Ex2-29

Use Theorem 2.17 to verify the formula for

∫
a
b

f
d
ϕ

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

 given in remark 3 in Section 2.3.

Remark 3: Let

ϕ
(
x
)

{\displaystyle \phi(x)}

 be a step function; that is, suppose there are points

# a

α
0

<

α
1

<
⋯
<

α
m

=
b

{\displaystyle a=\alpha\_0<\alpha\_1<\cdots<\alpha\_m=b}

 such that

ϕ

{\displaystyle \phi}

 is constant on each interval

(

α

i
−
1

,

α
i

)

{\displaystyle \left(\alpha\_{i-1}, \alpha\_i\right)}

.

Let

ϕ

(

α
i

)

=

lim

x
→

α
i

ϕ
(
x
)
,

# i
0
,
1
,
…
,
m
−
1
,

{\displaystyle
\phi\left(\alpha\_i+\right)=\lim \_{x \rightarrow \alpha\_i+} \phi(x), \quad i=0,1, \ldots, m-1,
}

and

ϕ

(

α
i

−

)

=

lim

x
→

α
i

−

ϕ
(
x
)
,

# i
1
,
…
,
m
,

{\displaystyle
\phi\left(\alpha\_i-\right)=\lim \_{x \rightarrow \alpha\_i-} \phi(x), \quad i=1, \ldots, m,
}

denote the limits from the right and left at

α

i

{\displaystyle \alpha\_{i}}

, and let

d
i

=
ϕ

(

α
i

)

−
ϕ

(

α
i

−

)

,
# i
1
,
…
,
m
−
1
,

d
0

=
ϕ

(

α
0

)

−
ϕ

(

α
0

)

{\displaystyle d\_i=\phi\left(\alpha\_i+\right)-\phi\left(\alpha\_i-\right), i=1, \ldots, m-1, d\_0=\phi\left(\alpha\_0+\right)-\phi\left(\alpha\_0\right)}

, and

d
m

=
ϕ

(

α
m

)

−
ϕ

(

α
m

−

)

{\displaystyle d\_m=\phi\left(\alpha\_m\right)-\phi\left(\alpha\_m-\right)}

 denote the jumps of

ϕ

{\displaystyle \phi}

. Then, for continuous

f

{\displaystyle f}

,

∫
a
b

f
d

# i
0

m

f

(

α
i

)

d
i

{\displaystyle
\int\_a^b f d \phi=\sum\_{i=0}^m f\left(\alpha\_i\right) d\_i
}

**Proof:** The step function

ϕ
:
[
a
,
b
]
→

R

{\displaystyle \phi:[a, b] \rightarrow \mathbb{R}}

 could be represented as:

ϕ
(
α
# )

{

c
0

 if
α
∈

(

α
0

,

α
1

)

c
1

 if
α
∈

(

α
1

,

α
2

)

⋮

c

m
−
1

 if
α
∈

(

α

m
−
1

,

α

m

)

{\displaystyle \begin{align}
\phi(\alpha)= \begin{cases}c\_0 & \text { if } \alpha \in\left(\alpha\_0, \alpha\_1\right) \\ c\_1 & \text { if } \alpha \in\left(\alpha\_1, \alpha\_2\right) \\ \vdots & \\ c\_{m-1} & \text { if } \alpha \in\left(\alpha\_{m-1}, \alpha\_{m}\right)\end{cases}
\end{align}}

with

f
(

α
1

)
,
f
(

α
2

)
,
…
,
f
(

α
n

)

{\displaystyle f(\alpha\_1), f(\alpha\_2),\ldots, f(\alpha\_n)}

 given arbitrary values.

By Theorem 2.17: If

∫
a
b

f
d
ϕ

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

 exists and

a
<
c
<
b

{\displaystyle a < c < b}

, then

∫
a
c

f
d
ϕ

{\displaystyle \int\_a^c f d \phi}

 and

∫
c
b

f
d
ϕ

{\displaystyle \int\_c^b f d \phi}

 both exist and

∫
a
b

f
d
# ϕ

∫
a
c

f
d
ϕ

∫
c
b

f
d
ϕ
.

{\displaystyle \begin{align}
\int\_a^b f d \phi=\int\_a^c f d \phi+\int\_c^b f d \phi .
\end{align}}

Since

∫
a
b

f
d
ϕ

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

 exist, then

∫

α

i
−
1

α
i

f
d
ϕ

{\displaystyle \int\_{\alpha\_{i-1}}^{\alpha\_i} f d \phi}

 exist for

# i
1
,
…
,
m
,

{\displaystyle \quad i=1, \ldots, m,}

∫
a
b

f
d

# i
1

m

∫

α

i
−
1

α
i

f
d
ϕ

{\displaystyle \begin{align}
\int\_a^b fd\phi = \sum\_{i=1}^m\int\_{\alpha\_{i-1}}^{\alpha\_i} f d \phi
\end{align}}

Consider partition

Γ
i

{\displaystyle \Gamma\_i}

 of the interval

[

α

i
−
1

,

α

i

]

{\displaystyle [\alpha\_{i-1},\alpha\_{i}]}

 such that

Γ
i

=
{

α

i
−
1

,
…
,

α

i

}

{\displaystyle \Gamma\_i=\{{\alpha\_{i-1}},\ldots,{\alpha\_{i}}\}}

 and

# i
1

m

Γ
i

{\displaystyle \Gamma = \sum\_{i=1}^{m}\Gamma\_i}

Since f is continuous, by using mean value theorem,

ξ
i

∈

(

α

i
−
1

,

α
i

)

{\displaystyle \xi\_i \in\left(\alpha\_{i-1}, \alpha\_i\right)}

, we can write the Riemann–Stieltjes sum as

R

Γ

=

∑

# i
1

m

f

(

ξ
i

)

⋅

(

ϕ

(

α
i

)

−
ϕ

(

α

i
−
1

)

)

{\displaystyle \begin{align}
R\_{\Gamma}= \sum\_{i=1}^{m}f\left(\xi\_i\right) \cdot\left(\phi\left(\alpha\_i\right)-\phi\left(\alpha\_{i-1}\right)\right)
\end{align}}

As

Γ

→
0

{\displaystyle \left|\Gamma\right| \rightarrow 0}

,

f
(

ξ
i

)
→
f
(

α

i

)

{\displaystyle f(\xi\_i) \rightarrow f(\alpha\_{i})}

.

Define

d
i

=
ϕ

(

α
i

)

−
ϕ

(

α
i

−

)

,
# i
1
,
…
,
m
−
1
,

d
0

=
ϕ

(

α
0

)

−
ϕ

(

α
0

)

{\displaystyle d\_i=\phi\left(\alpha\_i+\right)-\phi\left(\alpha\_i-\right), i=1, \ldots, m-1, d\_0=\phi\left(\alpha\_0+\right)-\phi\left(\alpha\_0\right)}

, and

d
m

=
ϕ

(

α
m

)

−
ϕ

(

α
m

−

)

{\displaystyle d\_m=\phi\left(\alpha\_m\right)-\phi\left(\alpha\_m-\right)}

, we get

∫
a
b

f
d
ϕ

=

lim

Γ

→
0

R

Γ

=

∑

# i
0

m

f

(

α
i

)

d

i

{\displaystyle \begin{align}
\int\_a^b f d \phi
&= \lim \_{|\Gamma| \rightarrow 0}R\_{\Gamma}\\
&= \sum\_{i=0}^{m}f\left(\alpha\_i\right) d\_{i}
\end{align}}

since the step function

ϕ

{\displaystyle \phi}

 will only step change at

α

i

{\displaystyle \alpha\_{i}}

 for

# i
1
,
…
,
m

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