1606 CHAPTER 50. RIEMANN STIELTJES INTEGRALS

γ

(−2h+ t +

2hb−a

(t−a))(

1+2h

b−a

)}and so γh ∈C1 ([a,b]) . The following lemma is significant.

Lemma 50.0.8 V (γh, [a,b])≤V (γ, [a,b]) .

Proof: Let a = t0 < t1 < · · ·< tn = b. Then using the definition of γh and changing thevariables to make all integrals over [0,2h] ,

n

∑j=1

∣∣γh (t j)− γh(t j−1

)∣∣=n

∑j=1

∣∣∣∣ 12h

∫ 2h

0

[γ

(s−2h+ t j +

2hb−a

(t j−a))−

γ

(s−2h+ t j−1 +

2hb−a

(t j−1−a

))]∣∣∣∣≤ 1

2h

∫ 2h

0

n

∑j=1

∣∣∣∣γ(s−2h+ t j +2h

b−a(t j−a)

)−

γ

(s−2h+ t j−1 +

2hb−a

(t j−1−a

))∣∣∣∣ds.

For a given s∈ [0,2h] , the points, s−2h+ t j +2h

b−a (t j−a) for j = 1, · · · ,n form an increas-ing list of points in the interval [a−2h,b+2h] and so the integrand is bounded above byV (γ, [a−2h,b+2h]) =V (γ, [a,b]) . It follows

n

∑j=1

∣∣γh (t j)− γh(t j−1

)∣∣≤V (γ, [a,b])

which proves the lemma.With this lemma the proof of the theorem can be completed without too much trouble.

Let H be an open set containing γ∗ such that H is a compact subset of Ω. Let 0 < ε <dist(γ∗,HC

). Then there exists δ 1 such that if h < δ 1, then for all t,

|γ (t)− γh (t)| ≤1

2h

∫ t+ 2h(b−a) (t−a)

−2h+t+ 2h(b−a) (t−a)

|γ (s)− γ (t)|ds

<1

2h

∫ t+ 2h(b−a) (t−a)

−2h+t+ 2h(b−a) (t−a)

εds = ε (50.0.11)

due to the uniform continuity of γ. This proves 50.0.8.From 50.0.2 and the above lemma, there exists δ 2 such that if ||P|| < δ 2, then for all

z ∈ K, ∣∣∣∣∣∣∣∣∫γ

f (·,z)dγ (t)−S (P)

∣∣∣∣∣∣∣∣< ε

3,

∣∣∣∣∣∣∣∣∫γh

f (·,z)dγh (t)−Sh (P)

∣∣∣∣∣∣∣∣< ε

3