Geometric Measure Theory

Transcription

1 Geometric Meaure Theory Lin, Fall 010 Scribe: Evan Chou Reference: H. Federer, Geometric meaure theory L. Simon, Lecture on geometric meaure theory P. Mittila, Geometry of et and meaure in Euclidean pace K. Falconer, Geometry of Fractal F.H.Lin, X.P. Yang, Geometric meaure theory: an introduction Coverage: About 80 page of [Lin,Yang], De Giorgi Theory, BV-Function, et of finite perimeter, and current time permitting. Week 1 (9/8/010) Background Notation and Definition: X a et, X denote the power et {A: A X} C X i a σ-algebra if 1. A C X \A C. A i C 3., X C i A i C, i A i C For (X, d) a metric pace, the Borel et are et in the σ-algebra generated by open et in (X, d). Meaure (denote outer meaure). µ: X [0, ] i a meaure if 1. µ( )=0. µ ( i A i ) i=1 µ(a i ) Meaurable et (Caratheodory definition) A i meaurable if µ(b)= µ(b A)+ µ(b A c ) for all et B A Borel meaure i a meaure for which all Borel et are meaurable. 1

2 A Borel regular meaure i a meaure µ for which µ(a) = inf µ(b) B Borel,B A Carathéodory criterion: Given a meaure µ on (X, d), µ i Borel if and only if µ(a B) = µ(a) + µ(b) for all A, B X.t. d(a, B)>0 If X i eparable and µ i σ-finite, then for A meaurable, µ(a) = inf µ(o) = O open,o A up K cloed,k A µ(k) A Radon meaure µ i a Borel regular meaure uch that µ(k)< for every K bounded. For a equence {µ k }, we ay that µ k µ weakly if When X = R n, can define Levi norm X k f(x)dµ k (x) f(x)dµ(x) for all f C 0 (K) X L(r) 4 {f: R n [0, ], upp f B r (0), f Lip 1} { F r (µ, ν)=up fdµ } fdν, f L(r) Then µ k µ weakly if and only if F r (µ k, µ) 0 for all r (0, ). Repreentation Theorem. Any continuou linear functional L on C 0 (X) i given by a Radon meaure, i.e. L(f)= f(x)dµ(x) for ome meaure µ and for all f C 0 (X) X Haudorff Meaure We will be looking at a generalization of Lebegue meaure. We will ue L n to denote the n-dimenional Lebegue meaure. We will focu on how to deal with lower dimenional object in R n. We define the approximate -dimenional Haudorff meaure H δ for 0 <, 0 δ to be α()= { ( ) H diam cj δ (A)=inf α(), c j R n.t. A } c j, diam(c j ) δ π/ Γ ( +1) o that vol(b 1 n )=α(n). For any ubet A R n, if A i bounded, then H δ (A)<. Can check that H δ i a meaure: For ubadditivity, given A i R n, we have H δ ( ) A k H δ (A k ) k=1 k=1

3 by definition of H δ. If we take a δ-cover c j (k) for A k, we can form a cover for the union by taking the union of the cover. H δ ( )=0 alo by definition, and thu H δ i a meaure. Alo, if δ 1 > δ, then H δ1 (A) H δ (A). Thi i alo by definition, ince a δ -cover of A i automatically a δ 1 -cover of A. We now define the Haudorff meaure by taking the limit: H (A)= lim δ 0 H δ (A) Exercie 1. If Γ i a C 1 cloed curve in R n, then H 1 (Γ) i the length of Γ.... Theorem 1. H i a Borel regular meaure for 0 n on R n Proof. Firt we how that H i a meaure. Subadditivity i preerved from taking limit, and trivially H ( )=0. To how that H i Borel, we ue the Carathéodory criteria, that H (A B)=H (A) +H (B) for d(a, B)>0. Let δ 0 =d(a, B)>0. Then for 0<δ < δ 0 we claim that H δ (A B)=H δ (A) +H δ (B) Given any δ-cover i c i of A B, each c i interect exactly one of A, B ince d(a, B) = δ 0. Thu the cover can be plit into two dijoint cover, one for A and one for B, and thu H δ (A B) H δ (A) +H δ (B) We already proved ubadditivity above, and hence we equality here. To how that H i Borel regular, take δ = 1, and for each k we find a 1 -cover c (k) k k j for which ( Note that H 1/k ( (k) diam c α() j Since A ( ) (k) k c j, we have that ( ( H 1/k (A) H 1/k j ) H1/k (A) +ε k ) (k) c j i bounded above by the left hand ide (c (k) j form a 1 -cover for the union). k k c j (k) )) H 1/k ( Taking the limit a k we conclude by the andwich above that H ( k ( c j (k) )) = H (A) c j (k) ) H 1/k (A)+ 1 k So we have found a Borel et containing A with the ame meaure a A, which how in particular that H (A)= inf B A,B Borel H (B) 3

4 Exercie. Let µ k µ be a equence of weakly converging Radon meaure. Then 1. limup k µ k (K) µ(k) for K compact.. liminf k µ k (O) µ(o) for O open. Note: If {µ k } i a equence of Radon meaure on R n uch that µ k = µ k (R n ) C < for all k, then there i a ubequence µ k uch that µ k µ weakly for ome Radon meaure µ.... Can ue continuou function that are 1 on K and 0 outide O for O K, (approximation of K by open et outide and approximation of O by compact et inide) Lemma. Obervation: 1. H 0 i imply the counting meaure. H (A)=0 for all A R n, >n 3. H (λa) =λ H (A), λ >0 µ k (K)= dµ k f K,O dµ k f K,O dµ= µ(k)+ε O K µ k (O)= dµ k f K,O dµ k f K,O dµ = µ(o) ε K O 4. H (QA)=H (A) where Q i a orthogonal linear tranformation. Proof. (1),(3), and (4) are immediate by definition. For (), by monotonicity it uffice to how that H (R n ) = 0 for > n. If we write R n = Q j a a union of unit cube, by countable ubadditivity it uffice to how that H (Q)=0 for Q=[0, 1] n (H i tranlation invariant). Now ( ) n ( n H 1/m (Q) m n α() = α() m ) m n cutting Q into cube of ide length 1/m. Since n < 0, a m, H 1/m (Q) 0 and hence H (Q) = 0, proving (). Theorem 3. Let A R n and 0 < t <. Then 1. If H (A) <, then H t (A) =0. If H t (A)>0, then H (A)= Proof. Since () i the contrapoitive of (1), we jut have to prove (1). Suppoe H (A) <. Then we know that H δ (A) C for all δ. Then for each δ there exit a cover c j A with diam c j <δ and ( ) diam cj α() C 4

5 Uing the ame cover, we have that ) t H t δ (A) ( diam cj α(t) ( ) t δ α(t) ( diam cj α() α() and thu taking δ 0 (and noting t >0) we have that H t (A) =0. ) Haudorff Dimenion With thi behavior of H ( ) a a function of in mind, we then define the Haudorff dimenion to be H dim (A) = inf { [0, ), H (A)=0} i.e. the dimenion where the meaure jump from 0 to. Week (9/15/010) Some note: H i not a Radon meaure, ince it i not locally finite: For 0 < n, H (B 1 n )=. One can define H for any metric pace (X, d), with the ame exact definition. We can alo define imilar meaure that retrict the cla of et ued for covering: Haudorff Spherical Meaure and { S δ (A) 4 inf α()ri, A } B ri (x i ), r i < δ S (A) 4 i=1 lim S δ (A) δ 0 which i the ame definition, except that we cover A with ball. We have the following relation: H (A) S (A) H (A) The firt inequality follow ince we are taking an infimum over more et for H and hence H i maller. The econd inequality follow ince any bounded c R n can be covered by a ball with twice the diameter: c B diam c (x 0 ) for any x 0 c. Net Meaure { N δ (A)=inf α()ri, A } Q j, dyadic cube with diameter δ i=1 N (A)= lim N δ (A) δ 0 5

6 What happen in the cae when = n? Theorem 4. H n =L n, the n-dimenional Lebegue meaure. Proof. (1) H n L n. We will how thi in two part. Firt we how that H n L n by howing that H n c(n)l n, and then we will improve the reult. For A R n, with L n (A) < (if = the inequality i trivial), we will cover with cube. Let ε, δ >0 be given. Then there exit {Q j } cube uch that 1. A. Q j L n (Q j ) L n (A)+ε with diam Q j < δ Thi i by definition of Lebegue meaure (here cube are very natural when working with the product meaure). Then we have that H n δ (A) ( ) n diam Qj α(n) ( ) n n α(n) L n (Q j ) n c(n)l n (A)+c(n)ε Letting ε, δ 0 how that H n c(n)l n, and c(n)>1, o we need to improve the reult. Now we will ue the reult that up to a et of meaure zero, we can cover each Q j by mutually dijoint ball contained in the interior: where B j,k Q j and L n (Z j ) =0. Thu Q j Z j B j,k (x j k ) k=1 A ( Q j Z j j,k=1 B j,k (x k j ) Since for each j, B j,k i mutually dijoint and contained in Q j, we have that and thu H δ n ( ) B j,k = k=1 k H δ n (B j,k ) = k ) L n (B j,k ) L n (Q j ) H δ n (A) H δ n (Z j )+ j L n (Q j ) H δ n (Z j )+L n (A)+ε Taking δ 0 give u H n (A) L n (A) + ε + H n (Z j )=0, and thu H n (Z j ). Since L n (Z j ) = 0, we have from before that H n (A) L n (A)+ε 6

7 Since ε i arbitrary, we have the reult. () L n H n To how thi, we will ue the following: Lemma 5. For A R n, ( diam A L n (A) α(n) ) n Note that the 1/ factor i important, otherwie the inequality i trivial (A B diam A (x) for all x A). Auming the Lemma for now, let u finih the reult. For any ε > 0, δ > 0, we can find c j with diam c j < δ A c j, and ( ) n diam cj α(n) H n δ (A)+ε Now applying the lemma, we have the computation L n (A) L n (c j ) (Lemma) ( diam cj α(n) H n δ (A)+ε ) n Letting ε, δ 0 give the reult L n (A) H n (A). Proof. (of Lemma 5) Firt, it uffice to prove the reult for the cae when A i cloed and bounded in R n, ince taking the cloure of A doe not affect the diameter and increae the meaure: We will ue a ymmetrization argument. ( ) n ( diam Ā diam A L n (A) L n (Ā) α(n) = α(n) ) n e A R n 1 e h(x) x x h(x ) S e (A) R n 1 Let e be a unit vector direction. The idea i to tranform A into a ymmetric et (with repect to ome direction e) while preerving the ame area, then we jut need to prove the reult for the ymmetrized et. Here i how to ymmetrize A. For a.e. x R n 1, let l x b the line {(x, y), y R}. Let h(x) = L 1 (A l x ). Then we add the point {(x, y): y [ h(x), h(x)]} to the ymmetrized et S e (A). Then we note that by Fubini, L n (A)= L 1 (A l x )dx= h(x)dx=l n (S e (A)) R n 1 R n 1 7

8 Furthermore, we how that diam(s e (A)) diam(a). Note that diam(a) = diam(s e (A))= up x,x Proj(A) up x,x Proj(A) where Proj(A) denote the projection of A to e. diam((a l x ) (A l x )) x x + h(x)+h(x ) Let a = inf Proj e (A l x ), b = up Proj e (A l x ), and a = inf Proj e (A l x ), b = up Proj e (A l x ). Aume that b a b a (otherwie witch x, x ). Note that diam((a l x ) (A l x )) x x + b a and b a (b a )+(b a) = (b a) +(b a ) h(x) +h(x ) = h(x)+h(x ). Thu diam((a l x ) (A l x )) x x + h(x)+h(x ) Now it uffice to prove the reult for S e (A), becaue once we have that, then ( ) n ( diam L n (A)=L n Se (A) diam A (S e (A)) α(n) α(n) Now tarting from A, we ymmetrize in the n tandard bai direction: A S e1 (A) S e (S e1 (A)) S en (S en 1 ( S e1 (A) ))=A denoting the reult by A. By what we jut proved it uffice to prove the reult for A. Thu we need to prove that ( ) diam A L n (A n ) α(n) Now for x A, ince A i ymmetric with repect to the tandard coordinate, x A and thu ) n o and L n (A ) L n (B d ia m A x B d i a m A (0) A B d i a m A (0) ) ( diam A (0) =α(n) ) n Exercie 3. Let Γ be a C 1 curve in R. Show that length(γ) N 1 (x)dx + N (y)dy R R 8

9 where N 1 (x) = # interection with l(x) and Γ where l(x) i vertical and N (y) i the correponding quantity for horizontal line l(y). There i ome generalization for higher dimenion a well. Recall (Lebegue Differentiation) If E R n i meaurable, then for a.e. x E, and E B lim r (x) = 1 for a.e. x E r 0 B r (x) E B lim r (x) = 0 for a.e. x E c r 0 B r (x) Thi let u define a meaure-baed boundary a the point where the limit above i between 0 and 1. What happen with our meaure? We will tudy thi next time. Covering Lemma Lemma 6. (Vitali Covering Lemma) Let E be a meaurable et in R n covered by a family of ball F = {B α } of bounded diameter. Then there i a countable dijoint ubfamily F of ball in F, i.e. F = {B i } i=1 with B i F, uch that B i E i=1 where B i = 5B i. Lemma 7. (Beicovitch Covering Lemma) Let E be a et in R n and let F be a family of ball with center at x E, F = {B rx (x): x E }. Then there i ome integer N > 0, depending only on the dimenion, and N ubfamilie of F, F 1, F N, uch that where ball in each F i are dijoint. N(n) E i=1 B F i B Remark. 1. The Vitali covering lemma only require metric pace tructure, independent of dimenion, etc. However, the lemma i uele unle the meaure ha ome tructure, i.e. doubling : µ(b) C 0 µ(b) for ome contant C 0. In Euclidean pace R n thi contant i n. Note that the ubcollection i dijoint, and to cover we then need to enlarge the et.. The Beicovitch covering lemma depend on Euclidean tructure, but i ueful for arbitrary meaure. The difference between the two covering lemma i that thi lemma allow for a bounded number of overlap when covering E, and that the et in the cover do not need to be enlarged. Proof. (Vitali Covering) Let { R F j = B F: j+1 diam(b) R } j 9

10 for j = 0, 1,, and R = up B F diam(b). Note if the diameter i arbitrarily large (i.e. R = ), then we can jut cover the whole pace eaily, taking a equence of B n whoe diameter go to. Thu we aume R <. We then contruct a equence a follow. Firt let β 0 be the maximal (countable) ubcollection of dijoint ball from F 0 (exitence i guaranteed by Haudorff Maximal Principle). Then let β j be the maximal ubcollection of dijoint ball from F j uch that they are alo dijoint from the previouly choen et. Then letting F = j=0 β j give the deired collection. ] k To prove thi, uppoe B F. Then B F k for ome k, and B [ j=0 β j, otherwie thi contradict the maximality of β k (can fit in one more et). Thu there exit S k j=0 β j uch that B S. Note diam(b) diam(s) by contruction. (Wort cae S β k, where diam(s) R diam(b)). Thu B j Ŝ = 5S (draw a picture). Thi implie that for any B F, B Ŝ for ome S F, and o E B F B B F B. Convex domain in R n Above we have been uing ball, which i ueful epecially for problem with linear tructure, uch a etimate involving harmonic function For other problem, for intance u = 0 det(u ij )=1 which ha affine tructure, it i more natural to ue convex et for etimate. Theorem 8. (F. John) Let Ω be convex in R n. Then B 1 A(Ω) B c(n) for ome affine tranformation A. (not ure if I got thi part correct...) 10