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6 Measure and Integration §2 Outer Lebesgue measure Deﬁnition (Outer Lebesgue measure) Let E be any subset of R. We call the nonnegative number ∞ ∞ λ∗ (E) = inf |Ij | : Ij are open intervals and Ij ⊃ E i=1 j=1 the outer Lebesgue measure of E. Proposition 2.1 (Basic properties of the outer Lebesgue measure). 1. λ∗ (∅) = 0 and for any x ∈ R, λ∗ ({x}) = 0. 2. If E1 ⊂ E2 , then λ∗ (E1 ) λ∗ (E2 ). ∞ ∞ 3. If (Ek )k 1 is a countable collection of sets, then λ∗ ( Ek ) λ∗ (Ek ). k=1 k=1 4. For any E ⊂ R and x ∈ R we have λ∗ (E + x) = λ∗ (E), where E + x = {y + x, y ∈ E} is the translation of E. Proof. 1. Indeed, take collection of empty intervals In = (0, 0), then In ⊃ ∅, so 0 ∗ λ (∅) 0. If x ∈ R is an arbitrary point, the interval Jε = (x − ε/2, x + ε/2) covers {x} and has length ε. Hence if I1 = Jε and Ik = (0, 0) = ∅, we get 0 λ∗ ({x}) ε for all ε > 0. Thus λ∗ ({x}) = 0. ∞ ∞ 2. If j=1 Ij ⊃ E2 , then j=1 Ij ⊃ E1 . Hence the inﬁmum for E1 is less than or equal to the inﬁmum for E2 . 3. The statement is obviously satisﬁed if there exists k such that λ∗ (Ek ) = +∞. We therefore assume λ∗ (Ek ) < +∞ for all k. Let ε > 0 be arbitrary. For each k, choose a (k) (k) (k) sequence of open intervals (Ij )∞ such that Ek ⊂ ∞ Ij and λ∗ (Ek ) j=1 j=1 ∞ j=1 |Ij | − (k) ε/2k . Then if we take all intervals {Ij ; j 1 and k 1}, we get a countable collection (k) of open intervals that cover ∞ Ek and ∞ λ∗ (Ek ) k=1 k=1 ∞ k=1 ∞ j=1 |Ij | − ε. The RHS of the latter inequality is at least λ∗ ( ∞ k=1 Ek ) − ε. Since ε > 0 is arbitrary, we get ∞ λ∗ (Ek ) λ∗ ( ∞ Ek . k=1 k=1 4. Any cover by open intervals (Ij ) of E corresponds to the cover by open intervals (x + Ij ) of x + E. Corollary 2.2. If E ⊂ R is ﬁnite or countable, then λ∗ (E) = 0. In particular, λ∗ (Q) = 0. §2. Outer Lebesgue measure 7 Proof. Let E = {q1 , q2 , . . . }. We know 0 λ∗ (E); by Proposition 2.1 we have λ∗ (E) ∞ n=1 λ∗ ({qn }) = 0 as λ∗ ({qn }) = 0 for every n by Proposition 2.1. Thus λ∗ (E) = 0. Lemma 2.3 (Outer Lebesgue measure of an interval). If I is an interval then λ∗ (I) coincides with its length. Proof. Assume ﬁrst I = [a, b] is a closed ﬁnite interval. Since I ⊂ (a − ε/2, b + ε/2) for any ε > 0, we conclude λ∗ (I) b − a + ε for every ε > 0 and so λ∗ (I) b − a = |I|. Assume now (Ij )j 1 is any collection of open intervals with union sontaining I. We are going to prove that the sum of their lengths ∞ |Ij | is at least |I|. This would imply that every j=1 element of the set in the right hand side of the deﬁnition of the outer Lebesgue measure of I is at least |I|, and hence λ∗ (I) |I|. By Heine-Borel lemma, we can choose ﬁnitely many of intervals among (Ij )j 1 with union containing I; thus ∃N such that N Ij ⊃ I = [a, b]. Denote j=1 F = {Ij , 1 j N}. Let J1 = (a1 , b1 ) ∈ F be an interval containing point a (then a1 < a). If b1 b then b1 ∈ [a, b] \ J1 . Therefore b1 ∈ Ij . Ij ∈F,Ij =J1 Let J2 ∈ F be an interval, such that b1 ∈ J2 . Assuming Jm = (am , bm ) ∈ F is deﬁned and bm b we conclude bm ∈ [a, b] \ m Ji ; choose Jm+1 ∈ F, such that bm ∈ Jm+1 . Since we i=1 have only ﬁnitely many intervals, this process must stop. We will therefore ﬁnd M such that bM > b. Note that J1 , . . . , JM ∈ F are all different intervals (1 i1 < i2 M implies bi1 < bi2 ), therefore, ∞ N M M M−1 |Ij | |Ij | |Ji | = (bi − ai ) = (bM − a1 ) + (bi − ai+1 ). j=1 j=1 i=1 i=1 i=1 Since ai+1 < bi for each 1 i M − 1, we conclude that the latter sum is greater than bM − a1 , which is bigger than b − a as bM > b and a1 < a. This proves ∞ |Ij | b − a = |I|. j=1 Assume now I is any bounded interval a, b (could be [a, b], (a, b), [a, b) or (a, b]) with a = b. Since [a + ε/2, b − ε/2] ⊂ I ⊂ [a − ε/2, b + ε/2] for any ε > 0, Proposition 2.1 implies |I| − ε λ∗ (I) |I| + ε for all ε > 0; thus λ∗ (I) = |I|. Finally, consider a case when I is an unbounded interval. In such a case, for any ∆ > 0 there exists an interval J of length ∆ such that J ⊂ I. Using again Proposition 2.1 we get λ∗ (I) ∆ for any ∆ > 0, thus λ∗ (I) = +∞. Remark. Property 3 from Proposition 2.1 implies that λ∗ (A ∪ B) λ∗ (A) + λ∗ (B). Our intuitive feeling is that in case A and B are disjoint, equality should hold in this relation. 8 Measure and Integration Unfortunately, this is not the case that equality holds for an arbitrary pair of disjoint sub- sets of R. However, we show that this desired equality relation is true in the case the sets A and B are at positive distance from each other. Lemma 2.4. If A, B ⊂ R are disjoint such that dist(A, B) > 0, then λ∗ (A∪B) = λ∗ (A)+λ∗ (B). Proof. We know λ∗ (A ∪ B) λ∗ (A) + λ∗ (B). We need to show λ∗ (A ∪ B) λ∗ (A) + λ∗ (B). Assume λ∗ (A ∪ B) < ∞ and d = dist(A, B) > 0. Let ε > 0. Consider any covering ∞ Ij ⊃ j=1 A ∪ B by open intervals, such that ∞ |Ij | < λ∗ (A ∪ B) + ε. We may cover each interval j=1 Ij by ﬁnitely many open intervals of length less than d with sum of their lengths less than |Ij | + ε/2j . Thus we may assume that we have a cover of A ∪ B by open intervals Uk with ∞ k=1 |Uk | < λ∗ (A ∪ B) + 2ε. Note that for each k, exactly one of the following three is satisﬁed: Uk ∩ A = ∅, Uk ∩ B = ∅, Uk ∩ (A ∪ B) = ∅. The sum of lengths of those Uk that intersect A is at least λ∗ (A), those that intersect B is at least λ∗ (B). Thus λ∗ (A ∪ B) + 2ε > ∞ k=1 |Uk | λ∗ (A) + λ∗ (B). Since ε > 0 is arbitrary, we are done. k k Corollary 2.5. If I1 , . . . , Ik are disjoint intervals, then λ∗ ( j=1 Ij ) = j=1 |Ij |. Proof. Follows from Lemma 2.3 and Lemma 2.4. Example (Non-additivity of outer Lebesgue measure) For x, y ∈ R, deﬁne x ∼ y if x − y is rational. This is an equivalence relationship on [0, 1]. For each equivalence class, pick an element out of that class (by the axiom of choice). Call the collection of such points A. Note λ∗ (A + q) = λ∗ (A) for any q by Proposition 2.1. Moreover, the sets A + q are disjoint for different rationals q as a1 , a2 ∈ A, q1 , q2 ∈ Q and a1 + q1 = a2 + q2 implies a1 − a2 = q2 − q1 ∈ Q, which is impossible unless a1 = a2 . In this case q1 = q2 . Let [−1, 1] ∩ Q = Q[−1,1] = {q1 , q2 , . . . }. We see q∈Q[−1,1] (A + q) ⊂ [−1, 2], so if λ∗ were additive, we would have for every n 1, ∞ n n 3 λ∗ ( (A + qi )) λ∗ ( (A + qi )) = λ∗ (A + qi ) = nλ∗ (A), i=1 i=1 i=1 which implies λ∗ (A) 3/n for every n 1 and so λ∗ (A) = 0. But [0, 1] ⊂ q∈Q[−1,1] (A + q), ∗ ∗ so using λ (A + q) = λ (A) = 0 for every q ∈ Q, we get by Proposition 2.1 ∞ ∞ 1 λ∗ ( (A + qi )) λ∗ (A + qi ) = 0, i=1 i=1 a contradiction. §2. Outer Lebesgue measure 9 Example (Calculating outer Lebesgue measure) 1. λ∗ ([0, 1] \ Q) = λ∗ ({irrational numbers between 0 and 1}) = 1. Proof. Let E = [0, 1] \ Q and Q1 = Q ∩ [0, 1]. Then 1 = λ∗ ([0, 1]) = λ∗ (E ∪ Q1 ) λ∗ (E) + λ∗ (Q1 ) = λ∗ (E) λ∗ ([0, 1]) = 1. Note that the set of irrational numbers between 0 and 1 contains no intervals, still its outer Lebesgue measure is strictly positive. 2. There exist uncountable sets of outer Lebesgue measure zero. Proof. Consider the Cantor set C = x ∈ [0, 1] | x = ∞ ak , ak ∈ {0, 2} for all k k=1 3 k 1 , ∞ ak let An = x ∈ [0, 1] | x = k=1 3k , ak ∈ {0, 1, 2} for all k 1 and ∃k n : ak = 1 . Clearly C ∩ An = ∅ for all n, therefore, λ∗ (C) λ∗ ([0, 1] \ An ) for all n. Note that [0, 1]\ An is a union of 2n disjoint closed intervals of equal length. This length is equal to (1/3)n , and so by Corollary 2.5, λ∗ ([0, 1] \ An ) = 2n /3n . Thus 0 λ∗ (C) (2/3)n for any n, so λ∗ (C) = 0. It remains to say that the Cantor set is uncountable.