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Unformatted text preview: The RiemannStieltjes Integral RiemannStieltjes integrals 7.1 Prove that a b d b a , directly from Definition 7.1. Proof : Let f 1 on a , b , then given any partition P a x ,..., x n b , then we have S P ,1, k 1 n f t k k , where t k x k 1 , x k k 1 n k b a . So, we know that a b d b a . 7.2 If f R on a , b and if a b fd 0 for every f which is monotonic on a , b , prove that must be constant on a , b . Proof : Use integration by parts, and thus we have a b df f b b f a a Given any point c a , b , we may choose a monotonic function f defined as follows. f 0 if x c 1 if x c . So, we have a b df c b . So, we know that is constant on a , b . 7.3 The following definition of a RiemannStieltjes integral is often used in the literature: We say that f is integrable with respect to if there exists a real number A having the property that for every 0, there exists a 0 such that for every partition P of a , b with norm P and for every choice of t k in x k 1 , x k , we have  S P , f , A  . (a) Show that if a b fd exists according to this definition, then it is also exists according to Definition 7.1 and the two integrals are equal. Proof : Since refinement will decrease the norm, we know that if there exists a real number A having the property that for every 0, there exists a 0 such that for every partition P of a , b with norm P and for every choice of t k in x k 1 , x k , we have  S P , f , A  . Then choosing a P with P , then for P P P . So, we have  S P , f , A  . That is, a b fd exists according to this definition, then it is also exists according to Definition 7.1 and the two integrals are equal. (b) Let f x x 0 for a x c , f x x 1 for c x b , f c 0, c 1. Show that a b fd exists according to Definition 7.1 but does not exist by this second definition. Proof : Note that a b fd exists and equals 0 according to Definition 7.1If a b fd exists according to this definition, then given 1, there exists a 0 such that for every partition P of a , b with norm P and for every choice of t k in x k 1 , x k , we have  S P , f ,  1. We may choose a partition P a x ,..., x n b with P and c x j , x j 1 , where j 0,. .., n 1. Then S P , f , f x x j 1 x j 1, where x c , x j 1...
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 Fall '09
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