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Unformatted text preview: that whenever we specify consistent probability measures μ n on increasing σalgebras A n , then we get a probability measure on the generated σalgebra A = σ ( S n A n ). But consider this example. Let the space be the open unit interval (0 , 1). For each n ∈ N , let A n be the smallest σalgebra that contains the set (0 , 1 n ) and the Borel subsets of [ 1 n , 1). On the measurable space ((0 , 1) , A n ) deﬁne the probability measure μ n by μ n { (0 , 1 n ) } = 1, μ n { [ 1 n , 1) } = 0. Show that the σalgebras A n generate the Borel σalgebra B (0 , 1) , but there is no probability measure μ on ((0 , 1) , B (0 , 1) ) such that, for all n , the restriction of μ to A n is μ n ....
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This note was uploaded on 11/18/2009 for the course MATH 241 taught by Professor Reed during the Spring '09 term at Duke.
 Spring '09
 Reed
 Probability

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