Lecture #1
MATH 321: Real Variables II
University of British Columbia
Lecture #1:
Instructor:
Scribe:
January 7, 2008
Dr. Joel Feldman
Peter Wong
From Last Time: Surely we had a merry Christmas and a happy New Year.
Annoucements
• Office = MATH 221 (Directly above the Mathematics Department Office.)
• Email = feldman@math.ubc.ca
• Text = “Baby Rudin” – Principles of Mathematical Analysis 3/e by Walter Rudin
• Course Website = “http://www.math.ubc.ca/∼feldman/m321/”
• Topics:
– Riemann-Stieltjes integral (§6)
– Sequences & Series of functions (§7)
– Power Series, Special Functions, Fourier Series (§8)
– Another topic to be determined.
• Grading:
(1) A midterm on Wed. Feb 27 (25%)
y = f(x)
(2) Weekly problem sets due each Wednesday (25%)
(t 3, f(t 3))
(3) Exam (50%)
(4) Grades will probably be scaled (up, usually)
Integration (Rudin §6)
Recall the definition of
Rb
a
f (x) dx from first-year calculus.
a = x0
Step 1: Slice up [a, b].
x1
x 2 t3 . . .
x n-1
x n= b
x
Definition. A partition of [a, b] is a finite set of points P = {x0 , x1 , . . . , xn } such that a = x0 < x1 < x2 <
· · · < xn−1 < xn = b.
Step 2: Pick a representation value of f for each subinterval.
Definition. A choice T for the partition P is a finite set of points t1 , . . . , tn obeying xi−1 ≤ ti ≤ xi for each
1 ≤ i ≤ n. We shall use f (ti ) as an approximate value of f on [xi−1 , xi ].
Step 3: Compute an approximate value of
Rb
a
f (x) dx.
Definition. A Riemann partial sum is a sum of the form
n
X
f (ti )[xi − xi−1 ] = S(P, T, f, x).
i=1
Step 4: Make the partition finer.
Definition. A partition P 0 is finer than the partition P if P 0 ⊃ P .
2
MATH 321: Lecture #1
Step 5: Hope the sums converge as the partition gets finer and finer.
Definition. A function f : [a, b] → R is said to be Riemann-integrable on [a, b] (denoted f ∈ R on [a, b])
if there is a number I ∈ R such that
∀ε > 0, ∃Pε such that for each partition P ⊃ Pε : |S(P, T, f, x) − I| < ε
for all choices T compatible with P . If so, we write
lim S(P, T, f, x) = I =
P
Z
b
f (x)dx.
a
Generalization of this definition
Let α : [a, b] → R and replace xi − xi−1 by α(xi ) − α(xi−1 ). The Riemann-Stieltjes partial sum is
S(P, T, f, α) =
n
X
f (ti )[α(xi ) − α(xi−1 )]
i=1
Definition. A function f : [a, b] → R is said to be Riemann-Stieltjes integrable with respect to α on [a, b]
(denoted by f ∈ R(α) on [a, b]) if there is a number I ∈ R such that
∀ε > 0, ∃Pε such that for each partition P ⊃ Pε : |S(P, T, f, α) − I| < ε
for all choices T compatible with P . If so, we write
lim S(P, T, f, α) = I =
P
Z
a
b
f (x) dα(x) =
Z
b
f dα.
a
[Examples and application shortly].
Remark.
1. The Riemann-Stieltjes integral reduces to the Riemann integral when α(x) = x.
2. We shall eventually prove that
(a) α monotonic, f continuous =⇒ f ∈ R(α).
(b) α continuous, f monotonic =⇒ f ∈ R(α). (From (a) and Integration by Parts)
(c) α strictly monotone, f unbounded on [a, b] =⇒ f ∈
/ R(α). (Use improper integrals.)
Lecture #2
MATH 321: Real Variables II
University of British Columbia
Lecture #2:
Instructor:
Scribe:
January 9, 2008
Dr. Joel Feldman
Peter Wong
From last time: We defined
Z
b
f (x) dα(x) = lim S(P, T, f, α).
P
a
Here, P = {x0 , x1 , · · · , xn } with a = x0 < x1 < x2 < · · · < xn = b,
a = x0
x1
x2
···
xn = b
and T = {t1 , · · · , tn } with ti ∈ [xi−1 , xi ].
S(P, T, f, α) =
n
X
f (ti )[α(xi ) − α(xi−1 )]
i=1
and
lim S(P, T, f, α) = I ⇐⇒ ∀ε > 0, ∃Pε such that P ⊃ Pε =⇒ |S(P, T, f, α) − I| < ε
P
And this time:
Remark. (Connection between
Rb
f dα) and Riemann Integral) Observe that
n
X
α(xi ) − α(xi−1 )
f (ti )
S(P, T, f, α) =
(xi − xi−1 )
xi − xi−1
i=1
|
{z
}
a
≈ α0 (ti )
We shall prove that if α has a continuous derivative, then
Z b
Z b
f dα =
f (x)α0 (x) dx
a
a
Definition. A step function on [a, b] is a function α : [a, b] → R such that
(i) α has finitely many points of discontinuity on [a, b]. Call si for 1 ≤ i ≤ n where a ≤ s1 < s2 < · · · < sn ≤ b.
(ii) α is constant on each subinterval.
[a, s1 ),
a
s1
s2
b = s3
(sj−1 , sj ), for 1 ≤ j ≤ n,
(sn , b].
Theorem. Let a < b. Let α : [a, b] → R be a step function with discontinuities at s1 < · · · < sn . Then f ∈ R(α)
on [a, b] and
Z b
n
X
−
f (sj )[α(s+
f dα =
j ) − α(sj )]
a
j=1
where
α(s+ )
α(s+
j ) = lim α(t),
t→sj
t>sj
α(s−
j ) = lim α(t)
t→sj
t<sj
α(s− )
s
2
MATH 321: Lecture #2
and by convention
s1 = a =⇒ α(a− ) = α(a)
sn = b =⇒ α(b+ ) = α(b)
Proof. Write I =
Pn
j=1
−
f (sj )[α(s+
j ) − α(sj )]. We prove that
∀ε > 0, ∃Pε such that P ⊃ Pε =⇒ |S(P, T, f, α) − I| < ε
Let ε > 0. Choose Pε obeying
(1) {s1 , . . . , sn } ⊂ Pε = {a = x̃0 < x̃1 < · · · < x̃m = b}
(2) The mesh (or norm) of Pε is
kPε k = max |x̃i − x̃i−1 | < δ
1≤i≤m
where δ = min{s2 − s1 , s3 − s2 , . . . , sn − sn−1 , δ0 }, and δ0 is given by
Insert (∗) here.
Let P = {x0 , x1 , . . . , xp } ⊃ Pε and T = {t1 , . . . , tp } be a choice for P and consider each term in
S(P, T, f, α) =
p
X
f (ti )[α(xi ) − α(xi−1 )]
i=1
Either
(1) neither xi−1 nor xi is on sj . In this case, α(xi ) = α(xi−1 ), so α(xi ) − α(xi−1 ) = 0, or
(2) xi = sj for some j. In this case, α(xi ) − α(xi−1 ) = α(sj ) − α(s−
j ), or
(3) xi−1 = sj for some j. In this case, α(xi ) − α(xi−1 ) = α(s+
j ) − α(sj ).
1
a
2
3
s1
1
2
3
s2
2
3
s3
1
b
Lecture #3
MATH 321: Real Variables II
University of British Columbia
Lecture #3:
Instructor:
Scribe:
January 11, 2008
Dr. Joel Feldman
Peter Wong
From last time: We were half-way through the proof of the Step Function Theorem. At the start of the class,
we have a brief recap of the first half of the proof. Please refer to page 1 of step.pdf.
Theorem. (The Step Function Theorem)
Proof. (Continued from last time)
S(P, T, f, α) =
p
X
f (ti )[α(xi ) − α(xi−1 )]
i=1




−
+
f (tij )[α(sj ) − α(sj )] + f (tij )[α(sj ) − α(sj )]
=

{z
} |
{z
}

j=1 |
n 

X
tij sj ti0j
case 2 terms
case 3 terms
Each of tij and ti0j lie in an interval of P with one endpoint: sj
Since
(
|sj − tij | < δ
kP k < δ =⇒
|sj − ti0j | < δ.
We are aiming for the integral to be
n
X
−
f (sj )[α(s+
j ) − α(sj )] =
n
X
j=1
j=1
+
f (sj )[α(sj ) − α(s−
j )] + f (sj )[α(sj ) − α(sj )]
But
n
X
+
− S(P, T, f, α) −
f (sj )[α(sj ) − α(sj )]
j=1
≤
n n
o
X
+
|f (tij ) − f (sj )||α(sj ) − α(s−
j )| + |f (ti0j ) − f (sj )||α(sj ) − α(sj )|
j=1
(*) for each 1 <≤ j ≤ n, f is continuous at sj implies ∃δj > 0 such that
|f (sj ) − f (t)| < Pn
k=1
ε
|α(sk ) −
α(s−
k )|
+ |α(s+
j ) − α(sk )|
for all |t − sj | < δj . Choose δ0 = min{δ1 , . . . , δn }.
Since kP k ≤ kPε k < δ0 , we have
n
X
+
− S(P, T, f, α) −
f
(s
)[α(s
)
−
α(s
)]
j
j
j < ε.
j=1
Remarks. (Applications of the Riemann-Stieltjes integral)
1. We now have a Dirac δ-function on a hand-waving level, δ(x) is defined by
(a) δ(x) < 0 for all x 6= 0
(b) δ(0) = +∞
δ(x)
2
(c)
Rb
a
MATH 321: Lecture #3
δ(x)dx = 1 for all a < 0, b > 0.
such that
Z
b
f (x)δ(x) dx =
a
Z
0
b
f (0)δ(x) dx,
∀a < 0, b > 0
(Not rigorous)
a
Notice that for x 6= 0, both integrals equal 0.
We can also define the Heaviside function
H(x) =
such that δ(x) = H 0 (x).
(
0, x < 0
1, x ≥ 0
(Not rigorous)
A more rigorous definition follows from the concept of Riemann-Stieltjes Integral:
Z
Z
a
b
f (x)δ(x) dx = f (0)
a
b
f (x)H 0 (x) dx = f (0)
Z
a
2. Application to Probability. (Next Lecture)
b
f dH = f (0)