MATH 321 - HOMEWORK #1
Due Friday, Jan 15.
P ROBLEM 1. Define an additive measure on R to be a function l from the set of all finite
intervals [s, t] to real numbers, satisfying
(1)l([s, t]) ≥ 0 for all s ≤ t.
(2)l([s, t]) = l([s, u]) + l([u, t]) for s ≤ u ≤ t.
Note that l is not a function on [s, t], it is a function that assigns a number to every interval.
We think of l([s, t]) as defining the length of [s, t] that may differ from the ordinary length,
but satisfies the basic properties (1) and (2).
Prove that if l is an additive measure on R then there exists a non-decreasing function
α : R → R, such that
l([s, t]) = α(t) − α(s)
for all s ≤ t.
The remaining problems ask you to prove that an integral exists and is zero. In all cases
the main problem is to show that the upper integral is zero. The upper sums are clearly
non-negative, so what is needed is a proof that the upper sums can get arbitrarily close to
0: for any ε > there exists a partition P, such that U(P, f, α) < ε.
P ROBLEM 2. Define functions f, α : [0, 2] → R
0 if x ≤ 1,
0
f(x) =
α(x) =
1 otherwise,
1
Prove that the integral
if x < 1,
otherwise.
Z2
fdα
0
exists and is equal to zero.
(Note that the two functions differ at x = 1. We showed in class that if f(x) is a continuous
R2
function, then 0 fdα = f(1). This is no longer true if f(x) is not continuous. For example,
R2
αdα does not even exist. You need to compute the upper and lower sums to prove the
0
statement.)
P ROBLEM 3. Define the function f : [0, 1] → R



1 if x = 0,
f(x) =
1
n


0
if x ∈ Q, x =
m
,
n
otherwise.
1
m, n ∈ N, gcd(m, n) = 1,
Prove that f(x) is Riemann-integrable, f(x) ∈ R, and
Z1
fdx = 0.
0
(Hint: Show that for any ε > 0, f(x) < ε except for a finite number of values x. Use this to
construct a partition P such that U(P, f) is arbitrarily close to zero.)
P ROBLEM 4. Recall the Cantor set C obtained from the interval [0, 1] by removing the
middle thirds ( 31 , 23 ), then removing the middle thirds of the remaining two intervals, and
so on. Let f : [0, 1] → R be the characteristic function of the Cantor set:
1 if x ∈ C,
f(x) =
0 otherwise.
Prove that f(x) is Riemann-integrable, f(x) ∈ R, and
Z1
fdx = 0.
0
(Hint: To construct partitions P such that U(P, f) is arbitrarily close to 0, you can use
the regular partitions Pn that have 3n parts of equal length. One can compute U(Pn , f)
inductively, relating it to U(Pn−1 , f). If we consider among the rectangles that compute
U(Pn , f) those that are based on the first third of the interval, [0, 1/3], then these are in
one-to-one correspondence with rectangles that compute U(Pn−1 , f); they are squeezed by
a factor of 1/3. This can be used to find a formula for U(Pn , f).)
Remark. The functions in the last two problems are interesting for the following reason. We
will prove in class that f(x) is integrable if it is continuous or if it has a finite number of
discontinuities. Theorem 11.33(b) in the textbook uses Lebesgue theory to prove that f(x)
is Riemann-integrable if and only if it is continuous “almost everywhere”, meaning the
points where it is not continuous have Lebesgue measure zero.
The function f(x) in Problem 3 is continuous at irrational points and discontinuous at
rational points. (How to prove it?) Thus, the points where f(x) is not continuous is a
countable set.
The function f(x) in Problem 4 is not continuous at any point of the Cantor set C, but it
is continuous outside C. The set C is not countable. To see these claims, write a number
x ∈ [0, 1] in the ternary form (base 3) x = a0 .a1 a2 a3 ..... Then x lies in C if and only if no
ai is equal to 1; in other words, points x ∈ C are sequences of zeros and twos. This set is
clearly not countable. However, any point of C is a limit of points not in C. For this, take
x ∈ C and change one of the digits ai to 1 for i large to find a point t very close to x, but
not in C.This means that f(x) = 1, but f(t) = 0 for t arbitrarily close to x, hence f is not
continuous at x.