Functions of Bounded Variation - Real Analysis, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Functions of Bounded Variation
Our main theorem concerning the existence of Riemann–Stietjes integrals assures us
that the integral
R
b
a
f(x) da(x) exists when f is continuous and a is monotonic. Our lin-
earity theorem then guarantees that the integral
R
b
a
f(x)da(x) exists when f is continuous
and a is the di?erence of two monotonic functions. In these notes, we prove that a is the
di?erence of two monotonic functions if and only if it is of bounded variation, where
De?nition 1
(a) The function a : [a,b]? IR is said to be of bounded variation on [a,b] if and only if
there is a constant M > 0 such that
n
X
i=1


a(x
i
)-a(x
i-1
)


=M
for all partitions IP ={x
0
,x
1
,···,x
n
} of [a,b].
(b) If a : [a,b]? IR is of bounded variation on [a,b], then the total variation of a on [a,b]
is de?ned to be
V
a
(a,b)= sup
n
n
P
i=1


a(x
i
)-a(x
i-1
)




 IP ={x
0
,x
1
,···,x
n
} is a partition of [a,b]
o
Example 2 If a : [a,b] ? IR is monotonically increasing, then, for any partition IP =
{x
0
,x
1
,···,x
n
} of [a,b]
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1

a(x
i
)-a(x
i-1
)
	
=a(x
n
)-a(x
0
) =a(b)-a(a)
Thus a is of bounded variation and V
f
(a,b)=a(b)-a(a).
Example 3 If a : [a,b] ? IR is continuous on [a,b] and di?erentiable on (a,b) with
sup
a<x<b
|a
'
(x)| = M, then, for any partition IP = {x
0
,x
1
,···,x
n
} of [a,b], we have, by
the Mean Value Theorem,
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1


a
'
(t
i
)[x
i
-x
i-1
]


=
n
X
i=1
M[x
i
-x
i-1
] =M(b-a)
Thus a is of bounded variation and V
f
(a,b)=M(b-a).
Page 2


Functions of Bounded Variation
Our main theorem concerning the existence of Riemann–Stietjes integrals assures us
that the integral
R
b
a
f(x) da(x) exists when f is continuous and a is monotonic. Our lin-
earity theorem then guarantees that the integral
R
b
a
f(x)da(x) exists when f is continuous
and a is the di?erence of two monotonic functions. In these notes, we prove that a is the
di?erence of two monotonic functions if and only if it is of bounded variation, where
De?nition 1
(a) The function a : [a,b]? IR is said to be of bounded variation on [a,b] if and only if
there is a constant M > 0 such that
n
X
i=1


a(x
i
)-a(x
i-1
)


=M
for all partitions IP ={x
0
,x
1
,···,x
n
} of [a,b].
(b) If a : [a,b]? IR is of bounded variation on [a,b], then the total variation of a on [a,b]
is de?ned to be
V
a
(a,b)= sup
n
n
P
i=1


a(x
i
)-a(x
i-1
)




 IP ={x
0
,x
1
,···,x
n
} is a partition of [a,b]
o
Example 2 If a : [a,b] ? IR is monotonically increasing, then, for any partition IP =
{x
0
,x
1
,···,x
n
} of [a,b]
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1

a(x
i
)-a(x
i-1
)
	
=a(x
n
)-a(x
0
) =a(b)-a(a)
Thus a is of bounded variation and V
f
(a,b)=a(b)-a(a).
Example 3 If a : [a,b] ? IR is continuous on [a,b] and di?erentiable on (a,b) with
sup
a<x<b
|a
'
(x)| = M, then, for any partition IP = {x
0
,x
1
,···,x
n
} of [a,b], we have, by
the Mean Value Theorem,
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1


a
'
(t
i
)[x
i
-x
i-1
]


=
n
X
i=1
M[x
i
-x
i-1
] =M(b-a)
Thus a is of bounded variation and V
f
(a,b)=M(b-a).
Example 4 De?ne the function a : [0,1]? IR by
a(x) =

0 if x = 0
xcos
p
x
if x6= 0
This function is continuous, but is not of bounded variation because it wobbles too much
near x = 0. To see this, consider, for each m? IN, the partition
IP
m
=

x
0
= 0, x
1
=
1
2m
, x
2
=
1
2m-1
, x
3
=
1
2m-2
,···,x
2m-2
=
1
3
, x
2m-1
=
1
2
, x
2m
= 1
	
The values of a at the points of this partition are
a(IP
m
) ={0,
1
2m
, -
1
2m-1
,
1
2m-2
,···, -
1
3
,
1
2
, -1}
x
y
y =a(x)
x
2m
= 1
x
2m-1
=
1
2
1
3
For this partition,
2m
X
i=1


a(x
i
)-a(x
i-1
)


=


1
2m
-0


+


-
1
2m-1
-
1
2m


+


1
(2m-2)
+
1
2m-1


+···+


-
1
3
-
1
4


+


1
2
+
1
3


+


-1-
1
2


=
1
2m
+0 +
1
2m-1
+
1
2m
+
1
(2m-2)
+
1
2m-1
+··· +
1
3
+
1
4
+
1
2
+
1
3
+ 1+
1
2
= 2

1
2m
+
1
2m-1
+···+
1
3
+
1
2


+1
The harmonic series
8
P
k=2
1
k
diverges. So given any M, there is an m ? IN for which the
partition IP
m
obeys
n
X
i=1


a(x
i
)-a(x
i-1
)


>M
Theorem 5
(a) If a,ß : [a,b]? IR are of bounded variation and c,d? IR, then ca+dß is of bounded
variation and
V
ca+dß
(a,b)=|c|V
a
(a,b)+|d|V
ß
(a,b)
Page 3


Functions of Bounded Variation
Our main theorem concerning the existence of Riemann–Stietjes integrals assures us
that the integral
R
b
a
f(x) da(x) exists when f is continuous and a is monotonic. Our lin-
earity theorem then guarantees that the integral
R
b
a
f(x)da(x) exists when f is continuous
and a is the di?erence of two monotonic functions. In these notes, we prove that a is the
di?erence of two monotonic functions if and only if it is of bounded variation, where
De?nition 1
(a) The function a : [a,b]? IR is said to be of bounded variation on [a,b] if and only if
there is a constant M > 0 such that
n
X
i=1


a(x
i
)-a(x
i-1
)


=M
for all partitions IP ={x
0
,x
1
,···,x
n
} of [a,b].
(b) If a : [a,b]? IR is of bounded variation on [a,b], then the total variation of a on [a,b]
is de?ned to be
V
a
(a,b)= sup
n
n
P
i=1


a(x
i
)-a(x
i-1
)




 IP ={x
0
,x
1
,···,x
n
} is a partition of [a,b]
o
Example 2 If a : [a,b] ? IR is monotonically increasing, then, for any partition IP =
{x
0
,x
1
,···,x
n
} of [a,b]
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1

a(x
i
)-a(x
i-1
)
	
=a(x
n
)-a(x
0
) =a(b)-a(a)
Thus a is of bounded variation and V
f
(a,b)=a(b)-a(a).
Example 3 If a : [a,b] ? IR is continuous on [a,b] and di?erentiable on (a,b) with
sup
a<x<b
|a
'
(x)| = M, then, for any partition IP = {x
0
,x
1
,···,x
n
} of [a,b], we have, by
the Mean Value Theorem,
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1


a
'
(t
i
)[x
i
-x
i-1
]


=
n
X
i=1
M[x
i
-x
i-1
] =M(b-a)
Thus a is of bounded variation and V
f
(a,b)=M(b-a).
Example 4 De?ne the function a : [0,1]? IR by
a(x) =

0 if x = 0
xcos
p
x
if x6= 0
This function is continuous, but is not of bounded variation because it wobbles too much
near x = 0. To see this, consider, for each m? IN, the partition
IP
m
=

x
0
= 0, x
1
=
1
2m
, x
2
=
1
2m-1
, x
3
=
1
2m-2
,···,x
2m-2
=
1
3
, x
2m-1
=
1
2
, x
2m
= 1
	
The values of a at the points of this partition are
a(IP
m
) ={0,
1
2m
, -
1
2m-1
,
1
2m-2
,···, -
1
3
,
1
2
, -1}
x
y
y =a(x)
x
2m
= 1
x
2m-1
=
1
2
1
3
For this partition,
2m
X
i=1


a(x
i
)-a(x
i-1
)


=


1
2m
-0


+


-
1
2m-1
-
1
2m


+


1
(2m-2)
+
1
2m-1


+···+


-
1
3
-
1
4


+


1
2
+
1
3


+


-1-
1
2


=
1
2m
+0 +
1
2m-1
+
1
2m
+
1
(2m-2)
+
1
2m-1
+··· +
1
3
+
1
4
+
1
2
+
1
3
+ 1+
1
2
= 2

1
2m
+
1
2m-1
+···+
1
3
+
1
2


+1
The harmonic series
8
P
k=2
1
k
diverges. So given any M, there is an m ? IN for which the
partition IP
m
obeys
n
X
i=1


a(x
i
)-a(x
i-1
)


>M
Theorem 5
(a) If a,ß : [a,b]? IR are of bounded variation and c,d? IR, then ca+dß is of bounded
variation and
V
ca+dß
(a,b)=|c|V
a
(a,b)+|d|V
ß
(a,b)
(b) If a : [a,b]? IR is of bounded variation on [a,b] and [c,d]? [a,b], then a is of bounded
variation on [c,d] and
V
a
(c,d)=V
a
(a,b)
(c) If a : [a,b]? IR is of bounded variation and c? (a,b), then
V
a
(a,b)=V
a
(a,c)+V
a
(c,b)
(d) If a : [a,b] ? IR is of bounded variation then the functions V(x) = V
a
(a,x) and
V(x)-a(x) are both increasing on [a,b].
(e) The function a : [a,b]? IR is of bounded variation if and only if it is the di?erence of
two increasing functions.
Proof: We shall use the shorthand notation
IP
X


?
i
a


for
n
X
i=1


a(x
i
)-a(x
i-1
)


where the partition IP ={x
0
,x
1
,···,x
n
}.
(a) follows from the observation that, for any IP partition of [a,b],
IP
X


?
i
(ca+dß)


=|c|
IP
X


?
i
a


+|d|
IP
X


?
i
ß


=|c|V
a
(a,b)+|d|V
ß
(a,b)
(b) follows from the observation that, for any partition IP of [c,d],
IP
X


?
i
a


=
IP?{a,b}
X


?
i
a


=V
a
(a,b)
(c) If IP =

x
0
,x
1
,···,x
n
	
is any partition of [a,b] and x
i-1
=c=x
i
, then


a(x
i
)-a(x
i-1
)


=


a(x
i
)-a(c)


+


a(c)-a(x
i-1
)


so that
IP
X

?
i
a

=
IP?{c}
X

?
i
a

=
(IP?{c})n[a,c]
X

?
i
a

+
(IP?{c})n[c,b]
X

?
i
a

=V
a
(a,c)+V
a
(c,b)
Page 4


Functions of Bounded Variation
Our main theorem concerning the existence of Riemann–Stietjes integrals assures us
that the integral
R
b
a
f(x) da(x) exists when f is continuous and a is monotonic. Our lin-
earity theorem then guarantees that the integral
R
b
a
f(x)da(x) exists when f is continuous
and a is the di?erence of two monotonic functions. In these notes, we prove that a is the
di?erence of two monotonic functions if and only if it is of bounded variation, where
De?nition 1
(a) The function a : [a,b]? IR is said to be of bounded variation on [a,b] if and only if
there is a constant M > 0 such that
n
X
i=1


a(x
i
)-a(x
i-1
)


=M
for all partitions IP ={x
0
,x
1
,···,x
n
} of [a,b].
(b) If a : [a,b]? IR is of bounded variation on [a,b], then the total variation of a on [a,b]
is de?ned to be
V
a
(a,b)= sup
n
n
P
i=1


a(x
i
)-a(x
i-1
)




 IP ={x
0
,x
1
,···,x
n
} is a partition of [a,b]
o
Example 2 If a : [a,b] ? IR is monotonically increasing, then, for any partition IP =
{x
0
,x
1
,···,x
n
} of [a,b]
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1

a(x
i
)-a(x
i-1
)
	
=a(x
n
)-a(x
0
) =a(b)-a(a)
Thus a is of bounded variation and V
f
(a,b)=a(b)-a(a).
Example 3 If a : [a,b] ? IR is continuous on [a,b] and di?erentiable on (a,b) with
sup
a<x<b
|a
'
(x)| = M, then, for any partition IP = {x
0
,x
1
,···,x
n
} of [a,b], we have, by
the Mean Value Theorem,
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1


a
'
(t
i
)[x
i
-x
i-1
]


=
n
X
i=1
M[x
i
-x
i-1
] =M(b-a)
Thus a is of bounded variation and V
f
(a,b)=M(b-a).
Example 4 De?ne the function a : [0,1]? IR by
a(x) =

0 if x = 0
xcos
p
x
if x6= 0
This function is continuous, but is not of bounded variation because it wobbles too much
near x = 0. To see this, consider, for each m? IN, the partition
IP
m
=

x
0
= 0, x
1
=
1
2m
, x
2
=
1
2m-1
, x
3
=
1
2m-2
,···,x
2m-2
=
1
3
, x
2m-1
=
1
2
, x
2m
= 1
	
The values of a at the points of this partition are
a(IP
m
) ={0,
1
2m
, -
1
2m-1
,
1
2m-2
,···, -
1
3
,
1
2
, -1}
x
y
y =a(x)
x
2m
= 1
x
2m-1
=
1
2
1
3
For this partition,
2m
X
i=1


a(x
i
)-a(x
i-1
)


=


1
2m
-0


+


-
1
2m-1
-
1
2m


+


1
(2m-2)
+
1
2m-1


+···+


-
1
3
-
1
4


+


1
2
+
1
3


+


-1-
1
2


=
1
2m
+0 +
1
2m-1
+
1
2m
+
1
(2m-2)
+
1
2m-1
+··· +
1
3
+
1
4
+
1
2
+
1
3
+ 1+
1
2
= 2

1
2m
+
1
2m-1
+···+
1
3
+
1
2


+1
The harmonic series
8
P
k=2
1
k
diverges. So given any M, there is an m ? IN for which the
partition IP
m
obeys
n
X
i=1


a(x
i
)-a(x
i-1
)


>M
Theorem 5
(a) If a,ß : [a,b]? IR are of bounded variation and c,d? IR, then ca+dß is of bounded
variation and
V
ca+dß
(a,b)=|c|V
a
(a,b)+|d|V
ß
(a,b)
(b) If a : [a,b]? IR is of bounded variation on [a,b] and [c,d]? [a,b], then a is of bounded
variation on [c,d] and
V
a
(c,d)=V
a
(a,b)
(c) If a : [a,b]? IR is of bounded variation and c? (a,b), then
V
a
(a,b)=V
a
(a,c)+V
a
(c,b)
(d) If a : [a,b] ? IR is of bounded variation then the functions V(x) = V
a
(a,x) and
V(x)-a(x) are both increasing on [a,b].
(e) The function a : [a,b]? IR is of bounded variation if and only if it is the di?erence of
two increasing functions.
Proof: We shall use the shorthand notation
IP
X


?
i
a


for
n
X
i=1


a(x
i
)-a(x
i-1
)


where the partition IP ={x
0
,x
1
,···,x
n
}.
(a) follows from the observation that, for any IP partition of [a,b],
IP
X


?
i
(ca+dß)


=|c|
IP
X


?
i
a


+|d|
IP
X


?
i
ß


=|c|V
a
(a,b)+|d|V
ß
(a,b)
(b) follows from the observation that, for any partition IP of [c,d],
IP
X


?
i
a


=
IP?{a,b}
X


?
i
a


=V
a
(a,b)
(c) If IP =

x
0
,x
1
,···,x
n
	
is any partition of [a,b] and x
i-1
=c=x
i
, then


a(x
i
)-a(x
i-1
)


=


a(x
i
)-a(c)


+


a(c)-a(x
i-1
)


so that
IP
X

?
i
a

=
IP?{c}
X

?
i
a

=
(IP?{c})n[a,c]
X

?
i
a

+
(IP?{c})n[c,b]
X

?
i
a

=V
a
(a,c)+V
a
(c,b)
which implies that V
a
(a,b) = V
a
(a,c) + V
a
(c,b). To prove the other inequality, we let
e > 0 and select a partition IP
1
of [a,c] for which
IP
1
P


?
i
a


= V
a
(a,c)-e and a partition
IP
2
of [c,b] for which
IP
2
P


?
i
a


= V
a
(c,b)-e. Then
IP
1
?IP
2
X


?
i
a


=
IP
1
X


?
i
a


+
IP
2
X


?
i
a


= V
a
(a,c)+V
a
(c,b)-2e
This assures that V
a
(a,b)=V
a
(a,c)+V
a
(c,b)-2e for all e> 0 and hence that V
a
(a,b)=
V
a
(a,c)+V
a
(c,b).
(d) Proof that V(x) is increasing: Let a=x
1
=x
2
= b. Then, by part (c),
V(x
2
)-V(x
1
) =V
a
(a,x
2
)-V
a
(a,x
1
) = V
a
(x
1
,x
2
)= 0
(d) Proof that V(x)-a(x) is increasing: Let a=x
1
=x
2
=b. By part (c),
{V(x
2
)-a(x
2
)}-{V(x
1
)-a(x
1
)} = V
a
(x
1
,x
2
)-{a(x
2
)-a(x
1
)}
= V
a
(x
1
,x
2
)-


a(x
2
)-a(x
1
)


= V
a
(x
1
,x
2
)-
{x
1
,x
2
}
X


?
i
a


= 0
(e) If a is of bounded variation then a(x) = V
a
(a,x)-

V
a
(a,x)-a(x)

expresses a as
the di?erence of two increasing functions. On the other hand if a is the di?erence ß-?
of two increasing functions, then ß and ? are of bounded variation by Example 2 and a is
of bounded variation by part (a).
Example 6 We know that if f is continuous and a is of bounded variation on [a,b], then
f ?R(a) on [a,b]. If f is of bounded variation and a is continuous on [a,b], then we have
f ?R(a) on [a,b] with
Z
b
a
fda =f(b)a(b)-f(a)a(a)-
Z
b
a
adf
byourintegrationbypartstheorem. Itispossibletohave f ?R(a)on[a,b]evenifneither
f nor a are of bounded variation on [a,b]. For example, we have seen, in Example 4, that
a(x) =

0 if x = 0
xcos
p
x
if x6= 0
Page 5


Functions of Bounded Variation
Our main theorem concerning the existence of Riemann–Stietjes integrals assures us
that the integral
R
b
a
f(x) da(x) exists when f is continuous and a is monotonic. Our lin-
earity theorem then guarantees that the integral
R
b
a
f(x)da(x) exists when f is continuous
and a is the di?erence of two monotonic functions. In these notes, we prove that a is the
di?erence of two monotonic functions if and only if it is of bounded variation, where
De?nition 1
(a) The function a : [a,b]? IR is said to be of bounded variation on [a,b] if and only if
there is a constant M > 0 such that
n
X
i=1


a(x
i
)-a(x
i-1
)


=M
for all partitions IP ={x
0
,x
1
,···,x
n
} of [a,b].
(b) If a : [a,b]? IR is of bounded variation on [a,b], then the total variation of a on [a,b]
is de?ned to be
V
a
(a,b)= sup
n
n
P
i=1


a(x
i
)-a(x
i-1
)




 IP ={x
0
,x
1
,···,x
n
} is a partition of [a,b]
o
Example 2 If a : [a,b] ? IR is monotonically increasing, then, for any partition IP =
{x
0
,x
1
,···,x
n
} of [a,b]
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1

a(x
i
)-a(x
i-1
)
	
=a(x
n
)-a(x
0
) =a(b)-a(a)
Thus a is of bounded variation and V
f
(a,b)=a(b)-a(a).
Example 3 If a : [a,b] ? IR is continuous on [a,b] and di?erentiable on (a,b) with
sup
a<x<b
|a
'
(x)| = M, then, for any partition IP = {x
0
,x
1
,···,x
n
} of [a,b], we have, by
the Mean Value Theorem,
n
X
i=1


a(x
i
)-a(x
i-1
)


=
n
X
i=1


a
'
(t
i
)[x
i
-x
i-1
]


=
n
X
i=1
M[x
i
-x
i-1
] =M(b-a)
Thus a is of bounded variation and V
f
(a,b)=M(b-a).
Example 4 De?ne the function a : [0,1]? IR by
a(x) =

0 if x = 0
xcos
p
x
if x6= 0
This function is continuous, but is not of bounded variation because it wobbles too much
near x = 0. To see this, consider, for each m? IN, the partition
IP
m
=

x
0
= 0, x
1
=
1
2m
, x
2
=
1
2m-1
, x
3
=
1
2m-2
,···,x
2m-2
=
1
3
, x
2m-1
=
1
2
, x
2m
= 1
	
The values of a at the points of this partition are
a(IP
m
) ={0,
1
2m
, -
1
2m-1
,
1
2m-2
,···, -
1
3
,
1
2
, -1}
x
y
y =a(x)
x
2m
= 1
x
2m-1
=
1
2
1
3
For this partition,
2m
X
i=1


a(x
i
)-a(x
i-1
)


=


1
2m
-0


+


-
1
2m-1
-
1
2m


+


1
(2m-2)
+
1
2m-1


+···+


-
1
3
-
1
4


+


1
2
+
1
3


+


-1-
1
2


=
1
2m
+0 +
1
2m-1
+
1
2m
+
1
(2m-2)
+
1
2m-1
+··· +
1
3
+
1
4
+
1
2
+
1
3
+ 1+
1
2
= 2

1
2m
+
1
2m-1
+···+
1
3
+
1
2


+1
The harmonic series
8
P
k=2
1
k
diverges. So given any M, there is an m ? IN for which the
partition IP
m
obeys
n
X
i=1


a(x
i
)-a(x
i-1
)


>M
Theorem 5
(a) If a,ß : [a,b]? IR are of bounded variation and c,d? IR, then ca+dß is of bounded
variation and
V
ca+dß
(a,b)=|c|V
a
(a,b)+|d|V
ß
(a,b)
(b) If a : [a,b]? IR is of bounded variation on [a,b] and [c,d]? [a,b], then a is of bounded
variation on [c,d] and
V
a
(c,d)=V
a
(a,b)
(c) If a : [a,b]? IR is of bounded variation and c? (a,b), then
V
a
(a,b)=V
a
(a,c)+V
a
(c,b)
(d) If a : [a,b] ? IR is of bounded variation then the functions V(x) = V
a
(a,x) and
V(x)-a(x) are both increasing on [a,b].
(e) The function a : [a,b]? IR is of bounded variation if and only if it is the di?erence of
two increasing functions.
Proof: We shall use the shorthand notation
IP
X


?
i
a


for
n
X
i=1


a(x
i
)-a(x
i-1
)


where the partition IP ={x
0
,x
1
,···,x
n
}.
(a) follows from the observation that, for any IP partition of [a,b],
IP
X


?
i
(ca+dß)


=|c|
IP
X


?
i
a


+|d|
IP
X


?
i
ß


=|c|V
a
(a,b)+|d|V
ß
(a,b)
(b) follows from the observation that, for any partition IP of [c,d],
IP
X


?
i
a


=
IP?{a,b}
X


?
i
a


=V
a
(a,b)
(c) If IP =

x
0
,x
1
,···,x
n
	
is any partition of [a,b] and x
i-1
=c=x
i
, then


a(x
i
)-a(x
i-1
)


=


a(x
i
)-a(c)


+


a(c)-a(x
i-1
)


so that
IP
X

?
i
a

=
IP?{c}
X

?
i
a

=
(IP?{c})n[a,c]
X

?
i
a

+
(IP?{c})n[c,b]
X

?
i
a

=V
a
(a,c)+V
a
(c,b)
which implies that V
a
(a,b) = V
a
(a,c) + V
a
(c,b). To prove the other inequality, we let
e > 0 and select a partition IP
1
of [a,c] for which
IP
1
P


?
i
a


= V
a
(a,c)-e and a partition
IP
2
of [c,b] for which
IP
2
P


?
i
a


= V
a
(c,b)-e. Then
IP
1
?IP
2
X


?
i
a


=
IP
1
X


?
i
a


+
IP
2
X


?
i
a


= V
a
(a,c)+V
a
(c,b)-2e
This assures that V
a
(a,b)=V
a
(a,c)+V
a
(c,b)-2e for all e> 0 and hence that V
a
(a,b)=
V
a
(a,c)+V
a
(c,b).
(d) Proof that V(x) is increasing: Let a=x
1
=x
2
= b. Then, by part (c),
V(x
2
)-V(x
1
) =V
a
(a,x
2
)-V
a
(a,x
1
) = V
a
(x
1
,x
2
)= 0
(d) Proof that V(x)-a(x) is increasing: Let a=x
1
=x
2
=b. By part (c),
{V(x
2
)-a(x
2
)}-{V(x
1
)-a(x
1
)} = V
a
(x
1
,x
2
)-{a(x
2
)-a(x
1
)}
= V
a
(x
1
,x
2
)-


a(x
2
)-a(x
1
)


= V
a
(x
1
,x
2
)-
{x
1
,x
2
}
X


?
i
a


= 0
(e) If a is of bounded variation then a(x) = V
a
(a,x)-

V
a
(a,x)-a(x)

expresses a as
the di?erence of two increasing functions. On the other hand if a is the di?erence ß-?
of two increasing functions, then ß and ? are of bounded variation by Example 2 and a is
of bounded variation by part (a).
Example 6 We know that if f is continuous and a is of bounded variation on [a,b], then
f ?R(a) on [a,b]. If f is of bounded variation and a is continuous on [a,b], then we have
f ?R(a) on [a,b] with
Z
b
a
fda =f(b)a(b)-f(a)a(a)-
Z
b
a
adf
byourintegrationbypartstheorem. Itispossibletohave f ?R(a)on[a,b]evenifneither
f nor a are of bounded variation on [a,b]. For example, we have seen, in Example 4, that
a(x) =

0 if x = 0
xcos
p
x
if x6= 0
is continuous but not of bounded variation on [0,1], because of excessive oscillation near
x = 0. Sof(x) =a(1-x) (still withthe a of Example4) iscontinuous but not ofbounded
variation on [0,1], because of excessive oscillation near x = 1. But f ? R(a) on [0,
1
2
],
by integration by parts, because f is of bounded variation on [0,
1
2
]. And f ? R(a) on
[
1
2
,1], because a is of bounded variation on [
1
2
,1]. So f ?R(a) on [0,1], by our linearity
theorem.