1873_solutions

# Is not closed and that f is a closed elementary

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Unformatted text preview: a 2  0 and a 2  2 we can go a step further and see that C is included in the set E 2  0, 1 Þ 2 , 3 Þ 6 , 7 Þ 8 , 1 9 99 99 9 0 1 9 2 9 1 3 2 3 7 9 8 9 1 and, in general, if n is any positive integer then C is included in the set E n that is the union of 2 n closed intervals of length 1/3 n and whose left endpoints are the numbers n j1 aj 3j where each number a j is either 0 or 2. It is not hard to show that C Ý  En. n1 297 Ý an n1 3 n Ý The value of the Cantor function  at each member Ý  n1 an 3n  n1 of the set C is defined by the equation an . 2 n 1 Thus 0 0 Ý 1  n1 2 3n Ý  n1 2 1 2 n 1  1  2  1 2 3 3 1  2  1  4 9 9 7   8  3.  4 9 9 The function  is a continuous strictly increasing function from C onto the interval 0, 1 . We now extend  to an increasing function from 0, 0 onto 0, 1 by making  constant on every component interval of the open set 0, 1 C. The graph of  is shown in the following figure: Graph of the Cantor Function A General Discussion of the integral Þ x p d x 1 0 The purpose of this document is to explore some integrals of the form Þ x p d x where p is a positive 0 integer. Note that such an integral always exists because the integrand is an increasing function. However, it is worth showing the existence of the integrals directly. We define f x  x p for each x 0, 1 . For each positive integer 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 1 0 1 9 2 9 1 3 2 3 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 p sn xj  Sn xj  xj for every j  0, 1, 2, , 2 n and by defining p sn x  xj 1 and p Sn x  xj whenever 298 7 9 8 9 1 xj xj xj−1 0  x  xj 1 1 We observe that s n f 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. When x j 1 , x j is a component interval of E n we have p p p1 p2 p3 p p1 x j x j 1  1n x j  x j x 1 1  x j x 2 1    x j 1  n j j 3 3 and so 2n Þ0 Sn 1 p s n d  xj 1 2n p xj 1 j1 2n  j1 1 xp 3n j 1 p2 p3 p1 1  xj x1 1  xj x2 1    xj j j 1 2n p  n. 3 Since the latter expression approaches 0 as n Ý we know that the pair of sequences s n and S n squeezes f with respect to  and so f is Riemann-Stieltjes integrable with respect to  on the interval 0, 1 . We deduce that Þ 0 s n d  Þ 0 x p d x 1 lim nÝ 1 We now look at the step functions s n more closely. We write the 2 n left endpoints of the component intervals of E n as c 1 , c 2 , c 3 , , c 2 n n we express c in the form and for each k  1, 2, 3, , 2 k n a kj . 3j ck  j1 Where each of the numbers a kj is eith...
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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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