Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 141–150

Failure of the condition N below W^{1,n}

Janne Kauhanen

University of Jyväskylä, Department of Mathematics and Statistics P.O. Box 35, FIN-40351 Jyväskylä, Finland; jpkau@math.jyu.fi

Abstract. In this paper we examine the size of the exceptional sets outside which the null sets for the Lebesgue measure are preserved under continuous Sobolev mappings. This persistence is called the condition N. For a W^{1,1} -mapping it is true that the condition N is satisfied outside a set of Lebesgue measure zero. On the other hand, for mappings in W^{1,n} this exceptional set can be chosen to have zero Hausdorff dimension. One might expect for some kind of interpolation results between these estimates on the Hausdorff dimension of exceptional sets. We show, however, that this is not the case by constructing examples of homeomorphisms that belong to ⋂_{p<n} W^{1,p} so that the exceptional set has to be n-dimensional.

1. Introduction

Suppose that f is a continuous mapping from an open bounded subset Ω of R^{n}, n ≥ 2, into R^{n}. We consider the following (Lusin) condition N: if E ⊂ Ω, ℒ^{n}(E) = 0, then ℒ^{n}(f(E)) = 0. Physically, this condition requires that there is no creation of matter under the deformation f of the n-dimensional body Ω. This is a natural requirement as the N-property with differentiability a.e. is sufficient for validity of various change-of-variable formulas, including the area formula, and the condition N holds for a homeomorphism f if and only if f maps measurable sets to measurable sets.

If f ∈ W^{1,1}(Ω, R^{n}), we can ask which integrability conditions on the derivative Df guarantee the condition N. According to a classical result by Marcus and Mizel [11], f satisfies the condition N if |Df| ∈ L^{p}(Ω) for some p > n. If |Df| ∈ L^{p}(Ω) with p ≤ n, the condition N may fail; see e.g. [10] or [15] for examples.

It is then interesting to ask for the size of the exceptional set, i.e. the set outside of which the condition N holds. It is worth pointing out that if f is any mapping in the Sobolev space W^{1,1}(Ω, R^{n}), then f satisfies the condition N outside a set of measure zero (see e.g. [14] for references). Malý and Martio proved in [10] (also see [8]), as a substantial improvement on this fact, that if f ∈ W^{1,n}(Ω, R^{n}), then the condition N holds for f outside of a set that has Hausdorff dimension zero.

It seems that very little is known about the size of the exceptional set below W^{1,n}. It would be natural to expect that the Hausdorff dimension of the exceptional set decreases as the exponent p increases to n. In this paper we will prove, however, that if f ∈ W^{1,1}(Ω, R^{n}) so that (1.1) holds, and thus in particular when |Df| ∈ ⋂_{p<n} L^{p}(Ω), one cannot, in general, find an exceptional set of Hausdorff dimension smaller than n.

2000 Mathematics Subject Classification: Primary 26B35, 46E35, 74B20.

Theorem 1.1. Let s ∈ (0, ∞). Then there exists a homeomorphism f: Q_{0} → Q_{0} from the unit cube Q_{0} = [0, 1]^{n} onto itself such that the following properties hold:

(a) f fixes the boundary ∂Q_{0} .

(b) f ∈ W^{1,1}(Q_{0}, R^{n}), f is differentiable almost everywhere, and

(1.1)  sup_{0<ε≤n−1} ε ∫_{Q_{0}} |Df(x)|^{n−ε} dx < ∞.

(c) The Jacobian determinant J(x, f) is strictly positive for almost every x ∈ Q_{0} and J( · , f) ∈ L^{1}(Q_{0}).

(d) If E ⊂ Q_{0} with ℋ ^{h}(E) = 0, where h(t) = t^{n} log^{s} log(4 + 1/t) (for small t), then there exists a compact set F ⊂ Q_{0} \ E of measure zero such that ℒ^{n}(f(F)) > 0.

If we put certain additional topological or analytical assumptions on f, then a higher degree of integrability of |Df| can be relaxed for the condition N. Reshetnyak proved in [16] that for a homeomorphism it suffices to assume that |Df| ∈ L^{n}(Ω). There are several generalizations of this result: instead of assuming that f is a homeomorphism it suffices to assume that f is continuous and open [10], or that J(x, f) > 0 almost everywhere [2] (see also [17] and [12]). For further results of this type see the survey paper [9]. Recently Kauhanen, Koskela and Malý [6] proved that, for a topologically sense-preserving (and thus in particular for homeomorphical) mapping f, it suffices to assume that

(1.2)  lim_{ε→0+} ε ∫_{Ω} |Df|^{n−ε} = 0.

The construction of f of Theorem 1.1 is a refinement of our construction in [6], where a similar mapping was constructed to show that the condition N may fail under condition (1.1). For the convenience of the reader we will present a complete construction here. The basic idea of the construction comes from Ponomarev [15]; also see [14]. Ponomarev gave an example of a homeomorphism f with |Df| ∈ ⋂_{p<n} L^{p}(Ω) such that f creates matter. He, however, did not consider the size of the exceptional set and worked only on the L^{p} -scale.

Failure of the condition N below W^{1,n} 143

2. About the function spaces

In this section we briefly review some function spaces that we dealt with in the introduction; for a more detailed discussion see [4]. Let us define BL^{n}(Ω) as the collection of all measurable functions u with

∥u∥_{n)} = sup_{0<ε≤n−1}(ε ∫_{Ω} |u|^{n−ε})^{1/(n−ε)} < ∞.

Then BL^{n}(Ω) is a Banach space and

V L^{n}(Ω) = {u ∈ BL^{n}(Ω) : lim_{ε→0^{+}} ε ∫_{Ω} |u|^{n−ε} = 0}

is a closed subspace. The following inclusions hold (see [3]):

L^{n}(Ω) ⊊ L^{n} log^{−1} L(Ω) ⊊ V L^{n}(Ω) ⊊ BL^{n}(Ω) ⊊ ⋂_{α<−1} L^{n} log^{α} L(Ω) ⊊ ⋂_{p<n} L^{p}(Ω),

where the space L^{n} log^{α} L(Ω) consists of all measurable functions u with

∫_{Ω} |u|^{n} log^{α}(e + |u|) < ∞.

One question that Theorem 1.1 and the other results mentioned in the introduction leave unanswered is the size of the exceptional set for continuous mappings f ∈ W^{1,1}(Ω, R^{n}) with |Df|, for example, in L^{n} log^{−1} L(Ω) or V L^{n}(Ω).

3. Proof of Theorem 1.1

Let us begin with some notation. Besides the usual euclidean norm |x| = (x_{1}^{2} + ⋯ + x_{n}^{2})^{1/2} we will use the cubic norm ∥x∥ = max_{i} |x_{i}|. Using the cubic norm, the x_{0} -centered closed cube with edge length 2r > 0 and sides parallel to coordinate axes can be represented in the form

Q(x_{0}, r) = {x ∈ R^{n} : ∥x − x_{0}∥ ≤ r}.

We then call r the radius of Q. c(a, b, …) denotes a positive constant depending only on a, b, … which might differ from occurrence to occurrence. We write x ≈ c(a, b, …)y if x ≤ c(a, b, …)y and y ≤ c(a, b, …)x.

We will be dealing with radial stretchings that map cubes Q(0, r) onto cubes. The following lemma can be verified by an elementary calculation.

Lemma 3.1. Let ϱ: (0, ∞) → (0, ∞) be a strictly monotone, differentiable function. Then for the mapping

f(x) = \frac{x}{∥x∥}ϱ(∥x∥), x ≠ 0,

we have for a.e. x

|Df(x)| ≈ c(n) max\{\frac{ϱ(∥x∥)}{∥x∥}, |ϱ′(∥x∥)|\} and J(x, f) ≈ c(n)\frac{ϱ′(∥x∥)ϱ(∥x∥)^{n−1}}{∥x∥^{n−1}} .

We will first give two Cantor set constructions in Q_{0} . Our mapping f will be defined as a limit of a sequence of piecewise continuously differentiable homeomorphisms f_{k}: Q_{0} → Q_{0} , where each f_{k} maps the k:th step of the first Cantor set construction onto the second one. Then f maps the first Cantor set onto the second one. Choosing the Cantor sets so that the Hausdorff h measure ℋ ^{h} of the first one is positive and finite and so that the second one has positive measure, we get the property (d). We will explain this argument later.

Let V ⊂ R^{n} be the set of all vertices of the cube Q(0, 1). Then the sets V^{k} = V ×⋯×V , k = 1, 2, …, will serve as the sets of indices for our construction (with the exception of the subscript 0). If w ∈ V^{k−1} , we denote

V^{k}[w] = {v ∈ V^{k} : v_{j} = w_{j}, j = 1, …, k − 1}.

Let z_{0} = (\frac{1}{2}, …, \frac{1}{2}), r_{0} = \frac{1}{2} and denote

ψ(k) = \frac{1}{2} \frac{log^{s/n} 2}{log^{s/n}(k + 2)}.

For v ∈ V^{1} = V let z_{v} = z_{0} + \frac{1}{4}v, P_{v} = Q(z_{v}, \frac{1}{4}) and Q_{v} = Q(z_{v}, ψ(1)2^{−1}). If k = 2, 3, 4, …, and Q_{w} = Q(z_{w}, r_{k−1}), w ∈ V^{k−1} , is a cube from the previous step of construction, then Q_{w} is divided into 2^{n} subcubes P_{v}, v ∈ V^{k}[w], with centers z_{v} and radius \frac{1}{2}r_{k−1} and inside them we pick concentric cubes Q_{v}, v ∈ V^{k}[w], with radius

r_{k} = ψ(k)2^{−k}.

Note that r_{k} < \frac{1}{2}r_{k−1} for all k. Thus, denoting v = (v_{1}, …, v_{k}), we have

\begin{aligned}z_{v} & = z_{w} + \frac{1}{2}r_{k−1}v_{k} = z_{0} + \frac{1}{2} ∑_{j=1}^{k} r_{j−1}v_{j}, \\ P_{v} & = Q(z_{v}, \frac{1}{2}r_{k−1}), Q_{v} = Q(z_{v}, r_{k}). \end{aligned}

Figure 1. Cubes Q_{v}, v ∈ V^{k}.

See Figure 1. The number of cubes in the family {Q_{v} : v ∈ V^{k}}, k = 1, 2, 3, …, is #V^{k} = 2^{nk}. If W is any subfamily of V_{k}, we have

(3.1)

∑_{w∈W} h(diam Q_{w}) ≈ c(n, s)#W r_{k}^{n} log^{s} log(1/r_{k})
≈ c(n, s)#W 2^{−nk}ψ(k)^{n} log^{s}(log 2^{k} + log ψ(k)^{−1})
≈ c(n, s)#W 2^{−nk} log^{−s} k log^{s} k ≈ c(n, s)#W 2^{−nk}.

This is behind the fact that the Hausdorff h measure of the resulting Cantor set

C = ⋂_{k=1}^{∞} ⋃_{v∈V^{k}} Q_{v}

is positive and finite. In lack of a convenient reference we give a short proof for this fact.

Lemma 3.2. We have 0 < ℋ ^{h}(C) < ∞.

Proof. We mimic the proof in [13, Section 4.10]. The estimate ℋ ^{h}(C) < ∞ directly follows from (3.1). Let us justify ℋ ^{h}(C) > 0.

Let {U_{j}} be an open covering of C. It suffices to show that

(3.2)  ∑_{j} h(diam U_{j}) ≥ c(n, s).

Since C is compact, we can assume that {U_{j}} is finite. For each j choose x_{j} ∈ U_{j} ∩ C (the intersection can be assumed to be non-empty) and denote B_{j} = B(x_{j}, diam U_{j}). Then there is k_{0} such that for all k ≥ k_{0} every cube Q_{v}, v ∈ V^{k}, is contained in some B_{j}, and because h(2t) ≤ 2^{n}h(t) for t ≥ 0, we have that

∑_{j} h(diam U_{j}) ≥ 2^{−n} ∑_{j} h(diam B_{j}).

Therefore it suffices to prove (3.2) for the family {B_{j}} in the place of {U_{j}}. We claim that for any open ball B with center in C and any fixed l,

(3.3)  ∑_{v∈V^{l}, Q_{v}⊂B} h(diam Q_{v}) ≤ c(n, s)h(diam B).

This gives (3.2), since for k ≥ k_{0} , by (3.1),

∑_{j} h(diam B_{j}) ≥ c(n, s) ∑_{j} ∑_{v∈V^{k}, Q_{v}⊂B_{j}} h(diam Q_{v})

≥ c(n, s) ∑_{v∈V^{k}} h(diam Q_{v}) ≈ c(n, s).

To prove (3.3), suppose that Q_{v} ⊂ B for some v ∈ V^{l}, and let m be the smallest integer for which Q_{w} ⊂ B for some w ∈ V^{m}. Then m ≤ l. Denote

W = {w ∈ V^{m} : Q_{w} ∩ B ≠ ∅}.

By considering the geometry of the construction of C, it is evident that #W ≤ c = c(n, s). Thus, using (3.1) twice,

h(diam B) ≥ \frac{1}{c} ∑_{w∈W} h(diam Q_{w}) ≈ c(n, s)#W 2^{−nm} = c(n, s)#W 2^{n(l−m)}2^{−nl}

= c(n, s)#{v ∈ V^{l} : Q_{v} ⊂ Q_{w}, w ∈ W }2^{−nl}

≈ c(n, s) ∑_{w∈W} ∑_{v∈V^{l}, Q_{v}⊂Q_{w}} h(diam Q_{v})

≥ c(n, s) ∑_{v∈V^{l}, Q_{v}⊂B} h(diam Q_{v}).

The second Cantor set construction is similar to the first one except that at this time we denote the centers by z′_{v} and the cubes by P′_{v}, Q′_{v}, v ∈ V^{k}, with

\begin{aligned}z′_{v} & = z′_{w} + \frac{1}{2}r′_{k−1}v_{k} = z_{0} + \frac{1}{2} ∑_{j=1}^{k} r′_{j−1}v_{j}, \\ P′_{v} & = Q(z′_{v}, \frac{1}{2}r′_{k−1}), Q′_{v} = Q(z′_{v}, r′_{k}). \end{aligned}

We set r′_{0} = \frac{1}{2} and

r′_{k} = φ(k)2^{−k}

for k = 1, 2, 3, …, where

φ(k) = \frac{1}{4}(1 + \frac{log log 4}{log log(k + 4)}).

Note that r′_{k} < \frac{1}{2}r′_{k−1} for each k. We have

ℒ^{n}( ⋂_{k=1}^{∞} ⋃_{v∈V^{k}} Q_{v}) = lim_{k→∞} ℒ^{n}( ⋃_{v∈V^{k}} Q_{v}) = lim_{k→∞} 2^{nk}(2r′_{k})^{n} = 2^{−n} > 0.

We are now ready to define the mappings f_{k}. Define f_{0} = id. We will give a mapping f_{1} that stretches each cube Q_{v}, v ∈ V^{1} , homogeneously so that f_{1}(Q_{v}) equals Q′_{v}. On the annulus P_{v} \Q_{v}, f_{1} is defined to be an appropriate radial map with respect to z_{v} and z′_{v} in the image in order to make f_{1} a homeomorphism. The general step is the following: If k > 1, f_{k} is defined as f_{k−1} outside the union of all cubes Q_{w}, w ∈ V^{k−1} . Further, f_{k} remains equal to f_{k−1} at the centers of cubes Q_{v}, v ∈ V^{k}. Then f_{k} stretches each cube Q_{v}, v ∈ V^{k}, homogeneously so that f(Q_{v}) equals Q′_{v}. On the annulus P_{v} \Q_{v}, f is defined to be an appropriate radial map with respect to z_{v} in preimage and z′_{v} in image to make f_{k} a homeomorphism (see Figure 2). Notice that the Jacobian determinant J(x, f_{k}) will be strictly positive almost everywhere in Q_{0} .

Figure 2. The mapping f_{k} acting on P_{v}, v ∈ V^{k}.

To be precise, let f_{0} = id | Q_{0} and for k = 1, 2, 3, … define

f_{k}(x) = \{ \begin{matrix}f_{k−1}(x) & if x ∉ ⋃_{v∈V^{k}} P_{v}, \\ f_{k−1}(z_{v}) + a_{k}(x − z_{v}) + b_{k} \frac{x − z_{v}}{∥x − z_{v}∥} & if x ∈ P_{v} \ Q_{v}, v ∈ V^{k}, \\ f_{k−1}(z_{v}) + c_{k}(x − z_{v}) & if x ∈ Q_{v}, v ∈ V^{k}, \end{matrix}

where a_{k}, b_{k} and c_{k} are chosen so that f_{k} maps each Q_{v} onto Q′_{v}, is continuous and fixes the boundary ∂Q_{0} :

(3.4)  a_{k}r_{k} + b_{k} = r′_{k},  \frac{1}{2}a_{k}r_{k−1} + b_{k} = \frac{1}{2}r′_{k−1},  c_{k}r_{k}

Clearly the limit f = lim_{k→∞} f_{k} is differentiable almost everywhere, its Jacobian determinant is strictly positive almost everywhere, and f is absolutely continuous on almost all lines parallel to coordinate axes. Continuity of f follows from the uniform convergence of the sequence (f_{k}): for any x ∈ Q_{0} and l ≥ j ≥ 1 we have

|f_{l}(x) − f_{j}(x)| ≤ c(n)r′_{j} → 0

as j → ∞.

It is easily seen that f is a one-to-one mapping of Q_{0} onto Q_{0} . Since f is continuous and Q_{0} is compact, it follows that f is a homeomorphism.

To finish the proof of the properties (b) and (c), we next estimate |Df(x)| and J(x, f) at x in the interior of the annulus P_{v} \ Q_{v}, v ∈ V^{k}, k = 1, 2, 3, …. Denote r = ∥x − z_{v}∥ ≈ c(n, s)r_{k}. In the annulus

f(x) = f_{k−1}(z_{v}) + (a_{k}∥x − z_{v}∥ + b_{k})\frac{x − z_{v}}{∥x − z_{v}∥}

whence denoting ϱ(r) = a_{k}r + b_{k} we have by Lemma 3.1 (it is easy to check that b_{k} > 0 for large k)

|Df(x)| ≈ c(n, s)(a_{k} + b_{k}/r_{k})

and

J(x, f) ≈ c(n, s)a_{k}(a_{k} + b_{k}/r_{k})^{n−1}.

From the equations (3.4) it follows that

a_{k} = \frac{\frac{1}{2}r′_{k−1} − r′_{k}}{\frac{1}{2}r_{k−1} − r_{k}} = \frac{φ(k − 1) − φ(k)}{ψ(k − 1) − ψ(k)} ≈ c(n, s)\frac{φ′(k)}{ψ′(k)} ≈ c(n, s)\frac{log^{s/n} k}{log^{2} log k}

and

a_{k} + b_{k}/r_{k} = r′_{k}/r_{k} = \frac{φ(k)}{ψ(k)} ≈ c(n)\frac{1}{ψ(k)} ≈ c(n, s) log^{s/n} k.

Therefore

|Df(x)| ≈ c(n, s) log^{s/n} k  and  J(x, f) ≈ c(n, s)\frac{log^{s} k}{log^{2} log k} .

The measure of ⋃_{v∈V^{k}}(P_{v} \ Q_{v}) is

2^{nk}(r_{k−1}^{n} − (2r_{k})^{n}) ≈ c(n)(ψ(k − 1)^{n} − ψ(k)^{n})

≈ c(n, s)(\frac{1}{log^{s}(k − 1)} − \frac{1}{log^{s} k} )

≈ −c(n, s)\frac{d}{dk} (\frac{1}{log^{s} k} ) ≈ c(n, s)\frac{1}{k log^{1+s} k}

and so for 0 < ε ≤ n − 1

(3.5)

ε ∫_{Q_{0}} |Df(x)|^{n−ε} dx ≈ c(n, s)ε ∑_{k=2}^{∞} \frac{1}{k log^{1+s} k} log^{(s/n)(n−ε)} k

= c(n, s)ε ∑_{k=2}^{∞} \frac{1}{k log^{1+(s/n)ε} k}

≈ c(n, s)ε ∫_{2}^{∞} \frac{dt}{t log^{1+(s/n)ε} t} ≈ c(n, s).

This proves (1.1), and it follows that f ∈ W^{1,1}(Q_{0}, R^{n}). Similarly we prove the property (c):

(3.6)

∫_{Q_{0}} J(x, f) dx ≈ c(n, s)ε ∑_{k=4}^{∞} \frac{1}{k log^{1+s} k} \frac{log^{s} k}{log^{2} log k}

≈ c(n, s) ∫_{4}^{∞} \frac{dt}{t log t log^{2} log t} < ∞.

Let us now prove the property (d). Since ℋ ^{h} is Borel regular, we find a Borel set E′ ⊃ E ∩ C such that ℋ ^{h}(E′) = 0. Now C \ E′ is a Borel set, whence there is a compact set F ⊂ C \ E′ such that ℋ ^{h}(F) > 0. Denote

W^{k} = {v ∈ V^{k} : Q_{v} ∩ F ≠ ∅}.

Then there exists a constant c > 0 such that

(3.7)  \frac{#W^{k}}{#V^{k}} ≥ c

for all k, since otherwice we would have by (3.1) that

∑_{v_{∈}W^{k}} h(diam Q_{v}) ≈ c(n, s)#W^{k} 2^{−nk} = c(n, s)\frac{#W^{k}}{#V^{k}} → 0

as k → ∞ for k in some subsequence (k) ⊂ N, whence we would have ℋ ^{h}(F) = 0, a contradiction. Since F is compact, we have that

F = ⋂_{k=1}^{∞} ⋃_{v∈W^{k}} Q_{v},

whence (3.7) yields

ℒ^{n}(f(F)) = ℒ^{n}(f( ⋂_{k=1}^{∞} ⋃_{v∈W^{k}} Q_{v})) = ℒ^{n}( ⋂_{k=1}^{∞} f( ⋃_{v∈W^{k}} Q_{v}))

= lim_{k→∞} ℒ^{n}( ⋃_{v∈W^{k}} Q′_{v}) ≈ c(n) lim_{k→∞} #W^{k}(φ(k)2^{−k})^{n}

≥ c(n) lim_{k→∞} #W^{k} 2^{−nk} = c(n) lim_{k→∞} \frac{#W^{k}}{#V^{k}} ≥ c(n)c > 0.

Acknowledgements. The author wishes to express his thanks to Professor Pekka Koskela for suggesting the problem and for many stimulating conversations.

The author is supported in part by the Academy of Finland, project 41933.

150 Janne Kauhanen

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Received 1 November 2000