Riemann Integral - Examples

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 (each is between and ). 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 choice!

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 and . 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 (that is, equal almost everywhere) 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 C be the Smith–Volterra–Cantor set, and let I_C be its indicator function. Because C is not Jordan measurable, I_C is not Riemann integrable. Moreover, no function g equivalent to I_C is Riemann integrable: g, like I_C, must be zero on a dense set, so as in the previous example, any Riemann sum of g has a refinement which is within ε of 0 for any positive number ε. But if the Riemann integral of g exists, then it must equal the Lebesgue integral of I_C, which is 1/2. Therefore g is not Riemann integrable.