→ ℝpbe continuous and of bounded variation. Let Ω be an open setcontainingγ∗and let f : Ω × K → ℝpbe continuous for K a compact set in ℝp, and let ε > 0 begiven. Then there existsη :
[a,b]
→ ℝpsuch thatη
(a)
= γ
(a)
,γ
(b)
= η
(b)
,η∈ C1
([a,b])
,and
∥γ − η ∥ < ε, (4.10)
(4.10)
||∫ ∫ ||
|| f (⋅,z)⋅dγ − f (⋅,z)dη|| < ε, all z ∈ K (4.11)
γ η
(4.11)
V (η,[a,b]) ≤ V (γ,[a,b]), (4.12)
(4.12)
where
∥γ − η ∥
≡ max
{|γ(t)− η(t)| : t ∈ [a,b]}
.
Proof: Extend γ to be defined on all ℝ according to γ
( ) ]|
γ s− 2h +t + -2h--(t − a) ||
j−1 b − a j−1 |
∫ n | ( )
≤ -1- 2h∑ ||γ s− 2h+ t + -2h--(t − a) −
2h 0 j=1| j b − a j
|
( -2h-- )||
γ s− 2h + tj−1 + b− a (tj− 1 − a) |ds.
For a given s ∈
[0,2h]
, the points, s− 2h + tj +
-2h
b−a
(t − a)
j
for j = 1,
⋅⋅⋅
,n form an increasing list of points in
the interval
[a − 2h,b + 2h]
and so the integrand is bounded above by V
(γ,[a − 2h,b+ 2h])
= V
(γ,[a,b])
.
It follows
∑n
|γh (tj)− γh (tj−1)| ≤ V (γ,[a,b])■
j=1
With this lemma the proof of Theorem 4.2.3 can be completed without too much trouble. Let H be an
open set containing γ∗ such that H is a compact subset of Ω. This is easily done as follows. For each
x ∈γ∗ let B
(x,rx)
be a ball centered at x which is contained in Ω. Since γ∗ is the continuous image of a
compact set, it is also compact and so there are finitely many of the balls
{B (x,rx∕2)}
whose union
contains γ∗. Denote these balls by
{B (xi, rxi)}
2
i=1n. Then let H be the union of the balls
{B(xi,rxi)}
i=1n Thus H is the union of the closures of these balls each of which is compact, so His also
compact and contained in Ω. Let 0 < ε <
1
2
min
{rxi : i ≤ n}
. Then there exists δ1 such that if h < δ1, then
for all t,
has values in H ⊆ Ω. Therefore, fix the partition P, and choose h small enough
that in addition to this, the following inequality is valid for all z ∈ K.
ε
|S (P )− Sh(P )| < -
3
This is possible because of 4.13 and the uniform continuity of f on H×K. It follows from 4.14 that for h
this small,
| |
||∫ ∫ ||
|| f (⋅,z)⋅dγ (t)− f (⋅,z)⋅dγh(t)|| ≤
γ γh
|∫ |
|| f (⋅,z)⋅dγ(t)− S(P )||+ |S(P )− S (P)|
| γ | h
| ∫ |
|| ||
+ ||Sh (P)− γ f (⋅,z)⋅dγh(t)|| < ε.
h
< ε + ε+ ε= ε ■
3 3 3
Of course the same result is obtained without the explicit dependence of f on z.
This is a very useful theorem because if γ is C1
([a,b])
, it is easy to calculate ∫γf ⋅ dγ and the above
theorem allows a reduction to the case where γ is C1. The next theorem shows how easy it is to compute
these integrals in the case where γ is C1.
Theorem 4.2.5If f : γ∗→ ℝpis continuous andγ :
[a,b]
→ ℝpis in C1
([a,b])
, then
∫ ∫
b ′
γ f ⋅dγ = a f (γ (t))⋅γ (t)dt. (4.16)
(4.16)
Proof: Let P be a partition of
[a,b]
, P =
{t0,⋅⋅⋅,tn}
and
||P ||
is small enough that whenever
|t− s|
<
||P ||
,
|f (γ(t)) − f (γ (s))| < ε (4.17)
(4.17)
and
| |
||∫ ∑n ||
|| f ⋅dγ − f (γ (τj))⋅(γ (tj)− γ(tj−1))||< ε.
|γ j=1 |
Now
n ∫ n
∑ f (γ(τ )) ⋅(γ (t)− γ (t )) = b∑ f (γ(τ ))X (s)⋅γ′(s) ds
j=1 j j j−1 a j=1 j [tj−1,tj]