Kenneth Kuttler

1604 CHAPTER 50. RIEMANN STIELTJES INTEGRALS

Clearly the extreme inequalities would be valid in 50.0.4 if Q had more than one extrapoint. You simply do the above trick more than one time. Let S (P) and S (Q) be RiemannSteiltjes sums for which ||P|| and ||Q|| are less than δ m and let R ≡P ∪Q. Then fromwhat was just observed,

|S (P)−S (Q)| ≤ |S (P)−S (R)|+ |S (R)−S (Q)| ≤ 2m

V (γ, [a,b]) .

and this shows 50.0.3 which proves 50.0.2. Therefore, there exists a unique complex num-ber, I ∈ ∩∞

m=1Fm which satisfies the definition of∫

γf dγ. This proves the theorem.

The following theorem follows easily from the above definitions and theorem.

Theorem 50.0.5 Let f ∈C (γ∗) and let γ : [a,b]→ C be of bounded variation and contin-uous. Let

M ≥max{|| f ◦ γ (t)|| : t ∈ [a,b]} . (50.0.5)

Then ∣∣∣∣∣∣∣∣∫γ

f dγ

∣∣∣∣∣∣∣∣≤MV (γ, [a,b]) . (50.0.6)

Also if { fn} is a sequence of functions of C (γ∗) which is converging uniformly to the func-tion, f on γ∗, then

limn→∞

∫γ

fndγ =∫

f dγ. (50.0.7)

Proof: Let 50.0.5 hold. From the proof of the above theorem, when ||P||< δ m,∣∣∣∣∣∣∣∣∫γ

f dγ−S (P)

∣∣∣∣∣∣∣∣≤ 2m

V (γ, [a,b])

and so ∣∣∣∣∣∣∣∣∫γ

f dγ

∣∣∣∣∣∣∣∣≤ ||S (P)||+ 2m

V (γ, [a,b])

≤n

∑j=1

M∣∣γ (t j)− γ

(t j−1

)∣∣+ 2m

V (γ, [a,b])

≤ MV (γ, [a,b])+2m

V (γ, [a,b]) .

This proves 50.0.6 since m is arbitrary. To verify 50.0.7 use the above inequality to write∣∣∣∣∣∣∣∣∫γ

f dγ−∫

fndγ

∣∣∣∣∣∣∣∣= ∣∣∣∣∣∣∣∣∫γ

( f − fn)dγ (t)∣∣∣∣∣∣∣∣

≤max{|| f ◦ γ (t)− fn ◦ γ (t)|| : t ∈ [a,b]}V (γ, [a,b]) .

Since the convergence is assumed to be uniform, this proves 50.0.7.It turns out to be much easier to evaluate such integrals in the case where γ is also

C1 ([a,b]) . The following theorem about approximation will be very useful but first here isan easy lemma.

1604 CHAPTER 50. RIEMANN STIELTJES INTEGRALSClearly the extreme inequalities would be valid in 50.0.4 if 2 had more than one extrapoint. You simply do the above trick more than one time. Let S(W) and S (2) be RiemannSteiltjes sums for which |||] and ||2]| are less than 6,, and let @ = AU Q. Then fromwhat was just observed,IS(P) —S(2)| <|S(P)—S(Z)| +|S(Z) —S(Q)| =v (y,(a,b)).and this shows 50.0.3 which proves 50.0.2. Therefore, there exists a unique complex num-ber, J € N7_1 Fim which satisfies the definition of Jy fdy. This proves the theorem.The following theorem follows easily from the above definitions and theorem.Theorem 50.0.5 Let f € C(y*) and let y: |a,b| — C be of bounded variation and contin-uous. LetM > max {||foy(t)||:¢ € [a,b]}. (50.0.5)Then/ fary| < MV (7,[a,b)). (50.0.6)YAlso if {fn} is a sequence of functions of C (y‘) which is converging uniformly to the func-tion, f on Y*, thenlim | frdy = / fay. (50.0.7)noe Jy YProof: Let 50.0.5 hold. From the proof of the above theorem, when || Y|| < bn,2[ fay-S(P)|| < =v (r.la.b)SY mand sofay} <|\S(P)\| + =V (7.[a,0))heey 2IAYM lve) —¥(t)-1) | + “vi, [a,b])j=< MV (y,[a,b)) +=V (1, (a).This proves 50.0.6 since m is arbitrary. To verify 50.0.7 use the above inequality to write| [rar [ nay] 7 | [u-mano |<max{||foy()—froyv()||:t € [a,b]} V (7, [a,5))-Since the convergence is assumed to be uniform, this proves 50.0.7.It turns out to be much easier to evaluate such integrals in the case where y is alsoC! ([a,b]) . The following theorem about approximation will be very useful but first here isan easy lemma.