SOLUTIONS OF HW7
MINGFENG ZHAO
April 04, 2011
(Problem 46, in Page 68) Let X = Y = [0, 1], M = N = B[0,1] , and µ=Lebesgue mea-
1.
If D = {(x, x) :
sure, and ν=counting measure.
RR
χD dµdν,
RR
χD dνdµ, and
R
x ∈ [0, 1]} is the diagonal in X × Y , then
χD d(µ × ν) are all unequal.
Proof. Consider the function f : [0, 1] × [0, 1] → R given by
f (x, y) = x − y.
So f is continuous on [0, 1] × [0, 1], so f is (B[0,1] × B[0,1] , BR )-measurable. In particular, D =
{(x, x) : x ∈ [0, 1]} = f −1 (0) is in B[0,1] × B[0,1] . By Proposition 2.34,in Page 65, we know for any
x, y ∈ [0, 1], Dx = {(x, x)}, Dy = {(y, y)} are in B[0,1] . Hence for any y ∈ [0, 1], we have
Z
Z
χD (x, y)dµ(x)
=
dµ(x)
{y}
[0,1]
=
µ({y})
=
0.
Therefore, we have
Z Z
Z
χD dµdν
Z
=
χD (x, y)dµ(x)dν(y)
[0,1]
[0,1]
Z
=
0dν(y)
[0,1]
=
0.
1
2
MINGFENG ZHAO
And also for any x ∈ [0, 1], we know
Z
Z
χD (x, y)dν(y)
=
dν(y)
{x}
[0,1]
= ν({x})
=
1.
Therefore, we have
Z Z
Z
Z
χD (x, y)dν(y)dµ(x)
χD dνdµ =
[0,1]
[0,1]
Z
=
1dµ(x)
[0,1]
=
For
R
1.
χD d(µ × ν), we know χD is the simple function on D, then
Z
χD d(µ × ν) = (µ × ν)(D).
By the definition of µ × ν, we know
(
(µ × ν)(D) = inf
∞
X
(µ × ν)(Ak × Bk ) : Ak , Bk ∈ B[0,1] ,
D⊂
k=1
For any D ⊂
S∞
k=1 (Ak ×Bk ),
∞
[
)
(Ak × Bk ) .
k=1
which means that for any 0 ≤ x ≤ 1, we have (x, x) ∈
S∞
k=1 (Ak ×Bk ).
In particular, we get
x∈
∞
[
x∈
Ak ,
k=1
So we get x ∈
S∞
k=1 (Ak
∞
[
Bk .
k=1
∩ Bk ). By the arbitrary of x ∈ [0, 1], we get
[0, 1] ⊂
∞
[
(Ak ∩ Bk ).
k=1
S∞
P∞
Hence µ([0, 1]) ≤ µ ( k=1 (Ak ∩ Bk )) ⊂ k=1 µ(Ak ∩ Bk ). So at least there exists some k0 such
that
µ(Ak0 ∩ Bk0 ) > 0.
SOLUTIONS OF HW7
3
Hence we get µ(Ak0 ) > 0, and µ(Bk0 ) > 0, which implies that Bk0 is an infinity set, so ν(Bk0 ) = ∞.
So we have
(µ× =
6 )(Ak0 × Bk0 ) = µ(Ak0 ) · ν(Bk0 ) = ∞.
So we get
P∞
k=1 (µ
× ν)(Ak × Bk ) = ∞. Since D ⊂
S∞
k=1 (Ak
× Bk ) is arbitrary, then by the
definition of (µ × ν)(D), we get
(µ × ν)(D) = ∞.
This implies that
Z
χD d(µ × ν) = (µ × ν)(D) = ∞.
2. (Problem 50, in Page 69) Suppose (X, M, µ) is a σ-finite measure space, and f ∈ L+ (X). Let
Gf = {(x, y) ∈ X × [0, ∞] : y ≤ f (x)}.
Then Gf is M ⊗ BR -measurable, and (µ × m)(Gf ) =
R
f dµ; the same is also true if the inequality
y ≤ f (x) in the definition of Gf is replaced by y < f (x). This is the definitive statement of the
familiar theorem from calculus, “ the integral of a function is the area under its graph”.
Proof. Without loss of generality, we assume 0 ≤ f (x) < ∞ for all x ∈ X. Consider the function
F : X × [0, ∞) −→ R2 given by
F (x, y) = (f (x), y).
For any A ∈ BR , since f is (BR , BR )-measurable, then f −1 (A) ∈ BR . So for any B ∈ B[0,∞) , we
get
F −1 (A × B)
= f −1 (A) × (B) ∈ BR × B[0,∞) ⊂ BR ⊗ B[0,∞)
Since BR ⊗ B[0,∞) = σ(BR × B[0,∞) ), then F is (M ⊗ B[0,∞) , R ⊗ B[0,∞) )-measurable.
4
MINGFENG ZHAO
Let the function G : R × [0, ∞) −→ R given by
G(x, y) = x − y.
Then G is continuous on R × [0, ∞), then G is (BR×[0,∞) , BR )-measurable. Therefore, we have
G ◦ F is (M ⊗ B[0,∞) , BR )-measurable. In particular, we have
Gf
=
{(x, y) ∈ X × [0, ∞) : y ≤ f (x)} = (G ◦ F )−1 ((−∞, 0]) ∈ M ⊗ B[0,∞)
G̃f
=
{(x, y) ∈ X × [0, ∞) : y ≤ f (x)} = (G ◦ F )−1 ((−∞, 0)) ∈ M ⊗ B[0,∞)
Consider χGf (x, y), χG̃f (x, y), so they are (M ⊗ B[0,∞) , BR )-measurable. Since M ⊗ B[0,∞) is σfinite, therefore we have
Z
(µ × m)(Gf )
χGf (x, y)d(µ × m)
=
X×[0,∞)
Z Z
=
χGf (x, y)dm(y)dµ(x)
X
By Tonelli’s theorem
[0,∞)
Z Z
=
dm(y)dµ(x)
X
[0,f (x)]
Z
=
f (x)dµ(x)
X
Z
(µ × m)(G̃f )
=
X×[0,∞)
χG̃f (x, y)d(µ × m)
Z Z
=
X
[0,∞)
χG̃f (x, y)dm(y)dµ(x)
By Tonelli’s theorem
Z Z
=
dm(y)dµ(x)
X
[0,f (x))
Z
=
f (x)dµ(x)
X
3. (Problem 7, in Page 88) Suppose that ν is a signed measure on (X, M) and E ∈ M. Show that
a. ν + (E) = sup {ν(F ) : E ∈ M, F ⊂ E}, and ν − (E) = − inf {ν(F ) : E ∈ M, F ⊂ E};
b. |ν|(E) = sup
nP
n
j=1
o
|ν(Ej )| : n ∈ N, E1 , · · · , En are disjoint, and ∪nj=1 Ej = E .
SOLUTIONS OF HW7
5
Proof. a. For any F ⊂ E, by the Jordan Decomposition Theorem, in Page 87, we know ν + and ν − are
two positive measure on X. So by the monotonicity of positive measure, we know ν + (F ) ≤ ν + (E),
and ν − (F ) ≤ ν − (E). Hence
ν(F )
ν(F )
=
ν + (F ) − ν − (F )
≤
ν + (F )
≤
ν + (E)
=
ν + (F ) − ν − (F )
≥
−ν − (F )
≥
−ν − (E)
By the arbitrary of F ⊂ E, we know
ν + (E) ≥
sup {ν(F ) : E ∈ M, F ⊂ E}
−ν − (E) ≤
inf {ν(F ) : E ∈ M, F ⊂ E}
On the other hand, by the Hahn Decomposition Theorem, in Page 86, we know that there exists
sets P and N such that ν(P ) ≥ 0, ν(N ) ≤ 0, and P ∪ N = X, P ∩ N = φ. And also by the definition
of ν + , ν − in the Jordan Decomposition Theorem, in Page 87, we know ν + (E) = ν(E ∩ P ), ν − (E) =
−ν(E ∩ N ). But E ∩ P ⊂ E, E ∩ N ⊂ E. Therefore, we get
ν + (E)
=
ν − (E)
= − inf {ν(F ) : E ∈ M, F ⊂ E}
sup {ν(F ) : E ∈ M, F ⊂ E}
6
MINGFENG ZHAO
b. For any E1 , · · · , En are disjoint, and ∪nj=1 Ej = E. By the Jordan Decomposition Theorem, in
Page 87, we know
n
X
n
X
|ν(Ej )| =
j=1
|ν + (Ej ) − ν − (Ej )|
j=1
n
X
≤
|ν + (Ej )| + |ν − (Ej )|
j=1
n
X
=
|ν|(Ej )
j=1
|ν|(E)
=
Since Ej ’s are disjoint, and |ν| is a measure on X
Hence we get
|ν|(E) ≥ sup

n
X

|ν(Ej )| : n ∈ N, E1 , · · · , En are disjoint, and ∪nj=1 Ej = E



j=1
On the other hand, by the Hahn Decomposition Theorem, in Page 86, we know that there exists
sets P and N such that ν(P ) ≥ 0, ν(N ) ≤ 0, and P ∪ N = X, P ∩ N = φ. So we can get
|ν|(E)
=
ν + (E) + ν − (E)
=
ν(E ∩ P ) − ν(E ∩ N )
By the definition of ν + , ν − in the Jordan Decomposition Theorem, in Page 87
=
|ν(E ∩ P )| + |ν(E ∩ N )|
Therefore, we get
|ν|(E) = sup

n
X

|ν(Ej )| : n ∈ N, E1 , · · · , En are disjoint, and ∪nj=1 Ej = E
j=1



Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu