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Chapter 1.6, Page 37

Problem 2:

(a) Prove that x is in the Cantor set iﬀ x has a ternary expansion that

uses only 0’s and 2’s.

(b) The Cantor-Lebesgue function is deﬁned on the Cantor set by writ-

ing x’s ternary expansion in 0’s and 2’s, switching 2’s to 1’s, and

re-interpreting as a binary expansion. Show that this is well-deﬁned

and continuous, F (0) = 0, and F (1) = 1.

(d) Show how to extend F to a continuous function on [0, 1].

Solution.

(a) The nth iteration of the Cantor set removes the open segment(s) con-

sisting of all numbers with a 1 in the nth place of the ternary expansion.

Thus, the numbers remaining after n iterations will have only 0’s and

2’s in the ﬁrst n places. So the numbers remaining at the end are pre-

cisely those with only 0’s and 2’s in all places. (Note: Some numbers

have a non-unique ternary representation, namely those that have a

representation that terminates. For these, we choose the inﬁnitely re-

peating representation instead; if it consists of all 0’s and 2’s, it is in

the Cantor set. This works because we remove an open interval each

time, and numbers with terminating representations are the endpoints

of one of the intervals removed.)

(b) First, we show that this is well-deﬁned. The only possible problem is

that some numbers have more than one ternary representation. How-

ever, such numbers can have only one representation that consists of

all 0’s and 2’s. This is because the only problems arise when one rep-

resentation terminates and another doesn’t. Now if a representation

terminates, it must end in a 2 if it contains all 0’s and 2’s. But then

the other representation ends with 12222... and therefore contains a

Next we show F is continuous on the Cantor set; given  > 0, choose

k such that

< . Then if we let δ =

, any numbers within δ will

agree in their ﬁrst k places, which means that the ﬁrst k places of their

images will also agree, so that their images are within

<  of each

other.

The equalities F (0) = 0 and F (1) = 1 are obvious; for the latter,

1 = 0.2222 . . . so F (1) = 0.1111 ··· = 1.

2’s, and re-interpret as a ternary expansion. By part (a), this will

produce a member of the Cantor set whose image is x. (Note: Their

may be more than one preimage of x, e.g. F (

) = F (

) =

(d) First, note that F is increasing on the Cantor set C. Now let

G(x) = sup{F (y) : y ≤ x, y ∈ C}.

Note that G(x) = F (x) for x ∈ C because F is increasing. G is

continuous at points not in C, because

C is open, so if z ∈

C, there

is a neighborhood of z on which G is constant. To show that G is

continuous on C, let x ∈ C and use the continuity of F (part b) to

since [0, 1] \ C

is a union of disjoint segments with this total length.

Then

m(C

) = 1 −

j=1

Now C

C, so by Corollary 3.3,

C) = lim

n→∞

m(C

) = 1 −

∞

k=1

(b) For k = 1, 2, . . . , let J

be the interval of C

which contains x. Let I

be the interval in

which is concentric with J

n−1

. (Thus, at the nth

step of the iteration, the interval I

is used to bisect the interval J

n−1

Let x

be the center of I

. Then x

∈

. Moreover, |x

−x| ≤ |J

n−1

since J

n−1

contains both x

and x. Since the maximum length of the

intervals in C

tends to 0, this implies x

→ x. Finally, x

∈ I

⊂

and I

⊂ J

n−1

⇒ |I

| → 0.

C is closed since it is the intersection of the closed sets C

. To

prove it contains no isolated points, we use the same construction from

the previous part. Let x ∈

C. This time, let x

be an endpoint of I

rather than the center. (We can actually take either endpoint, but for

speciﬁcity, we’ll take the one nearer to x.) Because I

is constructed

as an open interval, its endpoints lie in C

. Moreover, successive

iterations will not delete these endpoints because the kth iteration

only deletes points from the interior of C

k−1

. So x

∈

C. We also

have |x

− x| ≤ J

as before, so that x

→ x. Hence x is not an

isolated point. This proves that

C is perfect.

(d) We will construct an injection from the set of inﬁnite 0-1 sequences

into

C. To do this, we number the sub-intervals of C

in order from

left to right. For example, C

contains four intervals, which we denote

, I

, and I

. Now, given a sequence a = a

, a

, . . . of 0’s and

1’s, let I

denote the interval in C

whose subscript matches the ﬁrst

n terms of a. (For instance, if a = 0, 1, 0, 0, . . . then I

= I

0100

⊂ C

Finally, let

x ∈

∞

n=1

This intersection is nonempty because I

n+1

⊂ I

, and the intersection

of nested closed intervals is nonempty. On the other hand, it contains

only one point, since the length of the intervals tends to 0. Thus, we

have constructed a unique point in

C corresponding to the sequence a.

Since there is an injection from the uncountable set of 0-1 sequences

into

C is also uncountable.



Problem 5: Suppose E is a given set, and and O

is the open set

= {x : d(x, E) <

Show that

(a) If E is compact, then m(E) = lim

n→∞

m(O

(b) However, the conclusion in (a) may be false for E closed and un-

bounded, or for E open and bounded.

Proof. (a) First note that for any set E,

E =

∞

n=1

since the closure of E consists of precisely those points whose distance

to E is 0. Now if E is compact, it equals its closure, so

E =

∞

n=1

Note also that O

n+1

⊂ O

, so that O

& E. Now since E is bounded,

it is a subset of the sphere B

(0) for some N. Then O

⊂ B

N+1

(0),

so that m(O

) < ∞. Thus, by part (ii) of Corollary 3.3,

m(E) = lim

n→∞

m(O

(b) Suppose E = Z ⊂ R. Then m(O

) = ∞ for all n, since O

is a

collection of inﬁnitely many intervals of length

. However, m(Z) = 0.

This shows that a closed unbounded set may not work. To construct a

bounded open counterexample, we need an open set whose boundary

has positive measure. To accomplish this, we use one of the Cantor-

like sets

C from Problem 4, with the `

chosen such that m(

C) > 0.

Let E = [0, 1]\

C. Then E is clearly open and bounded. The boundary

of E is precisely

C, since

C contains no interval and hence has empty

interior. (This shows that the boundary of E contains

C; conversely,

it cannot contain any points of E because E is open, so it is exactly

equal to

C.) Hence

E = E ∪ ∂E = [0, 1]. Now



x ∈ R : d(x, E) <





x ∈ R : d(x,

E) <





−

, 1 +



Clearly m(O

) → 1, but m(E) = 1 − m(

C) < 1.



Problem 6: Using translations and dilations, prove the following: Let B be

a ball in R

of radius r. Then m(B) = v

, where v

= m(B

) and B

the unit ball {x ∈ R

: |x| < 1}.

Solution. Let  > 0. Choose a covering {Q

} of B

with total volume

less than m(B

) +



; such a covering must exist because the m(B

) is

the inﬁmum of the volumes of such cubical coverings. When we apply the

homothety x 7→ rx to R

, each Q

is mapped to a cube Q

whose side

length is r times the side length of Q

. Now {Q

} is a cubical covering of

with total volume less than r

m(B

)+. This is true for any  > 0, so we

must have m(B

) ≤ r

m(B

). Conversely, if {R

} is a cubical covering of

whose total volume is less than m(B

) + , we can apply the homothety

x 7→

x to get a cubical covering {R

} of B

with total volume less than

(m(B

) + ). This shows that m(B

) ≤

(m(B

) + ). Together, these

inequalities show that m(B

) = r

m(B

). 

Problem 7: If δ = (δ

, . . . , δ

) is a d-tuple of positive numbers with δ

> 0,

and E ⊂ R

, we deﬁne δE by

δE = {(δ

, . . . , δ

) : (x

, . . . , x

) ∈ E}.

Prove that δE is measurable whenever E is measurable, and

m(δE) = δ

. . . δ

m(E).

Solution. First we note that for an open set U , δU is also open. We could

see this from the fact that x → δx is an invertible linear transformation, and

therefore a homeomorphism. More directly, if p ∈ U , let B

(p) be a neigh-

borhood of p which is contained in U; then if we deﬁne

δ = min(δ

, . . . , δ

we will have B

δr

(δp) ⊂ δU.

Next, we note that for any set S, m

∗

(δS) = δ

. . . δ

∗

(S). The proof of

this is almost exactly the same as Problem 6: the dilation x 7→ δx and

its inverse map rectangular coverings of S to rectangular coverings of δS

and vice versa; but since the exterior measure of a rectangle is just its area

(Page 12, Example 4), the inﬁmum of the volume of rectangular coverings

is the same as the inﬁmum over cubical coverings. Hence a rectangular

covering within  of the inﬁmum for one set is mapped to a rectangular

covering within



...δ

for the other.

As a more detailed version of the preceding argument, suppose {Q

} is a

cubical covering of S with

| < m

∗

(S)+. Then {δQ

} is a rectangular

covering of δS with

|δQ

| < δ

. . . δ

∗

(S) + δ

. . . δ

. Now for each rec-

tangle δQ

we can ﬁnd a cubical covering {Q

} with

| < |δQ



Then ∩

j,k

is a cubical covering of δS with

j,k

| < δ

. . . δ

∗

(S)+

(1 + δ

. . . δ

). This implies that m

∗

(δS) ≤ δ

. . . δ

∗

(S). To get the re-

verse inequality we note that another δ-type transformation goes the other

direction, i.e. S = δ

(δS) where δ

= (1/δ

, . . . , 1/δ

Now let U ⊃ E be an open set with m

∗

(U \E) <



...δ

. Then δU ⊃ δE is

an open set. Moreover, δ(E\U) = δE\δU, so m

∗

(δU \δE) = δ

. . . δ

∗

(U \

E) < . Hence δE is also measurable. (Alternatively, we could prove that

δE is measurable by appealing to Problem 8.) 

Problem 8: Suppose L is a linear transformation of R

. Show that if E is a

measurable subset of R

, then so is L(E), by proceeding as follows:

(a) Note that if E is compact, so is L(E). Hence if E is an F

set, so is

L(E).

(b) Because L automatically satisﬁes the inequality

|L(x) − L(x

)| ≤ M |x −x

for some M, we can see that L maps any cube of side length ` into a

cube of side length c

M`, with c

= 2

√

d. Now if m(E) = 0, there is

a collection of cubes {Q

} such that E ⊂ ∩

, and

m(Q

) < .

Thus m

∗

(L(E)) ≤ c

, and hence m(L(E)) = 0. Finally, use Corol-

lary 3.5. (Problem 4 of the next chapter shows that m(L(E)) =

|det L|m(E).)

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