Math 321 - Riemann–Stieltjes Integrals

Because p we have sj tij sj and sj ti sj we

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Unformatted text preview: s2 Step Function α’s s3 b So S (P, T, f, α) = case (2) terms + case (3) terms n f (tij ) α(sj ) − α(sj −) + f (ti′ ) α(sj +) − α(sj ) j = (1) j =1 Here tij lies in the subinterval of P whose right hand end point is sj and ti′ lies in the j subinterval of P whose left hand end point is sj . Because P < δ , we have sj − δ < tij ≤ sj and sj ≤ ti′ < sj + δ . We may write the value of the integral given in the statement of j the theorem in a form quite like that of S (P, T, f, α): n n f (sj ) α(sj ) − α(sj −) + f (sj ) α(sj +) − α(sj ) f (sj ) α(sj +) − α(sj −) = j =1 j =1 (2) (The two f (sj )α(sj ) terms cancel.) Subtracting (2) from (1) and using the triangle inequality gives n S (P, T, f, α) − f (sj ) α(sj +) − α(sj −) j =1 n f (tij ) − f...
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This document was uploaded on 02/24/2014 for the course MATH 321 at University of British Columbia.

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