a basic introduction of geometric measure theory

7/23/2019 A BASIC INTRODUCTION OF GEOMETRIC MEASURE THEORY

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A BASIC INTRODUCTION OF GEOMETRIC MEASURE THEORY

QING HAN

Geometric measure theory studies properties of measures, functions and sets. In thisnote, we provide a basic introduction of geometric measure theory. We will study ele-mentary properties of Hausdorff measures, Lipschitz functions and countably rectifiablesets. To make this note easily accessible, we confine our discussions in Euclidean spaces.Important topics include the area and coarea formulae for Lipschitz functions and ap-

proximate tangent space properties of countably rectifiable sets.This note is prepared for lectures in the Special Lecture Series the author delivered in

Peking University, June 2006. He would like to thank Professor Gang Tian for providingsuch an opportunity. He would also like to thank Miss Xianghui Yu for helping himdealing various issues when he was visiting Peking University. Last but not least, theauthor would like to thank all participants of this series of lectures. Because of theiractive participation in class and numerous discussions with him after the class, the authorwas able to correct many mistakes in the early version of this lecture note.

Address: Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556, USA.Email: [email protected].

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2 Q. HAN

1. Prerequisite Knowledge

This section serves as a preparation for the rest of the note. In the first part, webriefly review some basic theory of outer measures. The second part includes the coveringtheory. Many techniques in geometric measure theory involve covering arguments.

Let X be a topological space. We denote byS the family of all subsets of X. AcollectionC of subsets ofX is said to be a -algebraif

(i), X C;(ii)i=1Ai C andi=1Ai C ifAi C for i = 1, 2, 3, ;(iii) X\ A C ifA C.LetB be the smallest -algebra containing all open subsets ofX. The elements ofB

are called Borel subsetsofX. Obviously,B also contains all closed subsets ofX.Now we recall the definition of (outer) measures.

LetXbe a topological space. If :S [0, ] is a function such that () = 0 and(A)

j=1

(Aj),

for any A j=1Aj, Aj S. Then is a measureon X.A subsetAX is -measurable if

(B) =(B\ A) + (B A) for any subsetBX.It is easy to show that the subfamily ofS consisting of all -measurable subsets is a-algebra.

A measure on Xis called aBorel measureif each Borel set is -measurable. A Borel

measure is calledBorel regularif for each subset AXthere exists a Borel setBAsuch that(B) =(A).

The following result is often referred to as the Caratheodory Criterion.

Theorem 1.1. Let be a measure on a metric space X. Then all open sets are -measurable if and only if

(A B)(A) + (B),for any subsetAX andBX with dist(A, B)> 0.Proof. The necessity is obvious. For the sufficiency, it suffices to prove that for any opensubsetOXand any subset T X

(T)(T\ O) + (T O),if(T) 0 and hence(T)(T Ck) + (T\ O).


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GEOMETRIC MEASURE THEORY 3

Now we claim

limk(T Ck) =(T C) =(T O),or limk (Ck) =(C).

Let C0 = and Rk = Ck Ck1 for k 1. By the condition of the sufficiency, weobtain for any positive integer k

(C2k)k

i=1

(R2i),

(C2k1)k

i=1

(R2i1).

Note that

C=

k=1

Ck = C2k+

i=k+1

R2i+

i=k+1

R2i1,

andi=1

(R2i)(T)


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4 Q. HAN

The proof is left as an exercise.

Suppose X is a locally compact and separable topological space. Then is a Radonmeasureif is Borel regular and finite on compact subsets ofX.LetUbe a bounded open subset in Rn. Denote byM(U) the space of signed Radon

measures on U with finite mass and by C0(U) the space of continuous (real-valued)functions onUwith compact support.

Definition 1.3. A sequence{k}k=1 M(U) converges weakly to M(U), denotedbyk weakly inM(U), if

Uf dk

U

f d as k , for each fC0(U).

Lemma 1.4. Assume thatk weakly inM(U). Thenlim sup

kk(C)(C),

for each compact setCU, and(O)lim inf

kk(O),

for each open setOU.Proof. (1) For any >0, there exist a >0 and a -neighborhood C ofC such that

(C)(C) + .By taking

f(x) = min{

1,1

dist(x, U

\C)

},

we have

(C) + >

f d = lim

k

f dklim sup

kk(C).

(2) For any m < (O), take a compact set A O with (A) > m and then take apositive number >0 such that the -neighborhood A O. We define fsimilarly asabove to obtain

m 0

Hs(A).We callHs the s-dimensional Hausdorff measureon Rn.

For the empty set, we simply defineHs() = 0 for any >0 andHs() = 0.Note that (s) is the volume of the unit ball in Rs ifs is a positive integer.

Remark 2.2. For > ,Hs(A) Hs(A). HenceHs() is a monotone decreasingfunction of(0, ]. In particular, we have for any subset A R

n

, >0 and s0Hs(A) Hs(A) Hs(A).We note that it is necessary to require 0 in order to force the coverings to followthe local geometry of the set A.

Observe thatLnBr(x)= (n)rn,

for any balls Br(x) Rn. Later on, we will prove that, ifs is an integer,Hs agrees withthe ordinary s-dimensional surface area on nice sets. This is the reason to include thenormalizing constant (s) in the definition.

Theorem 2.3. For anys

[0,

),

Hs is a Borel regular measure onRn.

Proof. (i)Hs is a measure, i.e., for any{Ai}i=1 Rn,

Hs(i=1

Ai)i=1

Hs(Ai).

By the definition ofHs for >0, we easily have

Hs(i=1

Ai)i=1

Hs(Ai).


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8 Q. HAN

Then take limit 0.(ii)H

s

is a Borel measure. We will proveHs(A B) =Hs(A) + Hs(B),for any Borel sets A, B Rn with dist(A, B)> 0. It is easy to prove by the definitionthat

Hs(A B) =Hs(A) + Hs(B),if dist(A, B)> 3. Then letting 0+, we obtain the desired result.

(iii)Hs is Borel regular. We need to show that, for anyA Rn withHs(A) 0.(iv)Hs(L(A)) =Hs(A) for any affine isometriesL: Rn Rn.

Proof. (iii) and (iv) are trivial. For (i), we note that (0) = 1 andH0({a}) = 1 for anya Rn.

For (ii), we need to proveHs(Q) = 0 for any unit cubeQ Rn.Fix an integerm1.Then the unit cube Q can be decomposed into mn cubes with the side length 1/m andthe diameter

n/m. Therefore,

Hsnm

(Q)mni=1

(s)(

n

m )s =(s)ns/2mns,

and the last term goes to zero as m ifs > n. HenceHs(Q) = 0, andHs(Rn) =0.

The next result yields an easy method to prove that a set has a zero measure.

Lemma 2.5. SupposeA Rn andHs(A) = 0 for a(0, ]. ThenHs(A) = 0.


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Proof. We will showHs(A) = 0 if Hs(A) = 0. It is obvious for s = 0. We considers > 0.For any >0, there exists a collection of sets{Cj}j=1 such that A j=1Cj and

j=1

(s)diamCj

2

s .In particular, for eachj1,

diamCj2

(s)

1/s ().Hence

Hs()(A).

Since ()0 as 0, thenHs

(A) = 0.

The next result asserts that, for any subset ARn, there is only one possible s0such thatHs(A) is meaningful.Lemma 2.6. LetA Rn and0s < t s andHt(A) = for any t < s. We note thatHs(A) may beany number between 0 and inclusive. Furthermore, dimH(A) need not be an integer.


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10 Q. HAN

Even if dimH(A) =k is an integer and 0


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It is easy to check by the Fubini Theorem that

(i) h(x) is a measurable function;(ii) 2

h(x)dx=Ln(A).Now we claim that diam(Se(A))diam(A). In fact, we prove

diam

Se(A)

= supx,x

(x x)2 + h(x) + h(x)2

supx,x

diam

(Lx A) (Lx A)

.

Set

a= inf(Lx A), b= sup(Lx A),a = inf(L

xA), b = sup(L

xA).

Assumeb a bawithout loss of generality. Sinceba 2h(x) andb a2h(x)by the definition ofh, we have

diam

(Lx A) (Lx A)|x x|2 + |b a|2

|x x|2 + h(x) + h(x)2.This finishes the proof of the claim.

Next, take an orthogonal basis{e1, , en} ofRn and setA =Sen Se1(A).

Then, diam(A)diam(A) andLn(A) =Ln(A). It is clear that A is symmetric withrespect to e1, , en. HenceA is symmetric with respect to the origin. Thus xAimpliesxA. So for any xA, we have 2|x| diamA, or xBdiamA/2(0). Thisimplies

A BdiamA/2(0).Then

Ln(A) =Ln(A)(n)diamA2

n (n)diamA2

n.

This finishes the proof.

Theorem 2.10.Hn

=Ln

onRn

.

Proof. (i) We first showLn(A) Hn(A) for any A Rn. For any > 0, consider anycollection of sets{Cj}j=1 such that A j=1Cj and diamCj . Then by Theorem2.9, we have

Ln(A)

j=1

Ln(Cj)n

j=1

(n)diamCj

2

n.

Taking the infimum, we haveLn(A) Hn (A), and thusLn(A) Hn(A).


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12 Q. HAN

(ii) We prove thatHn is absolutely continuous with respect toLn. To this end, wefirst observe that for any >0 andA

Rn

,

Hn (A)inf{i=1

(n)diamQj

2

n; A

j=1

Qj, Qj are cubes, diamQj}.

For each cube Q Rn, we have(n)

diamQ2

n=C(n)Ln(Q),

whereC(n) = 2n(n)

nn. Hence by the definition of Lebesgue measure,

Hn (A)C(n)Ln(A),or

Hn(A)C(n)Ln(A).(iii) We prove thatHn(A) Ln(A) for any A Rn. For any >0 and >0, there

exists a collection of cubes{Qj}j=1 such that diamQj < , A j=1Qj and

j=1

Ln(Qj) Ln(A) + .

For each cubeQi, choose disjoint balls{Bik}k=1Qisuch that diamBik < andLn(Qi \k=1 B

ik) = 0. Then by (ii),Hn(Qi\

k=1 B

ik) = 0. Hence we obtain

Hn (A)

i=1

Hn (Qi)

i=1

k=1

Hn (Bik)

i=1

k=1

(n)diamBik

2 n

i=1

k=1

Ln(Bik) =i=1

Ln(Qi) Ln(A) + .


3. Densities

In this section, we discuss densities for Hausdorff measures. We first define densitiesfor general measures.

Definition 3.1. Let be a measure on a metric space X. For any subset AX andany point x

X, the s-dimensional upper and lower densities ofx in A in the measure

, s(A, x) and s(A, x), are defined, respectively, bys(A, x) = lim sup

r0

1

(s)rs

A Br(x)

,

and

s(A, x) = lim infr0

1

(s)rs

A Br(x)

.

The s-dimensional density s(A, x) is defined as the common value if s(A, x)= s(A, x).


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It is easy to see that we may use closed balls in the above definition.

Remark 3.2. When =Hs, we write s(A, x), s(A, x) and s(A, x) instead ofs(HsA, x), s(HsA, x) and s(HsA, x). It is easy to show that s(A, x) ands(A, x) are Borel functions ofx Rn ifA Rn is Borel.

We first examine the densities of points inLn-measurable subsets of Rn. By theLebesgue differentiation theorem, we have for anLn-integral functionf

limr0

1

(n)rn

Br(x)

|f|=|f(x)|,

forLn-almost allx Rn. For anyLn-measurable subsetA ofRn, by taking f=A, weobtain

limr0

1(n)rn

LnA Br(x)= (x),forLn-almost allx Rn. Hence we conclude the following result.Lemma 3.3. LetA be anLn-measurable subset ofRn. Then

n(A, x) =

0 forLn-almost allx Rn \ A,1 forLn-almost allxA.

Now we prove some results concerning densities of Hausdorff measures.

Theorem 3.4. SupposeE Rn isHs-measurable withHs(E)0 and define

At=

x Rn \ E;limsupr0

1

(s)rsHsBr(x) E> t.

We prove thatHs(At) = 0 for each t >0.Note thatHsEis a Radon measure. For any >0, there exists a compact setKE

such that

Hs(E\ K).

Consider the open set U=Rn

\ K. We have AtU. Fix a >0 and considerB= Br(x); Br(x)U, r(0, ), 1

(s)rsHsBr(x) E> t.

ThenB is a fine cover ofAt. By Corollary 1.7, there exists a countable disjoint familyof balls{Bri(xi)}i=1 inB such that

Ati=1

B5ri(xi).


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14 Q. HAN

Then, we have

Hs10(At)i=1

(s)(5ri)s 5s

t

i=1

Hs(Bri(xi) E)

5s

tHs(U E) = 5

s

tHs(E\ K) 5

s

t.

Letting0, we haveHs(At)5st1. ThusHs(At) = 0 for any t >0. Theorem 3.5. Suppose thatE Rn isHs-measurable withHs(E)0 andt >1, and define

Et=

xE;limsupr01

(s)rsHs

Br(x) E> t.SinceHsE is a Radon measure, there exists an open set U containing Et with

Hs(U E) Hs(Et) + .Define

B={Br(x); Br(x)U, r(0, ), 1(s)rs

HsBr(x) E> t}.ThenBis a fine cover ofEt.By Corollary 1.7, there exists a countable disjoint family ofballs{Bri(xi)}i=1 inB such that

Et

k

i=1

Bri

(xi)

i=k+1

B5ri

(xi),

for eachk = 1, 2, . Then

Hs10(Et)k

i=1

(s)rsi +

i=k+1

(s)(5ri)s

1t

ki=1

Hs(Bri(xi) E) +5s

t

i=k+1

Hs(Bri(xi) E)

1tHs(U E) +5

s

tHs(

i=k+1Bri(xi) E),

for any k = 1, 2, . By taking k , we obtainHs10(Et)

1

tHs(U E) 1

t

Hs(Et) + .Letting0 and then 0, we have

Hs(Et) 1tHs(Et).

SinceHs(Et) Hs(E)1.


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Theorem 3.6. Suppose thatE Rn isHs-measurable. Then

s(HsE, x) 12s forHs-almost allxE.

Proof. Fix a t(0, 1) and define

At=

xE;limsupr0

1

(s)rsHs

Br(x) E

0 implies u / Yc. Hence there exists(a, r)P with

|a|2 r2c2


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18 Q. HAN

and

b=k

i=1

ig(xi), 1 =k

i=1

i,

where andbare as defined in Lemma 4.4. With the identity 2uv=|u|2 +|v|2|uv|2,we obtain

0 =2 k

i=1

i

g(xi) b2 = 2 k

i,j=1

ij

g(xi) b g(xj) b

=k

i,j=1

ij|g(xi) b|2 + |g(xj) b|2 |g(xi) g(xj)|2

k

i,j=1

ij

c2|xip|2 + c2|xjp|2 |xi xj|2

=k

i,j=1

ij

2c(xip) c(xjp) + (c2 1)|xi xj|2

=2c k

i=1

i(xip)2 + (c2 1) k

i,j=1

ij|xi xj|2.

Hence either k = 1 and c = 0 (because p=x1

T), or k >1 and c

1. We conclude

that c1, and hence|g(xi) b| |xip| for any i = 1, , k.

Next, we prove a theorem of Rademacher concerning the differentiability of Lipschitzfunctions on Euclidean spaces.

Theorem 4.5. Let f : Rm Rn be a locally Lipschitz function. Then f is differ-entiableLm-almost everywhere in Rm, i.e., forLm-almost every x Rm, Df(x) =(1f(x), , mf(x)) exists and

limyx

1

|y

x

|f(y) f(x) Df(x) (y x)= 0.

Proof. We assumen = 1 andfis Lipschitz. Fix a vectorvRm with|v|= 1, and define

vf(x) = limt0

1

t

f(x + tv) f(x) for anyx Rm,

if this limit exists. We define vf and vfsimilarly with lim replaced by lim sup andliminf.

Claim 1. For each vRm with|v|= 1, vf(x) exists forLm-almost allx Rm.


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To prove this, we first note that vf and vf are Borel functions, since, by the

continuity off,vf(x) = lim sup

t0

1

t

f(x + tv) f(x)

= limk

sup 1

k


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20 Q. HAN

where we used the Fubini theorem and the absolute continuity offon lines. This implies

Claim 2.Now choose a countable dense subset{vk}k=1 ofSm1. SetAk ={x Rm; vkf(x), Df(x) exist and vkf(x) =vk Df(x)},

and define A =k=1Ak. ThenLm(Rm \ A) = 0, vkf(x) =vk Df(x) for any xA andk= 1, 2, .

Claim 3. fis differentiable at each point xA.Fix anxA. For any vSm1 andt R with t= 0, define

Q(x,v,t) =1

t

f(x + tv) f(x) v Df(x).

Then for any v Sm1, we have

|Q(x,v,t) Q(x, v

, t)|1

t

f(x + tv) f(x + tv)+ |(v v) Df(x)|

Lip(f)|v v| + |Df||v v|(m + 1)Lipf|v v|.

By the compactness of Sm1, for any > 0, choose N large enough so that, for anyvSm1, there exists a k {1, 2, , N} satisfying

|v vk| 2(

m + 1)Lipf

.

Hence

limt0 Q(x, vk, t) = 0 for any k = 1, 2, , N,and there exists a >0 so that

|Q(x, vk, t)|< 2

for any 00, there exists aC1functiong: Rm R such that

Lm{x Rm; f(x)=g(x)} {x Rm; Df(x)=Dg(x)}< .


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We skip the proof as it is quite complicated.

5. Area and Coarea Formulae

In this section, we discuss the area formula and coarea formula for Lipschitz functionson Euclidean spaces. We first need a result on polar decompositions from the linearalgebra.

Lemma 5.1. LetL: Rm Rn be a linear map.(i) Ifmn, then there exist a symmetric map S: Rm Rm and an orthogonal map

O: Rm Rn such thatL= O S.(ii) Ifmn, then there exist a symmetric map S : Rn Rn and an orthogonal map

O: Rm Rn such thatL= S O.For case (i), Theorem 5.1 asserts that a linear map can be decomposed as a composition

of a dilation Sand an orthogonal injection O. A similar interpretation is immediate forcase (ii).

Proof. We need only prove (i). For (ii), we simply consider the adjoint map L : Rn Rm and apply (i).

Let A= LL. Then A is a symmetric nonnegative definite matrix. Let 1, , m0be the set of eigenvalues ande1, , em be a set of corresponding orthonormal eigenvec-tors, i.e., Aek =kek, k = 1, , m. Furthermore, let j =j and zj =L(ej/j) forj= 0. It is easy to see that < zk, zl >= kl for any k and l with k= 0 and l= 0.Note that{zj} is defined only for j with j= 0. Supplement those zj by unit vectorssuch that{zj}

m

j=1 forms an orthonormal set. DefineS :Rm R

m

bySej = jej anddefineO : Rm Rn byOej =zj, j = 1, , m. Then it is easy to check L = O S. We define the Jacobian ofL : Rm Rn by

[[L]] =|detS|.It is easy to see that

[[L]] =

det(LL), mn,det(LL), nm.

Hence [[L]] is independent of the decomposition in Lemma 5.1. Obviously, [[L]] = 0 ifrank(L) < min

{m, n

}. By the Cauchy-Binet formula, we can calculate the Jacobian of

L: Rm Rn, mn, by[[L]]2 =

(n,m)

det(P L)

2,

where (n, m) is the (n, m)-permutation group and P is the orthogonal projection tothe appropriate m-dimensional subspace in Rn.

Remark 5.2. LetL1, L2: Rm Rn be linear maps and t >0. If|L1v| t|L2v|for any

vRm, then [[L1]]tm[[L2]]. Such a simple fact will be employed frequently.


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22 Q. HAN

Now we introduce the Jacobian of differentiable maps. Supposef :ARm Rn isa map differentiable at xA. We define the Jacobianoff at x by

Jf(x) = [[Df(x)]].

Sometimes, in order to emphasize the maximal possible rank, we denote the JacobianbyJmf(x) for mn and Jnf(x) ifmn.

Let f : Rm Rn be a Lipschitz map and A be anLm-measurable subset ofRm. Weintend to calculate

AJf(x)dLm(x).

We will derive the area formula and the coarea formula for mnandmnrespectively.We first state state the area formula.

Theorem 5.3. Let f : Rm

Rn

be a Lipschitz map and m n. Then for anyLm

-measurable setA Rm,

AJmf(x)dx=

Rn

H0A f1(y)dHm(y).In particular, iff is one-to-one onA, then

AJmf(x)dx=Hm

f(A)

.

We note that the integral in the right hand side of the first formula is over f(A). Wewill prove that it is anHm-measurable set ofRn.

To prove Theorem 5.3, we need three preliminary lemmas.

Lemma 5.4. SupposeL: Rm Rn is linear, mn. ThenHm(L(A)) = [[L]]Lm(A).Proof. By Lemma 5.1, we have L = OS, where S : Rm Rm is symmetric andO: Rm Rn is orthogonal. Then

HmL(A)=HmO S(A)=HmS(A)=|detS|Lm(A) = [[L]]Lm(A).

Lemma 5.5. Letf : Rm Rn be Lipschitz, mn. IfA Rm isLm-measurable, then(i) f(A) is anHm-measurable subset ofRn;(ii) the functiony H0(A f1(y)) isHm-measurable onRn;(iii) Rn H

0(f1(y)

A)d

Hm(y)

(Lipf)m

Lm(A).

Proof. Without loss of generality, we assume that A is bounded. Otherwise, we simplydecompose it into a countable union of bounded sets.

(i) Fori = 1, 2, , we choose a sequence of compact setsKiA such thatLm(Ki)>Lm(A)1/i. Since f is continuous, f(Ki) is compact, which implies f(Ki) isHm-measurable, and so is their countable unioni=1f(Ki) =f(i=1Ki). Then, we have

Hmf(A) \ f( i=1

Ki) Hmf(A\

i=1

Ki)(Lipf)mLm(A \

i=1

Ki) = 0.


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This implies that f(A) isHm-measurable.(ii) We first decompose

Rm

into a set of pairwise disjoint small cubes. For any integerk1, letBk ={Q; Q= (a1, b1] (am, bm]}, whereai= ni/2k,bi= (ni + 1)/2k,nisare integers. Note that any element ofBk+1is a subset of some element ofBk. Obviously,Rm =QBkQ. Define

gk =

QBk

f(AQ) in Rn.

Then by (i), gk is anHm-measurable function in Rn for any k. Note thatgk(y) is thenumber of cubes Q Bk such that f1(y) A Q is nonempty. Therefore, gk increasesmonotonically toH0(f1(y) A) as k . By the Monotone Convergence Theorem,H0(f1(y) A) is also anHm-measurable function on Rn.

(iii) We first note that the result of (iii) is sharper than Lemma 4.2 for s = m. Letgk

be the functions defined in (ii). Then, by the Monotone Convergence Theorem, we haveRn

H0(f1(y) A)dHm(y) = limk

Rn

gk(y)dHm(y)

= limk

QBk

Hmf(A Q) limk

QBk

(Lipf)mLm(Q A)

=(Lipf)mLm(A).This finishes the proof.

Lemma 5.6. Letf : Rm Rn be Lipschitz, mn, t >1 andR={x; Df(x) exists andJmf(x)> 0}.

Then there is a countable collection{Ek}k=1 of Borel sets ofRm such that(i)R=k=1Ek;(ii) f|Ek is one-to-one;(iii) for eachk = 1, 2, , there is a symmetric nonsingular linear mapTk : Rm Rm

such that

Lip(f|EkT1k )t, Lip(Tk (f|Ek)1)t,and

t1|Tkv| |Df(p) v| t|Tkv| for anypEk, vRm,tm|detTk| (Jmf)|Ektm|detTk|.

Lemma 5.6 asserts that f can be locally approximated by a nonsingular symmetriclinear map as close as we wish.

Proof. We fix a small > 0 so that satisfies 1

t + < 1 < t . LetC be a countable

dense subset ofRand letSbe a countable dense subset of symmetric nonsingular linearmaps ofRm. Then we define, for each c C, T S, i= 1, 2, , a set E(c , T , i) to beall b R B1/i(c) satisfying

(5.1) (1

t + )|T v| |Df(b) v| (t )|T v| for anyvRm,


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24 Q. HAN

and

(5.2) |f(a) f(b) Df(b) (a b)| |T(a b)| for anyaB 2i

(b).

Note that for any b R, we have Df(b) =O S, for an orthogonal map O: Rm Rnand a nonsingular symmetric map S on Rm, and hence

|f(a) f(b) Df(b) (a b)| |S(a b)|,

if|a b| is small. SinceSis dense in the set of all symmetric nonsingular maps, we canreplace S by a T Sso that (5.1) and (5.2) are satisfied. ThusbE(c , T , i) for somec,T,i.

Also note that E(c , T , i) is a Borel set since Df is Borel measurable. By (5.1) and

(5.2), we have1

t|T(a b)| |f(a) f(b)| t|T(a b)|,

for any bE(c , T , i) and aB2/i(b). This shows in particular thatf is one-to-one onE(c , T , i). By (5.1), we have

1

t|T v| |Df(b) v| t|T v| for anyvRm,

and hence

tm|detT| Jmf(b)tm|detT|,for anybE(c , T , i). We relabel E(c , T , i) as{Ek}k=1 and this concludes the proof.

Now we are ready to prove Theorem 5.3.

Proof of Theorem 5.3. Since Df exists almost everywhere, by Lemma 5.5 (iii), we as-sume thatDfexists for all xA.

Case 1. Suppose A {Jmf(x)> 0}.Fix at >1 and choose Borel sets{Ei}i=1 as in Lemma 5.6. We assume that all ofEi

are pairwise disjoint. For each fixed integer k 1, we defineBk as in Lemma 5.5 andsetFij =Ej Qi A, Qi Bk. Then{Fij} is disjoint and A =i,jFij .

We claim that

(5.3) limk

i,j=1

Hmf(Fij )=Rn

H0f1(y) AdHm(y).To prove this, let gk =

i,jf(Fij )

in Rn. Thengk isHm-measurable by Lemma 5.5 andgk(y) is the number of sets{Fij} such that Fij f1(y)=for any y Rn. Thus gk(y)converges toH0(f1(y)A) monotonically. By applying the Monotone ConvergenceTheorem, we obtain (5.3).


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Next, we have by Lemma 5.6

Hm

f(Fi

j )

=Hm

f|EjT1

j Tj(Fi

j )

Lip(f|EjT1j )mHmTj(Fij )tmHmTj(Fij )= tm|detTj|Lm(Fij )t2m

Fij

Jmf(x)dx

t3m

Fij

|detTj|dxt3m|detTj|Lm(Fij )

t4mHmf(Fij ).Thus by dividing by t2m and summing over i and j , we obtain

t2mi,j

Hmf(Fij )

AJmf(x)dxt2m

i,j

Hmf(Fij ).By taking t1, we conclude the Case 1.

Case 2. Suppose A {Jmf(x) = 0}.Fix an >0. Define g : Rm Rn Rm byg (x) = (f(x), x) andP : Rn Rm Rn

by the projectionP(y, z) =y for any (y, z)Rn Rm. We then have f(x) =P g(x).We can check easily that 0 < Jmg(x)< C onA, for a constant C. Since LipP= 1, wehave by Case 1,

Hm(f(A)) Hm(g(A)) Rm+n H0

A g1(y, z)dH

m(y, z)

=

A

Jmg(x)dxC Lm(A)0 as 0.Thus we obtain by letting 0,

Hmf(A)= 0,and hence

Rn

H0A f1(y)dHm(y) = f(A)

H0A f1(y)dHm(y) = 0.Theorem 5.3 now is obtained by combining Cases 1 and 2.

Corollary 5.7. Supposef : Rm

Rn is Lipschitz andm

n. For every

Lm-measurable

setA, there exists a Borel set

B {xA; Jmf(x)> 0}such thatf|B is one-to-one andHm

f(A) \ f(B)= 0.

Proof. In caseLm(A) = 0, we take B =.In case A is a Borel set, we use Lemma 5.6 to obtain Borel sets Ei such that f|Ei is

one-to-one for i = 1, 2, , andP ={xA; Jmf(x)> 0} i=1Ei.


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26 Q. HAN

By setting

Fi=P Ei\ i1j=1

f1

f(P Ej)

for i = 1, 2, ,

B = i=1Fi,we conclude by Theorem 5.3 easily that B is Borel, f|B is one-to-one, f(B) =f(P) and

Hmf(A) \ f(B) Hmf(A \ P) A\P

Jmf dLm = 0.


We can also obtain the following change of variable formula.

Theorem 5.8. Letf : Rm

Rn be a Lipschitz function, m

n, and letu

L1(Rm).

Then

(5.4)

Rm

u(x)Jmf(x)dx=

Rn

xf1(y)

u(x)dHm(y).

In particular, iff is one-to-one, thenRm

u(x)Jmf(x)dx=

Rn

u

f1(y)

dHm(y).

Proof. We consider the case u0. The general case then follows easily.We first consider a decomposition ofu. Define

A1 =

{u

1

},

and inductively

Ak =

u 1k

+k1j=1

1

jAj

for k >1.

We claim

u=i=1

1

iAi .

In fact, it is easy to see that (i) uk

j=1

1

jAj and (ii), for 0 < u(x)


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where we used Theorem 5.3 in the last equality. This impliesi=1

1iH0f1(y) Ai

is finite forHm-almost allyRn. By the claim, it is equal toxf1(y)

u(x).


In the rest of this section section, we discuss the coarea formula for Lipschitz functionson Euclidean spaces. We first state the result.

Theorem 5.9. Letf :

Rm

Rn

be Lipschitz andmn. Then for anyLm

-measurablesubsetA Rm, A

Jnf(x)dx=

Rn

HmnA f1(y)dy.The co-area formula is the generalized version of Fubini Theorem. This can be seen

easily by choosing fto be an orthogonal projection.

Remark 5.10. If we take A ={x Rm; Jnf(x) = 0}, the set of critical points, weobtain forLn-almost all yRn,

Hmn{x Rm; Jnf(x) = 0} f1(y)= 0.This statement is weaker than the standard Sard Theorem.

Similar to Theorem 5.8, we also have the following change of variable formula for thecoarea formula.

Theorem 5.11. Letf : Rm Rn be a Lipschitz function, mn, and letuL1(Rm).Then

Rm

u(x)Jnf(x)dx=

Rn

f1(y)

udHmndy.

Now we begin to prove Theorem 5.9. First, we need a few lemmas.

Lemma 5.12. Let L : Rm Rn be a linear map, m n, and A Rm be anLm-measurable subset. Then

(i) the map y Hmn(A L1(y)) isLn-measurable;(ii)

Rn

Hmn(A L1(y))dy= [[L]]Lm(A).Proof. We divide the proof into the following three cases.

Case 1. Suppose dimL(Rm) < n, i.e., [[L]] = 0. Then AL1(y) = and henceHmn(A L1(y)) = 0 forLn-almost allyRn. Thus (i) and (ii) are trivial.

Case 2. Suppose L is an orthogonal projection P : Rm Rn. Then, by FubiniTheorem, y Lmn(AP1(y)) isLn-measurable and

RnLmn(AP1(y))dy =

Lm(A). So (i) and (ii) are proved for this case.


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28 Q. HAN

Case 3. Suppose L = SO where O : Rm Rn is an orthogonal projection andS:

Rn

Rn

is a symmetric nonsingular map as in Lemma 5.1. Then by Case 2 and bythe change of variables, we have

Lm(A) =Rn

HmnA O1(y)dy= detS1

Rn

HmnA O1 S1(z)dz (where z = Sy)= detS1

Rn

Hmn(A L1(z))dz.This concludes the proof.

Lemma 5.13. Let f : Rm Rn be Lipschitz for m n and A Rm be anLm-measurable set. Then(i) f(A) is anLn-measurable subset ofRn;

(ii) the functiony Hmn(A f1(y)) isLn-measurable onRn;(iii)

Rn

Hmn(A f1(y))dy(m n)(n)(m)

(Lipf)nLm(A).

Proof. The proof of (i) is the same as in the corresponding case for the area formula inLemma 5.5.

Next, we prove that A satisfies (iii) if it satisfies (ii). For each j = 1, 2, , thereis a collection of closed balls {Bji }i=1 such that A i=1Bji , diamBji 1/j and

i=1 Lm(Bji ) Lm(A) + 1/j. Define

gji (y) =(m n)12 diamBji mnf(Bji )(y).

By (i), gji is anLn-measurable function on Rn, and for any yRn

Hmn1j

A f1(y)

i=1

gji (y).

By Fatous Lemma and (ii), which we assumed to be true, we obtainRn

HmnA f1(y)dy= Rn

limj

Hmn1j

A f1(y)dy

Rn lim infj

i=1

gji (y)dylim infj

i=1Rn g

ji (y)dy

=liminfj

i=1

(m n)12

diamBjimnLnf(Bji )

liminfj

i=1

(m n)12

diamBjimn

(Lipf)n(n)1

2diamBji

n(n)(m n)

(m) (Lipf)nLm(A),


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where we used the isodiametric inequality

Lnf(Bji )(n)12 diamf(Bji )n.To prove (ii), we first consider the case when A is compact. For each fix t 0, we

claim

{yRn; HmnA f1(y)t}= i=1

Ui(t),

where, for each integeri1, Ui(t) is the collection of those pointsyRn such that thereare finitely many open sets S1, , Sl, A f1(y) lj=1Sj , diamSj 0. Given any integer i1,there exists a collection of sets{Sj}j=1 such that A f1(y) j=1Sj, diamSj < 1/iand

j=1 (m n)(diamSj/2)mn < t + 1/i. Furthermore, we can assume that Sj are

open. SinceA f1(y) is compact, we can choose a finite number ofSj .This shows thatyUi(t), i.e.,

{yRn; Hmn(A f1(y))t} i=1

Ui(t).

The other inclusion is immediate, and this proves the claim.Next, we treat the case when A is open. A can be written as a countable union of

compact sets, i.e., A =i=1Ki with compact sets KiKi+1, i = 1, 2, . ThenHmnA f1(y)= lim

iHmnKi f1(y).

ThusHmn(A f1(y)) is Borel measurable on Rn.To show (ii) for the general case, let Ki and Ui be compact and open sets such that

KiAUi andLm(Ui\Ki)< 1/i. Note (iii) holds for Ui\ Ki. Then we obtainRn

Hmn(Ui f1(y)) Hmn(Ki f1(y))dy=Rn H

mn

(Ui\Ki) f1(y)dy

(n)(m n)(m)

(Lipf)nLm(Ui\Ki)0,

as i . SinceHmnKi f1(y) HmnA f1(y) HmnUi f1(y),

we see thatHmn(A f1(y)) is a measurable function ofy . Now we begin to prove Theorem 5.9.


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30 Q. HAN

Proof of Theorem 5.9. First assume thatLm(A) 0}.

For a (m, m n), letP : Rm Rmn be the orthogonal projection defined byP(x) = (x1, , xmn). Thenf(x) = q h(x), where h : Rm Rm is defined byh(x) = (f(x), P(x)) andq: R

m = Rn Rmn Rn is defined by q(y, z) =y.Let A ={x A; Jmh(x) > 0}. Then A =(m,mn)A. So we assume that

A = A for a (m, mn) and denote h by h. Moreover, we assume that h isone-to-one and|h(x) h(y)| c|x y| for any x, yA. In fact, by Lemma 5.6, A maybe decomposed into a disjoint union of Borel sets{Fl}l=1 such that h is one-to-one ineach Fl and|h(x) h(y)| cl|x y| for any x, yFl. Soh1 is Lipschitz in h(Fl).

We apply Lemma 5.6 to h1 in h(A). For each fixed t > 1, there exists a disjointunion A =

k=1E

kand associated symmetric nonsingular linear maps S

k : Rm

Rm,

k= 1, 2, , such thatLip(S1k h|Ek)t, Lip((h|Ek)1 Sk)t,

and

t1|S1k v| |Dh1(y) v| t|S1k v| for anyyh(Ek) andvRm.Then we have

Rn

HmnEk f1(y)dy=Rn

HmnEk h1 q1(y)dy=Rn H

mnh1

Sk S1

k (h(Ek) q1

(y))

dy

tmnRn

HmnS1k h(Ek) S1k q1(y)dy=tmn[[q Sk]]Lm

S1k h(Ek)

(by Lemma 5.12(ii))

t2mn[[q Sk]]Lm(Ek)t2m

Ek

Jmf(x)dx,

where in the last step we used [[q Sk]]tnJnf(x) for any xEk. Similarly, we have

Rn

HmnEk f1(y)dyt2m Ek

Jnf(x)dx.

To conclude Case 1, we sum over k and let t1.Case 2. Suppose A {x Rm; Jnf(x) = 0}.We prove

Rn

Hmn(f1(y) A)dy= 0.Fix an >0 and define

g: Rm Rn Rn byg(x, z) =f(x) + z,P : Rm Rn Rn byP(x, z) =z.


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Then we see 0< Jng(x, z)< C. Apply Case 1 to g inA Bn1 Rm Rn, where Bn1 isthe unit ball in

Rn

. We getC Lm(A)Ln(Bn1 )

ABn

1

Jng(x, z)dLm+n(x, z)

=

Rn

Hm(A Bn1 ) g1(y)dLn(y).A slight modification of Lemma 5.13 shows that

Hm(A Bn1 ) g1(y)c(m, n)

Rn

Hmn(A Bn1 ) g1(y) P1(w)dLn(w).Thus

cLm(A)Ln(Bn1 )cRn

Rn

Hmn(A Bn1 ) g1(y) P1(w)dLn()dLn(y),(5.5)where

(x, z)A Bn1 , g(x, z) =y, p(x, z) =w,which is equivalent to xf1(y w) andwBn1 . Hence,

the right side of (5.5)

=c

Rn

Rn

HmnA f1(y w)dLn(w)Ln(y)=c

Bn1

Rn

HmnA f1(y w)dLn(y)dLn(w)=c(n)

Rn

HmnA f1(y)dLn(y).Letting0, we obtain

Rn

HmnA f1(y)dLn(y) = 0.Combining cases 1 and 2, we finish the proof.

6. Countably Rectifiable Sets

Countably rectifiable sets provide an appropriate notion of generalized submanifoldsin Euclidean spaces; they are the sets on which rectifiable currents and varifolds live.Throughout this section, m andn are two positive integers with 1mn 1.Definition 6.1. AnHm-measurable set E Rn is countably m-rectifiable if

EE0 j=1

fj(Rm)

,

whereHm(E0) = 0 and fj : Rm Rn is a Lipschitz function for j = 1, 2, .


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32 Q. HAN

By Theorem 4.3, this is equivalent to

E= E0 j=1

fj(Aj)

,

where Hm(E0) = 0 andfj :Aj Rn is a Lipschitz function on an Lm-measurable subsetAj Rm, j = 1, 2, .Definition 6.2. AnHm-measurable set E Rn is purely m-unrectifiable ifEcontainsno countably m-rectifiable subsets of positiveHm-measure.Remark 6.3. It is useful to remark that there is a dichotomy between rectifiable andpurely unrectifiable sets. Precisely speaking, for anyHm-measurable subset A Rn, byusing the Hausdorff maximal principle, there holds

A= B C, B C=,whereB is countably m-rectifiable and Cis purely m-unrectifiable in Rn.

Now, we introduce tangent cones. For any p Rn, vRn and >0, we defineX(p, v, ) ={x Rn; |r(x p) v|< for some r >0}.

Note that the requirement on x can be written as r(x p) v= y for some yB1, orx= p +

v + y

r .

It is easy to see that X(p, v, ) is a cone and X(p, sv, s) =X(p, v, ) for any s >0.

Definition 6.4. Suppose E is a subset ofRn and p is an arbitrary point in Rn. (Note

that p is not necessarily on E.) The tangent coneTpE ofEat p is defined byTpE={vRn; E X(p, v, ) B(p) \ {p} = for any >0}.

Vectors in TpEare called tangent vectorsofEat p.

Usually, we call TxEthe tangent space if it is a subspace ofRn.

Alternatively, tangent vectors can be defined in the following way: v TpE\ {0} ifand only if there exist xiE\ {p} and ri >0 such that xip and ri(x p)v. If,in addition,|v|= 1, we may take ri= 1|xip| .

Remark 6.5. We observe that(1) TpE is a closed subset ofR

n;

(2) TpEis a cone in Rn, i.e., vTpE ands[0, ) imply svTpE;(3) 0TpE (or TpE=) if and only ifp E.(4) TpE=TpE.

Lemma 6.6. Suppose E is a subset of Rn andf : Rn Rl is a map differentiable atp Rn. Then

Df(p)

TpETf(p)f(E).

Moreover, equality holds iff|E is one-to-one, pE, (f|E)1 is continuous atf(p), andDf(p) is a linear homeomorphism mappingRn onto a subspace ofRl.


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Proof. We assume p= 0, f(p) = 0 and write Df(p) =L. For each vT0E, there existxiE, ri(0, ) with xi0, rixiv as i . Hence

Lv rif(xi) =L(v rixi) + |rixi| Lxi f(xi)|xi| 0 as i ,

and Lv T0f(E). On the other hand, in case the additional hypotheses hold andw T0f(E), there exist xi E, ri (0, ) with f(xi) 0, rif(xi) w as i .Hencexi0 and

L(rixi) w=|rif(xi)| |xi||f(xi)|Lxi f(xi)

|xi| + rif(xi) w0 as i .

This implies that rixi converges to some vRn withLv = w, and hence vT0E. Of particular interest is the tangent space of an m-dimensional submanifold ofRn.

Proposition 6.7. LetM Rn be a subset. Then the following statements are equiva-lent.

(1) For any p M, there are neighborhoods U of p and V of 0 in Rn and a Ckdiffeomorphismf :UV such that

f(U M) =V (Rm {0}).(2) For anypM, there is a neighborhoodU ofp inRn and aCk maph : U Rnm

such thatUM=h1(0)andh is a submersion at each point ofU, i.e.,Dh is surjectiveat each point ofU.

(3) For anypM, there is a neighborhoodU ofp inRn, a neighborhood of 0 inRmand aCk map g :

Rn such that (, g) is a local parametrization ofU

M around

p, i.e., g : M U is a homeomorphism andg is an immersion at each point of,i.e., Dg is injective at each point of.

The subset M in Proposition 6.7 is called an m-dimensionalCk submanifoldin Rn.

Proposition 6.8. Let M Rn be an m-dimensional C1 submanifold and p M.SupposeUis a neighborhood ofp inRn.

(1) Supposef is a diffeomorphism fromUonto a neighborhoodVof 0 inRn such that

f(U M) =V (Rm {0}). ThenTpM=

Df(p)1

(Rm {0}).(2) Suppose h is a submersion of U to Rnm such that U M = h1(0). Then

TpM=K er

Dh(p)

.

(3) Suppose (, g) is a local parametrization of U

M such that g(u) = p. ThenTpM=

Dg(u)

(Rm).

Obviously, TpM is an m-dimensional suspace ofRn. For an m-dimensional Ck sub-

manifold M in Rn, tangent vectors can also be characterized by the following way:vTpMif and only if there is a Ck curve c onM such thatc(0) =p and c(0) =v.

The proof of Proposition 6.7 and Proposition 6.8 is left as an exercise.

Suppose Eis a subset ofRn and p is an arbitrary point in Rn. We define the cone

Tmp E={vRn; m

E X(p, v, ), p> 0 for any >0},


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34 Q. HAN

which is called the m-dimensional approximate tangent cone of E at p and whose el-

ements are called m-dimensional approximate tangent vectors of E at p. Obviously,Tmp E TpE. When m is clear from context, we simply say the approximate tangentcone ofEat p and approximate tangent vectors ofEat p.

It is easy to verify

Tmp E=

ERm

{TpE; m(E\ E, p) = 0}.

Remark 6.9. IfCis any compact subset ofRn \ Tmp E andT ={p + rv; r >0, vC},then m(E T, p) = 0. In fact,{a + v; vC} can be covered by a finite family of setsX(p, v, v) with

m(E X(p, v, v), p) = 0, with v depending onv, and this family alsocoversT.

Remark 6.9 asserts that the density of E at p in any closed cone away from theapproximate tangent cone Tmp Eis zero.

Example 6.10. LetEbe the union ofx1-x2 plane and the x3-axis in R3. ThenT0E is

Eitself andT20 E is the x1-x2 plane.

Now we start to discuss tangent cones of countably rectifiable sets. We first prove adecomposition of countably rectifiable sets similar to Lemma 5.6.

Lemma 6.11. SupposeEis a countablym-rectifiable subset ofRn andt(1, ). Thenthere exist compact subsets A1, A2, A3, of Rm and Lipschitz maps 1, 2, 3, :Rm Rn such that1(A1), 2(A2), 3(A3), are disjoint subsets ofE and that

E= E0i=1

i(Ai),

withHm(E0) = 0. Moreover, for eachi= 1, 2, Lip(i|Ai)t, i|Ai is one-to-one, Lip

(i|Ai)1

t,and

t1|v| |Di(p)v| t|v| for anypAi, vRm.Proof. By Definition 6.1, we write

E= E0i=1

fi(Ki),

where E0 is a subset ofRn

withHm

(E0) = 0 and, for each i = 1, 2, , Ki is a Borelsubset of Rm and fi : Ki Rn is Lipschitz. By Corollary 5.7, we assume that fi|Kiis one-to-one and Jmfi(x) > 0 for any x Ki. Furthermore, we assume by Theorem4.3 that fi is Liptschitz from R

m into Rn. For each Ki and fi, we apply Lemma 5.6to get a countable collection{Eij}j=1 of Borel sets ofRm and a sequence of symmetricnonsingular linear maps Tij : R

m Rm such that

Ki=

j=1

Eij,


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and for eachj = 1, 2, ,Lip(fi|Eij T1ij )t, Lip(Tij (fi|Eij)1)t,

and

t1|Tijv| |Dfi(p) v| t|Tijv| for anypEij, vRm.Now we set ij = fi T1ij and Aij = Tij(Eij). The result follows by a simple approxi-mation by compact subsets.

Theorem 6.12. Suppose E is a countably m-rectifiable subset of Rn. Then, forHm-almost allx inE,

m(E, x) = 1,

andTmx E is anm-dimensional subspace ofRn.

Proof. We first consider a special case. SupposeE= (A), where A is a subset ofRm

and : Rm Rn is a map such that |A is one-to-one and thatt(1, ) is a Lipschitzconstant for |A and (|A)1. Let p A be a point such that A hasLm-density 1 at

pA and has an injective differential at p. Now we claimt2m m

(A), (p)

m(A), (p)t2m,and

Tm(p)(A) =T(p)(A) = im D(p).

To prove this, we assume p = 0, (p) = 0 and write D(p) =L. First we note that

A B rt

1(A) Br, (A) Br(A Btr),and hence

t2mLm(A Br/t)(m)(r/t)m

Hm((A) Br)

(m)rm t2m L

m(A Btr)(m)(tr)m

,

for any r (0, ). Therefore, the first conclusion follows because A hasLm-density 1at 0. Next, it is easy to see that

T0

(A)imL.

To prove the oppositive inclusion, we suppose vRm and >0. By choosing >0 and >0 with

(|v| + ) + L ,we have for any r sufficiently small

|(x) Lx| |x| for anyxBmr/t,whereBmr/t is the ball with radius r/t and center at the origin in R

m. It implies

X(0, v , ) Br/tX0,L v,,

because, ifx Rm, s >0,|x| r/t and|sx v|< , then|s(x) Lv| s|(x) Lx| + |L(sx v)|

|sx|+ L |sx v|< (|v| + )+ L.


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36 Q. HAN

Therefore,

Hm(A) X(0,L v,) Br(m)rm

t2m Hm

A X(0, v , ) Br/t(m)(r/t)m

,

which approaches t2mLm(X(0, v , ) B1)/(m) as r approaches 0. We conclude thatm

(A) X(0,L v,), 0> 0,and hence LvTm0 ((A)). This finishes the proof of the claim.

Now we consider a general case. Assuming t (1, ), we choose Ai and i as inLemma 6.11. By Lemma 3.5 and Lemma 3.4, we have

m(E, x)1, mE\ i(Ai), x= 0,for

Hm-almost allx

i(Ai). Furthermore, Lemma 3.3, Theorem 4.5 and Lemma 5.5(iii)

imply that, forHm-almost all pointsxi(Ai), the hypotheses of the special case holdwithp = (|Ai)1x. We conclude that

t2m m (E, x)m(E, x)1,Tmx E= imDi(p),

atHm-almost allxi(Ai), withp = (|Ai)1x. To conclude this section, we give another characterization of countably rectifiable sets

and discuss its applications.

Lemma 6.13. AnHm-measurable setE Rn is countablym-rectifiable if and only if

E

j=0Nj ,

whereHm(N0) = 0 andNj is anm-dimensionalC1 submanifold ofRn for eachj1.Proof. The if part is essentially trivial and is omitted. The only if part is an easyconsequence of Theorem 4.6 as follows.

Without loss of generality, we assume E has the form E f(Rm) for a Lipschitzfunction f : Rm Rn. By Theorem 4.6, we choose C1 functions g1, g2, on Rm, suchthat

f(Rm)E0

i=1gi(R

m)

,

whereHm(E0) = 0. Then set

N0 = E0

i=1

gi(Ci)

,

where Ci ={x Rm; Jmgi(x) = 0} and Jmgi denotes the Jacobian ofgi. By Theorem5.3,Hm(gi(Ci)) = 0, and henceHm(N0) = 0. Now for eachx Rm \ Ci, we letUi(x) bean open subset ofRm \ Ci containing x such that gi|Ui(x) is one to one. By the InverseFunction Theorem, such a Ui(x) exists and Ni(x) gi(Ui(x)) is an m-dimensional C1


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submanifold ofRn. We choose a countable collection x1, x2, , ofRm \ Ci such thatgi(

Rm

\ Ci)

j=1Ni(xj). Hence, we have

f(Rm)N0

i,j=1

Ni(xj)

.


Another way to express Lemma 6.13 is to write

E= E0 j=1

Ej

,

whereHm(E0) = 0 and, for each j 1, there exist an open subset Aj Rm and a C1mapg

j :A

jRn such thatE

jg

j(A

j) and that g

jis one-to-one onA

j and Dg

j(x) is

injective for any xAj.By Lemma 6.13, anHm-measurable set E Rn is purely m-unrectifiable if and only

ifHm(E N) = 0 for any m-dimensional C1-manifoldN ofRn.As an application of Lemma 6.13, we give another characterization of approximate

tangent spaces of countably rectifiable sets. We show that approximate tangent spacescan be obtained by the weak convergence of suitable measures.

For any >0 and x Rn, we define x, : Rn Rn byx,(y) =

y x

for anyyRn.Theorem 6.14. SupposeEis a countablym-rectifiable subset inRn with0


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38 Q. HAN

whereJm is the Jacobian of . We note that the integration is over a bounded domain

for small in the last integral. Hence we may replace p,(A) byRm

. Letting 0and using Theorem 5.8 again, we obtain

L.H.S.Rm

f

D(p) vJm(p)dLm(v)=

D(p)(Rm)

f(y)dHm(y).

We note that D(1(x))(Rm) is the approximate tangent cone ofE atx.Now, we consider the general case. By Lemma 6.13, we assume

E= E0

j=1Ej

,

whereHm(E0) = 0 and, for each j 1, there exist an open subset Aj Rm and a C1map gj :Aj Rn such that Ejgj(Aj) and that gj is one-to-one in Aj and Dgj(x) isinjective for any xAj. Then for any fC0(Rn) and any j1, we have

x,(E)f dHm =

x,(E\Ej)

f dHm

x,(gj(Aj)\Ej)f dHm

+

x,(gj(Ej))

f dHm.

We assume suppfBR(0) for some R >0. Then, we have

x,(E\Ej)f dH

m=

1

m

E\Ejf x dHm

sup |f| 1m

Hm(E\ Ej) BR(0).By Theorem 3.4, we obtain forHm almost all xEj

x,(E\Ej)f dHm 0 as 0.

Similarly, we have x,(gj(Aj)\Ej)

f dHm 0 as 0.

By what we just proved, we obtainx,(gj(Aj))

f dHm

Dgj(g1j (x))(R

m)f dHm as 0.

Hence Dgj(g1j (x))(R

m) is the approximate tangent space ofE forHm almost all xEj.

Remark 6.9 and the density one property in Theorem 6.12 can also be obtained bychoosing appropriate f in (6.1).


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Corollary 6.15. SupposeE is a countablym-rectifiable subset inRn with0


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40 Q. HAN

7. Weak Approximate Tangent Space Property

Throughout this section, m and n are two positive integers with 1mn 1. Wefirst introduce some notations which will be used repeatedly.

LetGL(n, m) be the Grassmannian manifold of all m-dimensional linear subspaces ofRn. For any VGL(n, m), xRn, r(0, ) and(0, 1), a cone around the planeV + x with the opening is defined by

X(V, x) ={y; dist(y x, V)< |y x|},X(V,x,r) ={y; dist(y x, V)< |y x|, |y x| r}.

Finally, for any subsetA Rn, we write for r >0,A(r) =

{x

R

n; dist(x, A)< r

}.

In this section, we discuss the rectifiability by approximate tangent spaces. We firstprove a simple result.

Lemma 7.1. Suppose E Rn is anHm-measurable set such that for some V GL(n, m), (0, 1) andr(0, ),

E Br(x)X(V, x) for anyxE.ThenE is countablym-rectifiable.

Proof. Without loss of generality, we assume diam(E) < r, so that E Br(x) for anyx

E. LetPVbe the orthogonal projection ofR

n onto V. Then we have

|PVx PVy|

1 2|x y| for anyx, yE.Hence PV|E is injective. Taking f = (PV|E)1, we see that f : PV(E)(V) Rn is aLipschitz function with Lip(f)(1 2)1/2 and E=f(PVE). Thus,E is countablym-rectifiable.

In Lemma 7.1, the m-dimensional subspace V GL(n, m) is fixed for the entire setE. In fact, we may allow V, as well as and r, to change according to x.

We now discuss the rectifiability by approximate tangent spaces.

Theorem 7.2. Suppose E is anHm-measurable subset of Rn with 0


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Proof. First, we set

fi(x) = inf 0


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42 Q. HAN

wherePV andPVjare orthogonal projections ontoV and Vj respectively. Then we

obtain by (7.1)

HmE B2r(x) \ X(Vx, x) HmE Br4

(y)(m)r

4

m(0 ).

It follows (/8)m(0). This yields a contradiction if > 0 is chosen small,and hence proves (7.3). Now we apply Lemma 7.1 to conclude that Fj is countablym-rectifiable for each j = 1, , N.

In conclusion, there exists a positive number (m,,0) such that, for each 0, consider y E Br(x) \ X2(V, x). Then there exists a positiveintegeri such that r/2i r/2i1, and hence y

E

B(x,r/2i1)

\W(r/2i1). Therefore we have

E Br(x) \ X2(V, x)i=1

E Br/2i1(x) \ W(r/2i1)

.

Now (7.6) follows from (7.5) easily.

In the following, A(x,n,m) denotes the collection of all m-dimensional affine spacesofRn containingx.

To motivate our discussion, we first examine Corollary 7.5 and note that (7.5) can beformulated as follows. For any >0, there exists anr0> 0 such that for any r(0, r0)

1

rmHm

E Br(x) \ W(r)

< .

Here the affine subspace W depends only on the point x, independent of r. In manycases,Wis allowed to change according to r. We still intend to conclude the rectifiability.To do this, we need to strengthen the hypothesis on the density lower bound.

Definition 7.6. A subset E Rn has the weak approximate tangent space property atxEif, for any positive constants and , there exist positive contants r0 and suchthat, for any r(0, r0), there exists an affine m-plane W A(x,n,m) satisfying(7.7) HmE Br(y)rm for any yW Br(x),and

(7.8) HmE Br(x) \ W(r)< rm.

Such aWis called a weak approximate tangent space ofEat x at the scale r .

In Definition 7.6, the affine m-plane W is allowed to depend on r .

Remark 7.7. According to (7.8), most ofE lies nearW inBr(x). Clearly, (7.7) impliesthe positiveness of the lower density ofE, i.e., m (E, x)> 0 forHm almost all xE.Note that (7.7) also implies W Br(x)E(r). Hence, there are no big holes in EnearW Br(x).

The main result is the following theorem.


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44 Q. HAN

Theorem 7.8. SupposeE is anHm measurable subset of Rn withHm(E)


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Then (7.11) holds for anyxF11and rr11/2. If not, we take ayFBr(x)\W(r).On one hand, yBr(x) \ W(r) implies Br/2(y)B2r(x) \ W(r/2). Hence, we have

HmE Br2

(y) HmE B2r(x) \ W( r

2)

< r

2

m.

On the other hand, by yF, (7.10) impliesHmE B r

2(y)

> r

2

m.

This is a contradiction. To prove (7.12), we first observe that it holds with F replacedbyE by (7.7). Note that, when is fixed, = (x) in (7.7) depends only on x. Hence,we first take a compact subsetF F, withHm(F\F)< /4, such that(x)0> 0for any xF, namely, for any xF and r(0, r0)

HmE Br(y)0r

m for any y

W

Br(x).

By Theorem 3.4, m(E\ F, x) = 0 forHm almost all xF. Hence, we have a compactsetF12F, withHm(F \F12)< /4, and a constantr12r0such that for anyxF12andrr12

Hm(E\ F) B2r(x) 12

0rm.

This implies for any xF12 andrr12HmF Br(y) 1

20r

m for any yW Br(x).and hence (7.12) holds for any x F12 and r r12. Now we set F1 = F11 F12 andr1= min{r11/2, r12}.

By Theorem 3.4 and Theorem 3.5, m

(E, x)1 and m

(E\F1, x) = 0 for Hm

almostallxF1. Hence, as before, we see that forHm almost all xF1 there exists a positivenumber r2r1 such that for any r(0, r2) there is an affine m-space W A(x,n,m)for which

F1 Br(x) \ W(r) =,(7.13)W Br(x)F1(r),(7.14)HmE Br(x)< 2(m)rm,(7.15)Hm(E\ F1) Br(x)< 200m(m)trm,(7.16)

for a small t to be fixed. We note that r2 depends on xF1. Compare with r1, whichdepends only on F1.

Fix such x, r and W. For simplicity, we assume x = 0 and V = W GL(n, m).Let PVbe the orthogonal projection ofR

n onto V. We prove that if, for given ands(1/2, 1), the constant


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46 Q. HAN

Sinces


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Therefore, we take = (1 s)/4 and 50mtrm.

From now on, we only consideri= 1, , k. Recalling (7.13) and how the ballsBi(xi)were chosen, we see that there are points

(7.25) ziP1V

Bi(xi) V(r) F1 for i = 1, , k.

By (7.21), 1i r < r1, and we can apply (7.12) to zi and 1(1s)i/16 to getWiA(zi, n , m) such that

(7.26) AiB1(1s)i/16(zi) WiF 1

16(1 s)i

.

Then xi / PVAi. In fact, if there werexAi with PVx= xi, we could find by (7.26)a point yF with|x y| (1 s)i/16. Then PVyBi/2(xi) and by (7.25), (7.21)and


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V. Hence W is close to V. Therefore, most part ofE in Br(x) projects into a small

neighborhood ofPVx in V. The last assertion is reflected in (7.34) in the proof below.Proof of Theorem 7.8. We prove by a contradiction argument. By assuming E is purelym-unrectifiable, we proveHm(PVE) = 0 for anyVGL(n, m). This clearly contradicts(7.9).

Let (0, 1/2) and V GL(n, m). Then we find a compact subset F E andpositive numbers r0, and , withHm(E\ F)< and < , such that for any xFandr(0, r0) there is aW A(x,n,m) satisfying(7.31) HmE Br(x)> rm,and

(7.32) F

Br(x)

\W(r) =

.

The proof is similar to that of (7.10) and (7.11) in the proof of Lemma 7.9. Note thatwe also have

(7.33) HmPV(E\ F)< .Now we claim that, forHm almost allxF, there exists sufficiently small r such that

(7.34) HmPV(F Br(x))10mrm.To prove this, we first note by Lemma 7.1 that {x F; F X(V, x, 1/i) =} iscountablym-rectifiable for any integer i. Since F is purely m-unrectifiable, then

Hm

i=1{

x

F; F

X(V

, x,1

i

) =

}= 0.

Hence forHm almost all xF, there are points yFarbitrarily close to x such that(7.35) |PV(y x)|< |y x|.Let x F be such a point and take a y F such that x, y F satisfy (7.35) andr =|xy| < r0. Let W be the weak tangent plane at x at the scale r as in (7.32).Intuitively,Wis close toV. Hence the projection ofW, or evenW(r), which containsF, into V should be small. First by (7.32), we seeyW(r). By setting z =PWy, wehave

|z y| r, r2 |z x| r, |PV(z x)|< 2r.

We select an orthonormal basis{

e1,

, em}

for W {

x}

such thatPV(ei)

PV(ej) = 0for i=j . Then for an i {1, , m},

|PVei| 2r1|PV(z x)|< 4,because otherwise we should have

|PV(z x)|2 =m

j=1

|(z x) ej |2|PVej|2

>4r2|PV(z x)|2|z x|2 |PV(z x)|2.


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50 Q. HAN

It follows thatPV(WBr(x)) is contained in anm-dimensional rectangle with one side ofthe length 8r and the others of length 2r. Hence by (7.32), PV(FBr(x)) is containedin a rectangle with side lengths 10r, 2r+ 2r, , 2r+ 2r. Therefore, as

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