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MATH 650. HOMEWORK 5. SOLUTIONS Problem 1. Let X Be MATH 650. HOMEWORK 5. SOLUTIONS Problem 1. Let X be an uncountable set. Let R be the collection of all the finite subsets of X. Given A ∈ R let µ(A) be the number of elements in A. Show that R is a ring and µ is a measure on R. Solution. If A and B are in R, they are finite sets, and so is A ∪ B and A − B. If A, B ∈ R are disjoint, then the number of elements in A∪B equals the number of elements in A plus the number of elements in B, so µ(A ∪ B) = µ(A) + µ(B). ∞ For µ to be a measure, it must be countably additive. Let then {Ai}i=1 ⊂ R S∞ be a countable collection of mutually disjoint sets such that A = i=1 Ai is also in R. Then A and all Ai are finite, and so only finitely many Ai are non-empty. P∞ Therefore i=1 µ(Ai) is a finite sum and ∞ X µ(A) = µ(Ai) i=1 by additivity of µ. Problem 2. Let X be an infinite set and R the collection of all countable subsets of X. (1) Is R a σ-ring? (2) Let µ be a measure on R. Show that there is a function f : X → [0, ∞) such that X µ(A) = f(x), x∈A for all A ∈ R. (3) Show that f in (2) has the following properties: the set {x ∈ X | f(x) 6= 0} is countable, and P f(x) < ∞. (4) Conversely, show that if f : X → [0, ∞) has these two properties, then the formula X µ(A) = f(x) x∈A defines a measure on R. Solution. (1) Yes, a countable union of countable sets is a countable set. (2) Each one-point set {x} ∈ R. If A ∈ R, A is countable, and so it can be written S as a countable disjoint union of one-point sets: A = x∈A{x}. Thus, setting f(x) = µ({x}), the countably additivity of µ gives X µ(A) = f(x). x∈A 1 1 (3) The set {f(x) > 0} is countable. Indeed, let An be the interval An = ( n+1 , n ], ∞ n = 1, 2, ··· and A0 = (1, ∞). Then {An}n=0 is a countable partition of (0, ∞), and so Bn = {x ∈ X | f(x) ∈ An} is a countable partition of {f(x) > 0}. If {f(x) > 0} was uncountable, then at least one of the Bn must be also uncountable (for otherwise {f(x) > 0} would be a countable union of countable sets, hence 1 2 MATH 650. HOMEWORK 5. SOLUTIONS countable). Suppose then that Bn0 is uncountable. Then it contains an infinite countable set C = {x1, x2, ···}. Since C is countable, C ∈ R. But its measure ∞ ∞ X X 1 µ(C) = f(x ) ≥ = ∞, i n + 1 i=1 i=1 0 contradicting the fact that µ(A) is a non-negative number for every A ∈ R. Once we know that Y = {f(x) > 0} is countable, we know that Y ∈ R and so 0 ≤ µ(Y ) < ∞ and X X X f(x) = f(x) = f(x) = µ(Y ) < ∞. x∈X f(x)>0 x∈Y Problem 3. Let X be the real line and R = RLeb. Given A ∈ R, let µ(A) = 1 if, for some positive ε, A contains the interval (0, ε); otherwise let µ(A) = 0. Show that µ is an additive set function, but it is not countably additive. Solution. We have to show that if A, B ∈ R are disjoint, then µ(A ∪ B) = µ(A) + µ(B). A set in R is a finite union of intervals. Let A = A1 ∪ · · · ∪ Am and B = B1 ∪ · · · ∪ Bn, where the Ai’s and Bj’s are intervals. Assume that they are ordered so that the right endpoint of Ai is less than or equal to the left endpoint of Ai+1, i = 1, ··· , m − 1, and similarly for the Bj’s. Also assume that the right endpoint of A1 is smaller that the left endpoint of B1. (You should verify that this can be done.) There are two cases to consider. (1) µ(A∪B) = 1. This means that there exists ε > 0 such that (0, ε) ⊂ A∪B = A1 ∪ · · · ∪ Am ∪ B1 ∪ · · · Bn. But given that A1 is the left-most interval in A ∪ B, this implies that there exists a > 0 such that (0, a) ⊂ A1, and hence also that B ∩ (0, a) = ∅. It follows that µ(A) = 1 and µ(B) = 0. (2) µ(A ∪ B) = 0. This means that A ∪ B does contains no interval of the form (0, ε), and since A, B ⊂ A ∪ B, the same must be true for A and for B. Thus µ(A) = µ(B) = 0. 1 1 To show that µ is not countably additive, let An = ( n+1 , n ], n = 1, 2, ··· , so S∞ 1 1 that µ(An) = 0 for all n. Then A = i=1( n+1 , n ] = (0, 1] is in R, but ∞ X µ(A) = 1 6= µ(Ai) = 0 + 0 + 0 + ··· = 0. i=1 Problem 4. Given any collection C of subsets of X, show that there is a smallest ring R containing C. Solution. There is always a ring which contains C, namely the power set ring 2X . Let R be the intersection of all the rings which contain C. Then R is a non-empty collection of subsets of X. It is also a ring. For if A, B are in R, then they are in every ring which contains C. If S is a ring which contains C, then the union A ∩ B and the difference A − B are also in S. Thus A ∩ B and A − B are in every ring which contains C, that is, they are in R. .
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