MATH 321: Real Variables II
Lecture #4
University of British Columbia
Lecture #4:
Instructor:
Scribe:
January 14, 2008
Dr. Joel Feldman
Peter Wong
From last time: We discussed the possible application of Riemann-Stieltjes integral in the rigorous definition of
the Dirac δ-function.
Remarks. (Applications of the Riemann-Stieltjes integral)
2. Suppose you perforem a probability
experiment with possible outcomes s1 < . . . < sn having probabilities
P
pj . If a < s1 , b > sn , then
p1 , . . . , pn . Define α(x) =
··
·
1≤j≤n
sj ≤x
Pn
i=1 pi = 1
p2 {
a
p1 {
j=1
s1 s2 · · · sn
∞
Rb
By allowing α to both vary continuously and have jumps, we can use a f (x) dα(x) to compute expected value
for mixed continuous/discrete probability distributions.
(
1, if x ∈ Q,
Example. Let a < b. Let α(a) 6= α(b). Let f (x) =
Then f ∈ R(α) on [a, b]. For any partition
0, if x ∈ R \ Q.
P = {x0 , . . . , xn }
(
n
X
α(b) − α(a), if x ∈ T ⊂ Q,
f (tj )[α(xj ) − α(xj−1 )] =
S(P, T, f, α) =
0,
if x ∈ T ⊂ R \ Q.
j=1
Z
b
f (x) dα(x) =
n
X
f (sj )pj = mean/expected value of f .
Hence, f ∈
/ R(α) on [a, b].
Example. Let a = 0, b = 2. Let f (x) = α(x) =
(
1, if 0 ≤ x < 1 (notice the strict inequality)
Then f ∈
/ R(α)
0, if 1 ≤ x ≤ 2.
on [0, 2]. (See Problem Set 2, #1)
0
1
2
R1
if x = m
in
lowest
term
n
Example. Let a = 0, b = 1. Let α(x) = x and f (x) =
We claim that 0 f dx = 0.
0, if x ∈ R \ Q or x = 0.
1
Let ε > 0. Then 0 ≤ x ≤ 1 and f (x) > ε. This means that x = m
n with 1 ≤ m ≤ n and n > ε. This is equivalent
1
to saying that n < ε . Hence, there can only be finitely many values of x, say N of them, with f (x) > ε with
0 ≤ x ≤ 1. So, for any partition P = {x0 , . . . , xn },
(
S(P, T, f, α(x) = x) =
=
n
X
1
n,
f (tj )[xj − xj−1 ]
j=1
n
X
f (tj )[xj − xj−1 ] +
Choose Pε with kPε k ≤
ε
N.
N
{# terms}
f (tj )[xj − xj−1 ]
j=1
f (tj )<ε
j=1
f (tj )>ε
≤
n
X
×
1
{max f (tj )}
Then P ⊃ Pε =⇒ kP k ≤
ε
N
×
kP k
{max xj −xj−1 }
+ε
n
X
(xj − xj−1 )
j=1
|
=⇒ |S(P, T, f, α) − 0| < 2ε.
{z
1
}
2
MATH 321: Lecture #4
Properties of Riemann-Stieltjes Integrals
Theorem. (Linearity properties) Let a < b, and f, g, α, β : [a, b] → R.
(a) (Linearity with respect to f ) If f, g ∈ R(α) on [a, b] and c, d ∈ R, then cf + dg ∈ R(α) on [a, b] and
Rb
Rb
Rb
a cf + dg dα = c a f dα + d a g dα
Proof. Let ε > 0. Then f ∈ R(α) =⇒ ∃P1 such that P ⊃ P1 =⇒
Z b
f dα < See below (1)
S(P, T, f, α) −
a
Also, g ∈ R(α) =⇒ ∃P2 such that P ⊃ P2 =⇒
Z b
g dα < See below (2)
S(P, T, g, α) −
a
Choose Pε = See below (3) Then P ⊃ Pε implies
S(P, T, cf + dg, α) −
Hence, we choose (1):
c
Z
b
f dα + d
a
Z
b
a
!
g dα Z b
Z b
= c [S(P, T, f, α)] + d [S(P, T, g, α)] − c
f dα − d
g dα
a
a
Z b
Z b
ε ε
≤ |c| S(P, T, f, α) −
f dα + |d| S(P, T, g, α) −
g dα < + = ε.
2 2
a
a
ε
2|c|
and (2):
ε
2|d|
. Since P ⊃ P1 and P ⊃ P2 , why don’t we choose (3): Pε = P1 ∪ P2 ?
(b) (Linearity with respect to α) See Problem Set 1 #3.
R
(c) (Linearity with respect to on [a, b]) See Problem Set 1 #2.
MATH 321: Real Variables II
Lecture #5
University of British Columbia
Lecture #5:
Instructor:
Scribe:
January 16, 2008
Dr. Joel Feldman
Peter Wong
Properties of Riemann-Stieltjes Integrals
Theorem. (Linearity properties) Let a < b, and f, g, α, β : [a, b] → R.
(1) If f, g ∈ R(α) on [a, b] and c1 , c2 ∈ R, then
c1 f + c2 g ∈ R(α) on [a, b]
Z
and
b
c1 f + dg c2 α = c1
a
Z
b
f dα + c2
a
Z
b
g dα.
a
Proof. Done last time.
(2) If f ∈ R(α) ∩ R(β) on [a, b] and if c1 , c2 ∈ R, then
f ∈ R(c1 α + c2 β) on [a, b]
Z
and
b
f d(c1 α + c2 β) = c1
a
Z
b
f dα + c2
a
Z
b
f dβ
a
Proof. See Problem Set 1 #3.
(3) Suppose c ∈ (a, b). If f ∈ R(α) on [a, c] and on [c, b], then
f ∈ R(α) on [a, b]
and
Z
b
f dα =
a
Z
c
f dα +
a
Z
b
f dα
c
Proof. See Problem Set 1 #2.
(4) If [c, d] ⊂ [a, b] and f ∈ R(α) on [a, b], then f ∈ R(α) on [c, d].
Proof. See Problem Set 2 #1.
Theorem. (Integration by parts) Let a < b. Let f, α : [a, b] → R. If f ∈ R(α) on [a, b] , then
α ∈ R(f ) on [a, b]
and
Z
b
α df = f (b)α(b) − f (a)α(a) −
a
Z
b
f dα.
a
Another way to remember this formula is through one of its derivation as follows:
d(f α) = f dα + α df
Z b
Z b
α df
f dα +
d(f α) =
a
a
a
Z b
Z b
α df
f dα +
f (b)α(b) − f (a)α(a) =
Z
b
a
a
(integrate both sides on [a, b])
(apply telescoping sum to simplify LHS)
2
MATH 321: Lecture #5
Proof. Let P = {x0 , . . . , xn } be any partition and T = {t1 , . . . , tn } be any choice for P .
Z b
− S(P, T, α, f ) − f (b)α(b) − f (a)α(a) −
f
dα
a
Z b
n
n i=n:f (b)α(b)
n
X
i=1:f (a)α(a)
X
X
f dα
f (xi−1 )α(xi−1 ) −
f (xi )α(xi ) −
α(ti )(−f (xi ) + f (xi−1 )) +
=
a
i=1
i=1
i=1
Z
n
n
X
b
X
f dα
f ( xi )[α(xj ) − α(ti )] −
f ( xi−1 )[α(ti ) − α(ti−1 )] +
=
∈[ti ,xi ]
a
∈[xi−1 ,ti ]
i=1
i=1
Z b
···
t1
t2
tn
0
= S(P
∪ T}, T , f, α) −
f dα
| {z
a = x0
x1
x2 · · · xn−1 xn = b
a
=P 0
where T 0 = {x0 , x1 , x1 , x2 , x2 , . . . , xn }. Let ε > 0, f ∈ R(α) on [a, b] =⇒ ∃Pε such that P 0 ⊃ Pε =⇒
Rb
Rb
|S(P 0 , T 0 , f, α) − a f dα| < ε. Then P ⊃ Pε =⇒ P 0 = P ∩ T ⊃ Pε =⇒ |S(P 0 , T 0 , f, α) − a f dα| < ε
=|S(P,T,f,α)−desired answer|
Theorem. (Change of Variables)


Z b
Z c
x=g(y)


f (x) dα(x) =
f (g(y)) d α(g(y))

|
{z
}
{z
}
|
a
d
h(y)
β(y)
Let a < b, and α, f : [a, b] → R. Let f ∈ R(α) on [a, b]. Let g : [c, d] → [a, b] be continuous, strictly
monotone, and g(c) = a, g(d) = b. Define h(y) = f (g(y), β(y) = α(g(y)), then
h ∈ R(β) on [c, d]
and
Z
a
b
f dα =
Z
c
d
h dβ.
MATH 321: Real Variables II
Lecture #6
University of British Columbia
Lecture #6:
Instructor:
Scribe:
January 18, 2008
Dr. Joel Feldman
Peter Wong
Theorem. (Change of Variables x = g(y)) Let
• f ∈ R(α) on [a, b]
• g : [c, d] → [a, b] strictly monotonic and continuous where g(c) = a and g(d) = b
• h(y) = f (g(y))
• β(y) = α(g(y))
Then h ∈ R(β) on [c, d] and
Rd
c
h(y) dβ(y) =
Rb
a
f (x) dα(x).
Proof. Let P = {y0 = c < y1 < · · · < yn = d} be any partition of [c, d] and T = {s1 , · · · , sn } be any choice for P .
Z b
Z b
n
X
h(si )[β(yi ) − β(yi−1 )] −
f dα
f dα = S(P, T, h, α) −
a
a
i=1
n
Z b
X
f (g(si ))[α(g(yi )) − α(g(yi−1 ))] −
f dα
=
a
i=1
Z b
= S(g(P ), g(T ), f, α) −
f dα
a
where g(P ) = {x0 = g(y0 ), x1 = g(y1 ), . . . , xn = g(yn )} and g(T ) = {t0 = g(s0 ), t1 = g(s1 ), . . . , tn = g(sn )}. g(P )
is a partition for [a, b] and g(T ) is a choice for g(P ). Since x0 = g(y0 ) = g(c) = a and xn = g(yn ) = g(d) = b


(
)
)
(
yi−1 ≤ si ≤ yi ,




xi−1 ≤ ti ≤ xi ,
g(yi−1 ) ≤ g(si ) ≤ g(yi ),
=⇒
yi−1 < yi ,
.
=⇒


xi−1 < xi
g(yi−1 ) < g(yi )


g is strictly monotone
Rb
Let ε > 0, f ∈ R(α) on [a, b]. Then ∃Pε0 such that P 0 ⊃ Pε0 =⇒ |S(P, T, f, α) − a f dα|. Choose Pε = g −1 (Pε0 ). g −1
exists since g is bijective (strictly monotone and continuous.) Pε is a partition since g −1 is strictly monotonic. Let
P 0 = g(P ) with choice T 0 = g(T ). Then
Theorem. (Basic Bounds) Let
P ⊃ Pε =⇒ g(P ) ⊃ g(Pε ) = g(g −1 (Pε0 )) = Pε0
Z b
0
0
=⇒ S(P , T , f, α) −
f dα < ε
a
Z
b
=⇒ S(P, T, f, α) −
f dα < ε.
a
• a<b
• f, g : [a, b] → R
• f, g ∈ R(α) on [a, b] and α monotonic increasing
Then
2
MATH 321: Lecture #6
Rb
f α ≤ a g dα;
Rb
Rb
2. if |f (x)| ≤ g(x) for all a ≤ x ≤ b, then | a f α| ≤ a g dα.
1. if f (x) ≤ g(x) for all a ≤ x ≤ b, then
Rb
a
Proof. Let ε > 0. Then
Z b
f dα < ε,
∃Pf,ε such that P ⊃ Pf,ε =⇒ S(P, T, f, α) −
a
Z b
∃Pg,ε such that P ⊃ Pg,ε =⇒ S(P, T, f, α) −
g dα < ε.
a
Choose P = Pf,ε ∪ Pg,ε and let T be a choice for P .
1.
Z
b
f dα ≤ S(P, T, f, α) + ε =
a
n
X
f (ti )[α(xi ) − α(xi−1 )] + ε
i=1
=
n
X
g(ti )[α(xi ) − α(xi−1 )] + ε ≤ S(P, T, g, α) + ε ≤
Z
b
g dα + 2ε
a
i=1
2. Done similarly.
Theorem. (Reduction to Riemann Integral) Let a < b. Let f, α : [a, b] → R. Assume f is bounded, and α
has a continuous derivative on [a, b]. Then
f ∈ R(α) on [a, b]
Remark. If f = 1, the
Proof.
Rb
a
⇐⇒
α0 (x)dx =
f α ∈ R = R(β)
0
Rb
a
S(P, T, f, α) =
=
and
β=x
n
X
i=1
n
X
n
X
i=1
We finish the proof next time.
b
0
f (x)α (x) dx =
a
dα(x) = α(b) − α(a).
f (ti )[α(xi ) − α(xi−1 )]
f (ti )α0 (ui )[xi − xi−1 ] for some xi−1 ≤ ui ≤ xi
i=1
S(P, T, f α0 , x) =
Z
f (ti )α(ti )[xi − xi−1 ]
Z
a
b
f (x) dα(x)