Editing Riemann integral (section)

=== {{anchor|Riemann-integrable}} Riemann integral ===
Loosely speaking, the Riemann integral is the limit of the [[Riemann sum]]s of a function as the partitions get finer. If the limit exists then the function is said to be '''integrable''' (or more specifically '''Riemann-integrable'''). The Riemann sum can be made as close as desired to the Riemann integral by making the partition fine enough.<ref>{{Cite book|last=Taylor|first=Michael E. |author-link=Michael E. Taylor|title=Measure Theory and Integration| publisher=American Mathematical Society |year=2006 |isbn=9780821872468 |page=1|url=https://books.google.com/books?id=P_zJA-E5oe4C&pg=PA1}}</ref>

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 {{mvar|f}} exists and equals {{mvar|s}} if the following condition holds:

<blockquote>For all {{math|''ε'' > 0}}, there exists {{math|''δ'' > 0}} such that for any [[Partition of an interval|tagged partition]] {{math|''x''<sub>0</sub>, ..., ''x<sub>n</sub>''}} and {{math|''t''<sub>0</sub>, ..., ''t''<sub>''n'' − 1</sub>}} whose mesh is less than {{mvar|δ}}, we have
<math display="block">\left| \left( \sum_{i=0}^{n-1} f(t_i) (x_{i+1}-x_i) \right) - s\right| < \varepsilon.</math></blockquote>

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 {{mvar|f}} exists and equals {{mvar|s}} if the following condition holds:

<blockquote>For all {{math|''ε'' > 0}}, there exists a tagged partition {{math|''y''<sub>0</sub>, ..., ''y<sub>m</sub>''}} and {{math|''r''<sub>0</sub>, ..., ''r''<sub>''m'' − 1</sub>}} such that for any tagged partition {{math|''x''<sub>0</sub>, ..., ''x<sub>n</sub>''}} and {{math|''t''<sub>0</sub>, ..., ''t''<sub>''n'' − 1</sub>}} which is a refinement of {{math|''y''<sub>0</sub>, ..., ''y<sub>m</sub>''}} and {{math|''r''<sub>0</sub>, ..., ''r''<sub>''m'' − 1</sub>}}, we have
<math display="block">\left| \left( \sum_{i=0}^{n-1} f(t_i) (x_{i+1}-x_i) \right) - s\right| < \varepsilon.</math></blockquote>

Both of these mean that eventually, the Riemann sum of {{mvar|f}} with respect to any partition gets trapped close to {{mvar|s}}. Since this is true no matter how close we demand the sums be trapped, we say that the Riemann sums converge to {{mvar|s}}. These definitions are actually a special case of a more general concept, a [[net (mathematics)|net]].

As we stated earlier, these two definitions are equivalent. In other words, {{mvar|s}} works in the first definition if and only if {{mvar|s}} works in the second definition. To show that the first definition implies the second, start with an {{mvar|ε}}, and choose a {{mvar|δ}} that satisfies the condition. Choose any tagged partition whose mesh is less than {{mvar|δ}}. Its Riemann sum is within {{mvar|ε}} of {{mvar|s}}, and any refinement of this partition will also have mesh less than {{mvar|δ}}, so the Riemann sum of the refinement will also be within {{mvar|ε}} of {{mvar|s}}.

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 {{mvar|ε}}, and choose a partition {{math|''y''<sub>0</sub>, ..., ''y<sub>m</sub>''}} such that the lower and upper Darboux sums with respect to this partition are within {{math|''ε''/2}} of the value {{mvar|s}} of the Darboux integral. Let
<math display="block"> r = 2\sup_{x \in [a, b]} |f(x)|.</math>

If {{math|''r'' {{=}} 0}}, then {{mvar|f}} is the zero function, which is clearly both Darboux and Riemann integrable with integral zero. Therefore, we will assume that {{math|''r'' > 0}}. If {{math|''m'' > 1}}, then we choose {{mvar|δ}} such that
<math display="block">\delta < \min \left \{\frac{\varepsilon}{2r(m-1)}, \left(y_1 - y_0\right), \left(y_2 - y_1\right), \cdots, \left(y_m - y_{m-1}\right) \right \}</math>

If {{math|''m'' {{=}} 1}}, then we choose {{mvar|δ}} to be less than one. Choose a tagged partition {{math|''x''<sub>0</sub>, ..., ''x<sub>n</sub>''}} and {{math|''t''<sub>0</sub>, ..., ''t''<sub>''n'' − 1</sub>}} with mesh smaller than {{mvar|δ}}. We must show that the Riemann sum is within {{mvar|ε}} of {{mvar|s}}.

To see this, choose an interval {{math|[''x<sub>i</sub>'', ''x''<sub>''i'' + 1</sub>]}}. If this interval is contained within some {{math|[''y<sub>j</sub>'', ''y''<sub>''j'' + 1</sub>]}}, then
<math display="block"> m_j \leq f(t_i) \leq M_j</math>
where {{mvar|m<sub>j</sub>}} and {{mvar|M<sub>j</sub>}} are respectively, the infimum and the supremum of ''f'' on {{math|[''y<sub>j</sub>'', ''y''<sub>''j'' + 1</sub>]}}. 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 {{mvar|s}}. This is the case when {{math|''m'' {{=}} 1}}, so the proof is finished in that case.

Therefore, we may assume that {{math|''m'' > 1}}. In this case, it is possible that one of the {{math|[''x<sub>i</sub>'', ''x''<sub>''i'' + 1</sub>]}} is not contained in any {{math|[''y<sub>j</sub>'', ''y''<sub>''j'' + 1</sub>]}}. Instead, it may stretch across two of the intervals determined by {{math|''y''<sub>0</sub>, ..., ''y<sub>m</sub>''}}. (It cannot meet three intervals because {{mvar|δ}} is assumed to be smaller than the length of any one interval.) In symbols, it may happen that
<math display="block">y_j < x_i < y_{j+1} < x_{i+1} < y_{j+2}.</math>

(We may assume that all the inequalities are strict because otherwise we are in the previous case by our assumption on the length of {{mvar|δ}}.) This can happen at most {{math|''m'' − 1}} times.

To handle this case, we will estimate the difference between the Riemann sum and the Darboux sum by subdividing the partition {{math|''x''<sub>0</sub>, ..., ''x<sub>n</sub>''}} at {{math|''y''<sub>''j'' + 1</sub>}}. The term {{math|''f''(''t<sub>i</sub>'')(''x''<sub>''i'' + 1</sub> − ''x<sub>i</sub>'')}} in the Riemann sum splits into two terms:
<math display="block">f\left(t_i\right)\left(x_{i+1}-x_i\right) = f\left(t_i\right)\left(x_{i+1}-y_{j+1}\right)+f\left(t_i\right)\left(y_{j+1}-x_i\right).</math>

Suppose, [[without loss of generality]], that {{math|''t<sub>i</sub>'' ∈ [''y<sub>j</sub>'', ''y''<sub>''j'' + 1</sub>]}}. Then
<math display="block">m_j \leq f(t_i) \leq M_j,</math>
so this term is bounded by the corresponding term in the Darboux sum for {{mvar|y<sub>j</sub>}}. To bound the other term, notice that
<math display="block">x_{i+1}-y_{j+1} < \delta < \frac{\varepsilon}{2r(m-1)},</math>

It follows that, for some (indeed any) {{math|''t''{{su|b=''i''|p=*}} ∈ [''y''<sub>''j'' + 1</sub>, ''x''<sub>''i'' + 1</sub>]}},
<math display="block">\left|f\left(t_i\right)-f\left(t_i^*\right)\right|\left(x_{i+1}-y_{j+1}\right) < \frac{\varepsilon}{2(m-1)}.</math>

Since this happens at most {{math|''m'' − 1}} times, the distance between the Riemann sum and a Darboux sum is at most {{math|''ε''/2}}. Therefore, the distance between the Riemann sum and {{mvar|s}} is at most&nbsp;{{mvar|ε}}.