MA2224 (Lebesgue integral) Tutorial sheet 4
[February 12, 2016]
Name: Solutions
1. Suppose that A is a σ-algebra of subsets of R and µ : A → [0, ∞] is a measure. (So that
means µ(∅) = 0 and µ is countably additive.)
(a) Show that µ is finitely additive.
Solution: Start with E1 , E2 ∈ A disjoint and define Ej = ∅ for j > 2. Then we have
a (pairwise) disjoint sequence E1 , E2 , . . . ∈ A and we know
!
∞
∞
X
[
µ(En )
µ
En =
n=1
n=1
by countable additivity. This reduces to
µ(E1 ∪ E2 ) = µ(E1 ) + µ(E2 ) + 0 = µ(E1 ) + µ(E2 )
[If you prefer to state finite additivity using any finite number of disjoint sets E1 , E2 , . . . , En
in A, you can define Ej = ∅ for j > n and continue more or less as before. But of
course 2 sets is enough because induction gives it for any finite number.]
(b) Show that µ is monotone: E1 , E2 ∈ A, E1 ⊆ E2 implies µ(E1 ) ≤ µ(E2 )
Solution: Write E2 = E1 ∪ (E2 ∩ E1c ) (where E1 , E2 ∈ A, E1 ⊆ E2 ). Using finite
additivity (and disjointness of E 1 and E2 ∩ E1c , which are both in A)
µ(E2 ) = µ(E1 ∪ (E2 ∩ E1c )) = µ(E1 ) + µ(E2 ∩ E1c )) ≥ µ(E1 )
P∞
S
E
)
≤
(c) Show that µ is countably subadditive: E1 , E2 , . . . ∈ A ⇒ µ ( ∞
n
n=1 µ(En ).
n=1
Solution: Starting with E1 , E2 , . . . ∈ A define F1 = E1 and for n > 1
!
n−1
[
Fn = En \
Ej
j=1
to get a disjoint sequence F1 , F2 , . . . ∈ A such that
Fn ⊆ En and
n
[
Fj =
j=1
n
[
Ej
j=1
S
S∞
(for all n ≥ 1). Moreover ∞
j=1 Fj =
j=1 Ej . So by countable additivity of µ
!
!
∞
∞
∞
∞
[
X
X
[
µ
Ej = µ
Fj =
µ(Fj ) ≤
µ(Ej )
j=1
j=1
j=1
j=1
(using monotonicty of µ at the last inequality). This shows countable subadditivity.
2. Show that the collection Σ of all subsets S ⊆ R satisfying one of m∗ (S) = 0 or m∗ (R \
S) = 0 is a σ-algebra (of subsets of R).
Solution: S = ∅ is included because m∗ (∅) = 0 and Σ is closed under complements since
m∗ (S) = 0 ⇒ T = S c has m∗ (R \ T ) = m∗ (S) = 0 ⇒ T ∈ Σ
and
m∗ (R \ S) = 0 ⇒ T = S c has m∗ (T ) = 0 ⇒ T ∈ Σ
S
For countable unions ∞
n=1 En with En ∈ Σ∀n, there are two cases
!
∞
∞
∞
[
X
[
∗
∗
∗
m (En ) = 0∀n ⇒ m
En ≤
m (En ) = 0 ⇒
En ∈ Σ
n=1
n=1
n=1
or
∗
∗
∗
∃m with m (Em ) 6= 0 ⇒ m (R \ Em ) = 0 ⇒ m
R\
∞
[
!
En
≤ m∗ (R \ Em ) = 0
n=1
since m∗ is monotone and Em ⊆
in this case also.
S∞
n=1
En ⇒ R \
Richard M. Timoney
2
S∞
n=1
En ⊆ R \ Em . Thus
S∞
n=1
En ∈ Σ