→ ℝ^{p}be continuous and of bounded variation. Let Ω be an open setcontainingγ^{∗}and let f : Ω × K → ℝ^{p}be continuous for K a compact set in ℝ^{p}, and let ε > 0 begiven. Then there existsη :
[a,b]
→ ℝ^{p}such thatη
(a)
= γ
(a)
,γ
(b)
= η
(b)
,η∈ C^{1}
([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 + t_{j} +
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=1}^{n}. Then let H be the union of the balls
{B(xi,rxi)}
_{i=1}^{n} 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 δ_{1} such that if h < δ_{1}, 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 C^{1}
([a,b])
, it is easy to calculate ∫_{γ}f ⋅ dγ and the above
theorem allows a reduction to the case where γ is C^{1}. The next theorem shows how easy it is to compute
these integrals in the case where γ is C^{1}.
Theorem 4.2.5If f : γ^{∗}→ ℝ^{p}is continuous andγ :
[a,b]
→ ℝ^{p}is in C^{1}
([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]