First we need to define what is meant by finite total variation.
Definition 9.1.1Suppose g is an arbitrary function defined on
[a,b]
. ForP_{[a,x]
}≡
{x0,⋅⋅⋅,xn}
a partition of
[a,x]
,define V
( )
P[a,x],g
by
∑n
|g (xi)− g (xi−1)|.
i=1
Define the total variation of g on
[a,x]
by
{ ( ) }
V[a,x](g) ≡ sup V P [a,x],g : P[a,x] is a partition of [a,x ] .
Then g is said to be ofbounded variation on
[a,b]
if V_{[a,b]
}
(g)
is finite.
Then with this definition, one has an important proposition which pertains to the case of
principal interest here in which the functions are all real valued. The above definition of finite
total variation works for functions which have values in some normed linear space
however.
Proposition 9.1.2Every real valued function g of bounded variation can be written asthedifference of two increasing function, one of which is the function
x → V[a,x](g)
Furthermore, the functions of bounded variation are exactly those functions which are thedifference of two increasing functions.
Proof: Let g be of bounded variation. It is obvious from the definition that x → V_{}
[a,x]
(g)
is an increasing function. Also
( )
g (x ) = V[a,x](g)− V[a,x](g)− g (x)
The first part of the proposition is proved if I can show x → V_{}
[a,x]
(g)
− g
(x)
is increasing.
Let x ≤ y. Is it true that
V (g)− g (x ) ≤ V (g)− g(y)?
[a,x] [a,y]
This is true if and only if
g (y) − g (x) ≤ V[a,y](g)− V[a,x](g) (9.1)
(9.1)
To show this is so, first note that
( ) ( )
V P [a,x],g ≤ V Q[a,x],g
whenever the partition Q_{}
[a,x]
⊇ P_{[a,x]
}. You demonstrate this by adding in one point at a time
and using the triangle inequality. Now let P_{y} and P_{[a,x]
} be partitions of
[a,y]
and
[a,x]
respectively such that
( )
V P[a,x],g + ε > V [a,x](g),V (Py,g)+ ε > V[a,y](g)
Without loss of generality P_{y} contains x because from what was just shown you could add in
the point x and the approximation of V
(Py,g)
to V_{[a,y]
}
(g)
would only be better. Then from
the definition,
( ( ) )
V[a,y](g)− V[a,x](g) ≥ V (Py,g)− V P[a,x],g + ε
≥ |g(y)− g(x)|− ε
≥ g (y) − g (x)− ε
and since ε is arbitrary, this establishes 9.1. This proves the first part of the proposition.
Now suppose
g(x) = g1(x)− g2(x)
where each g_{i} is an increasing function. Why is g of bounded variation? Letting x < y
is defined as I. I will denote this Riemann Stieltjes sumapproximating I as∑_{P}f
(zi)
(g(xi)− g(xi−1))
. When f is Riemann Stieltjes integrable on
[a,b]
with respect to g as just described, we write f ∈ R
([a,b],g)
Lemma 9.1.4The integral∫_{a}^{b}f
(x)
dg
(x)
is well defined in the sense that if thereis such a number I then there is only one.
Proof:Suppose you have two of them I,Î and that P,
ˆP
are corresponding partitions
such that
∥P∥
,
∥ ∥
∥∥ˆP∥∥
are both small enough that
|| || || ||
||I − ∑ f (zi)(g (xi)− g (xi−1))|| < ε,||ˆI − ∑ f (ˆzi)(g(xi)− g(xi− 1))||< ε
| P | || ˆP ||
whenever z_{i} or ẑ_{i} are in
[xi− 1,xi]
. Let Q ≡ P ∪
ˆP
and choose z_{i} and ẑ_{i} to be the left
endpoint of the sub intervals defined by the partition Q. Then
∥Q∥
≤ min
( ∥ ∥)
∥P∥ ,∥∥Pˆ∥∥
and
so
| |
|I − S| < ε,||ˆI − S|| < ε
where S = ∑_{Q}f
(xi−1)
(g(xi)− g(xi−1))
. Then
| | | |
||I − ˆI|| ≤ |I − S|+ ||S − ˆI|| < 2ε
Since ε is arbitrary, I = Î. ■
Next is a fairly obvious theorem which says essentially that things which hold for sums
typically hold for integrals also, provided the integrals exist.
Theorem 9.1.5Assuming all integrals make sense, the following relation exists forf,g functions and a,b scalars.
∫ ∫ ∫
b( ˆ) b b ˆ
a af + bf dg = a a fdg+ b a fdg.
Assuming all integrals make sense and g is increasing, it follows that
| |
∫ b ||∫ b ||
a |f |dg ≥ || a f dg||.
Also, if a < c < b and all integrals make sense for I =
[a,c]
,
[a,b]
,
[c,b]
, then
∫ b ∫ c ∫ b
f dg = fdg + f dg
a a c
Proof:Consider the first claim. Since all is assumed to make sense, (It is shown in the
next theorem that continuity of the f functions and bounded variation of the g is sufficient.)
there exists δ > 0 such that if
||P||
< δ, then
||∫ b ( ) ( )|| || ∫ b || || ( ) ∫ b ||
|| af + bˆf dg− SP af + bˆf ||,||aSP (f) − a fdg||,||bSP ˆf − b ˆfdg||< ε
| a | | a | | a |
where S_{P}
(h )
denotes a suitable Riemann sum with respect to such a partition.
Choose the same point in
[x ,x]
i−1 i
for each function in the list. Then from the
above,
|| ( ∫ ∫ ∫ ) ||
||bS (fˆ) + aS (f) − S (af + bˆf)− a bfdg +b bˆfdg− b(af + bfˆ)dg ||< 3ε
| P P P a a a |
However, bS_{P}
( ˆ)
f
+ aS_{P}
(f )
− S_{P}
( ˆ)
af + bf
= 0 from properties of sums. Therefore,
| |
|| ∫ b ∫ bˆ ∫ b( ˆ) ||
||a a f dg+ b a fdg− a af + bf dg|| < 3ε
and since ε is arbitrary, this shows that the expression inside
|⋅|
equals 0.
If g is increasing, then the Riemann sums are all nonnegative if f ≥ 0. Thus
∫ b ∫ b
(|f |− f )dg, (|f|+ f)dg ≥ 0
a a
and so
∫ b (∫ b ∫ b )
|f |dg ≥ max fdg,− fdg
a a a
so
| |
∫ b ||∫ b ||
|f|dg ≥ || fdg||
a a
For the last claim, let δ > 0 be such that when
||P||
< δ, all integrals are approximated within
ε be a Riemann sum based on P. Without loss of generality, letting P_{I} be such a partition for
I each of the intervals needed, we can assume P_{[a,b]
} contains c since adding it in will not
increase
||P||
. Also we can let each of the other two P_{I} be the restriction of P_{[a,b]
} to
[a,c]
or
[c,b]
. Then
| | | |
||∫ c || ||∫ b || ||∫ b ||
||a fdg − SP[a,c] (f)|| < ε,||c fdg − SP[c,b]|| < ε,|| a fdg− SP[a,b]|| < ε
we can also pick the same intermediate point in each of these sums. Then S_{P}
[a,b]
= S_{P[a,c]
} +S_{P[c,b]
}
and so, from the triangle inequality,
| ( ) |
||∫ b ∫ c ∫ b ||
|| a fdg− a fdg+ c fdg || < 3ε
and since ε is arbitrary, the desired relation follows. ■
When does the integral make sense? The main result is the next theorem. We
have in mind the case where f and g have real values but there is no change in the
argument if they have complex values or even more general situations such as where g
has values in a complete normed linear space and f has scalar values or when f
has values in a normed linear space and g has scalar values or even more general
situations. You simply change the meaning of the symbols used in the following
argument.
Theorem 9.1.6Let f be continuous on
[a,b]
and let g be of finite totalvariation on
[a,b]
. Then f is Riemann Stieltjes integrable in the sense of Definition9.1.3, f ∈ R
([a,b],g)
.
Proof: Since f is continuous and
[a,b]
is sequentially compact, it follows from Theorem
6.7.2 that f is uniformly continuous. Thus if ε > 0 is given, there exists δ > 0 such that if
|x − y|
< δ, then
|f (x)− f (y)| <-(---ε-----).
2 V[a,b](g)+ 1
Let P =
{x0,⋅⋅⋅,xn}
be a partition such that
||P||
< δ. Now if you add in a point z on the
interior of I_{j} and consider the new partition,
x0 < ⋅⋅⋅ < xj−1 < z < xj < ⋅⋅⋅ < xn
denoting it by P^{′},
′ j−∑ 1 ′
S (P,f)− S (P ,f) = (f (ti)− f (ti))(g(xi)− g(xi−1))
i=1
( )
+f (tj)(g (xj)− g (xj−1))− f t′j (g(z)− g(xj−1))
( ′ ) ∑n ( (′ ))
− f tj+1 (g (xj)− g (z))+ f (ti)− f ti+1 (g(xi)− g (xi−1))
i=j+1
The term, f
(tj)
(g(xj)− g(xj−1))
can be written as
f (tj)(g(xj)− g(xj−1)) = f (tj)(g (xj)− g (z))+ f (tj)(g(z)− g(xj−1))
and so, the middle terms can be written as
f (tj)((g) (xj)− g (z))+ f (tj)((g(z))− g(xj− 1))
− f t′j (g(z)− g(xj−1))− f t′j+1 (g (xj) − g (z))
( ( ))
= f (tj) − f t′j+1 (g(xj)− g(z))
+ (f (t)− f (t′))(g(z)− g(x ))
j j j−1
The absolute value of this is dominated by
< --(---ε------) (|g (xj)− g (z)|+ |g (z) − g (xj− 1)|)
2 V[a,b](g)+ 1
This is because the various pairs of values at which f is evaluated are closer than δ. Similarly,
||j−1 ||
||∑ (f (ti) − f (t′))(g(xi) − g (xi−1))||
|i=1 i |
j−∑1
≤ |f (ti)− f (t′i)||g (xi)− g(xi−1)|
i=1
j−∑1
≤ -(----ε-----) |g(xi)− g(xi− 1)|
i=1 2 V[a,b](g)+ 1
and
|| n ||
||∑ ( (′ )) ||
||i=j+1 f (ti)− f ti+1 (g(xi) − g (xi−1))||
n
∑ --(---ε------)
≤ 2 V[a,b](g)+ 1 |g(xi)− g (xi−1)|.
i=j+1
Similar reasoning would apply if you added in two new points in the partition or
more generally, any finite number of new points. You would just have to consider
more exceptional terms. Therefore, if
||P ||
< δ and Q is any partition, then from
what was just shown, you can pick the points on the intervals any way you like
and