The Riemann Stieltjes Integral Riemann Stieltjes integrals b 7 1 Prove that d b a directly from Definition 7 1 a Proof Let f 1 on a b then given any partition P a x 0 x n b then we have n S P 1 f t k k where t k x k 1 x k k 1 n k k 1 b a b So we know that d b a a 7 2 If f R on a b and if ab fd 0 for every f which is monotonic on a b prove that must be constant on a b Proof Use integration by parts and thus we have b a df f b b f a a Given any point c a b we may choose a monotonic function f defined as follows f 0 if x c 1 if x c So we have b a df c b So we know that is constant on a b 7 3 The following definition of a Riemann Stieltjes integral is often used in the literature We say that f is integrable with respect to if there exists a real number A having the property that for every 0 there exists a 0 such that for every partition P of a b with norm P and for every choice of t k in x k 1 x k we have S P f A b a Show that if fd exists according to this definition then it is also exists according a to Definition 7 1 and the two integrals are equal Proof Since refinement will decrease the norm we know that if there exists a real number A having the property that for every 0 there exists a 0 such that for every partition P of a b with norm P and for every choice of t k in x k 1 x k we have S P f A Then choosing a P with P then for P P P So we have S P f A b That is fd exists according to this definition then it is also exists according to a Definition 7 1 and the two integrals are equal b Let f x x 0 for a x c f x x 1 for c x b b f c 0 c 1 Show that fd exists according to Definition 7 1 but does not exist a by this second definition b b Proof Note that fd exists and equals 0 according to Definition 7 1If fd exists a a according to this definition then given 1 there exists a 0 such that for every partition P of a b with norm P and for every choice of t k in x k 1 x k we have S P f 1 We may choose a partition P a x 0 x n b with P and c x j x j 1 where j 0 n 1 Then S P f f x x j 1 x j 1 where x c x j 1 which contradicts to S P f 1 7 4 If f R according to Definition 7 1 prove that ab f x dx also exists according to definition of Exercise 7 3 Contrast with Exercise 7 3 b b Hint Let I f x dx M sup f x x a b Given 0 choose P so that a U P f I 2 notation of section 7 11 Let N be the number of subdivision points in P and let 2MN If P write U P f M k f x k S 1 S 2 where S 1 is the sum of terms arising from those subintervals of P containing no points of P and S 2 is the sum of remaining terms Then S 1 U P f I 2 and S 2 NM P NM 2 and hence U P f I Similarly L P f I if P for some Hence S P f I if P min Proof The hint has proved it Remark There are some exercises related with Riemann integrals we write thme as references b 1 Suppose that f 0 and f is continuous on a b and f x dx 0 Prove that a f x 0 on a b Proof Assume that there is a point c a b such that f c 0 Then by continuity of f c f we know that given 2 0 there is a 0 such that as x c x a b we have f c f x f c 2 which implies that f c f x if x c c a b I 2 So we have b f c 0 I f x dx f x dx 0 where 0 I the length of I 2 I a which is absurb Hence we obtain that f x 0 on a b 2 Let f be a continuous function defined on a b Suppose that for every continuous function g defined on a b which satisfies that b a g x dx 0 we always have b a f x g x dx 0 Show that f is a constant function on a b b Proof Let f x dx I and define g x f x a I b a then we have b a g x dx 0 which implies that by hypothesis b a f x g x dx 0 which implies that b a f x c g x dx 0 for any real c So we have b a g x 2 dx 0 if letting c I b a which implies that g x 0 forall x a b by 1 That is f x I b a on a b 3 Define 0 if x 0 1 Q h x 1 n if x is the rational number m n in lowest terms 1 if x 0 Then h R 0 1 Proof Note that we have shown that h is continuous only at irrational numbers on 0 1 Q We use it to show that h is Riemann integrable i e h R 0 1 Consider the upper sum U P f as follows Given 0 there exists finitely many points x such that f x 2 Consider a partition P x 0 a x n b so that its subintervals I j x j 1 x j for some j containing those points and I j 2 So we have n U P f M k x k k 1 1 2 2 2 where 1 1 M j I j and 2 is the sum of others So we have shown that f satisfies the Riemann condition with respect to x x Note 1 The reader can show this by Theorem 7 48 Lebesgue s Criterion for Riemann Integrability Also compare Exercise 7 32 and Exercise 4 16 with this 2 In Theorem 7 19 if we can make sure that there is a partition P such that U P f L P f then we automatically have for any finer P P …