35

Proof: If not, then there exists ε > 0 and xn,yn, |xn− yn|< 1/n but

| f (xn)− f (yn)| ≥ ε.

By Corollary 3.0.3, there is a subsequence{

xnk

}and point x∈ [a,b] such that limk→∞ xnk =

x. Then it follows that also limk→∞ ynk = x also because∣∣ynk − x∣∣≤ ∣∣ynk − xnk

∣∣+ ∣∣xnk − x∣∣

and both of the terms on the right converge to 0. But then by continuity of f ,

0 = f (x)− f (x) = limk→∞

(f(xnk

)− f

(ynk

))which is impossible because

∣∣ f (xnk

)− f

(ynk

)∣∣≥ ε for all k. ■With this preparation, here is the major result on the existence of the integral of a

continuous function.

Theorem 3.0.9 Let f : [a,b]→ R be continuous. Then∫ b

a f (x)dx exists. In fact, thereexists a sequence δ m converging to 0 such that if ∥P∥< δ m, and if ∑P f is a Riemann sumfor P, then ∣∣∣∣∣∑P f −

∫ b

af dx

∣∣∣∣∣< 2m(b−a)

δ m is defined to be such that if |x− y| < δ m, then | f (x)− f (y)| < 1m and the sequence is

decreasing.

Proof: Consider a partition P given by a = x0 < x1 < · · ·< xn = b. Then you could addin another point as follows:

a = x0 < x1 < · · ·< xi−1 < x∗ < xi < · · ·< xn = b

Denote this one by Q. Then if you have a Riemann sum,

∑P

f =n

∑j=1

f (y j)(x j− x j−1

)You could write this sum in the following form.

i−1

∑j=1

f (y j)(x j− x j−1

)+ f (yi)(x∗− xi−1)+ f (yi)(xi− x∗)+

n

∑j=i+1

f (y j)(x j− x j−1

)In fact, you could continue adding in points and doing the same trick and thereby write theoriginal sum in terms of any partition containing P. If R is a partition containing P and ifδ m corresponds to ε = 1/m in the above Lemma with · · ·> δ m > δ m+1 · · · 3.0.8, then onecan conclude that if ∥P∥< δ m, then∣∣∣∣∣∑P f −∑

Rf

∣∣∣∣∣≤ 1m(b−a)