f :

Feige's question.
In the 90's Uriel Feige asked a fascinating question [22, Problem 2.12] 5 , see also [21,Problem 5.5]: Question 5.1. Is there a constant L > 0 such that for every n ∈ N and every set S ⊂ Z d such that |S| = n d there is an L-Lipschitz bijection f : S → [n] d ?
We provide a negative solution to Question 5.1 in a strong sense in all dimensions d ∈ N, d ≥ 2. For d = 1 the answer is trivially positive. First, we transform Feige's question into a question about densities of measures supported on I d . In order to do so, we adapt a construction of Burago and Kleiner [4] that encodes densities into separated nets in R d . Then we prove that if we plug in any of the positive non-realisable functions whose existence is ensured by Theorem 4.1, the sequence of discrete sets that arise from the chosen function provide a negative solution to Question 5.1.
We provide an equivalent version of Question 5.1 that fits better the tools we have. The equivalence of Question 5.1 and Question 5.2 is easy to see. Thus, we provide its formal justification only in Appendix D in Observation D.1.
The following theorem, in conjunction with Theorem 4.1, proves that the answer to Question 5.2 is negative. In other words, it proves Theorem 1.2 stated in the introduction.
Assume that the answer to Question 5.2 is positive. Then for every measurable function ρ : Before we start a presentation of the proof of Theorem 5.3, let us add a convenient observation, which asserts that it is sufficient to prove Theorem 5.3 only for densities ρ with average value 1. Proof. We write f : I d → R d for a Lipschitz regular mapping satisfying f (ρL) = L| f (I d ) . Let us consider a mapping φ α (x) := d √ α · x and observe that φ α • f is the sought after solution: 5 Feige asked the question for d = 2, as it was announced in the introduction in Question 1.1.

for any measurable set
The proposed proof of Theorem 5.3 relies on the discretisation procedure of Burago and Kleiner [4], which they used to encode a bilipschitz non-realisable density ρ into a separated net S in R d that cannot be bijectively mapped onto Z d in a bilipschitz way. We will use their procedure, with a small technical modification, to encode a given bounded, measurable function ρ : I d → (0, ∞) into a sequence of separated sets S i in R d such that each S i has cardinality n d i for some n i ∈ N. Burago and Kleiner [4] showed that a bilipschitz bijection S → Z d would yield a bilipschitz solution to the equation f (ρL) = L| f (I d ) . We will show that if there are L-Lipschitz bijections f i : S i → [n i ] d for some L > 0 and every i ∈ N, then there is also a Lipschitz regular solution to the equation f (ρL) = L| f (I d ) . This part of our proof is different than that of Burago and Kleiner, although we follow their overall idea 6 .
The preceding explanation implies that if we use any of the positive continuous functions ρ ∈ C(I d , R)\E, where E is the set from Theorem 4.1, we get a sequence of separated sets that will provide a counterexample to Question 5.2, and thus, also a negative answer to the question of Feige.
In the next paragraph, we describe our modified version of the Burago-Kleiner construction [4].
Encoding a bounded, positive function into a sequence of separated sets. To begin with, let us describe the main ideas of the construction informally. Each set S i in the sequence of sets 'discretising' a given bounded, measurable function ρ : I d → (0, ∞) represents 'a picture' of ρ taken with a resolution increasing with i. More precisely, to define S i we first blow up the domain of ρ by some factor l i and then subdivide it into m d i cubes of the same size. Let T be one of these cubes. The set S i ∩ T will be then formed by regularly spaced points inside T of number proportional to the average value of the blow up of ρ by the factor l i over T . If we choose l i and m i so that l i m i goes to infinity with i, the set S i will capture more and more precisely variations of ρ on small scales. This idea is illustrated in Figure 8.
Since we want to use the sets S i in Question 5.2, we need to make sure that the total number of points forming each S i is a d-th power of some natural number. That's the technical difference in comparison to the original construction of Burago and Kleiner [4]. Now, we will write everything formally.
We assume that we are given a measurable function ρ : I d → (0, ∞) such that 0 < inf ρ ≤ sup ρ < ∞. We choose two sequences {l i } i∈N ⊂ R + and {m i } i∈N ⊂ N. We put several conditions on them, which we describe a bit later. First, let us introduce some notation.
We write ϕ i (x) := l i · x for a blow up by factor l i . It is clear that Figure 8: An illustration of the construction that encodes a given density ρ into a sequence of separated sets {S i } ∞ i=1 . The position and the number of points inside S i approximates ρ, after rescaling by ϕ −1 i , with a precision increasing with i.
This number will stand for |S i ∩ T i,k |. The possibility to choose n i,k among two different values will allow us to ensure that |S i | = n d i for some n i ∈ N. The change of variables formula implies Using the bounds on ρ we infer that Now we can state the required conditions on l i , m i and n i,k : 1. l i → ∞, m i → ∞ and l i m i → ∞ as i → ∞. 2. for every i ∈ N we choose each n i,k from the two possibilities so that there is We show that these conditions can be satisfied at once. We set l i := m 1+p i for a suitable p > 0 and choose {m i } ∞ i=1 ⊂ N as an increasing sequence. This will satisfy the first condition.
In order to satisfy the second condition, it is sufficient to make sure that the interval i dL contains a d-th power of a natural number. If we denote by a i the largest integer such that a d i < thus we see that m i satisfies m d i > C(d)a d−1 i provided we choose p < 1 d−1 and m i sufficiently large.
After setting up the parameters l i and m i properly, we can construct the separated sets S i . We first form sets S i,k by placing n i,k distinct points inside each T i,k and then set But instead of providing an explicit formula for S i,k , it will be enough to consider any sufficiently separated set of n i,k points inside T i,k and argue that the separation constant can be chosen independently of i, k.
Since each T i,k has a side of length l i m i , given r > 0 satisfying we may define S i,k as any r-separated set of n i,k points inside T i,k that, in addition, satisfies dist(S i,k , ∂T i,k ) ≥ r/2; an example is depicted in Figure 9. The last condition ensures that the set S i is r-separated as well. It remains to verify the existence of the separation constant r > 0 satisfying the inequality (5.3) for all i, k. Using the inequality (5.2) and standard estimates we obtain that Proof of Theorem 5.3. Using Observation 5.4, we may assume that − I d ρ = 1. We use the construction described above on ρ and get sequences of r-separated sets S i and S i,k together with the parameters l i , m i , n i , n i,k , mappings ϕ i and cubes T i,k .
Assuming the positive answer to Question 5.2, we get L > 0 and a sequence of L- We pull each f i back to I d in the following way. We write X i for the set ϕ −1 i (S i ) and define a mapping g i : . It is not hard to see that the mappings g i are uniformly Lipschitz: for every x, y ∈ X i . Thus, we need to examine the behaviour of the sequence l i From the definition of n i,k we have the following bounds on n d i = m d i k=1 n i,k : Using the identity − I d ρ = 1 and the fact that Jac(ϕ i ) = l d i we immediately obtain that Since m i l i → 0, the sequence n i l i i∈N is bounded and converges to 1, and hence, l i n i → 1 as well. Consequently, for any L > L we can find i 0 ∈ N such that for every i ≥ i 0 the mappings g i are L -Lipschitz. By trimming off the initial segment of the sequence (g i ) i∈N up to i 0 , we can assume that all mappings g i are L -Lipschitz for any chosen L > L.
We extend each g i , by Kirszbraun's extension theorem, to a mappingḡ i : I d → R d such that Lip(ḡ i ) = Lip(g i ). By the Arzelà-Ascoli theorem we know that the sequence (ḡ i ) ∞ i=1 subconverges to a limit f , which is also L -Lipschitz 7 . By passing to a convergent subsequence, we may assume thatḡ i ⇒ f .
Before we prove that f transforms measure according to the equation we have to establish additional notation and present two lemmas about weak convergence of measures 8 . For i ≥ 1 we define a measure µ i on I d by In order to show that f (ρL) = L| f (I d ) we first prove that µ i converges weakly to ρL on I d . Moreover, this will be shown to imply that (ḡ i ) (µ i ) converges weakly to f (ρL).
Finally, we prove that (ḡ i ) (µ i ) also converges weakly to L| f (I d ) , and hence, f (ρL) and L| f (I d ) must be the same by the uniqueness of weak limits.
To this end we use the following two lemmas, which are probably a part of a common knowledge in measure theory, but the authors were unable to find a proper reference. Since their proofs are straightforward, we include them only in Appendix D.
Lemma 5.5. Let ν and (ν n ) ∞ n=1 be finite Borel measures on a compact metric space K. Moreover, assume that there is, for each n ∈ N, a finite collection Q n of Borel subsets of K that cover ν-almost all of K and, at the same time, Then ν n converges weakly to ν.
Lemma 5.6. Let K be a compact space and (ν n ) n∈N be a sequence of finite, Borel measures on K converging weakly to a finite Borel measure ν. Let X be a metric space and h n : K → X be a sequence of continuous mappings converging uniformly to h. Then (h n ) (ν n ) converges weakly to h (ν).
Equipped with the two lemmas above, we resume proving Theorem 5.3. Proof. We verify that the measures µ i and ρL satisfy the assumptions of Lemma 5.5. The only non-trivial assumption in this case is the existence of the collection Q i . We take Hence, µ i and ρL satisfy the assumptions of Lemma 5.5.
Proof. We know thatḡ i ⇒ f and that µ i converge weakly to ρL by Claim 5.3.1. Thus, it is sufficient to directly apply Lemma 5.6. Proof. We aim to use Lemma 5.5 again, but first we need to show that f ( is the support of the measure (ḡ i ) (µ i ), is precisely the set 1 n i , 2 n i , . . . , n i n i d , that is, a regular grid with n d i points inside I d . The situation is depicted in Figure 10. Figure 10: Weak convergence of (ḡ i ) (µ i ) to L| I d .
Now we observe that I d ⊆ f (I d ). Let y ∈ I d . Each g i (X i ) forms a √ d n i -net in the cube I d . Therefore, we get that dist(y,ḡ i (I d )) ≤ dist(y, g i (X i )) ≤ √ d n i . Sinceḡ i ⇒ f , we infer that dist(y, f (I d )) = 0. Because f (I d ) is closed, the point y has to lie in f (I d ).
Next, we need to argue that f (I d ) ⊆ I d . To this end, we observe that there is a sequence (s i ) i∈N ⊂ R + converging to zero such that each g i (X i ) forms an s i -net of f (I d ).
By the inequality (5.2) we know that Since l i m i → ∞, we can assume that for every i ∈ N and every k ∈ [m d i ] we have n i,k ≥ 1. This in turn implies that each set X i has at least one point in each cube Now let y ∈ f (I d ). By the discussion above, for every i ∈ N we can find y i ∈ g i (X i ) ⊂ I d such that y − y i 2 ≤ s i . This means that y i → y, and hence, y ∈ I d as well. This implies that f (I d ) ⊆ I d .
Finally, we can construct the collection Q i . It comprises of half-open cubes The cubes in Q i tile I d ; see again Figure 10. Observe that each cube Q ∈ Q i contains precisely one point of the set g i (X i ), and conversely, each point of g i (X i ) is contained in some cube Q ∈ Q i . Therefore, for every i ∈ N and Q ∈ Q i we get that Since we have already shown that f (ρL) = L| f (I d ) , we get for every measurable set inf ρ . Applying Lemma 2.4 we conclude that f is Lipschitz regular. This finishes the proof of Theorem 5.3.
A. Appendix to Section 2: Lipschitz regular mappings.
. Given ε > 0, we may choose δ i > 0 sufficiently small so that On the other hand, from the above inequality, we can also deduce that where the final inequality is obtained by choosing ε sufficiently small. Therefore, for all δ ∈ (0, δ))). Applying Proposition 2.7, part (iii) we infer that deg(f, B(x i , δ), y) = deg(g, B(x i , δ), y) = sign(Jac(g)(x i )) = sign(Jac(f )(x i )) for every δ ∈ (0, δ i ]. In the above we used the formula (2.1) for the degree function of mappings in C 1 (U , R d ). Next, we choose δ < min {δ 1 , . . . , δ n } sufficiently small so that the sets (B(x i , δ)) n i=1 are pairwise disjoint subsets of U . In the case n = 1 we choose a (possibly empty) open set U 2 ⊆ U \ B(x 1 , δ) and use y ∈ R d \ f (U 2 ), Proposition 2.7, part (ii) and property (d2) with U 1 = B(x 1 , δ) to obtain the desired result. When n > 1, we iteratively apply property (d2) to get sign(Jac(f (x))). Lemma 2.9. Let a ∈ R d , r > 0 and f : B(a, r) → R d be a continuous mapping (Fréchet) differentiable at the point a with rank(Df (a)) = d. Then there is δ 0 > 0 such that for every δ ∈ (0, δ 0 ] we have B(a, δ)).
Proof. Without loss of generality, we assume that a = 0 and that f (0) = 0. We write z(y) := f (y) − Df (0)(y). For every y ∈ R d we have z(y) 2 ∈ o( y 2 ). We pick β > 0 small, whose precise value will be set later, and choose δ 0 small enough so that z(y) 2 ≤ β y 2 for every y ∈ B(0, δ 0 ). Since the linear mapping Df (0) has full rank, its inverse Df (0) −1 is well defined and has finite norm. We get that Df (0) −1 • f (y) − y = Df (0) −1 (z(y)). We fix δ ∈ (0, δ 0 ]. Let us write g δ (y) := 1 δ Df (0) −1 • f (δy). The mapping g δ is defined on the ball B(0, 1) and continuous. We also get that We set β := . Applying Lemma 2.10 we get that g δ (B(0, 1)) ⊇ B 0, 1 2 ; hence, we infer that non-empty open balls and f : U 1 ∪ U 2 → R d be a continuous mapping such that f | U i is a homeomorphism for i = 1, 2 and f (U 1 ) = f (U 2 ). Then there exists s > 0 such that for any continuous mapping h : Proof. Fix y ∈ f (U 1 ) = f (U 2 ). The multiplication theorem for the topological degree of a composition, see e.g. [14,Theorem 2.10], implies that the degree of the composition of two injective mappings is equal to the 'chain rule type product' of the degrees of the two mappings. Applying this to (f | U i ) −1 • f | U i = id and using the property (d1) of the degree we have that deg(f, U i , y) ∈ {−1, 1} for i = 1, 2.
B. Appendix to Section 3: Geometric properties of bilipschitz mappings. such that for all c > 0, n ∈ N, N ∈ N, N ≥ 2 and all L-Lipschitz mappings h : [0, c] → R n at least one of the following statements holds: Proof. Let M ∈ N and ϕ ∈ (0, 1) be parameters to be determined later in the proof. Let c > 0, n ∈ N, N ∈ N and h : [0, c] → R n be an L-bilipschitz mapping. The assertion of the Lemma holds for h if and only if the assertion holds for ρ•h, where ρ : R n → R n is any distance preserving transformation. Therefore, we may assume that h(0) = (0, 0, . . . , 0) and h(c) = (A, 0, . .
Assume that the second statement does not hold for h. In other words we have that We complete the proof, by verifying that the first statement holds for h.
For later use, we point out that (B.1) implies , t = t(L, ε) ∈ (ϕ, 1) be some parameter to be determined later in the proof and For x ∈ P we have This inequality follows from the definition of P , the inequality (B.2) and t > ϕ. For the remaining co-ordinate functions we have Combining the two inequalities above we deduce can only intersect in the endpoints. Therefore the set is a finite union of intervals with endpoints in c N M Z ∩ [0, c] and with total length c − |Γ|c N . Using Γ ∩ P = ∅ and (B.2) we deduce that Rearranging this inequality, we obtain where, for the last inequality, we apply N ≥ 2. It follows that the set c Letting . Moreover for any i ∈ Ω and x ∈ S i , we can find x ∈ P with |x − x| ≤ c/N M . This allows us to apply (B.3) to get We are now ready to specify the parameters t, M and ϕ, so that the inequalities obtained above verify statement 1. First, we prescribe that t ∈ (0, 1) is sufficiently small and M ∈ N is sufficiently large so that √ 5tL + 2L M < ε. Finally we demand that ϕ ∈ (0, t) is small enough so that 6ϕ ϕ+t < ε.
Proof. In this proof we will sometimes add the superscript d or d − 1 to objects such as the Lebesgue measure L or vectors e i , 0 in order to emphasise the dimension of the Euclidean space to which they correspond. For d ≥ 2, we will express points in R d in the form x = (x 1 , x 2 , . . . , x d ). Given x = (x 1 , . . . , x d ) ∈ R d and s ∈ R we let Thus, we get that for each s ∈ [0, c/N ] at least one of the following statements holds: To deduce the fourth inequality in the sequence above we use the lower bilipschitz bound on h. In fact, this is the only place in the proof of Lemma 3.3 where we use that the mapping h is bilipschitz and not just Lipschitz. The final inequality is ensured by taking Therefore, by Fubini's theorem, For each i ∈ [N − 1] we let S i := (i−1)c N , ic N × 0, c N d−1 , define and observe that Combining the two inequalities derived above for L d (A), we deduce Moreover, for any i ∈ Ω and any cube Q ⊆ S i with side length (2 √ θL d (S i )) 1 d we have A ∩ Q = ∅. Therefore, for any i ∈ Ω and any x ∈ S i we can find x ∈ A ∩ S i with Using this approximation, we obtain Thus, statement 1 is verified when we prescribe that θ > 0 is sufficiently small and N 0 is sufficiently large so that (1 − √ θ) ≥ 1 − ε and 4L √ dθ 1/2d + θ + 2L N 0 < ε.
such that statement 1 of Lemma 3.3 is valid for the mapping g p : [0, c p ] × [0, c p /N ] d−1 → R n defined by Proof. The appropriate condition on the parameter r = r(d, L, ε) ∈ N will be determined later in the proof. Let ϕ = ϕ(d, L, ε) be the parameter given by the conclusion of Lemma 3.3. We implement the following algorithm.
1. If statement 1 of Lemma 3.3 holds for h = g i and c = c i then stop. If not proceed to step 2. Therefore, whenever the algorithm does not terminate in step 1, we have that such a point z i+1 required by step 2 exists by Lemma 3.3.
To complete the proof, it suffices to verify that Algorithm B.1 terminates after at most r iterations. This is clear, after rewriting (B.5) in the form where the latter inequality follows by induction and the lower bilipschitz inequality for g = g 1 . If Algorithm B.1 completes r + 1 iterations then, for r > 2 log L log(1+ϕ) , the above inequality with i = r contradicts the L-Lipschitz condition on g.
The next lemma is an intuitively clear statement: It says that if two L-bilipschitz mappings defined on a cube S ∈ Q d λ are close in supremum norm, then the volumes of their images are also close.
Since the above argument is completely symmetric with respect to f 1 and f 2 , we also have L(f 2 (S)) − L(f 1 (S)) ≤ 2dL d+1 εL(S).
C. Appendix to Section 4: Realisability in spaces of functions. Proof. Let r ∈ N and the finite, tiled families S 1 , S 2 , . . . , S r of cubes in U be given by the conclusion of Lemma 3.1 applied to d, k, L, ζ and η ∈ (0, 1), where η is a parameter to be determined later in the proof. We will now define a sequence ψ 0 , ψ 1 , ψ 2 , . . . , ψ r of continuous functions on U . The sought after function ψ will then be defined on U by ψ| U = ψ r . We begin by setting ψ 0 = 0. If i ≥ 0 and ψ i is already constructed, we define ψ i+1 as any continuous function on U with the following properties: (iii) For every cube S ∈ S i , − S ψ i ∈ {−8ε/9, 8ε/9}.
(iv) For every pair of e 1 -adjacent cubes S, S ∈ S i , − S ψ i = − S ψ i . It is clear that such a continuous function exists; see Figure 11 for an example. Note that conditions (iii) and (iv) just prescribe that the average values of ψ i on the cubes in S i follow a 'chessboard' pattern. The final function ψ r clearly satisfies −ε ≤ ψ r ≤ ε and ψ r = 0 outside of r i=1 S i . Moreover, for any i ∈ [r] and e 1 -adjacent cubes S, S ∈ S i we will prove that − S ψ r − − S ψ r ≥ ε. (C.1) Fix i ∈ [r], e 1 -adjacent cubes S, S ∈ S i and combine (iii) and (iv) to obtain − S ψ i − − S ψ i = 16ε/9. Letting T := S ∩ r j=i+1 S j and T := S ∩ r j=i+1 S j , we have from conclusion 1 of Lemma 3.1 that max {L(T ), L(T )} ≤ ηL(S). Moreover, condition (i) in the construction above guarantees that ψ r = ψ i on (S \ T ) ∪ (S \ T ). We conclude that Thus, setting η = 1/9, we verify (C.1). To complete the proof, it now only remains to extend the function ψ to the whole of I d by setting ψ = 0 outside of U so that supp(ψ) ⊆ U .
Proof. We define the function ψ on I d inductively as follows. Pick any S 1 ∈ S such that the first co-ordinate projection map π 1 : S → R attains its minimum on S 1 . This ensures that S 1 = T + λe 1 for any cube T ∈ S. We set ψ = ρ on S 1 . If n ≥ 1, distinct cubes S 1 , . . . , S n ∈ S and the function ψ| n i=1 S i are defined, we extend ψ as follows: If S \ {S 1 , . . . , S n } = ∅, we complete the construction of ψ by setting ψ = ρ on I d \ S. Otherwise, we choose, if possible, S n+1 ∈ S \ {S 1 , . . . , S n } such that S n+1 = S n + λe 1 . If this is not possible we choose S n+1 ∈ S \ {S 1 , . . . , S n } arbitrarily such that the mapping π 1 : S \ n i=1 S i → R attains its minimum on S n+1 . In the first case, we define ψ on S n+1 by In the latter case, we may simply take ψ = ρ on S n+1 . It is now readily verified that the function ψ ∈ L ∞ (I d ) produced by this construction possesses all of the required properties.
Symmetrically, we derive the lower bound Summing over all elements of Q n and using the assumption that elements of Q n form ν-almost disjoint cover of K we get K ψ dν n − K ψ dν ≤ 2εν(K) + ε(max |ψ| + ε).
Since ψ is continuous on a compact K it is also bounded and the right hand side of the last inequality tends to zero with ε.
Lemma 5.6. Let K be a compact space and (ν n ) n∈N be a sequence of finite, Borel measures on K converging weakly to a finite Borel measure ν. Let X be a metric space and h n : K → X be a sequence of continuous mappings converging uniformly to h. Then (h n ) (ν n ) converges weakly to h (ν).
Proof. We take any ψ ∈ C(X, R) with compact support. We can bound As n → ∞ the first term in the final sum goes to zero since ψ • h n converges uniformly to ψ • h. Moreover, the second term converges to zero as well, because ν n converges weakly to ν.