1603

Proof: The function, f ◦ γ , is uniformly continuous because it is defined on a compactset. Therefore, there exists a decreasing sequence of positive numbers, {δ m} such that if|s− t|< δ m, then

| f (γ (t))− f (γ (s))|< 1m.

LetFm ≡ {S (P) : ||P||< δ m}.

Thus Fm is a closed set. (The symbol, S (P) in the above definition, means to include allsums corresponding to P for any choice of τ j.) It is shown that

diam(Fm)≤2V (γ, [a,b])

m(50.0.2)

and then it will follow there exists a unique point, I ∈ ∩∞m=1Fm. This is because X is

complete. It will then follow I =∫

γf (t)dγ (t) . To verify 50.0.2, it suffices to verify that

whenever P and Q are partitions satisfying ||P||< δ m and ||Q||< δ m,

|S (P)−S (Q)| ≤ 2m

V (γ, [a,b]) . (50.0.3)

Suppose ||P|| < δ m and Q ⊇P . Then also ||Q|| < δ m. To begin with, suppose thatP ≡

{t0, · · · , tp, · · · , tn

}and Q ≡

{t0, · · · , tp−1, t∗, tp, · · · , tn

}. Thus Q contains only one

more point than P . Letting S (Q) and S (P) be Riemann Steiltjes sums,

S (Q)≡p−1

∑j=1

f (γ (σ j))(γ (t j)− γ

(t j−1

))+ f (γ (σ∗))(γ (t∗)− γ (tp−1))

+ f (γ (σ∗))(γ (tp)− γ (t∗))+n

∑j=p+1

f (γ (σ j))(γ (t j)− γ

(t j−1

)),

S (P)≡p−1

∑j=1

f (γ (τ j))(γ (t j)− γ

(t j−1

))+

= f(γ(τ p))(γ(tp)−γ(tp−1))︷ ︸︸ ︷f (γ (τ p))(γ (t∗)− γ (tp−1))+ f (γ (τ p))(γ (tp)− γ (t∗))

+n

∑j=p+1

f (γ (τ j))(γ (t j)− γ

(t j−1

)).

Therefore,

|S (P)−S (Q)| ≤p−1

∑j=1

1m

∣∣γ (t j)− γ(t j−1

)∣∣+ 1m

∣∣γ (t∗)− γ (tp−1)∣∣+

1m

∣∣γ (tp)− γ (t∗)∣∣+ n

∑j=p+1

1m

∣∣γ (t j)− γ(t j−1

)∣∣≤ 1m

V (γ, [a,b]) . (50.0.4)