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Rudin_6 - The Riemann-Stieltjes Integral Written by Men-Gen...

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The Riemann-Stieltjes Integral Written by Men-Gen Tsai email: [email protected] 1. Suppose α increases on [ a, b ], a x 0 b , α is continuous at x 0 , f ( x 0 ) = 1, and f ( x ) = 0 if x = x 0 . Prove that f ∈ R ( α ) and that fdα = 0. Proof: Note that L ( P, f, α ) = 0 for all partition P of [ a, b ]. Thus 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 n > 1. Thus U ( P, f, α ) = n i =1 M i Δ α i 2( b - a ) n for all N n > 1. Thus 0 inf U ( P, f, α ) 2( b - a ) n for all n N . Thus inf U ( P, f, α ) = 0 . Hence a b fdα = 0; thus, a b fdα = 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 δ > 0 such that | f ( x ) - f ( p ) | < whenever x ( x - δ, x + δ ) [ a, b ], that is, 0 < 1 2 f ( p ) < f ( x ) < 3 2 f ( p ) 1

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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 > 0 , 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 counter-example is shown in Exercise 6.1). 3. 4. If f ( x ) = 0 for all irrational x , f ( x ) = 1 for all rational x , prove that f / R on [ a, b ] for any a < b . Proof: Take any partition P of [ a, b ], say a = x 0 x 1 .. x n - 1 x n = b. By P we can construct the new partition P without repeated points, and U ( P, f ) = U ( P f ), L ( P, f ) = L ( P , f ). Say P a = y 0 < y 1 < ... < y m - 1 < y m = b.
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