REAL ANALYSIS I - HOMEWORK 4 - Spring 2015 - Selected Solutions
(1.) R-F, page 28, Number 50
Show that a Lipschitz function is uniformly continuous but there are uniformly continuous functions that are not Lipschitz.
(i.) Show that a Lipschitz function is uniformly continuous.
Proof. Suppose that f is Lipschitz. Then there exists M > 0 such that
|f (x) − f (y)| ≤ M |x − y| for all x, y. Let > 0, and choose δ = M . Then if
|x − y| < δ, we have |f (x) − f (y)| ≤ M |x − y| < M · δ = M · M = . Therefore f is
uniformly continuous.
(ii.) Let us give a function that is uniformly continuous but not Lipschitz.
√
Example: Let f (x) = 3 x on [−1, 1]. This function is continuous on [−1, 1], a
compact subset of R. Thus, f is uniformly continuous on [−1, 1]. Now, let n ∈ N
and choose x1 = n13 , x2 = − n13 . Note:
|f (x1 ) − f (x2 )|
=
|x1 − x2 |
q
q
| 3 n13 − 3 − n13 |
| n13 − − n13 |
=
2
n
2
n3
= n2 .
Clearly there is no M for which M > n2 for all n.
(2.) R-F, page 34, Number 5
By using properties of outer measure, prove that the interval [0, 1] is not countable.
(3.) R-F, page 34, Number 6
Let A be the set of irrational numbers in the interval [0, 1]. Prove that m∗ (A) = 1.
(4.) R-F, page 34, Number 7 (This solution is to a problem stated a little differently than in
the fourth edition)
Prove: Given any set A and any > 0, there is an open set O such that A ⊆ O and
m∗ (O) ≤ m∗ (A) + . Also, there is a G ∈ Gδ such that A ⊆ G and m∗ (A) = m ∗ (G).
Proof:
Case 1: Suppose m∗ (A) = ∞. Then, taking O = R, we get O ⊇ A and ∞ = m∗ (R) =
m∗ (A) + .
Case 2: Suppose m∗ (A) < ∞. By definition of outer measure and infimum,
for any
S
> 0, there is a countable collection of open intervals {In }n≥1 such that n≥1 In ⊇ A
X
S
and
l(In ) < m∗ (A) + . Now, take O = n≥1 In . Then O is open and A ⊆ O, and
n≥1
we have

m∗ (O) = m∗ 

[
n≥1
In  ≤
X
n≥1
m∗ (In ) =
X
l(In ) < m∗ (A) + .
n≥1
So, we have an open set O such that A ⊆ O and m∗ (O) < m∗ (A) + .
To prove the second part of the proposition, first recall that a Gδ set is an intersection of
countably many open sets. Take the first part of this proof with = n1 . For each n ∈ N,
we can find an open set On with A ⊆ On such that
m∗ (On ) ≤ m∗ (A) +
Let
G=
∞
\
1
.
n
On .
n=1
Clearly G is a Gδ set. Since each On ⊇ A, we have G ⊇ A. Then by the monotone
property of outer measure, m∗ (G) ≥ m∗ (A). It remains to show that m∗ (G) ≤ m∗ (A).
But, for every n, G ⊆ On , so that for every n ∈ N,
m∗ (G) ≤ m∗ (On ) ≤ m∗ (A) +
1
.
n
Letting n go to ∞, we get m∗ (G) ≤ m∗ (A). Therefore m∗ (G) = m∗ (A) and we are done.
(5.) R-F, page 34, Number 8
Let A be the set of rational numbers
P between 0 and 1. Let {In } be a finite collection of
open intervals covering A. Then
l(In ) ≥ 1.
Proof:
Q, and let {In }N
n=1 be a finite collection of open intervals covering A.
N
N
[
[
This means that
In ⊇ A, so that O =
In is an open set containing A. We
Let A = (0, 1)
T
n=1
n=1
know that O can be written as the union of a countable collection of pairwise disjoint
open intervals (Proposition 8, page 42). So we may as well assume that we have {In =
(an , bn )}N
n=1 pairwise disjoint (if not, change the collection into one in which the intervals
are pairwise disjoint). Order things so that
0 = b0 ≤ a1 < b1 ≤ a2 < b2 . . . ≤ an−1 < bn−1 ≤ an < bn ≤ an+1 = 1.
Suppose that for some k, 1 ≤ k ≤ n + 1, we have bk−1 < ak . Then [bk−1 , ak ] is a closed
interval of positive length contained in [0, 1] ∼ O ⊂ [0, 1] ∼ A. But [bk−1 , ak ] must contain
a rational number because
between any
T
T two distinct
T ˜ real numbers is a rational number.
This
means
that
A
{[0,
1]
∼
A}
=
A
{[0,
1]
A} 6= ∅, which is a contradiction, since
T
T
T ˜
A {[0, 1] ∼ A} = A {[0, 1] A} = ∅. Thus, we have bk−1 = ak for all k, 1 ≤ k ≤ n. It
follows that [0, 1] ∼ O is a finite number of points, which has outer measure zero. Then
!
N
N
N
X
X
[
∗
∗
l(Ii ) =
m (Ii ) ≥ m
Ii
i=1
i=1
i=1
m∗ (O)
=
= m∗ (O) + 0
= m∗ (O)S+ m∗ ([0, 1] ∼ O)
≥ m∗ (O [0, 1] ∼ O)
= m∗ ([0, 1])
= 1.
(6.) R-F, page 34, Number 9
Prove that if m∗ (A) = 0, then m∗ (A ∪ B) = m∗ (B).
For this proof, use the properties of monotonicity and countable subadditivity.
(7.) R-F, page 34, Number 10
Let A and B be bounded sets for which there is an α > 0 such that |a − b| ≥ α for all
a ∈ A, b ∈ B. Prove that m∗ (A ∪ B) = m∗ (A) + m∗ (B).
Proof:
∗ ∗ ∗Consider an open cover G of the set A such that G is a collection of bounded, open
intervals such that the elements of the union of the sets in G are all within a distance of α2
from the elements of A. Let G be the union of all the sets in the cover G. So |a − c| < α2
when a ∈ A, c ∈ G.
Let G1 be the union of the sets in such a cover for A and let G2 be the union of the
sets in such a cover for B. We first claim that G1 and G2 are disjoint. Let us show by
contradiction that G1 ∩ G2 = ∅. Suppose that c ∈ G1 ∩ G2 . Then c ∈ G1 and c ∈ G2 .
Let a ∈ A and b ∈ B. Note:
|a − b| = |a − c + c − b| ≤ |a − c| + |c − b| <
α α
+ = α.
2
2
This contradicts the fact that the sets A and B are at least a distance of α from each
other, so it must be the case that G1 and G2 are disjoint.
Also note that for such sets G1 and G2 , it is clearly true that G1 ∪ G2 ⊃ A ∪ B so that
the collection of sets G1 ∪ G2 is a cover of A ∪ B satisfying condition ∗ ∗ ∗. Conversely, a
cover G of A ∪ B satisfying condition ∗ ∗ ∗ can be viewed as the disjoint union of such a
cover of A and such a cover for B.
In the definition of outer measure, we may simply consider all open covers that satisfy
condition ∗ ∗ ∗. If we have a cover that doesn’t satisfy such a condition, we can always
“do better.”
∗
m (A ∪ B) = inf
(∞
X
)
l(Ik ) :
{Ik }∞
k=1
is a cover of A ∪ B satisfying condition ∗ ∗∗
k=1
= inf
and
(∞
X
l(Ik ) +
k=1
{Ii }∞
i=1
(∞
X
= inf
∞
X
l(Ii ) : {Ik }∞
k=1 is a cover of A satisfying condition ∗ ∗∗
i=1
is a cover of B satisfying condition ∗ ∗∗}
l(Ik ) :
)
{Ik }∞
k=1
is a cover of A satisfying condition ∗ ∗∗
k=1
+ inf
(∞
X
)
l(Ii ) : {Ii }∞
i=1 is a cover of B satisfying condition ∗ ∗∗
i=1
∗
= m (A) + m∗ (B).