This preview shows page 1. Sign up to view the full content.
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
n1 n1 , 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
n1 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
n1 var , A n .
n1 The desired result follows at once from the fact that 293 Ý . 2n n1 4. Given that E is an elementary set and that A n is a sequence of elementary sets and that
Ý An, E n1 prove that
Ý var , A n . var , E
n1 To obtain a contradiction, assume that
Ý var , E var , A n .
n1 Choose 0 such that Ý Ý var , A n var , E n1 Ý var , A n
n1 n1 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
n1 2n n1 var , U n
n1 which, in view of Exercise 2, is impossible because
Ý Un. E n1 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.
n1 Prove that
Ý var , A n . var , E
n1 Given any postiive integer N we see that
N N var , E var , A n n1 var , A n .
n1 Therefore
Ý N var , E var , A n lim NÝ n1 var , A n ,
n1 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 j1 where each number a j is either 0 or 2. The right endpoint of I is
n
j1 Ý n aj
1n
3
3j aj
2
3 j jn1 3 j j1 and we observe that
n
j1 Ý aj
2
3 j jn1 3 j n
j1 Ý n aj
3j 1
2 j1 Ý
jn1 aj
1
2 j jn1 2 j n 1
2 j1 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 RiemannStieltjes integrable with respect to the Cantor function . Can you give an example of a
function f that is RiemannStieltjes 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...
View
Full
Document
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.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details