Chapter 2 Lebesgue measure 飞光飞光，劝尔一杯酒 香不识李委气整 a idea of the first step isto appr ts:open cibe f th owever,this appro smore pl is the from extha encompas the property of the disjoint additivity,one has to disregard some sets of highly irrega (non-measurable sets).Therefore a satisfactory measure theory does not include all subsets of R". 2.1 Exterior measure As said above,measure is a generalization of 'length,area,volume,'.So the very first agreement s that the meas re c the n-dimensional open cube C an) and me any su lizatio of volume,su monotone:if A C B.then A's measure is not greater than B's measure: .disjoint additivity:A's measure is the sum of A'measure if A are disjoint .translation invariant 。Scaling property. many Definition 2.1.Given EC R",the exterior measure of E is defined as m():gc where is a sequence of countable open cubes that cover E and is the volume ofI. 17
Chapter 2 Lebesgue measure ú1ú1ß!ò À Æÿ£ìUpßë/˛ ))oÂ5£º·6 In this chapter, we shall generalize ’length, area, volume, ...’ of regular regions to the measure of arbitrary sets. There are two steps involved. The idea of the first step is to approximate a general set by familiar regular sets: open cubes. However, this approximation is more plausible from exterior of a set, which leads to the definition of the exterior measure. The second step is the discovery that to encompass the property of the disjoint additivity, one has to disregard some sets of highly irregular (non-measurable sets). Therefore a satisfactory measure theory does not include all subsets of R n. 2.1 Exterior measure As said above, measure is a generalization of ’length, area, volume, ...’ . So the very first agreement is that the measure of the n-dimensional open cube C = (a1, b1) × · · ·(an, bn) is its volume (b1 − a1) × · · ·(bn − an), and measure of regular regions are their volume. Moreover, geometric intuition echoes that any such generalization should inherit nice properties of volume, such as • monotone: if A ⊂ B, then A’s measure is not greater than B’s measure; • disjoint additivity: ∪ n i=1Ai ’s measure is the sum of Ai ’ measure if Ai are disjoint; • translation invariant; • Scaling property. We use the covering of cubes to define the measure for a general set, and we shall allow countable many cubes for the covering. Definition 2.1. Given E ⊂ R n, the exterior measure of E is defined as m∗ (E) := inf E⊂∪∞k=1Ik X∞ k=1 |Ik|, where {Ik}∞ k=1 is a sequence of countable open cubes that cover E and |Ik| is the volume of Ik. 17
CHAPTER 2.LEBESGUE MEASURE the exterior measure. Erample 9.Let A be a set consists of countable many points,then m"(A)=0 Proof.This proof is a common trick in real analysis,which relies on ◇ Erample 10.m*(C)=0,where C is the Cantor set. Remark 2.2.The definition builds on the volume of n-dimensional cubes.Therefore it can't distin- guish sets of 'lower dimension'.For example,a line segment in R2 has exterior measure(area)zero s length.The more intrinsic way to encode the dimension information of sets is The next theorem shows that the exterior measure has all the nice properties we could expect. Theorem 2.3.The erterior measure satisfies the following 。nonnegativity:m*(E)≥0 ·monotone:fACB,then m'(A)≤m'(B) ·sub-additivity::m'(U21A)≤∑1m*(Ak)i 。translation invariant:m'(E+{zo）=m'(E)月 .scaling:m*(AE)=X"m"(E);>0. sts follo that m(A≤∑kl<m'(A)+ =1 Clearly i is a countable union of open cubes that covers UA,thus mug1Ae)s∑∑IM<∑m'a)+ k=1i=1 arbitrary,we get the desired sub-additivity. There is still one unsatisfied issue:the exterior measure only has subadditivity,and is lack of additivity for disjoint sets.That is m1A)=∑m(4) whenever A are disjoint.Here is an example
18 CHAPTER 2. LEBESGUE MEASURE The reason we call it exterior measure rather than measure will be clear momentarily. Before that we shall get used to this definition by exploring several simple yet important facts and properties of the exterior measure. Example 9. Let A be a set consists of countable many points, then m∗ (A) = 0. Proof. This proof is a common trick in real analysis, which relies on X∞ n=1 2 n = . Example 10. m∗ (C) = 0, where C is the Cantor set. Remark 2.2. The definition builds on the volume of n-dimensional cubes. Therefore it can’t distinguish sets of ’lower dimension’. For example, a line segment in R 2 has exterior measure (area) zero, but it certainly has length. The more intrinsic way to encode the dimension information of sets is the notion called Hausdorff measure. The next theorem shows that the exterior measure has all the nice properties we could expect. Theorem 2.3. The exterior measure satisfies the following • nonnegativity: m∗ (E) ≥ 0; • monotone: if A ⊂ B, then m∗ (A) ≤ m∗ (B); • sub-additivity: m∗ (∪∞ k=1Ak) ≤ P∞ k=1 m∗ (Ak); • translation invariant: m∗ (E + {x0}) = m∗ (E); • scaling: m∗ (λE) = λ nm∗ (E); ∀λ > 0. Proof. We only prove the sub-additivity. The rests follow more or less directly from definition and thus are left to the reader. ∀ > 0, there exists a covering of open cubes {Ik,i} for each Ak, such that m∗ (Ak) ≤ X∞ i=1 |Ik,i| < m∗ (Ak) + 2 k . Clearly ∪∞ i,k=1Ii,k is a countable union of open cubes that covers ∪∞ k=1Ak, thus m∗ (∪ ∞ k=1Ak) ≤ X∞ k=1 X∞ i=1 |Ik,i| < X∞ k=1 m∗ (Ak) + . Since is arbitrary, we get the desired sub-additivity. There is still one unsatisfied issue: the exterior measure only has subadditivity, and is lack of additivity for disjoint sets. That is m∗ (∪ ∞ k=1Ak) = X∞ k=1 m∗ (Ak) whenever Ak are disjoint. Here is an example
2.2.MEASURE 19 ruct a set N C [0,1].First the disjoint union of different equivalent classes: 0,1=UE We pick ar ea in each quivnt cWe cae disioint ent class. m心N)=∑mw Clearly, UN.c-1.2. and thus 1≤∑m(N)≤3 (2.1) k=1 In view of the translation invariant,m"(N)=m"(N).Vk.No value for m"(N)would justify (2.1) Remark 2.4.We shall point out,the definition of N,namely the pick of one elemer nt from encl equivalent class requires the Ariom of choice.Formally,it states that for every indexed family nents such that ri e 5 2.2 Measure dy is t attention to the Definition 2.5.Let ACR",A is called a measurable set if m'(T)=m'(TOA)+m*(TnA),VT CR". (2.2) is that t if(22 m*(T)2m'(Tn4)+m'(Tn4) Suppose m"(A)=0,then m'(TnA)=0 and m'(TnA)s m'(T),we infer that all sets with e sets is denoted by M.We prove the following Theorem2.6.1.0∈M; 2.A∈M,then A∈M:
2.2. MEASURE 19 Example 11. [A non-measurable set] We shall construct a set N ⊂ [0, 1]. First, we define an equivalent relation, say x ∼ y if x − y ∈ Q. Under this equivalent relation, [0, 1] can be written as the disjoint union of different equivalent classes: [0, 1] = [ α∈Λ Eα. We pick a representative rα ∈ Eα in each equivalent class and set N := {rα}α∈Λ. Denote all rational numbers in [−1, 1] as {q1, q2, · · · , }. We claim Nk := N + qk are disjoint. Suppose Nk ∩ Nl 6= ∅, then there exists x, y ∈ N, such that x + qk = y + ql , which means x ∼ y. This contradicts the only one pick from each equivalent class. If Nk satisfied the disjoint additivity, we would have m∗ ( [∞ k=1 Nk) = X∞ k=1 m∗ (Nk). Clearly, [0, 1] ⊂ [∞ k=1 Nk ⊂ [−1, 2], and thus 1 ≤ X∞ k=1 m∗ (Nk) ≤ 3. (2.1) In view of the translation invariant, m∗ (Nk) = m∗ (N), ∀k. No value for m∗ (N) would justify (2.1). Remark 2.4. We shall point out, the definition of N, namely the pick of one element from each equivalent class requires the Axiom of choice. Formally, it states that for every indexed family (Si)i∈I of nonempty sets there exists an indexed family (xi)i∈I of elements such that xi ∈ Si for every i ∈ I. The reader is referred to https://en.wikipedia.org/wiki/Axiom of choice for more details. 2.2 Measure The example 11 shows in general we do not have disjoint additivity of exterior measure for all subsets of R n. A remedy is to restrict our attention to those sets, for which the disjoint additivity hold. Carath´eodory made the following convenient criterion for the sets we shall be concerned with. Definition 2.5. Let A ⊂ R n, A is called a measurable set if m∗ (T) = m∗ (T ∩ A) + m∗ (T ∩ A c ), ∀T ⊂ R n . (2.2) A useful observation is that to verify (2.2), one just needs to show m∗ (T) ≥ m∗ (T∩A)+m∗ (T∩Ac ) Since m∗ (T) ≤ m∗ (T ∩ A) + m∗ (T ∩ Ac ) always holds by the sub-additivity. Suppose m∗ (A) = 0, then m∗ (T ∩ A) = 0 and m∗ (T ∩ Ac ) ≤ m∗ (T), we infer that all sets with zero exterior measure are measurable. The collection of all measurable sets is denoted by M. We prove the following Theorem 2.6. 1. ∅ ∈ M; 2. if A ∈ M, then Ac ∈ M;
CHAPTER 2.LEBESGUE MEASURE .fAk∈M fork-l,2,…,hemU1Ak∈M,moreover me-2w） whenever Akare disjoint. A41,A2日 rable. m(T)=m*(TOA)+m(TnAf) m(TOA0A)+m(TnA0A5)+m(TOAfO A2)+m(TnAf 0A5). Notice Tn(A1UA2)=(TnA0A2)U(TnAnA5)U(Tn AfnA2),by sub-additivity,we have m*(T∩(A.UA2)<m"(T∩A.nA2)+m(TnA,∩As)+m*(TnA9nA2). and thus m(T)>m(Tn(AI UA2))+m"(TnAinA5)m'(Tn(A UA2))+m'(T(AIU A2)). This implies that A UA2EM. eaegT=Ukm忉=mTn+mTnA, m'(41UA2)=m'(4)+m*(42. (2.3) Setting T of the form Tn(A UA2)we also have m'(Tn(AIU A))=m'(TnA)+m'(TOA2). (2.4) A holds,i.e. m(U214)=∑m(4), and mTn(ugA》=∑m(TnA wher ountable union,first suppose A1.....An..M are all disjoint.Let S:=UAn and Sk=U=1An.Using Sk E M,we have for any T that m'(T)=m'(TnSk)+m'(TnSk) =∑m'(TnA)+m(Tnsk)≥∑m*Tn4n)+m*(Tns9). Above inequality holds for all k,letting we obtain m'(T≥ Cm(TOAn)+m'(TnS)>m'(Tns)+m"(TOS)
20 CHAPTER 2. LEBESGUE MEASURE 3. if Ak ∈ M for k = 1, 2, · · · , then ∪ ∞ k=1Ak ∈ M, moreover m∗ (∪ ∞ k=1Ak) = X∞ k=1 m∗ (Ak) whenever Ak are disjoint. Proof. Notice (2.2) is symmetric about A and Ac , 2 of the theorem immediately follows. To show 3, we first show if A1, A2 ∈ M, then A1 ∪ A2 ∈ M. Using A1, A2 are measurable, we have for any T, m∗ (T) = m∗ (T ∩ A1) + m∗ (T ∩ A c 1 ) = m∗ (T ∩ A1 ∩ A2) + m∗ (T ∩ A1 ∩ A c 2 ) + m∗ (T ∩ A c 1 ∩ A2) + m∗ (T ∩ A c 1 ∩ A c 2 ). Notice T ∩(A1 ∪ A2) = (T ∩ A1 ∩ A2)∪(T ∩ A1 ∩ Ac 2 )∪(T ∩ Ac 1 ∩ A2), by sub-additivity, we have m∗ (T ∩ (A1 ∪ A2)) ≤ m∗ (T ∩ A1 ∩ A2) + m∗ (T ∩ A1 ∩ A c 2 ) + m∗ (T ∩ A c 1 ∩ A2), and thus m∗ (T) ≥ m∗ (T ∩ (A1 ∪ A2)) + m∗ (T ∩ A c 1 ∩ A c 2 ) = m∗ (T ∩ (A1 ∪ A2)) + m∗ (T ∩ (A1 ∪ A2) c ). This implies that A1 ∪ A2 ∈ M. Moreover suppose A1 ∩ A2 = ∅, then setting T = A1 ∪ A2 in m∗ (T) = m∗ (T ∩ A1) + m∗ (T ∩ Ac 1 ), we get the additivity for two disjoint sets: m∗ (A1 ∪ A2) = m∗ (A1) + m∗ (A2). (2.3) Setting T of the form T ∩ (A1 ∪ A2) we also have m∗ (T ∩ (A1 ∪ A2)) = m∗ (T ∩ A1) + m∗ (T ∩ A2). (2.4) Iterate this process finite many times together with the property 2, we infer that if A1, · · · An ∈ M, then any union or intersection among them is still measurable, and finite disjoint additivity holds, i.e., m∗ (∪ n i=1Ai) = Xn i=1 m∗ (Ai), and m∗ (T ∩ (∪ n i=1Ai)) = Xn i=1 m∗ (T ∩ Ai), whenever Ai are all disjoint. For countable union, first suppose A1, · · · , An, · · · ∈ M are all disjoint. Let S := ∪∞ n=1An and Sk = ∪ k n=1An. Using Sk ∈ M, we have for any T that m∗ (T) = m∗ (T ∩ Sk) + m∗ (T ∩ Sk c ) = X k n=1 m∗ (T ∩ An) + m∗ (T ∩ Sk c ) ≥ X k n=1 m∗ (T ∩ An) + m∗ (T ∩ S c ). Above inequality holds for all k, letting k → ∞ we obtain m∗ (T) ≥ X∞ n=1 m∗ (T ∩ An) + m∗ (T ∩ S c ) ≥ m∗ (T ∩ S) + m∗ (T ∩ S c )
2.2.MEASURE te心gair,肥 m'(TnS)≥∑m'(TnAn) On the other hand,m(T)m(TA)always holds by sub-additivity.Therefore m'(TnS)=∑m(TnAn by taking T=we t the disioint additivity Finally,if fAM are not necessarily disjoint from each other,then we make the following change: B=A1,B=(U1A:)\(U=A)k≥2. It follows (Bx}are disjoint and UA=UB E M. From now on we shall write simply m(A)for the exterior measure of a measurable set A.Our task of defining the measure for suitable subsets ofis now completed. We conclude this section with two useful facts about interchanging measure with limit operation. Proposition 2.7.Let ACA+be a sequence of increasing measurable sels,set A=UnAn,then m(A)=lim m(An). Proof.If m(A.)=oo for some n.then the desired equality holds.Therefor m(A）< for all n.Set B=A1,B2=A2\A1,B=An An-1,then Ba are all disjoint.Using countable disjoint additivity,we get mv.B)m(B) We obtain the desired equality as U B=Un An and m(An)=m(B) For decreasing sequence,we have 2k1kaae可tgmnsA=nA m(A)lim m(An). (2.5) Proof.We view A as the ambient set and take complement with respect to A.We then have 0C4A5C…CA…, Applying Proposition 2.7,we have m(UnA)lim m(A ) (2.6) Since m(Af)+m(An)m(Ai)and m(nA)+m(A)=m(A:), plugging back to (2.6),we get (2.5). 0
2.2. MEASURE 21 Hence S ∈ M. Using T ∩ S in the above inequality, we get m∗ (T ∩ S) ≥ X∞ n=1 m∗ (T ∩ An). On the other hand, m∗ (T ∩ S) ≤ P∞ n=1 m∗ (T ∩ An) always holds by sub-additivity. Therefore m∗ (T ∩ S) = X∞ n=1 m∗ (T ∩ An), by taking T = R n, we get the disjoint additivity. Finally, if {An} ∈ M are not necessarily disjoint from each other, then we make the following change: B1 = A1, Bk = (∪ k i=1Ai) \ ((∪ k−1 i=1 Ai)) ∀k ≥ 2. It follows {Bk} are disjoint and ∪∞ n=1An = ∪∞ k=1Bk ∈ M. From now on we shall write simply m(A) for the exterior measure of a measurable set A. Our task of defining the measure for suitable subsets of R n is now completed. We conclude this section with two useful facts about interchanging measure with limit operation. Proposition 2.7. Let An ⊂ An+1 be a sequence of increasing measurable sets, set A = ∪nAn, then m(A) = limn→∞ m(An). Proof. If m(An) = ∞ for some n, then the desired equality holds. Therefore we assume m(An) < ∞ for all n. Set B1 = A1, B2 = A2 \ A1, Bn = An \ An−1, then Bn are all disjoint. Using countable disjoint additivity, we get m(∪nBn) = X∞ k=1 m(Bk). We obtain the desired equality as ∪nBn = ∪nAn and m(An) = Pn k=1 m(Bk). For decreasing sequence, we have Proposition 2.8. Let An ⊃ An+1 be a sequence of decreasing measurable sets, set A = ∩nAn, assume m(A1) < ∞ then m(A) = limn→∞ m(An). (2.5) Proof. We view A1 as the ambient set and take complement with respect to A1. We then have ∅ ⊂ A c 2 ⊂ · · · ⊂ A c n · · · , Applying Proposition 2.7, we have m(∪nA c n ) = limn→∞ m(A c n ). (2.6) Since m(A c n ) + m(An) = m(A1) and m(∪nA c n ) + m(A) = m(A1), plugging back to (2.6), we get (2.5). Remark 2.9. The assumption m(A1) < ∞ is necessary. For example, let An = (n, ∞), then ∩nAn = ∅ and (2.5) fails