1607

for all h. Here S (P) is a Riemann Steiltjes sum of the form

n

∑i=1

f (γ (τ i) ,z)(γ (ti)− γ (ti−1))

and Sh (P) is a similar Riemann Steiltjes sum taken with respect to γh instead of γ. Becauseof 50.0.11 γh (t) has values in H ⊆ Ω. Therefore, fix the partition, P , and choose h smallenough that in addition to this, the following inequality is valid for all z ∈ K.

|S (P)−Sh (P)|< ε

3

This is possible because of 50.0.11 and the uniform continuity of f on H×K. It follows∣∣∣∣∣∣∣∣∫γ

f (·,z)dγ (t)−∫

γh

f (·,z)dγh (t)∣∣∣∣∣∣∣∣≤

∣∣∣∣∣∣∣∣∫γ

f (·,z)dγ (t)−S (P)

∣∣∣∣∣∣∣∣+ ||S (P)−Sh (P)||

+

∣∣∣∣∣∣∣∣Sh (P)−∫

γh

f (·,z)dγh (t)∣∣∣∣∣∣∣∣< ε.

Formula 50.0.10 follows from the lemma. This proves the theorem.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

∫γ

f dγ

and the above theorem allows a reduction to the case where γ is C1. The next theoremshows how easy it is to compute these integrals in the case where γ is C1. First note that iff is continuous and γ ∈C1 ([a,b]) , then by Lemma 50.0.6 and the fundamental existencetheorem, Theorem 50.0.4,

∫γ

f dγ exists.

Theorem 50.0.9 If f : γ∗→ X is continuous and γ : [a,b]→ C is in C1 ([a,b]) , then∫γ

f dγ =∫ b

af (γ (t))γ

′ (t)dt. (50.0.12)

Proof: Let P be a partition of [a,b], P = {t0, · · · , tn} and ||P|| is small enough thatwhenever |t− s|< ||P|| ,

| f (γ (t))− f (γ (s))|< ε (50.0.13)

and ∣∣∣∣∣∣∣∣∣∣∫

γ

f dγ−n

∑j=1

f (γ (τ j))(γ (t j)− γ

(t j−1

))∣∣∣∣∣∣∣∣∣∣< ε.

Nown

∑j=1

f (γ (τ j))(γ (t j)− γ

(t j−1

))=∫ b

a

n

∑j=1

f (γ (τ j))X[t j−1,t j ] (s)γ′ (s)ds