# (Some) Solutions to Homework # 2

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