Editing Riemann integral (section)

== Examples ==
Let <math>f:[0,1]\to\R</math> be the function which takes the value 1 at every point. Any Riemann sum of {{mvar|f}} on {{math|[0, 1]}} will have the value 1, therefore the Riemann integral of {{mvar|f}} on {{math|[0, 1]}} is 1.

Let <math>I_{\Q}:[0,1]\to\R</math> be the [[indicator function]] of the rational numbers in {{math|[0, 1]}}; that is, <math>I_{\Q}</math> 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 {{math|''x''<sub>0</sub>, ..., ''x<sub>n</sub>''}} and {{math|''t''<sub>0</sub>, ..., ''t''<sub>''n'' − 1</sub>}} be a tagged partition (each {{mvar|t<sub>i</sub>}} is between {{mvar|x<sub>i</sub>}} and {{math|''x''<sub>''i'' + 1</sub>}}). Choose {{math|''ε'' > 0}}. The {{mvar|t<sub>i</sub>}} have already been chosen, and we can't change the value of {{mvar|f}} at those points. But if we cut the partition into tiny pieces around each {{mvar|t<sub>i</sub>}}, we can minimize the effect of the {{mvar|t<sub>i</sub>}}. Then, by carefully choosing the new tags, we can make the value of the Riemann sum turn out to be within {{mvar|ε}} of either zero or one.

Our first step is to cut up the partition. There are {{mvar|n}} of the {{mvar|t<sub>i</sub>}}, and we want their total effect to be less than {{mvar|ε}}. If we confine each of them to an interval of length less than {{math|''ε''/''n''}}, then the contribution of each {{mvar|t<sub>i</sub>}} to the Riemann sum will be at least {{math|0 · ''ε''/''n''}} and at most {{math|1 · ''ε''/''n''}}. This makes the total sum at least zero and at most {{mvar|ε}}. So let {{mvar|δ}} be a positive number less than {{math|''ε''/''n''}}. If it happens that two of the {{mvar|t<sub>i</sub>}} are within {{mvar|δ}} of each other, choose {{mvar|δ}} smaller. If it happens that some {{mvar|t<sub>i</sub>}} is within {{mvar|δ}} of some {{mvar|x<sub>j</sub>}}, and {{mvar|t<sub>i</sub>}} is not equal to {{mvar|x<sub>j</sub>}}, choose {{mvar|δ}} smaller. Since there are only finitely many {{mvar|t<sub>i</sub>}} and {{mvar|x<sub>j</sub>}}, we can always choose {{mvar|δ}} sufficiently small.

Now we add two cuts to the partition for each {{mvar|t<sub>i</sub>}}. One of the cuts will be at {{math|''t<sub>i</sub>'' − ''δ''/2}}, and the other will be at {{math|''t<sub>i</sub>'' + ''δ''/2}}. If one of these leaves the interval [0, 1], then we leave it out. {{mvar|t<sub>i</sub>}} will be the tag corresponding to the subinterval
<math display="block">\left [t_i - \frac{\delta}{2}, t_i + \frac{\delta}{2} \right ].</math>

If {{mvar|t<sub>i</sub>}} is directly on top of one of the {{mvar|x<sub>j</sub>}}, then we let {{mvar|t<sub>i</sub>}} be the tag for both intervals:
<math display="block">\left [t_i - \frac{\delta}{2}, x_j \right ], \quad\text{and}\quad \left [x_j,t_i + \frac{\delta}{2} \right ].</math>

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 {{math|1 − ''ε''}}. 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 {{mvar|ε}}.

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 {{mvar|s}}, so this function is not Riemann integrable. However, it is [[Lebesgue integral|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. <math>I_{\Q}</math> 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 {{mvar|C}} be the [[Smith–Volterra–Cantor set]], and let {{math|''I<sub>C</sub>''}} be its indicator function. Because {{mvar|C}} is not [[Jordan measure|Jordan measurable]], {{math|''I<sub>C</sub>''}} is not Riemann integrable. Moreover, no function {{mvar|g}} equivalent to {{math|''I<sub>C</sub>''}} is Riemann integrable: {{mvar|g}}, like {{math|''I<sub>C</sub>''}}, must be zero on a dense set, so as in the previous example, any Riemann sum of {{mvar|g}} has a refinement which is within {{mvar|ε}} of 0 for any positive number&nbsp;{{mvar|ε}}. But if the Riemann integral of {{mvar|g}} exists, then it must equal the Lebesgue integral of {{math|''I<sub>C</sub>''}}, which is {{math|1/2}}. Therefore, {{mvar|g}} is not Riemann integrable.