MATH 202A HOMEWORK 12
Exercise 4
.
Proof.
Let ˜
g
(
x
) =
R
b
0
χ
{
x
≤
t
}

f
(
t
)

t
dt
. It is clear that the function in the integral is mea
surable and thus by Tonelli Theorem, we have
R
b
0
˜
g
(
x
) =
R
b
0
±
R
b
0
χ
{
x
≤
t
}

f
(
t
)

t
dt
²
dx
=
R
b
0
±
R
b
0
χ
{
x
≤
t
}

f
(
t
)

t
dx
²
dt
=
R
b
0

f
(
t
)

t
±
R
b
0
χ
{
x
≤
t
}
dx
²
dt
=
R
b
0

f
(
t
)

t
±
R
t
0
dx
²
dt
=
R
b
0

f
(
t
)

t
t
dt
=
R
b
0

f
(
t
)

dt
<
∞
.
Hence, ˜
g
is integrable and thus
g
(
x
) is also integrable. Now we can use Fubini
Theorem and similarly, by replacing

f
(
t
)

with
f
(
t
), we have
R
b
0
g
(
x
)
dx
=
R
b
0
f
(
t
)
dt
.
±
Exercise 5
. (a)
Proof.
Since
F
is closed in
R
,
∃
x
0
,y
0
, s.t.

x

x
0

=
δ
(
x
)
,

y

y
0

=
δ
(
y
). Hence,

x

x
0
 ≤ 
x

y
0

,

y

y
0
 ≤ 
y

x
0

.
By triangle inequality, we have

x

y
0

=

x

y
+
y

y
0
 ≤ 
x

y

+

y

y
0

and thus

x

x
0

y

y
0
 ≤ 
x

y
0

y

y
0
 ≤ 
x

y

. Similarly, we have

y

y
0

x

x
0
 ≤ 
x

y

.
Therefore, we know

δ
(
x
)

δ
(
y
)

=

x

x
0
  
y

y
0
 ≤ 
x

y

.
±
(b)
Proof.
F
c
=
t
n
∈
N
U
n
, where
U
n
are disjoint open sets and the expression is unique.
∀
x
∈
F
c
, suppose
x
∈
(
a,b
). Let
I
= (
x
+
a
2
,
x
+
b
2
). Hence,
∀
y
∈
I,δ
(
y
)
≥
min
{
a

x
2
,
b

x
2
}
,
r
.
Thus, we have
I
(
x
)
≥
R
I
δ
(
y
)

x

y

2
dy
≥
R
I
r

x

y

2
dy
≥
r
R
(
a

x
2
,
b

x
2
)
1
z
2
dz
. Since 0
∈
(
a

x
2
,
b

x
2
), we have
R
(
a

x
2
,
b

x
2
)
1
z
2
dz
=
∞
and thus
I
(
x
)
≥ ∞
, which implies
I
(
x
) =
∞
,
∀
x
∈
F
c
.
±
(c)
Proof.
By hint, we investigate
R
F
I
(
x
)
dx
.
Since
δ
(
x
) is continuous,
δ
(
y
)

x

y

2
is also continuous and thus measurable. Hence,
I
(
x
) is nonnegative and measurable function. By extended Fubini Theorem and part
(a), we have
R
F
I
(
x
)
dx
=
R
F
R
F
c
δ
(
y
)

x

y

2
dydx
=
R