A partition P of [a, b] is a finite collection of points of [a, b] such that P contains a, b

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

(1.1) m ≤ f (x) ≤ M, 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< · · · < x_n= b}.

Let P be a partition on [a, b] and f : [a, b] → R be a bounded function. For each 1 ≤ i ≤ n, let M_i and m_i be the following real numbers

Mi = sup

x∈[xi−1,xi]

f (x), mi= inf

x∈[xi−1,xi]f (x).

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

i=1

Mi∆xi, L(f, P ) =

n

X

i=1

mi∆xi. Here ∆xi= xi− x_i−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 a

f (x)dx = inf

P U (f, P ), Z b

a

f (x)dx = sup

P

L(f, P )

where P runs through all partition P of [a, b]. It follows from the definition that Z b

a

f (x)dx ≤ Z b

a

f (x)dx.

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 a

f (x)dx = Z b

a

f (x)dx.

In this case, we define Z b

a

f (x)dx = Z b

a

f (x)dx = Z b

a

f (x)dx.

Example 1.1. Let f : [0, 1] → R be the function (it is called the Dirichlet function) f (x) =

(1 if x is a rational number in [0, 1];

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

[xi−1,xi]

f (x) = 1, mi= inf

[xi−1,xi]f (x) = 0.

Then L(f, P ) = 0 and U (f, P ) =

n

X

i=1

M_i∆x_i=

n

X

i=1

(x_i− x_i−1) = x_n− x₀ = 1.

This gives us

Z 1 0

f (x)dx = 1, Z 1

0

f (x)dx = 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 A(Ω) =

Z b a

f (x)dx.

Proof. Let P = {a = x0 < · · · < xn= b} be a partition of [a, b] and Mi = sup

[xi−1,xi]

f (x), mi = inf

[xi−1,xi]f (x)

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

i=1

Mi∆xi, L(f, P ) =

n

X

i=1

mi∆xi

where ∆x_i = x_i− x_i−1 for 1 ≤ i ≤ n. Let R⁰_i and R_i be the rectangles

R⁰_i= {(x, y) : x_i−1≤ x ≤ x_i, 0 ≤ y ≤ M_i}, R_i= {(x, y) : x_i−1≤ x ≤ x_i, 0 ≤ y ≤ m_i}.

Then S = R1∪ · · · ∪ R_n is a simple region contained in Ω and S⁰ = R⁰₁∪ · · · ∪ R⁰_nis a simple region containing Ω. By definition,

A(S⁰) =

n

X

i=1

Mi∆xi = U (f, P ) and

A(S) =

n

X

i=1

m_i∆x_i = L(f, P ).

Since S⁰ contains Ω, A(S⁰) ≥ 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 a

f (x)dx ≤ A−(Ω) ≤ A₊(Ω) ≤ Z b

a

f (x)dx.

3

If f is Riemann integrable, Z b

a

f (x)dx = Z b

a

f (x)dx = Z b

a

f (x)dx.

This shows that

Z b a

f (x)dx ≤ A−(Ω) ≤ A+(Ω) ≤ Z b

a

f (x)dx.

We conclude that

A−(Ω) = A+(Ω) = Z b

a

f (x)dx.

Hence Ω has an area and its area equals Z b

a

f (x)dx.