SOLUTIONS OF HW2
MINGFENG ZHAO
February 14, 2011
1. (Page 39, 28) Let F be increasing and right continuous, and let µF be the associated measure. Then
µF ({a}) = F (a) − F (a−), µF ([a, b)) = F (b−) − F (a−), µF ([a, b]) = F (b) − F (a−), and µF ((a, b)) =
F (b−) − F (a).
Proof. Let F be increasing and right continuous, by 1.16 Theorem in Page 35, we know that we can
find an associated Borel measure µF such that
µF ((a, b]) = F (b) − F (a)
for all a, b ∈ R.
Since F is increasing, so F (a−) exists for all a ∈ R. We look at En = a − n1 , a , then En ⊃ En+1
for all n ≥ 1, and
T∞
n=1
En = {a}, and µF (En ) = F (a) − F a −
1
n
. And also we know µF (E1 ) =
F (a)−F (a−1) < ∞. Since µF is a measure, then {a} is µF -measurable. By the continuity of measure
from above, we get
µF ({a})
=
=
=
=
lim µF (En )
n→∞
1
lim F (a) − F a −
n→∞
n
1
F (a) − lim F a −
n→∞
n
F (a) − F (a−).
S
For µF ([a, b]), and a < b, since [a, b] = (a, b] {a}, and (a, b], {a} are µF -measurable, then [a, b] is
µF -measurable.And we have
µF ([a, b])
= µF ((a, b]
[
{a})
1
2
MINGFENG ZHAO
=
µF ((a, b]) + µF ({a})
By additivity of µF
=
F (b) − F (a) + F (a) − F (a−)
=
F (b) − F (a−).
S
Since [a, b) {b} = [a, b], so [a, b) = [a, b]\{b}. Since [a, b], {b} are µF -measurable, then [a, b) is
µF -measurable. And by additivity of µF , we have
µF ([a, b]) = µF ([a, b)) + µF ({b})
That is, F (b) − F (a−) = µF ([a, b)) + F (b) − F (b−), so
µF ([a, b))
= F (b) − F (a−) − (F (b) − F (b−))
= F (b) − F (a−) − F (b) + F (b−)
= F (b−) − F (a−).
S
Since (a, b) {b} = (a, b], so (a, b) = (a, b]\{b}. Because (a, b], {b, } are µF -measurable, then (a, b)
is µF -measurable. And by additivity of µF , we have
µF ((a, b]) = µF ((a, b)) + µF ({b}).
So we get
µF ((a, b))
= µF ((a, b]) − µF ({b})
= F (b) − F (a) − (F (b) − F (b−))
= F (b) − F (a) − F (b) + F (b−)
= F (b−) − F (a).
SOLUTIONS OF HW2
3
2. (Page 39, 29) Let E be a Lebesgue measurable set.
a. If E ⊂ N , where N is the nonmeasurbale set described in 1.1, then m(E) = 0.
b. If m(E) > 0, then E contains a nonmeasurbale set.
T
Proof. a. If m(E) > 0, consider Er = r + E = {r + e : e ∈ E}, for any r ∈ Q [0, 1]. From the
construction of N , we know that Er
T
T
Es = φ if r 6= s, r, s ∈ Q [0, 1]. On the other hand, from
T
the definition of m, we know Er is also m-measurable for all r ∈ Q [0, 1]. And by the translation
invariant of m, we get m(Er ) = m(E) > 0 for all r ∈. But since N ⊂ [0, 1], then Er ⊂ [0, 2] for all
r∈Q
T
[0, 1] , which implies that
S
r∈Q
T
[0,1]
Er ⊂ [0, 2], so m
S
r∈Q
T
[0,1]
Er ≤ 2.
T
Since Q [0, 1] is countable, then by the additivity of measure, hence we have

m

[
r∈Q
T
Er 
X
=
[0,1]
r∈Q
r∈Q
= ∞
S
Since Er
\
Es = φ if r 6= s, r, s ∈ Q
m(E)
T
[0,1]
Since m(E) > 0, and Q
T
r∈Q [0,1]
\
[0, 1] is countable.
Er ≤ 2. Therefore, m(E) = 0.
b. For any E which is m-measurable, and m(E) > 0. For any n ∈ Z, let En = E
En
T
\
[0, 1]
T
[0,1]
X
=
This contradicts with m
m(Er )
T
[n, n + 1). Then
El = φ if n 6= l, n, l ∈ Z, and En is m-measurable.
Claim I: There exists some k ∈ Z such that m(Ek ) > 0.
If m(En ) = 0 for all n ∈ Z.By the additivity of m, we have
!
m(E)
= m
[
En
n∈Z
=
X
m(En )
n∈Z
=
X
0
n∈Z
=
0,
which contradicts with m(E) > 0. Therefore, there exists some k ∈ Z such that m(Ek ) > 0.
4
MINGFENG ZHAO
Claim II: There exists some subset F ⊂ Ek ⊂ E which is not m-measurable.
If all subsets of Ek are measurable. Since Ek ⊂ [k, k + 1), then Ek − k ⊂ [0, 1). For any r ∈
[−1, 1]
T
Q, we let
Fr = (Ek − k)
Since [0, 1] ⊂
S
r∈[−1,1]
T
Q
\
Nr = Ek
Nr , so Ek − k =
S
\
r∈[−1,1]
Since all subsets of Ek are m-measurable, then Ek
m, we know Fr = Ek
T
Nr+k − k ⊂ [0, 1].
T
T
Q
Fr .
Nr+k is m-measurable. By the definition of
Nr+k − k is m-measurable. So Fr − r ⊂ Nr − r = N is m-measurable, by the
result of part a, we know m(Fr − r) = 0, so m(Fr ) = 0. Since [0, 1] ⊂

m(Ek − k)
=
X
r∈[−1,1]
Q
Nr , then
T
Q
m(Fr )
T
Q
T
Q
X
r∈[−1,1]
=
T
Fr 
r∈[−1,1]
=
r∈[−1,1]

[
m
=
S
0
0,
which contradicts with m(Ek − k) = m(Ek ) > 0. Therefore, There exists some subset F ⊂ Ek ⊂ E
which is not m-measurable.
Remark 1. By the same idea as book, we can get another proof of part b.
For any n ∈ N, let En = E
T
[−n, n]. Then En is m-measurable, and En ⊂ En+1 for all n ∈ N.
Since m(E) > 0, by the continuity of measure from below, we know for
such that whenever n ≥ N , we have m(En ) > m(E) −
m(E)
2
=
m(E)
2
m(E)
2
> 0, there exists N ∈ N
> 0.
Let G = EN , then m(G) > 0.
Now we will find a subset F ⊂ G = EN ⊂ E such that F is not m-measurable. (Follow the same
idea before.)
1. Firstly, we define an equivalence relation ∼ on G: We say x ∼ y if x − y ∈ Q.
SOLUTIONS OF HW2
5
Claim I: ∼ is an an equivalence relation ∼ on G.
i. (Reflexivity) For any x ∈ G, we know x − x = 0 ∈ Q, so x ∼ x.
ii. (Symmetry) For any x, y ∈ G, and x ∼ y, that is x − y ∈ Q. Since Q is a field, then
y − x = −(x − y) ∈ Q. So y ∼ x.
iii. (Transitivity) For any x, y, z ∈ G, and x ∼ y, y ∼ z, that is x − y ∈ Q, y − z ∈ Q. Since Q is a
field, then x − z = (x − y) + (y − z) ∈ Q. So x ∼ z.
By i, ii, iii, we know that ∼ is an an equivalence relation ∼ on G.
Let [x] = {y ∈ G : x ∼ y}.
T
Claim II: For any x, y ∈ G, if [x] [y] 6= φ, then [x] = [y].
T
T
Since [x] [y] 6= φ, we take z ∈ [x] [y]. Then x ∼ z, and y ∼ z. By the transitivity of ∼, we know
x ∼ y.
For any t ∈ [x], then t ∼ x. Since x ∼ y, by the transitivity of ∼, then t ∼ y, which implies t ∈ [y].
On the other hand, for any t ∈ [y], then t ∼ y. Since x ∼ y, By the transitivity of ∼, we know t ∼ x,
which implies t ∈ [x]. Therefore, we get [x] = [y].
Let P = {[x] : x ∈ G}. For any A ∈ P, we know A = [x] for some x ∈ G. Since x ∈ [x], so A 6= φ.
Claim III: P is a partition of G.
For any x ∈ G, we know x ∈ [x], and [x] ∈ P, so x ∈
For any A, B ∈ G, if A
T
S
A∈P
A.
B 6= φ. From the definition of P, we know A = [x], B = [y] for some
T
x, y ∈ G, then [x] [y] 6= φ. By Claim II, we know [x] = [y], that is A = B.
Therefore, P is a partition of G.
Now we assume P = {Bα : α ∈ A}, where A is an index set such that α1 6= α2 , then Bα1
T
Bα2 =
φ. For each Bα , we take one point xα in Bα . Let F = {xα : α ∈ A}. Hence F ⊂ G. For any
T
r ∈ Q [−2R, 2R], let Fr = F + r = {f + r : f ∈ F }.
Claim IV: Fr
T
T
Fs = φ if r 6= s, r, s ∈ Q [−2R, 2R].
6
MINGFENG ZHAO
If r 6= s, r, s ∈ Q
T
[−2R, 2R], but Fr
T
Fs 6= φ. We say z ∈ Fr
T
Fs . So z − r ∈
T
F , and
z − s ∈ F , let z − r = f1 , z − s = f2 , which implies z = r + f1 = s + f2 . So r − s = f2 − f1 ∈ Q.
By the definition of F , we know f1 = f2 , so r = s, contradiction. Therefore, we have Fr
r 6= s, r, s ∈ Q
T
Claim V: G ⊂
S
T
Fs = φ if
[−2R, 2R].
r∈Q
T
[−2R,2R]
Fr ⊂ [−4R, 4R].
For any x ∈ G, by the definition of F , there exists y ∈ F such that x ∼ y, that is x − y ∈ Q. Since
T
x, y ∈ G ⊂ [−R, R], then −2R ≤ x − y ≤ 2R, so x − y ∈ Q [−2R, 2R], hence x ∈ Fx−y . By the
arbitrary of x ∈ G, we conclude that G ⊂
S
On the other hand, since for any x ∈
S
r∈Q
T
r∈Q
[−2R,2R]
Fr .
[−2R,2R]
Fr , we have x = y + r for some y ∈ G, r ∈
T
T
Q [−2R, 2R]. But we know y, r ∈ [−2R, 2R], then x = y + r ∈ [−4R, 4R]. By the arbitrary of
x∈
S
r∈Q
T
[−2R,2R] ,
therefore, we get
S
r∈Q
T
[−2R,2R]
Fr ⊂ [−4R, 4R].
Claim VI: F is not m-measurable.
T
If F is m-measurable. By the definition of F , we know for any r ∈ Q [−2R, 2R], Fr is measurable,
and m(F ) = m(Fr ).
By Claim V, we know 0 <
m(E)
2
≤ m(G) ≤ m(F ) ≤ 8R, and
S
r∈Q
T
[−2R,2R]
Fr ⊂ [−4R, 4R].
Hence

m([−4R, 4R]) ≥

[
m
r∈Q
T
X
=
r∈Q
T
r∈Q
T
m(Fr )
By Claim IV
[−2R,2R]
X
=
Fr 
[−2R,2R]
m(F )
[−2R,2R]
T
Since Q [−2R, 2R] is countable, and m([−4R, 4R]) < ∞, then m(F ) = 0, which contradicts with
m(F ) ≥ m(G) ≥
m(E)
2
> 0.
Therefore, F is not m-measurable subset of E.
SOLUTIONS OF HW2
7
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu