5.6The Riemann Integral

Definition 5.6.1 Let f be a function that is defined and bounded on a

be any arbitrary selection of points of [a, b]. Then the Riemann Sum that is associated with P and C is denoted R(P ) and is defined by

R(P ) = f(c1)(x1 − x0) + f(c2)(x2 − x1) + · · · + f(cn)(xn + xn−1)

n

X

=f(ci)(xi − xi−1).

i=1

lim f(ci)Δxi = I

|| ||→0

i=1

if and only if for each > 0 there exists some δ > 0 such that

whenever || || < δ for all partitions P and all selections C that define the Riemann Sum.

If the limit I exists as a finite number, we say that f is (Riemann) integrable and write

Z b

I = f(x) dx.

a

Next we will show that if f is continuous, the Riemann integral of f is the definite integral defined by lower and upper sums and it exists. We first prove two results that are important.

Definition 5.6.2 A function f is said to be uniformly continuous on its domain D if for each > 0 there exists δ > 0 such that if |x1 − x2| < δ, for any x1 and x2 in D, then

|f(x1) − f(x2)| < .

244 CHAPTER 5. THE DEFINITE INTEGRAL

Definition 5.6.3 A collection C = {Uα : Uα is an open interval} is said to cover a set D if each element of D belongs to some element of C.

Theorem 5.6.1 If C = {Uα : Uα is an open interval} covers a closed and bounded interval [a, b], then there exists a finite subcollection B = {Uα1 , Uα2 , · · · , Uαn } of C that covers [a, b].

Proof. We define a set A as follows:

A = {x : x [a, b] and [a, x] can be covered by a finite subcollection of C}.

Since a A, A is not empty. A is bounded from above by b. Then A has a least upper bound, say lub(A) = p. Clearly, p ≤ b. If p < b, then some Uα in C contains p. If Uα = (aα, bα), then aα < p < bα. Since p = `ub(A), there exists some point a of A between aα and p. There exists a subcollection

B = {Uα1 , · · · , Uαn } that covers [a, a ]. Then the collection

B1 = {Uα1 , · · · , Uαn , Uα} covers [a, bα). By the definition of A, A must contain all points of [a, b] between p and bα. This contradicts the assump-

tion that p = `ub(A). So, p = b and b A. It follows that some finite subcollection of C covers [a, b] as required.

Theorem 5.6.2 If f is continuous on a closed and bounded interval [a, b], then f is uniformly continuous on [a, b].

and pj. We note that

|pi − pj| = |(pi − x1) + (x1 − x2) + (x2 − pj)| ≤ |pi − xi| + |x1 − x2| + |x2 − pj|

≤ max{δpi , δpj }.

It follows that both pi and pj are either in Upi or Upj . Suppose that pi and pj are both in Upi . Then

|x2 − pi| = |(x2 − x1)| + (x1 − pi)| ≤ |x2 − x1| + |x1 − pi|

1

< δ + 3 δpi < δpi .

So, x1, x2, pi and pj are all in Upi . Then

|f(x1) − f(x2)| = |(f(x1) − f(pi)) + (f(pi) − f(x2))| ≤ |f(x1) − f(pi)| + |f(pi) − f(x2)|

<3 + 3

<.

By Definition 5.6.2, f is uniformly continuous on [a, b].

Theorem 5.6.3 If f is continuous on [a, b], then f is (Riemann) integrable and the definite integral and the Riemann integral have the same value.

Proof. Let P = {a = x0 < x1 < x2 < . . . < xn = b} be a partition of [a, b] and C = {ci : xi−1 ≤ ci ≤ xi, i = 1, 2, . . . , n} be an arbitrary selection. For each i = 1, 2, . . . , n let

mi = absolute minimum of f on [xi−1, xi] obtained at ci , f(ci ) = mi; Mi = absolute maximum of f on [xi−1, xi] obtained at ci ) = Mi;

m = absolute minimum of f on [a, b];

M = absolute maximum of f on [a, b];

n

≤f(ci )(xi − xi−1) ≤ M(b − a).

i=1

We note that L(P ) and U(P ) are also Riemann sums and for every partition P , we have

L(P ) ≤ R(P ) ≤ U(P ).

To prove the theorem, it is su cient to show that lub{L(P )} = glb{U(P )}.

Since f is uniformly continuous, by Theorem 5.6.2, for each > 0 there is

some δ > 0 such that |f(x)−f(y)| < b − a whenever |x−y| < δ for x and y in [a, b]. Consider all partitions P , selections C = {ci}, C = {ci }, C = {ci } such that

m

X

≤|f(ci ) − f(ci )| xi