Ex2-15

Question: Suppose

f

{\displaystyle f}

 is continuous and

ϕ

{\displaystyle \phi}

 is of bounded variation on

[
a
,
b
]

{\displaystyle [a, b]}

. Show that:

* ψ
(
x
# )

∫

a

x

f

d
ϕ

{\displaystyle \psi(x)=\int\_{a}^{x}f \,d\phi}

 is of bounded variation.
* If

g

{\displaystyle g}

 is continuous, then

∫

a

b

g

d
# ψ

∫

a

b

g
f

d
ϕ
.

{\displaystyle \int\_{a}^{b}g\,d\psi=\int\_{a}^{b}gf\,d\phi.}

Solution:

**First part**

First note that for any subinterval

[

a
′

,

b
′

]
⊆
[
a
,
b
]

{\displaystyle [a',b']\subseteq [a,b]}

,

∫

a
′

b
′

f

d
ϕ

{\displaystyle \int\_{a'}^{b'}f \,d\phi}

 exists since

f

{\displaystyle f}

 is continuous and

ϕ

{\displaystyle \phi}

 is of bounded variation on that subinterval. Let

# Γ
{

x
i

}

# i
0

n

{\displaystyle \Gamma=\{x\_i\}\_{i=0}^{n}}

 be a partition of

[
a
,
b
]

{\displaystyle [a, b]}

. Then,

S

Γ

=

∑

# i
1

n

ψ
(

x
i

)
−
ψ
(

x

i
−
1

)

=

∑

# i
1

n

∫

x

i
−
1

x
i

f

d
ϕ

.

=\sum\_{i=1}^{n}\Big|\int\_{x\_{i-1}}^{x\_i}f\,d\phi\Big|.
}

Each of the summands is bounded since

∫

x

i
−
1

x
i

f

d
ϕ

≤
(

sup

[

x

i
−
1

.

x
i

]

f
)

V
(
ϕ
;
[

x

i
−
1

,

x
i

]
)

≤
(

sup

[
a
,
b
]

f
)

V
(
ϕ
;
[

x

i
−
1

,

x
i

]
)
.

{\displaystyle \begin{align}
\Big|\int\_{x\_{i-1}}^{x\_i}f\,d\phi\Big|&\leq (\sup\_{[x\_{i-1}.x\_i]} f)\,V(\phi;[x\_{i-1},x\_i])\\
&\leq (\sup\_{[a,b]}f)\,V(\phi;[x\_{i-1},x\_i]).
\end{align}}

Since variation is additive, we see that

S

Γ

=

∑

# i
1

n

ψ
(

x
i

)
−
ψ
(

x

i
−
1

)

=

∑

# i
1

n

∫

x

i
−
1

x
i

f

d
ϕ

≤
(

sup

[
a
,
b
]

f
)

∑

# i
1

n

V
(
ϕ
;
[

x

i
−
1

,

x
i

]
)

=
(

sup

[
a
,
b
]

f
)

V
(
ϕ
;
[
a
,
b
]
)
.

{\displaystyle \begin{align}
S\_{\Gamma}&=\sum\_{i=1}^{n}|\psi(x\_i)-\psi(x\_{i-1})|\\
&=\sum\_{i=1}^{n}\Big|\int\_{x\_{i-1}}^{x\_i}f\,d\phi\Big| \\
&\leq (\sup\_{[a,b]}f)\,\sum\_{i=1}^{n}V(\phi;[x\_{i-1},x\_i])\\
&=(\sup\_{[a,b]}f)\,V(\phi;[a,b]).
\end{align}}

This shows that

V
(
ψ
;
[
a
,
b
]
)

{\displaystyle V(\psi;[a,b])}

 exists and is finite. Therefore,

ψ

{\displaystyle \psi}

 is of bounded variation.
**Second part**

∫

a

b

g

d
ψ
,

∫

a

b

g
f

d
ϕ

{\displaystyle \int\_{a}^{b}g\,d\psi,\int\_{a}^{b}gf\,d\phi}

 both exist since

g
,
f

{\displaystyle g, f}

 are continuous and

ϕ
,
ψ

{\displaystyle \phi, \psi}

 are of bounded variation. We first introduce this mean value theorem for Riemann-Stieltjes Integral:

*Mean Value Theorem:* If

f

{\displaystyle f}

 is continuous on

[
a
,
b
]

{\displaystyle [a, b]}

 and

ϕ

{\displaystyle \phi}

 is bounded and increasing on

[
a
,
b
]

{\displaystyle [a, b]}

, then there exists

ξ
∈
[
a
,
b
]

{\textstyle \xi \in [a,b]}

 such that

∫

a

b

f

d
# ϕ
f
(
ξ
)
[
ϕ
(
b
)
−
ϕ
(
a
)
]
.

{\displaystyle \int\_{a}^{b}f\,d\phi=f(\xi)[\phi(b)-\phi(a)].}

The proof can be found in page 33 of the textbook.
Due to the properties of Riemann-Stieltjes Integral and the fact that every function of bounded variation can be written as a difference of two bounded increasing function, it suffices to prove for the case where

ϕ

{\displaystyle \phi}

 is bounded and increasing.

Since

f

{\displaystyle f}

 is continuous on

[
a
,
b
]

{\displaystyle [a, b]}

, it is uniformly continuous and bounded. Let

M
>
0

{\displaystyle M > 0}

 be an upper bound of

f

{\displaystyle f}

 and

# V
ϕ
(
b
)
−
ϕ
(
a
)

{\displaystyle V=\phi(b)-\phi(a)}

 be the variation of

ϕ

{\displaystyle \phi}

. Let

# K
M
(
V
1
)

{\displaystyle K=M(V+1)}

.

Given

ε
>
0

{\displaystyle \varepsilon > 0}

, there exists

δ
1

>
0

{\displaystyle \delta\_1>0}

 such that if

x
−

x
′

<

δ
1

{\displaystyle |x-x'|<\delta\_1}

, then

f
(
x
)
−
f
(

x
′

)

<

ε

2
K

{\displaystyle |f(x)-f(x')|<\frac{\varepsilon}{2K}}

.

Since

∫

a

b

g
f

d
ϕ

{\displaystyle \int\_{a}^{b}gf\,d\phi}

 exists, there exists

δ
2

>
0

{\displaystyle \delta\_2>0}

 such that if

Γ

<

δ
1

{\displaystyle |\Gamma|<\delta\_1}

, then

R
Γ

−

∫

a

b

g
f

d
ϕ

<
ε

/

{\displaystyle \Big|R\_\Gamma -\int\_{a}^{b}gf\,d\phi\Big|<\varepsilon/2.}

Now, let

# Γ
{

x
i

}

# i
0

n

{\displaystyle \Gamma=\{x\_i\}\_{i=0}^{n}}

 be a partition of

[
a
,
b
]

{\displaystyle [a, b]}

. Then the Riemann-Stieltjes sum for

g
,
ψ

{\displaystyle g,\psi}

 is

R

Γ

=

∑

# i
1

n

g
(

ξ
i

)

∫

x

i
−
1

x
i

f

d
ϕ
.

{\displaystyle R\_{\Gamma}=\sum\_{i=1}^{n}g(\xi\_i)\int\_{x\_{i-1}}^{x\_i}f\,d\phi.}

By the Mean value theorem, for each i, we can find

c
i

∈
[

x

i
−
1

,

x
i

]

{\displaystyle c\_i\in [x\_{i-1},x\_i] }

 such that

∫

x

i
−
1

x
i

f

d
# ϕ
f
(

c
i

)
[
ϕ
(

x
i

)
−
ϕ
(

x

i
−
1

)
]
.

{\displaystyle \int\_{x\_{i-1}}^{x\_i}f\,d\phi=f(c\_i)[\phi(x\_i)-\phi(x\_{i-1})].}

Thus,

R

Γ

=

∑

# i
1

n

g
(

ξ
i

)
f
(

c
i

)
[
ϕ
(

x
i

)
−
ϕ
(

x

i
−
1

)
]
.

{\displaystyle R\_{\Gamma}=\sum\_{i=1}^{n}g(\xi\_i)f(c\_i)[\phi(x\_i)-\phi(x\_{i-1})].}

Now, given

ε
>
0

{\displaystyle \varepsilon > 0}

, take

# δ
min
{

δ
1

,

δ
2

}

{\displaystyle \delta = \min\{ \delta\_1, \delta\_2 \}}

. If

Γ

{\displaystyle \Gamma}

 is a partition having gap smaller than

δ

{\displaystyle \delta}

, then

R

Γ

−

∫

a

b

g
f

d
ϕ

≤

R

Γ

−

∑

# i
1

n

g
(

ξ
i

)
f
(

ξ
i

)
[
ϕ
(

x
i

)
−
ϕ
(

x

i
−
1

)
]

∑

# i
1

n

g
(
ξ
)
f
(
ξ
)
[
ϕ
(

x
i

)
−
ϕ
(

x

i
−
1

)
]
−

∫

a

b

g
f

d
ϕ

<
M
×
V
(

ε

2
K

)

ε
2

<
ε
.

{\displaystyle \begin{align}
&<M\times V (\frac{\varepsilon}{2K})+\frac{\varepsilon}{2}\\
&<\varepsilon .
\end{align}}

This proves the second part.