For (b): For
p
=
∞
,
the conclusion fails. Take
E
=
R
and
f
k
=
χ
(
−
k,k
)
, f
= 1
.
Then
f
k
, f
∈
L
∞
(
R
)
, f
k
→
f
a.e. on
R
,
and
k
f
k
k
∞
=
k
f
k
∞
= 1
.
But we do not have
k
f
k
−
f
k
∞
→
0
as
k
→ ∞
.
Another example is
E
= (0
,
1)
, f
k
=
χ
(1
/k,
1)
, f
= 1
, k
∈
N
.
Then we have the same conclusion.
¤
7. (15 points) Let
E
= (0
,
∞
)
×
(0
,
1)
⊂
R
2
and
f
(
x, y
) =
ye
−
xy
sin
x,
(
x, y
)
∈
E.
(a) (5 points) Show that
f
(
x, y
)
is
integrable
on
E,
i.e.,
f
∈
L
(
E
)
.
(b) (10 points) Evaluate the integral
ZZ
E
f
(
x, y
)
dxdy.
3
solution:
For (a), note that
ZZ
E

f
(
x, y
)

dxdy
≤
ZZ
E
ye
−
xy
dxdy
and by Tonelli Theorem we have
ZZ
E
ye
−
xy
dxdy
=
Z
1
0
∙
Z
∞
0
ye
−
xy
dx
¸
dy
=
Z
1
0
dy
= 1
which implies that
f
∈
L
(
E
)
.
For (b), we apply Fubini Theorem and get
ZZ
E
f
(
x, y
)
dxdy
=
Z
1
0
y
μZ
∞
0
e
−
xy
sin
xdx
¶
dy.
Recall the elementary formula
Z
e
ax
sin
bxdx
=
e
ax
(
a
sin
bx
−
b
cos
bx
)
a
2
+
b
2
+
C,
a, b
are constants
we see that for each
y >
0
Z
∞
0
e
−
xy
sin
xdx
=
e
−
xy
(
−
y
sin
x
−
cos
x
)
y
2
+ 1
¯
¯
¯
¯
x
=
∞
x
=0
=
1
y
2
+ 1
and so
Z
1
0
y
μZ
∞
0
e
−
xy
sin
xdx
¶
dy
=
Z
1
0
y
y
2
+ 1
dy
=
1
2
log 2
.
¤
8. (20 points) Answer
Yes
or
No
to each statement. Just provide the answers.
(a) If
x
is a point of density of a measurable set
E,
then
x
∈
E.
ANS: _________
(b) Let
0
< p <
∞
.
If
k
f
k
−
f
k
p,E
→
0
as
k
→ ∞
,
then there exists a subsequence
f
k
j
→
f
almost
everywhere in
E
as
j
→ ∞
.
ANS: _________
(c) Let
f
and
g
be two
nonnegative
measurable functions on measurable set
E
satisfying
{
x
∈
E
:
f
(
x
)
> y
}
=
{
x
∈
E
:
g
(
x
)
> y
}
for all
y >
0
.
Then we must have
R
E
f
=
R
E
g.
ANS: _________
(d)
k
f
k
−
f
k
∞
,E
→
0
as
k
→ ∞
if and only if
f
k
→
f
uniformly
on
E
except on a set
Z
⊂
E
of
measure zero.
ANS: _________
solution:
The answers to the five problems are: NO, YES, YES, YES.
¤
4
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 Fall '08
 Akhmedov,A
 dx, lim sup