Editing Riemann integral (section)

== 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, {{math|''t<sub>i</sub>'' {{=}} ''x<sub>i</sub>''}} for all {{mvar|i}}, and in a right-hand Riemann sum, {{math|''t<sub>i</sub>'' {{=}} ''x''<sub>''i'' + 1</sub>}} for all {{mvar|i}}. 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 {{mvar|t<sub>i</sub>}}. In more formal language, the set of all left-hand Riemann sums and the set of all right-hand Riemann sums is [[cofinal (mathematics)|cofinal]] in the set of all tagged partitions.

Another popular restriction is the use of regular subdivisions of an interval. For example, the {{mvar|n}}th regular subdivision of {{math|[0, 1]}} consists of the intervals
<math display="block">\left [0, \frac{1}{n} \right], \left [\frac{1}{n}, \frac{2}{n} \right], \ldots, \left[\frac{n-1}{n}, 1 \right].</math>

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]] <math>I_{\Q}</math> will appear to be integrable on {{math|[0, 1]}} 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:
<math display="block">\int_0^{\sqrt{2}-1} I_\Q(x) \,dx + \int_{\sqrt{2}-1}^1 I_\Q(x) \,dx = \int_0^1 I_\Q(x) \,dx.</math>

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 <math>I_{\Q}.</math> The Lebesgue integral is defined in such a way that all these integrals are 0.