Axioms for Integration - math.bard.edu

Axioms for Integration

These notes develop the theory of definite integrals axiomatically. This approach is fairly
simple, but has the disadvantage that we don’t really prove that the integral exists.

To begin, we assume that there exists a certain operation called the definite integral.
This operation takes as input a continuous function f : [a, b] → R, where [a, b] ⊆ R is a
closed bounded interval, and outputs a real number, which is denoted

b

f (x) dx.

a

This operation satisfies the following axioms:

Axioms. Let [a, b] ⊆ R be a closed bounded interval, let f, g : [a, b] → R be continuous
functions, and let k ∈ R.

1. (Linearity) The definite integral is linear, i.e. bb

b bb kf (x) dx = k f (x) dx.

f (x) + g(x) dx = f (x) dx + g(x) dx and aa

a aa

cb b

2. (Additivity) If c ∈ [a, b], then f (x) dx + f (x) dx = f (x) dx.

ac a

b

3. (Positivity) If f (x) ≥ 0 for all x ∈ [a, b], then f (x) dx ≥ 0.

a

b

4. (Normalization) k dx = k(b − a).

a

Though it may not be obvious, these axioms completely determine the definite integral.
The following proposition lists some easy consequences of the axioms.

Proposition. Let [a, b] be a closed bounded interval, and let f, g : [a, b] → R be continuous

functions.

b bb

1. f (x) − g(x) dx = f (x) dx − g(x) dx.

a aa

bb

2. If f (x) ≤ g(x) for all x ∈ [a, b], then f (x) dx ≤ g(x) dx.

aa

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Proof.
(1) By the linearity axiom,

bb bb

f (x) − g(x) dx = f (x) + (−1)g(x) dx = f (x) dx + (−1)g(x) dx

aa aa

b b bb

= f (x) dx + (−1) g(x) dx = f (x) dx − g(x) dx.

a a aa

(2) Suppose that f (x) ≤ g(x) for all x ∈ [a, b]. Then g(x) − f (x) ≥ 0 for all x ∈ [a, b].

Therefore, by the positivity axiom and part (1) of the proposition, we have

b b b

g(x) dx − f (x) dx = g(x) − f (x) dx ≥ 0,

a a a

and the result follows.

Theorem (Triangle Inequality for Integrals). Let [a, b] ⊆ R be a closed bounded inter-
val, and let f : [a, b] → R be a continuous function. Then

bb

f (x) dx ≤ |f (x)| dx.

aa

Proof. Let f+, f− : [a, b] → R be the functions defined by

f+(x) = |f (x)| + f (x) and f−(x) = |f (x)| − f (x)

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for all x ∈ [a, b]. Note that:

1. f+ and f− are continuous.

2. f = f+ − f− and |f | = f+ + f−.

3. f+(x) ≥ 0 and f−(x) ≥ 0 for all x ∈ [a, b].

Let I+ = b f+(x) dx, and let I− = b f−(x) dx. By positivity, I+, I− ≥ 0. Then
a a

bb

f (x) dx = |I+ − I−| ≤ |I+| + |I−| = I+ + I− = |f (x)| dx.

aa

Fundamental Theorem of Calculus, Part I. Let [a, b] ⊆ R be a closed bounded interval,

and let f : [a, b] → R be a continuous function. Define a function F : [a, b] → R by the

formula x

F (x) = f (t) dt.

a

Then F is continuous, and F (x) = f (x) for all x ∈ (a, b).

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Proof. We first prove that F is continuous. Let > 0. Since f is continuous on [a, b], there
exists an M > 0 so that |f (x)| ≤ M for all x ∈ [a, b]. Let δ = /M , let x, y ∈ [a, b], and
suppose that |x − y| < δ. If x ≤ y, then

yy y

|F (x) − F (y)| = f (t) dt ≤ |f (t)| dt ≤ M dx = M (y − x) < M δ = ,

xx x

and the x > y case is similar. We conclude that F is (uniformly) continuous on [a, b].
We now prove that F (c) = f (c) for all c ∈ (a, b). Let c ∈ (a, b), and let > 0. Since f is

continuous at c, there exists a δ > 0 so that

t ∈ [a, b] and |t − c| < δ ⇒ |f (t) − f (c)| < .

Let x ∈ [a, b] − {c}, and suppose that |x − c| < δ. If x ≥ c, then

F (x) − F (c) − f (c) = F (x) − F (c) − (x − c)f (c)
x − c x−c

= x f (t) dt − x f (c) dt ≤ cx|f (t) − f (c)| dt ≤ x dt = (x − c)
c c x−c c x−c = ,

x−c x−c

and the x < c case is similar. We conclude that F (c) = lim F (x) − F (c) = f (c).
x→c
x−c

Definition. Let [a, b] ⊆ R be a closed bounded interval, and let f : [a, b] → R be a continuous
function. An antiderivative of f is a continuous function F : [a, b] → R such that F (x) =
f (x) for all x ∈ (a, b).

This definition is slightly different from Definition 4.4.8 in the textbook, since the domain
interval [a, b] is closed.

Proposition. Let [a, b] ⊆ R be a closed bounded interval, let f : [a, b] → R be a continuous
function, and let F, G : [a, b] → R be antiderivatives of f . Then there exists a C ∈ R such
that F (x) = G(x) = C for all x ∈ [a, b].

Proof. By Corollary 4.4.9 in the textbook, there exists a C ∈ R such that F (x) = G(x) + C
for all x ∈ (a, b). Since F and G are continuous, it follows that

F (a) = lim F (x) = lim G(x) = G(a) + C
x→a+ x→a+

and similarly F (b) = G(b) + C, and hence F (x) = G(x) + C for all x ∈ [a, b].

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We are now ready for the second part of the Fundamental Theorem of Calculus, which
lets us evaluate integrals using antiderivatives.

Fundamental Theorem of Calculus, Part II. Let [a, b] ⊆ R be a closed bounded interval,

let f : [a, b] → R be a continuous function, and let F : [a, b] → R be an antiderivative of f .

Then b

f (x) dx = F (b) − F (a).

a

Proof. Let G : [a, b] → R be the function defined by the formula

t

G(x) = f (t) dt

a

and note that b f (x) dx = G(b) − G(a). By the previous theorem, G is also an antiderivative
a

of f . Therefore, there exists a constant C so that F (x) = G(x) + C for all x ∈ [a, b]. Then

b

F (b) − F (a) = G(b) + C − G(a) + C = G(b) − G(a) = f (x) dx.

a

Note that this theorem proves our claim that the four axioms completely determine the
definite integral. In particular, we have proven that any operation satisfying the axioms for
a definite integral is given by the formula

b

f (x) dx = F (b) − F (a),

a

and is therefore equal to the definite integral we are familiar with from Calculus I.

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