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B if a b r then every upper bound of b is also an

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(b) If A B R , then every upper bound of B is also an upper bound of A ; in particular sup B is an upper bound of A , hence sup B sup A . Thus also μ ( B ) = max(sup B, 0) max(sup A, 0) = μ ( A ) . (c) Let { A n } be a sequence of subsets of R . Let x S n =1 A n . Then there is m N such that x A m , hence x sup A m μ ( A m ) sup n N μ ( A n ) X n =1 μ ( A n ) . Thus n =1 μ ( A n ) is an upper bound of S n =1 A n , hence sup S n =1 A n n =1 μ ( A n ). Moreover, n =1 μ ( A n ) 0, thus μ ˆ [ n =1 A n ! = max ˆ sup [ n =1 A n , 0 ! X n =1 μ ( A n ) . Simple, isn’t it? 2
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• Spring '11
• Speinklo
• Topology, Empty set, Metric space, Open set, Topological space, Closed set

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