Royden, Real Analysis 3rd ed.
Chapter 4
Problem 42
a.
Let
h
be the upper envelop of
f
.
It is known from
Problem 251 that
h
(
y
) = max
μ
f
(
y
)
,
lim
x
→
y
f
(
x
)
¶
.
(1)
If
ϕ
≥
f
is a step function, it is clear from
(1)
that
ϕ
(
y
)
≥
lim
x
→
y
f
(
x
)
≥
h
(
y
)
(2)
at each continuous point
y
of
ϕ
.
Hence
(2)
holds except at the jump points
of
ϕ
,
which are
fi
nite. As a result,
Z
b
a
h
≤
Z
b
a
ϕ
and thus
R
Z
b
a
f
≥
Z
b
a
h.
(3)
By Problem 250
g
, there exists step functions
ϕ
n
such that
ϕ
n
↓
h.
By
Bounded Convergence Theorem,
R
Z
b
a
f
≥
lim
Z
b
a
ϕ
n
= lim
Z
b
a
ϕ
n
=
Z
b
a
h.
(4)
The conclusion follows from
(3)
and
(4)
.
b.
Let
g
be the lower envelop of
f.
Similar to part
a,
we have
R
Z
b
a
f
=
Z
b
a
g.
Hence
f
is Riemann integrable if and only if
Z
b
a
g
=
Z
b
a
h.
(5)
Because
g
≤
f
≤
h,
(5)
implies
g
(
y
) =
h
(
y
)
a.e. (See Problem 43). The
conclusion follows from Problem 251
a
.
Problem 43
Let
A
n
=
{
x
:
f
(
x
)
>
1
/n
}
for
n
= 1
,
2
, . . .
. Since
{
x
:
f
(
x
)
6
= 0
}
=
{
x
:
f
(
x
)
>
0
}
=
∪
∞
n
=1
A
n
,
it su
ﬃ
ces to show that
m
(
A
n
) = 0
for each
n
≥
1
,
which is shown as below
0 =
Z
f
≥
Z
f
·
χ
A
n
≥
Z
n
−
1
·
χ
A
n
=
n
−
1
m
(
A
n
)
.
1
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Problem 44
a.
It is clear that
f
∧
n
↑
f.
But
f
∧
n
is not simple.
Let
E
n,k
=
{
x
:
k
·
2
−
n
< f
(
x
)
≤
(
k
+ 1)
·
2
−
n
}
.
De
fi
ne
g
n
(
x
) =
2
2
n
X
k
=
−
2
2
n
k
·
2
−
n
χ
E
n,k
(
x
)
.
Then

g
n

≤
2
n
and
g
n
↑
f.
There is still no guarantee that
m
{
x
:
g
n
(
x
)
6
= 0
}
<
∞
.
This can be remedied by introducing
ϕ
n
(
x
) =
g
n
(
x
)
·
χ
[
−
n,n
]
(
x
)
.
b.
By Proposition 8(iii),
Z
f
≥
sup
½Z
ϕ
:
ϕ
is a simple function with
ϕ
≤
f
¾
.
Applying the Monotone Convergence Theorem to
{
ϕ
n
}
in part
a.
sup
Z
ϕ
≥
lim
Z
ϕ
n
=
Z
f.
The conclusion follows.
Problem 45
Note that
F
(
x
) =
Z
x
−∞
f
(
t
)
dt
=
Z
f
(
t
)
χ
(
−∞
,x
]
(
t
)
dt.
Since
χ
(
−∞
,x
n
]
→
χ
(
−∞
,x
]
everywhere except possibly at
x
for any
x
n
→
x,
we have from Theorem 10,
lim
n
→∞
F
(
x
n
) =
F
(
x
)
for any
x
n
↑
x.
(1)
That is
F
(
x
−
) =
F
(
x
)
.
In order to show
F
(
x
+) =
F
(
x
)
,
it su
ﬃ
ces to
verify
(1)
still holds when
x
n
↓
x.
Suppose all
x
n
≤
x
+ 1
.
Then by Theorem
10,
F
(
x
+ 1)
−
F
(
x
n
) =
Z
f
(
t
)
χ
(
x
n
,x
+1]
↑
Z
f
(
t
)
χ
(
x,x
+1]
=
F
(
x
+ 1)
−
F
(
x
)
.
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 Spring '08
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 Calculus, lim fn, lim F, Dominated convergence theorem, Monotone convergence theorem

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