Ex3-29

***Question***: Use Theorem 2.17 to verify the formula for

∫
f

d
ϕ

{\displaystyle \int f\ d\phi }

 given in remark 3 in Section 2.3.

***Answer***: We will prove the statement directly (i.e., without appealing to Theorem 2.17). In particular, we want to prove that *if

ϕ
:
[
a
,
b
]
→

R

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

 is a step function with jump discontinuities at

α

1

,
…
,

α

n

{\displaystyle \alpha \_{1},\dots ,\alpha \_{n}}

 with

α

k

<

α

k
1

{\displaystyle \alpha \_{k}<\alpha \_{k+1}}

 for

1
≤
k
≤
n

{\displaystyle 1 \leq k \leq n }

 and

f
:
[
a
,
b
]
→

R

{\displaystyle f:[a,b]\to\R}

 is a function that is continuous at

α
k

{\displaystyle \alpha\_k}

 for all

1
≤
k
≤
n

{\displaystyle 1 \leq k \leq n }

, then

∫

a

b

f

d
ϕ

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

 exists, and*

*∫

a

b

f

d

# k
1

n

f
(

α

i

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

−
)
,

{\displaystyle \int \_{a}^{b}f\ d\phi =\sum \_{k=1}^{n}f(\alpha \_{i})(\phi (\alpha \_{k}+)-\phi (\alpha \_{k}-),}*

where

*ϕ
(
α
# )

lim

x
→

α

ϕ
(
x
)

and

ϕ
(
α
−
# )

lim

x
→

α

−

ϕ
(
x
)
;

{\displaystyle \phi (\alpha +)=\lim \_{x\to \alpha ^{+}}\phi (x)\qquad {\text{and}}\qquad \phi (\alpha -)=\lim \_{x\to \alpha ^{-}}\phi (x);}*

*if

α

1

=
a

{\displaystyle \alpha \_{1}=a}

 then

ϕ
(

α

1

−
# )
ϕ
(
a
)

{\displaystyle \phi (\alpha \_{1}-)=\phi (a)}

; and if

α

n

=
b

{\displaystyle \alpha \_{n}=b}

 then

ϕ
(

α

n

# )
ϕ
(
b
)

{\displaystyle \phi (\alpha \_{n}+)=\phi (b)}

.*

To begin, given

ε
>
0

{\displaystyle \varepsilon > 0}

. Let

# E
{

α

1

,
…
,

α

n

}

{\displaystyle E=\{\alpha \_{1},\dots ,\alpha \_{n}\}}

. Let

Γ

ε

=
{
# a

x

0

¯

,

x

1

¯

,
…
,

x

m

¯

=
b
}

{\displaystyle \Gamma \_{\varepsilon }=\{a={\overline {x\_{0}}},{\overline {x\_{1}}},\dots ,{\overline {x\_{m}}}=b\}}

 be a partition of

[
a
,
b
]

{\displaystyle [a, b]}

 so that

α

k

∈

Γ

ε

{\displaystyle \alpha \_{k}\in \Gamma \_{\varepsilon }}

 for all

1
≤
k
≤
n

{\displaystyle 1 \leq k \leq n }

, and

g
a
p

(

Γ

ε

)

<
δ

{\displaystyle \mathrm {gap} \left(\Gamma \_{\varepsilon }\right)<\delta }

, where

# δ
min
{

α

2

−

α

1

,
…
,

α

n

−

α

n
−
1

,

δ

0

}
,

with

δ

0

{\displaystyle \delta\_{0}}

 given by (1).

Let

# Γ
{

x

i

}

# i
0

r

{\displaystyle \Gamma =\{x\_{i}\}\_{i=0}^{r}}

 be a refinement of

Γ

ε

{\displaystyle \Gamma \_{\varepsilon }}

. Then

g
a
p

(
Γ
)

≤

g
a
p

(

Γ

ε

)

<
δ

{\displaystyle \mathrm {gap} \left(\Gamma \right)\leq \mathrm {gap} \left(\Gamma \_{\varepsilon }\right)<\delta }

. Let

{

ξ

i

}

# i
1

r

{\displaystyle \{\xi \_{i}\}\_{i=1}^{r}}

 be a sequence of intermediate points of the subintervals of

Γ

{\displaystyle \Gamma}

. Consider the Riemann-Stieltjes sum

R

Γ

(
f
,
ϕ

# i
1

r

f
(

ξ

i

)
(
ϕ
(

x

i

)
−
ϕ
(

x

i
−
1

)
)
.

{\displaystyle R\_{\Gamma }(f,\phi )=\sum \_{i=1}^{r}f(\xi \_{i})(\phi (x\_{i})-\phi (x\_{i-1})).}

For each

1
≤
i
≤
r

{\displaystyle 1\leq i\leq r}

, we have that

x

i
−
1

{\displaystyle x\_{i-1}}

 and

x
i

{\displaystyle x\_i }

 are neighboring points in

Γ

{\displaystyle \Gamma}

. Since

α

k

∈
Γ

{\displaystyle \alpha \_{k}\in \Gamma }

, we find that

α

k

∉
(

x

i
−
1

,

x

i

)

{\displaystyle \alpha \_{k}\notin (x\_{i-1},x\_{i})}

 for all

1
≤
k
≤
n

{\displaystyle 1 \leq k \leq n }

. Moreover, since

g
a
p

(
Γ
)

=
max
{

x

i

−

x

i
−
1

∣
1
≤
i
≤
p
}
<

min

1
≤
k
≤
n

α

k

−

α

k
−
1

,

it follows that we cannot have both

x

i
−
1

,

x

i

∈
E

{\displaystyle x\_{i-1},x\_{i}\in E}

. Thus, we have either on of the following cases.

**(Case 1:

x

i
−
1

,

x

i

∉
E

{\displaystyle x\_{i-1},x\_{i}\notin E}

)** In this case, since no discontinuity

α
k

{\displaystyle \alpha\_k}

 lies between

x

i
−
1

{\displaystyle x\_{i-1}}

 and

x
i

{\displaystyle x\_i }

, we have

ϕ
(

x

i

)
−
ϕ
(

x

i
−
1

# )

{\textstyle \phi (x\_{i})-\phi (x\_{i-1})=0.}

**(Case 2:

x

i

=

α

k

{\displaystyle x\_{i}=\alpha \_{k}}

 for some

1
≤
k
≤
n

{\displaystyle 1 \leq k \leq n }

)** In this case,

x

i
−
1

≠

α

k
′

{\displaystyle x\_{i-1}\neq \alpha \_{k'}}

 (since

g
a
p

(
Γ
)

<

α

k

−

α

k
−
1

{\displaystyle \mathrm {gap} \left(\Gamma \right)<|\alpha \_{k}-\alpha \_{k-1}|}

) and no

α

k
′

{\displaystyle \alpha \_{k'}}

 can lie between

x

i
−
1

{\displaystyle x\_{i-1}}

 and

α
k

{\displaystyle \alpha\_k}

, so it follows that

ϕ
(

x

i

)
−
ϕ
(

x

i
−
1

# )
ϕ
(

α

k

)
−
ϕ
(

α

k

−
)
.

{\textstyle \phi (x\_{i})-\phi (x\_{i-1})=\phi (\alpha \_{k})-\phi (\alpha \_{k}-).}

**(Case 3:

x

i
−
1

=

α

k

{\displaystyle x\_{i-1}=\alpha \_{k}}

 for some

1
≤
k
≤
n

{\displaystyle 1 \leq k \leq n }

)** In this case,

x

i

≠

α

k
′

{\displaystyle x\_{i}\neq \alpha \_{k'}}

 (since

g
a
p

(
Γ
)

<

α

k
1

−

α

k

{\displaystyle \mathrm {gap} \left(\Gamma \right)<|\alpha \_{k+1}-\alpha \_{k}|}

) and no

α

k
′

{\displaystyle \alpha \_{k'}}

 can lie between

α
k

{\displaystyle \alpha\_k}

 and

x
i

{\displaystyle x\_i }

, so it follows that

ϕ
(

x

i

)
−
ϕ
(

x

i
−
1

# )
ϕ
(

α

k

)
−
ϕ
(

α

k

)
.

{\textstyle \phi (x\_{i})-\phi (x\_{i-1})=\phi (\alpha \_{k}+)-\phi (\alpha \_{k}).}

All the three cases are illustrated below.

Hence, we find that

R

Γ

(
f
,
ϕ
)

=

∑

# i
1

r

f
(

ξ

i

)
(
ϕ
(

x

i

)
−
ϕ
(

x

i
−
1

)
)

=

{

terms in (Case 1)

}

⏟

=
0

{

terms in (Case 2)

}
{

terms in (Case 3)

}

=

∑

# k
1

n

[

f
(

ξ

i

k

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

−
)
)
f
(

ξ

i

k

′

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

)
)

]

.

{\displaystyle {\begin{aligned}&R\_{\Gamma }(f,\phi )\\&=\sum \_{i=1}^{r}f(\xi \_{i})(\phi (x\_{i})-\phi (x\_{i-1}))\\&=\underbrace {\{{\text{terms in (Case 1)}}\}} \_{=0}+\{{\text{terms in (Case 2)}}\}+\{{\text{terms in (Case 3)}}\}\\&=\sum \_{k=1}^{n}\left[f(\xi \_{i\_{k}})(\phi (\alpha \_{k})-\phi (\alpha \_{k}-))+f(\xi \_{i\_{k}}')(\phi (\alpha \_{k}+)-\phi (\alpha \_{k}))\right].\end{aligned}}}

Here we remark that

ξ

i

k

{\displaystyle \xi \_{i\_{k}}}

 is a point in the subinterval of

Γ

{\displaystyle \Gamma}

 whose right-endpoint is

α
k

{\displaystyle \alpha\_k}

 and thus is within a distance

g
a
p

(
Γ
)

<
δ

{\displaystyle \mathrm {gap} \left(\Gamma \right)<\delta }

 of

α
k

{\displaystyle \alpha\_k}

. Likewise,

ξ

i

k

′

{\displaystyle \xi \_{i\_{k}}'}

 is a point in the subinterval of

Γ

{\displaystyle \Gamma}

 whose left-endpoint is

α
k

{\displaystyle \alpha\_k}

 and thus is within a distance

g
a
p

(
Γ
)

<
δ

{\displaystyle \mathrm {gap} \left(\Gamma \right)<\delta }

 of

α
k

{\displaystyle \alpha\_k}

. Now, we have

R

Γ

(
f
,
ϕ
)
−

∑

# k
1

n

f
(

α

i

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

−
)

=

∑

# k
1

n

[

f
(

ξ

i

k

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

−
)
)
f
(

ξ

i

k

′

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

)
)

]

−

∑

# k
1

n

[

f
(

α

k

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

−
)
)
f
(

α

k

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

)
)

]

≤

∑

# k
1

n

[

f
(

ξ

i

k

−
f
(

α

k

)

ϕ
(

α

k

)
−
ϕ
(

α

k

−
)

f
(

ξ

i

k

′

−
f
(

α

k

)

ϕ
(

α

k

)
−
ϕ
(

α

k

)

]

.

Now since

f

{\displaystyle f}

 is continuous at

α
k

{\displaystyle \alpha\_k}

 for all

1
≤
k
≤
n

{\displaystyle 1 \leq k \leq n }

, there exists

δ

k

>
0

{\displaystyle \delta \_{k}>0}

 for

1
≤
k
≤
n

{\displaystyle 1 \leq k \leq n }

 such that

f
(
x
)
−
f
(

α

k

)

<

ε

∑

# k
1

n

(

ϕ
(

α

k

)
−
ϕ
(

α

k

−
)

ϕ
(

α

k

)
−
ϕ
(

α

k

)

)

for all

x
−

α

k

<

δ

k

{\displaystyle |x-\alpha \_{k}|<\delta \_{k}}

. Let

δ

0

=
min
{

δ

1

,
…
,

δ

n

}
.

(
1
)

{\displaystyle \delta \_{0}=\min\{\delta \_{1},\dots ,\delta \_{n}\}.\qquad \qquad (1)}

Then it follows that

R

Γ

(
f
,
ϕ
)
−

∑

# k
1

n

f
(

α

i

)
(
ϕ
(

α

k

)
−
ϕ
(

α

k

−
)

<
ε
,

{\displaystyle \left|R\_{\Gamma }(f,\phi )-\sum \_{k=1}^{n}f(\alpha \_{i})(\phi (\alpha \_{k}+)-\phi (\alpha \_{k}-)\right|<\varepsilon ,}

and this completes the proof.