Ex2-30

**Chapter 2, Question 30**

Let

f

{\displaystyle f}

and

ϕ

{\displaystyle \phi }

 be real-valued functions on

[
a
,
b
]

{\displaystyle [a,b]}

.

(a) If

∫

a

b

f
d
ϕ

{\displaystyle \int \_{a}^{b}fd\phi }

 exists and

ϕ

{\displaystyle \phi }

 is not constant on any subinterval of

[
a
,
b
]

{\displaystyle [a,b]}

, show that

f

{\displaystyle f}

 is bounded on

[
a
,
b
]

{\displaystyle [a,b]}

.

Proof: We assume

f

{\displaystyle f}

is unbounded on

[
a
,
b
]

{\displaystyle [a,b]}

,

ϕ

{\displaystyle \phi }

 is not constant on any subinterval of

[
a
,
b
]

{\displaystyle [a,b]}

 and

∫

a

b

f
d
ϕ

{\displaystyle \int \_{a}^{b}fd\phi }

 exists, then get a contradiction. Without loss of generality, we assume

f

{\displaystyle f}

is unbounded above, then there is a sequence

{

x

n

}

{\displaystyle \{x\_{n}\}}

in

[
a
,
b
]

{\displaystyle [a,b]}

such that

{
f
(

x

n

)
}

{\displaystyle \{f(x\_{n})\}}

is strictly increasing and unbounded. By closedness of

[
a
,
b
]

{\displaystyle [a,b]}

 and the fact that every sequence in

R

{\displaystyle \mathbb {R} }

 has a monotone subsequence [see for example, Theorem 2.32 in Advanced Calculus, Fitzpatrick], we find that

{

x

n

}

{\displaystyle \{x\_{n}\}}

 has a monotone subsequence

{

x

n

k

}

{\displaystyle \{x\_{n\_{k}}\}}

that converges to a point

x
∈
[
a
,
b
]

{\displaystyle x\in [a,b]}

. For the sake of simplicity, we rename the subsequence as

{

y

k

}

{\displaystyle \{y\_{k}\}}

and without loss of generality, assume

x
>
a

{\displaystyle x>a}

 and

{

y

k

}

{\displaystyle \{y\_{k}\}}

 is an increasing sequence. For the case where

# x
a

{\displaystyle x=a}

 and

{

y

k

}

{\displaystyle \{y\_{k}\}}

 decreasing, one can repeat the proof below by changing just a few words.

From assumptions, we know

lim

Γ

→
0

R

Γ

{\displaystyle \lim \_{|\Gamma |\to 0}R\_{\Gamma }}

exists and is equal to

∫

a

b

f
d
ϕ

{\displaystyle \int \_{a}^{b}fd\phi }

, so there is

δ
>
0

{\displaystyle \delta >0}

 such that

∫

a

b

f
d
ϕ
−

R

Γ

<
1

 whenever

Γ

 is a partition such that

Γ

<
δ
.

Since

ϕ

{\displaystyle \phi }

 is not constant on any subinterval, we can choose

x
′

<
x
,
x
−

x
′

<
δ

{\displaystyle x'<x,x-x'<\delta }

 such that

ϕ
(
x
)
≠
ϕ
(

x
′

)

{\displaystyle \phi (x)\neq \phi (x')}

 (this can be done, or otherwise

ϕ

{\displaystyle \phi }

 will be constant on the interval

(
x
−
δ
,
x
]

{\displaystyle (x-\delta ,x]}

). Let

Γ

{\displaystyle \Gamma }

 be a partition satisfying

Γ

<
δ

{\displaystyle |\Gamma |<\delta }

 that contains the two adjacent partition points

x
′

,
x

{\displaystyle x',x}

. Then we fix

ξ

1

,
…
,

ξ

n

{\displaystyle \xi \_{1},\ldots ,\xi \_{n}}

to be the inputs of

f

{\displaystyle f}

 in

R

Γ

{\displaystyle R\_{\Gamma }}

 except the one in

[

x
′

,
x
]

{\displaystyle [x',x]}

, say

# ξ

ξ

i

{\displaystyle \xi =\xi \_{i}}

, which can be changed. Now we treat

R

Γ

=

R

Γ

(
ξ
)

{\displaystyle R\_{\Gamma }=R\_{\Gamma }(\xi )}

 as a function of

ξ
∈
[

x
′

,
x
]

{\displaystyle \xi \in [x',x]}

 and investigate its behavior. Using the triangle inequality we should have

R

Γ

(
ξ
)
−

R

Γ

(

ξ
′

)

<
2

 for all

ξ
,

ξ
′

∈
[

x
′

,
x
]
.

{\displaystyle |R\_{\Gamma }(\xi )-R\_{\Gamma }(\xi ')|<2\quad {\text{ for all }}\xi ,\xi '\in [x',x].}

We choose

K

{\displaystyle K}

 a positive integer large enough such that

y

k

∈
[

x
′

,
x
]

{\displaystyle y\_{k}\in [x',x]}

 for

k
≥
K

{\displaystyle k\geq K}

. Then for

k
≥
K

{\displaystyle k\geq K}

 we have

2
>

R

Γ

(
x
)
−

R

Γ

(

y

k

)

=

f
(
x
)
−
f
(

y

k

)

ϕ
(
x
)
−
ϕ
(

x
′

)

,

which is a contradiction since

{
f
(

y

k

)

}

# k
K

∞

{\displaystyle \{f(y\_{k})\}\_{k=K}^{\infty }}

is unbounded above and

ϕ
(
x
)
−
ϕ
(

x
′

)

>
0

{\displaystyle |\phi (x)-\phi (x')|>0}

.