Serge Ballif
MATH 501 Homework 3
October 10, 2007
In problems below
L
stands for the
σ
algebra of Lebesgue measurable subsets of
d
,
m
for Lebesgue measure on
L
, (
X,
M
, μ
) stands for the measure space, and
L
1
=
L
1
(
X,
M
, μ
).
1.
Let
E
∈ L
be such that
m
(
E
)
<
∞
and let
(
f
k
)
be a sequence of measurable functions defined on
E
. Assume also
that for every
x
∈
E
there exists a constant
M
x
such that

f
k
(
x
)
 ≤
M
x
for all
k
≥
1
. Prove that given
>
0
there
exists a closed set
F
⊂
E
and a constant
M
such that
m
(
E
\
F
)
<
and

f
k
(
x
)
 ≤
M
for all
x
∈
F
and all
k
≥
1
.
Each function
f
k
satisfies the criteria for Lusin’s Theorem:
1.
f
k
is measurable,
2.
f
k
is finite valued on
E
, and
3.
E
has finite measure.
Hence, for every
>
0, there exists a closed set
F
, with
F
⊂
E
and
m
(
E

F
)
≤
/
2 and such that
f

F
is continuous.
The fact that
m
(
E
)
<
∞
also tells us that there exists a compact set
K
such that
K
⊂
E
and
m
(
E

K
)
≤
/
2.
The set
F
∩
K
is compact as an
intersection of a closed set with a compact set. We also have
F
∩
K
⊂
E
and
m
(
E

(
F
∩
K
)) =
m
((
E

F
)
∪
(
E

K
))
≤
m
(
E

F
) +
m
(
E

K
) =
.
Let
F
=
F
∩
K
. Then
f
k
is continuous on
F
since it is continuous on the
larger set
F
.
Hence, the function

f
k

is continuous on
F
as a composition of
continuous functions
 ·  ◦
f
k
. Therefore, as a continuous function on a compact
set,

f
k

always attains a maximum value
M
k
.
We need an
M
that bounds all
M
k
. hmmmmmmmmmmmm
2.
Let
f
:
d
→
be a Lebesgue measurable function. Show there exists a Borel measurable function
g
:
d
→
such that
f
=
g
a.e.
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