REAL ANALYSIS I HOMEWORK 2
CİHAN BAHRAN
The questions are from Stein and Shakarchi’s text, Chapter 1.
1. Prove that the Cantor set C constructed in the text is totally disconnected and
perfect. In other words, given two distinct points x, y ∈ C, there is a point z ∈
/C
that lies in between x and y, and yet C has no isolated points.
[Hint: If x, y ∈ C and |x − y| > 1/3k , then x and y belong to two different intervals
in Ck . Also, given any x ∈ C there is an end-point yk of some interval in Ck that
satisfies x 6= yk and |x − yk | ≤ 1/3k .]
Given two distinct points x, y ∈ C, since |x − y| > 0 there exists k ∈ N such that
|x − y| > 1/3k . Because Ck is a union of closed intervals of length 1/3k and x, y ∈ Ck ,
there exists z in between x and y which does not lie in Ck . Thus z ∈
/ C.
And given x ∈ C, for every k ∈ N, x lies in one of the 2k intervals that make up Ck .
Thus there exists an end-point yk in Ck such that |x − yk | ≤ 1/3k . We may assume
yk 6= x here (if x happens to be an end-point of an interval in Ck itself, choose the
other end-point of the interval to be yk ). But we know that the end-points survive
the Cantor intersection, that is they lie in C. Hence [x − 1/3k , x + 1/3k ] − {x} intersects C for every k. Since x ∈ C was arbitrary, we conclude that C has no isolated points.
2. The Cantor set C can also be described in terms of ternary expansions.
(a) Every number in [0, 1] has a ternary expansion
x=
∞
X
ak 3−k , where ak = 0, 1 or 2.
k=1
k
Note that this decomposition is not unique since, for example, 1/3 = ∞
k=2 2/3 .
Prove that x ∈ C if and only if x has a representation as above where every ak is
either 0 or 2.
P
The question states that there is a well-defined surjective map
f : {0, 1, 2}N → [0, 1]
(an )n∈N 7→
X
an 3−n .
n∈N
The
asks us to prove f ({0, 2}N ) = C. In the last homework, I showed that
Ä question
ä
f {0, 2}N ⊆ C (I also showed that f maps {0, 2}N injectively to C and from here
deduced that C is uncountable).
Ä
ä
Ä
ä
So it’s enough to show that C ⊆ f {0, 2}N , or equivalently [0, 1]rf {0, 2}N ⊆ [0, 1]rC.
Ä
ä
So let x ∈ [0, 1]r f {0, 2}N . Since f is surjective there exists a sequence (an )n∈N with
an ∈ {0, 1, 2} such that x = n∈N an 3−n . By assumption, the set {n ∈ N : an = 1}
must be nonempty. So pick the least element k from that set. Thus ak = 1 but
P
1
REAL ANALYSIS I HOMEWORK 2
2
−i
a1 , . . . , ak−1 ∈ {0, 2}. Let y = k−1
i=1 ai 3 . Note that y is an end-point in Ck−1 . More
precisely, [y, y + 3−k+1 ] is one of the intervals in Ck−1 . Note that on one hand
P
x−y =
∞
X
−i
ai 3
−k
=3
∞
X
+
i=k
ai 3−i ≥ 3−k
i=k+1
and on the other hand
x − y ≤ 3−k + 2
∞
X
3−i = 3−k + 2 ·
i=k+1
3−k−1
= 3−k + 3−k = 2 · 3−k
1 − 3−1
which implies
î
(y + 3−k+1 ) − x = y + 3 · 3−k − x ≥ 3−k .
ó
Thus we get x ∈ y + 3−k , (y + 3−k+1 ) − 3−k , which is the middle third of the interval
P
−i
[y, y Ä+ 3−k+1 ]. Moreover, y andä y + 3−k+1 = y + ∞
lie in f ({0, 2}N ), thus
i=k 2 · 3
x ∈ y + 3−k , (y + 3−k+1 ) − 3−k and so x ∈
/ Ck as Ck loses this interval while being
constructed from Ck−1 . In particular x ∈
/ C.
(b) The Cantor-Lebesgue function is defined on C by
F (x) =
bn 2−n
X
n∈N
−n
if x = n∈N an 3 where bn = an /2. In this definition we choose the expansion of
x in which ak = 0 or 2.
P
Show that F is well-defined and continuous on C, and moreover F (0) = 0 as well
as F (1) = 1.
In (a), we showed that there is a bijection
f : {0, 2}N → C
(an )n∈N 7→
X
an 3−n .
n∈N
Note that there is a surjection
g : {0, 1}N → [0, 1]
(bn )n∈N 7→
X
bn 2−n
n∈N
since every number in [0, 1] has a binary expansion like 0.0011110010110001... . Now
we can define
F = g ◦ ι ◦ f −1 : C → [0, 1]
where
ι : {0, 2}N → {0, 1}N
(an )n∈N 7→ (an /2)n∈N .
If we let 0 to be simply the sequence consisting entirely of 0’s, we have
(g ◦ ι)(0) = g(0) = 0 = f (0)
thus
F (0) = (g ◦ ι)(f −1 (0)) = (g ◦ ι)(0) = 0 .
REAL ANALYSIS I HOMEWORK 2
3
And if we let 2 to be the sequence consisting entirely of 2’s and 1 to be the sequence
of 1’s, we have
(g ◦ ι)(2) = g(1) =
X −n
2
=1=
n∈N
X
2 · 3−n = f (2)
n∈N
thus
F (1) = (g ◦ ι)(f −1 (1)) = (g ◦ ι)(2) = 1 .
For continuity, consider the discrete topology on {0, 1} and then endow the set of
sequences {0, 1}N with the product topology. Do the same for {0, 2}N . Then being
continuous coordinate-wise, ι is definitely continuous. We will show that f and g are
also continuous. From here it follows that F is continuous, and since the domain {0, 2}N
of f is compact (Tychonoff’s theorem!) and the codomain C of f is Hausdorff, the
bijective continuous map f is a homeomorphism; hence F is a composite of continuous
maps. And Proposition 2 yields that f and g are continuous (g by taking d = 2 and f
by taking d = 3 and restricting to {0, 2}N ).
Lemma 1. Let d be a natural number greater than 1 and write D = {0, . . . , d − 1}. Let
(an )n∈N and (bn )n∈N be two distinct sequences with elements in D. Let k = min{n ∈
N : an 6= bn }. Then
X
an d−n <
n∈N
X
bn d−n
n∈N
if and only if one of the three possibilities below holds:
(1) ak < bk − 1.
(2) ak = bk − 1 and bl > 0 for some l > k.
(3) ak = bk − 1 and al < d − 1 for some l > k.
Proof. Omitted. This is a standard base-d expansion result.
Proposition 2. Let d be a natural number greater than 1. Put the discrete topology
on the set D = {0, . . . , d − 1} and consider DN with the product topology. Then the
surjective map
h : DN → [0, 1]
(an )n∈N 7→
X
an d−n
n∈N
is continuous.
Proof. It is enough to show that for every b ∈ [0, 1] the set
h−1 ((−∞, b) ∩ [0, 1]) = {(an )n∈N :
X
an d−n < b}
n∈N
is open in DN . Pick (bn )n∈N such that
X
bn d−n = b. If bn ’s eventually become 0, replace
n∈N
them by a sequence which is eventually d − 1 which still is an expansion of b. Write
Uk = {b1 } × · · · {bk−1 } × {c ∈ D : c < bk − 1} × D × D × D × · · ·
Vk,l = {b1 } × · · · {bk−1 } × {bk − 1} × D × D × · · · × D × (D − {d − 1}) × D × D × · · ·
REAL ANALYSIS I HOMEWORK 2
4
for every k, l ∈ N with l > k where the D − {d − 1} term in Vk,l occurs at the l-th
coordinate. Note that every Uk and Vk,l is open in DN . Now by Lemma 1 we get that
{(an )n∈N :
X
an d−n < b} =
[
n∈N
[ [
Uk ∪
k∈N
Vk,l
k∈N l>k
which is open.
(c) Prove that F : C → [0, 1] is surjective.
We know that f is bijective and g is surjective. ι is also clearly surjective by its definition. Thus F = g ◦ ι ◦ f −1 is also surjective.
(d) One can also extend F to be a continuous function on [0, 1] as follows. Note
that if (a, b) is an open interval of the complement of C, then F (a) = F (b). Hence
we may define F to have constant value on that interval.
A connected component of the complement of C is of the form
n
X
ai
n
X
3−i + 3−n ,
a 3−i
i
i=1
!
+ 2 · 3−n
i=1
for some a1 , . . . , an ∈ {0, 2}. Write r =
n
X
ai 3−i + 3−n so the interval is (r, r + 3−n ).
i=1
Note that
r=
n
X
∞
X
ai 3−i +
i=1
2 · 3−i ∈ C
i=n+1
and
F (r) =
=
=
n
X
∞
X
i=1
n
X
i=n+1
(ai /2)2−i +
2−i
(ai /2)2−i + 2−n
i=1
n−1
X
(ai /2)2−i +
i=1
=F
=F
n−1
X
i=1
n
X
an + 2 −n
·2
2
!
−i
ai 3
+ (an + 2)3
−n
!
−i
ai 3
−n
+2·3
i=1
= F (r + 3−n )
as desired.
5. Suppose E is a given set, and On is the open set
On = {x ∈ Rd : d(x, E) < 1/n}.
Show:
(a) If E is compact, then m(E) = limn→∞ m(On ).
REAL ANALYSIS I HOMEWORK 2
5
Note that since E is closed, it is measurable so writing m(E) is OK. Also note that
(On ) is a decreasing family of open sets.
Let O be an arbitrary open set containing E and write C = Rd r O. So E and C are
disjoint, moreover since E is compact and C is closed they are distant by a previous
exercise. Hence d(C, E) ≥ 1/n for some n. That is, every point in C is at least 1/n
away from E hence C ∩ On = ∅. This implies that On ⊆ O.
So we showed the set {On : n ∈ N} is cofinal in {O ⊆ Rd : E ⊆ O-open} with respect
to the reverse containment ⊇. Thus by the monotonicity of m the set {m(On ) : n ∈ N}
is cofinal in {m(O) : E ⊆ O-open} with respect to ≥. Thus their infima coincide:
lim On = inf{m(On ) : n ∈ N} = inf{m(O) : E ⊆ O-open} = m(E) .
n→∞
(b) However, the conclusion in (a) may be false for E closed and unbounded; or E
open and bounded.
For the first part, take E = N as a subset of R, which is closed and unbounded. In this
case,
[
On =
(N − 1/2n, N + 1/2n) .
N ∈N
Since the union is disjoint, by countable additivity of m we have
X
m(On ) =
1/n = ∞
N ∈N
whereas m(E) = 0, being a countable set.
For the second part, let (qn )n∈N be an enumeration of Q ∩ [0, 1] and take
E=
[
(qn − 4−n , qn + 4−n ).
n∈N
E is certainly open and bounded. Now by countable subadditivity we have
m(E) ≤
X
2 · 4−n = 2 ·
n∈N
4−1
1/4
=2·
= 2/3 .
−1
1−4
3/4
However, E is dense in [0, 1]. So given x ∈ [0, 1], for every n ∈ N there exists q ∈ E
such that d(x, q) < 1/n so x ∈ On . Thus [0, 1] ⊆ On for every n and hence
1 = m([0, 1]) ≤ m(On )
for every n. Therefore
lim m(On ) ≥ 1.
n→∞
11. Let A be the subset of [0, 1] which consists of all numbers which do not have
the digit 4 appearing in their decimal expansion. Find m(A).
This corresponds to dividing [0, 1] into 10 equal pieces, labeling them from 0 to 9 and
taking away the piece with label 4 (since the numbers between 0.4 and 0.5 should be
deleted). Then this process is applied to the 9 remaining pieces. So at the first step
REAL ANALYSIS I HOMEWORK 2
6
we take away an interval of length 1/10, at the second step we take away 9 intervals of
length 1/100 and so on. So the complement of A has measure
Ç
X n−1
9
n∈N
1
1 X 9
· n =
·
10
10 n∈N 10
ån−1
=
1
1
·
9 = 1
10 1 − 10
hence A has measure 0.
13. The following deals with Gδ and Fσ sets.
(a) Show that a closed set is a Gδ and an open set an Fσ .
[Hint: If F is closed, consider On = {x : d(x, F ) < 1/n}.]
Let’s use the notation of the hint. Since F is closed, d(x, F ) = 0 if and only if x ∈ F
therefore
F =
\
On .
n∈N
Let G be an open set. Then Gc is closed and hence is a Gδ set. Therefore G is an Fσ
set by de Morgan.
(b) Give an example of an Fσ that is not a Gδ .
[Hint: This is more difficult; let F be a denumerable set that is dense.]
We can definitely choose an F as in the hint for any Rd because Rd is a separable
topological space (points with rational coordinates form a dense set). Since F is a
countable union of singletons, it is an Fσ . To show that it is not Gδ we will refer to the
Baire category theorem.
Definition 3. A topological space X is called a Baire space if for any countable
collection
{An } of closed sets of X each of which has empty interior in X, their union
[
An also has empty interior in X.
Theorem 4. (Baire category theorem) Complete metric spaces are Baire spaces.
Now suppose F is Gδ to get a contradiction. Then F =
\
Un where Un ’s are a count-
n∈N
able collection of open sets. Since F is dense, each Un is dense. Write Cn = Rd r Un .
Then since Int(Cn ) is disjoint from the dense set Un , we have Int(Cn ) = ∅. So
[
Rd r F =
Cn
n∈N
is a countable union of closed sets with empty interiors. But F is also a countable union
of closed sets with empty interiors, namely singletons. Thus Rd = (Rd r F ) ∪ F is a
countable union of closed sets with empty interiors. But by Baire category theorem Rd
is a Baire space hence Rd has empty interior, nonsense.
Therefore F is not Gδ .
(c) Give an example of a Borel set which is not a Gδ nor an Fσ .
Lemma 5. Let X be any topological space and A, B ⊆ X .
(1) If A and B are Fσ sets, so are A ∩ B and A ∪ B.
REAL ANALYSIS I HOMEWORK 2
7
(2) If A and B are Gδ sets, so are A ∩ B and A ∪ B.
Proof. (2) follows from (1) by taking complements. To prove (1), write A =
S
and B = n∈N Ln where Fn , Ln are closed. Then
A∪B =
[
[
(Fn ∪ Ln ), A ∩ B =
n∈N
S
n∈N
Fn
(Fn ∩ Lm )
n,m∈N
are Fσ sets.
Let F = [0, ∞) ∩ Q and G = (−∞, 0] ∩ (R r Q) in R. Note that closed sets in R are
trivially Fσ and also Gδ by part (a). By Lemma 5, F is an Fσ set and G is a Gδ set.
Let E = F ∪ G. Being the union of two Borel sets, E is certainly Borel. Suppose
that E is Gδ . Then by Lemma 5 the set E ∩ [0, ∞) = F is Gδ . But then the set
F 0 = (−∞, 0] ∩ Q is also Gδ because F 0 is the image of F under the homeomorphism
x 7→ −x of R. Thus F ∪ F 0 = Q is Gδ . This contradicts what we’ve shown in part (b).
Thus E is not Gδ .
Now suppose that E is Fσ . Then E ∩ (−∞, 0] = G is Fσ . Similar to above, from here
it follows that Rr Q is Fσ which then implies that Q is Gδ , again a contradiction.
Thus E is neither Gδ nor Fσ .
16. The Borel-Cantelli lemma. Suppose {Ek }∞
k=1 is a countable family of
d
measurable subsets of R such that
∞
X
m(Ek ) < ∞.
k=1
Let
E = {x ∈ Rd : x ∈ Ek , for infinitely many k}
= lim sup(Ek ).
k→∞
(a) Show that E is measurable.
[Hint: Write E =
T∞
n=1
S
k≥n
Ek ].
Let’s verify the hint. Assume x ∈ Ek for infinitely many k. This implies that for
every n ∈ N, there exists k ≥ n such that x ∈ Ek . That is, for every n ∈ N,
S
T
S
T
S
x ∈ k≥n Ek and hence x ∈ n∈N k≥n Ek . Conversely, assume that x ∈ n∈N k≥n Ek .
S
S
Since x ∈ k≥1 Ek , there exists k1 ≥ 1 such that x ∈ Ek1 . And since x ∈ k≥2 Ek ,
there exists k2 > max{2, k1 } such that x ∈ Ek2 . Going on like this, we can construct
a sequence k1 < k2 < k3 · · · such that x ∈ Ekn for every n. Thus x lies in infinitely
many Ek . Having proved the claim, we are done since the collection of measurable sets
is closed under countable unions and countable intersections.
(b) Prove m(E) = 0.
Since
P∞
k=1
m(Ek ) < ∞, for every ε > 0 there exists n ∈ N such that
Ñ
ε>
X
k≥n
m(Ek ) ≥ m
é
[
k≥n
Ek
≥ m(E) .
REAL ANALYSIS I HOMEWORK 2
8
Thus m(E) = 0.
17. Let {fn } be a sequence of measurable functions on [0, 1] with |fn (x)| < ∞ for
a.e. x. Show that there exists a sequence cn of positive real numbers such that
fn (x)
→ 0 a.e. x
cn
[Hint: Pick cn such that m({x : |fn (x)/cn | > 1/n}) < 2−n , and apply the BorelCantelli lemma.]
Fix n. Since |fn | < ∞ a.e., for every n there exists dn > 0 such that
m({x : |fn (x)| > dn }) < 2−n .
Now let cn = ndn . Clearly (cn ) is a sequence of positive real numbers. If we write
®
´
|fn (x)|
> 1/n ,
En = {x : |fn (x)| > dn } = x :
cn
we have
X
m(En ) <
X −n
2
= 1 < ∞.
n∈N
n∈N
So by the Borel-Cantelli lemma, the E := lim supn→∞ (En ) is a null set. Now by
definition, every x ∈ E c lies in only finitely many En ’s. That is, given x ∈ E c , there
exists N ∈ N such that for every n ≥ N we have
|fn (x)|
≤ 1/n .
cn
Therefore
fn (x)
→ 0 for every x ∈ E c .
cn
29. Suppose E is a measurable subset of R with m(E) > 0. Prove that the
difference set of E, which is defined by
{x − y : x, y ∈ E}
contains an open interval centered at the origin.
It is enough to prove the claim for a measurable subset of E with positive measure, so
we do some reductions by finding some nice subsets of E like this and replacing E with
these subsets.
First, since the collection {E ∩ (n, n + 1] : n ∈ Z} of disjoint measurable sets cover E,
by countable additivity there exists n ∈ N such that m(E ∩ (n, n + 1]) > 0. So we may
assume that E has finite measure.
Second, since 0 < m(E) < ∞, by an exercise in the previous homework there exists a
compact set K contained in E such that (choosing ε = m(E)/2)
m(E r K) ≤ m(E)/2 .
Then by addivity of the measure we get m(K) ≥ m(E)/2 > 0. So we may assume that
E is compact.
REAL ANALYSIS I HOMEWORK 2
9
Third, since 0 < m(E) < ∞ there exists an open set U containing E such that
m(U r E) ≤ m(E)/2 .
Hence m(U ) ≤ 3/2 · m(E) < 2m(E). Now E and U c are disjoint sets where E is
compact and U c is closed. Therefore δ := d(E, U c ) > 0. We claim that (−δ, δ) lies in
the difference set of E. So let t ∈ (−δ, δ). Then by the definition of δ, the set
E + t = {x + t : x ∈ E}
does not intersect U c , therefore E + t ⊆ U and hence (E + t) ∪ E ⊆ U . Note that
E + t is a measurable set with m(E + t) = m(E). Suppose (E + t) ∩ E = ∅. Then by
additivity, we get
m(U ) ≥ m((E + t) ∪ E) = 2m(E)
which is a contradiction. Thus there exists x, y ∈ E such that x + t = y, so t lies in the
difference set of E.
30. If E and F are measurable subsets of R and m(E) > 0, m(F ) > 0, prove that
E + F = {x + y : x ∈ E, y ∈ F }
contains an interval.
We show that the difference set ∆(E, F ) = {x − y : x ∈ E, y ∈ F } contains an interval
under the same assumptions. This implies the desired conclusion because then the set
−F = {−y : y ∈ F } is also measurable and m(−F ) = m(F ) > 0, so E +F = ∆(E, −F )
contains an interval.
By Exercise 28, for every α ∈ (0, 1) there exists an open interval I and an open interval
J such that m(E ∩ I) ≥ 34 m(I) and m(F ∩ J) ≥ 43 m(J). WLOG, we may assume
m(I) ≥ m(J). Then there exists a ∈ R such that J + a ⊆ I.
Write δ = 14 m(J). Observe that for any 0 < |c| < δ, the intervals I and J + a + c
intersect in an interval K of length more than 34 m(J). Writing E0 = E ∩ K and
F0 = (F + a + c) ∩ K, we have
1
m(E ∩ I) = m(E0 ) + m(E ∩ (I r K)) < m(E0 ) + m(J)
4
hence
3
1
3
1
1
m(E0 ) > m(I) − m(J) > m(J) − m(J) = m(J) .
4
4
4
4
2
And
m(F ∩ J) = m((F + a + c) ∩ (J + a + c))
= m(F0 ) + m((F + a + c) ∩ (J + a + c)r K)
1
< m(F0 ) + m(J)
4
so
3
1
1
m(F0 ) > m(J) − m(J) = m(J) .
4
4
2
Note that E0 ∪ F0 is contained in K and hence m(E0 ∪ F0 ) ≤ m(K) ≤ m(J). But both
E0 and F0 has measure strictly greater than 12 m(J), so by the addivity of the measure
we conclude that E0 ∩ F0 6= ∅. Thus E ∩ (F + a + c) 6= ∅, so there exists x ∈ F such
REAL ANALYSIS I HOMEWORK 2
10
that x + a + c ∈ E. Therefore a + c ∈ ∆(E, F ). Since this is true whenever 0 < |c| < δ,
the interval (a − δ, a + δ) is contained in ∆(E, F ).
33. Let N denote the non-measurable set constructed in the text. Recall from the
exercise above that measurable subsets of N have measure zero.
Show that the set N c = I − N satisfies m∗ (N c ) = 1, and conclude that if E1 = N
and E2 = N c , then
m∗ (E1 ) + m∗ (E2 ) 6= m∗ (E1 ∪ E2 )
although E1 and E2 are disjoint.
[Hint: To prove that m∗ (N c ) = 1, argue by contradiction and pick a measurable
set U such that U ⊆ I, N c ⊆ U and m∗ (U ) < 1 − ε.]
Suppose m∗ (N c ) < 1, so there exists ε > 0 such that m∗ (N c ) < 1 − ε. Since
m∗ (N c ) = inf{m(U ) : N c ⊆ U -open}
there exists an open, hence measurable set U containing N c such that m(U ) < 1 − ε.
Note that U ∩ I is also a measurable containing N c with m(U ∩ I) < 1 − ε, so we may
assume U ⊆ I. Since N c = I − N ⊆ U ⊆ I, we have I − U ⊆ N . But I − U is a
measurable set so by additivity of the measure we have
m(I − U ) = m(I) − m(U ) = 1 − m(U ) > ε .
So I − U is a measurable subset of N with positive measure; a contradiction. Therefore
m∗ (N c ) ≥ 1 but on the other hand m∗ (N c ) ≤ m∗ (I) = 1 hence m∗ (N c ) = 1.
We know that sets with outer measure zero are measurable, so m∗ (N ) > 0. Thus
m∗ (N ) + m∗ (N c ) > 1 = m∗ (I) = m∗ (N ∪ N c ) .
34. Let C1 and C2 be any two Cantor sets (constructed in Exercise 3). Show that
there exists a function F : [0, 1] → [0, 1] with the following properties:
(i) F is continuous and bijective.
(ii) F is monotonically increasing.
(iii) F maps C1 surjectively onto C2 .
[Hint: Copy the construction of the standard Cantor-Lebesgue function.]
Let C be a Cantor set of constant dissection as in Exercise 3. By construction, C is the
intersection of a family {Cn }n∈N of closed sets where each Cn is a disjoint union of 2n
closed intervals. So we can label these 2n intervals from left to right by bit strings of
length n, that is, words of length n consisting of 0’s and 1’s. So for example C1 = I0 ∪I1
where I0 is the interval on the left hand side in C1 and I1 is the one on the right. Keeping
the labeling in a lexicographic order, we have C2 = I00 ∪ I01 ∪ I10 ∪ I11 and in general
Cn is the union of Ib ’s where b’s vary over length n bit strings.
Note that Ib ⊆ Ic if and only if c can be truncated from the right to obtain b. For
example I0 ⊇ I01 ⊇ I010 ⊇ I0100 . In general given an infinite sequence a = (an ) of 0’s
and 1’s, if we write a|n for its n-truncation (a1 , . . . , an ) there is a decreasing sequence
Ia|1 ⊇ Ia|2 ⊇ Ia|3 · · ·
REAL ANALYSIS I HOMEWORK 2
11
By compactness, the intersection
\
Ia|n
n∈N
which lies in C, is nonempty. Yet the diameter of the intersection is zero, hence it must
be a singleton. Therefore every infinite sequence a of 0’s and 1’s uniquely determines a
point in C. So we get a map
f : {0, 1}N → C
which is surjective since points in C by definition survives the intersection of Cn ’s hence
lie in infinitely many (hence in an infinite decreasing chain of) Ib ’s. If two infinite
sequences are distinct, they have different truncations so as the intervals get finer, the
two points these sequences determine will fall into different intervals. Hence f is a
bijection.
Two points in C lie in the same Ib where b is a finite bit string if and only if their
inverse images under f both start with b. It follows from this observation (as we did
for the middle thirds Cantor set in Exercise 2) that f is continuous. And since f goes
from a compact space to a Hausdorff space, f is a homeomorphism.
Also observe that if we order {0, 1}N by lexicographic ordering, then f preserves the
order. Because if a sequence beats another sequence lexicographically, then at some
point it will lie to the right side of a dissection while the other lies on the left side.
So if C1 and C2 are two Cantor sets, we have order preserving homeomorphisms f1 :
{0, 1}N → C1 and f2 : {0, 1}N → C2 ; thus f2 ◦ f1−1 gives an order preserving homeomorphism from C1 to C2 .
35. Give an example of a measurable function and a continuous function Φ so that
f ◦ Φ is non-measurable.
[Hint: Let Φ : C1 → C2 as in Exercise 34, with m(C1 ) > 0 and m(C2 ) = 0. Let
N ⊂ C1 be non-measurable, and take f = χΦ(N ) .]
Use the construction in the hint to show that there exists a Lebesgue measurable
set that is not a Borel set.
Let’s do as the hint commands. We know such N exists by Exercise 32(b). Since
Φ(N ) ⊆ C2 and m(C2 ) = 0, we have m∗ (Φ(N )) = 0 and so Φ(N ) is a measurable set.
Therefore f is a measurable function. However,
(f ◦ Φ)−1 (1) = Φ−1 (f −1 ({1})) = Φ−1 (Φ(N )) = N
is not measurable, hence f ◦ Φ is not a measurable function. Also the measurable set
Φ(N ) cannot be Borel because the inverse images of Borel sets under continuous functions are Borel, but although Φ is continuous, Φ−1 (Φ(N )) = N is not even measurable.