Lecture #7
MATH 321: Real Variables II
University of British Columbia
Lecture #7:
Instructor:
Scribe:
January 21, 2008
Dr. Joel Feldman
Peter Wong
See handout: Reduction to the Riemann Integral
Theorem. (Reduction to the Riemann Integral Theorem)
Proof. (Continued)
|S(P, T, f α0 , x) − S(P, T, f, α)| =
n
X
|f (ti )| |α0 (ti ) − α0 (vi )| |xi − xi−1 |
i=1
Let ε > 0, f bounded. Then
∃M > 0 such that |f (ti )| ≤ M, ∀i
and
α0 is continuous on [a, b]
=⇒ α0 is uniformly continuous on [a, b]
ε
2M (b − a)
ε
=⇒ ∃δ such that kP k < δ =⇒ |α0 (ti ) − α0 (vi )| <
2M (b − a)
=⇒ ∃δ such that |t − v| < δ =⇒ |α0 (t) − α0 (v)| <
=⇒ ∃δ such that kP k < δ =⇒ |S(P, T, f α0 , x) − S(P, T, f, α)| ≤
n
X
M
i=1
ε
2M (b − a)
[xi − xi−1 ] =
ε
.
2
Suppose limP S(P, T, f α0 , x) = I. Then
∃Pε0 such that P ⊃ Pε0 =⇒ |S(P, T, f α0 , x) − I| <
ε
.
2
Choose Pε = Pε0 ∪ {finitely many points chosen such that kPε k < δ}. Then
P ⊃ Pε =⇒ |S(P, T, f, α) − I| ≤ |S(P, T, f, α) − S(P, T, f α0 , x)| + |S(P, T, f α0 , x) − I| <
Existence (Rudin p.120–127)
b
U (P, f, α)
S(P, T, f, α)
b
L(P, f, α)
Strategy.
L(P, f, α) ≤ S(P, T, f, α) ≤ U (P, f, α)
We shall show that as P gets finer, L(P, f, α) % and U (P, f, α) & ; and if f is continuous,
U (P, f, α) − L(P, f, α) → 0.
Until further notice, assume f is bounded and α is monotonic increasing.
ε ε
+ = ε.
2 2
2
MATH 321: Lecture #7
Definition. For any partition P = {x0 , x1 , . . . , xn } on [a, b]. Set
L(P, f, α) =
U (P, f, α) =
n
X
i=1
n
X
mi [α(xi ) − α(xi−1 )] where mi =
inf
f (x)
sup
f (x)
xi−1 ≤x≤xi
Mi [α(xi ) − α(xi−1 )] where Mi =
xi−1 ≤x≤xi
i=1
Lemma.
1. For any partition P
inf f (x)[α(b) − α(a)] ≤ L(P, f, α) ≤ S(P, T, f, α) ≤ U (P, f, α) ≤ sup f (x)[α(b) − α(a)]
a≤x≤b
a≤x≤b
2. If P 0 ⊃ P , then
L(P, f, α) ≤ L(P 0 , f, α) and U (P, f, α) ≥ U (P 0 , f, α)
3. If P, P 0 are partitions (not necessarily related), then
L(P, f, α) ≤ U (P 0 , f, α)
Proof.
1. Trivial since
inf f (x) ≤ mi ≤ f (ti ) ≤ Mi ≤ sup f (x),
x∈[a,b]
x∈[a,b]
and we can multiply this by α(xi ) − α(xi−1 ) and sum.
2. Easy: P 0 = P ∪ {a finite # of points}. It suffices, by induction, to consider P 0 = P ∪ {y} with y ∈ (xj−1 , xj ).
Then
n
X
Mi [α(xi ) − α(xi−1 )].
U (P 0 , f, α) =
i=1
i6=j
More on this next time. . .
Lecture #8
MATH 321: Real Variables II
University of British Columbia
Lecture #8:
Instructor:
Scribe:
January 23, 2008
Dr. Joel Feldman
Peter Wong
Existence Theory
Until further notice, we assume f is bounded and α is non-decreasing.
Notation.
P = {x0 , x1 , . . . , xn },
L(P, f, α) =
n
X
mi =
inf
xi−1 ≤x≤xi
mi [α(xi ) − α(xi−1 )]
f (x), Mi = supxi−1 ≤x≤xi f (x)
and U (P, f, α) =
n
X
Mi [α(xi ) − α(xi−1 )]
i=1
i=1
Lemma.
(1) inf a≤x≤b f (x)[α(b) − α(a)] ≤ L(P, f, α) ≤ S(P, T, f, α) ≤ U (P, f, α) ≤ supa≤x≤b f (x)[α(b) − α(a)]
(2) P 0 ⊃ P =⇒ L(P, f, α) ≤ L(P 0 , f, α) and U (P, f, α) ≥ U (P 0 , f, α)
(3) For any P, P 0 , L(P, f, α) ≤ U (P 0 , f, α).
Proof.
(1) Done last class.
(2) It suffices to consider P 0 = {y} for y ∈ (xj−1 ,j ). For U ,
U (P 0 , T, α) =
≤
≤
n
X
i=1
i6=j
n
X
i=1
i6=j
n
X
Mi ∆αi +
sup
f (x)[α(y) − α(xj−1 )] + sup f (x)[α(xj ) − α(y)]
xj−1 ≤x≤y
y≤x≤xj
Mi ∆αi + Mj ∆αj
Mi ∆xi = U (P, T, α).
i=1
The proof for L is similar.
(2)
(1)
(2)
(3) Set P ∗ = P ∪ P 0 . Then L(P, f, α) ≤ L(P ∗ , f, α)) ≤ U (P ∗ , f, α) ≤ U (P 0 , f, α).
Definition. Define
Z
b
f dα = inf{ U (P, T, α | P is a partition of [a, b] }
a
Z b
f dα = sup{ L(P, T, α | P is a partition of [a, b] }
a
Note. For all bounded f ’s, increasing α’s,
Z
a
b
f dα and
Z
Theorem. The following are equivalent:
a
b
f dα always exist, and
Z
a
b
f dα ≥
Z
a
b
f dα.
2
MATH 321: Lecture #8
(i) f ∈ R(α) on [a, b]. (In-class definition)
(ii) ∀ε > 0, ∃Pε such that 0 ≤ U (Pε , f, α) − L(Pε , f, α) < ε.
(iii)
Rb
af
dα =
Rb
af
dα. (Rudin’s definition)
Proof. (i)=⇒(ii) Assume f ∈ R(α). Let ε > 0. Choose Pε so that
Z b
ε
f dα < .
S(Pε , T, f, α) −
3
a
Since
we have
Z b
ε
f dα <
U (Pε , f, α) = sup S(Pε , T, f, α) =⇒ U (Pε , f, α) −
3
T
a
Z
ε
b
L(Pε , f, α) = inf S(Pε , T, f, α) =⇒ L(Pε , f, α) −
f dα < ,
T
3
a
0 ≤ U (Pε , f, α) − L(Pε , f, α)
"
#
Z
"
b
= U (Pε , f, α) −
f dα − L(Pε , f, α) −
a
≤
Z
b
f dα
a
ε ε
+ < ε.
3 3
(ii)=⇒(iii) Let ε > 0. We know that
L(P, f, α) ≤
Z
f dα ≤
Z
f dα ≤ U (P, f, α)
≤ ε
=⇒ 0 ≤
=⇒
Z
Z
f dα ≤
f dα =
Z
Z
f dα < ε,
f dα
#
∀ε > 0
Lecture #9
MATH 321: Real Variables II
University of British Columbia
Lecture #9:
Instructor:
Scribe:
January 25, 2008
Dr. Joel Feldman
Peter Wong
Until further notice, f is bounded and α is monotone increasing.
Theorem. The following are equivalent:
(i) f ∈ R(α) on [a, b]. (The in-class definition)
(ii) ∀ε > 0, ∃Pε such that 0 ≤ U (Pε , f, α) − L(Pε , f, α) < ε.
(iii)
Rb
a
f dα =
Rb
a
f dα. (Rudin’s definition of integrability)
Proof.
(i)=⇒(ii): Done last class.
(ii)=⇒(iii): Done last class.
(iii)=⇒(i): Let ε > 0. Recall that we have
b
Z
f dα = inf U (P, f, α) =⇒ ∃Pε0 such that 0 ≤ U (Pε0 , f, α) −
P
a
Z
b
f dα < ε
a
b
f dα = sup L(P, f, α) =⇒
P
a
Z
∃Pε00
such that 0 ≤
Z
b
f dα − L(Pε00 , f, α) < ε
a
Choose Pε = Pε0 ∪ Pε00 . Let P ⊃ Pε . Then
b
≤
Z
b
S(P, T, f, α) ≥ L(P, f, α) ≥ L(Pε00 , f, α) ≥
Z
S(P, T, f, α) ≤ U (P, f, α) ≤
U (Pε0 , f, α)
f dα + ε
a
f dα − ε
a
Hence, f ∈ R(α) on [a, b].
Theorem. If f is continuous and α is monotone increasing on [a, b], then f ∈ R(α) on [a, b].
Proof. Let ε > 0. Given that f is continuous on a compact set [a, b], it follows that f is uniformly continuous on
[a, b], which means that ∃δ > 0 such that for any a ≤ x, y ≤ b,
|x − y| < δ
=⇒
|f (x) − f (y)| < See Below (1)
Choose Pε to be See Below (2). Then
U (Pε , f, α) − L(Pε , f, α) =
n
X
(Mi − mi )∆αi
i=1
where
M i − mi =
sup
xi−1 ≤x≤xi
f (x)
−
inf
xi−1 ≤x≤xi
f (x) =
sup
xi−1 ≤x,y≤xi
[f (x) − f (y)] ≤ See Below (1)
2
MATH 321: Lecture #9
Choose (2) any partition with kPε k < δ. Then
U (Pε , T, α) − L(Pε , T, α) =
n
X
(Mi − mi )∆αi ≤
n
X
See Below (1) ∆αi
i=1
i=1
= See Below (1)
n
X
∆αi < ε
i=1
| {z }
α(b)−α(a)
Choosing (1)
ε
2[α(b)−α(a)]
completes the proof.
Theorem. Let f : [a, b] → [−M, M ], α : [a, b] → R be increasing, and ϕ : [−M, M ] → R be continuous. If f ∈ R(α)
on [a, b], then ϕ ◦ f ∈ R(α) on [a, b]. Hence ϕ ◦ f (x) = ϕ(f (x)).
Before we prove this theorem, we would like to see two corollaries.
Corollary 1. If f ∈ R(α) on [a, b], then |f |, f n ∈ R(α) for all n ∈ N.
Corollary 2. If f, g ∈ R(α) on [a, b], then f g ∈ R(α) on [a, b].
Proof. Suppose f, g ∈ R(α). By linearity of integrals, f + g ∈ R(α). By Corollary 1, (f + g)2 , f 2 , and g 2 ∈ R(α).
Note that f g = 21 [(f + g)2 − f 2 − g 2 ]. So by linearity, f g ∈ R(α).
Proof of Main Theorem. Let ε > 0. Given that
1. ϕ is continuous on the compact set [−M.M ]. This means that ϕ is bounded, so ∃Mϕ such that |ϕ(y)| < M ϕ
for all |y| ≤ M . Set ε0 = See Below (3), then ∃δ > 0 such that |x − y|δ =⇒ |ϕ(x) − ϕ(y)| < ε0 , for all
|x| ≤ M.
2. f ∈ R(α) on [a, b]. Set η = See Below (4). Then ∃Pη such that U (Pη , f, α) − L(Pη , f, α) < η.
Choose Pε = Pη . We will show that
U (Pε , ϕ ◦ f, α) − L(Pε , ϕ ◦ f, α) < ε.
We know that
U (Pη , f, α) − L(Pη , f, α) =
n
X
(Mi − mi )∆αi < η
i=1
where Mi − mi =
sup
[f (x) − f (y)].
xi−1 ≤x,y≤xi
We wish to show that
U (Pη , ϕ ◦ f, α) − L(Pη , ϕ ◦ f, α) =
n
X
i=1
(Mi∗ − m∗i )∆αi < ε
where Mi∗ − m∗i =
sup
[ϕ(f (x)) − ϕ(f (y))].
xi−1 ≤x,y≤xi
We know that if Mi − mi < δ, then Mi∗ − m∗i < ε0 . Next lecture, we shall reveal that (3) ε0 =
(4) η =
εδ
4Mϕ
ε
2[α(b)−α(a)]
and