Singularity of maps of several variables and a problem of Mycielski concerning prevalent homeomorphisms

S. Banach pointed out that the graph of the generic (in the sense of Baire category) element of $\text{Homeo}([0,1])$ has length $2$. J. Mycielski asked if the measure theoretic dual holds, i.e., if the graph of all but Haar null many (in the sense of Christensen) elements of $\text{Homeo}([0,1])$ have length $2$. We answer this question in the affirmative. We call $f \in \text{Homeo}([0,1]^d)$ singular if it takes a suitable set of full measure to a nullset, and strongly singular if it is almost everywhere differentiable with singular derivative matrix. Since the graph of $f \in \text{Homeo}([0,1])$ has length $2$ iff $f$ is singular iff $f$ is strongly singular, the following results are the higher dimensional analogues of Banach's observation and our solution to Mycielski's problem. We show that for $d \ge 2$ the graph of the generic element of $\text{Homeo}([0,1]^d)$ has infinite $d$-dimensional Hausdorff measure, contrasting the above result of Banach. The measure theoretic dual remains open, but we show that the set of elements of $\text{Homeo}([0,1]^d)$ with infinite $d$-dimensional Hausdorff measure is not Haar null. We show that for $d \ge 2$ the generic element of $\text{Homeo}([0,1]^d)$ is singular but not strongly singular. We also show that for $d \ge 2$ almost every element of $\text{Homeo}([0,1]^d)$ is singular, but the set of strongly singular elements form a so called Haar ambivalent set (neither Haar null, nor co-Haar null). Finally, in order to clarify the situation, we investigate the various possible definitions of singularity for maps of several variables, and explore the connections between them.

It turns out that this result is closely related to the following notions. Definition 1. 4. We say that f ∈ Homeo([0, 1] d ) is singular if there exists a Borel set F ⊆ [0, 1] d such that λ d (F ) = 1 and λ d (f (F )) = 0, and strongly singular if f is differentiable at x and det f (x) = 0 for almost every x ∈ [0, 1] d .
In Theorem 3.7 below we will prove that the graph of f ∈ Homeo([0, 1]) has length 2 iff f is singular iff f is strongly singular. Therefore the following problems are all natural generalizations of Banach's observation and Mycielski's problem. We show that for d ≥ 2 the generic element of Homeo([0, 1] d ) is singular but not strongly singular, and also that for d ≥ 2 the prevalent element of Homeo([0, 1] d ) is singular, but the set of strongly singular elements form a so called Haar ambivalent set (neither Haar null, nor prevalent).
Moreover, in Section 3, in order to clarify the situation, we investigate the various possible definitions of singularity for maps of several variables, and explore the connections between them.
Finally, in the Appendix we prove an approximation result communicated to us by J. Luukkainen stating that the set of somewhere smooth homeomorphisms is dense in Homeo([0, 1] d ). These types of problems are really delicate, dimension 4 is particularly difficult, since for example there exists a homeomorphism f of (0, 1) 4 into R 4 that cannot be uniformly approximated even by bi-Lipschitz homeomorphisms, let alone smooth ones [9, page 183]. Moreover, J. Luukkainen [20] pointed out to us that we can even find such an f ∈ Homeo([0, 1] 4 ).
We note here that several problems we consider in this paper have well-known analogues for continuous maps instead of homeomorphisms. For example it is classical that the generic f ∈ C([0, 1] d , R d ) is nowhere differentiable, (the case of d = 1 dates back to Banach [4], for the general case see e.g. Claim 3.6), hence not strongly singular, but singular (see Theorem 3.2). The prevalent f ∈ C([0, 1], R) is nowhere differentiable [16], hence not strongly singular, and also not singular, since quite the opposite holds, namely, the occupation measure λ • f −1 is absolutely continuous [3,Theorem 4.2]. (It will be shown in Theorem 3.2 that f is singular iff the occupation measure is a singular measure.) But it should be noted that the case of homeomorphisms is very different and much more difficult. First, as we have just mentioned above, the problem of approximating homeomorphism with more regular homeomorphisms is very subtle. Second, Homeo([0, 1] d ) is not commutative, and the theory of Haar null sets is notoriously complex for non-abelian groups (the group structure on C([0, 1] d , R d ) is pointwise addition, hence commutative).
As our final remark, we mention that there is a completely different approach to define random homeomorphisms, see Graf, Mauldin, and Williams [15].

Preliminaries and notations
Let G be a Polish group. A set A ⊆ G is called compact catcher if for every compact set K ⊆ G there exist g, h ∈ G such that gKh ⊆ A. The following fact is well known, see e.g. [10, Lemma 6.6.1].
Fact 2.1. If A is compact catcher then it is not Haar null.
We use λ d to denote the d-dimensional Lebesgue measure on R d , and we simply write λ instead of λ 1 . Let | · | stand for the Euclidean norm on R d . Let B(x, r) and U (x, r) denote the closed and open balls of radius r around x, respectively. We say that K ⊆ R d is a Cantor set if it is perfect and totally disconnected. For a set A let diam A, ∂A, and int A stand for the diameter, the boundary, and the interior of Let Homeo([0, 1] d ) denote the group of homeomorphisms of the d-dimensional cube [0, 1] d (under composition, as the group operation), and let us equip this group with the maximum metric. It is well known that this is a Polish group [17, 9.B.8], i.e., a completely metrizable separable topological group. Note that the maximum metric itself is not complete, but instead the topology it generates can also be generated by the complete metric (f, g) = f − g + f −1 − g −1 , where · denotes the maximum norm. We will also often work with the Polish space of continuous functions f : [0, 1] d → R d endowed with the maximum metric as well. Let B(f, r) denote the closed ball of radius r around f with respect to the maximum metric.
We use B(X) to denote the Borel sets of a topological space X and P(X) to denote the space of the Borel probability measures on a separable metrizable space X. The classical result [17,Theorem 17.19] states that P(X) is also a separable metrizable space when equipped with the weak topology.
For A ⊆ R d and s ≥ 0 we define s-dimensional Hausdorff measure as The length of a curve γ : [0, 1] → R d is defined as The length of the graph of a function f : . Since γ is one-to-one, the following well-known fact follows e.g. from [ Let X be a completely metrizable topological space. A subset of X is somewhere dense if it is dense in a non-empty open set, otherwise it is called nowhere dense. We say that M ⊆ X is meager if it is a countable union of nowhere dense sets, and a set is called co-meager if its complement is meager. It is not difficult to show that a set is co-meager iff it contains a dense G δ set. We say that the generic element x ∈ X has property P if {x ∈ X : x has property P} is co-meager. See e.g. [17] for more on these concepts. Let The above matrix L is not necessarily unique. We say that f is not differentiable at x 0 if there exists no L satisfying (2.1). Actually, we will always prove in this case the stronger property that lim sup Note that if f is differentiable at x 0 and either H = [0, 1] d or x 0 is a Lebesgue density point of H then there is a unique L satisfying (2.1), in which case we define f (x 0 ) = L.

Versions of singularity
In this section we explore the connections between the various possible definitions of singularity of maps. The special cases when the map is one-to-one, or differentiable almost everywhere, or d = 1 are also examined.
We say that f is strongly singular if f is differentiable at x and det f (x) = 0 for almost every x ∈ H, and f is singular if there exists a Borel set F ⊆ H such that λ d (H \ F ) = 0 and λ d (f (F )) = 0. (1) f is strongly singular, = 0 for almost every x ∈ H, = 0 for almost every y ∈ R d , (4) the pushforward measure λ d • f −1 is singular with respect to λ d , (5) f is singular.
and then construct the counterexamples.
(1) ⇒ (2): Assume that f exists at x ∈ H and det f (x) = 0 and let For ε > 0 fixed, let r 0 > 0 be small enough so that is contained in the εr-neighborhood of a (d − 1)-dimensional ball of radius Rr for any r < r 0 . Using the formula for the volume of the d-dimensional ball, λ d (B(0, r)) = C d r d , it follows that for all r < r 0 , where c x does not depend on r, r 0 or ε. Therefore the lim sup in question is indeed 0.
(2) ⇒ (5): Let For a fixed ε > 0, let Clearly, F ⊆ B. The 5r-covering theorem implies that there is a countable family B ⊆ B consisting of pairwise disjoint balls such that 5B def = {5B : B ∈ B } covers B, where 5B = B(x, 5r) for a ball B = B(x, r). Then where we used the disjointness of the balls in B and the fact that B ⊆ [−1, 2] d . It follows that λ d (F ) = 0, finishing the proof.
(3) ⇔ (4): The equivalence of a measure, in this case λ d • f −1 being singular with respect to λ d and lim sup r↓0 = 0 almost everywhere is a wellknown fact in the theory of differentiation of measures. In particular, it follows from [27,Theorem 7.14].
We note first that the sets defined throughout the proof could be proved to be Borel sets using standard methods. However, as the proof works without using that the sets in question are Borel, the reader is encouraged to think of λ d as an outer measure.
Using ( Using that for each x ∈ A the derivative of f at x exists with non-zero determinant, it is easy to check that ∞ n=1 ∞ k=1 A n,k = A. It follows that λ d (A n,k ) > 0 for some n and k. Let us fix such an n and k. Using the definition of A n,k , one easily checks that f is one-to-one on A n,k and that f −1 | f (A n,k ) is Lipschitz with Lipschitz constant at most n. Using also that A n,k ⊆ F , we obtain that (3) ⇒ (2): To show that the generic continuous function satisfies (3), it is enough to show that it satisfies (5). It is probably widely known, but we include its short proof for the sake of completeness.

Proof. Take a union of Cantor sets
∞ n=1 F i,n gives our desired co-meager set. The set F i,n is clearly open by the outer regularity of Lebesgue measure, so it suffices to check that it is dense. Let g ∈ C([0, 1] d , R d ) and ε > 0 be given, we need to find f ∈ F i,n ∩ B(g, ε). By the uniform continuity of g there exists δ > 0 such that |g(x) − g(y)| < ε whenever |x − y| ≤ δ. As C i is totally disconnected, there is a finite family of pairwise disjoint open sets {U 1 , . . . , U k } such that diam(U j ) < δ for all 1 ≤ j ≤ k and C i ⊆ k j=1 U j . We can define f ∈ B(g, ε) such that f | Ci∩Uj is constant for all 1 ≤ j ≤ k. Indeed, it is enough that f | Ci ∈ B(g| Ci , ε), then Tietze Extension Theorem guarantees the existence of a continuous extension f ∈ B(g, ε). Clearly f ∈ F i,n , and the proof is complete. Now we prove that the generic continuous map f : [0, 1] d → R d does not satisfy (2). Moreover, we prove the following more general claim. First we need a known fact.
Proof. For the sake of simplicity we denote f (B(x, r) ∩ [0, 1] d )) by f (B(x, r)). Fix n ∈ N + , it is enough to show that contains a dense open set, since then ∞ n=1 F n will give our desired co-meager set. Assume that g ∈ C([0, 1] d , R d ) and ε > 0 are given, we will find f ∈ B(g, 2ε) such that B(f, θ) ⊆ F n for some θ > 0. By the uniform continuity of g we can choose a positive δ < ε/(6n) such that |g(x) − g(y)| ≤ ε whenever |x − y| ≤ δ.
Composing f with a homothety of ratio 1 6n and applying Fact 3.4 to it with β = δ and α = δ Therefore h ∈ F n , and the proof is complete.
Then we have λ d (f (K)) = 0, for the well-known argument see the proof of Claim 3.3, so f automatically satisfies (2). As K is totally disconnected, it is widely known that the generic f ∈ C(K, R d ) is one-to-one, see e.g. [2,Lemma 2.6]. Hence f is a homeomorphism from K to f (K). Therefore, it is enough to prove that the generic f ∈ C(K, R d ) is nowhere differentiable. We prove the following more general statement.
Claim 3.6. Let (K, ρ) be a compact, perfect metric space. Then for the generic Proof. For all k, n ∈ N + let Clearly the functions in ∞ k=1 ∞ n=1 F k,n satisfy (3.1), so it is enough to prove that the F k,n are dense open sets.
Fix k, n ∈ N + . First we prove that F k,n is open. Assume that f i ∈ F c k,n and f i → f uniformly as i → ∞, we need to prove that f ∈ F c k,n . Fix m ≥ n arbitrarily. Then there exist x i,m ∈ K for all i ∈ N + such that diam f i (B(x i,m , 1/m)) ≤ k/m. As K is compact, by choosing a subsequence we may assume that x i,m → x m as i → ∞. Then clearly diam f (B(x m , 1/m)) ≤ k/m. As this holds for all m ≥ n, we obtain that f ∈ F c k,n . Finally, we prove that F k,n is dense. Let ε > 0 and a continuous g : K → R d be given. We will construct an f ∈ B(g, 2ε) ∩ F k,n . Choose m ≥ n with k/m < ε and finitely many distinct points Then |f (z i ) − g(z i )| ≤ 2ε, so applying Tietze Extension Theorem for the finite set {x i , z i : i ≤ N } we obtain a continuous map f ∈ B(g, 2ε) with the above property. We need to check that f ∈ F k,n . Indeed, fix x ∈ K arbitrarily. Then there exists This completes the proof.
Indeed, consider finitely many points 0 = a 0 < a 1 < · · · < a k = 1. Then for the corresponding approximation of the length of graph(f ) we obtain There is a Borel nullset N ⊆ [0, 1] with λ(f (N )) = 1. As the H 1 measure cannot increase under an orthogonal projection, projecting graph(f | N ) to the y-axis implies that Similarly, projecting graph(f | [0,1]\N ) to the x-axis yields Adding up the above two inequalities implies (3.4). Here we used that the two parts of the graphs are continuous one-to-one images of Borel sets, so they are Borel (in particular, H 1 measurable) by [17,Corollary 15.2]. Now we prove that (6) ⇒ (1): Let f ∈ Homeo([0, 1]) be a homeomorphism satisfying (6). Since the length of the graph of f is 2, for each n ∈ N + we can choose points 0 = a n 0 < a n 1 < · · · < a n kn = 1 such that (3.5) a n i − a n i−1 < 1 n for all 1 ≤ i ≤ k n and the sum of the lengths of the corresponding line segments satisfies Define S n = (a n i−1 , a n i ) : 1 ≤ i ≤ k n and f (a n i ) − f (a n i−1 ) a n i − a n i−1 It is enough to prove that λ(S n ) → 1 as n → ∞. Indeed, let S = lim sup n S n , then clearly λ(S) = 1. By the definition of S and (3.5) at each point x ∈ S the lower derivative Finally, we prove that λ(S n ) → 1 as n → ∞. For each n ∈ N + define Fix n ∈ N + , we clearly have For all 1 ≤ i ≤ k n define w n i = (a n i − a n i−1 ) + (f (a n i ) − f (a n i−1 )) − |(a n i − a n i−1 , f (a n i ) − f (a n i−1 )|. For every i ∈ I n we obtain by elementary calculation that (3.8) w n i ≥ (a n i − a n i−1 ) 1 + 1/n − 1 + 1/n 2 ≥ a n i − a n i−1 n + 1 .

Singularity of generic homeomorphisms
In this section we prove that for each d ≥ 2 the generic f ∈ Homeo([0, 1] d ) is not strongly singular, but for all d ≥ 1 it is singular. This answers the generic case of Questions 1.6 and 1.7. The next result implies that the generic f ∈ Homeo([0, 1] d ) is not strongly singular.
We will only need the full power of the following theorem in the next section. Here we only use it for the one-element compact sets K = {g}.
be a compact set and let ε > 0. Then there exists a homeomorphism f ∈ B(id, 2ε) such that g • f is nowhere differentiable for all g ∈ K. Moreover, for all g ∈ K and x ∈ [0, 1] d , First we prove the following elementary lemma. We will only use its consequence Corollary 4.4 in this section, the more general statement will be needed in Section 6.  Proof of Lemma 4.3. We call a closed interval 2 m ] for some ∈ N, and we call it dyadic if it is m-dyadic for some m ∈ N. Let us fix a strictly increasing sequence a m of natural numbers such that for all m ∈ N. We construct piecewise linear continuous functions ϕ m : [0, 1] → [0, 1] such that ϕ 0 ≡ 0 and for all m ≥ 1 we have First we show that this is indeed enough. By (1) the functions ϕ m uniformly converge, let ϕ = lim m→∞ ϕ m be their limit. Then clearly ϕ : To check that ϕ satisfies the lemma with some q n , let I and J be intervals of length 2 −n and s n for some n, respectively. Clearly, if n < a 0 , then we can set q n = 1 to satisfy the conclusion of the lemma. Otherwise, let m be chosen so that a m ≤ n < a m+1 . Then one can find a sequence of a m+1 -dyadic intervals A s n is decreasing, (4.2) yields Hence, we can find a sequence of consecutive 2 m+1 -dyadic intervals J 1 , J 2 , . . . , Using (4.4), applying (2) to each pair of intervals I i and J j , and (4.3) yield that We can then define q n = 18 · 2 −2 m for the largest m with a m ≤ n. One can easily check that q n ↓ 0, which will finish the proof of the lemma.
Finally, we construct the functions ϕ m satisfying (1) and (2). If m = 0 then ϕ 0 ≡ 0 is continuous and linear, while (1) and (2) are vacuous. Now let m ≥ 1, and suppose that ϕ 0 , . . . , ϕ m−1 are already constructed satisfying our conditions. Our goal is to construct ϕ m . Using that ϕ m−1 is piecewise affine, we can decompose Then it suffices to define ϕ m on each such interval separately, with the condition that ϕ m and ϕ m−1 coincide on the boundary of such intervals.
So fix such intervals I and To see that (iv) can indeed be satisfied, note that there are exactly 2 2 m−1 many 2 m -dyadic intervals inside J i . By (i) and (ii) our function ϕ m is a well defined, piecewise linear, continuous function. Then ϕ m satisfies (1) because of (iii), and (2) follows from (iv). This completes the proof. onto ∂B(0, r) for all 0 ≤ r ≤ 1. Therefore, it is enough to prove (4.1) for T . For all n ∈ N + define s n = min{0 ≤ s ≤ 1 : |g(t 1 ) − g(t 2 )| ≥ 1/n whenever g ∈ K and |t 1 − t 2 | ≥ s}.
By the compactness of K we have s n > 0 for all n, and s n ↓ 0 as n → ∞. By Now we can define f : T → T by f (r, α, y) = (h(r), α + ϕ(r) mod 2π, y).
Proof of Theorem 4.1. Define The elements of G are clearly nowhere differentiable. Theorem 4.2 yields that G is dense in Homeo Fix n ∈ N + , it is sufficient to show that G c n is closed. Let f k ∈ G c n be a sequence such that f k → f uniformly as k → ∞, we need to prove that f ∈ G c n . By definition, for each k there exists x k ∈ [0, 1] d such that |f k (x k ) − f k (y)| ≤ n|x k − y| for all y ∈ U (x k , 1/n). By choosing a convergent subsequence we may assume that x k → x as k → ∞. Let y ∈ U (x, 1/n) be arbitrarily fixed. Then y ∈ U (x k , 1/n) for all large enough k, so Thus f ∈ G c n , and the proof is complete.
Proof. By choosing h to be a 'radial' homeomorphism fixing x and moving the points away from x (but fixing the boundary) we can obtain that h([0, 1] d \ B(x, r)) is in an arbitrarily small neighborhood of the boundary of [0, 1] d , hence its measure can be arbitrarily small. Proof. Choose n ∈ N + such that the cubes in the 1 n -grid are of diameter less than δ > 0. In every cube in this 1 n -grid, since K is nowhere dense, we can fix a ball that is disjoint from K. In every such cube apply the (scaled version of the) previous lemma to this ball. Then the obtained homeomorphisms together form a homeomorphism h that clearly works. Proof. This set is clearly open (by the outer regularity of Lebesgue measure), so it suffices to check that it is dense. Let g ∈ Homeo([0, 1] d ) and δ > 0 be given. Apply the previous lemma with K = g(C). Then f = h • g is in the δ-neighborhood of g, Proof of Theorem 4.6. Let (C n ) n∈N + be a sequence of nowhere dense compact sets in [0, 1] d such that M ⊆ ∞ n=1 C n . Applying the previous lemma for every C n and for every ε = 1 k (k ∈ N + ) we obtain that λ d (f (M )) = 0 holds for the generic f ∈ Homeo([0, 1] d ).

Singularity of prevalent homeomorphisms
In this section we prove that for all d ≥ 2 the strongly singular homeomorphisms f ∈ Homeo([0, 1] d ) form a Haar ambivalent set and also that for every d ≥ 1 the prevalent f ∈ Homeo([0, 1] d ) is singular. These results answer the prevalent case of Questions 1.6 and 1.7.    In the nth step we remove the middle open intervals of length 2 −2n from each of them, and obtain 2 n disjoint, nth level elementary intervals. We continue this procedure for all n ∈ N + , and the limit set is the Smith-Volterra-Cantor set K. A closed interval is an elementary interval of K if it is an nth level elementary interval for some n ≥ 1. We can define elementary intervals analogously for similar copies of K. Then For all n ≥ 1 the length of the nth level elementary intervals equals ϕ(r) = max{0 ≤ s ≤ 1 : |g(x) − g(y)| ≤ r 3 whenever g ∈ K and |x − y| ≤ s}.
Clearly, ϕ is increasing, and by the compactness of K we obtain that ϕ(r) = 0 iff r = 0. We will define a homeomorphism f : [0, 1] → [0, 1] and families of nonoverlapping closed intervals {I n } n≥1 such that for each n and I ∈ I n we have First we prove that this suffices. Let G = lim inf n I n and for each x ∈ G and large enough n choose I n (x) ∈ I n such that x ∈ I n (x). Let P = {x ∈ G : B(x, 4 −n ) ⊆ I n (x) for all large enough n}.
First we show that λ(P ) = 1. Indeed, for every n ≥ 1 define Therefore λ(S) = 0 by the Borel-Cantelli Lemma, hence λ(P ) = λ(G) = 1. Now we define F ∈ Homeo([0, 1] d ) by We show that for all g ∈ K we have (g•F ) (x) = 0 d×d at each x ∈ P d . Fix x ∈ P d and g ∈ K and, then I n (x i ) are defined such that B(x i , 4 −n ) ⊆ I n (x i ) for all large enough n and 1 ≤ i ≤ d. Fix such an n and let B n = B(x, Then the definition of F , (2), and (1) imply that diam F (B n ) ≤ ϕ(2 −n ). Thus the definition of ϕ yields that Finally, we construct I n and f satisfying (1) We will construct an increasing homeomorphism f n : [0, 1] → [0, 1] and a compact set K n ⊆ [0, 1] such that K n−1 ⊆ K n and f n | Kn−1 = f n−1 | Kn−1 . We will obtain K n by filling up the complementary intervals of K n−1 with similar copies of K. More precisely, enumerate the complementary intervals {(u i , v i )} i≥1 of K n−1 and let us decompose each [u i , v i ] into finitely many non-overlapping closed intervals {J i,j = [w i,j−1 , w i,j ]} 1≤j≤ki such that w i,0 = u i , w i,ki = v i , and for all 1 ≤ j ≤ k i we have w i,j − w i,j−1 = 2 −N for some large enough N = N (i) ∈ N + for which the oscillation of f n−1 satisfies Let K i,j be the similar copy of K with endpoints {w i,j−1 , w i,j }, more precisely, let Let us modify f n−1 on each interval J i,j simultaneously as follows. Fix i, j, we will define f n | Ji,j as a limit of uniformly convergent, strictly increasing functions for m = 1, 2. We can define h k for all k ∈ N. Since the length of a (k − 1)st level elementary interval is at most 2 −(k−1) , for all k ≥ 1 we obtain where the last inequality comes from id ∈ K. Inequality (5.4) yields that h k uniformly converges to some continuous h = h i,j . We show that h is strictly increasing. Let x, y ∈ J i,j such that x < y, we need to prove that h(x) < h(y). Choose a complementary interval U of K i,j such that U ⊆ (x, y). By our construction h| U = (h k )| U for a large enough k and h k is strictly increasing, so h(x) < h(y). By the construction for every k ≥ 1 and kth level elementary interval Define the homeomorphism f n : Note that f n is indeed strictly increasing, as f n−1 and h i,j are strictly increasing as well. We have defined f n for all n ∈ N. The construction and (5.2) imply f n − f n−1 ≤ 2 −n for all n ≥ 1, so f n converges to a homeomorphism f : [0, 1] → [0, 1]. Indeed, it is easy to show that f is strictly increasing: Let 0 ≤ x < y ≤ 1 be arbitrary, we need f (x) < f (y). As ∞ n=1 K n is dense in [0, 1], we can choose n ∈ N and x n , y n ∈ K n such that x < x n < y n < y. As f | Kn = (f n )| Kn by our construction and f n is strictly increasing, we obtain proving that f is a homeomorphism. Let I be the family of elementary intervals I of all the copies of K of the form K i,j that we used during the construction. We show that for all I ∈ I we have (5.6) diam f (I) ≤ ϕ(diam I).
Indeed, each I ∈ I is an elementary interval of K i,j ⊆ K n for some i, j, n.
We need to show that f and I n satisfy (1), (2), and (3). Property (1) follows from the definition of I n , and (5.6) yields (2). Hence it is enough to show (3). Let We need to show that G ⊆ lim inf n I n (actually equality holds). Indeed, if z ∈ G, then z is in a similar copy C of K with similarity ratio 2 −m for some m ∈ N. For each k ≥ 1 the kth elementary interval I k of C containing z satisfies |I k | = 2 −m b k , so 2 −(m+k)−1 ≤ |I k | < 2 −(m+k) by (5.1). Thus I k ∈ I k+m , and z ∈ I k+m for all k ≥ 1. Thus z ∈ lim inf n I n , and the proof is complete.
Before proving the main result of this section we need some preparation. We also need [17,Theorem 17.25], which states the following.
Theorem 5.6. Let (X, S) be a measurable space, Y a separable metrizable space, and A ⊆ X × Y a measurable set. Then the map is the set of probability measures on Y endowed with the weak topology, and B(·) stands for the Borel σ-algebra.
Now we are ready to prove the following.
we need to prove that F is prevalent. First we prove that F is a Borel set. For each n ∈ N + define

By Theorem 3.2 we obtain that
Notice that the set is Borel. Indeed, it can be written as To conclude the proof, we will construct a measure µ which witnesses that F is prevalent. Fix a singular homeomorphism f 0 ∈ Homeo([0, 1] d ) and a subset F ⊆ [0, 1] d such that λ d (F ) = 1 and λ d (f 0 (F )) = 0. For s = (s 1 , s 2 , . . . , s d ), where s i ∈ [1, 2] for each 1 ≤ i ≤ d, let us define ψ s ∈ Homeo([0, 1] d ) as It is easy to see that this defines a Borel probability measure on Homeo([0, 1] d ). We prove that µ is indeed a witness measure for F. Assume that g, h ∈ Homeo([0, 1] d ) are arbitrarily fixed, it is enough to show that µ(g −1 • F • h −1 ) = 1.
According to the definition of µ, This means that we want to prove that for almost every pair (s, t) the composition g • ψ s • f 0 • ψ t • h is an element of F. Using the definition of F, this is equivalent to the condition that for almost every pair (s, t) there exists a set C s,t ⊆ [0, 1] d such that λ d (C s,t ) = 1 and Suppose that x ∈ (0, 1) d . Since h is a homeomorphism, h(x) ∈ (0, 1) d as well. By definition t → ψ t (h(x)) is a one-to-one, bi-Lipschitz map from [1,2] Now assume x ∈ (0, 1) d and t ∈ [1, 2] d . By Lemma 5.5 there exists a Borel set R g ⊆ [0, 1] d such that λ d (R g ) = 1 and λ d (g(N )) = 0 for every nullset N ⊆ R g . As f 0 • ψ t • h is still a homeomorphism, we have (f 0 • ψ t • h)(x) ∈ (0, 1) d . We may repeat the previous argument to show that Applying (5.8), (5.9), and Fubini's theorem we obtain that almost every triple Applying Fubini's theorem in the other direction yields that for almost every pair (s, t) there exists a set C s,t ⊆ [0, 1] d with λ d (C s,t ) = 1 satisfying We only need to show that C s,t satisfies (5.7). By (5.10) we obtain Since ψ s is Lipschitz, it maps measure zero sets to measure zero sets, so Then (5.11) and the definition of R g imply (5.7), and the proof is complete.
6. Solution to the problem of Mycielski and H d -measure of graphs in higher dimensions In this section first we answer Mycielski's problem, then formulate the generalizations of this question and Banach's result to higher dimensions.
First, Theorems 5.7 and 3.7 immediately yield the following, answering Question 1.3 of Mycielski. New we turn to Question 1.5, the natural generalization to higher dimensions. First we answer this question in the generic case, then we partially answer it in the prevalent case. Somewhat surprisingly, the H d -measure tend to be infinite. We need to prove two lemmas first. Lemma 6.4. Let d ≥ 2 and C, ε ∈ R + be given. Then for every f ∈ Homeo([0, 1] d ) that is Lipschitz on a cube Q ⊆ [0, 1] d there exists g ∈ Homeo([0, 1] d ) that is also Lipschitz on Q such that g ∈ B(f, ε) and H d (graph(g| Q )) > C.
Proof. By shrinking Q if necessary, we can assume that the distance between Q and the boundary of [0, 1] d is positive. Then, since f is a homeomorphism, there exists δ ∈ R + such that First we check that λ d ({(x 1 , . . . , x d ) ∈ Q : det f (x 1 , . . . , x d ) = 0}) > 0 (note that f is Lipschitz on Q, and therefore differentiable almost everywhere on Q by the Rademacher Theorem [12, Theorem 3.1.6]). Indeed, if the determinant were 0 almost everywhere on Q, then by the implication (1) ⇒ (5) of Theorem 3.2 we could find a Borel set F ⊆ Q with λ d (Q \ F ) = 0 and λ d (f (F )) = 0, but using that f is Lipschitz on Q we would also have λ d (f (Q \ F )) = 0, hence λ d (f (Q)) = 0, which is absurd since f is a homeomorphism.
This implies that in particular f 2 cannot be the zero vector almost everywhere on Q (here f 2 is the second coordinate function of f ). Hence we can find a set P ⊆ Q with λ d (P ) > 0 such that f 2 exists and f 2 = 0 on P . By partitioning P into d many pieces and picking the one with positive measure we can assume that there is a fixed coordinate j such that ∂f2 ∂xj exists and ∂f2 ∂xj = 0 on P . And finally, by partitioning P into countably many pieces and picking the one with positive measure we can assume that ∃P ⊆ Q with λ d (P ) > 0 ∃j ∈ {1, . . . , d} ∃ ∈ R + such that ∂f 2 (x) ∂x j exists and ∂f 2 (x) ∂x j ≥ for every x ∈ P. (1) |ϕ| < min{ε, δ}, (2) the points of non-differentiability of ϕ are in D, where C is given in the statement of the theorem.
Next we slightly modify P . Since f is differentiable almost everywhere, we can assume by removing a nullset from P (i.e. (6.2) will still hold) that (6.3) f is differentiable at every point of P.
Similarly, for the finitely many points a of non-differentiability of ϕ we remove the set f −1 2 (a) ∩ P from P . By (2) these sets are of measure zero, hence (6.2) still holds, but now we also have that (6.4) ϕ is differentiable at f 2 (x) for every x ∈ P.
Clearly, g ∈ Homeo([0, 1] d ), and (since Φ is piecewise affine, and hence Lipschitz), g is Lipschitz on Q. So all that remains to check is H d (graph(g| Q )) > C.
The Hausdorff measure of the graph of a function can be computed by the Area Formula [12,Theorem 3.2.3]. Let ).
Then clearly graph(g| Q ) = range(G| Q ). Since G is Lipschitz and one-to-one on Q, we can apply the Area Formula which states that ) is the Jacobian of G, which exists almost everywhere on Q. In order to show that it suffices to prove that (6.5) JG(x) exists and JG(x) > C λ d (P ) at every point of the set P ⊆ Q from (6.2) above.
As G is Lipschitz on Q, by throwing away a final nullset from P we can assume that G , and hence also JG exist at every point of P .
Let x ∈ P be arbitrary. The Cauchy-Binet Formula [8, page 9] implies that det G (x) T G (x) is the sum of the squares of the determinants of the d × d sized minors of G (x). Hence, in order to obtain (6.5) it suffices to find a single minor of size d × d with the absolute value of its determinant greater than C λ d (P ) . We claim that the minor obtained by taking the first d rows (which form the d × d identity matrix) and replacing the jth row by the (d + 1)st works (here j comes from (6.2)). Some easy linear algebra shows that the determinant of this minor is the jth entry of its diagonal, that is, ∂g1(x1,...,x d ) ∂xj . Therefore, the proof will be complete if we show that at every point x = (x 1 , . . . , x d ) ∈ P we have .
By the definition of Φ and by (6.1) for every x = (x 1 , ..., x d ) ∈ Q (and even on a neighborhood of Q) we have that ) on a neighborhood of Q.
This completes the proof of (6.6), and hence the proof of the lemma.
Remark. Note that the construction of the map Φ above was inspired by the so called 'slides' from [1].
The following lower semicontinuity result seems to be known, the case of d = 2 is due to Besicovitch [5, page 21]. As we were not able to find a full reference (and for the reader's convenience), we include its proof here.
Before proving Lemma 6.5 we need some preparation. Let (K(R m ), d H ) be the non-empty compact subsets of R m endowed with the Hausdorff metric, that is, for all compact sets K 1 , K 2 ⊆ R m we have d H (K 1 , K 2 ) = min {r : K 1 ⊆ B(K 2 , r) and K 2 ⊆ B(K 1 , r)} , where B(A, r) = {x ∈ R m : ∃y ∈ A such that |x − y| ≤ r}. Then (K(R m ), d H ) is a Polish space, see [17] for more on this concept.
Let K n ⊆ R m be compact sets such that K n converges to K in the Hausdorff metric. We say that {K n } n≥1 is almost uniformly concentrated in dimension s if for every ε > 0 there is an s-almost Vitali covering B of K such that for each B ∈ B, For the proof of the following claim see [24,Theorem 10.14] after the straightforward modifications. Claim 6.7. Assume that {K n } n≥1 is an almost uniformly concentrated sequence of compact sets in R m in dimension s > 0 and K n converges to K in the Hausdorff metric. Then lim inf n→∞ H s (K n ) ≥ H s (K).
Proof of Lemma 6.5. Let f ∈ Homeo([0, 1] d ) be Lipschitz on a cube Q ⊆ [0, 1] d and assume that {f n } n≥1 is an arbitrary sequence of homeomorphisms such that f n → f uniformly as n → ∞. Then we need to prove that It is enough to show that K n def = graph(f n | Q ) is an almost uniformly concentrated sequence in dimension d. Indeed, we may assume that Q is closed, so the compact sets K n converge to K def = graph(f | Q ) in the Hausdorff metric. Therefore, applying Claim 6.7 for K n , K, and s = d will finish the proof of (6.8).
By the Rademacher Theorem λ d (Q \ D) = 0. As f | Q is Lipschitz, we have thus B is really a d-almost Vitali covering of K. Fix z ∈ D and B = B((z, f (z)), r) ∈ B with some 0 < r < δ(z). Let T = T z and V = {(x, T (x)) : x ∈ R d } be the tangent plane at z. Choose N ∈ N + such that for all n ≥ N and x ∈ [0, 1] d we have It is enough to prove that for all n ≥ N . Fix an arbitrary n ≥ N . By (6.10) and (6.9) for all x ∈ B(z, r) we obtain 2δ)). Note that if y ∈ B then there is a unique x ∈ B(z, r) such that y = (x, T (x)). Consider the continuous map S : B → V such that for y = (x, T (x)) ∈ B we have (6.13) S(y) = pr V ((x, f n (x))), where pr V is the orthogonal projection to V . By (6.12) if (x, T (x)) ∈ B then (x, f n (x)) ∈ K n ∩ B. Therefore, since projections cannot increase the Hausdorff measure, we obtain (6.14) (S(B )).
As K is compact, it is easy to see that s n ↓ 0. By Lemma 4.3 there exist a sequence q n ↓ 0 and a continuous function ϕ : [0, 1] → [0, 1 4 ] such that for all n ∈ N if I, J are intervals of length 2 −n and s n , respectively, then (6.17) λ({z ∈ I : ϕ(z) ∈ J}) ≤ q n 2 −n .
Fix an arbitrary f ∈ K, we will show that Fix an arbitrary n ∈ N and cubes Q 1 ⊆ T and Q 2 ⊆ [0, 1] d of edge length 2 −n , it is enough to prove that (6.19) µ(Q 1 × Q 2 ) ≤ q n 2 −nd .
Indeed, then the Mass Distribution Principle [22,Theorem 4.19] implies that with some c d > 0 we have so (6.18) holds. In order to prove (6.19), define the interval J as the orthogonal projection of f −1 (Q 2 ) to the first coordinate axis. Since diam Q 2 < d2 −n , the definition of s n implies that diam f −1 (Q 2 ) ≤ s n , so diam J ≤ s n as well. Let Q 1 = I 1 × · · · × I d , by (6.17) for all x 1 ∈ I 1 we obtain (6.20) λ({x 2 ∈ I 2 : ϕ(x 2 ) ∈ J − x 1 }) ≤ q n 2 −n .
It is straightforward that Hence (6.21) and (6.20) easily yield Thus (6.19) holds, and the proof is complete.

Concluding remarks and open problems
In Theorem 6.2 we showed that for d ≥ 2 the d-dimensional Hausdorff measure of the graph of the generic f ∈ Homeo([0, 1] d ) is infinite. The next question asks weather this can be slightly improved.
Question 7.1. Let d ≥ 2 be an integer. Is it true that H d | graph(f ) is non-σ-finite for the generic f ∈ Homeo([0, 1] d )?
Since graph(f ) ⊆ [0, 1] 2d , it may even be possible that dim H graph(f ) > d for the generic f ∈ Homeo([0, 1] d ), where dim H is the Hausdorff dimension, see e.g. [21] for the definition. In fact, we can show that if d = 4 and d = 5 then the generic f ∈ Homeo([0, 1] d ) satisfies dim H graph(f ) = d, we plan to publish this in a forthcoming paper. As {f ∈ Homeo([0, 1] d ) : dim H graph(f ) = d} is G δ , the real question is the following. For the case of prevalent homeomorphisms, we do not know whether the graph has infinite d-dimensional Hausdorff measure. The answer to the following question may easily be known, but we have been unable to find it in the literature. By setting Q = [0, 1] d in Lemma 6.5 one obtains a form of lower semi-continuity result for the d-dimensional Hausdorff measure of a graph of a homeomorphism f ∈ Homeo([0, 1] d ): if f is Lipschitz with H d (graph(f )) > c then there exists ε > 0 such that H d (graph(g)) > c for all g ∈ B(f, ε). This observation motivates the following question. Since the case d = 1 is straightforward, we assume d ≥ 2. Remark. Instead of using Hausdorff measures in the above problems, one may consider the notion of surface area introduced by Lebesgue [19]. The Lebesgue area L(f ) of a continuous map f : [0, 1] d → R d is obtained by taking the infimum of limit inferiors of surface areas of piecewise affine functions converging uniformly to f . Then L is lower semi-continuous, and L(f ) = H d (graph(f )) whenever f is Lipschitz, see e.g. [11] for more on this concept. Using our above results concerning the H d measure of the generic graph, it is straightforward to obtain that L(f ) = ∞ holds for the generic f ∈ Homeo([0, 1] d ).
8. Appendix (an approximation result in differential topology) We establish for convenience a known result needed above. Recall that a map f is called somewhere smooth if it is smooth on a non-empty open set (that may depend on f ). Theorem 8.1. Let d ≥ 2 be an integer, f ∈ Homeo((0, 1) d ) and ε : (0, 1) d → (0, ∞) be continuous. Then there exists a somewhere smooth g ∈ Homeo((0, 1) d ) such that |f (x) − g(x)| < ε(x) for every x ∈ (0, 1) d .
In fact, an easy and elementary argument shows that the statement also holds for d = 1, but we will not need this here.
Proof. First we show that, if d = 4, then the above g can even be a PL (piecewise affine) homeomorphism of (0, 1) d .