Serge Ballif
MATH 501 Homework 4
October 22, 2007
In problems 1 and 5, (
X,
M
,μ
) stands for the measure space and
L
1
=
L
1
(
X,
M
,μ
).
1.
Let
(
X,
M
,μ
)
be a measure space such that
μ
(
X
)
<
∞
. Assume that
(
f
n
)
is a sequence of integrable functions
f
n
:
X
→
C
converging uniformly to the function
f
on
X
. Prove that
R
f
n
→
R
f
.
The sequence (
f
n
) is uniformly convergent with limit
f
if for every
± >
0, there
exists a natural number
N
such that for all
x
in
X
and all
n
≥
N
, we have

f
n
(
x
)

f
(
x
)

< ±
.
Claim 1.
The function
f
is in
L
1
.
Proof.
We need to show that
R

f

<
∞
. Using the triangle inequality, together
with the facts that
μ
(
X
)
<
∞
and
R

f
n

<
∞
we can calculate
Z

f

=
Z

f

f
n
+
f
n

≤
Z

f

f
n

+

f
n

=
Z

f

f
n

+
Z

f
n

≤
Z
±
+
Z

f
n

=
±μ
(
X
) +
Z

f
n

<
∞
.
Thus,
f
is in
L
1
.
To see that
R
f
k
→
R
f
we must show that for any
± >
0, there is an
N >
0
such that for
n
≥
N
,

R
f
n

R
f

< ±
. Using the uniform convergence of the
sequence (
f
n
) we can choose
N
such that for
n > N
we have

f
n

f

<
±
μ
(
X
)
.
Now,
±
±
±
±
Z
f
n

Z
f
±
±
±
±
=
±
±
±
±
Z
f
n

f
±
±
±
±
≤
Z

f
n

f

≤
Z
±
μ
(
X
)
=
±.
Hence,
R
f
k
→
R
f
.
2.
Let
f
n
(
x
) =
n
1+
n
6
x
2
. Compute
lim
n
→∞
R
[
a,
∞
)
f
n
dm
when
a <
0
,
a
= 0
, and
a >
0
.
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View Full DocumentSerge Ballif
MATH 501 Homework 4
October 22, 2007
Case 1: Suppose
a >
0. Then

f
n
(
x
)

=

n
1+
n
6
x
2

<
1
x
2
for all
x
. Since the positive
function
1
x
2
is Riemann integrable for
a >
0, it is Lebesgue integrable. Therefore,
by the dominated convergence theorem, lim
n
→∞
R
[
a,
∞
)
f
n
dm
=
R
[
a,
∞
)
lim
n
→∞
f
n
dm
. We
calculate the limit lim
n
→∞
f
n
(
x
) = 0 for all
x
except
x
= 0, where lim
n
→∞
f
n
(0) =
∞
.
Thus, lim
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 Fall '08
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 Math, lim, Dominated convergence theorem, Lebesgue integration, Serge Ballif, fn dm

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