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Math 321 - Riemann–Stieltjes Integrals

Math 321 - Riemann–Stieltjes Integrals -...

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Riemann–Stieltjes Integrals with α a Step Function Theorem. Let α : [ a, b ] IR be a step function with discontinuities at s 1 < ··· < s n , where a s 1 and s n b . Let f : [ a, b ] be continuous at each s j , 1 j n . Then f ∈ R ( α ) on [ a, b ] and i b a f dα = n s j =1 f ( s j ) b α ( s j +) α ( s j ) B where α ( s + ) = lim t s t>s α ( t ) α ( s - ) = lim t s t<s α ( t ) α ( s ) α ( s ) α ( s +) and, by convention, if s 1 = a α ( s 1 ) = α ( a ) if s n = b α ( s n +) = α ( b ) Proof: Let ε > 0. Choose the partition P ε = ± a = ˜ x 0 < ˜ x 1 < < ˜ x m = b ² so that (1) { s 1 , , s n } ⊂ P ε (2) the norm or mesh of P ε = b P ε b = max 1 i m v v ˜ x i ˜ x i - 1 v v < δ with δ = min ± | s 2 s 1 | , , | s n s n - 1 | , δ 0 ² and δ 0 is given by Insert ( ) here . Now let P = { a = x 0 , x 1 , , x p = b } ⊃ P ε be any partitition Fner than P ε and T = { t 1 , , t p } be any choice for P and consider each term in S ( P, T, f, α ) = p s i =1 f ( t i ) b α ( x i ) α ( x i - 1 ) B ±or each 1 i p , either (1) neither x i nor x i - 1 is in { s 1 , , s n } , in which case both x i and x i - 1 lie in a
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Math 321 - Riemann&ndash;Stieltjes Integrals -...

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