MATH2060B TUTORIAL 5

-
For Riemann Integration Theory, we will follow closely to the
notes uploaded in the course webpage instead of the textbook.

Construction of Riemann Integral:

We always consider bounded functions f, g, h, ... etc defined on a closed bounded
interval [a, b], and let m and M be an lower and upper bound of f respectively.
i.e., m < f(x) < M for any x [a, b].
ˋ ˋ E

* A partition P of the interval [a, b] is a finite set of points x , x , ..., x such that
0 1 n

a = x < x < ...... < x = b.

0 1 n

*
For any partition P of [a, b], denote
·
Δx i = xi - x for i = 1, 2, ..., n
·
|| P || = max Δx ii
* For any partition P of [a, b] and function f defined on [a, b], denote
m (f, P) = inf { f(x) : x [x , x ] }.
ii iiiitlnis
M (f, P) = sup { f(x) : x [x , x ] }.
Always exist because
f is bounded!
·
ω (f, P) = M i (f, P) - m (f, P) = sup { |f(x) - f(y)| : x, y [x , x ] }.
i i
E H i

4
Why?

(Lower sum) L(f, P) = Σ m (f, P) Δx

0
We always have
:
(Upper sum) U(f, P) = Σ Mi (f, P) Δxi
:= m(b-a) < L(f, P) < U(f, P) < M(b-a)
ˋ ˋ

for any partition P.

ˋ

* For any function f defined on [a, b], denote

\
(Lower integral)
iaof = sup { L(f, P) : P is a partition of [a, b] }
.

Always exists by

a
(Upper integral) i f = inf { U(f, P) : P is a partition of [a, b] } ←
this observation!

* If f has equal upper and lower integral, we say that f is Riemann integrable.
We write f R[a, b] in this case and
Eiii f= f= f (integral of f)

Prepared by Ernest Fan

Example 1: Visualize the notations in a picture.

te
P = { a, x , x , x , x , b }
y L(f, P) = Σ m (f, P) Δx
M (f, P) U(f, P) = Σ M (f, P) Δx
M
ω (f, P)

y = f(x)
m (f, P)

m
Δx || P || = Δx Δx
x
a=x x x x x x =b

Example 2: Show that the function f(x) = x is Riemann integrable on [0, 1].
Solution: We need to show that f has the same lower and upper integrals.
For each n EN , consider the partition Pn of [0, 1] defined by
Pn = {0, 1/n, 2/n, ..., 1}.
On each subinterval [(i-1)/n, i/n], where i = 1, 2, ..., n, we have
Δxi = 1/n, m (f, Pn) = (i-1)/n and M (f, Pn) = i/n
i

iii
Compute the corresponding upper and lower sum:
1 1 n(n - 1) 1
L(f, Pn) = Σm Δx = n Σ(i - 1) = n = (1 - 1/n)
2 2
: ==
U(f, Pn) = ΣM Δx =
1
n Σ i =
1 n(n + 1)
n 2
=
1
2
(1 + 1/n)
:-(

1 1
It follows that
2
_

(1 - 1/n) < ˋ
!! f < ǕEN
ˋ
f < 2 (1 + 1/n), for all n

Letting n _ , we conclude that f is Riemann integrable with integral 1/2. *

Prepared by Ernest Fan

 Example 2: Show that the Dirichlet’s function is not Riemann integrable on [0, 1].
Solution: Recall that the Dirichlet’s function is defined by
1, if x is rational;
g(x) = { 0, if x is irrational.
We need to show that it has unequal upper and lower integrals.
Fix any partition P of [0, 1]. On each subinterval [x , x ], we have i -
1 i

m (g, P) = 0
i and M (g, P) = 1
i (We don’t know about Δx i !)

Then L(g, P) = Σ m Δx = 0, and U(g, P) = Σ M Δx = Σ Δx i = 1.

i i i i

Since the partition P is arbitrary, it follows that

if = 0 and i f = 1. *

Useful Propositions:
·
Let f be a function defined on [a, b] and P, Q be partitions of [a, b].
*
If P E Q, then L(f, P) < L(f, Q) < U(f, Q) < U(f, P).
- ˋ ˋ

*
L(f, P) <
、
1点 f 亦
< f < U(f, Q)

•
Let f be a bounded function defined on [a, b]. Then f R[a, b] if and only if E

for any ε > 0, there exists a partition P of [a, b] such that

U(f, P) - L(f, P) = Σ ω (f, P)Δx < ε. i i

.
C[a, b] E R[a, b].
Remark: Let me say more about the proof of Lemma 1.2 (i) in the notes. It claims that
it suffices to show the case that Q = P {c}. i.e., Q contains exactly one more U

point than P. Here is why: Suppose in general that Q contains k more points
than P. i.e., Q = P {c , c , ..., c }. If we write
U
1 2 K

Q = P {c }, Q = Qf {c }, ..., Q = Q = Q
1 U
1 2 2 K k
-
1
U {c }.
K

Then by applying the special case k times, we have

L(f, P) < L(f, Q ) < L(f, Q ) < ...... < L(f, Q ) = L(f, Q).
ˋ 1 ˋ
2 ˋ ˋ
K

Prepared by Ernest Fan

 Exercises:
1. Suppose that f is a continuous and non-negative function defined on [a, b]. If the
integral of f is 0, show that f is constantly zero.
Solution: We show the assertion by contradiction.
Suppose on a contrary that f(c) > 0 for some c [a, b]. Since f is continuous
E

at c, there exists δ > 0 such that whenever | x - c | < δ,

| f(x) - f(c) | < f(c)/2. (This act as ε)

t.li
i.e., 0 < f(c)/2 < f(x) < 3f(c)/2 on (c-δ, c+δ)
Now consider the partition P of [a, b] defined by P = { a, u, v, b }, where
a < c-δ < u < v < c+δ < b 3f(c)/2

兩
f(c)
Then we can compute the lower sum: f(c)/2

L(f, P) = m (f, P)Δx + m (f, P)Δx + m (f, P)Δx , c-δ c c+δ

+ m:(f, P)Δxi +
.

u v
> 0 0
> 0 (> 0 is not enough in this proof!)

It is a contradiction because

0= ii.f=
!
f > L(f, P) > 0
*
ǜf 名
Remark: Can the continuity of f be dropped?

Prepared by Ernest Fan