SOLUTION OF HW1
MINGFENG ZHAO
February 05, 2011
1. (Page 27, 11) A finitely additive measure µ is a measure iff it is continuous from below as in
Theorem 1.8c. If µ(X) < ∞, µ is a measure iff it is continuous from above as in Theorem 1.8d.
Proof. (a) (⇐=) Assume µ is a finitely additive measure, and continuous from below. We want to
show µ is a measure, it suffices to show µ is countablely additive.
For any {Ek }∞
k=1 of µ-measurable subsets of X such that Ei
Sn
k=1
Ek , then Fn ⊂ Fm if n ≤ m, and
∞
[
µ
S∞
n=1
Fn =
n=1
∞
[
!
En
S∞
=µ
n=1
T
Ej = φ if i 6= j. Define Fn =
En . By the continuity from below of µ, then
!
Fn
= lim µ(Fn )
n→∞
n=1
Since µ is finitely additive, and E1 , · · · , En are disjoint, then
µ(Fn ) = µ
n
[
!
Ek
=
n
X
µ(Ek ).
k=1
k=1
Hence
µ
∞
[
∞
[
!
En
= µ
n=1
!
Fn
n=1
=
=
=
lim µ(Fn )
n→∞
lim
n
X
n→∞
∞
X
µ(Ek )
k=1
µ(Ek )
Since µ(Ek ) ≥ 0
k=1
Therefore, µ is countablely additive.
(=⇒) If µ is a measure, by Theorem 1.8c, we know µ is continuous from below.
1
2
MINGFENG ZHAO
(b) (⇐=) Assume µ is a finitely additive measure such that µ(X) < ∞, and µ is continuous from
above. Since µ(X) < ∞, then for all µ-measurable subset E of X, we have µ(E) ≤ µ(X) < ∞. We
want to show µ is a measure, by the result of part (a), it suffices to show µ is also continuous from
below.
For any {Ek }∞
k=1 of µ-measurable subsets of X such that Ei ⊂ Ej if i ≤ j.
S∞
k=1
Ek , X\ (
S∞
k=1
Then
Ek ) are µ-measurable, and X\Ek is measurable for all k ≥ 1, and X\Ei ⊃ X\Ej
if i ≤ j. Since µ is continuous from above, then
(1)
∞
\
µ
!
X\Ek
= lim µ(X\Ek ).
k→∞
k=1
T∞
By De Morgan’s laws, we have
k=1
S∞
X\Ek = X\ ( k=1 Ek ). Since µ is fintedly additive, and
µ(X) < ∞, then µ(X\Ek ) = µ(X) − µ(Ek ) for all k ≥, and
µ
∞
\
∞
[
!
X\Ek
= µ X\
k=1
!!
Ek
= µ(X) − µ
k=1
∞
[
!
Ek
.
k=1
By (1), we know
µ(X) − µ
∞
[
!
= lim [µ(X) − µ(Ek )] = µ(X) − lim µ(Ek ).
Ek
k→∞
k=1
k→∞
Since µ(X) < ∞, then we get
µ
∞
[
k=1
!
Ek
= lim µ(Ek ).
k→∞
So µ is also continuous from below.
(=⇒) If µ is a measure, and µ(X) < ∞, by Theorem 1.8d, we know µ is continuous from below.
2. (Page 27, 12) Let (X, M, µ) be a finite measure space.
a If E, F ∈ M and µ(E∆F ) = 0, then µ(E) = µ(F ).
b Say that E ∼ F if µ(E∆F ) = 0, then ∼ is an equivalence relation on M.
SOLUTION OF HW1
3
c For E, F ∈ M, define ρ(E, F ) = µ(E∆F ). Then ρ(E, G) ≤ ρ(E < F ) + ρ(F < G), and hence
ρ defines a metric on the space M/ ∼ of equivalence classes.
Proof. a. Assume E, F ∈ M, and µ(E∆F ) = 0.
Claim: E ⊂ F
S
(E∆F ).
In fact, for any x ∈ E, if x ∈ F , then x ∈ F
Since x is arbitrary in E, then E ⊂ F
µ(E) ≤ µ(F
S
[
S
(E∆F ); if x ∈
/ F , then x ∈ E\F , so x ∈ F
S
(E∆F ).
(E∆F ). Therefore, we have
(E∆F )) ≤ µ(F ) + µ(E∆F ) = µ(F ).
By the similar argument as above, we can get µ(F ) ≤ µ(E). Therefore µ(E) = µ(F ).
b. 1.(Reflexivity) For any E ∈ M, then E∆E = φ. So µ(E∆E) = µ(φ) = 0, which implies E ∼ E.
2.(Symmetry) If E, F ∈ M, and E ∼ F .
Claim: E∆F = F ∆E.
S
S
In fact, E∆F = (E\F ) (F \E) = (F \E) (E\F ) = F ∆E.
So µ(E∆F ) = µ(F ∆E) = 0, which implies F ∼ E.
3. (Transitivity) If E, F, G ∈ M, and E ∼ F , and F ∼ G. So µ(E∆F ) = µ(F ∆G) = 0.
Claim: E∆G ⊂ (E∆F
S
F ∆G).
For any x ∈ E∆G = E\G
S
G\E. Case I: x ∈ E\G. If x ∈ E\G, and x ∈ F , then x ∈ F \G, which
implies x ∈ F ∆G, so x ∈ (E∆F
x ∈ E∆F , so x ∈ (E∆F
we can get x ∈ (E∆F
S
S
S
F ∆G); if x ∈ E\G, and x ∈
/ F , then x ∈ E\F , which implies
F ∆G). By the similar argument as above for the case II: x ∈ G\E, then
F ∆G).
Anyway we know x ∈ (E∆F
S
F ∆G), so E∆G ⊂ (E∆F
S
F ∆G).
Then
µ(E∆G) ≤ µ(E∆F
[
F ∆G) ≤ µ(E∆F ) + µ(F ∆G) = 0.
Hence µ(E∆G) = 0, which implies E ∼ G.
4
MINGFENG ZHAO
By 1,2,3, we know ∼ is an equivalence relation on M.
c. i. (Positive-definite) For any E, F ∈ M such that ρ(E, F ) = µ(E∆F ) = 0, by the result of part b,
we know E ∼ F , that is E = F in M/ ∼.
ii. (Symmetry) For any E, F ∈ M, then ρ(E, F ) = µ(E∆F ) = µ(F ∆E) = ρ(F, E).
iii. (Triangle inequality) For any E, F, G ∈ M, by the proof of part b, we know that
[
E∆G ⊂ E∆F
F ∆G .
Therefore, µ(E∆G) ≤ µ (E∆F
S
F ∆G) ≤ µ(E∆F ) + µ(F ∆G), that is, ρ(E, G) ≤ ρ(E, F ) +
ρ(F, G).
By i,ii,iii, we know ρ is a metric on M/ ∼.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu