1873_solutions

Elementary set that is not closed and that f is a

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Unformatted text preview: nn1 32 For each n we see easily that Þ0 l Pn, f 1 Þ0 u Pn, f 1 0 and 1 n and therefore Þ0 l Pn, f 1 lim nÝ nÝ lim Þ0 u Pn, f 1  0. 2. Suppose that f is defined on the interval in such a way that whenever x 0, 1 and x has the form 1 for some n 1 positive integer n we have f x  0 and whenever x belongs to an interval of the form n1 , 1 for some n positive integer n we have f x  1  1 n. Draw a rough sketch of the graph of this function and explain why it is integrable on the interval 0, 1 . Solution: 0 11 1 65 4 1 2 1 3 1 For each positive integer n we define P n to be the following partition of the interval 0, 1 : P n  0, 1 , 1 , , 1 , 1 , 1 . nn1 32 For each n we see easily that Þ0 l Pn, f 1 n1  j1 1 j1 1 j 1 1 j and Þ0 u Pn, f  2  n 1 n1 j1 and so 271 1 j 1 j1 1 1 j Þ0 u Pn, f Þ0 l Pn, f 1 lim nÝ 1 nÝ2 0 lim n and we have shown that f is integrable on 0, 1 . Incidentally, we have also shown that n1 lim Þ0 f  n Ý 1 1 j j1 1 j1 1 1 j n nÝ lim 2 2j 2j  1 j1 n nÝ lim 1 j 2j  1 j1 In the chapter on infinite series you will learn how to show that the latter limit is 2 2 log 2. 3. Given that f is a bounded function on an interval a, b , prove that the following conditions are equivalent: a. The function f is integrable on the interval a, b . b. For every number  0 there exist step functions s and S on the interval a, b such that s Þa S b c. For every number such that if S and f S and s.  0 there exist step functions s and S on the interval a, b such that s E x f a, b Sx sx , we have m E  . To show that condition a implies condition b we assume that condition a holds. In other words, we assume that f is integrable on a, b . Suppose that  0. Using the first criterion for integrability we choose a partition P of a, b such that Þ a w P, f b . We define S  u P, f and s  l P, f and observe that the functions s and S have the desired properties. The proof that condition a implies condition c is very similar. This time, the partition P is chosen using the second criterion for integrability. To prove that condition b implies condition a we assume that condition b holds. What we shall show is that the first criterion for integrability holds. Suppose that  0. Using condition b we choose step functions s and S on the interval a, b such that s f S and Þa S b s. Choose a partition P of a, b such that both s and S step within P and observe that since s l P, f u P, f S we have Þ a w P, f b  Þ a u P, f b Þa S b l P, f s. The proof that condition c implies condition a is very similar. This time we show that f is integrable by showing that the second criterion for integrability holds. 4. Suppose that f is a bounded function on an interval a, b and that for every number  0 there exists an elementary subset E of a, b such that m E  and such that the function f 1  E is Riemann integrable on a, b . Prove that f must be Riemann integrable on the interval a, b . 272 Solution: In order to show that f is integrabl...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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