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Unformatted text preview: The RiemannStieltjes Integral Written by MenGen Tsai email: b89902089@ntu.edu.tw 1. Suppose increases on [ a,b ], a x b , is continuous at x , f ( x ) = 1, and f ( x ) = 0 if x = x . Prove that f R ( ) and that fd = 0. Proof: Note that L ( P,f, ) = 0 for all partition P of [ a,b ]. Thus a b fd = 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 n > 1. Thus U ( P,f, ) = n i =1 M i i 2( b a ) n for all N n > 1. Thus inf U ( P,f, ) 2( b a ) n for all n N . Thus inf U ( P,f, ) = 0 . Hence a b fd = 0; thus, a b fd = 0. 2. Suppose f 0, f is continuous on [ a,b ], and 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 > 0 such that  f ( x ) f ( p )  < whenever x ( x , x + ) [ 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 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)....
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 Spring '08
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