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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 n1
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 RiemannStieltjes 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
j1 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 n1 where a n is a sequence in the set 0, 2 . The least member of the Cantor set is
Ý 0 0
3n n1 and the greatest member is
Ý The least member Ý
an
n0 3 n 2 1.
3n n1 of C for which a 1 2 is
Ý
n2 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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 Fall '08
 STAFF
 Math, Calculus

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