Riemann integral

Overview

Definition

Partitions of an interval

Riemann sums

Riemann integral

For all, there exists such that for any tagged partition and whose mesh is less than, we have
Unfortunately, this definition is very difficult to use. It would help to develop an equivalent definition of the Riemann integral which is easier to work with. We develop this definition now, with a proof of equivalence following. Our new definition says that the Riemann integral of equals if the following condition holds:

For all, there exists a tagged partition and such that for any tagged partition and which is a refinement of and, we have
Both of these mean that eventually, the Riemann sum of with respect to any partition gets trapped close to. Since this is true no matter how close we demand the sums be trapped, we say that the Riemann sums converge to. These definitions are actually a special case of a more general concept, a net.
As we stated earlier, these two definitions are equivalent. In other words, works in the first definition if and only if works in the second definition. To show that the first definition implies the second, start with an, and choose a that satisfies the condition. Choose any tagged partition whose mesh is less than. Its Riemann sum is within of, and any refinement of this partition will also have mesh less than, so the Riemann sum of the refinement will also be within of.
To show that the second definition implies the first, it is easiest to use the Darboux integral. First, one shows that the second definition is equivalent to the definition of the Darboux integral; for this see the article on Darboux integration. Now we will show that a Darboux integrable function satisfies the first definition. Fix, and choose a partition such that the lower and upper Darboux sums with respect to this partition are within of the value of the Darboux integral. Let
If, then is the zero function, which is clearly both Darboux and Riemann integrable with integral zero. Therefore, we will assume that. If, then we choose such that
If, then we choose to be less than one. Choose a tagged partition and with mesh smaller than. We must show that the Riemann sum is within of.
To see this, choose an interval. If this interval is contained within some, then
where and are respectively, the infimum and the supremum of f on. If all intervals had this property, then this would conclude the proof, because each term in the Riemann sum would be bounded by a corresponding term in the Darboux sums, and we chose the Darboux sums to be near. This is the case when, so the proof is finished in that case.
Therefore, we may assume that. In this case, it is possible that one of the is not contained in any. Instead, it may stretch across two of the intervals determined by. In symbols, it may happen that
This can happen at most times.
To handle this case, we will estimate the difference between the Riemann sum and the Darboux sum by subdividing the partition at. The term in the Riemann sum splits into two terms:
Suppose, without loss of generality, that. Then
so this term is bounded by the corresponding term in the Darboux sum for. To bound the other term, notice that
It follows that, for some ,
Since this happens at most times, the distance between the Riemann sum and a Darboux sum is at most. Therefore, the distance between the Riemann sum and is at most .

Examples

Let be the function which takes the value 1 at every point. Any Riemann sum of on will have the value 1, therefore the Riemann integral of on is 1.
Let be the indicator function of the rational numbers in ; that is, takes the value 1 on rational numbers and 0 on irrational numbers. This function does not have a Riemann integral. To prove this, we will show how to construct tagged partitions whose Riemann sums get arbitrarily close to both zero and one.
To start, let and be a tagged partition. Choose. The have already been chosen, and we can't change the value of at those points. But if we cut the partition into tiny pieces around each, we can minimize the effect of the. Then, by carefully choosing the new tags, we can make the value of the Riemann sum turn out to be within of either zero or one.
Our first step is to cut up the partition. There are of the, and we want their total effect to be less than. If we confine each of them to an interval of length less than, then the contribution of each to the Riemann sum will be at least and at most. This makes the total sum at least zero and at most. So let be a positive number less than. If it happens that two of the are within of each other, choose smaller. If it happens that some is within of some, and is not equal to, choose smaller. Since there are only finitely many and, we can always choose sufficiently small.
Now we add two cuts to the partition for each. One of the cuts will be at, and the other will be at. If one of these leaves the interval , then we leave it out. will be the tag corresponding to the subinterval
If is directly on top of one of the, then we let be the tag for both intervals:
We still have to choose tags for the other subintervals. We will choose them in two different ways. The first way is to always choose a rational point, so that the Riemann sum is as large as possible. This will make the value of the Riemann sum at least. The second way is to always choose an irrational point, so that the Riemann sum is as small as possible. This will make the value of the Riemann sum at most.
Since we started from an arbitrary partition and ended up as close as we wanted to either zero or one, it is false to say that we are eventually trapped near some number, so this function is not Riemann integrable. However, it is Lebesgue integrable. In the Lebesgue sense its integral is zero, since the function is zero almost everywhere. But this is a fact that is beyond the reach of the Riemann integral.
There are even worse examples. is equivalent to a Riemann integrable function, but there are non-Riemann integrable bounded functions which are not equivalent to any Riemann integrable function. For example, let be the Smith–Volterra–Cantor set, and let be its indicator function. Because is not Jordan measurable, is not Riemann integrable. Moreover, no function equivalent to is Riemann integrable:, like, must be zero on a dense set, so as in the previous example, any Riemann sum of has a refinement which is within of 0 for any positive number . But if the Riemann integral of exists, then it must equal the Lebesgue integral of, which is. Therefore, is not Riemann integrable.

Similar concepts

It is popular to define the Riemann integral as the Darboux integral. This is because the Darboux integral is technically simpler and because a function is Riemann-integrable if and only if it is Darboux-integrable.
Some calculus books do not use general tagged partitions, but limit themselves to specific types of tagged partitions. If the type of partition is limited too much, some non-integrable functions may appear to be integrable.
One popular restriction is the use of "left-hand" and "right-hand" Riemann sums. In a left-hand Riemann sum, for all, and in a right-hand Riemann sum, for all. Alone this restriction does not impose a problem: we can refine any partition in a way that makes it a left-hand or right-hand sum by subdividing it at each. In more formal language, the set of all left-hand Riemann sums and the set of all right-hand Riemann sums is cofinal in the set of all tagged partitions.
Another popular restriction is the use of regular subdivisions of an interval. For example, the th regular subdivision of consists of the intervals
Again, alone this restriction does not impose a problem, but the reasoning required to see this fact is more difficult than in the case of left-hand and right-hand Riemann sums.
However, combining these restrictions, so that one uses only left-hand or right-hand Riemann sums on regularly divided intervals, is dangerous. If a function is known in advance to be Riemann integrable, then this technique will give the correct value of the integral. But under these conditions the indicator function will appear to be integrable on with integral equal to one: Every endpoint of every subinterval will be a rational number, so the function will always be evaluated at rational numbers, and hence it will appear to always equal one. The problem with this definition becomes apparent when we try to split the integral into two pieces. The following equation ought to hold:
If we use regular subdivisions and left-hand or right-hand Riemann sums, then the two terms on the left are equal to zero, since every endpoint except 0 and 1 will be irrational, but as we have seen the term on the right will equal 1.
As defined above, the Riemann integral avoids this problem by refusing to integrate The Lebesgue integral is defined in such a way that all these integrals are 0.

Properties

Linearity

The Riemann integral is a linear transformation; that is, if and are Riemann-integrable on and and are constants, then
Because the Riemann integral of a function is a number, this makes the Riemann integral a linear functional on the vector space of Riemann-integrable functions.

Integrability

A bounded function on a compact interval is Riemann integrable if and only if it is continuous almost everywhere. This is known as the or Lebesgue's criterion for Riemann integrability or the Riemann–Lebesgue theorem. The criterion has nothing to do with the Lebesgue integral. It is due to Lebesgue and uses his measure zero, but makes use of neither Lebesgue's general measure or integral.
The integrability condition can be proven in various ways, one of which is sketched below.
One direction can be proven using the oscillation definition of continuity: For every positive, Let be the set of points in with oscillation of at least. Since every point where is discontinuous has a positive oscillation and vice versa, the set of points in, where is discontinuous is equal to the union over for all natural numbers.
If this set does not have a zero Lebesgue measure, then by countable additivity of the measure there is at least one such so that does not have a zero measure. Thus there is some positive number such that every countable collection of open intervals covering has a total length of at least. In particular this is also true for every such finite collection of intervals. Note that this remains true also for less a finite number of points.
For every partition of, consider the set of intervals whose interiors include points from. These interiors consist of a finite open cover of, possibly up to a finite number of points. Thus these intervals have a total length of at least. Since in these points has oscillation of at least, the infimum and supremum of in each of these intervals differ by at least. Thus the upper and lower sums of differ by at least. Since this is true for every partition, is not Riemann integrable.
We now prove the converse direction using the sets defined above. Note that for every, is compact, as it is bounded and closed:

• For every series of points in that is converging in, its limit is in as well. This is because every neighborhood of the limit point is also a neighborhood of some point in, and thus has an oscillation of at least on it. Hence the limit point is in.

Now, suppose that is continuous almost everywhere. Then for every, has zero Lebesgue measure. Therefore, there is a countable collections of open intervals in which is an open cover of, such that the sum over all their lengths is arbitrarily small. Since is compact, there is a finite subcover – a finite collections of open intervals in with arbitrarily small total length that together contain all points in. We denote these intervals, for, for some natural.
The complement of the union of these intervals is itself a union of a finite number of intervals, which we denote .
We now show that for every, there are upper and lower sums whose difference is less than, from which Riemann integrability follows. To this end, we construct a partition of as follows:
Denote and, where and are the infimum and supremum of on. Since we may choose intervals with arbitrarily small total length, we choose them to have total length smaller than.
Each of the intervals has an empty intersection with, so each point in it has a neighborhood with oscillation smaller than. These neighborhoods consist of an open cover of the interval, and since the interval is compact there is a finite subcover of them. This subcover is a finite collection of open intervals, which are subintervals of . In each of these, oscillates by less than. Since the total length of these is not larger than, they together contribute at most to the difference between the upper and lower sums of the partition.

• The intervals. These have total length smaller than, and oscillates on them by no more than. Thus together they contribute less than to the difference between the upper and lower sums of the partition.

In total, the difference between the upper and lower sums of the partition is smaller than, as required.
In particular, any set that is at most countable has Lebesgue measure zero, and thus a bounded function with only finitely or countably many discontinuities is Riemann integrable.
An indicator function of a bounded set is Riemann-integrable if and only if the set is Jordan measurable. The Riemann integral can be interpreted measure-theoretically as the integral with respect to the Jordan measure.
If a real-valued function is monotone on the interval it is Riemann-integrable, since its set of discontinuities is at most countable, and therefore of Lebesgue measure zero.
If a real-valued function on is Riemann-integrable, it is Lebesgue-integrable. That is, Riemann-integrability is a stronger condition than Lebesgue-integrability.
If is a uniformly convergent sequence on with limit, then Riemann integrability of all implies Riemann integrability of, and
However, the Lebesgue monotone convergence theorem does not hold. In Riemann integration, taking limits under the integral sign is far more difficult to logically justify than in Lebesgue integration.

Generalizations

It is easy to extend the Riemann integral to functions with values in the Euclidean vector space for any. The integral is defined component-wise; in other words, if then
In particular, since the complex numbers are a real vector space, this allows the integration of complex valued functions.
The Riemann integral is only defined on bounded intervals, and it does not extend well to unbounded intervals. The simplest possible extension is to define such an integral as a limit, in other words, as an improper integral:
This definition carries with it some subtleties, such as the fact that it is not always equivalent to compute the Cauchy principal value
For example, consider the sign_function which is 0 at, 1 for, and −1 for. By symmetry,
always, regardless of. But there are many ways for the interval of integration to expand to fill the real line, and other ways can produce different results; in other words, the multivariate limit does not always exist. We can compute
In general, this improper Riemann integral is undefined. Even standardizing a way for the interval to approach the real line does not work because it leads to disturbingly counterintuitive results. If we agree that the improper integral should always be
then the integral of the translation is −2, so this definition is not invariant under shifts, a highly undesirable property. In fact, not only does this function not have an improper Riemann integral, its Lebesgue integral is also undefined.
Unfortunately, the improper Riemann integral is not powerful enough. The most severe problem is that there are no widely applicable theorems for commuting improper Riemann integrals with limits of functions. In applications such as Fourier series it is important to be able to approximate the integral of a function using integrals of approximations to the function. For proper Riemann integrals, a standard theorem states that if is a sequence of functions that converge uniformly to on a compact set, then
On non-compact intervals such as the real line, this is false. For example, take to be on and zero elsewhere. For all we have:
The sequence converges uniformly to the zero function, and clearly the integral of the zero function is zero. Consequently,
This demonstrates that for integrals on unbounded intervals, uniform convergence of a function is not strong enough to allow passing a limit through an integral sign. This makes the Riemann integral unworkable in applications, because there is no other general criterion for exchanging a limit and a Riemann integral, and without such a criterion it is difficult to approximate integrals by approximating their integrands.
A better route is to abandon the Riemann integral for the Lebesgue integral. The definition of the Lebesgue integral is not obviously a generalization of the Riemann integral, but it is not hard to prove that every Riemann-integrable function is Lebesgue-integrable and that the values of the two integrals agree whenever they are both defined. Moreover, a function defined on a bounded interval is Riemann-integrable if and only if it is bounded and the set of points where is discontinuous has Lebesgue measure zero.
An integral which is in fact a direct generalization of the Riemann integral is the Henstock–Kurzweil integral.
Another way of generalizing the Riemann integral is to replace the factors in the definition of a Riemann sum by something else; roughly speaking, this gives the interval of integration a different notion of length. This is the approach taken by the Riemann–Stieltjes integral.
In multivariable calculus, the Riemann integrals for functions from are multiple integrals.