Hausdorff measures and the Morse-Sard theorem

Publ. Mat. 45 (2001), 149–162 HAUSDORFF MEASURES AND THE MORSE-SARD THEOREM Carlos Gustavo T. de A. Moreira Abstract Let F:U⊂Rn→Rmbe a differentiable function and p<man integer. If k≥1 is an integer, α∈[0,1] and F∈Ck+(α), if we set Cp(F)={x∈U|rank(Df(x)) ≤p}then the Hausdorff measure of dimension (p+n−p k+α)ofF(Cp(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:Rn→Rmof class Cn−m+1 has zero Lebesgue measure in Rm. 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 Xand a positive real number α, we define the Hausdorff measure of dimension αassociated to a covering U=(Uλ)λ∈L of Xby bounded sets Uλby mα(U)=λ∈L(diam Uλ)α, where diam Uλ denotes the diameter of Uλ, and, if we define the norm of a covering U by ||U|| = supU∈U(diam U), then the Hausdorff measure of dimension α of Xis mα(X) = lim inf Ucovering of X ||U||→0 mα(U). It is not difficult to see that there is a unique d∈[0,+∞] such that if α>dthen mα(X) = 0 and if α<dthen mα(X)=+∞. This number d is called the Hausdorff dimension of X. It is easy to see that if X⊂Rn then its Hausdorff dimension d=: HD(X) belongs to [0,n]. 150 C. G. T. de A. Moreira Sard himself proved that if Cp(F)={x∈Rn|rank(DF(x)) ≤p} then for any ε>0 there is k∈Nsuch that if Fis Ckthen F(Cp(F)) has zero Hausdorff measure of dimension p+ε([S2]). This result was made more precise by Federer ([F]), who proved that if k∈Nthen the Hausdorff measure of dimension p+n−p kof F(Cp(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(Cp(F)) is at most p+n−p k+α, provided that F∈Ck+α, where k∈N and 0 ≤α<1. More recently, Bates ([B2]) proved that if F∈Ck+α with k∈N,0<α≤1 and p+n−p k+α=mthen F(Cp(F)) has zero Lebesgue measure in Rm(this in particular improves the hypothesis of the classical Morse-Sard theorem from F∈Cn−m+1 to F∈Cn−m+Lips., i.e., F∈Cn−mand Dn−mFLipschitz). 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⊂Rn→Rmis of class Ck+(α)at a subset Aof Uif F is Ckin Uand for each x∈Athere are εx>0, Kx>0 such that |y−x|<ε x⇒|DkF(y)−DkF(x)|≤Kx|y−x|α(this is less restrictive than supposing F∈Ck+α). Our main result is the following Theorem. Let F:U⊂RnCk −→ Rmand let p<mbe an integer. If Cp(F):={x∈U|rank(DF(x)) ≤p}and if Fis of class Ck+(α)at Cp(F)then the Hausdorff (p+n−p k+α)-measure of F(Cp(F)) is zero. In particular, if k+α=n−p m−p, we recover the result of [B2], with a weaker hypothesis. We remark that if p+n−p k+α<m, the Hausdorff (p+ n−p k+α)-measure is not the Lebesgue measure or a product measure in Rm, 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. Hausdorff Measures and the Morse-Sard Theorem 151 Theorem 2.1. Let k≥1,α∈[0,1],n>pand A⊂U⊂Rn, where Uis an open set. Then there are sets A1,A 2... ⊂Asuch that A= ∞ i=1 Ai, where for each i=1,2,... there is a function ψi:Bi×Vi C1 −→ U where Biis a ball in some Rri,ri≥0and Viis a ball in Rpsuch that ψi(x, y)=( ψi(x, y),y), and |ψi(x1,y 1)−ψi(x2,y 2)|≥|(x1,y 1)−(x2,y 2)|, ∀(x1,y 1),(x2,y 2)∈Bi×Viand Ai⊂ψi(Bi×Vi), with the following property: We can write Ai=A i∪A iso that ψ−1 i(A i)has measure zero in Bi×Vi, and if f:U→Rvanishes in Aand fis Ck+(α)at Awe have: •lim sup (x,y0)→(x0,y0) f(ψi(x, y0)) |x−x0|k+α<+∞,∀(x0,y 0)∈Bi×Visuch that ψi(x0,y 0)∈Ai, •lim (x,y0)→(x0,y0) f(ψi(x, y0)) |x−x0|k+α=0,∀(x0,y 0)∈Bi×Visuch that ψi(x0,y 0)) ∈A i. Proof: Let us consider first the case k= 1 and df (x)·v=0∀x∈A, v∈Rn−p×{0}. In this case we take A=(A∩A)∪A where Ais the set of density points of Ain the direction of Rn−p×{0}((x, y)∈ A⇒lim ε→0 m((Bε(x)×{y})∩A) m(Bε(x)) = 1, where mis the (n−p)-dimensional measure). The measure of A =A−Ais zero, since it is zero in each plane Rn−p×{y}. For (x0,y 0)∈Atake B((x0,y 0),ε(x0,y 0)) a ball contained in Uand ψ=Id|B((x0,y0),ε(x0,y0)). We have lim sup (x,y0)→(x0,y0) f(x, y0) |x−x0|1+α<+∞, since f(x, y0)=f(x, y0)−f(x0,y 0)=df (tx0+(1−t)x)(x−x0), t∈ (0,1) ⇒|f(x, y0)|≤Kx0|x−x0|1+α. For (x0,y 0)∈A, lim δ→0 1 vol(Sn−p−1)Sn−p−11 δδ 0 χA(x0+tv, y0)dtdv =1, so ∀ε>0∃δ0>0 s.t. |x−x0|<δ 0⇒∃v∈Sn−p−1with v−x−x0 |x−x0|<ε and  1 |x−x0||x−x0| 0 χA(x0+tv, y0)dt −1<ε, 152 C. G. T. de A. Moreira so, if ˜x=x0+|x−x0|v, |f(x, y0)−f(x0,y 0)|≤|f(x, y0)−f(˜x, y0)|+|f(˜x, y0)−f(x0,y 0)|, but f(x, y0)−f(˜x, y0)=df (θt +(1−θ)˜x, y0)·(x−˜x),θ∈(0,1) ⇒|f(x, y0)−f(˜x, y0)|≤Kx0|x−x0|α·ε|x−x0|=εKx0|x−x0|1+α and f(˜x, y0)−f(x0,y 0) =|˜x−x0| 0 df (x0+tv, y0)·vdt ≤Kx0|x−x0|α·mt∈[0,|˜x−x0|]|∂f ∂x(x0+tv, y0)=0  ≤Kx0|x−x0|α·ε|x−x0|=εKx0|x−x0|1+α. So |f(x, y0)|=|f(x, y0)−f(x0,y 0)|≤2εKx0|x−x0|1+α ⇒lim (x,y0)→(x0,y0) f(x, y0) |x−x0|1+α=0. We can take a countable subcovering of Aby the B((x0,y 0),ε(x0,y 0)) to finish the proof in this case. Consider now the case k≥1, narbitrary. We have A=A∗∪A∗∗ where A∗={x∈A|∃g:UCk −→ R,g|A≡0, ∃v∈Rn−p×{0}, dg(x)·v=0}.A∗∗ =A\A∗.If(x0,y 0)∈A∗there is gas above, so there is ε>0 such that g−1(0) ∩Bε(x0,y 0) is contained in the image of ψ:B×VCk −→ Uwhere Bis a ball in Rn−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.Ifk= 1, the result was yet proved for A∗∗.Ifk>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⊂ψi(Bi×Vi),ψ i∈C1, Hausdorff Measures and the Morse-Sard Theorem 153 ψi(x0,y 0)∈A∗∗ i⇒lim sup x→x0 ||df (ψi(x, y0))|Rn−p×{0}|| |x−x0|k−1+α<+∞ ⇒lim sup x→x0 |f(ψi(x, y0))| |x−x0|k+α<+∞ and ψi(x0,y 0)∈(A∗∗ i)⇒lim x→x0 ||df (ψi(x, y0))|Rn−p×{0}|| |x−x0|k−1+α=0 ⇒lim x→x0 f(ψi(x, y0)) |x−x0|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>pand A⊂U⊂Rn, where Uis an open set. Then there are sets A1,A 2... ⊂Asuch that A= ∞ i=1 Ai, where for each i=1,2,... there is a function ψi:Bi×Vi C1 −→ U where Biis a ball in some Rri,ri≥0and Viis a ball in Rpsuch that ψi(x, y)=( ψi(x, y),y), and |ψi(x1,y 1)−ψi(x2,y 2)|≥|(x1,y 1)−(x2,y 2)|, ∀(x1,y 1),(x2,y 2)∈Bi×Viand Ai⊂ψi(Bi×Vi), with the following property: We can write Ai=A i∪A iso that ψ−1 i(A i)has measure zero in Bi×Vi, and if f:U→Ris Ck+(α)at Aand Dxf≡0in Awe have: •lim sup (x,y0)→(x0,y0) |f(ψi(x, y0)) −f(ψi(x0,y 0))| |x−x0|k+α<+∞,∀(x0,y 0)∈Bi× Visuch that ψi(x0,y 0)∈Ai, •lim (x,y0)→(x0,y0)|f(ψi(x, y0)) −f(ψi(x0,y 0))| |x−x0|k+α=0,∀(x0,y 0)∈Bi×Vi such that ψi(x0,y 0)) ∈A i. Proof: If k≥2 this is an immediate consequence of Theorem 2.1 applied to Dxfand 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∈Bis.t. ψi(x)∈Aithere are εx>0,Kx>0such that |y−x|<ε x⇒|f(ψi(y)) −f(ψi(x))|≤Kx|y−x|k+α, and for any ε>0 there is a δ>0so that λ(ψ−1 i(Ai)∩Br(x)) λ(Br(x)) >1−δ⇒|f(ψi(y))−f(ψi(x))|≤ εKxrk+α,ifr≤εxand |y−x|≤r(δdepends only on εand n, but not on for on x). 154 C. G. T. de A. Moreira 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 f(ψi(y)) |y−x|k+α= 0, for each x∈Bisuch that ψi(x)∈A i. 3. The main results Lemma 3.1. Let A⊂Rmwith λ(A)<∞and let Ube a family of balls Br(x),x∈Asuch that for each x∈Athere is an εx>0such that r≤εx⇒Br(x)∈U. Then for each ε>0there are xn∈A,rn>0 with Brn(xn)∈Uand A⊂∞ n=1 Brn(xn)such that ∞ n=1 λ(Brn(xn)) < λ(A)+ε. Proof: This lemma is essentially the Vitali covering theorem from measure theory. Take U⊃Aan open set with λ(U)<λ(A)+ε 2.Ifwe choosed B˜r1(x1),... ,B˜rn(xn), define sn= sup{r>0|∃x∈As.t. r< εx 5,Br(x)⊂Uand Br(x)∩(B˜r1(x1)∪···∪B˜rn(xn)) = ∅}. Choose B˜rn+1 (xn+1) such that ˜rn+1 >sn 2,˜rn+1 <εxn+1 5,B˜rn+1 (xn+1)⊂Uand B˜rn+1 (xn+1)∩(B˜r1(x1)∪···∪B˜rn(xn)) = ∅. Since the B˜ri(xi) are disjoint and contained in Uwe have ∞ i=1 λ(B˜ri(xi)) <λ(A)+ε 2, and so there is an0∈Nsuch that ∞ i=n0λ(B5˜ri(xi)) <ε 2. We take Bri(xi)=B˜ri(xi), i<n 0and Bri(xi)=B5˜ri(xi), i≥n0. Clearly we have ∞ i=1 λ(Bri(xi)) <λ(A)+ε. To prove that A⊂ ∞ n=1Brn (xn), take x∈Aand r=min{˜rn0,ε x/5,d(x, Uc∪i<n0Bri (xi))}. If r>0, take n≥n0such that sn<r≤sn−1(we have r≤˜rn0≤sn0−1), and note that sn<r⇒Br(x)∩(B˜r1(x1)∪···∪B˜rn(xn)) =∅⇒∃i≤n such that Br(x)∩B˜ri(xi)=∅. We have n≥n0since r≤d(x, B˜ri(xi)), and ˜ri>sn−1 2≥r 2, since i≤n. Therefore, we have x∈B5˜ri(xi). If r=0 then x∈Bri(xi) for some i<n 0. This proves that A⊂∞ n=1 Brn(xn). Taking ˜rn=( λ(A)+ε ∞ i=1 λ(Bri(xi)) )1/2m·rn, we have A⊂∞ n=1 B˜rn(xn), with ∞ n=1 λ(B˜rn(xn)) = (λ(A)+ε)1/2(∞ i=1 λ(Bri(xi)))1/2<λ(A)+ε. Remark 3.1.In the Lemma 3.1 we can replace a family of balls Br(x)by a family of cubes Cr(x)=m i=1[xi−r, xi+r], where x=(x1,... ,x m), using the same proof. Lemma 3.2. Let F:U⊂Rn→Rmbe a function, A⊂Uand d>0 such that for any x∈Athere are εx>0,Kx>0such that md(F(Bε(x)∩ A)) ≤Kx.λ(Bε(x)),∀ε<ε x, where mdis the Hausdorff measure of dimension d, and there is A⊂Asuch that λ(A\A)=0and lim ε→0 md(F(Bε(x)∩A)) λ(Bε(x)) =0,∀x∈A. Then md(F(A)) = 0. Hausdorff Measures and the Morse-Sard Theorem 155 Remark 3.2.The same result is true if we replace Bε(x)byCε(x). Remark 3.3.We can replace the condition “md(F(Bε(x)∩A)) ≤Kxλ(Bε(x)),∀ε<ε x” by “F(Bε(x)∩A) can be covered by balls Bδi(yi),i∈N, with ∞  i=1 δd i≤Kxλ(Bε(x)),∀ε<ε x”, and the condition “ lim ε→0 md(F(Bε(x)∩A) λ(Bε(x)) =0,∀x∈A” by “F(Bε(x)∩A) can be covered by balls Bδ(ε) i (yi),i∈N with lim ε→0∞ i=1(δ(ε) i)d λ(Bε(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(Bε(x)∩A) can be covered by balls Bδi(yi), i∈N, with ∞ i=1 δd i≤kλ(Bε(x)), ∀ε<ε x(note that here kdoes not depend on x), and λ(A)<∞”, then we can conclude, using the same proof, that md(F(A)) ≤kλ(A). Proof: We may suppose that Ahas finite Lebesgue measure, since Ais 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 Ak, where Ak={x∈A|Kx≤k}, we may suppose Kx≤K,∀x∈A. Let Cbe the Lebesgue measure of A. Let ε>0. For each x∈Atake δx>0 such that Bδx(x)⊂Uand r≤δx⇒md(F(Br(x)∩A)) λ(Br(x)) ≤ε 2(C+1) . By the Lemma 3.1 we can cover A by ∞ n=1 Brn(xn) with ∞  n=1 λ(Brn(xn)) <C+1 156 C. G. T. de A. Moreira and rn≤δxn⇒∞  n=1 md(F(Brn(xn)∩A)) ≤ε 2(C+1)·(C+1)= ε 2⇒md(F(A)) ≤ε 2. By Lemma 3.1 we can cover A\Aby ∞ n=1 B˜rn(xn) such that B˜rn(xn)⊂ Uand ˜rn<ε xn,∀n∈N, with ∞  n=1 λ(B˜rn(xn)) <ε 2K⇒∞  n=1 md(F(B˜rn(xn))) ≤ε 2K·K=ε 2⇒md(F(A\A)) ≤ε 2⇒md(F(A)) ≤ε 2+ε 2=ε. Since ε>0 is arbitrary we have md(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⊂Rn→Rnbe 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∈Athen lim r→0 λ(F(Br(x))) λ(Br(x)) = 0. Indeed, x∈A⇒F(x+h)=F(x)+DF(x).h + r(h), where lim h→0 r(h) |h|= 0. Let K=||DF(x)||, and let ε∈(0,1). Let δ>0 such that |h|≤δ⇒|r(h)| |h|<ε 2(K+1)n−1. Then, if |h|≤δ, F(x+h)−F(x) belongs to an ε.|h| 2(K+1)n−1neighbourhood of a ball of radius K|h|in a subspace of Rnof dimension n−1 (a fixed subspace of Rnof 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 ε|h| 2(K+1)n−1. Therefore, λ(F(Br(x)) ≤ ε.r.rn−1(K+1)n−1 (K+1)n−1vn−1=εrnvn−1, where vn−1is the volume of the unitary ball in Rn−1, and, since ε>0 is arbitrary, lim r→0 λ(F(Br(x))) λ(Br(x)) =0. Theorem 3.4. Let F:U⊂RnCk −→ Rmbe a function of class Ck+(α)(α∈ (0,1]) at Cp(F):={x∈U|rank(DF(x)) ≤p}. Then the Hausdorff measure of dimension d=p+n−p k+αof F(Cp(F)) is zero, ∀p<min{m, n}. Hausdorff Measures and the Morse-Sard Theorem 157 Proof: Since Cp(F)=p r=0{x∈U|rank(DF(x)) = r}, and r+n−r k+α≤ p+n−p k+αfor 0 ≤r≤p, we may restrict our attention to  Cp(F)={x∈ U|rank(DF(x)) = p}.Ifx0∈Cp(F), we have, after a change of coordinates of class Ck,F(z,y)=(z,G(z,y)), with (z,y)∈Rp×Rn−p and G(z,y)∈Rm−p, in a neighbourhood Vof x0=(z0,y 0). We shall restrict our attention to this neighbourhood. We have x=(z,y)∈ Cp(F)⇔DyG(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 DyG, and obtain the decomposition A=∞ i=1 Ai,Ai⊂ψi(Vi×Bi), where A={(z,y)∈V|DyG(z,y)=0}. Let us fix such an Ai. Since ψ−1 i(Ai)=m∈N{x∈ψ−1 i(Ai)|εx≥1 m,K x≤m},wemay suppose εx≥1 M,Kx≤M,∀x∈ψ−1 i(Ai), for some fixed Mand also that Vhas finite Lebesgue measure λ(V). With these assumptions, we shall prove that there is a constant K0 such that for any X⊂V,ν>0, we can cover F(Ai∩X) by balls Bδi(pi) so that ∞ i=1 δd i≤K0(λ(X)+ν). For this, given a point x∈Ai∩X and an ε< 1 2√nM , we can divide the cube Cε(x)=Cε(z)×Cε(y) into ([ε1−(k+α)]+1) pboxes Cδ(zi)×Cε(y), δ<ε k+α. If there is some point (zi,y i)in(Cδ(zi)×Cε(y)) ∩(Ai∩X), then for any point (z i,y i)in (Cδ(z1)×Cε(y)) ∩(Ai∩X), we have |F(z i,y)−F(zi,y i)|≤|F(z i,y)− F(zi,y)|+|F(zi,y)−F(zi,y i)|≤Kδ+|F(zi,y i)−F(zi,y i)|(where K is √ptimes a Lipschitz constant of F|Vwhich we may suppose to exist) ≤Kδ+|G(zi,y i)−G(zi,y i)|. Observe now that (zi,y i)=(zi, ψi(p1)) and (z i,y i)=(zi, ψi(p2)), for some p1,p2in {zi}×Biwith |p1−p2|≤|yi−y i|≤2ε√n. Let γ:[0,1] → Vi×Bibe a straight path joining p1and p2. Then G(zi,y i)−G(zi,y i)= 1 0 ∂G ∂y (γ(t)) ·γ(t)dt, where γ:= ψi◦γ. We have ∂G ∂y (γ(0)) = 0, so     ∂G ∂y (γ(t))   =    ∂G ∂y (γ(t)) −∂G ∂y (γ(0))    ≤M|p1−p2|k+α−1 ≤M(2ε√n)k+α−1⇒    ∂G ∂y (γ(t))   |γ(t)| ≤Kεk+α, for some constant K. Indeed, |γ(t)|is limited by a constant multiple of |p1−p2|≤2√nε.So |G(z1,y i)−G(zi,y i)|≤1 0 ∂G ∂y (γ(t)) ◦γ(t)dt ≤Kεk+α