Math 515
Professor Lieberman
September 20, 2004
HOMEWORK #4 SOLUTIONS
Chapter 3
28. (a) From Exercise 2.48, f1 is continuous and increasing, so f is continuous and strictly
increasing. (To see that f is strictly increasing, we note that x > y implies that f (x) =
f1 (x) + x ≥ f1 (y) + x > f1 (y) + y = f (y).) Since f (0) = 0 and f (1) = 2, it follows from
the intermediate value theorem that f is onto, and f is one-to-one because it’s strictly
increasing. Exercise 2.46 also tells us that f is a homeomorphism.
(b) Let hIn i be the sequence of “middle third” intervals from the definition of C. Then
∞
[
[0, 2] = f (C) ∪
f (In )
n=1
and this is a disjoint union because f is one-to-one and C ∩ In = Im ∩ In = ∅ if m 6= n.
Hence
∞
X
2 = m(f (C)) +
m(f (In )).
n=1
But f (In ) is an interval with the same length as In (because f is constant on each of these
intervals), so
∞
∞
X
X
m(f (In )) =
`(In ) = 1.
n=1
n=1
Combining the two displayed equations yields m(f (C)) = 1.
(c) Let P be a nonmeasurable subset of F (from Exercise 3.16) and set A = f −1 (P ).
Then A is a subset of C and hence measurable because mC = 0 and Lebesgue measure is
complete. However, g −1 [A] = f (A) = P is not measurable.
(d) Take g = f −1 and h = χA . Then (h ◦ g)−1 (1/2, ∞) = g −1 (A) is not measurable, so h ◦ g
is not measurable.
(e) The set A from part (c) is measurable but it isn’t a Borel set because of Exercise 3.26.
31. From Proposition 34, for each positive integer n, there is a continuous function fn and a
measurable set An such that mAn < 2−n−1 δ and |fn − f | < 1/n on Ãn . (To accomplish
this, we must apply the proposition with ε = min{1/n, 2−n δ}.) Then set A = ∪An and
B = [a, b] ∼ A. Then Proposition 28 gives a closed set F such that F ⊂ B and m(B ∼
F ) < δ/2. Egoroff’s Theorem implies that hfn i converges uniformly on F to a function g,
which is therefore continuous (and only defined on F ). Problem 2.40 now gives a function
ϕ which is continuous on (−∞, ∞), and hence on [a, b] such that ϕ = g on F . On the
other hand, because fn → f on B, it follows that f = g on F and hence f = ϕ on F . But
{x : f (x) 6= ϕ(x)} ⊂ (B ∼ F ) ∪ A, so
m{x : f (x) 6= ϕ(x)} ≤ m(B ∼ F ) + mA < δ.
1
2
Chapter 4
1. (a) Because every interval with positive length contains a rational number and an irrational
number, we see that (in the notation of the textbook) Mi = 1 and mi = 0 for any partition.
Hence
n
n
X
X
S=
(ξi − ξi−1 )Mi =
(ξi − ξi−1 ) = 1
i=1
i=1
for any partition. Therefore
Z
b
R
f (x) dx = 1.
a
Similarly s = 0 for any partition, so
Z b
R f (x) dx = 0.
a
newline (b) Let hrn i be an enumeration of the rationals and define gn (x) = χ{rn } , and
fn (x) =
n
X
gn (x).
i=1
Then fn is nonnegative and Riemann integrable because it only has a finite number of
discontinuities. Since each gn is nonnegative, it also follows that fn increases monotonically
to its limit, which is clearly f .
The conclusion is that changing the order of integration and the limiting process is not
always allowable.
Chapter 11
13. (a) For each positive integer ν, there is a positive integer nν such that
m{|fn − f | ≥ 2−ν } < 2−ν
for any n ≥ nν . We now set
Eν = {x : |fnν (x) − f (x)| < 2−ν },
S
−ν for any ν ≥ k, so f
and Gk = ∞
nν → f on
ν=k Eν . If xT∈ G̃k , then |fnν (x) − f (x)| < 2
∞
G̃k . Finally, we set A = k=1 Gk . Then fnν → f on à and mA = lim mGk = 0.
(b) From Proposition 35 (applied with A in place of X) with ε = δ, we see that hfn i
converges to f in measure.
(c) Since the sequence is Cauchy in measure, the argument from part (a) shows that there is
a subsequence hfnν i which is Cauchy a.e. Write F for the set of measure zero on which this
subsequence is not Cauchy and we take f = lim fnν on F̃ with f = 0 on F . By Theorem
and , f is measurable. Let ε > 0 be given and let N and E be the quantities given by the
definition of “Cauchy in measure”. Then for all sufficiently large ν (for example ν > N ),
we have
|fn (x) − fnν (x)| < ε
on Ẽ if n ≥ N . Now we set A = E ∪ F and we send ν to ∞ to see that
|fn (x) − f (x)| < ε
on à if n ≥ N . Since µA = µE < ε, it follows that hfn i converges to f in measure.