Jump to content

Real Analysis/Riemann integration

From Wikibooks, open books for an open world
Real Analysis
Riemann integration

Definition

[edit | edit source]

Riemann integration is the formulation of integration most people think of if they ever think about integration. It is the only type of integration considered in most calculus classes; many other forms of integration, notably Lebesgue integrals, are extensions of Riemann integrals to larger classes of functions. The Riemann integral was developed by Bernhard Riemann in 1854 and was, when invented, the first rigorous definition of integration applicable to not necessarily continuous functions.

We will first define some preliminary ideas.

Partitions

[edit | edit source]

Definition

[edit | edit source]

Let

A Partition is defined as the ordered -tuple of real numbers such that

Norm of a Partition

[edit | edit source]

Let be a partition given by

Then, the Norm (or the "mesh") of is defined as

Tagged Partition

[edit | edit source]

Let be a partition

A Tagged Partition is defined as the set of ordered pairs such that . The points are called Tags.

Riemann

[edit | edit source]
Riemann sum of a function

Riemann Sums

[edit | edit source]

Let

Let be a tagged partition of

The Riemann Sum of over with respect to is given by

Riemann Integral

[edit | edit source]

Let

Let

We say that is Integrable on if and only if, for every there exists such that for every partition satisfying , we have that

is said to be the integral of over , and is written as

or as

Properties

[edit | edit source]

Theorem (Uniqueness)

[edit | edit source]

Let be integrable on

Then the integral of is unique

Proof

[edit | edit source]

Assume, if possible that are both integrals of over . Consider

As are integrals, there exist such that for all that satisfy and for all that satisfy

Let . Hence, if is a partition satisfying , then we have and that

That is, , which is an obvious contradiction. Hence the integral of is unique.


We now state (without proof) two seemingly obvious properties of the integral.

Theorem

[edit | edit source]

Let be integrable and let

Then:

(i)

(ii)

Theorem (Boundedness Theorem)

[edit | edit source]

Let be Riemann integrable. Then is bounded over

Proof

[edit | edit source]

Assume if possible that is unbounded. For every divide the interval into parts. Hence, for every , is unbounded on at least one of these parts. Call it .

Now, let be given. Consider an arbitrary . Let be a tagged partition such that and , where is taken so as to satisfy .

Thus we have that . But as is arbitrary, we have a contradiction to the fact that is Riemann integrable.

Hence, is bounded.

Integrability

[edit | edit source]

We now study classes of Riemann integrable functions. The first "constraint" on Riemann integrable functions is provided by the Cauchy Integrability Criterion.

Theorem (Cauchy Criterion)

[edit | edit source]

Let

Then,

(i) is Riemann integrable on if and only if

(ii) For every , there exists such that if are two partitions satisfying then

Proof

[edit | edit source]

()Let and let be given.

Then, there exists such that for every partition satisfying ,we have

Now, let partitions be such that .

Thus we have that , that is

() For every , consider such that for all partitions satisfying , we have .

Without loss of generality, we can assume that when . For every , let be a partition such that

The sequence is a Cauchy sequence, and hence it has a limit .

Now, for every , we have a such that implies .

Thus

Theorem (Squeeze Theorem)

[edit | edit source]

Let

Then,

(i) is Riemann integrable on if and only if

(ii) For every , there exist Riemann integrable functions such that

for all and

Proof

[edit | edit source]

()Take . It is easy to see that

()Let . Then, there exist functions such that . Further, if and , then there exist such that if a partition satisfies then and then

Now let be a partition satisfying .

Now, we can easily see that . Hence, is a Cauchy sequence, with a limit , and as in the previous proof, we can show that