Real Variables: Solutions to Homework 3
September 23, 2011
Exercise 0.1. Chapter 3, # 2: Show that the cantor set C consists of all x such that x has
some triadic expansion for which every ck is either 0 or 2.
P
ck
Proof. We are being asked to prove that x ∈ C if and only if x = ∞
k=1 3k for ck = {0, 2}.
First suppose that x ∈ C. We proceed by induction on n in that we will show that the left
endpoint of the intervals in Cn are x which have triadic expansions with ck is either 0 or 2.
Consider first C1 = [0, 13 ] ∪ [ 32 , 1]. Then either x = 0 or x = 23 , which are both the correct
P
ck
triadic expansions as claimed. Now suppose this is true P
for Cn−1 , that is x = n−1
k=1 3k with
n
ck
ck = {0, 2} is a left endpoint of Cn−1 . Now for Cn , let x = k=1 3k with ck = {0, 2}. If an = 0
the x is the left endpoint of the left subinterval and if x = 2 then x is the left endpoint of
the right subinterval. This is the construction.
Now, suppose that x =
P∞
ck
k=1 3k
with ck = {0, 2}. Then we have the inequalities
n
∞
n
X
X
X
ck
2
ck
≤x≤
+
k
k
3
3
3k
k=1
k=n+1
k=1
n
X
ck
1
=
+ n.
k
3
3
k=1
P
This shows us that if In ⊂ Cn and nk=1 3ckk is the left endpoint of the In , then x ∈ Cn . This
holds for all n so therefore x ∈ C. Say that x ∈ C1 , then c1 = 0 if x is in the left interval and
c1 = 2 if x is in the right interval. By the same construction suppose that x ∈ Cn−1 and that
ai have
been defined for i = 1, 2, . . . , n − 1. If x ∈ In−1 , an interval in Cn−1 then it is written
Pn−1
x = k=1 a3kk ∈ In−1 . Either x in the left or right interval of In−1 . Again, if it is the left say
that an = 0 and that an = 2 if it is in the right. But then we see that
x−
so
n
∞
X
X
ck
2
≤
,
k
k
3
3
k=1
k=n+1
∞
X
ck
x=
with ck = {0, 2}.
k
3
k=1
1
S
N
Exercise 0.2.
Chapter
3,
#
7:
If
{I
}
is
a
finite
collection
of
non-overlapping,
Ik is
k
k=1
S P
measurable
Ik = |Ik |.
S
Proof. From the book, we know that Ik is measurable follows from (3.13). Now, to complete the proof, let ˚
Ik denote the interior of sets Ik . These interiors are disjoint so the enjoy
some nice properties like additivity in measure. Then
[ X
|Ik |
Ik ≤
by subadditivity. We seek to show the other direction for the inequality. Consider
[ [ X X
˚ Ik =
|Ik | .
Ik ≥ ˚
Ik =
Thus
[ X
Ik =
|Ik |.
Exercise 0.3. Chapter 3, # 8: Show that the Borel σ-algebra B in Rn is the smallest
σ-algebra containing the closed sets in Rn .
Proof. By definition, B is the smallest σ-algebra containing the open sets. In particular,
B is a field and is therefore closed under set complementation. Because B contains all the
open sets in Rn , it also contains the compliments of all such set, which are exactly all closed
sets in Rn . To see that it is the smallest such σ-algebra, imagine that there is one smaller
that contains all the closed sets, call it A . Again, because A is a σ-algebra it is a field
and is closed under complementation so in particular A contains all the open sets in Rn .
However this is a contradiction because by definition, B is the smallest σ-algebra containing
the open sets in Rn and we are done.
P∞
Exercise 0.4. Chapter 3, # 9: If {En }∞
n=1 is a collection of sets with
n=1 |En |e < ∞, show
that lim sup E and lim inf E have measure zero.
P
Proof. Now, since ∞
k=1 |En |e < ∞, there must be some m such that
∞
X
|En |e < .
n=m
Then,
lim sup En =
∞
\
∞
[
j=1
n=j
!
En
⊂
∞
[
En .
n=m
Proceed to calculate,
∞
\
| lim sup En |e = j=1
∞
[
n=j
!
∞
∞
[
X
En ≤ En =
|En |e < .
n=m n=m
e
2
e
Also
lim inf En =
∞
[
∞
\
j=1
n=j
∞
[
!
En
⊂
En ,
n=m
and the same argument follows. Thus
| lim sup En |e = | lim inf En |e = 0.
Exercise 0.5. Chapter 3, # 10: If E1 and E2 are measurable show that |E1 ∪E2 |+|E1 ∩E2 | =
|E1 | + |E2 |.
Proof. We know since E1 is measurable that
|E1 ∪ E2 | = |(E1 ∪ E2 ) ∩ E1 | + |(E1 ∪ E2 ) ∩ E1c |
= |E1 | + |E2 ∩ E1c |
Furthermore we know that
|E1 ∩ E2 | + |E2 ∩ E1c | = |E2 |,
so
|E1 ∪ E2 | + |E1 ∩ E2 | = |E1 | + |E2 |.
Exercise 0.6. Chapter 3, # 17: Our example will be the Cantor-Lebesgue function:
F (x) =
∞
X
ak
bk
where
b
=
k
2k
2
k=1
P
ak
if x = ∞
k=1 3k . We first must go through the tedious task of showing it is continuous, that
F (0) = 0 and F (1) = 1, and that F is surjective. We begin:
Proof. The ternary expansion for zero is ak = 0 for all k. Looking at the function,
F (0) =
X
0×
k
1
=0
2k
The ternary expansion for 1 is ak = 2 for all k.
F (1) =
X
2×
k
1
=1
2k
Thus F (0) = 0 and F (1) = 1.
Now we need to prove continuity. To do this, we need to show that for all > 0 there
exists δ > 0 s.t. |F (a) − F (b)| < when |a − b| < δ for a, b ∈ [0, 1]. Let > 0 and pick
3
and n such that 2−n < . Pick δ = 3−(n+1) . Pick a, b s.t. |a − b| < δ. Let a have a ternary
expansion with ak ∈ {0, 2} and let b have ternary expansion with bk ∈ {0, 2}. We need to
show that ak = bk for k < n. To do so, pick the minimal k where ak 6= bk , k 0 .
∞ ∞
X
X
ak
bk ak0
bk 0
ak
bk −
− k0 +
−
≤
3k 3k 3k0
3
3k 3k 0
k=1
k=k +1
Again, we conclude that
∞
a0
X
0
1
a
b
b
k
k
k
k ≤ k0 − k0 +
− k k
3
3k0
3
3
3
k=k0 +1
giving us that k 0 > n. So for the case where ak = bk , k < n and ck = ak /2, dk = bk /2, then
∞ ∞
∞ X
ck
dk X ck
dk X −k
2 = 2−n < − k =
− k ≤
|F (a) − F (b)| = k
k
2
2
2
2
k=1
k=1
k=n+1
This shows the function F (x) to be continuous on C.
We need to prove F to be surjective. We start with the fact that every element x ∈ [0, 1]
has a binary expansion. That is
∞
X
bn
x=
.
2n
n=1
Define ak = 2bk . Because ak ∈ {0, 2} then by Exercise 1 in this homework, for any element
c ∈ C,
∞
X
an
c=
3n
n=1
and F (c) = x. Therefore F is surjective.
Finally, we can get to the example:
Proof. Take A ⊂ [0, 1] to be non measurable. Now, we know that F (C) = [0, 1]. Every
subset of a set of measure zero is measurable so then in particular
B = F −1 (A) ∩ C
is measurable. However,
F (B) = F (F −1 (A) ∩ C) = A
is non measurable.
4
Exercise 0.7. Chapter 3, # 23: let Z be a subset of R1 with measure zero. Show that the
set Z 0 = {x2 : x ∈ Z} also has measure zero.
Proof. Either Z is bounded or it is infinite. Suppose first that Z is bounded, that is x ≤
M ∈ R for all x ∈ Z. BecausePZ has measure zero we can cover it will a collection of
intervals In ⊂ BM (0) such that n |In | < . Now, Z 0 is covered by a union of intervals In2 .
But clearly |In2 | ≤ diamBM (0)|In | = 2M |In |. So
X
X
|Z 0 | ≤
|In2 | ≤ 2M
|In | < 2M .
n
n
Thus, taking → 0 shows that |Z 0 | = 0.
Now, in the unbounded case, define Z restricted to a bounded interval on the real line as
0
ZM = Z ∩ BM (0) and ZM
= Z 0 ∩ BM (0).
By construction
Z=
[
ZM and Z 0 =
M
[
0
ZM
.
M
0
|
|ZM
= 0 (this is simply the bounded case we did above. Using the property
Now, for any M ,
of subadditivity of the measure
[
X
0
0 |ZM
| = 0.
|Z 0 | = ZM
≤
M
M
We have shown that Z 0 = {x2 : x ∈ Z} has measure zero if Z has measure zero.
5