5-16

Suppose that

f
≥
0

{\displaystyle f\ge0}

and is measurable on

E

{\displaystyle E}

 and that

ω

{\displaystyle \omega}

 is finite on

(
0
,
∞
)

{\displaystyle (0, \infty)}

. Show that Theorem 5.51 holds without any further restrictions (i.e. we don't need to assume

f
∈

L
p

(
E
)

{\displaystyle f \in L^p(E)}

 and we don't need to assume

E

{\displaystyle |E|}

 is finite) if we interpret

∫
0

∞

α

p
−
1

ω
(
α
)
d
# α

lim

a
→
0
,
b
→
0
∞

∫
a
b

α
p

d
ω
(
α
)

{\displaystyle \int\_0^{\infty} \alpha^{p-1}\omega(\alpha) d\alpha=\lim\_{a\to 0+, b\to 0+\infty} \int\_a^b \alpha^p d \omega(\alpha )}

.

Answer: Given that

f
≥
0

{\displaystyle f\ge0}

 and

E

{\displaystyle |E|}

 is assumed not necessarily finite.

E

{\displaystyle |E|}

could be finite of infinite.

And from theorem 5.46, if

ϕ
(
x
# )

x

p

{\displaystyle \phi(x)=|x|^ p }

, then we have

∫

E

a
b

ϕ
(
f
# )
−

∫
a
b

ϕ
(
α
)
d
ω
(
α
)

{\displaystyle \int\_{E\_{ab}}\phi(f) = - \int\_a^b \phi(\alpha)d \omega(\alpha)}

 is finite.

We write down

f

E

a
b

{\displaystyle f|\_{E \_ {ab}}}

 as the increasing limit of simple measurable functions over partitions of

[
a
,
b
]

{\displaystyle [a, b]}

, and the norms of partitions tends to be 0 (This can be shown in theorem 4.13 (textbook page 68) , Every function f can be written as the limit of a sequence

f
k

{\displaystyle { f\_k }}

 of simple functions.).

And for the Riemman-Stietljes form,

ϕ

{\displaystyle \phi}

 is monotone increasing on nonnegative real numbers, and we have

φ

(

f

k

)

↗
φ
(
f
)

{\displaystyle \varphi\left(f\_{k}\right) \nearrow \varphi(f)}

, so by the Monotone Covergence theorem, we obtain

∫

E

a
b

φ

(

f

k

)

→

∫

E

a
b

φ
(
f
)

{\displaystyle \int\_{E\_{a b}} \varphi\left(f\_{k}\right) \rightarrow \int\_{E\_{a b}} \varphi(f)}

.

φ
(

f
k

)

{\displaystyle \varphi(f \_ k)}

 is a sequence of simple functions , we can get

∑

j

φ

(

α

j
−
1

(
k
)

)

[

ω

(

α

j

(
k
)

)

−
ω

(

α

j
−
1

(
k
)

)

]

≤

∫

E

a
b

φ

(

f

k

)

≤

∑

j

φ

(

α

j

(
k
)

)

[

ω

(

α

j

(
k
)

)

−
ω

(

α

j
−
1

(
k
)

)

]

{\displaystyle \sum\_{j} \varphi\left(\alpha\_{j-1}^{(k)}\right)\left[\omega\left(\alpha\_{j}^{(k)}\right)-\omega\left(\alpha\_{j-1}^{(k)}\right)\right] \leq \int\_{E\_{a b}} \varphi\left(f\_{k}\right) \leq \sum\_{j} \varphi\left(\alpha\_{j}^{(k)}\right)\left[\omega\left(\alpha\_{j}^{(k)}\right)-\omega\left(\alpha\_{j-1}^{(k)}\right)\right]}

.

where

α
j

(
k
)

{\displaystyle \alpha\_j^{(k)}}

 is the j th partition element of

[
a
.
b
]

{\displaystyle [a.b]}

 for the

f
k

{\displaystyle f\_k}

 The norms of partitions approaches 0 as

k
→
∞

{\displaystyle k \rightarrow \infty}

, and

∫

E

a
b

φ

(

f

k

)

→
−

∫

a

b

φ
(
α
)
d
ω
(
α
)

{\displaystyle \int\_{E\_{a b}} \varphi\left(f\_{k}\right) \rightarrow-\int\_{a}^{b} \varphi(\alpha) d \omega(\alpha)}

, we let

a
→
0

{\displaystyle a \rightarrow 0}

 and

b
→
∞

{\displaystyle b \rightarrow +\infty}

, then by monotone covergence theorem it shows that

∫

E

φ

(

f

k

)

→
−

∫

0

∞

φ
(
α
)
d
ω
(
α
)

{\displaystyle \int\_{E} \varphi\left(f\_{k}\right) \rightarrow-\int\_{0}^{ \infty} \varphi(\alpha) d \omega(\alpha)}

.

**Case1 (FINITE CASE) :** Suppose that

∫
0
∞

α
p

d
ω
(
α
)

{\displaystyle \int \_0 ^ \infty \alpha ^ p d \omega ( \alpha )}

 is finite, then

f
∈

L
p

(
E
)

{\displaystyle f \in L^p(E)}

, we using the integrate by parts and we can get

∫

a

b

α

p

d
ω
(
α
# )

α

p

ω
(
α
)

a

b

−
p

∫

a

b

α

p
−
1

ω
(
α
)
d
α
→
−
p

∫

0

∞

α

p
−
1

ω
(
α
)
d
α

{\displaystyle \int\_{a}^{b} \alpha^{p} d \omega(\alpha)=\left.\alpha^{p} \omega(\alpha)\right|\_{a} ^{b}-p \int\_{a}^{b} \alpha^{p-1} \omega(\alpha) d \alpha \rightarrow- p \int\_{0}^{\infty} \alpha^{p-1} \omega(\alpha) d \alpha }

Conversely,

∫
0
∞

α

p
−
1

d
ω
(
α
)

{\displaystyle \int \_0 ^ \infty \alpha ^ {p-1} d \omega ( \alpha )}

 is finite, and integrating by parts again, and

l
i
m

a
→
0

a
p

ω
(
a
)

{\displaystyle {lim}\_{ a \rightarrow 0 } a ^p \omega(a)}

 equal to 0 by exercise 15.

α

p

ω
(
α
)

a

b

−
p

∫

a

b

α

p
−
1

ω
(
α
)
d
# α

∫

a

b

α

p

d
ω
(
α
)
.

{\displaystyle \left.\alpha^{p} \omega(\alpha)\right|\_{a} ^{b}-p \int\_{a}^{b} \alpha^{p-1} \omega(\alpha) d \alpha= \int\_{a}^{b} \alpha^{p} d \omega(\alpha) .}

As

a
→
0

{\displaystyle a \rightarrow 0}

 and

b
→
∞

{\displaystyle b \rightarrow +\infty}

 in both sides all get what we want,

∫

E

f

p

=
−

∫

0

∞

α

p

d
ω
(
α
# )
p

∫

0

∞

α

p
−
1

ω
(
α
)
d
α

{\displaystyle \int\_{E} f^{p}=-\int\_{0}^{\infty} \alpha^{p} d \omega(\alpha)=p \int\_{0}^{\infty} \alpha^{p-1} \omega(\alpha) d\alpha}

**Case II,** if

∫
0
∞

α
p

d
ω
(
α
)

{\displaystyle \int \_0 ^ \infty \alpha ^ p d \omega ( \alpha )}

 or

∫
0
∞

α

p
−
1

d
ω
(
α
)

{\displaystyle \int \_0 ^ \infty \alpha ^ {p-1} d \omega ( \alpha )}

 is not finite, then they are both equal to

∞

{\displaystyle +\infty }

, as f is nonnegative in our assuption. Obviously

∫

E

f

p

=
−

∫

0

∞

α

p

d
ω
(
α
# )
p

∫

0

∞

α

p
−
1

ω
(
α
)
d
# α
∞

{\displaystyle \int\_{E} f^{p}=-\int\_{0}^{\infty} \alpha^{p} d \omega(\alpha)=p \int\_{0}^{\infty} \alpha^{p-1} \omega(\alpha) d\alpha = + \infty}