M ATH 531
R EAL A NALYSIS I
FALL 2009
(Some) Solutions to Homework # 2
Definition: Let (X, M, µ) be a measure space. If for each E ∈ M with µ(E) = ∞ there exists F ∈ M with F ⊆ E and
0 < µ(F ) < ∞, µ is called semifinite.
Folland, p.27, Exercise 14: If µ is a semifinite measure and µ(E) = ∞, for any C > 0 there exists F ⊆ E with
C < µ(F ) < ∞.
Solution. Set
α := sup {µ(F ) : F ∈ I(E)}
where I(E) = {F ∈ M : F ⊆ E, µ(F ) < ∞} .
Since µ is semifinite, α > 0. Let us see that α = ∞. For otherwise, for each n ∈ N, there exists Fn ∈ I(E)
such that µ(Fn ) ≥ α − 1/n. Since µ is subadditive, for all n ∈ N, ∪n
j=1 Fn ∈ I(E) and thus
´
` n
α ≥ µ ∪j=1 Fj ≥ µ(Fn ) ≥ α − 1/n.
Accordingly, for F = ∪n∈N Fn ⊂ E, the continuity of µ from below implies that
´
`
µ(F ) = lim µ ∪n
j=1 Fj = α
n→∞
and F ∈ I(E). But then µ(E\F ) = ∞, and since µ is semifinite, there exists F ′ ⊆ E\F with 0 <
µ(F ′ ) < ∞. This provides a contradiction since F ∪ F ′ ∈ I(E) and µ(F ∪ F ′ ) = α + µ(F ′ ) > α.
Folland, p.27, Exercise 15: Given a measure µ on (X, M), define µ0 on M by
µ0 (E) = sup {µ(F ) : F ∈ I(E)}
where I(E) = {F ∈ M : F ⊆ E, µ(F ) < ∞} .
a. µ0 is a semifinite measure. It is called the semifinite part of µ.
Solution. Clearly µ0 (∅) = 0. Let us now see that µ0 is countably-additive. To this end, let {En }n∈N ⊆ M
be a sequence of disjoint sets. Set E = ∪n∈N En . Note that
F ∈ I(E)
⇒
⇒
⇒
⇒
F ⊆ E & µ(F ) < ∞
F ∩ En ⊆ E ∩ En = En
F ∩ En ∈ I(En )
µ(F ∩ En ) ≤ µ0 (En )
&
µ(F ∩ En ) ≤ µ(F ) < ∞
so that, since {F ∩ En }n∈N is a sequence of disjoint sets, we get
X
X
µ0 (En );
µ(F ∩ En ) ≤
µ(F ) = µ(F ∩ E) = µ (∪n∈N F ∩ En ) =
n∈N
n∈N
this shows that
µ0 (E) ≤
X
µ0 (En ).
n∈N
The reverse inequality clearly holds when µ0 (E) = ∞. We may thus assume that µ0 (E) < ∞. Since
I(En ) ⊆ I(E), we then have that µ0 (E) ≤ µ0 (E) < ∞. Now given n ∈ N and ε > 0, choose
−j
Fj ∈ I(EP
< µ(Fj )[≤ µ0 (Ej ) < ∞]. Then F = ∪n
j ) (j = 1,. . . ,n) such that µ0 (Ej ) − ε2
j=1 Fj ⊆ E and
Pn
µ(F ) = n
µ(F
)
≤
µ
(E
)
<
∞
so
that
F
∈
I(E).
Accordingly,
j
0
j
j=1
j=1
µ0 (E) ≥ µ(F ) =
n
X
j=1
µ(Fj ) >
n “
X
j=1
n
” X
µ0 (Ej ) − ε2−j >
µ0 (Ej ) − ε.
j=1
Letting first ε → 0, and then n → ∞, we thus get
µ0 (E) ≥
X
µ0 (En ).
n∈N
Therefore, µ0 is a measure on (X, M). As is clear from its definition, µ0 is semifinite.
b. If µ is semifinite, then µ = µ0 . (Use Exercise 14.)
Solution. Since µ is monotone, we have that µ0 (E) = µ(E) when µ(E) < ∞. On the other hand, if
µ(E) = ∞, then Exercise 14 implies that µ0 (E) = ∞.
c. There is a measure ν on M (in general, not unique) which assumes only the values 0 and ∞ such that µ = µ0 + ν.
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M ATH 531
R EAL A NALYSIS I
FALL 2009
Solution. Let us agree to say that a set E ∈ M is semifinite for µ if for all F ⊆ E with µ(F ) = ∞, there
exists F ′ ⊆ F such that 0 < µ(F ′ ) < ∞. Define ν : M → [0, ∞] by setting

0,
if E is semifinite for µ,
ν(E) =
∞, otherwise.
Note that any E ∈ M with µ(E) < ∞ is semifinite for µ. Thus in particular ν(∅) = 0. Let us see that ν
is countably additive. To this end, let {En }n∈N ⊆ M be a sequence of disjoint sets, and let E = ∪n∈N En .
We have two cases:
(i) For some n ∈ N, En fails to be semifinite for µ: Clearly, in this case, E cannot be semifinite for µ.
Accordingly, we have
X
ν(En ).
ν(E) = ∞ =
n∈N
(ii) Each En is semifinite for µ: In this case, we have to show that E too is semifinite so that the identity
X
ν(En )
ν(E) = 0 =
n∈N
holds. Indeed, if F ⊆ E is any set with µ(F ) = ∞, then the identity
X
∞ = µ(F ) = µ(F ∩ E) =
µ(F ∩ En )
n∈N
shows that µ(F ∩ En0 ) > 0 for at least one n0 ∈ N. If µ(F ∩ En0 ) < ∞, we set F ′ = F ∩ En0 .
Otherwise, F ∩ En0 ⊆ En0 and µ(F ∩ En0 ) = ∞ so that, since En0 is semifinite, there exists
F ′ ⊆ F ∩ En0 with 0 < µ(F ′ ) < ∞. Therefore E is semifinite for µ as claimed.
We have thus shown that ν is a measure on (X, M).
Let us now see that µ = µ0 + ν. For let E ∈ M. If µ(E) < ∞, then µ0 (E) = µ(E), and E is semifinite
for µ so that ν(E) = 0. Thus, in this case, µ(E) = µ0 (E) + ν(E). If µ(E) = ∞ and E is semifinite for µ,
then Exercise 14 shows that µ0 (E) = ∞, and this entails µ(E) = µ0 (E)+ν(E). If µ(E) = ∞ and E is not
semifinite for µ, then the identity µ(E) = µ0 (E) + ν(E) holds trivially.
Exercise: Show that ν is not the unique such measure.
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