测度论讲义

Measure Theory V. Liskevich 1998 1 Introduction We always denote by X our universe, i.e. all the sets we shall consider are subsets of X. Recall some standard notation. 2X everywhere denotes the set of all subsets of a given set X. If A ∩ B = ∅ then we often write A unionsq B rather than A ∪ B, to underline the disjointness. The complement (in X) of a set A is denoted by Ac. By A4B the symmetric difference of A and B is denoted, i.e. A4B = (A \B) ∪ (B \A). Letters i, j, k always denote positive integers. The sign � is used for restriction of a function (operator etc.) to a subset (subspace). 1.1 The Riemann integral Recall how to construct the Riemannian integral. Let f : [a, b]→ R. Consider a partition pi of [a, b]: a = x0 < x1 < x2 < . . . < xn−1 < xn = b and set ∆xk = xk+1 − xk, |pi| = max{∆xk : k = 0, 1, . . . , n − 1}, mk = inf{f(x) : x ∈ [xk, xk+1]},Mk = sup{f(x) : x ∈ [xk, xk+1]}. Define the upper and lower Riemann— Darboux sums s(f, pi) = n−1∑ k=0 mk∆xk, s¯(f, pi) = n−1∑ k=0 Mk∆xk. One can show (the Darboux theorem) that the following limits exist lim |pi|→0 s(f, pi) = sup pi s(f, pi) = ∫ b a fdx lim |pi|→0 s¯(f, pi) = inf pi s¯(f, pi) = ∫ b a fdx. 1 Clearly, s(f, pi) ≤ ∫ b a fdx ≤ ∫ b a fdx ≤ s¯(f, pi) for any partition pi. The function f is said to be Riemann integrable on [a, b] if the upper and lower integrals are equal. The common value is called Riemann integral of f on [a, b]. The functions cannot have a large set of points of discontinuity. More presicely this will be stated further. 1.2 The Lebesgue integral It allows to integrate functions from a much more general class. First, consider a very useful example. For f, g ∈ C[a, b], two continuous functions on the segment [a, b] = {x ∈ R : a 6 x 6 b} put ρ1(f, g) = max a6x6b |f(x)− g(x)|, ρ2(f, g) = ∫ b a |f(x)− g(x)|dx. Then (C[a, b], ρ1) is a complete metric space, when (C[a, b], ρ2) is not. To prove the latter statement, consider a family of functions {ϕn}∞n=1 as drawn on Fig.1. This is a Cauchy sequence with respect to ρ2. However, the limit does not belong to C[a, b]. 2 - 6 � � � � � � � � �� L L L L L L L L LL −1 2 + 1 n 1 2 − 1 n −1 2 1 2 Figure 1: The function ϕn. 2 Systems of Sets Definition 2.1 A ring of sets is a non-empty subset in 2X which is closed with respect to the operations ∪ and \. Proposition. Let K be a ring of sets. Then ∅ ∈ K. Proof. Since K 6= ∅, there exists A ∈ K. Since K contains the difference of every two its elements, one has A \ A = ∅ ∈ K. � Examples. 1. The two extreme cases are K = {∅} and K = 2X . 2. Let X = R and denote by K all finite unions of semi-segments [a, b). Definition 2.2 A semi-ring is a collection of sets P ⊂ 2X with the following properties: 1. If A,B ∈ P then A ∩B ∈ P; 3 2. For every A,B ∈ P there exists a finite disjoint collection (Cj) j = 1, 2, . . . , n of sets (i.e. Ci ∩ Cj = ∅ if i 6= j) such that A \B = n⊔ j=1 Cj. Example. Let X = R, then the set of all semi-segments, [a, b), forms a semi-ring. Definition 2.3 An algebra (of sets) is a ring of sets containing X ∈ 2X . Examples. 1. {∅, X} and 2X are the two extreme cases (note that they are different from the corresponding cases for rings of sets). 2. Let X = [a, b) be a fixed interval on R. Then the system of finite unions of subin- tervals [α, β) ⊂ [a, b) forms an algebra. 3. The system of all bounded subsets of the real axis is a ring (not an algebra). Remark. A is algebra if (i) A,B ∈ A =⇒ A ∪B ∈ A, (ii) A ∈ A =⇒ Ac ∈ A. Indeed, 1) A ∩B = (Ac ∪Bc)c; 2) A \B = A ∩Bc. Definition 2.4 A σ-ring (a σ-algebra) is a ring (an algebra) of sets which is closed with respect to all countable unions. Definition 2.5 A ring (an algebra, a σ-algebra) of sets, K(U) generated by a collection of sets U ⊂ 2X is the minimal ring (algebra, σ-algebra) of sets containing U. In other words, it is the intersection of all rings (algebras, σ-algebras) of sets containing U. 4 3 Measures Let X be a set, A an algebra on X. Definition 3.1 A function µ: A −→ R+ ∪ {∞} is called a measure if 1. µ(A) > 0 for any A ∈ A and µ(∅) = 0; 2. if (Ai)i>1 is a disjoint family of sets in A ( Ai ∩ Aj = ∅ for any i 6= j) such that⊔∞ i=1Ai ∈ A, then µ( ∞⊔ i=1 Ai) = ∞∑ i=1 µ(Ai). The latter important property, is called countable additivity or σ-additivity of the measure µ. Let us state now some elementary properties of a measure. Below till the end of this section A is an algebra of sets and µ is a measure on it. 1. (Monotonicity of µ) If A,B ∈ A and B ⊂ A then µ(B) 6 µ(A). Proof. A = (A \B) unionsqB implies that µ(A) = µ(A \B) + µ(B). Since µ(A \B) ≥ 0 it follows that µ(A) ≥ µ(B). 2. (Subtractivity of µ). If A,B ∈ A and B ⊂ A and µ(B) < ∞ then µ(A \ B) = µ(A)− µ(B). Proof. In 1) we proved that µ(A) = µ(A \B) + µ(B). If µ(B) <∞ then µ(A)− µ(B) = µ(A \B). 3. If A,B ∈ A and µ(A ∩B) <∞ then µ(A ∪B) = µ(A) + µ(B)− µ(A ∩B). Proof. A ∩B ⊂ A, A ∩B ⊂ B, therefore A ∪B = (A \ (A ∩B)) unionsqB. Since µ(A ∩B) <∞, one has µ(A ∪B) = (µ(A)− µ(A ∩B)) + µ(B). 5 4. (Semi-additivity of µ). If (Ai)i≥1 ⊂ A such that ⋃∞ i=1Ai ∈ A then µ( ∞⋃ i=1 Ai) 6 ∞∑ i=1 µ(Ai). Proof. First let us proove that µ( n⋃ i=1 Ai) 6 n∑ i=1 µ(Ai). Note that the family of sets B1 = A1 B2 = A2 \ A1 B3 = A3 \ (A1 ∪ A2) . . . Bn = An \ n−1⋃ i=1 Ai is disjoint and ⊔n i=1Bi = ⋃n i=1Ai. Moreover, since Bi ⊂ Ai, we see that µ(Bi) ≤ µ(Ai). Then µ( n⋃ i=1 Ai) = µ( n⊔ i=1 Bi) = n∑ i=1 µ(Bi) ≤ n∑ i=1 µ(Ai). Now we can repeat the argument for the infinite family using σ-additivity of the measure. 3.1 Continuity of a measure Theorem 3.1 Let A be an algebra, (Ai)i≥1 ⊂ A a monotonically increasing sequence of sets (Ai ⊂ Ai+1) such that ⋃ i≥1 ∈ A. Then µ( ∞⋃ i=1 Ai) = lim n→∞ µ(An). Proof. 1). If for some n0 µ(An0) = +∞ then µ(An) = +∞∀n ≥ n0 and µ( ⋃∞ i=1Ai) = +∞. 2). Let now µ(Ai) <∞ ∀i ≥ 1. 6 Then µ( ∞⋃ i=1 Ai) = µ(A1 unionsq (A2 \ A1) unionsq . . . unionsq (An \ An−1) unionsq . . .) = µ(A1) + ∞∑ k=2 µ(Ak \ Ak−1) = µ(A1) + lim n→∞ n∑ k=2 (µ(Ak)− µ(Ak−1)) = lim n→∞ µ(An). 3.2 Outer measure Let a be an algebra of subsets of X and µ a measure on it. Our purpose now is to extend µ to as many elements of 2X as possible. An arbitrary set A ⊂ X can be always covered by sets from A, i.e. one can always find E1, E2, . . . ∈ A such that ⋃∞ i=1Ei ⊃ A. For instance, E1 = X,E2 = E3 = . . . = ∅. Definition 3.2 For A ⊂ X its outer measure is defined by µ∗(A) = inf ∞∑ i=1 µ(Ei) where the infimum is taken over all A-coverings of the set A, i.e. all collections (Ei), Ei ∈ A with ⋃ iEi ⊃ A. Remark. The outer measure always exists since µ(A) > 0 for every A ∈ A. Example. Let X = R2, A = K(P), -σ-algebra generated by P, P = {[a, b) × R1}. Thus A consists of countable unions of strips like one drawn on the picture. Put µ([a, b)× R1) = b − a. Then, clearly, the outer measure of the unit disc x2 + y2 6 1 is equal to 2. The same value is for the square |x| 6 1, |y| 6 1. Theorem 3.2 For A ∈ A one has µ∗(A) = µ(A). In other words, µ∗ is an extension of µ. Proof. 1. A is its own covering. This implies µ∗(A) 6 µ(A). 2. By definition of infimum, for any ε > 0 there exists a A-covering (Ei) of A such that∑ i µ(Ei) < µ ∗(A) + ε. Note that A = A ∩ ( ⋃ i Ei) = ⋃ i (A ∩ Ei). 7 - 6 � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � � � � � � � �� � � � � �� � � � � � � � � � � � �� � ��� ��� � � � � � �� � � � � � � � � � � �� � � � �� � � � � � � � � �� �� a b Using consequently σ-semiadditivity and monotonicity of µ, one obtains: µ(A) 6 ∑ i µ(A ∩ Ei) 6 ∑ i µ(Ei) < µ ∗(A) + ε. Since ε is arbitrary, we conclude that µ(A) 6 µ∗(A). � It is evident that µ∗(A) > 0, µ∗(∅) = 0 (Check !). Lemma. Let A be an algebra of sets (not necessary σ-algebra), µ a measure on A. If there exists a set A ∈ A such that µ(A) <∞, then µ(∅) = 0. Proof. µ(A \ A) = µ(A)− µ(A) = 0. � Therefore the property µ(∅) = 0 can be substituted with the existence in A of a set with a finite measure. Theorem 3.3 (Monotonicity of outer measure). If A ⊂ B then µ∗(A) 6 µ∗(B). Proof. Any covering of B is a covering of A. � Theorem 3.4 (σ-semiadditivity of µ∗). µ∗( ⋃∞ j=1Aj) 6 ∑∞ j=1 µ ∗(Aj). 8 Proof. If the series in the right-hand side diverges, there is nothing to prove. So assume that it is convergent. By the definition of outer measur for any ε > 0 and for any j there exists an A-covering⋃ k Ekj ⊃ Aj such that ∞∑ k=1 µ(Ekj) < µ ∗(Aj) + ε 2j . Since ∞⋃ j,k=1 Ekj ⊃ ∞⋃ j=1 Aj, the definition of µ∗ implies µ∗( ∞⋃ j=1 Aj) 6 ∞∑ j,k=1 µ(Ekj) and therefore µ∗( ∞⋃ j=1 Aj) < ∞∑ j=1 µ∗(Aj) + ε. � 3.3 Measurable Sets Let A be an algebra of subsets of X, µ a measure on it, µ∗ the outer measure defined in the previous section. Definition 3.3 A ⊂ X is called a measurable set (by Carathe`odory) if for any E ⊂ X the following relation holds: µ∗(E) = µ∗(E ∩ A) + µ∗(E ∩ Ac). Denote by A˜ the collection of all set which are measurable by Carathe`odory and set µ˜ = µ∗ � A˜. Remark Since E = (E ∩ A) ∪ (E ∩ Ac), due to semiadditivity of the outer measure µ∗(E) ≤ µ∗(E ∩ A) + µ∗(E ∩ Ac). Theorem 3.5 A˜ is a σ-algebra containing A, and µ˜ is a measure on A˜. 9 Proof. We devide the proof into several steps. 1. If A,B ∈ A˜ then A ∪B ∈ A˜. By the definition one has µ∗(E) = µ∗(E ∩B) + µ∗(E ∩Bc). (1) Take E ∩ A instead of E: µ∗(E ∩ A) = µ∗(E ∩ A ∩B) + µ∗(E ∩ A ∩Bc). (2) Then put E ∩ Ac in (1) instead of E µ∗(E ∩ Ac) = µ∗(E ∩ Ac ∩B) + µ∗(E ∩ Ac ∩Bc). (3) Add (2) and (3): µ∗(E) = µ∗(E ∩ A ∩B) + µ∗(E ∩ A ∩Bc) + µ∗(E ∩ Ac ∩B) + µ∗(E ∩ Ac ∩Bc). (4) Substitute E ∩ (A ∪B) in (4) instead of E. Note that 1) E ∩ (A ∪B) ∩ A ∩B = E ∩ A ∩B 2) E ∩ (A ∪B) ∩ Ac ∩B = E ∩ Ac ∩B 3) E ∩ (A ∪B) ∩ A ∩Bc = E ∩ A ∩Bc 4) E ∩ (A ∪B) ∩ Ac ∩Bc = ∅. One has µ∗(E ∩ (A ∪B)) = µ∗(E ∩ A ∩B) + µ∗(E ∩ Ac ∩B) + µ∗(E ∩ A ∩Bc). (5) From (4) and (5) we have µ∗(E) = µ∗(E ∩ (A ∪B)) + µ∗(E ∩ (A ∪B)c). 2. If A ∈ A˜ then Ac ∈ A˜. The definition of measurable set is symmetric with respect to A and Ac. Therefore A˜ is an algebra of sets. 3. Let A,B ∈ A, A ∩B = ∅. From (5) µ∗(E ∩ (A unionsqB)) = µ∗(E ∩ Ac ∩B) + µ∗(E ∩ A ∩Bc) = µ∗(E ∩B) + µ∗(E ∩ A). 10 4. A˜ is a σ-algebra. From the previous step, by induction, for any finite disjoint collection (Bj) of sets: µ∗(E ∩ ( n⊔ j=1 Bj)) = n∑ j=1 µ∗(E ∩Bj). (6) Let A = ⋃∞ j=1Aj, Aj ∈ A. Then A = ⋃∞ j=1Bj, Bj = Aj \ ⋃j−1 k=1Ak and Bi ∩Bj = ∅ (i 6= j). It suffices to prove that µ∗(E) > µ∗(E ∩ ( ∞⊔ j=1 Bj)) + µ ∗(E ∩ ( ∞⊔ j=1 Bj) c). (7) Indeed, we have already proved that µ∗ is σ-semi-additive. Since A˜ is an algebra, it follows that ⊔n j=1Bj ∈ A˜(∀n ∈ N) and the following inequality holds for every n: µ∗(E) > µ∗(E ∩ ( n⊔ j=1 Bj)) + µ ∗(E ∩ ( n⊔ j=1 Bj) c). (8) Since E ∩ (⊔∞j=1Bj)c ⊂ E ∩ (⊔nj=1Bj)c, by monotonicity of the mesasure and (8) µ∗(E) ≥ n∑ j=1 µ∗(E ∩Bj) + µ∗(E ∩ Ac). (9) Passing to the limit we get µ∗(E) ≥ ∞∑ j=1 µ∗(E ∩Bj) + µ∗(E ∩ Ac). (10) Due to semiadditivity µ∗(E ∩ A) = µ∗(E ∩ ( ∞⊔ j=1 Bj)) = µ ∗( ∞⊔ j=1 (E ∩Bj)) ≤ ∞∑ j=1 µ∗(E ∩Bj). Compare this with (10): µ∗(E) ≥ µ∗(E ∩ A) + µ∗(E ∩ Ac). Thus, A ∈ A˜, which means that A˜ is a σ-algebra. 5. µ˜ = µ∗ � A˜ is a measure. 11 We need to prove only σ-additivity. Let E = ⊔∞ j=1Aj. From(10) we get µ∗( ∞⊔ j=1 Aj) > ∞∑ j=1 µ∗(Aj). The oposite inequality follows from σ-semiadditivity of µ∗. 6. A˜ ⊃ A. Let A ∈ A, E ⊂ X. We need to prove: µ∗(E) > µ∗(E ∩ A) + µ∗(E ∩ Ac). (11) If E ∈ A then (11) is clear since E ∩ A and E ∩ Ac are disjoint and both belong to A where µ∗ = µ and so is additive. For E ⊂ X for ∀ε > 0 there exists a A-covering (Ej) of E such that µ∗(E) + ε > ∞∑ j=1 µ(Ej). (12) Now, since Ej = (Ej ∩ A) ∪ (Ej ∩ Ac), one has µ(Ej) = µ(Ej ∩ A) + µ(Ej ∩ A) and also E ∩ A ⊂ ∞⋃ j=1 (Ej ∩ A) E ∩ Ac ⊂ ∞⋃ j=1 (Ej ∩ Ac) By monotonicity and σ-semiadditivity µ∗(E ∩ A) 6 ∞∑ j=1 µ(Ej ∩ A), µ∗(E ∩ Ac) 6 ∞∑ j=1 µ(Ej ∩ Ac). Adding the last two inequalities we obtain µ∗(E ∩ A) + µ∗(E ∩ Ac) ≤ ∞∑ j=1 µ∗(Ej) < µ∗(E) + ε. Since ε > 0 is arbitrary, (11) is proved. � The following theorem is a direct consequence of the previous one. 12 Theorem 3.6 Let A be an algebra of subsets of X and µ be a measure on it. Then there exists a σ-algebra A1 ⊃ A and a measure µ1 on A1 such that µ1 � A = µ. Remark. Consider again an algebra A of subsets of X. Denot by Aσ the generated σ-algebra and construct the extension µσ of µ on Aσ. This extension is called minimal extension of measure. Since A˜ ⊃ A therefore Aσ ⊂ A˜. Hence one can set µσ = µ˜ � Aσ. Obviously µσ is a minimal extension of µ. It always exists. On can also show (see below) that this extension is unique. Theorem 3.7 Let µ be a measure on an algebra A of subsets of X, µ∗ the corresponding outer measure. If µ∗(A) = 0 for a set A ⊂ X then A ∈ A˜ and µ˜(A) = 0. Proof. Clearly, it suffices to prove that A ∈ A˜. Further, it suffices to prove that µ∗(E) > µ∗(E ∩ A) + µ∗(E ∩ Ac). The latter statement follows from monotonicity of µ∗. Indeed, one has µ∗(E ∩ A) 6 µ∗(A) = 0 and µ∗(E ∩ Ac) 6 µ∗(E). � Definition 3.4 A measure µ on an algebra of sets A is called complete if conditions B ⊂ A, A ∈ A, µ(A) = 0 imply B ∈ A and µ(B) = 0. Corollary. µ˜ is a complete measure. Definition 3.5 A measure µ on an algebra A is called finite if µ(X) < ∞. It is called σ-finite if the is an increasing sequence (Fj)j≥1 ⊂ A such that X = ⋃ j Fj and µ(Fj) <∞ ∀j. Theorem 3.8 Let µ be a σ-finite measure on an algebra A. Then there exist a unique extension of µ to a measure on A˜. Proof. It suffices to sjow uniqueness. Let ν be another extension of µ (ν � A = µ � A). First, let µ (and therefore ν, µ∗) be finite. Let A ∈ A˜. Let (Ej) ⊂ A such that A ⊂ ⋃j Ej. We have ν(A) ≤ ν( ∞⋃ j=1 Ej) ≤ ∞∑ j=1 ν(Ej) = ∞∑ j=1 µ(Ej). Therefore ν(A) ≤ µ∗(A) ∀A ∈ A˜. 13 Since µ∗ and ν are additive (on A˜) it follows that µ∗(A) + µ∗(Ac) = ν(A) + ν(Ac). The terms in the RHS are finite and ν(A) ≤ µ∗(A), ν(Ac) ≤ µ∗(Ac). From this we infer that ν(A) = µ∗(A) ∀A ∈ A˜. Now let µ be σ-finite, (Fj) be an increasing sequence of sets from A such that µ(Fj) < ∞ ∀j and X = ⋃∞j=1 Fj. From what we have already proved it follows that µ∗(A ∩ Fj) = ν(A ∩ Fj) ∀A ∈ A˜. Therefore µ∗(A) = lim j µ∗(A ∩ Fj) = lim j ν(A ∩ Fj) = ν(A). � Theorem 3.9 (Continuity of measure). Let A be a σ-algebra with a measure µ, {Aj} ⊂ A a monotonically increasing sequence of sets. Then µ( ∞⋃ j=1 Aj) = lim j→∞ µ(Aj). Proof. One has: A = ∞⋃ j=1 Aj = ∞⊔ j=2 (Aj+1 \ Aj) unionsq A1. Using σ-additivity and subtractivity of µ, µ(A) = ∞∑ j=1 (µ(Aj+1)− µ(Aj)) + µ(A1) = lim j→∞ µ(Aj). � Similar assertions for a decreasing sequence of sets in A can be proved using de Morgan formulas. Theorem 3.10 Let A ∈ A˜. Then for any ε > 0 there exists Aε ∈ A such that µ∗(A4 Aε) < ε. Proof. 1. For any ε > 0 there exists an A cover ⋃ Ej ⊃ A such that∑ j µ(Ej) < µ ∗(A) + ε 2 = µ˜(A) + ε 2 . 14 On the other hand, ∑ j µ(Ej) > µ˜( ⋃ j EJ). The monotonicity of µ˜ implies µ˜( ∞⋃ j=1 EJ) = lim n→∞ µ˜( n⋃ j=1 Ej), hence there exists a positive integer N such that µ˜( ∞⋃ j=1 Ej)− µ˜( N⋃ j=1 Ej) < ε 2 . (13) 2. Now, put Aε = N⋃ j=1 Ej and prove that µ∗(A4 Aε) < ε. 2a. Since A ⊂ ∞⋃ j=1 Ej, one has A \ Aε ⊂ ∞⋃ j=1 Ej \ Aε. Since Aε ⊂ ∞⋃ j=1 Ej, one can use the monotonicity and subtractivity of µ˜. Together with estimate (13), this gives µ˜(A \ Aε) ≤ µ˜( ∞⋃ j=1 Ej \ Aε) < ε 2 . 2b. The inclusion Aε \ A ⊂ ∞⋃ j=1 Ej \ A implies µ˜(Aε \ A) 6 µ˜( ∞⋃ j=1 Ej \ A) = µ˜( ∞⋃ j=1 Ej)− µ˜(A) < ε 2 . 15 Here we used the same properties of µ˜ as above and the choice of the cover (Ej). 3. Finally, µ˜(A4 Aε) 6 µ˜(A \ Aε) + µ˜(Aε \ A). � 16 4 Monotone Classes and Uniqueness of Extension of Measure Definition 4.1 A collection of sets, M is called a monotone class if together with any monotone sequence of sets M contains the limit of this sequence. Example. Any σ-ring. (This follows from the Exercise 1. below). Exercises. 1. Prove that any σ-ring is a monotone class. 2. If a ring is a monotone class, then it is a σ-ring. We shall denote by M(K) the minimal monotone class containing K. Theorem 4.1 Let K be a ring of sets, Kσ the σ-ring generated by K. ThenM(K) = Kσ. Proof. 1. Clearly,M(K) ⊂ Kσ. Now, it suffices to prove thatM(K) is a ring. This follows from the Exercise (2) above and from the minimality of Kσ. 2. M(K) is a ring. 2a. For B ⊂ X, set KB = {A ⊂ X : A ∪B,A ∩B,A \B,B \ A ∈M(K)}. This definition is symmetric with respect to A and B, therefore A ∈ KB implies B ∈ KA. 2b. KB is a monotone class. Let (Aj) ⊂ KB be a monotonically increasing sequence. Prove that the union, A = ⋃ Aj belongs to KB. Since Aj ∈ KB, one has Aj ∪B ∈ KB, and so A ∪B = ∞⋃ j=1 (Aj ∪B) ∈M(K). In the same way, A \B = ( ∞⋃ j=1 Aj) \B = ∞⋃ j=1 (Aj \B) ∈M(K); 17 B \ A = B \ ( ∞⋃ j=1 Aj) = ∞⋂ j=1 (B \ Aj) ∈M(K). Similar proof is for the case of decreasing sequence (Aj). 2c. If B ∈ K then M(K) ⊂ KB. Obviously, K ⊂ KB. Together with minimality of M(K), this implies M(K) ⊂ KB. 2d. If B ∈M(K) then M(K) ⊂ KB. Let A ∈ K. Then M(K) ⊂ KA. Thus if B ∈M(K), one has B ∈ KA, so A ∈ KB. Hence what we have proved is K ⊂ KB. This implies M(K) ⊂ KB. 2e. It follows from 2a. — 2d. that if A,B ∈M(K) then A ∈ KB and so A ∪ B, A ∩ B, A \B and B \ A all belong to M(K). � Theorem 4.2 Let A be an algebra of sets, µ and ν two measures defined on the σ- algebra Aσ generated by A. Then µ � A = ν � A implies µ = ν. Proof. Choose A ∈ Aσ, then A = limn→∞An, An ∈ A, for Aσ =M(A). Using continuity of measure, one has µ(A) = lim n→∞ µ(An) = lim n→∞ ν(An) = ν(A). � Theorem 4.3 Let A be an algebra of sets, B ⊂ X such that for any ε > 0 there exists Aε ∈ A with µ∗(B 4 Aε) < ε. Then B ∈ A˜. Proof. 1. Since any outer measure is semi-additive, it suffices to prove that for any E ⊂ X one has µ∗(E) > µ∗(E ∩B) + µ∗(E ∩Bc). 2a. Since A ⊂ A˜, one has µ∗(E ∩ Aε) + µ∗(E ∩ Acε) 6 µ∗(E). (14) 2b. Since A ⊂ B ∪ (A 4 B) and since the outer measure µ∗ is monotone and semi- additive, there is an estimate |µ∗(A)− µ∗(B)| 6 µ∗(A4 B) for any A,B ⊂ X. (C.f. the proof of similar fact for measures above). 2c. It follows from the monotonicity of µ∗ that |µ∗(E ∩ Aε)− µ∗(E ∩B)| 6 µ∗((E ∩ Aε)4 (E ∩B)) 6 µ(Aε ∩B) < ε. 18 Therefore, µ∗(E ∩ Aε) > µ∗(E ∩B)− ε. In the same manner, µ∗(E ∩ Acε) > µ∗(E ∩Bc)− ε. 2d. Using (14), one obtains µ∗(E) > µ∗(E ∩B) + µ∗(E ∩Bc)− 2ε. � 19 5 The Lebesgue Measure on the real line R1 5.1 The Lebesgue Measure of Bounded Sets of R1 Put A for the algebra of all finite unions of semi-segments (semi-intervals) on R1, i.e. all sets of the form A = k⋃ j=1 [aj, bj). Define a mapping µ : A −→ R by: µ(A) = k∑ j=1 (bj − aj). Theorem 5.1 µ is a measure. Proof. 1. All properties including the (finite) additivity are obvious. The only thing to be proved is the σ-additivity. Let (Aj) ⊂ A be such a countable disjoint family that A = ∞⊔ j=1 Aj ∈ A. The condition A ∈ A means that ⊔Aj is a finite union of intervals. 2. For any positive integer n, n⋃ j=1 Aj ⊂ A, hence n∑ j=1 µ(Aj) 6 µ(A), and ∞∑ j=1 µ(Aj) = lim n→∞ n∑ j=1 µ(Aj) 6 µ(A). 3. Now, let Aε a set obtained from A by the following construction. Take a connected component of A. It is a semi-segment of the form [s, t). Shift slightly on the left its right-hand end, to obtain a (closed) segment. Do it with all components of A, in such a way that µ(A) < µ(Aε) + ε. (15) 20 Apply a similar procedure to each semi-segment shifting their left end point to the left Aj = [aj,