Ex2-11

Chapter 2 Question 11

Question: Show that

∫
a
b

f
d
ϕ

{\displaystyle \int\_a^b fd\phi}

 exists if and only if given

ε
>
0

{\displaystyle \varepsilon > 0}

, there exists

δ
>
0

{\displaystyle \delta > 0}

 such that

R
Γ

−

R

Γ
′

<
ε

{\displaystyle |R\_\Gamma-R\_{\Gamma'}|<\varepsilon}

 if

Γ

,

Γ
′

<
δ

.

Proof: First assume

∫
a
b

f
d
ϕ

{\displaystyle \int\_a^b fd\phi}

 exists, then for partition

# Γ
{

x
0

,

x
1

,
…
,

x
n

}

{\displaystyle \Gamma=\{x\_0,x\_1,\ldots,x\_n\}}

 of

[
a
,
b
]

{\displaystyle [a, b]}

 and

n

{\displaystyle n}

points

ξ
1

,
…
,

ξ
n

{\displaystyle \xi\_1,\ldots,\xi\_n}

 satisfying

x

i
−
1

≤

ξ
i

≤

x
i

{\displaystyle x\_{i-1}\leq \xi\_i\leq x\_i}

 for all

# i
1
,
…
,
n

{\displaystyle i=1,\ldots,n}

, the equality below holds

lim

Γ

→
0

R
Γ

=

∫
a
b

f
d
ϕ
,

{\displaystyle \lim\_{|\Gamma|\to 0}R\_\Gamma=\int\_a^b fd\phi,}

where

R
Γ

{\displaystyle R\_\Gamma}

is the Riemann-Stieltjes sum for

Γ

{\displaystyle \Gamma}

. Given

ε
>
0

{\displaystyle \varepsilon > 0}

 we may choose

δ
>
0

{\displaystyle \delta > 0}

 so that if

Γ

<
δ

{\displaystyle |\Gamma|<\delta}

then

∫
a
b

f
d
ϕ
−

R
Γ

<

ε
2

{\displaystyle \left|\int\_a^b fd\phi-R\_\Gamma\right|<\dfrac\varepsilon2\quad}

Therefore, by triangle inequality, for

Γ

,

Γ
′

<
δ

 we have

R
Γ

−

R

Γ
′

≤

∫
a
b

f
d
ϕ
−

R
Γ

∫
a
b

f
d
ϕ
−

R

Γ
′

<
ε
.

To prove the converse, assume that given

ε
>
0

{\displaystyle \varepsilon > 0}

, there exists

δ
>
0

{\displaystyle \delta > 0}

 such that

R
Γ

−

R

Γ
′

<
ε

{\displaystyle |R\_\Gamma-R\_{\Gamma'}|<\varepsilon}

 if

Γ

,

Γ
′

<
δ

. We define

# U

lim sup

Γ

→
0

R
Γ

 and

# L

lim inf

Γ

→
0

R
Γ

.

We claim both

U
,
L

{\displaystyle U,L}

 are finite. We will only prove that

U

{\displaystyle U}

is finite, since the proof of the finiteness of

L

{\displaystyle L}

 is similar. If

U

{\displaystyle U}

is not finite, then it is either

∞

{\displaystyle +\infty }

 or

−
∞

{\displaystyle -\infty }

. If

U

{\displaystyle U}

 is

∞

{\displaystyle +\infty }

, we can construct a sequence of partitions

{

Γ
n

}

{\displaystyle \{\Gamma\_n\}}

 (with some intermediate choices of points

ξ
i

{\displaystyle \xi\_i}

) such that

Γ
n

<

1
n

 and

R

Γ
n

>

R

Γ

n
−
1

1

 for
n
≥

{\displaystyle |\Gamma\_n|<\frac1n\quad\text{ and }\quad R\_{\Gamma\_n}>R\_{\Gamma\_{n-1}}+1\quad\text{ for }n\geq 2.}

However, we know from assumption that there is

δ
>
0

{\displaystyle \delta > 0}

 such that

R

Γ

−

R

Γ
′

<
1

{\displaystyle |R\_{\Gamma}-R\_{\Gamma'}|<1}

 for

Γ

,

Γ
′

<
δ

, this is contradictory once we choose

# Γ

Γ
n

,

Γ
′

=

Γ

n
1

{\displaystyle \Gamma=\Gamma\_n, \Gamma'=\Gamma\_{n+1}}

 with

1
n

<
δ

{\displaystyle \frac1n<\delta}

. If on the other hand,

# U
−
∞

{\displaystyle U=-\infty}

, then given a real number

A

{\displaystyle A}

 there is

δ
>
0

{\displaystyle \delta > 0}

 such that

Γ

<
δ

{\displaystyle |\Gamma|<\delta}

 implies

R
Γ

<
A

{\displaystyle R\_\Gamma<A}

. Therefore, we can choose another sequence of partitions

{

Γ
n
′

}

{\displaystyle \{\Gamma\_n'\}}

 such that

Γ
n
′

<

1
n

 and

R

Γ
n
′

<

R

Γ

n
−
1

′

−
1

 for all
n
≥

{\displaystyle |\Gamma\_n'|<\frac1n \quad\text{ and }\quad R\_{\Gamma\_n'}<R\_{\Gamma\_{n-1}'}-1\quad\text{ for all }n\geq 2.}

Using the assumption again, we obtain a similar contradiction. Thus

U

{\displaystyle U}

 is finite, as well as

L

{\displaystyle L}

.
Now we know both

U
,
L

{\displaystyle U,L}

 are finite, in particular,

L
≤
U

{\displaystyle L\leq U}

. It suffices to prove

# L
U

{\displaystyle L=U}

 to complete the proof, because we could conclude from

L
≤

lim

Γ

→
0

R
Γ

≤
U

{\displaystyle L\leq \lim\_{|\Gamma|\to 0}R\_\Gamma\leq U}

. Assume the contrary that

L
<
U

{\displaystyle L<U}

, let

# ε
(
U
−
L
)

/

3

{\displaystyle \varepsilon=(U-L)/3}

, then we can choose

δ
>
0

{\displaystyle \delta > 0}

 such that

R
Γ

−

R

Γ
′

<
ε

∀

Γ

,

Γ
′

<
δ
.

By definitions of

lim sup

{\displaystyle \limsup}

and

lim inf

{\displaystyle \liminf}

, we can choose partitions with

Γ

,

Γ
′

<
δ

 such that

R
Γ

<
L
ε
,

R

Γ
′

>
U
−
ε

{\displaystyle R\_\Gamma<L+\varepsilon, R\_{\Gamma'}>U-\varepsilon}

. However, this lead us to

R

Γ
′

>
U
−
# ε
L
2
ε
>

R
Γ

ε

⟹

R

Γ
′

−

R

Γ

>
ε
,

{\displaystyle R\_{\Gamma'}>U-\varepsilon=L+2\varepsilon>R\_\Gamma+\varepsilon\implies |R\_{\Gamma'}-R\_{\Gamma}|>\varepsilon,}

a contradiction.