1873_solutions

# A b such that f x 0 whenever a number x lies outside

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Unformatted text preview: f  is the Cantor function then 1 Solution: The solutions to Exercises 3, 4, and 5 are given after the solution to Exercise 6 below. 6. Prove that if  is the Cantor function then Þ 0  x d x 1 295  1. 2 Once again we recall that C is the intersection of the family of sets E n that were defined in our discussion of the Cantor set and that each set E n is the union of 2 n closed intervals. Look, once again, at the set E 2 : 1 9 1 3 2 3 8 7 1 9 9 The function  takes the value 1 at each of the points 1 and 2 and takes the value 1 at each of 4 9 9 2 the points 1 and 2 and takes the value 3 at each of the points 7 and 8 . Therefore, if I is any one 4 9 9 3 3 of the four component intervals of E 2 then we have var , I  1 . 4 More generally we may see that for each positive integer n, if I is any one of the 2 n component intervals of E n then var , I  1n . 2 For each natural number n we define P n to be the partition of 0, 1 whose points are the endpoints of the component intervals of the set E n . For example, P 2  0, 1 , 2 , 1 , 2 , 7 , 8 , 1 . 993399 For each n, if the partition P n is expressed as P  x 0 , x 1 , , x 2 n1 then we define two step functions s n and S n on 0, 1 by defining sn xj  Sn xj   xj n and by defining for every j  0, 1, 2, , 2 sn x   xj 1 and Sn x   xj whenever xj 1  x  xj 0 2 9 xj xj−1 0 1 We observe that s n  S n . Now given any two consecutive points x j 1 and x j of the partition P n there are two possibilities: Either the interval x j 1 , x j is a component interval of the set E n ; in which case var , x j 1 , x j  1n 2 or the interval x j 1 x j is a gap between two component intervals of E n ; in which case var , x j 1 , x j  0. We deduce that Þ0 Sn 1 1 1  1. 4n 2n 2n Since the latter expression approaches 0 as n Ý we know that the pair of sequences s n and S n squeezes  with respect to  and so  is Riemann-Stieltjes integrable with respect to itself on 1 the interval 0, 1 . To find the value of the integral Þ d we observe that for each n we have s n d  0 Þ 0 S n d  1 2n j1 j 2n 1 2n and so 296  1n 4 2n 2n  1 2 Þ 0 S n d  1 lim nÝ 1. 2 Therefore Þ 0 d  1 1. 2 1 The fact that this integral is 2 becomes obvious after one has studied the integration by parts identity. As a matter of fact, if  is any increasing continuous function on an interval a, b then 2 2 b Þ a d   b 2  a . Integrating with Respect to the Cantor Function Solutions to Exercises 3, 4 and 5 Quick Review of the Cantor set and the Cantor function The Cantor set C is the set of all those real numbers that can be expressed in the form Ý an 3n n1 where a n is a sequence in the set 0, 2 . The least member of the Cantor set is Ý 0 0 3n n1 and the greatest member is Ý The least member Ý an n0 3 n 2  1. 3n n1 of C for which a 1  2 is Ý n2 and the greatest member for which a 1  0 is 2 3 2  1. 3 3n Thus the set C is included in the set E 1  0, 1 3 Þ 2 ,1 3 1 3 0 2 3 1 By looking at the two cases...
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