exam_1_sol

# exam_1_sol - Solutions to Exam 1 1. Let A Z. Since (a) = 0,...

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Solutions to Exam 1 1. Let A Z . Since μ * ( a ) = 0, it follows that inf n μ ( E n ) = 0, where the inf is taken collections of elementary sets { E n } n> 0 , s.t. n E n is an open cover of A . Thus for any m> 0, there is an open cover of elementary sets { E n } n> 0 for A s.t. n μ ( E n ) < 1 m . Let F m = n E n . Since A F m , the symmetric diFerence S ( A,F m ) F m . But μ * ( S ( A,F m )) μ * ( F m ) < 1 m This approaches 0 as m →∞ , thus F m A . Thus A is measurable. Since A Z was arbitrary, any element of Z is measurable. 2. We need to show that if A,B Z , then A - B Z , and if { A n } n> 0 Z , then n A n Z If A,B Z , then μ ( A ) = 0. ±urthermore, A - B A , so μ ( A - B ) μ ( A ) = 0. Thus μ ( A - B ) = 0, so A - B Z ±or the second property, by subadditivity, μ ( n A n ) ± n μ ( A n )= ± n 0 = 0 Thus Z is a sigma ring. 3. Since Q is countable, let { q n } n> 0 be an enumeration of it. A set con- sisting of any point q n is measurable and has measure 0. By countable additivity, m ( Q )= ± n> 0 m ( { q n } )= ± n> 0 0 = 0 4. We need to show that the set { x | f ( x ) >a } is measurable for all real numbers a . Suppose a> 0. Then since

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## This note was uploaded on 07/15/2008 for the course MATH 118 taught by Professor Ponce during the Spring '08 term at UCSB.

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exam_1_sol - Solutions to Exam 1 1. Let A Z. Since (a) = 0,...

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