The Riemann–Darboux Integral

Throughout this section,f denotes an arbitrary bounded, real-valued function on a closed and bounded interval[a, b].

The first step in the development of the Riemann–Darboux integral is to partition the interval [a, b] into finitely many subintervals, which are used to formupper andlower sumsoff. Under suitable conditions, the sums converge to the integral.

5.1.1 Definition. A partition of [a, b] is a set P = {x0, x1, . . . , xn−1, xn}, where

x0:=a < x1<ã ã ã< xn−1< xn:=b.

The points x1, . . . , xn−1 are called the interior points of the partition. The mesh of the partitionis defined as

kPk:= max

1≤j≤n∆xj, where ∆xj:=xj−xj−1, 1≤j ≤n.

A refinement of P is a partition containing P. The common refinement of

partitionsP andQis the partitionP ∪ Q. ♦

5.1.2 Example. Letp∈N. Then, for eachn∈N, Pn :={j/pn : j= 0,1, . . . , pn}

is a partition of [0,1],kPnk=p−n, andPn+1 is a refinement ofPn. ♦ 5.1.3 Definition. Thelowerandupper(Darboux) sums off over a partition P of[a, b] are defined, respectively, by

S(f,P) :=

j=1

mj∆xj and S(f,P) :=

j=1

Mj∆xj, where

mj=mj(f) := inf

xj−1≤x≤xj

f(x) and Mj =Mj(f) := sup

xj−1≤x≤xj

f(x). ♦ 107

108 A Course in Real Analysis

A geometric interpretation of the upper and lower sums for a positive continuous function is given in Figure 5.1. The lower (upper) sum is the total area of the smaller (larger) rectangles.

a x1 x2 x3 x4 b x f

FIGURE 5.1: Upper and lower sums off.

The following proposition asserts that refinements increase lower sums and decrease upper sums.

5.1.4 Proposition. If Q is a refinement ofP, then S(f,P)≤S(f,Q)≤S(f,Q)≤S(f,P).

Proof. The middle inequality is clear. To prove the rightmost inequality, let P ={x0=a < x1<ã ã ã< xn−1< xn =b}and assume first thatQ=P ∪ {c}. Choosekso thatxk−1< c < xk and set

Mk0 = sup

xk−1≤x≤c

f(x) and Mk00= sup

c≤x≤xk

f(x). ThenMk0, Mk00≤Mk, hence

S(f,Q) =

k−1

j=1

Mj∆xj+

j=k+1

Mj∆xj+Mk0(c−xk−1) +Mk00(xk−c)

≤

k−1

j=1

Mj∆xj+

j=k+1

Mj∆xj+Mk(c−xk−1) +Mk(xk−c)

=S(f,P).

For the general case, observe that any refinement QofP may be obtained by successively adding points toP. At each step, the upper sum is decreased so that ultimately one obtains the desired inequality. The proof for lower sums is similar.

5.1.5 Corollary. For any partitionsP andQ of[a, b],

S(f,Q)≤S(f,P). (5.1) Proof. By 5.1.4,S(f,Q)≤S(f,P ∪ Q)≤S(f,P ∪ Q)≤S(f,P).

Riemann Integration on R 109 5.1.6 Definition. Thelower andupper(Darboux)integrals off on[a, b] are defined, respectively, by

Z b a

f = Z b a

f(x)dx:= sup

S(f,P) and Z b a

f = Z b a

f(x)dx:= inf

P S(f,P), where the supremum and infimum are taken over all partitions P of [a, b]. In each case,f is called theintegrand andxtheintegration variable. ♦ 5.1.7 Proposition. For any partition P of [a, b],

S(f,P)≤ Z b

f ≤ Z b

f ≤S(f,P).

Proof. The left and right inequalities are immediate from the definition of lower and upper integrals. The middle inequality follows by taking the infimum overQand then the supremum overP in (5.1).

5.1.8 Proposition. The following statements are equivalent:

(a) Z b a

f = Z b a

(b) For each ε >0 there exists a partitionPε of [a, b]such that S(f,Pε)−S(f,Pε)≤ε.

Proof. (a)⇒(b): Givenε >0, there exist partitionsP0 andP00such that Z b

f−ε/2< S(f,P0) and S(f,P00)<

Z b a

f+ε/2.

By 5.1.4, the inequalities still hold ifP0 and P00 are each replaced by their common refinementPε:=P0∪ P00. Subtracting the resulting inequalities and applying (a) yields (b).

(b)⇒(a): If the inequality in (b) holds then, by 5.1.7, 0≤

Z b a

f− Z b

f ≤S(f,Pε)−S(f,Pε)< ε.

Sinceεis arbitrary, the integrals must be equal.

5.1.9 Definition. The functionf is said to be (Darboux)integrable on[a, b] if one (hence both) of the conditions (a), (b) of 5.1.8 hold. In this case, the common value of the integrals in (a) is called the (Riemann–Darboux)integral off on [a, b] and is denoted by

Z b a

f =Z b a

f(x)dx.

110 A Course in Real Analysis Also, define

Z a b

f =− Z b

f and Z a a

f = 0.

The collection of all integrable functions on [a, b] is denoted byRba. ♦ The following theorem guarantees a rich supply of integrable functions.

5.1.10 Theorem. Iff is continuous on[a, b] except possibly at finitely many points, then f ∈ Rba.

Proof. Denote the points of discontinuity off, if any, byd1<ã ã ã< dn. For convenience, we assume that these lie in (a, b); only a minor modification of the proof is needed if d1 = a ordn = b. Let ε > 0. For each j, remove an open interval of widthrcentered atdj, the value ofr to be determined. Since f is continuous on each of the resulting n+ 1 closed intervals I0, . . . , In, it is uniformly continuous there. (If f is continuous on [a, b], then n= 0 and I0= [a, b].) Thus there exists aδ >0 such that for eachj,

|f(x)−f(y)|< ε/2(b−a) for allx, y∈Ij with|x−y|< δ.

Now, the endpoints of the intervalsIj form a partitionP of [a, b]. If necessary, refineP by inserting points (marked by∗in Figure 5.2) into these intervals so that the distance between consecutive points is less thanδ. The subintervals of

a b

P I0 I1 I2

β β β

β α β α

r r

∗ ∗

d1 d2

FIGURE 5.2: The partitionsP andQ.

the resulting partitionQare of two types: those that contain some dj, which we mark by α, and those that do not, which we mark by β. Thus, in the obvious notation,

S(f,Q)−S(f,Q) =X

(Mj−mj)∆xj+X

(Mj−mj)∆xj.

In the first sum, ∆xj< rand in the second,Mj−mj≤ε/2(b−a). Since the first sum hasnterms (corresponding to thendiscontinuitiesdj),

S(f,Q)−S(f,Q)<2M nr+ε/2,

where M is a bound for |f| on [a, b]. Choosing r < ε/4M n, we then have S(f,Q)−S(f,Q)< ε, which shows thatf is integrable on [a, b].

Riemann Integration on R 111 The set of discontinuities of an integrable function can be infinite but may not be too large. We make this precise in Section 5.8. In the meantime, we offer the following examples to illustrate the basic idea. In the first example, the function is discontinuous only on a countably infinite set, while in the second the function is discontinuous everywhere.

0 x11/n x2 x3 1/(n−1) x4 ã ã ã x2n−3 1/2 x2n−2 1

FIGURE 5.3: The partitionPn of Example 5.1.11.

5.1.11 Example. Letf be any bounded function on [0,1] such thatf(x) = 0 ifx6∈ {1/n : n= 2,3. . .}. We claim that f is integrable and thatR1

0 f = 0.

The idea is to enclose the points of discontinuity off in small intervals, as in the proof of 5.1.10. Fixnand let

Pn={x0= 0, x1, x2, . . . , x2n−2, x2n−1= 1}, where

x2j−1<1/(n−j+ 1)< x2j< x2j+1, j= 1,2, . . . , n−1, and

∆x2j =x2j−x2j−1<1/n2, j= 1,2, . . . , n.

(See Figure 5.3.) Let |f| ≤ M on [0,1]. Since f = 0 on [x2j, x2j+1] and mj≥ −M,

S(f,Pn) =m1x1+m2(x2−x1) +ã ã ã+m2n−2(x2n−2−x2n−3)

≥ −M

x1+ (x2−x1) + (x4−x3) +ã ã ã+ (x2n−2−x2n−3)

≥ −M(1/n+ (n−1)/n2) =−M(2/n−1/n2).

A similar calculation shows thatS(f,Pn)≤M(2/n−1/n2). Therefore, limn S(f,Pn) = lim

n S(f,Pn) = 0,

hencef is integrable with zero integral. ♦

5.1.12 Example. The Dirichlet functiond(x) (3.1.7) is not integrable on any (nondegenerate) interval [a, b]. Indeed, every upper sum ofd(x) has the value

b−aand every lower sum has the value 0. ♦

A useful characterization of integrability may be given in terms of the limits ofS(f,P) andS(f,P) askPk →0.

5.1.13 Definition. LetL∈R. We writeL= limkPk→0S(f,P) if, givenε >0, there existsδ >0 such that|S(f,P)−L|< εfor all partitionsP withkPk< δ. The limit limkPk→0S(f,P) is defined analogously. ♦

112 A Course in Real Analysis

5.1.14 Lemma. LetP0={x00=a < x01<ã ã ãx0n < x0n+1=b} be a partition of[a, b] and let|f| ≤M on[a, b]. Then

S(f,P)≤S(f,P0) + 3nMkPk for all partitionsP of[a, b]with kPk< δ0 := minj∆x0j.

P γ γ α γ α γ γ

P0 a x01 x02 b

P00

β β β β

γ γ γ γ γ

FIGURE 5.4: The partitionsP0,P, andP00.

Proof. SincekPk<∆x0j, no interval ofP can contain more that one interior point ofP0. Mark the intervals ofP that contain exactly one interior point of P0 byαand mark those that contain no interior point ofP0 by γ. Consider the common refinementP00=P ∪ P0 ofP andP0. Some of the intervals of P00 were formed from an interval of P of typeα; we mark those byβ. The remaining intervals ofP00, intervals that were not formed from an interval ofP of typeα, are precisely the intervals markedγinP. Thus the terms ofS(f,P) andS(f,P00) corresponding to intervals of typeγare identical, hence cancel under substraction of upper sums. Therefore, in the obvious notation,

S(f,P)−S(f,P00) =X

Mj(f)∆xj−X

Mj00(f)∆x00j

≤Mh X

∆xj+X

∆x00ji

≤M nkPk+ 2nkP00k ,

the last inequality because there are at mostnintervals of typeαand at most 2nintervals of typeβ. SinceP00 is a refinement ofP0 andP,

S(f,P)−S(f,P0)≤S(f,P)−S(f,P00)≤3nMkPk.

5.1.15 Theorem. For any bounded functionf on [a, b], Z b

f = lim

kPk→0S(f,P) and Z b a

f = lim

kPk→0S(f,P). (5.2) Thus f is integrable on[a, b] iff the limits in(5.2)are equal, in which case

Z b a

f = lim

kPk→0S(f,P) = lim

kPk→0S(f,P). (5.3)

Riemann Integration on R 113 Proof. Givenε >0, choose a partitionP0 such that

S(f,P0)<

Z b a

f+ε/2.

In the notation of 5.1.14, for any partitionP withkPk< δ0, S(f,P)≤S(f,P0)|+ 3nMkPk<

Z b a

f+ε/2 + 3nMkPk.

Hence ifkPk<min{δ0, ε/6nM}, then Z b

f ≤S(f,P)<

Z b a

f +ε.

Sinceεwas arbitrary, the first limit in (5.2) is established. The second follows from the first by considering−f and using Exercise 5.1.3.

Equation (5.3) represents the integral as a limit of upper and lower sums.

It is also possible to represent the integral as a limit of intermediate sums, calledRiemann sums.

5.1.16 Definition. LetP ={x0=a < x1<ã ã ã< xn=b}be a partition of [a, b] and letξ= (ξ1, . . . , ξn), whereξj ∈[xj−1, xj]. The sum

S(f,P,ξ) :=

j=1

f(ξj)∆xj

is called theRiemann sum off determined by P andξ. ♦ Figure 5.5 illustrates a Riemann sum for a positive continuous function f. In this caseS(f,P,ξ) is the total area of the rectangles with heightsf(ξj) and bases ∆xj.

aξ1x1 ξ2x2 ξ3x3 ξ4x4 ξ5 b f

x FIGURE 5.5: A Riemann sum.

5.1.17 Definition. LetP ={x0=a < x1<ã ã ã< xn=b}be a partition of [a, b] and letξ= (ξ1, . . . , ξn), whereξj ∈[xj−1, xj]. We write

L= lim

kPk→0S(f,P,ξ)

114 A Course in Real Analysis

if for each ε > 0 there exists δ > 0 such that |S(f,P,ξ)−L| < ε for all partitionsP withkPk< δ and all choices of ξ. Similarly, we write

L= lim

P S(f,P,ξ)

if for eachε >0 there exists a partitionPεsuch that|S(f,P,ξ)−L|< εfor

all refinementsP ofPεand all choices of ξ. ♦

We may now give Riemann’s characterization of integrability.

5.1.18 Theorem. The following statements are equivalent:

(a) f ∈ Rba. (b) lim

kPk→0S(f,P,ξ)exists in R. (c) lim

P S(f,P,ξ)exists inR.

If these conditions hold, then Z b

f = lim

kPk→0S(f,P,ξ) = lim

P S(f,P,ξ). Proof. (a)⇒(b): LetL=Rb

af. For any partitionP and anyξ, we have S(f,P)−L≤S(f,P,ξ)−L≤S(f,P)−L,

hence (b) follows from 5.1.15.

(b) ⇒(c): Let L:= limkPk→0S(f,P,ξ). Givenε >0, choose δ >0 such that

|S(f,P,ξ)−L|< ε for all partitionsP withkPk< δ and allξ. (5.4) Choose any partition Pε withkPεk< δ. If P is any refinement ofPε, then kPk ≤ kPεk< δ, hence (5.4) holds for P.

such that

|S(f,P,ξ)−L|< ε for all refinementsP ofPε and allξ. (5.5) For such a partitionP, by the approximation property of suprema there exists for eachja sequence{ξj,k}∞k=1in [xj−1, xj] such that limkf(ξj,k) =Mj(f). It follows that

limk S(f,P,ξk) =S(f,P), where ξk = (ξ1k, ξ2k, . . . , ξnk). From (5.5), Rb

a f −L ≤ S(f,P)−L ≤ ε. Sinceε was arbitrary, Rb af ≤L.

Similarly, Rb

af ≥L.Therefore Rb af = Rb

a f.

Riemann Integration on R 115 Exercises

1. Prove that ifk is a constant, then Rb a k= Rb

a k=k(b−a).

2. Let a ≤ c < d ≤ b. Define f on [a, b] by f(x) = 1 if x ∈ [c, d] and f(x) = 0 otherwise. Show thatf ∈ Rba and evaluateRb

af. 3.S ⇓1Prove that

(a) S(−f,P) =−S(f,P) and Rb

a(−f) =−Rb af.

(b) f ∈ Rba⇒ −f ∈ Rba andRb

a(−f) =−Rb af.

4. ⇓2 Prove that a monotone function is integrable.

5.S Letf ∈ Rba and letg: [a, b]→Rbe any function that differs fromf at finitely many points in [a, b]. Prove that g∈ Rba and thatRb

a f =Rb a g. Does the same result hold if gdiffers from f at countably many points?

6. Letf ∈ Rba. Prove:

(a) If infa≤x≤bf(x)>0, then 1/f∈ Rba. (b) Iff(x)≥0 for allx∈[a, b], then√

f ∈ Rba. (c)S sin(f)∈ Rba.

7.S LetF(P) be a real-valued function of partitionsP on an interval [a, b].

Write

L= lim

P F(P)

if, givenε >0, there exists a partitionPεsuch that|F(P)−L|< εfor all partitionsP refiningPε.

(a) Show that the limit is linear, that is, limP

αF(P) +βG(P)

=αlim

P F(P) +βlim

P G(P), provided the right side exists.

(b) Letf be a bounded function on [a, b]. With this definition, show that Z b

f = lim

P S(f,P) and Z b a

f = lim

P S(f,P).

8. Letf ∈ R10 and setg(x) =xq, whereq >0. Prove thatf ◦g∈ R10.

1This exercise will be used in 5.2.2.

2This exercise will be used in 5.9.8.

116 A Course in Real Analysis

Definition of Derivative and Examples

A Deeper Look at Riemann Integrability