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Analysis 2
Homework 03
1. Let (Ω
,
F
) be a measurable space on which the measures
μ
and
ν
satisfy
μ
(Ω) =
ν
(Ω)
<
∞
. Must the collection
E
=
{
F
∈ F
:
μ
(
F
) =
ν
(
F
)
}
be a
σ
algebra? Proof or counterexample.
2. Let Ω =
N
=
{
1
,
2
,...
}
and on
F
=
P
(Ω) deﬁne the (probability) measure
μ
by
μ
(
{
n
}
) =
1
n
(
n
+ 1)
.
For
F
n
=
{
n,n
+ 1
,...
}
calculate
∑
∞
n
=1
μ
(
F
n
) and
μ
(lim sup
n
→∞
F
n
). Any
comments?
Solutions
1. Answer: ‘No; it need not be’. Notice that
E
is closed under complementa
tion and countable pairwisedisjoint unions. The problem is that
σ
algebras
must be closed under
all
countable unions, but the
σ
additivity of a measure
only controls
pairwisedisjoint
countable unions; a counterexample comes by
arranging nonempty overlap. Thus, let Ω =
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 Spring '09
 LARSON

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