Ex2-7

**Chapter 2 Question 7**

Question: Suppose

f

{\displaystyle f}

 is of bounded variation on

[
a
,
b
]

{\displaystyle [a, b]}

. If

f

{\displaystyle f}

 is continuous at a point

x
0

{\displaystyle x\_0}

 , show that

V
(
x
)
,
P
(
x
)
,

{\displaystyle V(x), P(x), }

 and

N
(
x
)

{\displaystyle N(x)}

 are also continuous at

x
0

{\displaystyle x\_0}

. In particular, if

f

{\displaystyle f}

 is continuous on

[
a
,
b
]

{\displaystyle [a, b]}

, then so are

V
(
x
)
,
P
(
x
)
,

{\displaystyle V(x), P(x), }

 and

N
(
x
)

{\displaystyle N(x)}

.

Proof: Given a partition

Γ

{\displaystyle \Gamma}

 with two adjacent points

x
0

<
x

{\displaystyle x\_0<x}

, we prove the inequality below:

V
[

x
0

,
x
]
≤

f
(
x
)
−
f
(

x
0

)

V
[
a
,
b
]
−

S
Γ

.

{\displaystyle V[x\_0,x]\leq |f(x)-f(x\_0)|+V[a,b]-S\_\Gamma.}

Indeed, by applying Theorem 2.2(ii) twice, we have

V
[
a
,
b
]
−
V
[

x
0

,
x
# ]
V
[
a
,

x
0

]
V
[
x
,
b
]

{\displaystyle V[a,b]-V[x\_0,x]=V[a,x\_0]+V[x,b]}

. Moreover, we can write

S
Γ

{\displaystyle S\_\Gamma}

 as

S
Γ

=

S

Γ
1

f
(
x
)
−
f
(

x
0

)

S

Γ
2

,

{\displaystyle S\_\Gamma=S\_{\Gamma\_1}+|f(x)-f(x\_0)|+S\_{\Gamma\_2},}

where

Γ
1

=
Γ
∩
[
a
,

x
0

]

{\displaystyle \Gamma\_1=\Gamma\cap [a,x\_0]}

 and

Γ
2

=
Γ
∩
[
x
,
b
]

{\displaystyle \Gamma\_2=\Gamma\cap [x,b]}

. Basically, we have split

Γ

{\displaystyle \Gamma}

 into three parts:

Γ
1

,
{

x
0

,
x
}
,

{\displaystyle \Gamma\_1, \{x\_0,x\}, }

 and

Γ
2

{\displaystyle \Gamma\_2}

. Since

Γ

1

{\displaystyle \Gamma\_{1}}

 is a partition of

[
a
,

x
0

]

{\displaystyle [a,x\_0]}

, we have

V
[
a
,

x
0

]
≥

S

Γ
1

{\displaystyle V[a,x\_0]\geq S\_{\Gamma\_1}}

. Similarly,

V
[
x
,
b
]
≥

S

Γ
2

{\displaystyle V[x,b]\geq S\_{\Gamma\_2}}

. Finally, we have

S
Γ

≤
V
[
a
,

x
0

]

f
(
x
)
−
f
(

x
0

)

V
[
x
,
b
# ]

f
(
x
)
−
f
(

x
0

)

V
[
a
,
b
]
−
V
[

x
0

,
x
]
.

Now we prove the statement. Given

ε
>
0
,

x
0

∈
[
a
,
b
]

{\displaystyle \varepsilon>0, x\_0\in[a,b]}

, we choose

δ
>
0

{\displaystyle \delta > 0}

 such that

f
(
x
)
−
f
(

x
0

)

<

ε
2

 whenever

x
−

x
0

<
δ
.

Next, we choose a partition

Γ

{\displaystyle \Gamma}

 such that

V
[
a
,
b
]
−

S
Γ

<
ε

/

2

{\displaystyle V[a,b]-S\_\Gamma<\varepsilon/2}

. Without loss of generality, we might assume

x
0

∈
Γ

{\displaystyle x\_0 \in \Gamma}

 since a refinement of partition will increase the value of

S
Γ

{\displaystyle S\_\Gamma}

. We can also assume two points adjacent to

x
0

{\displaystyle x\_0}

 (if

x
0

{\displaystyle x\_0}

 is not boundary point) or the one point adjacent to

x
0

{\displaystyle x\_0}

 (if

x
0

{\displaystyle x\_0}

 is boundary point) is contained in the open interval

(

x
0

−
δ
,

x
0

δ
)

{\displaystyle (x\_0 - \delta, x\_0 + \delta)}

. If this is not true, we simply add more points in the partition.
Let

x
1

{\displaystyle x\_1}

 be the point in

Γ

{\displaystyle \Gamma}

 after

x
0

{\displaystyle x\_0}

, we have

V
(

x
1

)
−
V
(

x
0

)

=
V
[

x
0

,

x
1

]

≤

f
(

x
1

)
−
f
(

x
0

)

V
[
a
,
b
]
−

S

Γ

<

ε
2

ε
2

=
ε
.

{\displaystyle \begin{align}|V(x\_1)-V(x\_0)|&=V[x\_0,x\_1]\\
&\leq |f(x\_1)-f(x\_0)|+V[a,b]-S\_{\Gamma}\\
&< \dfrac\varepsilon2 +\frac\varepsilon2\\
&=\varepsilon.\end{align}}

Similar result holds for

x
2

{\displaystyle x\_2}

 being the point in

Γ

{\displaystyle \Gamma}

 before

x
0

{\displaystyle x\_0}

.
By monotonicity of

V
(
x
)

{\displaystyle V(x)}

, we conclude

V
(
x
)
−
V
(

x
0

)

<
ε

{\displaystyle |V(x)-V(x\_0)|<\varepsilon}

 for all

x
∈
[

x
2

,

x
1

]

{\displaystyle x\in[x\_2,x\_1]}

 where

[

x
2

,

x
1

]

{\displaystyle [x\_2,x\_1] }

 has

x
0

{\displaystyle x\_0}

 as an interior point. Thus

V

{\displaystyle V}

 is continuous at

x
0

{\displaystyle x\_0}

.

By Theorem 2.6, we find that since

V
(
x
)
,
f
(
x
)

{\displaystyle V(x),f(x)}

 are both continuous at

x
0

{\displaystyle x\_0}

, so are

P
(
x
# )

1
2

[
V
(
x
)
f
(
x
)
−
f
(
a
)
]

 and

N
(
x
# )

1
2

[
V
(
x
)
−
f
(
x
)
f
(
a
)
]
.

{\displaystyle P(x)=\frac12[V(x)+f(x)-f(a)]\quad\text{ and }\quad N(x)=\frac12[V(x)-f(x)+f(a)].}