HAUSDORFF MEASURES AND THE MORSE–SARD THEOREM

Carlos Gustavo T. de A. Moreira IMPA Estrada Dona Castorina, 110 22460-320, Rio de Janeiro, RJ e-mail: gugu@impa.br

Abstract Let F: U ⊂ ℝ^{n} → ℝ^{m} be a differentiable function and p < m an integer. If k ≥ 1 is an integer, α ∈ [0, 1] and F ∈ C^{k+(α)}, if we set C_{p}(F) = {x ∈ U | rank(Df(x)) ≤ p} then the Hausdorff measure of dimension (p + \frac{n−p}{k+α}) of F (C_{p}(F)) is zero.

1 Introduction

The Morse-Sard theorem is a fundamental theorem in analysis that is in the basis of transversality theory and differential topology. The classical Morse-Sard theorem states that the image of the set of critical points of a function F: ℝ^{n} → ℝ^{m} of class C^{n−m+1} has zero Lebesgue measure in ℝ^{m}. It was proved by Morse ([M]) in the case m = 1 and by Sard ([S1]) in the general case.

Due to its theoretical importance, the Morse-Sard theorem was generalized in many directions. Many of these generalizations are related with Hausdorff measures and Hausdorff dimensions.

Given a metric space X and a positive real number α, we define the Hausdorff measure of dimension α associated to a covering U = (U_{λ})_{λ∈L} of X by bounded sets U_{λ} by m_{α}(U) = ∑_{λ∈L}(diam U_{λ})^{α}, where diam U_{λ} denotes the diameter of U_{λ}, and, if we define the norm of a covering U by ||U|| = sup_{U∈U}(diam U), then the Hausdorff measure of dimension α of X is m_{α}(X) = lim inf_{U covering of X ||U||→0} m_{α}(U).

It is not difficult to see that there is a unique d ∈ [0, +∞] such that if α > d then m_{α}(X) = 0 and if α < d then m_{α}(X) = +∞. This number d is called the Hausdorff dimension of X. It is easy to see that if X ⊂ ℝ^{n} then its Hausdorff dimension d =: HD(X) belongs to [0, n].

Sard himself proved that if C_{p}(F) = {x ∈ ℝ^{n} | rank(DF (x)) ≤ p} then for any ε > 0 there is k ∈ ℕ such that if F is C^{k} then F (C_{p}(F)) has zero Hausdorff measure of dimension p+ε ([S2]). This result was made more precise by Federer ([F]), who proved that if k ∈ ℕ then the Hausdorff measure of dimension p + \frac{n−p}{k} of F (C_{p}(F)) is zero. We should also mention the works of Church ([Ch1], [Ch2]), which gave more results about the structure of the set of critical values of differentiable maps. Later, Yomdin ([Y]) proved that the Hausdorff dimension of F (C_{p}(F)) is at most p + \frac{n−p}{k+α}, provided that F ∈ C^{k+α}, where k ∈ ℕ and 0 ≤ α < 1. More recently, Bates ([B2]) proved that if F ∈ C^{k+α} with k ∈ ℕ, 0 < α ≤ 1 and p + \frac{n−p}{k+α} = m then F (C_{p}(F)) has zero Lebesgue measure in ℝ^{m} (this in particular improves the hypothesis of the classical Morse-Sard theorem from F ∈ C^{n−m+1} to F ∈ C^{n−m+Lips.}, i.e., F ∈ C^{n−m} and D^{n−m}F Lipschitz.

The aim of this work is to generalize the mentioned results by proving a general version of the Morse-Sard Theorem involving Hausdorff measures. Let k ≥ 1 be an integer and α ∈ [0, 1]. We say that a function F: U ⊂ ℝ^{n} → ℝ^{m} is of class C^{k+(α)} at a subset A of U if F is C^{k} in U and for each x ∈ A there are ε_{x} > 0, K_{x} > 0 such that |y − x| < ε_{x} ⇒ |D^{k} F (y) − D^{k} F (x)| ≤ K_{x}|y − x|^{α} (this is less restrictive than supposing F ∈ C^{k+α}). Our main result is the following

Theorem: Let F: U ⊂ ℝ^{n} ⟶^{C^{k}} ℝ^{m} and let p < m be an integer. If C_{p}(F) := {x ∈ U | rank (DF (x)) ≤ p\} and if F is of class C^{k+(α)} at C_{p}(F) then the Hausdorff (p + \frac{n−p}{k+α})-measure of F (C_{p}(F)) is zero.

In particular, if k + α = \frac{n−p}{m−p} , we recover the result of [B2], with a weaker hypothesis. We remark that if p + \frac{n−p}{k+α} < m, the Hausdorff (p + \frac{n−p}{k+α})-measure is not the Lebesgue measure or a product measure in ℝ^{m}, and so we can not use Fubini’s Theorem. This difficulty is solved in the present paper by replacing the use of Fubini’s theorem by a careful decomposition of the critical set, combined with a parametrized strong version of the main lemma of Morse’s paper([M]) (Theorem 2.1).

We shall also give examples that show that our result is quite sharp, by giving counterexamples to slight changes of the hypothesis or of the conclusion.

2 Functions whose zeros include a given set

We shall prove here a version of Theorem 3.6 of [M] and Lemma 3.4.2 of [F], which will be fundamental for the later results.

Theorem 2.1. Let k ≥ 1, α ∈ [0, 1], n > p and A ⊂ U ⊂ ℝ^{n}, where U is an open set. Then there are sets A_{1}, A_{2} …, ⊂ A such that A = ⋃_{i=1}^{∞} A_{i}, where for each i = 1, 2, … there is a function ψ_{i}: B_{i} × V_{i} ⟶^{C^{1}} U where B_{i} is a ball in some ℝ^{r_{i}}, r_{i} ≥ 0 and V_{i} is a ball in ℝ^{p} such that ψ_{i}(x, y) = (ψ̃_{i}(x, y), y), and |ψ_{i}(x_{1}, y_{1}) − ψ_{i}(x_{2}, y_{2})| ≥ |(x_{1}, y_{1}) − (x_{2}, y_{2})|, ∀ (x_{1}, y_{1}), (x_{2}, y_{2}) ∈ B_{i} × V_{i} and A_{i} ⊂ ψ_{i}(B_{i} × V_{i}), with the following property: We can write A_{i} = A′_{i} ∪ A′′_{i} so that ψ_{i}^{−1}(A′′_{i}) has measure zero in B_{i} × V_{i}, and if f: U → ℝ vanishes in A and f is C^{k+(α)} at A we have:

• lim sup_{(x,y_{0})→(x_{0},y_{0})} \frac{f(ψ_{i}(x, y_{0}))}{|x − x_{0}|^{k+α}} < +∞, ∀ (x_{0}, y_{0}) ∈ B_{i} × V_{i} such that ψ_{i}(x_{0}, y_{0}) ∈ A_{i},

• lim_{(x,y_{0})→(x_{0},y_{0})} \frac{f(ψ_{i}(x, y_{0}))}{|x − x_{0}|^{k+α}} = 0, ∀ (x_{0}, y_{0}) ∈ B_{i} × V_{i} such that ψ_{i}(x_{0}, y_{0})) ∈ A′_{i}.

Proof. Let us consider first the case k = 1 and df(x) · v = 0 ∀ x ∈ A, v ∈ ℝ^{n−p} × {0}. In this case we take A = (A′ ∩ A) ∪ A′′ where A′ is the set of density points of \overline{A} in the direction of ℝ^{n−p} × \{0\} ((x, y) ∈ A′ ⇒ lim_{ε→0} \frac{m((B_{ε}(x) × \{y\}) ∩ \overline{A})}{m(B_{ε}(x))} = 1, where m is the (n − p)-dimensional measure). The measure of A′′ = A − A′ is zero, since it is zero in each plane ℝ^{n−p} × {y}.

For (x_{0}, y_{0}) ∈ A take B((x_{0}, y_{0}), ε(x_{0}, y_{0})) a ball contained in U and ψ = Id|_{B((x_{0},y_{0}),ε(x_{0},y_{0}))}. We

have lim sup_{(x,y_{0})→(x_{0},y_{0})} \frac{f(x, y_{0})}{|x − x_{0}|^{1+α}} < +∞, since f(x, y_{0}) = f(x, y_{0}) − f(x_{0}, y_{0}) = df(tx_{0} + (1 − t)x)(x − x_{0}),

t ∈ (0, 1) ⇒ |f(x, y_{0})| ≤ K_{x_{0}}|x − x_{0}|^{1+α}. For (x_{0}, y_{0}) ∈ A′,

lim_{δ→0} \frac{1}{vol(S^{n−p−1})} ∫_{S^{n−p−1}} (\frac{1}{δ} ∫_{0}^{δ} χ\overline{A}(x_{0} + tv, y_{0})dt)dv = 1,

so ∀ ε > 0 ∃δ_{0} > 0 s.t. |x − x_{0}| < δ_{0} ⇒ ∃ v ∈ S^{n−p−1} with |v − \frac{x − x_{0}}{|x − x_{0}|} | < ε and

| \frac{1}{|x − x_{0}|} ∫_{0}^{|x−x_{0}|} χ\overline{A}(x_{0} + tv, y_{0})dt − 1| < ε, so, if x̃ = x_{0} + |x − x_{0}|v,

|f(x, y_{0}) − f(x_{0}, y_{0})| ≤ |f(x, y_{0}) − f(x̃, y_{0})| + |f(x̃, y_{0}) − f(x_{0}, y_{0})|, but f(x, y_{0}) − f(x̃, y_{0}) = df(θt + (1 − θ)x̃, y_{0}) · (x − x̃), θ ∈ (0, 1) ⇒ |f(x, y_{0}) − f(x̃, y_{0})| ≤ K_{x_{0}}|x − x_{0}|^{α} · ε|x − x_{0}| = εK_{x_{0}}|x − x_{0}|^{1+α} and f(x̃, y_{0}) − f(x_{0}, y_{0}) = ∫_{0}^{|x̃−x_{0}|} df(x_{0} + tv, y_{0}) · v dt ≤

K_{x_{0}}|x − x_{0}|^{α} · m\{t ∈ [0, |x̃ − x_{0}|]|\frac{∂f}{∂x}(x_{0} + tv, y_{0}) ≠ 0\} ≤ K_{x_{0}}|x − x_{0}|^{α} · ε|x − x_{0}| = εK_{x_{0}}|x − x_{0}|^{1+α}.

So |f(x, y_{0})| = |f(x, y_{0}) − f(x_{0}, y_{0})| ≤ 2εK_{x_{0}}|x − x_{0}|^{1+α} ⇒ lim_{(x,y_{0})→(x_{0},y_{0})} \frac{f(x, y_{0})}{|x − x_{0}|^{1+α}} = 0.

We can take a countable subcovering of A by the B((x_{0}, y_{0}), ε(x_{0}, y_{0})) to finish the proof in this case.

Consider now the case k ≥ 1, n arbitrary. We have A = A^{∗} ∪ A^{∗∗} where A^{∗} = {x ∈ A |

∃ g: U ⟶^{C^{k}} ℝ, g|_{A} ≡ 0, ∃ v ∈ ℝ^{n−p} × {0}, dg(x) · v ≠ 0}. A^{∗∗} = A\A^{∗}. If (x_{0}, y_{0}) ∈ A^{∗} there is g as

above, so there is ε > 0 such that g^{−1}(0) ∩ B_{ε}(x_{0}, y_{0}) is contained in the image of ψ: B × V ⟶^{C^{k}} U where B is a ball in ℝ^{n−p−1}, as in the statement, and A ⊂ g^{−1}(0). Taking a countable subcovering of A^{∗} by these balls we reduce the proof in this case to a case with smaller n. If k = 1, the result was yet proved for A^{∗∗}. If k > 1 , and assuming by induction the result for k − 1, we have

A^{∗∗} = ⋃_{i=1}^{∞} A_{i}^{∗∗}, A_{i}^{∗∗} = (A_{i}^{∗∗})′ ∪ (A_{i}^{∗∗})′′, A_{i}^{∗∗} ⊂ ψ_{i}(B_{i} × V_{i}), ψ_{i} ∈ C^{1},

ψ_{i}(x_{0}, y_{0}) ∈ A_{i}^{∗∗} ⇒ lim sup_{x→x_{0}} \frac{||df(ψ_{i}(x, y_{0}))|_{ℝ^{n−p}×\{0\}}||}{|x − x_{0}|^{k−1+α}} < +∞ ⇒ lim sup_{x→x_{0}} \frac{|f(ψ_{i}(x, y_{0}))|}{|x − x_{0}|^{k+α}} < +∞

and

ψ_{i}(x_{0}, y_{0}) ∈ (A_{i}^{∗∗}) ⇒ lim_{x→x_{0}} \frac{||df(ψ_{i}(x, y_{0}))|_{ℝ^{n−p}×\{0\}}||}{|x − x_{0}|^{k−1+α}} = 0 ⇒ lim_{x→x_{0}} \frac{f(ψ_{i}(x, y_{0}))}{|x − x_{0}|^{k+α}} = 0,

both by the mean value theorem, and the proof is finished by induction.

Corollary 2.2. Let k ≥ 1, α ∈ [0, 1], n > p and A ⊂ U ⊂ ℝ^{n}, where U is an open set. Then there are sets A_{1}, A_{2} …, ⊂ A such that A = ⋃_{i=1}^{∞} A_{i}, where for each i = 1, 2, … there is a function ψ_{i}: B_{i} × V_{i} ⟶^{C^{1}} U where B_{i} is a ball in some ℝ^{r_{i}}, r_{i} ≥ 0 and V_{i} is a ball in ℝ^{p} such that ψ_{i}(x, y) = (ψ̃_{i}(x, y), y), and |ψ_{i}(x_{1}, y_{1}) − ψ_{i}(x_{2}, y_{2})| ≥ |(x_{1}, y_{1}) − (x_{2}, y_{2})|, ∀ (x_{1}, y_{1}), (x_{2}, y_{2}) ∈ B_{i} × V_{i} and A_{i} ⊂ ψ_{i}(B_{i} × V_{i}), with the following property: We can write A_{i} = A′_{i} ∪ A′′_{i} so that ψ_{i}^{−1}(A′′_{i}) has measure zero in B_{i} × V_{i}, and if f: U → ℝ is C^{k+(α)} at A and D_{x}f ≡ 0 in A we have:

• lim sup_{(x,y_{0})→(x_{0},y_{0})} \frac{|f(ψ_{i}(x, y_{0})) − f(ψ_{i}(x_{0}, y_{0}))|}{|x − x_{0}|^{k+α}} < +∞, ∀ (x_{0}, y_{0}) ∈ B_{i} × V_{i} such that ψ_{i}(x_{0}, y_{0}) ∈ A_{i},

• lim_{(x,y_{0})→(x_{0},y_{0})} \frac{|f(ψ_{i}(x, y_{0})) − f(ψ_{i}(x_{0}, y_{0}))|}{|x − x_{0}|^{k+α}} = 0, ∀ (x_{0}, y_{0}) ∈ B_{i} × V_{i} such that ψ_{i}(x_{0}, y_{0})) ∈ A′_{i}.

Proof. If k ≥ 2 this is an immediate consequence of Theorem 2.1 applied to D_{x}f and of the mean value theorem. If k = 1 this can be proved exactly as the case k = 1 of the Theorem 2.1.

Corollary 2.3. In the statements of Theorem 2.1 and Corollary 2.2, for any x ∈ B_{i} s.t. ψ_{i}(x) ∈ A_{i} there are ε_{x} > 0, K_{x} > 0 such that |y − x| < ε_{x} ⇒ |f(ψ_{i}(y)) − f(ψ_{i}(x))| ≤ K_{x}|y − x|^{k+α}, and for any ε > 0 there is a δ > 0 so that \frac{λ(ψ_{i}^{−1}(A_{i})∩B_{r}(x))}{λ(B_{r}(x))} > 1 − δ ⇒ |f(ψ_{i}(y)) − f(ψ_{i}(x))| ≤ εK_{x}r^{k+α}, if r ≤ ε_{x} and |y − x| ≤ r (δ depends only on ε and n, but not on f or on x).

Proof. This is only a more precise formulation of the results proved in the demonstration of the theorem.

Remark 2.1. For k = 0 we have the same results, except the statement lim_{y→x} \frac{f(ψ_{i}(y))}{|y−x|^{k+α}} = 0, for each x ∈ B_{i} such that ψ_{i}(x) ∈ A′_{i}.

3 The main results

Lemma 3.1. Let A ⊂ ℝ^{m} with λ(A) < ∞ and let U be a family of balls B_{r}(x), x ∈ A such that for each x ∈ A there is an ε_{x} > 0 such that r ≤ ε_{x} ⇒ B_{r}(x) ∈ U. Then for each ε > 0 there are x_{n} ∈ A, r_{n} > 0 with B_{r_{n}}(x_{n}) ∈ U and A ⊂ ⋃_{n=1}^{∞} B_{r_{n}}(x_{n}) such that ∑_{n=1}^{∞} λ(B_{r_{n}}(x_{n})) < λ(A) + ε.

Proof. This lemma is essentially the Vitali covering theorem from measure theory. Take U ⊃ A an open set with λ(U) < λ(A) + \frac{ε}{2}. If we choosed Br̃_{1}(x_{1}), …, Br̃_{n}(x_{n}), define s_{n} = sup\{r > 0 | ∃ x ∈ A s.t. r < \frac{ε_{x}}{5} , B_{r}(x) ⊂ U and B_{r}(x) ∩ (Br̃_{1}(x_{1}) ∪ ⋯ ∪ Br̃_{n}(x_{n})) = ∅\}. Choose Br̃_{n+1}(x_{n+1}) such that r̃_{n+1} > \frac{s_{n}}{2} , r̃_{n+1} < \frac{ε_{x_{n+1}}}{5} , Br̃_{n+1}(x_{n+1}) ⊂ U and Br̃_{n+1}(x_{n+1}) ∩ (Br̃_{1}(x_{1}) ∪ ⋯ ∪ Br̃_{n}(x_{n})) = ∅. Since the Br̃_{i}(x_{i}) are disjoint and contained in U we have ∑_{i=1}^{∞} λ(Br̃_{i}(x_{i})) < λ(A) + \frac{ε}{2} , and so there is a n_{o} ∈ ℕ such that ∑_{i=n_{o}}^{∞} λ(B_{5r̃_{i}}(x_{i})) < \frac{ε}{2} . We take B_{r_{i}}(x_{i}) = Br̃_{i}(x_{i}), i < n_{o} and B_{r_{i}}(x_{i}) = B_{5r̃_{i}}(x_{i}), i ≥ n_{o}.

Clearly we have ∑_{i=1}^{∞} λ(B_{r_{i}}(x_{i})) < λ(A) + ε. To prove that A ⊂ ⋃_{n=1}^{∞} \overline{B_{r_{n}}(x_{n})}, take x ∈ A and r = min{r̃_{n_{o}}, ε_{x}/5, d(x, U^{c} ∪ ⋃_{i<n_{o}} B_{r_{i}}(x_{i}))}. If r > 0, take n ≥ n_{o} such that s_{n} < r ≤ s_{n−1} (we have r ≤ r̃_{n_{o}} ≤ s_{n_{o}−1}), and note that s_{n} < r ⇒ B_{r}(x) ∩ (Br̃_{1}(x_{1}) ∪ ⋯ ∪ Br̃_{n}(x_{n})) ≠ ∅ ⇒ ∃ i ≤ n such that B_{r}(x) ∩ Br̃_{i}(x_{i}) ≠ ∅. We have n ≥ n_{o} since r ≤ d(x, Br̃_{i}(x_{i})), and r̃_{i} > \frac{s_{n−1}}{2} ≥ \frac{r}{2}, since i ≤ n. Therefore, we have x ∈ B_{5r̃_{i}}(x_{i}). If r = 0 then x ∈ \overline{B_{r_{i}}(x_{i})} for some i < n_{o}. This proves that A ⊂ ⋃_{n=1}^{∞} \overline{B_{r_{n}}(x_{n})}. Taking r̃_{n} = (\frac{λ(A)+ε}{∑_{i=1}^{∞} λ(B_{r_{i}}(x_{i}))})^{1/2m} · r_{n}, we have A ⊂ ⋃_{n=1}^{∞} Br̃_{n}(x_{n}), with ∑_{n=1}^{∞} λ(Br̃_{n}(x_{n})) = (λ(A) + ε)^{1/2}(∑_{i=1}^{∞} λ(B_{r_{i}}(x_{i})))^{1/2} < λ(A) + ε.

Remark 3.1. In the Lemma 3.1 we can replace a family of balls B_{r}(x) by a family of cubes C_{r}(x) = ∏_{i=1}^{m} [x_{i} − r, x_{i} + r], where x = (x_{1}, …, x_{m}), using the same proof.

Lemma 3.2. Let F: U ⊂ ℝ^{n} → ℝ^{m} be a function, A ⊂ U and d > 0 such that for any x ∈ A there are ε_{x} > 0, K_{x} > 0 such that m_{d}(F (B_{ε}(x) ∩ A)) ≤ K_{x}.λ(B_{ε}(x)), ∀ ε < ε_{x}, where m_{d} is the Hausdorff measure of dimension d, and there is A′ ⊂ A such that λ(A\A′) = 0 and lim_{ε→0} \frac{m_{d}(F (B_{ε}(x)∩A))}{λ(B_{ε}(x))} = 0, ∀ x ∈ A′. Then m_{d}(F (A)) = 0.

Remark 3.2. The same result is true if we replace B_{ε}(x) by C_{ε}(x).

Remark 3.3. We can replace the condition “m_{d}(F (B_{ε}(x) ∩ A)) ≤ K_{x} λ(B_{ε}(x)), ∀ ε < ε_{x}” by “F (B_{ε}(x) ∩ A) can be covered by balls B_{δ_{i}}(y_{i}), i ∈ ℕ, with ∑_{i=1}^{∞} δ_{i}^{d} ≤ K_{x} λ(B_{ε}(x)), ∀ ε < ε_{x}”, and the condition “lim_{ε→0} \frac{m_{d}(F (B_{ε}(x)∩A)}{λ(B_{ε}(x))} = 0, ∀x ∈ A′ ” by “F (B_{ε}(x) ∩ A) can be covered by balls B_{δ_{i}^{(ε)}}(y_{i}), i ∈ ℕ with lim_{ε→0} \frac{∑_{i=1}^{∞}(δ_{i}^{(ε)})^{d}}{λ(B_{ε}(x))} = 0, ∀ x ∈ A′ ”. The proof remains essentially the same, and Remark 3.2 is still valid. Remark 3.4. If we replace the conditions of this lemma by “F (B_{ε}(x) ∩ A) can be covered by balls B_{δ_{i}}(y_{i}), i ∈ ℕ, with ∑_{i=1}^{∞} δ_{i}^{d} ≤ k λ(B_{ε}(x)), ∀ ε < ε_{x} (note that here k does not depend on x), and λ(A) < ∞”, then we can conclude, using the same proof, that m_{d}(F (A)) ≤ k λ(A).

Proof. We may suppose that A has finite Lebesgue measure, since A is a countable union of sets with finite measure, and a countable union of sets with Hausdorff d-measure zero has Hausdorff d-measure zero. Moreover, since A = ⋃_{k=1}^{∞} A_{k}, where A_{k} = {x ∈ A | K_{x} ≤ k}, we may suppose K_{x} ≤ K, ∀ x ∈ A. Let C be the Lebesgue measure of A.

Let ε > 0. For each x ∈ A′ take δ_{x} > 0 such that B_{δ_{x}}(x) ⊂ U and r ≤ δ_{x} ⇒ \frac{m_{d}(F (B_{r}(x)∩A))}{λ(B_{r}(x))} ≤ \frac{ε}{2(C+1)}. By the Lemma 3.1 we can cover A′ by ⋃_{n=1}^{∞} B_{r_{n}}(x_{n}) with ∑_{n=1}^{∞} λ(B_{r_{n}}(x_{n})) < C + 1 and r_{n} ≤ δ_{x_{n}} ⇒ ∑_{n=1}^{∞} m_{d}(F (B_{r_{n}}(x_{n}) ∩ A)) ≤ \frac{ε}{2(C+1)} · (C + 1) = \frac{ε}{2} ⇒ m_{d}(F (A′)) ≤ \frac{ε}{2} . By Lemma 3.1 we can cover A\A′ by ⋃_{n=1}^{∞} Br̃_{n}(x̃_{n}) such that Br̃_{n}(x̃_{n}) ⊂ U and r̃_{n} < εx̃_{n}, ∀ n ∈ ℕ, with ∑_{n=1}^{∞} λ(Br̃_{n}(x̃_{n})) < \frac{ε}{2K} ⇒ ∑_{n=1}^{∞} m_{d}(F (Br̃_{n}(x̃_{n}))) ≤ \frac{ε}{2K} ·K = \frac{ε}{2} ⇒ m_{d}(F (A\A′)) ≤ \frac{ε}{2} ⇒ m_{d}(F (A)) ≤ \frac{ε}{2} + \frac{ε}{2} = ε. Since ε > 0 is arbitrary we have m_{d}(F (A)) = 0.

We first use Lemma 3.2 to prove the following strong version of Constantin’s result ([Co]), that does not suppose continuity of the derivatives. Here we do not suppose differentiability in every point, but only in the set of critical points under consideration.

Theorem 3.3. Let F: X ⊂ ℝ^{n} → ℝ^{n} be a function, and let A = {x ∈ X | DF (x) exists and is not surjective}. Then λ(F (A)) = 0.

Proof. It is a simple consequence of Lemma 3.2, since if x ∈ A then lim_{r→0} \frac{λ(F (B_{r}(x)))}{λ(B_{r}(x))} = 0. Indeed, x ∈ A ⇒ F (x + h) = F (x) + DF (x).h + r(h), where lim_{h→0} \frac{r(h)}{|h|} = 0. Let K = ||DF (x)||, and let ε ∈ (0, 1). Let δ > 0 such that |h| ≤ δ ⇒ \frac{|r(h)|}{|h|} < \frac{ε}{2(K+1)^{n−1}} . Then, if |h| ≤ δ, F (x+h)−F (x) belongs to an \frac{ε.|h|}{2(K+1)^{n−1}} neighbourhood of a ball of radius K|h| in a subspace of ℝ^{n} of dimension n − 1 (a fixed subspace of ℝ^{n} of dimension n − 1 which contains the image of DF), and thus belongs to the orthogonal product of a ball of radius (K + 1)|h| in this subspace by an interval of radius \frac{ε|h|}{2(K+1)^{n−1}} . Therefore, λ(F (B_{r}(x)) ≤ \frac{ε.r.r^{n−1}(K+1)^{n−1}}{(K+1)^{n−1}} v_{n−1} = εr^{n}v_{n−1}, where v_{n−1} is the volume of the unitary ball in ℝ^{n−1}, and, since ε > 0 is arbitrary, lim_{r→0} \frac{λ(F (B_{r}(x)))}{λ(B_{r}(x))} = 0.

Theorem 3.4. Let F: U ⊂ ℝ^{n} ⟶^{C^{k}} ℝ^{m} be a function of class C^{k+(α)}(α ∈ (0, 1]) at C_{p}(F) := {x ∈ U | rank (DF (x)) ≤ p\}. Then the Hausdorff measure of dimension d = p + \frac{n−p}{k+α} of F (C_{p}(F)) is zero, ∀ p < min{m, n}.

Proof. Since C_{p}(F) = ⋃_{r=0}^{p} \{x ∈ U | rank (DF (x)) = r\}, and r + \frac{n−r}{k+α} ≤ p + \frac{n−p}{k+α} for 0 ≤ r ≤ p, we may restrict our attention to C̃_{p}(F) = {x ∈ U | rank (DF (x)) = p}. If x_{o} ∈ C_{p}(F), we have, after a change of coordinates of class C^{k}, F (z, y) = (z, G(z, y)), with (z, y) ∈ ℝ^{p} ×ℝ^{n−p} and G(z, y) ∈ ℝ^{m−p}, in a neighbourhood V of x_{o} = (z_{o}, y_{o}). We shall restrict our attention to this neighbourhood. We have x = (z, y) ∈ C_{p}(F) ⇔ D_{y} G(z, y) = 0. We can apply the results of the section 2 (Theorem 2.1, Corollary 2.3 and Remark 2.1) to the function D_{y} G, and obtain the decomposition A = ⋃_{i=1}^{∞} A_{i}, A_{i} ⊂ ψ_{i}(V_{i} × B_{i}), where A = {(z, y) ∈ V | D_{y}G(z, y) = 0}. Let us fix such an A_{i}.

Since ψ_{i}^{−1}(A_{i}) = ⋃_{m∈ℕ} \{x ∈ ψ_{i}^{−1}(A_{i}) | ε_{x} ≥ \frac{1}{m}, K_{x} ≤ m\}, we may suppose ε_{x} ≥ \frac{1}{M}, K_{x} ≤ M, ∀ x ∈ ψ_{i}^{−1}(A_{i}), for some fixed M and also that V has finite Lebesgue measure λ(V).

With these assumptions, we shall prove that there is a constant K_{o} such that for any X ⊂ V, ν > 0, we can cover F (A_{i} ∩ X) by balls B_{δ_{i}}(p_{i}) so that ∑_{i=1}^{∞} δ_{i}^{d} ≤ K_{o}(λ(X) + ν). For this, given a point x ∈ A_{i}∩X and an ε < \frac{1}{2\sqrt{n} M} , we can divide the cube C_{ε}(x) = C_{ε}(z)×C_{ε}(y) into ([ε^{1−(k+α)}]+1)^{p} boxes C_{δ}(z̃_{i}) × C_{ε}(y), δ < ε^{k+α}. If there is some point (z_{i}, y_{i}) in (C_{δ}(z̃_{i}) × C_{ε}(y)) ∩ (A_{i} ∩ X), then for any point (z′_{i}, y′_{i}) in (C_{δ}(z̃_{1})×C_{ε}(y))∩(A_{i} ∩X), we have |F (z′_{i}, y′)−F (z_{i}, y_{i})| ≤ |F (z′_{i}, y′)−F (z_{i}, y′)|+ |F (z_{i}, y′) − F (z_{i}, y_{i})| ≤ K′δ + |F (z_{i}, y′_{i}) − F (z_{i}, y_{i})| (where K′ is \sqrt{p} times a Lipschitz constant of F |_{V} which we may suppose to exist) ≤ K′δ + |G(z_{i}, y′_{i}) − G(z_{i}, y_{i})|.

Observe now that (z_{i}, y_{i}) = (z_{i}, ψ̃_{i}(p_{1})) and (z′_{i}, y′_{i}) = (z_{i}, ψ̃_{i}(p_{2})), for some p_{1}, p_{2} in {z_{i}} × B_{i} with |p_{1} − p_{2}| ≤ |y_{i} − y′_{i}| ≤ 2ε\sqrt{n}. Let γ: [0, 1] → V_{i} × B_{i} be a straight path joining p_{1} and p_{2}. Then G(z_{i}, y′_{i}) − G(z_{i}, y_{i}) = ∫_{0}^{1} \frac{∂G}{∂y}(γ̃(t)) · γ̃′(t) dt, where γ̃ := ψ_{i} ∘ γ. We have \frac{∂G}{∂y}(γ̃(0)) = 0, so

∥\frac{∂G}{∂y}(γ̃(t))∥ = ∥\frac{∂G}{∂y}(γ̃(t)) − \frac{∂G}{∂y}(γ̃(0))∥

≤ M|p_{1} − p_{2}|^{k+α−1} ≤ M(2ε\sqrt{n})^{k+α−1} ⇒ ∥\frac{∂G}{∂y}(γ̃(t))∥ |γ′(t)|

≤ K′′ε^{k+α}, for some constant K′′. Indeed, |γ̃′(t)| is limited by a constant multiple of |p_{1} − p_{2}| ≤ 2\sqrt{n}ε. So

|G(z_{1}, y′_{i}) − G(z_{i}, y_{i})| ≤ ∫_{0}^{1} | \frac{∂G}{∂y}(γ̃(t)) ∘ γ̃(t)| dt ≤ K′′ε^{k+α}

and

|F (z′_{i}, y′) − F (z_{i}, y_{i})| ≤ K′δ + K′′ε^{k+α} ≤ K′_{0} · ε^{k+α},

where K′_{0} = K′ + K′′.

Therefore, F (C_{δ}(z̃_{i}) × C_{ε}(y)) is contained in some ball B_{δ_{i}}(q_{i}), with δ_{i} ≤ K′_{0} · ε^{k+α}, so

∑_{i} δ_{i}^{d} ≤ ([ε^{1−(k+α)}] + 1)^{p}(K′_{o} ε^{k+α})^{d}

= (K′_{o})^{d}([ε^{1−(k+α)}] + 1)^{p}(ε^{k+α})^{p+ \frac{n−p}{k+α}} ≤ K̃_{o} ε^{n}

for some constant K̃_{o}. So, F (C_{ε}(x)) can be covered by balls B_{δ_{i}}(q_{i}) with ∑_{i} δ_{i}^{d} ≤ K_{o}.λ(C_{ε}(x)), and by the Lemma 3.2, remarks 3.2 and 3.4, we can conclude that m_{d}(F (A_{i} ∩ X)) ≤ K_{o} λ(X) where K_{0} = 2^{n}K̃_{0}, and so we can cover F (A_{i} ∩ X) by balls B_{δ_{i}}(p_{i}) so that ∑_{i=1}^{∞} δ_{i}^{d} ≤ K_{o}(λ(X) + ν), as we stated.

We shall prove now that there is an A′_{i} ⊂ A_{i} ⊂ V with λ(A_{i}\A′_{i}) = 0 such that

F (C_{ε}(x) ∩ A_{i}) can be covered by balls B_{δ_{i}^{(ε)}}(W_{i}), i ∈ ℕ with lim_{ε→0} \frac{∑_{i}(δ_{i}^{(ε)})^{d}}{λ(C_{ε}(x))} = 0, ∀x ∈ A′_{i}. This

will imply our theorem, by the Lemma 3.2, remarks 3.2 and 3.3, since we have proved above that m_{d}(F (C_{ε}(x)∩A_{i}) ≤ K_{o} λ(C_{ε}(x)), ∀ ε < \frac{1}{2\sqrt{n} M} . For this, since A_{i} ⊂ ψ_{i}(V_{i} ×B_{i}), B_{i} ⊂ ℝ^{r_{i}}, r_{i} ≤ n−p, we may suppose r_{i} = n−p and ψ_{i} = identity, because r_{i} < n−p ⇒ λ(A_{i}) = 0 and we can take A′_{i} = A_{i}. Let us take A′_{i} equal to the set of the density points of A_{i}. Given a point x ∈ A′_{i}, and an η′ > 0, we want to find an ε_{o} > 0 such that ε < ε_{o} ⇒ F (C_{ε}(x) ∩ A_{i}) can be covered by balls B_{δ_{i}^{(ε)}}(W_{i}), i ∈ ℕ such that ∑_{i}(δ_{i}^{(ε)})^{d} ≤ η′ λ(C_{ε}(x)). Let η, η̃ > 0, ε̃ < \frac{1}{2\sqrt{n} M} such that \frac{λ(C_{ε}(x)∩A_{i})}{λ(C_{ε}(x))} > 1 − η̃^{2}, ∀ ε ≤ ε̃. Divide the cube C_{ε}(x) = C_{ε}(z̃) × C_{ε}(ỹ), ε < ε̃ into N = ([ε^{1−(K+α)}η^{−1}] + 1)^{p} boxes C_{δ}(z̃_{i}) × C_{ε}(ỹ), δ < η ε^{k+α}, 1 ≤ i ≤ N. Then for at least (1 − η̃)N values of i, there is a z_{i} ∈ C_{δ}(z̃_{i}) such that λ({y ∈ C_{ε}(ỹ) | (z_{i}, y) ∈ A_{i}}/λ(C_{ε}(ỹ)) > 1 − η̃ (here λ is the Lebesgue measure in ℝ^{n−p}), because λ(C_{ε}(x) ∩ A_{i}) > (1 − η̃^{2})λ(C_{ε}(x)). For such an i, take an y_{i} such that (z_{i}, y_{i}) ∈ A_{i}. Then, applying Theorem 2.1, Corollaries 2.2 and 2.3, given η we can choose η̃ so that |F (z_{i}, y) − F (z_{i}, y_{i})| < η · ε^{k+α} ⇒ |F (z, y) − F (z_{i}, y_{i})| ≤ 2K′\sqrt{n}.ηε^{K+α} + ηε^{K+α} ≤ K′′ηε^{k+α}, for some constant K′′ and for any z ∈ C_{ε}(z̃_{i}), where K′ is a Lipschitz constant for F ⇒ F (C_{δ}(z̃_{i}) × C_{ε}(ỹ)) is contained in a ball B_{δ_{i}}(Q_{i}), with ∑_{i} δ_{i}^{d} over these values of i less than

([ε^{1−(K+α)}η^{−1}] + 1)^{p}(K′′ηε^{k+α})^{d} = (K′′)^{d}([ε^{1−(k+α)}.η^{−1}] + 1)^{p}(ηε^{k+α})^{p+ \frac{n−p}{k+α}} ≤ K̃_{o}η^{\frac{n−p}{k+α}} .ε^{n}

for some constant K̃_{o}. The union of the remaining (at most η̃N) boxes has volume at most η̃ε^{n} ⇒ the union of the image of the intersection of A_{i} with the union of these boxes by F is contained in a union of balls B_{δ_{i}}(Q_{i}) with ∑_{i} δ_{i}^{d} ≤ 2K_{o}η̃ε^{n}, by the statement proved before, and so F (C_{ε}(x)) can be covered by balls B_{δ_{i}}(Q̃_{i}) with Σδ_{i}^{d} ≤ (K̃_{o}η^{\frac{n−p}{k+α}} + 2K_{o}η̃)ε^{n}. Choosing η, η̃ so small that K̃_{o}η^{\frac{n−p}{k+α}} + 2K_{o}η̃ ≤ η′, we obtain the desired result with ε_{o} = ε̃.

Remark 3.5. In the cases of functions of class C^{k} (C^{k+α} with α = 0) we have the same result. If k ≥ 2, it follows from the case of class C^{k−1+(1)} of the theorem. If k = 1, p + \frac{n−p}{k+α} = n, and the proof of the Theorem 3.3 shows that if F: ℝ^{n} → ℝ^{m} is a function and C(F) = {x ∈ U | DF (x) exists and rank DF (x) < n} then m_{n}(F (C(F))) = 0, where m_{n} is the Hausdorff measure of dimension n.

4 Examples

In this section we give some examples which show that the previous results are quite sharp. In all these examples we shall use a certain kind of functions of the real line that we shall describe below.

Definition 4.1. Let (λ_{n})_{n∈ℕ} be a sequence of real numbers with 0 < λ_{i} < \frac{1}{2} , ∀ i ∈ ℕ. The central Cantor set K\underline{λ} is the Cantor set constructed as follows: We remove from the interval [0,1] the central open interval U_{1,1} of proportion 1 − 2λ_{1}, then we remove from the two remaining intervals the central open intervals U_{2,1} and U_{2,2} of proportion 1 − 2λ_{2}, and so on. After the r-th step of the construction there will remain 2^{r} intervals of lenght λ_{1}λ_{2} … λ_{r}. The intersection of all these sets is the central Cantor set K\underline{λ}. The open intervals removed in the r-th step of the construction have lenght λ_{1}λ_{2} ⋯ λ_{r−1}(1 − 2λ_{r}).

Let ψ: ℝ ⟶^{C^{∞}} ℝ be a fixed function such that ψ(ℝ) ⊂ [0, 1], ψ(x) = 0, ∀ x ≤ 0, ψ(x) = 1, ∀x ≥ 1. Given two central Cantor sets K\underline{λ} and K\underline{μ}, we construct the function f_{\underline{λ},\underline{μ}}: ℝ → ℝ as follows: f_{\underline{λ},\underline{μ}}(x) = 0, ∀ x ≤ 0, f_{\underline{λ},\underline{μ}}(x) = 1, ∀x ≥ 1, and if U_{i,j} = (a, b) and v_{i,j} = (c, d) are corresponding removed intervals in the constructions of K\underline{λ} and K\underline{μ}, respectively, we define f_{\underline{λ},\underline{μ}}(x) = c + (d − c)ψ (\frac{x−a}{b−a}), ∀ x ∈ (a, b). We extend f_{\underline{λ},\underline{μ}} to K\underline{λ} by continuity, obtaining f_{\underline{λ},\underline{μ}}(K\underline{λ}) = K\underline{μ}. It is easy to check that if g_{r} := λ_{1}λ_{2} … λ_{r−1}(1 − 2λ_{r}) and g̃_{r} := μ_{1}μ_{2} … μ_{r−1}(1 − 2μ_{r}) satisfy lim_{r→∞} \frac{g̃_{r}}{g_{r}^{k}} = 0 then f_{\underline{λ},\underline{μ}} is C^{k} (if k ≥ 1 is an integer). Moreover, if q > 1, and →_{r} sup \frac{g̃_{r}}{g_{r}^{q}} < ∞ then f_{\underline{λ},\underline{μ}} is C^{q−1,1} if q is integer and is C^{q} (i.e., it is C^{[q]+{q}}, where {q} = q − [q] ∈ (0, 1)) otherwise. See [BMPV] for more details.

Example 4.1. Let λ_{n} = \frac{1}{2}−\frac{1}{2n}, μ_{n} ≡ a. Then lim_{n→∞} \frac{g̃_{n}}{g_{n}^{q}} = 0, ∀ q < \frac{− log a}{log 2}, and so f_{\underline{λ},\underline{μ}} is C^{q}, ∀ q < \frac{− log a}{log 2}. On the other hand, m_{d}(K\underline{μ}) = 1 where d = \frac{− log 2}{log a} (see [PT]). Moreover, since a ∈ (0, \frac{1}{2}), lim_{n→∞} \frac{g̃_{n}}{g_{n}} = 0, and so f′_{\underline{λ},\underline{μ}}(x) = 0, ∀ x ∈ K\underline{λ}. If F: ℝ^{n+p} → ℝ^{n+p} is given by F (x_{1}, x_{2}, …, x_{n}, x_{n+1}, …, x_{n+p}) = (f_{\underline{λ},\underline{μ}}(x_{1}), f_{\underline{λ},\underline{μ}}(x_{2}) …, f_{\underline{λ},\underline{μ}}(x_{n}), x_{n+1} …, x_{n+p}), then F (C_{p}(F)) = F (K\underline{λ} × K\underline{λ} × ⋯ × K\underline{λ} × ℝ^{p}) = K\underline{μ} × K\underline{μ} × ⋯ × K\underline{μ} × ℝ^{p} that is a set with positive (nd + p)-measure. This shows that given q > 1, p > 0 and n > p there is a map F: ℝ^{n} → ℝ^{n} such that m_{d}(F (C_{p}(F))) > 0, where d = p + \frac{n−p}{q} , and F is of class C^{q′} for each q′ < q. Remark 4.1. If a = \frac{1}{2^{n}}, F (x_{1}, x_{2}, …, x_{n}) = f_{\underline{λ},\underline{μ}}(x_{1})+2f_{\underline{λ},\underline{μ}}(x_{2})+⋯+2^{n−1}f_{\underline{λ},\underline{μ}}(x_{n}) gives an example of a function F: ℝ^{n} → ℝ which is of class C^{q}, ∀ q < n (q ∈ ℝ) such that F (ℝ^{n}) contains an open set, since K\underline{μ} + 2K\underline{μ} + ⋯ + 2^{n−1}K\underline{μ} = [0, 2^{n} − 1], which can be proved easily using representation in basis 2^{n}. Example 4.2. Let λ_{n} = \frac{1}{2} − \frac{1}{3n^{2}} , μ_{n} = a − \frac{a}{3n}, a ∈ (0, 1/2]. Then lim_{n→∞} \frac{g̃_{n}}{g_{n}^{q}} = 0, where q = \frac{− log a}{log 2} , and so f_{\underline{λ},\underline{μ}} is C^{q}. On the other hand we have HD(K\underline{μ}) ≥ \frac{− log 2}{log a} . Indeed, if b < a and θ_{n} ≡ b, f_{\underline{μ},\underline{θ}} is clearly C^{1}, and f_{\underline{μ},\underline{θ}}(K\underline{μ}) = K\underline{θ} ⇒ HD(K\underline{μ}) ≥ HD(K\underline{θ}) = \frac{− log 2}{log b} , ∀ b < a. If F: ℝ^{n+p} → ℝ^{n+p} is given by F (x_{1}, x_{2}, …, x_{n}, x_{n+1}, …, x_{n+p}) = (f_{\underline{λ},\underline{μ}}(x_{1}), …, f_{\underline{λ},\underline{μ}}(x_{n}), x_{n+1}, …, x_{n+p}) then F (C_{p}(F)) = K\underline{μ} × ⋯ × K\underline{μ} × ℝ^{p}, that is a set with Hausdorff dimension nd + p, where d = \frac{− log 2}{log a} . This shows that given q ≥ 1, p > 0 and n > p there is a map F: ℝ^{n} → ℝ^{n} such that HD(F (C_{p}(F))) = p + \frac{n−p}{q} , and F is of class C^{q}.

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