and Properties of Riemann Integral Dr.Hamed Al-Sulami
THE INTEGRABILITY OF MONOTONE AND CONTINUOUS FUNCTIONS AND PROPERTIES OF RIEMANN INTEGRAL
DR.HAMED AL-SULAMI
Let [a, b]⊆Rbe a closed and bounded interval,P ={x0, x1, . . . , xn}be a partition of [a, b], and f : [a, b]→Rbe bounded function. We use the familiar notation
Mk = sup{f(x)|x∈[xk−1, xk]},andmk= inf{f(x)|x∈[xk−1, xk]}.
Note that
U(f, P)−L(f, P) = Xn k=1
(Mk−mk)4xk
Theorem 0.1(Integrability of Monotone Functions). Let f : [a, b]→Rbe monotone function on [a, b]. Then f is integrable on[a, b].
Discussion: Suppose f is increasing on [a, b]. Let P = {x0, x1, . . . , xn} be a partition of [a, b] such that xk−xk−1 = b−an for k = 1,2, . . . , n. Since f is increasing on [xk−1, xk], then mk = f(xk−1) and Mk =f(xk).Hence
U(f, P)−L(f, P) = Xn k=1
(Mk−mk)4xk
= Xn
k=1
[f(xk)−f(xk−1)]
·b−a n
¸
=b−a n
Xn
k=1
[f(xk)−f(xk−1)]
=b−a n
£(©©f(x©1)−f(x0)) + (©©f(x©2)−©©f(x©1)) +. . .+ (f(xn)−»»»f(xn−1»))¤
=b−a
n [f(xn)−f(x0)]
=b−a
n [f(b)−f(a)]. So we need to choosen∈Nsuch that b−a
n [f(b)−f(a)]< ε.
Proof. Suppose f is increasing on [a, b]. Choose n ∈ N such that b−a
n [f(b)−f(a)] < ε. Let P = {x0, x1, . . . , xn} be a partition of [a, b] such thatxk−xk−1=b−an fork= 1,2, . . . , n.Now,
U(f, P)−L(f, P) =b−a
n [f(b)−f(a)]< ε.
Hencef is integrable. ¤
Theorem 0.2 (Integrability of Continuous Functions). Let f : [a, b]→Rbe continuous function
and Properties of Riemann Integral Dr.Hamed Al-Sulami Proof. Sincef is continuous on [a, b], then f is uniformly continuous on [a, b]. Hence givenε >0 there existδ >0 such that ifx, y∈[a, b] and|x−y|< δ⇒ |f(x)−f(y)|< b−aε .Now, choosen∈Nsuch that
b−a
n < δ.LetP ={x0, x1, . . . , xn}be a partition of [a, b] such thatxk−xk−1= b−an fork= 1,2, . . . , n.
Sincef is continuous on [xk−1, xk],thenf takes its maximum and minimum at some points in [xk−1, xk].
Hencemk=f(uk) andMk=f(vk) for some uk, vk ∈[xk−1, xk].
Hence|vk−uk| ≤ |xk−xk−1|= b−an < δ⇒ |f(vk)−f(uk)|< b−aε . Therefore
U(f, P)−L(f, P) = Xn
k=1
(Mk−mk)4xk
= Xn k=1
[f(vk)−f(uk)]
·b−a n
¸
< b−a n
Xn
k=1
ε b−a
= b−a n
nε b−a
=ε.
Hencef is integrable. ¤
Theorem 0.3. Let f : [a, b]→Rbe integrable on [a, b]. Ifc∈R, thencf is integrable on[a, b], and Z b
a
cf=c Z b
a
f.
Proof. Case 1: c >0.
LetI⊆[a, b].Now, since
inf{(cf)(x) :x∈I}=cinf{f(x) :x∈I} and sup{(cf)(x) :x∈I}=csup{f(x) :x∈I}, then
U(cf, P) =cU(f, P) andL(cf, P) =cL(f, P) for any partitionP of [a, b].
Letε >0 be given. Sincef is integrable ∃a partitionPε such thatU(f, Pε)−L(f, Pε)<εc. Now,
U(cf, Pε)−L(cf, Pε) =cU(f, Pε)−cL(f, Pε)
=c[U(f, Pε)−L(f, Pε)]
< cε c
=ε.
Hencecf is integrable.
and Properties of Riemann Integral Dr.Hamed Al-Sulami Now,
cL(f, Pε) =L(cf, Pε)≤ Z b
a
cf ≤U(cf, Pε) =cU(f, Pε) (1) and
L(f, Pε)≤ Z b
a
f ≤U(f, Pε).
If we multiply the last inequality byc >0 we get cL(f, Pε)≤c
Z b
a
f ≤cU(f, Pε).
Now, usingU(cf, P) =cU(f, P) andL(cf, P) =cL(f, P) in the last inequality we have L(cf, Pε)≤c
Z b
a
f ≤U(cf, Pε).
Hence
−U(cf, Pε)≤ −c Z b
a
f ≤ −L(cf, Pε) (2) Adding (1) and (2) we get
−[U(cf, Pε)−L(cf, Pε)]≤ Z b
a
cf−c Z b
a
f ≤U(cf, Pε)−L(cf, Pε)⇒
¯¯
¯¯
¯ Z b
a
cf−c Z b
a
f
¯¯
¯¯
¯< U(cf, Pε)−L(cf, Pε)< ε.
Hence ¯
¯¯
¯¯ Z b
a
cf−c Z b
a
f
¯¯
¯¯
¯= 0⇒ Z b
a
cf−c Z b
a
f = 0⇒ Z b
a
cf=c Z b
a
f Case 2: c= 0.
The functioncf = 0 which is integrable and Z b
a
cf= Z b
a
0 = 0 = 0.
Z b
a
f =c Z b
a
f.
Case 3: c <0.
Try to do this case.
¤ Theorem 0.4. Let f, g: [a, b]→R be integrable on[a, b]. Thenf+g is integrable on [a, b],and
Z b
a
f+g= Z b
a
f+ Z b
a
g.
Proof. LetI⊆[a, b].Now, since
inf{(f+g)(x) :x∈I} ≥inf{f(x) :x∈I}+ inf{g(x) :x∈I}
and
sup{(f +g)(x) :x∈I} ≤sup{f(x) :x∈I}+ sup{g(x) :x∈I}, then
and Properties of Riemann Integral Dr.Hamed Al-Sulami Letε >0 be given. Sincef andgare integrable∃a partitionsP1andP2such thatU(f, P1)−L(f, P1)< ε2 andU(f, P2)−L(f, P2)<2ε.
Now, let
Pε=P1∪P2⇒U(f+g, Pε)≤U(f, Pε) +U(g, Pε) andL(f, Pε) +L(g, Pε)≤L(f+g, Pε) also,
Pε=P1∪P2⇒U(f, Pε)≤U(f, P1), L(f, P1)≤L(f, Pε), U(g, Pε)≤U(g, P2), and,L(g, P2)≤L(g, Pε).
Now,
U(f+g, Pε)−L(f+g, Pε)≤U(f, Pε) +U(g, Pε)−[L(f, Pε) +L(g, Pε)]
≤[U(f, Pε)−L(f, Pε)] + [U(g, Pε)−L(g, Pε)]
≤[U(f, P1)−L(f, P1)] + [U(g, P2)−L(g, P2)]
< ε 2+ε
2
=ε.
Hencef +gis integrable. Now, Z b
a
f +g≤U(f+g, Pε)≤U(f, Pε) +U(g, Pε)≤L(f, Pε) +L(g, Pε) +ε≤ Z b
a
f+ Z b
a
g+ε.
Hence Z b
a
f+g−( Z b
a
f+ Z b
a
g)≤ε (1).
Also, Z b
a
f+ Z b
a
g≤U(f, Pε) +U(g, Pε)≤L(f+g, Pε) +ε≤ Z b
a
f+g+ε.
Hence
−ε≤ Z b
a
f +g−( Z b
a
f+ Z b
a
g) (2).
By (1) and (2) we have
−ε≤ Z b
a
f +g−( Z b
a
f+ Z b
a
g)≤ε.
Hence
¯¯
¯¯
¯ Z b
a
f +g−( Z b
a
f+ Z b
a
g)
¯¯
¯¯
¯≤ε⇒
¯¯
¯¯
¯ Z b
a
f +g−( Z b
a
f+ Z b
a
g)
¯¯
¯¯
¯= 0⇒ Z b
a
f+g= Z b
a
f+ Z b
a
g.
¤
Theorem 0.5. Let f : [a, b]→Rbe integrable on [a, b]. Then |f|is integrable on [a, b],and
| Z b
a
f| ≤ Z b
a
|f|.
and Properties of Riemann Integral Dr.Hamed Al-Sulami Proof. LetI⊆[a, b].Now, since We have
sup{|f(x)|:x∈I} −inf{|f(x)|:x∈I} ≤sup{f(x) :x∈I} −inf{f(x) :x∈I}.
Letε >0 be given. Sincef is integrable ∃a partitionPε such thatU(f, Pε)−L(f, Pε)< ε
Now,
U(|f|, Pε)−L(|f|, Pε)≤U(f, Pε)−L(f, Pε)< ε.
Hence|f| is integrable. Now, f(x)≤ |f(x)| ⇒
Z b
a
f(x)≤ Z b
a
|f(x)| −f(x)≤ |f(x)| ⇒ − Z b
a
f(x)≤ Z b
a
|f(x)|.
Hence
− Z b
a
|f(x)| ≤ Z b
a
f(x)≤ Z b
a
|f(x)| ⇒
¯¯
¯¯
¯ Z b
a
f(x)
¯¯
¯¯
¯≤ Z b
a
|f(x)|.
¤
Theorem 0.6. Let f : [a, b]→Rbe integrable on [a, b]. Then f2 is integrable on[a, b].
Proof. Sincef is bounded ,then there existM >0 such that|f(x)| ≤M for allx∈[a, b].
Now,letI⊆[a, b].We have
sup{f2(x) :x∈I} −inf{f2(x) :x∈I} ≤2M[sup{|f(x)|:x∈I} −inf{|f(x)|:x∈I}]
Letε >0 be given. Since|f| is integrable∃a partitionPεsuch thatU(|f|, Pε)−L(|f|, Pε)< 2Mε
Now,
U(f2, Pε)−L(f2, Pε)≤2M U(|f|, Pε)−L(|f|, Pε)<2M ε 2M =ε.
Hencef2 is integrable. ¤
