2
INTRODUCTION TO LEBESGUE INTEGRATION
14
2.5
Convergence Theorems
Example.
1. If
f
n
is measurable and
f
n
!
f
, it is not necessarily true that
R
f
n
!
R
f
. Let
f
n
(
x
) be
n
for
x
2
(0
,
1
n
) and 0
otherwise. That is
f
n
=
n
χ
(0
,
1
n
)
. Note that
f
n
!
0 pointwise, but
R
[0
,
1]
f
n
= 1 for all
n
.
2. Let
f
n
= 1 on [
n, n
+ 1] and 0 otherwise. Then
f
n
!
0, but
R
R
f
n
= 1 for all
n
.
Theorem
(Monotone Convergence Theorem)
.
Suppose
f
n
≥
0 (allowing
1
) are measurable functions with
f
n
(
x
)
%
f
(
x
) for all
x
. Then
R
E
f
= lim
n
!1
R
E
f
(
n
) for all measurable
E
.
Note.
Note that this fails for Riemann integrability: Let
Q
\
[0
,
1] =
{
r
n
}
1
1
. Define
f
n
(
x
) = 1 if
x
=
r
1
, . . . , r
n
and
0 otherwise.
f
n
≥
0 an measurable for all
n
, and
f
n
(
x
)
%
f
=
χ
Q
\
[0
,
1]
. Then
R
R
1
0
f
n
= 0, but
R
R
1
0
f
does not exist.
However,
R
[0
,
1]
f
n
= 0 =
R
[0
,
1]
f
.
Lemma.
Let
φ
be positive and simple,
A
n
be measurable sets with
A
n
✓
A
n
+1
for all
n
. Put
A
=
S
1
n
=1
A
n
. Then
R
A
n
φ
!
R
A
φ
, or equivalently
R
φχ
A
n
!
R
φχ
A
.
Proof.
This is a special case of MCT with
f
n
=
φχ
A
n
and
f
=
φχ
A
. Let
φ
=
P
N
i
=1
a
i
χ
E
i
. Then
Z
A
n
φ
=
N
X
i
=1
a
i
m
(
A
n
\
E
i
)
.
Note that
E
i
\
A
n
✓
E
i
\
A
n
+1
8
n
and
S
1
n
=1
E
i
\
A
n
=
E
i
\
A
. By continuity of measure,
m
(
E
i
\
A
n
)
!
m
(
E
i
\
A
)
for each
i
. Hence
N
X
i
=1
a
i
m
(
E
i
\
A
n
)
!
N
X
i
=1
a
i
m
(
E
i
\
A
) =
Z
A
φ
.
Proof of MCT.
Recall that
f
n
measurable,
f
n
!
f
implies
f
measurable. Since
f
n
f
, by monotonicity,
R
E
f
n
R
E
f
.
lim
n
!1
R
E
f
n
exists by monotonicity and lim
n
!1
R
E
f
n
R
E
f
. We must prove that
R
E
f
lim
n
!1
R
E
f
n
. Recall
that
R
E
f
= sup
0
φ
f
φ
simple
R
E
φ
, thus it will be enough to prove that
R
φ
lim
n
!1
R
f
n
for all simple
φ
with 0
φ
f
.
Fix such a
φ
, and take any 0
<
↵
<
1. Define
A
n
=
{
x
2
E
:
f
n
(
x
)
≥
↵φ
(
x
)
}
=
E
\
(
f
n

↵φ
)

1
[0
,
1
], which are
measurable. Also note that
A
n
✓
A
n
+1
since the
f
n
s are increasing. Now we claim that
S
A
n
=
E
. Let
x
2
E
. If
φ
(
x
) = 0 then
f
n
(
x
)
≥
↵φ
(
x
)
8
n
, so
x
2
A
n
8
n
. Otherwise
φ
(
x
)
>
0, in which case
↵φ
(
x
)
<
φ
(
x
)
f
(
x
). Since
f
n
(
x
)
!
f
(
x
), for large enough
n
we have
f
n
(
x
)
>
↵φ
(
x
), so
x
2
A
n
, and hence
x
2
S
A
n
. Hence
E
✓
S
A
n
, and
clearly
S
A
n
✓
E
, so
S
A
n
=
E
. Now by the lemma,
↵
Z
A
n
φ
!
↵
Z
S
A
j
φ
=
↵
Z
E
φ
.
On the other hand,
↵
Z
A
n
φ
=
Z
A
n
↵φ
Z
A
n
f
n
=
Z
f
n
χ
A
n
Z
f
n
χ
E
=
Z
E
f
n
!
lim
n
!1
Z
E
f
n
So
↵
R
E
φ
lim
n
!1
R
E
f
n
for every
↵
2
(0
,
1). Let
↵
!
1 to get
R
E
φ
lim
R
E
f
n
.
Proposition.
If
f, g
≥
0 are measurable, then
R
f
+
g
=
R
f
+
R
g
.