MATH 321:201: Real Variables II (Term 2, 2010)
Home work assignment #2
Due date: Friday, Jan 22, 2010 (hand-in in class)
Problem 1 : Is the converse of [Rudin, Theorem 6.12 (c)] true? Namely, let a < c < b and suppose
f ∈ R(α) on [a, c] and on [c, b]. Is f ∈ R(α) on [a, b] as well? Justify your answer rigorously.
Problem 2 : Do [Rudin, Ch. 6. Exercise 11].
Problem 3 : Do [Rudin, Ch. 6. Exercise 12].
Problem 4 : Find a monotonically increasing functin α : [0, 1] → R with α(0) = 0, α(1) = 4/3
with the following property: for any continuous function f : [0, 1] → R,
Z
0
Here
R
1
Z 2−n+1
∞ h
i
X
1
1
1
f( ) + n
f dα =
f dx .
2n 2n
2 2−n
n=1
f dx denotes the ordinary Riemann integral. Justify your answer.
The following are suggested exercises (Please DO NOT hand-in) :
Problem: Do Rudin, Ch. 6, Exercises #7, #8. #15.
Problem: Let f, g be two Riemann-Stieltjes integrable functions with respect to α on the interval
[a, b]: i.e. f, g ∈ R(α). Define two functions max[f, g](x) = max[f (x), g(x)], min[f, g](x) =
min[f (x), g(x)]. Show both max[f, g], min[f, g] ∈ R(α).
Problem: This exercise concerns [Rudin, Theorem 6.19 (Change of variable)]. We use the same
notation as in there.
• (a) Assume instead only that ϕ is strictly increasing (so remove the continuity assumption).
Can we still say g ∈ R(β)? Justify your answer.
• (b) Assume that ϕ is a monotonically increasing continuous function, but let us NOT assume strictness of the monotonicity of ϕ. Show that the same conclusion holds as in
Theorem 6.19.
Problem: This exercise answers some questions raised in the class. (Thank you for the questions!)
Let f : [a, b] → R with m ≤ f ≤ M and φ : [m, M ] → R.
• (a) Suppose both f and φ are Riemann integrable. Is φ ◦ f also Riemann integrable on
[a, b]?
The answer is NO. This is done in the following counterexample.
– Example Let a = 0, b = 1. Let
(
1/n if x = m
in the lowest term (e.g. f (8//12) = 1/3),
n
f (x) =
0
if x ∈ R \ Q or x = 0..
First show that this function f is Riemann integrable on [0, 1]. Let φ : [0, 1] → R be
the step function (thus it is Riemann integrable):
(
0 if x ≤ 0,
φ(x)
1 if x > 0.
Show φ ◦ f is NOT Riemann integrable on [0, 1].
• (b)* This is a hard and long exercise. Suppose f is continuous and φ is Riemann integrable. Is φ ◦ f Riemann integrable? The answer is again NO. Finding a counterexample
to this is not so easy. This is a bit beyond the scope of this course. If you are interested in,
you may want to consider the following famous example:
– Example: Let P be the Cantor set:
∞
X
P = {x| x =
aj 3−j ; aj = 0 or 2 for all j}
j=1
Construct the Cantor function f : [0, 1] → [0, 1] defined as the following rule:
∞
∞
X
X
aj −j
2 if x ∈ P and x =
aj 3−j ∈ P ;
f (x) =
2
j=1
j=1
and for x ∈ [0, 1] \ P , f (x) = f (y) where y ∈ P is the nearest point in P from the
right to x.
(i) Show that this function is continuous (moreover it is monotone) and the image
f (P ) = [0, 1].
Consider the function g : [0, 1] → [0, 2] with g(x) = x + f (x).
(ii) Show g is 1-1 and let h : [0, 2] → [0, 1] be the inverse map h = g −1 .
(iii) Show h is continuous.
Define a function φ : [0, 1] → R as
(
0 if x ∈ P ,
φ(x) =
1 otherwise.
(iv) Show φ is Riemann integrable.
(v) Show φ ◦ h : [0, 2] → R is not Riemann integrable
2