1873_solutions

# 1873_solutions

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Unformatted text preview: s Given any closed subset H of E we know from Exercise 1 that Ý var , U n . var , H Since Ý var n1 n1 , U n is an upper bound of the set var , H H is elementary and closed and H E and since var , E is the least upper bound of this set we have Ý var , U n . var , E n1 3. Given that A n is a sequence of elementary sets and that  0 and that the series var , A n is convergent, and given that for each positive integer n the set U n is an open elementary set that includes A n and satisfies the inequality var , U n  var , A n  n , 2 prove that Ý Ý var , U n  n1 var , A n  . n1 The desired result follows at once from the fact that 293 Ý . 2n n1 4. Given that E is an elementary set and that A n is a sequence of elementary sets and that Ý  An, E n1 prove that Ý var , A n . var , E n1 To obtain a contradiction, assume that Ý var , E  var , A n . n1 Choose  0 such that Ý Ý var , A n  var , E   n1 Ý var , A n  n1 n1 2n . For each n, choose an open elementary set U n that includes A n such that var , U n  var , A n  n . 2 We see that Ý Ý Ý var , A n  var , E  n1 2n n1  var , U n n1 which, in view of Exercise 2, is impossible because Ý  Un. E n1 5. Suppose that E is an elementary set and that A n is a sequence of elementary sets with the property that whenever i and j are positive integers and i j we have Ai Aj  . Suppose that E is an elementary set and that E Ý  An. n1 Prove that Ý var , A n . var , E  n1 Given any postiive integer N we see that N N var , E var ,  A n  n1 var , A n . n1 Therefore Ý N var , E var , A n lim NÝ n1  var , A n , n1 and the desired result therefore follows from Exercise 4. Some Exercises on Integrals with Respect to the Cantor 294 Function 1. Prove the claim that was made earlier that if  is the Cantor function and I is one of the component intervals of the elementary set E n then var , I  1n . 2 Solution: We can express the left endpoint of I in the form n aj 3j j1 where each number a j is either 0 or 2. The right endpoint of I is n j1 Ý n aj  1n  3 3j aj 2  3 j jn1 3 j j1 and we observe that n  j1 Ý aj 2  3 j jn1 3 j n  j1 Ý n aj 3j 1 2 j1 Ý  jn1 aj 1  2 j jn1 2 j n 1 2 j1 aj 2j 1  1. 2n 2j 2. The preceding examples show that a function f can be Riemann integrable on the interval 0, 1 even though it fails to be Riemann-Stieltjes integrable with respect to the Cantor function . Can you give an example of a function f that is Riemann-Stieltjes integrable with respect to  on 0, 1 but fails to be Riemann integrable on 0, 1 ? We can make use of the fact that the Cantor function is constant in the interval 1 , 2 . We define 33 0 if x is irrational fx  0 if x 0, 1 1 3 , 1 if x is rational and x 2 3 1 3 , 2 3 3. Prove that if  is the Cantor function then Þ 0 x 2 d x  3. 8 Þ 0 x 3 d x  5. 16 Þ 0 x 3 d x  39 . 160 1 Solution: 4. Prove that if  is the Cantor function then 1 Solution: 5. Prove that i...
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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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