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Unformatted text preview: Math 507/420: Measure Theory and Integration (2010) SOLUTIONS Homework Assignment #1 Due: Friday, Sept. 24, at beginning of class. You may use any result from Chapter 0 or Sections 1.1., 1.2. or 1.3 of Folland or estab lished in class. 1. True or False (justify your answer with a proof or counterexample): (a) The intersection of two algebras is an algebra. Solution: Same proof as for σalgebras. (b) The union of two algebras is an algebra. Solution: Let X = { 1 , 2 , 3 , 4 } . Let A be algebra generated by {{ 1 , 2 } , { 3 , 4 }} . Let B be algebra generated by {{ 1 , 3 } , { 2 , 4 }} . Then { 1 , 2 } and { 1 , 3 } are in the union algebra but their union { 1 , 2 , 3 } is not. 2. Let A be an algebra. Show that the following are equivalent: (a) A is a σalgebra (b) A is closed under countable intersections (c) A is closed under countable increasing unions (i.e., if A 1 ⊆ A 2 ⊆ A 3 ... and each A i ∈ A then ∪ i A i ∈ A ) Solution: ( a ) ⇒ ( b ): ∩ n A n = ( ∪ n A c n ) c ( b ) ⇒ ( c ): ∪ i A i = ( ∩ i A c i ) c which belongs to A since A is closed under complements (since A is an algebra) and countable intersections (by assumption). ( c ) ⇒ ( a ): Since A is an algebra, it suffices to show that A is closed under countable unions. Let A i ∈ A and define B n = ∪ n i =1 A i . Since A is an algebra, each B n ∈ A and { B n } is an increasing sequence. Thus, ∪ ∞ i =1 A i = ∪ ∞ n =1 B n ∈ A . 3. Verify that ([0 , 1) , ⊕ ) is a commutative group 1 x ⊕ y = x + y if x + y < 1 x + y 1 if x + y ≥ 1 Solution: The operation is clearly commutative. Closure: In both cases x ⊕ y ∈ [0 , 1). Associativity: 6 cases need to be checked explicitly: 1 this is one of the very few pure abstract algebra problems to be assigned in this course; in case you have forgotten the definition or have never seen it, ask me. 1 – Assume x + y < 1 and y + z ≥ 1. Then 1 ≤ x + y + z < 2. Thus, ( x ⊕ y ) ⊕ z = ( x + y ) ⊕ z = x + y + z 1 and x ⊕ ( y ⊕ z ) = x ⊕ ( y + z 1) = x + y + z 1 – The case x + y ≥ 1 and y + z < 1 follows by symmetry. – The case x + y < 1 and y + z < 1 breaks into two subcases according to whether or not x + y + z < 1 – The case x + y ≥ 1 and y + z ≥ 1 breaks into two subcases according to whether or not x + y + z < 2 Identity: x ⊕ 0 = x Inverse: x ⊕ (1 x ) = 0 4. Show that a σalgebra must be either finite or uncountable (i.e., cannot be countably infinite)....
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 Spring '05
 feritöztürk
 Math

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