Assignment 6 of Math 5302

Assignment 6 of Math 5302

1. Assume f is a real-valued function defined on [a, b] and f is Lipschitz continuous on [a, b]. Show
that f is absolutely continuous on [a, b].
2. If f is continuous and α is of bounded variation on [a, b]. Then f is Riemann-Stieltjes integrable
with respect to α on [a, b]. Let β(x) = V
x
a
(α) and γ(x) = β(x) − α(x), x ∈ [a, b]. Show that
(a)
 
 
 
 
Z b
a
f(x)dα(x)
 
 
 
 
≤
Z b
a
|f(x)|dβ(x) ≤ max
x∈[a,b]
|f|V
b
a
(α).
(b) The function α is Riemann-Stieltjes integrable with respect to f on [a, b].
3. Given a positive integer n and numbers c0, c1, c2, ..., cn, let α be the step function defined on [0, 1]
by
α(0) = 0,
α(x) = c0 for 0 < x <
1
n
,
α(x) = X
k−1
i=0
ci for k − 1
n
≤ x <
k
n
, k = 2, 3, ..., n,
α(1) = Xn
i=0
ci
.
Show that V
1
0
(α) ≤
Pn
i=0
|ci
|. (Hint: Use Riemann-Stieltjes integral to estimate the variation.)
4. Let
f(x) = 
x
2
if −1 ≤ x ≤ 0;
x
3
if 0 < x ≤ 1; and α(x) =



1 if x = −1;
2x
2
if −1 < x < 1;
−1 if x = 1.
Evaluate the Darboux-Stieltjes integral R 1
−1
f(x)dα(x).
5. Let C be the Cantor set in [0, 1]. The Cantor set C is created by iteratively deleting the open middle
third from a set of non-overlapping closed intervals. One starts by deleting the open middle third (
1
3
,
2
3
)
from the interval [0, 1], leaving two closed intervals: [0,
1
3
] and [
2
3
, 1]. Next, the open middle third of
1
each of these remaining intervals is deleted, leaving four closed intervals: [0,
1
9
], [
2
9
,
1
3
], [
2
3
,
7
9
], and [
8
9
, 1].
Continue this process for ever. The Cantor set contains all points in the interval [0, 1] that are not deleted
at any step in this infinite process. Let D be the open set deleted. Then C = [0, 1] ∼ D.
A continuous function f is defined to be zero on C and on each component interval (α, β) of D to
have its graph as shown in the figure. The exact equation is not important, but on (α, β), f
0
is continuous,
f
0
(α
+) = f
0
(β
−) = 0, maxx∈(α,β)
|f
0
(x)| = 1, and maxx∈(α,β) f(x) ≤ (β −α)
2
. Show that the Riemann
integral R 1
0
f
0
(x)dx doesn’t exist even though f
0
(x) exists and are bounded on [0, 1].
2