1. Riemann sum and Riemann integral
A function f : [a, b] → R on [a, b] is bounded if there exist real numbers M and m such
that
m ≤ f (x) ≤ M,
(1.1)
for any a ≤ x ≤ b.
Definition 1.1. A partition P of [a, b] is a finite collection of points of [a, b] such that P
contains a, b.
P is often written as
P = {a = x0 < x1 < · · · < xn = b}.
Let P be a partition on [a, b] and f : [a, b] → R be a bounded function. For each
1 ≤ i ≤ n, let Mi and mi be the following real numbers
Mi =
sup
f (x),
mi =
x∈[xi−1 ,xi ]
inf
f (x).
x∈[xi−1 ,xi ]
The upper Riemann sum U (f, P ) of f with respect to P and the lower Riemann sum
L(f, P ) of f with respect to P are defined respectively by
U (f, P ) =
n
X
Mi ∆xi ,
L(f, P ) =
i=1
n
X
mi ∆xi .
i=1
Here ∆xi = xi − xi−1 . For any partition P,
L(f, P ) ≤ U (f, P ).
The Riemann upper integral of f and the Riemann lower integral of f (over [a, b]) are
defined by
Z b
Z b
f (x)dx = inf U (f, P ),
f (x)dx = sup L(f, P )
P
a
P
a
where P runs through all partition P of [a, b]. It follows from the definition that
Z b
Z b
f (x)dx ≤
f (x)dx.
a
a
Definition 1.2. Let f : [a, b] → R be a bounded function. We say that f is Riemann
integrable (over [a, b]) if
Z b
Z b
f (x)dx =
f (x)dx.
a
a
In this case, we define
Z
b
Z
f (x)dx =
a
b
Z
f (x)dx =
a
b
f (x)dx.
a
Example 1.1. Let f : [0, 1] → R be the function (it is called the Dirichlet function)
(
1 if x is a rational number in [0, 1];
f (x) =
0 if x is an irrational number in [0, 1].
Then f is not Riemann integrable.
1
2
Proof. Let P = {0 = x0 < · · · < xn = 1} be a partition of [0, 1]. For each 1 ≤ i ≤ n,
Mi =
sup f (x) = 1,
mi =
[xi−1 ,xi ]
inf
f (x) = 0.
[xi−1 ,xi ]
Then L(f, P ) = 0 and
U (f, P ) =
n
X
n
X
Mi ∆xi =
(xi − xi−1 ) = xn − x0 = 1.
i=1
i=1
This gives us
1
Z
Z
f (x)dx = 1,
1
f (x)dx = 0.
0
0
This implies that f is not Riemann integrable.
Theorem 1.1. Let f : [a, b] → R be a nonnegative bounded function and
Ω = {(x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f (x)}.
If f is Riemann integrable, then Ω has an area and
Z b
A(Ω) =
f (x)dx.
a
Proof. Let P = {a = x0 < · · · < xn = b} be a partition of [a, b] and
Mi =
sup f (x),
[xi−1 ,xi ]
mi =
inf
f (x)
[xi−1 ,xi ]
for 1 ≤ i ≤ n. The upper Riemann sum U (f, P ) and the lower Riemann sum L(f, P ) of f
with respect to P are given respectively by
U (f, P ) =
n
X
Mi ∆xi ,
L(f, P ) =
i=1
mi ∆xi
i=1
where ∆xi = xi − xi−1 for 1 ≤ i ≤ n. Let
Ri0
n
X
Ri0
= {(x, y) : xi−1 ≤ x ≤ xi , 0 ≤ y ≤ Mi },
and Ri be the rectangles
Ri = {(x, y) : xi−1 ≤ x ≤ xi , 0 ≤ y ≤ mi }.
Then S = R1 ∪ · · · ∪ Rn is a simple region contained in Ω and S 0 = R10 ∪ · · · ∪ Rn0 is a simple
region containing Ω. By definition,
A(S 0 ) =
n
X
Mi ∆xi = U (f, P )
i=1
and
A(S) =
n
X
mi ∆xi = L(f, P ).
i=1
Since S 0 contains Ω, A(S 0 ) ≥ A+ (Ω). Since S is contained in Ω, and A(S) ≤ A− (Ω). We
obtain that for any partition P of [a, b],
L(f, P ) ≤ A− (Ω) ≤ A+ (Ω) ≤ U (f, P ).
This implies that
Z
b
Z
f (x)dx ≤ A− (Ω) ≤ A+ (Ω) ≤
a
b
f (x)dx.
a
3
If f is Riemann integrable,
Z
b
b
Z
f (x)dx =
a
This shows that
Z
Z
b
f (x)dx.
f (x)dx =
a
a
b
Z
f (x)dx ≤ A− (Ω) ≤ A+ (Ω) ≤
a
b
f (x)dx.
a
We conclude that
Z
b
f (x)dx.
A− (Ω) = A+ (Ω) =
a
Z b
Hence Ω has an area and its area equals
f (x)dx.
a