This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: The RiemannStieltjes Integral Written by MenGen Tsai email: [email protected] 1. Suppose α increases on [ a,b ], a ≤ x ≤ b , α is continuous at x , f ( x ) = 1, and f ( x ) = 0 if x 6 = x . Prove that f ∈ R ( α ) and that R fdα = 0. Proof: Note that L ( P,f,α ) = 0 for all partition P of [ a,b ]. Thus Z a b fdα = 0 . Take a partition P such that P = { a,a + 1 n ( b a ) ,...,a + k n ( b a ) ,...,a + n 1 n ( b a ) ,b } for all N 3 n > 1. Thus U ( P,f,α ) = n X i =1 M i Δ α i ≤ 2( b a ) n for all N 3 n > 1. Thus ≤ inf U ( P,f,α ) ≤ 2( b a ) n for all n ∈ N . Thus inf U ( P,f,α ) = 0 . Hence R a b fdα = 0; thus, R a b fdα = 0. 2. Suppose f ≥ 0, f is continuous on [ a,b ], and R b a f ( x ) dx = 0. Prove that f ( x ) = 0 for all x ∈ [ a,b ]. (Compare this with Exercise 1.) Proof: Suppose not, then there is p ∈ [ a,b ] such that f ( p ) > 0. Since f is continuous at x = p , for = f ( p ) / 2, there exist δ > 0 such that  f ( x ) f ( p )  < whenever x ∈ ( x δ,x + δ ) T [ a,b ], that is, < 1 2 f ( p ) < f ( x ) < 3 2 f ( p ) 1 for x ∈ B r ( p ) ⊂ [ a,b ] where r is small enough. Next, consider a partition P of [ a,b ] such that P = { a,p r 2 ,p + r 2 ,b } . Thus L ( P,f ) ≥ r · 1 2 f ( p ) = rf ( p ) 2 . Thus sup L ( P,f ) ≥ L ( P,f ) ≥ rf ( p ) 2 > , a contradition since R b a f ( x ) dx = sup L ( P,f ) = 0. Hence f = 0 for all x ∈ [ a,b ]. Note: The above conclusion holds under the condition that f is con tinuous. If f is not necessary continuous, then we cannot get this conclusion. (A counterexample is shown in Exercise 6.1)....
View
Full
Document
This note was uploaded on 11/07/2009 for the course MATH 410 taught by Professor Ddd during the Spring '09 term at Punjab Engineering College.
 Spring '09
 ddd

Click to edit the document details