(b) If
E
1
∪
F
1
, E
2
∪
F
2
∈
M
and
E
1
∪
F
1
=
E
2
∪
F
2
where
E
i
∈ M
and
F
i
is a subset
of a set
N
i
of measure 0, then by monotonicity and subadditivity
μ
(
E
1
∪
F
1
) =
μ
(
E
2
∪
F
2
)
≤
μ
(
E
2
∪
N
2
)
≤
μ
(
E
2
)+
μ
(
N
2
) =
μ
(
E
2
)+0. Hence
μ
(
E
2
∪
F
2
)
≤
μ
(
E
2
).
Monotonicity gives us the reverse inequality, so
μ
(
E
1
∪
F
1
) =
μ
(
E
2
∪
F
2
) =
μ
(
E
2
).
By a symmetric argument
μ
(
E
1
∪
F
1
) =
μ
(
E
2
∪
F
2
) =
μ
(
E
1
).
(c) We must show that
μ
(
∅
) = 0 and that if
{
E
i
∪
F
i
}
∞
i
=1
is a sequence of dis
joint sets in
M
, then
μ
(
S
∞
i
=1
E
i
) =
∑
∞
i
=1
μ
(
E
i
).
Since
∅ ∈ M
we know that
μ
(
∅
) =
μ
(
∅
) = 0. It follows immediately from part (b) that
μ
satisfies the rules
for monotonicity and subadditivity.
Applying the subadditivity of
μ
, part (b),
additivity of
μ
, and monotonicity of
μ
(in that order) we get
μ
∞
[
i
=1
(
E
i
∪
F
i
)
!
≤
∞
X
i
=1
μ
(
E
i
∪
F
i
) =
∞
X
i
=1
μ
(
E
i
) =
μ
∞
[
i
=1
E
i
!
≤
μ
∞
[
i
=1
(
E
i
∪
F
i
)
!
.
Therefore,
μ
(
S
∞
i
=1
E
i
) =
∑
∞
i
=1
μ
(
E
i
), so
μ
is a measure. To see that
μ
is complete
we note that if
μ
(
E
∪
F
) = 0, then
μ
(
E
) = 0 =
μ
(
F
). If
E
0
∪
F
0
⊂
E
∪
F
where
E
0
⊂
E
and
F
0
⊂
F
, then each of the sets (
E
∪
F
)
⊃
(
E
∪
F
0
)
⊃
(
E
0
∪
F
0
) is in
M
by definition.
3.Let(fn)be a sequence of measurable functionsfn:X→. Prove that:
(a) The set
L
=
{
x
∈
X

lim
n
→∞
f
n
(
x
)
exists
}
belongs to
M
.
(b) If
(
X,
M
, μ
)
is complete and
f
n
→
f μ
a.e., then
f
is measurable.
(a) We will break up the proof into two claims that combine to prove that
L
is
measurable.
Claim 1.
{
x
∈
X

lim
n
→∞
f
n
(
x
) =
∞}
and
{
x
∈
X

lim
n
→∞
f
n
(
x
) =
∞}
are measurable sets.
Claim 2.
If we assume that each function
f
n
is finite
valued, then
L
is measurable.
Proof of Claim 1.
Both sets
{
x
∈
X

lim
n
→∞
f
n
(
x
) =
∞}
=
T
∞
N
=1
S
∞
n
=1
{
x

f
n
(
x
)
>
N
}
, and
{
x
∈
X

lim
n
→∞
f
n
(
x
) =
∞}
=
T
∞
N
=1
S
∞
n
=1
{
x

f
n
(
x
)
<

N
}
are count
able intersections of countable unions of measurable sets. Hence they are mea
surable.
Page 3 of 5
Serge Ballif
MATH 501 Homework 2
October 3, 2007
Proof of Claim 2.
We note first that the real valued sequence lim
n
→∞
f
n
(
x
) exists if
and only if
f
n
(
x
) is a Cauchy sequence. That is, for
>
0 there exists
N
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 Math, measure, Lebesgue measure, Lebesgue integration, lim sup Ej, Hj Ej