208 CHAPTER 9. INTEGRATION

9.6 Uniform Convergence and the IntegralIt turns out that uniform convergence is very agreeable in terms of the integral. The follow-ing is the main result.

Theorem 9.6.1 Let g be of bounded variation and let fn be continuous and converg-ing uniformly to f on [a,b]. Then f is also integrable and

∫ ba f dg = limn→∞

∫ ba fndg.

Proof: The uniform convergence implies f is also continuous. See Theorem 6.9.7.Therefore,

∫ ba f dg exists. Now let n be given large enough that

∥ f − fn∥ ≡ maxx∈[a,b]

| f (x)− fn (x)|< ε

Next pick δ > 0 small enough that if ||P||< δ , then∣∣∣∣∣∫ b

af dg−

n

∑k=1

f (tk)(g(xk)−g(xk−1))

∣∣∣∣∣ < ε∣∣∣∣∣∫ b

afndg−

n

∑k=1

fn (tk)(g(xk)−g(xk−1))

∣∣∣∣∣ < ε

for any choice tk ∈ [xk−1,xk]. Pick such a P and the same tk for both sums. Then∣∣∣∣∫ b

af dg−

∫ b

afndg

∣∣∣∣≤∣∣∣∣∣∫ b

af dg−

n

∑k=1

f (tk)(g(xk)−g(xk−1))

∣∣∣∣∣+

∣∣∣∣∣ n

∑k=1

( f (tk)− fn (tk))(g(xk)−g(xk−1))

∣∣∣∣∣+∣∣∣∣∣∫ b

afndg−

n

∑k=1

fn (tk)(g(xk)−g(xk−1))

∣∣∣∣∣< ε +

n

∑k=1

ε |g(xk)−g(xk−1)|+ ε ≤ 2ε +V[a,b] (g)ε

Since ε is arbitrary, this shows that limn→∞

∣∣∣∫ ba f dg−

∫ ba fndg

∣∣∣= 0.

9.7 A Simple Procedure for Finding IntegralsSuppose f is a continuous function and F is an increasing integrator function. How do youfind

∫ ba f (x)dF? Is there some sort of easy way to do it which will handle lots of simple

cases? It turns out there is a way. It is based on Lemma 9.4.3. First of all

F (x+)≡ limy→x+

F (y) , F (x−)≡ limy→x−

F (y)

For an increasing function F , the jump of the function at x equals F (x+)−F (x−).

Procedure 9.7.1 Suppose f is continuous on [a,b] and F is an increasing functiondefined on [a,b] such that there are finitely many intervals determined by the partition