In this section we characterize Riemann integrability of a function in terms of the size of its set of discontinuities.
5.8.1 Definition. A setAof real numbers is said to have (Lebesgue)measure zeroif for eachε >0 there exists a finite or infinite sequence of intervalsIn with total lengthP
n|In|< εsuch that the sequencecovers A, that is, every
member ofAis contained in someIn. ♦
Any countable set has measure zero. Indeed, ifA={a1, a2, . . .} andε >0, then the intervalsIn= (an−ε/2n+2, an+ε/2n+2) obviously coverAand have total length< ε. In particular, the set of rational numbers has measure zero.
An uncountable set of measure zero is constructed in Example 10.3.4.
The following result will be proved in Chapter 11.
5.8.2 Theorem. Let f be bounded on [a, b]. Then f ∈ Rba iff its set of discontinuities has measure zero.
152 A Course in Real Analysis
Examples 5.1.11 and 5.1.12 are relevant here: The function in the first example, shown to be integrable, has a countable set of discontinuities. The function in the second example, shown not to be integrable, has [0,1] as its set of discontinuities, certainlynot a set of measure zero.
Theorem 5.8.2 allows simple proofs of many of the properties discussed in this chapter. For example, iff andgare integrable with sets of discontinuity AandB, respectively, thenf+g andf g have sets of discontinuity contained inA∪B, a set of measure zero (Exercise 2), and hence are integrable.
Exercises
1. Show that ifB has measure zero andA⊆B, thenAhas measure zero.
2.SProve: IfAnhas measure zero for everyn∈N, then so doesA1∪A2∪ã ã ã. 3. LetAhave measure zero. Prove thatA+Qhas measure zero.
4. Letf : [a, b]→[c, d] be integrable and g: [c, d]→Rcontinuous. Prove that g◦f is integrable.
5. A setAof real numbers has (Jordan)content zeroif for eachε >0 there exist finitely many intervals of total length< εthat cover A. Show that (a) a convergent sequence has content zero.
(b) [0,1]∩Qdoes not have content zero.
6.S Prove that the functionf in Exercise 3.3.10 is integrable on [a, b] and find its integral.
*5.9 Functions of Bounded Variation
5.9.1 Definition. Let P ={a=x0< x1<ã ã ã < xn =b} be a partition of [a, b]. Forf : [a, b]→Rdefine
VP(f) =
n
X
j=1
|f(xj)−f(xj−1)|.
Thetotal variation off on [a, b] is the extended real number Vab(f) := sup
P
VP(f).
The functionf is said to havebounded variation on [a, b] ifVab(f)<+∞. The set of all functions with bounded variation on [a, b] is denoted byBVba. ♦
Riemann Integration on R 153 5.9.2 Proposition. Let f : [a, b]→R.
(a) If f ∈ BVba, thenf is bounded.
(b) If f has a bounded derivative on [a, b], thenf ∈ BVba. (c) If f is monotone on [a, b], thenVab(f) =|f(b)−f(a)|. (d) If g∈ Rba andf(x) =Rx
a g(t)dt, thenVab(f)≤(b−a) sup[a,b]|g|.
(e) IfP is a partition of[a, b]andQis a refinement ofP, thenVP(f)≤VQ(f).
(f) If f, g∈ BVba and c∈R, thenf+g, cf, f g∈ BVba. Proof. (a) Leta < x < bandP ={a, x, b}. Then
2|f(x)| ≤ |f(x)−f(a)|+|f(x)−f(b)|+|f(a)|+|f(b)|
=VP(f) +|f(a)|+|f(b)|
≤Vab(f) +|f(a)|+|f(b)|.
(b) Let|f0| ≤C on [a, b]. By the mean value theorem, given a partitionP, there exists for eachj a pointtj∈(xj−1, xj) such that
VP(f) =X
P
|f(xj)−f(xj−1)|=X
P
|f0(tj)|(xj−xj−1)≤C(b−a). Therefore,Vab(f)≤C(b−a).
(c) Iff is increasing, then X
P
|f(xj)−f(xj−1)|=X
P
f(xj)−f(xj−1)
=f(b)−f(a).
(d) LetM := supa≤t≤b|g(t)|. Then, for any partitionP, VP(f)≤X
P
Z xj
xj−1
|g(t)|dt≤MX
P
(xj−xj−1) =M(b−a).
(e) Let P = {a = x0 < x1 < ã ã ã < xn = b} and P0 = P ∪ {c}, where c∈[xi−1, xi]. Then
VP(f) =X
j6=i
|f(xj)−f(xj−1)|+|f(xi)−f(xi−1)|
≤X
j6=i
|f(xj)−f(xj−1)|+|f(xi)−f(c)|+|f(c)−f(xi−1)|
=VP0(f).
Adding points successively, yields (e).
154 A Course in Real Analysis (f) Let|f|,|g| ≤M on [a, b]. The inequality
|(f g)(xj)−(f g)(xj−1)| ≤M|g(xj)−g(xj−1)|+M|f(xj)−f(xj−1)| shows thatf g∈ BVba. The proofs of the remaining parts of (f) are similar.
5.9.3 Example. Forα >0, define a continuous functionfαon [0,1] by fα(x) :=
(xαsin(1/x) if 0< x≤1,
0 ifx= 0.
We show that ifα≤1, thenfαdoes not have bounded variation on [0,1]. Set ak := 1
2kπ+π/2 = 2
(4k+ 1)π and bk := 1 2kπ and note that
fα(bk) = 0 and fα(ak) =aαk = c
(4k+ 1)α, where c:= 2α πα. Sincebk+1< ak< bk, for sufficiently smallε >0 we may form the partition
Pε={ε < ap< bp< ap−1<ã ã ã< ak < bk<ã ã ã< bq+1< aq < bq <1} of [ε,1], where p and q are, respectively, the largest and smallest integers satisfyingε < ap< bq <1, equivalently,
1
2π < q < p < 2−πε 4πε . Fromfα(ak)−fα(bk) = c
(4k+ 1)α, V01(fα)≥Vε1(fα)≥c
p
X
k=q
1 (4k+ 1)α.
Sinceεmay be chosen arbitrarily small, the upper limitpof the sum on the right may be made arbitrarily large. Since the seriesP∞
k=1(4k+ 1)−αdiverges,
V01(fα) = +∞. ♦
5.9.4 Theorem. If f0∈ Rba, then f ∈ BVba and Vab(f) =Z b
a
|f0(x)|dx. (5.30)
Proof. LetP be a partition of [a, b]. By the mean value theorem, there exists ξj ∈[xj−1, xj] such that
VP(f) =X
P
|f0(ξj)| |xj−xj−1|=S(|f0|,P,ξ),
Riemann Integration on R 155 By 5.1.18,
Z b a
|f0|= lim
P S(f0,P,ξ) = lim
P VP(f). (5.31) On the other hand, givenr < Vab(f), we may choose a partitionProf [a, b] such thatr < VPr(f)≤Vab(f). By 5.9.2(e),r < VP(f)≤Vab(f) for all refinements P of Pr. Thus, by (5.31),r≤Rb
a |f0| ≤Vab(f). Since r was arbitrary, (5.30) follows.
5.9.5 Corollary. If f is continuous ata andf0 is locally integrable on(a, b], then (5.30) holds, where the integral is improper. Thus f ∈ BVba iff Rb
a|f0| converges.
Proof. By the theorem and the definition of improper integral, it suffices to show that
Vab(f) = lim
t→a+Vtb(f)
= sup
a<t≤b
Vtb(f) .
Clearly, we may assume thatVab(f)>0. Let 0< s < r < Vab(f) and choose a partitionPr={x0=a < x1<ã ã ã< xn=b}such that r < VPr(f)≤Vab(f).
Next, chooset∈(a, x1) so that|f(t)−f(a)|< r−s. For sucht, letPt=Pr∪{t}. Then
Vtb(f)≥ |f(x1)−f(t)|+
n−1
X
j=1
|f(xj+1)−f(xj)|
=VPt(f)− |f(t)−f(a)|
> VPr(f)−(r−s)> s.
It follows that lim
t→a+Vtb(f)≥s. Sinceswas arbitrary, the assertion follows.
5.9.6 Example. We use 5.9.5 to show that the function fα in 5.9.3 has bounded variation on [0,1] ifα >1. We have
|fα0(x)|=|αxα−1sin(1/x)−xα−2cos(1/x)| ≤αxα−1+xα−2. Ifα >1, the integralR1
0 xα−2dxconverges, henceR1
0 |fα|converges. ♦ 5.9.7 Theorem. If f ∈ BVba, then there exist monotone increasing functions g andhon[a, b] such thatf =g−h.
Proof. Forx∈[a, b], defineg(x) :=Vax(f) andh(x) :=g(x)−f(x). Clearly, g is increasing. To see thathis increasing, letx < y, letPxbe an arbitrary partition of [a, x], and letPy=Px∪ {y}. Then
VPx(f) +f(y)−f(x) =VPy(f)≤g(y).
Taking suprema over all partitions Px yieldsg(x) +f(y)−f(x)≤g(y), that is,h(x)≤h(y).
From Exercise 5.1.4 we have 5.9.8 Corollary. BVba⊆ Rba.
156 A Course in Real Analysis
*5.10 The Riemann–Stieltjes Integral
In this section we describe the main features of the Riemann-Stieltjes integral, a generalization of the Riemann integral. These integrals have many of the properties of Riemann integrals; however, as we shall see, there are some striking differences.
Definition and General Properties
5.10.1 Definition. Let f andw be bounded, real-valued functions on an interval [a, b]. IfP ={x0=a < x1<ã ã ã< xn =b}andξj∈[xj−1, xj], then
Sw(f,P,ξ) :=
n
X
j=1
f(ξj)∆wj, ∆wj:=w(xj)−w(xj−1), ξ:= (ξ1, . . . , ξn), is called a Riemann-Stieltjes sum off with respect to w. The function f is said to beRiemann-Stieltjes integrable with respect towif for someI∈Rand eachε >0, there exists a partitionPεsuch that
|Sw(f,P,ξ)−I|< ε for all refinementsP ofPεand all choices ofξ.
In this caseI is called theRiemann-Stieltjes integral with respect to wand is denoted by
Z b a
f dw=Z b a
f(x)dw(x) = lim
P Sw(f,P,ξ). (5.32) The functionf is called theintegrand andwtheintegrator. The collection of all functions that are Riemann-Stieltjes integrable with respect towis denoted
byRba(w). ♦
It follows from 5.1.18 that, for the integrator w(x) = x, the Riemann- Stieltjes integral reduces to the Riemann integral.
It is clear that constant functions are Riemann-Stieltjes integrable. The following example shows that, in contrast to the Riemann integral, iff has a simple discontinuity, thenRb
a f dwmay not exist.
5.10.2 Example. Letf : [0,1]→Rand define w(x) :=
(0 if 0≤x <1, 1 ifx= 1 We show thatf ∈ R10(w) ifff is continuous at 1.
LetP ={x0= 0< x1<ã ã ã< xn= 1} be any partition of [0,1]. Then Sw(f,P,ξ) =f(ξn)[w(1)−w(xn−1)] =f(ξn).
Riemann Integration on R 157 Hence iff ∈ R10(w) andξis chosen so that firstξn= 1 and secondξn<1, we see thatf is continuous at 1 andR1
0 f dw=f(1).
Conversely, if f is not continuous at 1, then there exists a sequence{am} andr >0 such thatam↑1 and|f(am)−f(1)| ≥rfor everym. LetPmdenote the refinementP ∪ {am}of P, where am∈(xn−1,1]. Ifξconsists of the left endpoints of the intervals ofPm, then Sw(f,Pm,ξ) =f(am), hence
|Sw(f,Pm,ξ)−f(1)|=|f(am)−f(1)| ≥r.
SinceP was arbitrary, f 6∈ R10(w). ♦
5.10.3 Theorem. Iff, g∈ Rba(w)andα, β∈R, thenαf+βg∈ Rba(w)and Z b
a
(αf+βg)dw=α Z b
a
f dw+β Z b
a
g dw.
Proof. This follows from the identity
Sw(αf+βg,P,ξ) =αSw(f,P,ξ) +βSw(g,P,ξ)
and the linearity of the limit in (5.32), as is readily established by a standard argument.
5.10.4 Theorem. Let w:=αu+βv, whereα, β∈Randu, v : [a, b]→Rare bounded. Iff ∈ Rba(u)∩ Rba(v), thenf ∈ Rba(w)and
Z b a
f dw=α Z b
a
f du+β Z b
a
f dv.
Proof. This follows from
Sw(f,P,ξ) =αSu(f,P,ξ) +βSv(f,P,ξ) and the linearity of the limit in (5.32).
5.10.5 Theorem. Leta < c < b. Iff|[a,c] ∈ Rca(w)and f|[c,b]∈ Rbc(w), then f ∈ Rba(w)and
Z b a
f dw=Z c a
f dw+Z b c
f dw.
Proof. Givenε >0, choose partitionsPε0 of [a, c] andPε00of [c, b] such that the following hold:
Z c a
f dw−Sw(f,P0,ξ0)
< ε/2 for all refinements P0 ofPε0 and allξ0,
Z b c
f dw−Sw(f,P00,ξ00)
< ε/2 for all refinementsP00ofPε00 and allξ00.
158 A Course in Real Analysis
ThenPε:=Pε0∪ Pε00 is a partition of [a, b] containingc. Moreover, ifP is a refinement ofPε, thenP0:=P ∩[a, c] andP00=P ∩[c, b] are refinements of Pε0 andPε00, respectively. From
Sw(f,P,ξ) =Sw(f,P0,ξ0) +Sw(f,P00,ξ00) and the above inequalities we see that
Z c a
f dw+Z b c
f dw−Sw(f,P,ξ)
< ε/2 +ε/2 =ε.
This establishes the existence ofRb
af dw as well as the desired equality.
5.10.6 Example. Consider the floor function integratorw(x) =bxc. A slight modification of the argument in 5.10.2 shows thatRn
0 f(x)dbxcexists iff f is left continuous at the integers 1,2, . . . , n, in which caseRk
k−1f(x)dbxc=f(k).
For such a function, 5.10.5 implies that Z n
0
f(x)dbxc=
n
X
k=1
Z k k−1
f(x)dbxc=
n
X
1
f(k). ♦
The preceding example suggests that improper Riemann-Stieltjes integra- tion could be used to provide a unified theory that includes both improper Riemann integrals and infinite series. This is indeed possible; however, it turns out that Lebesgue integration is a more efficient approach. Lebesgue theory onRn is developed in Chapter 11.
The following theorem reveals a remarkable symmetry between integrand and integrator.
5.10.7 Integration by Parts Formula. Iff ∈ Rba(w), thenw∈ Rba(f)and Z b
a
f dw+Z b a
w df =f(b)w(b)−f(a)w(a). Proof. For any partitionP{x0=a, x1, . . . , xn−1, xn =b},
f(b)w(b)−f(a)w(a) =
n
X
j=1
f(xj)w(xj)−
n
X
j=1
f(xj−1)w(xj−1) and
Sf(w,P,ξ) =
n
X
j=1
w(ξj)f(xj)−
n
X
j=1
w(ξj)f(xj−1). Subtracting we obtain
f(b)w(b)−f(a)w(a)−Sf(w,P,ξ)
=
n
X
j=1
f(xj−1)[w(ξj)−w(xj−1)] +
n
X
j=1
f(xj)[w(xj)−w(ξj)]
=Sw(f,Q,ζ),
Riemann Integration on R 159 where ζ = (a, x1, x1, x2, x2, . . . , xn−1, xn−1, b) and Q is the refinement of P obtained by adding the coordinates ofξtoP. Therefore,
a x1 x2 x3 x4 b
ξ1 ξ2 ξ3 ξ4 ξ5
P ξ
a ξ1 x1 ξ2 x2 ξ3 x3 ξ4 x4 ξ5 b Q
ζ
FIGURE 5.10: The partitionQ.
f(b)w(b)−f(a)w(a)− Z b
a
f dw−Sf(w,P,ξ) =
Sw(f,Q,ζ)− Z b
a
f dw . Since f ∈ Rba(w), the right side may be made arbitrarily small, Therefore, Rb
aw df exists and equalsf(b)w(b)−f(a)w(a)−Rb a f dw.
The next result shows that under certain general conditions the Riemann- Stieltjes integral reduces to a Riemann integral.
5.10.8 Theorem. Let f ∈ Rba(w). If w is continuously differentiable, then f w0∈ Rba and
Z b a
f dw=Z b a
f(x)w0(x)dx.
Proof. For any partitionP of [a, b] and anyξ, Sw(f,P,ξ)−S(f w0,P,ξ) =
n
X
j=1
f(ξj)∆wj−
n
X
j=1
f(ξj)w0(ξj)∆xj. By the mean value theorem, for eachj there existstj ∈(xj−1, xj) such that
∆wj =w(xj)−w(xj−1) =w0(tj)∆xj. Therefore,
Sw(f,P,ξ)−S(f w0,P,ξ) =
n
X
j=1
f(ξj)
w0(tj)−w0(ξj)∆xj. (5.33) Let|f| ≤M on [a, b]. By uniform continuity of w0, givenε >0, there exists a δ >0 such that
|w0(x)−w0(y)|< ε
2M(b−a) whenever|x−y|< δ. (5.34)
160 A Course in Real Analysis
LetPε0 be a partition of [a, b] withkPε0k< δ. From (5.33) and (5.34),
|Sw(f,P,ξ)−S(f w0,P,ξ)| ≤ ε 2(b−a)
n
X
j=1
∆xj = ε
2 (5.35)
for all refinementsP ofPε0 and allξ. Next, choose a partitionPε00such that
Z b a
f dw−Sw(f,P,ξ)
< ε/2 for allξand all refinementsP ofPε00. (5.36) IfP is a refinement ofPε0 ∪ Pε00, then both (5.35) and (5.36) hold, hence, by the triangle inequality,
Z b a
f dw−S(f w0,P,ξ) < ε.
This shows that f w0∈ Rba and establishes the equality.
Monotone Increasing Integrators
Ifw: [a, b]→Ris monotone increasing, then the Riemann-Stieltjes integral may be characterized in terms of upper and lower sums, as in the Darboux theory. This fact will lead to an important existence theorem for integrators of bounded variation and continuous integrands.
Letf : [a, b]→Rbe bounded and let P be a partition of [a, b]. Define the upper and lower Darboux–Stieltjes sums off with respect towby
Sw(f,P) =
n
X
j=1
Mj∆wj and Sw(f,P) =
n
X
j=1
mj∆wj, where
Mj=Mj(f) := sup
xj−1≤x≤xj
f(x) and mj=mj(f) := inf
xj−1≤x≤xj
f(x). The upper andlower Darboux–Stieltjes integrals of f with respect to w are defined, respectively, by
Z b a
f dw:= inf
P Sw(f,P) and Z b a
f dw:= sup
P
Sw(f,P).
As in the Darboux theory, if Q is a refinement of P then, because w is increasing,
Sw(f,P)≤Sw(f,Q)≤ Z b
a
f dw≤ Z b
a
f dw≤Sw(f,Q)≤Sw(f,P). Here is the analog of 5.1.8 for Riemann–Stieltjes integrals.
Riemann Integration on R 161 5.10.9 Theorem. The following statements are equivalent:
(a) f ∈ Rba(w).
(b) For each ε >0, there exists a partition Pε such that Sw(f,P)−Sw(f,P)< ε.
(c) Z b a
f dw= Z b a
f dw.
If these conditions hold, then Z b a
f dw= Z b a
f dw= Z b a
f dw.
Proof. That (b) and (c) are equivalent is proved exactly as in 5.1.8.
Assume that (a) holds. Given ε >0, choose a partitionPεsuch that
Z b a
f dw−Sw(f,P,ξ)
< ε/3 for all refinementsP ofPε and allξ. (5.37) For such a partition P and for each j, there exists a sequence{ξj,k}∞k=1 in [xj−1, xj] such that limkf(ξj,k) =Mj(f). It follows that
limk Sw(f,P,ξk) =Sw(f,P), where ξk = (ξ1,k, . . . , ξn,k). From (5.37),
Z b a
f dw−Sw(f,P)
≤ε/3. Similarly,
Z b a
f dw−Sw(f,P) ≤ε/3. Part (b) now follows from the triangle inequality.
Now assume that (c) holds. LetI denote the common value of the integrals in (c). Givenε >0, choose partitionsPε0 andPε00 such that
I−ε < Sw(f,Pε0) and Sw(f,Pε00)< I+ε.
The inequalities still hold ifPε0 andPε00 are replaced by any refinementP of Pε:=Pε0 ∪ Pε00. Thus
−ε < Sw(f,P)−I≤Sw(f,P,ξ)−I≤Sw(f,P)−I < ε.
This shows that f ∈ Rba(w) andRb
af dw=I.
162 A Course in Real Analysis Integrators of Bounded Variation
Recall that a function of bounded variation may be expressed as the difference of two monotone increasing functions (5.9.7). This, together with 5.10.9, allows for a simple proof of the following existence theorem.
5.10.10 Theorem. If f : [a, b] → R is continuous and w : [a, b] → R has bounded variation, thenf ∈ Rba(w).
Proof. By the remark preceding the theorem and by 5.10.4, we may assume thatwis increasing. By uniform continuity off, givenε >0, there exists a δ >0 such that
|f(x)−f(y)|< ε
w(b)−w(a) + 1 for allx, y with|x−y|< δ.
LetPε be a partition withkPεk< δ. For any refinementP of Pε, kPk< δ, hence
Mj(f)−mj(f)≤ ε
w(b)−w(a) + 1. Therefore,
Sw(f,P)−Sw(f,P) =
n
X
j=1
Mj(f)−mj(f)∆wj≤ε, which shows thatf ∈ Rba(w).
The conclusion of the theorem does not necessarily hold if wfails to have bounded variation, even ifw is continuous:
5.10.11 Example. Letf =w=f1/2, wherefαis defined as in Example 5.9.3.
We show thatR1
0 f dw does not exist. Referring to that example, letPεbe the partition
ε < ap< bp< ap−1<ã ã ã< bk+1< ak < bk<ã ã ã< bq+1< aq < bq <1, of [ε,1], and letξconsist of left endpoints ofPε. Then
Sw(f,P,ξ) =f(ε)
w(aq)−w(ε)
+f(bq)
w(1)−w(bq) +
p
X
k=q
f(ak)
w(bk)−w(ak) +
p−1
X
k=q
f(bk+1)
w(ak)−w(bk+1) . Sincef1/2(bk) = 0 andf1/2(ak) =√
ak, Sw(f,Pε,ξ) =f(ε)√
aq−w(ε)
−
p
X
k=q
ak.
Since the sums diverge asε→0, limε→0Sw(f,Pε,ξ) =−∞. ♦
Chapter 6
Numerical Infinite Series
An infinite series is the limit of a sequence of expanding finite sums. The terms of these sums may be real numbers or functions. In this chapter we examine the convergence behavior of series of the former type; series whose terms are functions are treated in the next chapter. In the first section, we give examples of series that may be summed, that is, for which an explicit numerical value may be calculated. The remaining sections describe various tests for convergence of general series. Additional methods of summing series may be found in Section 7.4.