If a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?

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1 2 Lebesgue Measure I Chapter 1 we defied the cocept of a set of measure zero, ad we have observed that every coutable set is of measure zero. Here are some atural questios: If a subset E of R cotais a ope iterval (of ozero legth), the ca it be of measure zero? If a subset E of R cotais o ope iterval, is it of zero measure? For istace, is the set of irratioals i [0, 1] is of measure zero? We shall fid aswer to these questios after defiig the cocept of Lebesgue outer measure. I fact we shall see that if Lebesgue outer measure of a set is zero, the it of measure zero. NOTATION: I the followig, by sayig that {A } is a coutable family of sets, we mea a family {A : Λ} of sets A, Λ, where Λ is a coutable set, that is, fiite or deumerable. The, {A : Λ} ad {A : Λ} will be writte as A ad A, respectively. Also, for a coutable set {a R : Λ}, the series a will be writte as a. Λ By a iterval we mea a subset I of R such that (i) for every a, b I with a < b, (a, b) I ad (ii) there exists x I ad ε > 0 such that (x ε, x + ε) I. Thus, itervals are of the forms (a, b), [a, b), (a, b], [a, b], (a, ), [a, ), (, a), (, a] for a, b R with a < b. The itervals (a, b), [a, b), (a, b], [a, b] are bouded itervals with edpoits a ad b, ad the itervals (a, ), [a, ), (, a), (, a], (, ) are ubouded itervals. 11

2 12 Lebesgue Measure Legth of a iterval I is deoted by l(i). For a bouded iterval I of ed poits a, b, l(i) is defied as b a. If I is a ubouded iterval, the we say that its legth is ifiity, ad write l(i) =. Although is ot a umber, i certai cotexts such as discussio ivolvig ubouded itervals, we shall use the covetio that a + = for every a R ad + =. 2.1 Lebesgue Outer Measure Defiitio For E R, the Lebesgue outer measure of E is defied as m (E) := if l(i ), I E where the ifimum is take over the collectio I E of all coutable family {I } of ope itervals which covers E, that is, E I. Note that m (E) 0, ad m (E) ca take the value as well. Thus, m ca be thought of as a fuctio from the family of all subsets of R ito the set [0, ]. Exercise 2.1 If (I ) is a sequece of ope itervals, the the iequality m ( I ) l(i ) holds Why? Theorem Let E R. For every ε > 0 there exists a coutable family {I } of ope itervals such that E I ad l(i ) m (E) + ε. Strict iequality occurs i the above if m (E) <. Proof. If m (E) =, the the coclusio is true trivially. So, assume that m (E) <. The the result follows from the defiitio of ifimum of a subset of R. By Theorem 2.1.1, for E R, m (E) = 0 if ad oly if there exists a coutable family {I } of ope itervals such that E I ad l(i ) ε. Thus, E is of measure zero if ad oly if its outer measure is zero.

3 Lebesgue Outer Measure 13 Corollary Let E R. For every ε > 0, there exists a ope set G i R such that E G ad m (G) m (E) + ε. Proof. Let ε > 0 be give. By Theorem 2.1.1, there exists a coutable family {I } of ope itervals such that E I ad l(i ) m (E) + ε. Take G = I. The, by the defiitio of m, m (G) l(i ). Thus, we have m (G) m (E) + ε. Throughout the text, we cosider the topology o R as the usual topology. Thus, a set G R is ope if ad oly if for every x G, there exists r > 0 such that {y R : x y < r} G, a set F R is closed if ad oly if its complemet is ope. Remark It ca be see that, if we assume m (E) < i Corollary 2.1.2, the strict iequality holds i the coclusio as well. Theorem The followig hold. (i) If I is a ope iterval of fiite legth l(i), the m (I) l(i). (ii) m ( ) = 0. (iii) If E is a coutable subset of R, the m (E) = 0. Proof. (i) Let I be a ope iterval of fiite legth l(i). Takig the sigleto family {I}, we obtai from the defiitio that m (I) l(i). (ii) For every ε > 0, we have ( ε, ε). Hece by (i), m ( ) 2ε. This is true for every ε > 0. Hece, m ( ) = 0. (iii) First let E be a fiite set, say E = {a 1,..., a k } R. The for every ε > 0, E k I i, where I i = (a i ε, a i + ε). Hece, m (E) 2kε. Sice this is true for every ε > 0, m (E) = 0. Next suppose that E is a coutably ifiite set, say E = {a i : i N}. The takig we have E N I so that I := (a ε/2 +1, a + ε/2 +1 ), m (E) N l(i ) = N(ε/2 ) ε. Sice this is true for every ε > 0, we obtai m (E) = 0.

4 14 Lebesgue Measure NOTATION: Sets cosidered i this chapter are are subsets of R. Also, for E R, we shall use the otatio I E to deote the class of all coutable family of ope itervals {I } such that E I. For subsets A ad B of R, we defie A + B = {x + y : x A, y B}. Also for E R ad a R, we deote, E + a = E + {a} so that E + a := {x + a : x R}. Theorem We have the followig: (i) A B m (A) m (B). (ii) m (A B) m (A) + m (B). (iii) If E A ad m (E) = 0, the m (A \ E) = m (A). (iv) If E R ad x R, the m (E + x) = m (E). Proof. (i) Let A B ad let {I } I B. The {I } I A. Hece m (A) l(i ). Now, takig ifimum over all {I } I B, we have m (A) m (B). (ii) If at least oe of m (A) ad m (B) is ifiity the the result holds. Next, assume that both m (A) ad m (B) are fiite. Hece, by Theorem 2.1.3(iv), give ε > 0 there exist {I } a {J } i I A ad I B, respectively, such that l(i ) < m (A) + ε 2, l(j ) < m (B) + ε 2. The, the collectio {I } {J k} k=1 of ope itervals cover A B. Therefore, m (A B) l(i ) + l(j ) ( m (A) + ε ) ( + m (B) + ε ) 2 2 = m (A) + m (B) + ε. This is true for all ε > 0, so that m (A B) m (A) + m (B). (iii) If E A ad m (E) = 0, the by (i) ad (ii), m (A) m (E) + m (A \ E) = m (A \ E) m (A).

5 Lebesgue Outer Measure 15 Hece, m (A \ E) = m (A). (iv) Suppose E R ad x R. Give ε > 0, let {I } i I E be such that l(i ) m (E) + ε. Note that each I + x is a ope iterval ad E + x (I + x). Hece, m (E + x) l(i + x) = l(i ) = m (E) + ε. This is true for every ε > 0. Hece, m (E + x) m (E). Sice E = (E + x) + ( x), it follows from the above that m (E) m (E + x). Thus the proof is complete. The property (i) i Theorem is called the mootoicity property of m, ad the property (iv) is called the traslatio ivariace of m. Makig use of the mootoicity of m, we deduce the followig corollary from Corollary Corollary Let E R. The there exists a set G which is a coutable itersectio of ope sets i R such that E G ad m (G) = m (E). Proof. By Corollary 2.1.2, for each N, there exists a ope set G i R such that E G ad m (G ) m (E) + 1. Take G = G. The, E G ad m (E) m (G) m (G ) for every N, so that m (G) m (E) + 1 N. Lettig ted to ifiity, we obtai, m (G) m (E). Thus, we have proved m (G) = m (E). Defiitio A subset of R is said to be a G δ set if it is a coutable itersectio of of ope sets, ad a subset of R is said to be a F δ set if it is a coutable uio of closed sets. Sice a subset of R is closed if ad oly if its complemet is ope, usig De Morga s law, it follows that a subset of R is G δ set if ad oly if its complemet is a F σ set.

6 16 Lebesgue Measure Recall that for a give {A α : α Λ} of sets, where Λ is some idex set, the De Morga s law states that ( ) c ( ) c A α = ad A α = A c α. α Λ α Λ A c α I particular, for a coutably ifiite family, that is, if Λ = N, the we have α Λ α Λ ( ) c A i = A c i ad ( ) c A i = A c i. The above relatios icludes the case for fiite family {A i : i = 1,..., } also, as we ca take A i = for i >. Note that a G δ -set eed ot be ope. For example, [0, 1) is a G δ -set as [0, 1) = ( 1/, 1). I fact, for every a, b R with a < b, the itervals [a, b), (a, b], [a, b] are G δ -sets which are ot ope sets (verify!). The followig result geeralizes Theorem 2.1.4(ii). Theorem Suppose A k R for k N. The ) A k k=1 m (A k ). Proof. If m (A k ) = for some k N, the the result holds trivially. Hece, assume that m (A k ) < for all k N. The for each k N ad ε > 0, there exists {I k, } I Ak such that l(i k,) < m (A k ) + ε/2 k. Note that k=1 A k k=1 I k,. Therefore, ) A k k=1 k=1 l(i k, ) k=1 m (A k ) + ε. k=1 This is true for all ε > 0. Hece the result follows. Remark Note that Theorem icludes the result ) A k k=1 m (A k ) k=1 for ay fiite family {A 1,..., A } as we ca take A k = for k >.

7 Lebesgue Outer Measure 17 The property of m i Theorem is called the coutable subadditivity of m. Now a result that we are waitig for. Theorem The Lebesgue outer measure of ay iterval is its legth. Proof. Let I be a iterval. Case 1: Suppose I = [a, b] with < a < b <, ad let ε > 0. Let I ε := (a ε, b + ε). Clearly, m (I) m (I ε ) b a + 2ε. This is true for all ε > 0. Hece, m (I) b a. Thus, it remais to show that m (I) b a. For this, it is eough to show that b a l(i ) {I } I I, ( ) because, i that case we ca take ifimum over all such {I } I I ad obtai b a m (I). So, let {I } I I. Without loss of geerality, we may assume that each I is of fiite legth. By the compactess of I, there exists a fiite sub-collectio {I 1,,..., I k } of {I } such that I k I i. Let I i := (a i, b i ) for i {1,..., k}. Without loss of geerality, assume that The (a i, b i ) [a, b] ad a i+1 < b i, i {1,..., k 1}. k l(i i ) = Thus, k k 1 (b i a i ) = b k + (b i a i+1 ) a 1 b k a 1 b a. b a k l(i i ) l(i ). Thus, we have proved ( ). This completes the proof of m (I) = b a. Case 2: Suppose I is a iterval of fiite legth with ed poits a ad b with a < b. The for sufficietly small ε > 0, we have [a + ε, b ε] I [a ε, b + ε]. Hece, Thus, by case (i), we have m ([a + ε, b ε]) m (I) m ([a ε, b + ε]). b a 2ε m (I) b a + 2ε.

8 18 Lebesgue Measure Sice this is true for every ε > 0, it follows that m (I) = b a. Case 3: Suppose I is of ifiite legth. The for every M > 0 there exists a closed iterval I M of legth M such that I M I. Hece M = m (I M ) m (I). Thus, M m (I) for all M > 0 so that m (I) =. Corollary Every o-degeerate iterval is a ucoutable set. Now, let us address the followig questio: Ca a ucoutable set be of measure 0? The aswer is i affirmative as the followig example of Cater s terary set or simply Cater set shows. Example Let us first recall how the Cater set is costructed. Cosider the uit iterval [0, 1]. Let C 1 be the set obtaied from I after removig its middle third J 1 := ( 1 3, 2 3 ) from C 0[0, 1]. That is C 1 = [ 0, 1 ] [ 2 ] 3 3, 1. Net, let C 2 be the set obtaied from C 1 after removig the middle thirds from each of the two subitervals i C 1. Let the removed set be J 2. Thus, [ C 2 = 0, 1 ] [ 2 9 9, 1 [ 2 3] 3 9], 7 [ 8 ] 9, 1. Cotiue this procedure to obtai C 3, C 4 ad so o. At the th stage, we obtai C = C 1 \ J, where J is the uio of the middle thirds of each of the subitervals i C 1. Now Cater set C is defied by Note that C = C. C 1 C 2 C 3. ad m (C 1 ) 2/3, m (C 2 ) (2/3) 2, m (C 3 ) (2/3) 3, etc., ad more geerally, ( 2 ), m (C ) N. 3 Hece, k ( 2 ) k m (C) C = m (C k ) k N. 3

9 Lebesgue Measurable Sets 19 Thus, m (C) = 0. To see that C is a ucoutable set, first we recall that every umber a [0, 1] ca be writte as a series a 3, where a {0, 1, 2}. It ca be see that a {0, 2} a J so that a J N a {0, 2} N. Hece, C = { b = b } 3 : b {0, 2}, N. Thus, C is i oe-oe correspodece with the family of all sequece (b ) with b {0, 2}, ad hece, C is a ucoutable set. 2.2 Lebesgue Measurable Sets Suppose A 1 ad A 2 are disjoit subsets of R. We may expect that m (A 1 A 2 ) = m (A 1 ) + m (A 2 ). (1) Is it true for ay two disjoit sets A 1 ad A 2? Suppose for a momet that (1) is true for ay two disjoit sets A 1 ad A 2. The we also have ) A i = m (A i ) (2) for ay pairwise disjoit sets A 1,..., A. Now, cosider a deumerable disjoit family {A }. The usig the subadditivity (Theorem 2.1.6) ad mootoicity (Theorem (i)) ad the equality (2) above, we obtai m (A ) ) A ) A i = m (A i ) N. Thus, m (A i ) ) A m (A ) for all N. Takig limit as, we obtai ) A = m (A ). (3)

10 20 Lebesgue Measure We ow show that (3) eed ot hold for every deumerable disjoit family {A }, so that (1) eed ot be true for every disjoit sets A 1, A 2. For this, cosider a relatio o R by defiig x y x y Q. It ca be easily see that is a equivalece relatio o R. Hece, R is the disjoit uio of equivalece classes. Let E be the subset of [0, 1] such that its itersectio with each equivalece class is a sigleto set. Such a set E exists by usig the axiom of choice o the collectio E := {[x] [0, 1] : x [0, 1]}, where [x] is the equivalece class of x. We ote that if x y, the the ratioal umber r := x y satisfies 1 r 1. Let {r 1, r 2,...} be the set of all ratioal umbers i [ 1, 1]. Let E := E + r for N. Note that {E } is a disjoit family ad so that [0, 1] E [ 1, 2]. 1 ) E 3. Therefore, if (3) is true, the the above relatio implies 1 m (E ) 3. This is ot possible, sice m (E ) = m (E + r ) = m (E) for all N. Hece, we coclude that oe of the relatios (3), (2), (1) hold for all possible disjoit family of sets ivolved. Thus, we have proved the followig result. Theorem There exist disjoit subsets A 1 ad A 2 of R such that m (A 1 A 2 ) m (A 1 ) + m (A 2 ). I establishig the above theorem what we have actually proved is the followig. Theorem There exists a subset E of R such that ) E m (E ), where E := E + r with {r : N} = [ 1, 1] Q.

11 Lebesgue Measurable Sets 21 The above discussio motivates us to cosider a family of sets i which the relatio (1), ad hece (2) ad (3) hold for all possible disjoit family of sets ivolved. Defiitio A set E R is said to be Lebesgue measurable if for every A R, m (A) = m (A E) + m (A E c ). We deote the set of all Lebesgue measurable sets by M. The proof of the followig theorem is easy ad hece left as exercise. Theorem The followig results hold. (i) E M m (A) m (A E) + m (A E c ) A R. (ii) M. (iii) E M E c M. (iv) m (E) = 0 E M. I view of Theorem 2.2.3(iv), M cotais all coutable sets. I particular, Q M, ad hece by (iii) Q c M. Theorem Let A 1, A 2 subsets of R such that A 1 A 2 =. If oe of A 1, A 2 belogs to M, the m (A 1 A 2 ) = m (A 1 ) + m (A 2 ). Proof. Suppose A 1 A 2 =. Assume A 1 M. The m (A 1 A 2 ) = m ((A 1 A 2 ) A 1 ) + m ((A 1 A 2 ) A c 1). But, (A 1 A 2 ) A 1 ) = A 1 ad (A 1 A 2 ) A c 1 ) = A 2. Hece the result The proof of the followig theorem is alog the same lies as we have deduced (2.3) from (2.2). However, we give its details here as well. Theorem Let {A : N} be a disjoit family i M. The ) A = m (A ).

12 22 Lebesgue Measure Proof. If {A } is a fiite family, the by Theorem 2.2.4, ) A i = m (A i ) for every N. Hece, by usig the mootoicity of m, we have m (A ) ) A ) A i = for all N. Lettig ted to ifiity, the result follows. m (A i ) Remark Note that Theorem icludes the equality ) A i = m (A i ) for ay fiite disjoit family {A 1,..., A } i M, as i this case we ca take A k = for k >. The property of m give i Theorem is called coutable additivity of m o M. Remark I view of the coutable additivity of m o M, Theorem shows that there exists E R which does ot belog to M. From this, it also follows that M is ot the whole of 2 R, the power set of R. Our ext attempt is to show that M cotais ot oly coutable sets but a lot may subsets of R. Theorem Let A 1, A 2 M. The A 1 A 2 M. More geerally, if {A 1,..., A } M, the A i M. Proof. Let A R. We have to show that m (A) m (A (A 1 A 2 )) + m (A A c 1 A c 2). ( ) Note that A (A 1 A 2 ) = (A A 1 ) (A A 2 A c 1 ), so that m (A (A 1 A 2 )) m (A A 1 ) + m (A A 2 A c 1). Therefore, the right had side of ( ) is less tha or equal to m (A A 1 ) + m (A A 2 A c 1) + m (A A c 1 A c 2). Now, sice A 2 M, we get m (A A 2 A c 1) + m (A A c 1 A c 2) = m (A A c 1).

13 Lebesgue Measurable Sets 23 Hece, usig the fact that A 1 M, m (A (A 1 A 2 )) + m (A A c 1 A c 2) m (A A 1 ) + m (A A c 1) = m (A) which completes the proof of ( ). The last part follows by repeated applicatio of the first part. By the above theorem we ca say that M is closed uder fiite uios. Next we show that M is closed uder coutable uios. For this purpose we shall make use of the followig lemma which is more geeral tha the Theorem Lemma Let {A 1,..., A } be a disjoit family i M. The for ay A R, ) A A i = m (A A i ). Proof. Let = 2. Sice A 1 M, m (A (A 1 A 2 )) = m (A (A 1 A 2 ) A 1 ) + m (A (A 1 A 2 ) A c 1). But, A (A 1 A 2 ) A 1 = A A 1, A (A 1 A 2 ) A c 1 = A A 2. Hece, we have m (A (A 1 A 2 )) = m (A A 1 ) + m (A A 2 ). Thus, the result is proved for = 2. The result for geeral follows by iductio. Now we prove the result that we had promised. Theorem If E M for N, the E M. Proof. Let E = E, where E M, N ad let A R. We have to show that m (A) m (A E) + m (A E c ). ( ) We write E as a disjoit uio E = A where A 1 = E 1 ad for 2, A = E \ 1 E i. Let F = A i. Note that F M. Hece, m (A) m (A F ) + m (A F c ).

14 24 Lebesgue Measure Now, Lemma implies m (A F ) = m (A A ) ad the relatio F c E c implies m (A F c ) m (A E c ). Thus, m (A) m (A A ) + m (A E c ). This is true for all N. Lettig ted to ifiity, m (A) m (A A ) + m (A E c ) Therefore, usig the fact that (A A i) = A A i ad the subadditivity of m, m (A) Thus the proof is complete. m (A A ) + m (A E c ) m (A A i ) + m (A E c ) = m (A E) + m (A E c ). Corollary If E M for N, the E M. Proof. Let E M for N. By De Morga s law, ( c. E = E) c Hece, the result follows from Theorem together with the fact that E M implies E c M. Defiitio The fuctio m restricted to M is called the Lebesgue measure o R, ad it is deoted by m. For E M, m(e) := m (E) is called the Lebesgue measure of E. Let us list some of the importat properties of the family M that we have proved: M E M E c M E i M, i N i E i M & m ( i E i) = i m (E i )

15 Lebesgue Measurable Sets 25 The followig theorem shows that the class M is really large. Theorem The followig results hold. (i) For ay a R, (a, ) M. (ii) For ay a R, [a, ) M. (iii) For ay a R, (, a) M. (iv) For ay a R, (, a] M. (v) For ay a, b R with a < b, (a, b) M. (vi) For ay a, b R with a < b, [a, b] M. (vii) Ope subsets of R belog to M. (viii) Closed subsets of R belog to M. (ix) Coutable uios ad coutable itersectios of ope subsets of R belog to M. (x) Coutable uios ad coutable itersectios of closed subsets of R belog to M. Proof. We first proved (i) ad the deduce other results (ii)-(x) by usig some of the properties of m. For provig (i), let a R ad E = (a, ). Let A R ad ε > 0. By Theorem 2.1.1, there exists a coutable family {I } of ope itervals such that A I, l(i ) < m (A) + ε. Note that A E (I (a, ), A E c (I (, a]. Hece, takig I := I (a, ) ad I := I (, a], we have m (A E) + m (A E c ) m (I ) + m (I ) = [m (I ) + m (I )]. Note that I ad I are itervals such that I I = ad I I = I so that m (I ) + m (I ) = l(i ) + l(i ) = l(i ).

16 26 Lebesgue Measure Thus, we have proved that m (A E) + m (A E c ) l(i ) m (A) + ε. This is true for ay ε > 0, so that we get m (A E) + m (A E c ) l(i ) m (A). Hece, (a, ) = E M, provig (i). Next we observe that for ay a R, [a, ) = (a 1, ), (, a) = R \ [a, ), ad for ay a, b R with a < b, (a, b) = (a, ) (, b), (, a] = R \ (a, ), [a, b] = [a, ) (, b]. Hece, usig the facts that M is closed uder coutable uios, coutable itersectios ad complemetatio, ad the fact that every ope subset of R is a coutable uio of ope itervals, the results listed i (ii)-(x) follow. By the above theorem, M cotais most of the sets that we ecouter i aalysis. Still, we kow that it is ot whole of the power set of R. Now, we prove a compaio result to Corollary ad Corollary Theorem Let E R. The the followig are equivalet. (i) E M. (ii) For every ε > 0, there exists a ope set G i R such that E G ad m (G \ E) ε. (iii) There exists a G δ -set set G i R such that E G ad m (G \ E) = 0. Proof. (i) (ii): Suppose E M ad ε > 0 be give. First, let us assume that m (E) <. By Corollary 2.1.2, there exists a ope set G i R such that E G ad m (G) m (E) + ε.

17 Lebesgue Measurable Sets 27 Sice E ad G are i M ad G \ E = G E c is also i M. Therefore, m (E) + m (G \ E) = m (G) m (E) + ε. Sice m (E) <, we obtai m (G \ E) ε. For the case whe m (E) is ot ecessarily fiite, we write E = E, where m (E ) < for every N. For example, we ca take E := E [, ] for N. Sice m (E ) < for each N, by the above part, there exists ope set G E such that m (G \ E ) < ε/2. Takig G = G, we have G is ope, G E ad Therefore, Thus, (ii) holds. G \ E = [ G ] \ [ E ] (G \ E ). m (G \ E) m (G \ E ) ε 2 = ε. (ii) (iii): Assume (ii). The, for every N, there exists a ope set V i R such that E V ad m (V \ E) 1. Let G = V. The E G V ad G \ E V \ E for all N so that m (G \ E) 1 N. Lettig ted to ifiity, we obtai m (G \ E) = 0. Thus, (iii) holds. (iii) (i): Assume (iii). The there exists a G δ -set set G i R such that E G ad m (G \ E) = 0. I particular, G \ E M. Also, This completes the proof. E = G \ (G \ E) = G (G \ E) c M. Corollary Let E R. The E M if ad oly if there exist a G δ -set G ad a F σ -set F such that F E G ad m (G \ F ) = 0. Proof. Suppose E M. The by Theorem , there exists a G δ set G such that E G ad m (G \ E) = 0.

18 28 Lebesgue Measure Also, by takig E c i place of E, there exists a G δ -set H such that E c H ad m (H \ E c ) = 0. Hece, H \ E c M, F := H c E, H \ E c = E \ H c = E \ F. Thus, F is a closed set such that F E ad m (E \ F ) = 0. Sice G \ F = (G \ E) (E \ F ) we obtai m (G \ F ) = 0. Coversely, suppose that there exists a G δ -set G ad a F σ -set F such that F E G ad m (G \ F ) = 0. I particular, G \ E G \ F so that m (G \ E) = 0 ad hece G \ E M. Therefore, E = G \ (G \ E) M. Usig Theorem , we obtai the followig theorem. The details of its proof is left as a exercise. Theorem Let E R. The the followig are equivalet. (i) E M. (ii) For every ε > 0, there exists a closed set F i R such that F E ad m (E \ F ) ε. (iv) There exists a F σ -set set F i R such that F E ad m (E \ F ) = Problems 1. Prove that, i Defiitio 2.1.1, m (E) remais the same if we take I E to be the collectio of all (a) coutable family {I } of itervals of the form J = (a, b ), (b) coutable family {I } of itervals of the form J = [a, b ), (c) coutable family {I } of itervals of the form J = (a, b ], (d) coutable family {I } of itervals of the form J = [a, b ], (e) sequeces (I ) of itervals of the form J = (a, b ), (f) sequeces (I ) of itervals of the form J = [a, b ), (g) sequeces (I ) of itervals of the form J = (a, b ], (h) sequeces (I ) of itervals of the form J = [a, b ],

19 Problems 29 with a, b R. 2. Show that, if E A ad m (E) = 0, the m (A E) = m (A). 3. From Theorem 2.1.6, deduce that outer measure of every coutable set is Justify the statemet: There exists a coutable disjoit family {E } of subsets of R such that ) E m (E ). 5. If E is a subset of a iterval I such that m (E) = 0, the prove that E c is dese i I. 6. Prove Theorem Suppose {A 1,..., A } is a disjoit family of subsets of R such that A 1,..., A 1 belog to M. The show that ) A i = m (A i ). 8. Let M = M {E 0 }, where E 0 M. Prove that if {A } is a coutable disjoit family i M, the ) A i = m (A i ). 9. Show that if E M, the m(e) = if{m(g) : G ope E G}. 10. Supply details of the proof of Theorem If E R such that with m (E) <. Prove that the followig are equivalet: (a) E M. (b) There exists a G δ set G E such that E = G \ E 0, where m (E 0 ) = 0. (c) There exists a F σ set F E such that E = F E 0, where m (E 0 ) = 0.

20 30 Lebesgue Measure 12. If A, B M with A B ad if m(b) <, the show that 13. If A, B M, prove that m(b \ A) = m(b) m(a). m(a B) + m(a B) = m(a) + m(b). 14. Supply details of the proof for the results listed i Theorem Fid a ope dese subset A of R with fiite m(a) <. [Hit: Use desity of the set of ratioal umbers.] 16. Fid a closed subset B of R with empty iterior ad with m(b) =. 17. Give ε > 0 fid a closed set F [0, 1] such that m(f ) 1 ε. [Hit: Fid a ope subset A of [0, 1] such that m(a) < ε.] 18. Fid a subset E of [0, 1] which is a coutable uio of closed owhere dese sets such that m(e) = 1. [Hit: Use last exercise. Recall that a set A is said to be owhere dese if its closure has empty iterior.]

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