Riemann integral

In the branch of mathematics known as real analysis, the Riemann integral, created by Bernhard Riemann, was the first rigorous definition of the integral of a function on an interval. It was presented to the faculty at the University of Göttingen in 1854, but not published in a journal until 1868. For many functions and practical applications, the Riemann integral can be evaluated by the fundamental theorem of calculus or approximated by numerical integration, or simulated using Monte Carlo integration.

Overview

Imagine you have a curve on a graph, and the curve stays above the x-axis between two points, a and b. The area under that curve, from a to b, is what we want to figure out. This area can be described as the set of all points on the graph that follow these rules: a ≤ x ≤ b and 0 < y < f. Mathematically, this region can be expressed in set-builder notation as
To measure this area, we use a Riemann integral, which is written as:
This notation means “the integral of f from a to b,” and it represents the exact area under the curve f and above the x-axis, between x = a and x = b.
The idea behind the Riemann integral is to break the area into small, simple shapes, add up their areas, and then make the rectangles smaller and smaller to get a better estimate. In the end, when the rectangles are infinitely small, the sum gives the exact area, which is what the integral represents.
If the curve dips below the x-axis, the integral gives a signed area. This means the integral adds the part above the x-axis as positive and subtracts the part below the x-axis as negative. So, the result of can be positive, negative, or zero, depending on how much of the curve is above or below the x-axis.

Definition

Partitions of an interval

A partition of an interval is a finite sequence of numbers of the form
Each is called a sub-interval of the partition. The mesh or norm of a partition is defined to be the length of the longest sub-interval, that is,
A tagged partition of an interval is a partition together with a choice of a sample point within each sub-interval: that is, numbers with for each. The mesh of a tagged partition is the same as that of an ordinary partition.
Suppose that two partitions and are both partitions of the interval. We say that is a refinement of if for each integer, with, there exists an integer such that and such that for some with. That is, a tagged partition breaks up some of the sub-intervals and adds sample points where necessary, "refining" the accuracy of the partition.
We can turn the set of all tagged partitions into a directed set by saying that one tagged partition is greater than or equal to another if the former is a refinement of the latter.

Riemann sum

Let be a real-valued function defined on the interval. The Riemann sum of with respect to a tagged partition of is
Each term in the sum is the product of the value of the function at a given point and the length of an interval. Consequently, each term represents the area of a rectangle with height and width. The Riemann sum is the area of all the rectangles.
Closely related concepts are the lower and upper Darboux sums. These are similar to Riemann sums, but the tags are replaced by the infimum and supremum of on each sub-interval:
If is continuous, then the lower and upper Darboux sums for an untagged partition are equal to the Riemann sum for that partition, where the tags are chosen to be the minimum or maximum of on each subinterval. The Darboux integral, which is similar to the Riemann integral but based on Darboux sums, is equivalent to the Riemann integral.

Riemann integral

Loosely speaking, the Riemann integral is the limit of the Riemann sums of a function as the partitions get finer. If the limit exists then the function is said to be integrable. The Riemann sum can be made as close as desired to the Riemann integral by making the partition fine enough.
One important requirement is that the mesh of the partitions must become smaller and smaller, so that it has the limit zero. If this were not so, then we would not be getting a good approximation to the function on certain subintervals. In fact, this is enough to define an integral. To be specific, we say that the Riemann integral of exists and equals if the following condition holds:

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 exists and 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 Darboux integral article. 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.