Riemann integral and (un)equal partitions In the original definition of the Riemann integral we use partitions of the interval in which the sub-intervals do not need to be of equal length. Is there any example in which the sub-intervals of equal length will cause some problems (proofs, computations)? With many thanks.
 A: As long as the partitions get smaller and smaller any partitioning will do and they will converge to the same limit if it exists. Sometimes by choosing a clever partition you can make an otherwise difficult integral very easy if say, the function is always zero at the partition points.
You may be interested in conditionally convergent sequences which can be made to take on any value by rearranging the order of the terms as proven by the Riemann series theorem. They will give you some idea of how complications can arise if we treat infinite series carelessly.
A: Consider Thomae's function on $[0,1]$.
$$
f(x)=
\begin{cases}
\frac1q, \text{ if }x=\frac{p}{q}, \text{ with } p\in \mathbb{Z} \text{ and }q\in\mathbb{N} \text{ coprime};\\
0, \text{ if }x\notin \mathbb{Q}.
\end{cases}
$$
The Riemann integrability of $f$ on $[0,1]$ can be shown by definition, which is that the supremum of lower Riemann sum equals to the infimum of upper Riemann sum.
For any partition $P$ of $[0,1]$, the lower Riemann sum of $f$ at $P$ would be $0$.
The reason is as follows.
Every interval on $\mathbb{R}$ contains irrational and rational numbers. The value of $f$ on rational numbers must be greater than $0$, so the infimum within every interval would be $0$, which is the value of $f$ at every irrational point.
So, the supremum of lower Riemann sum is $0$.
Showing that the infimum of upper Riemann sum is more complicating, but I am sure that you can find some resource at other question.
Now, the Riemann integral of Thomae function on $[0,1]$ at equal-length partitions is $0$, the value of definite Riemann integral $\int_0^1f(x)dx$. That is,
$$
\lim_{n\rightarrow\infty}\sum_{i=1}^nf(\frac{i}{n})\cdot\frac1n=\int_0^1f(x)dx=0
$$
Keep this in mind, because we are going to change the Thomae function.
We select a real number $a\neq0$. This would be the value of new function on irrational numbers. See the following definition.
Define $\hat{f}:[0,1]\rightarrow\mathbb{R}$ by
$$
\hat{f}(x)=
\begin{cases}
\frac1q, \text{ if }x=\frac{p}{q}, \text{ with } p\in \mathbb{Z} \text{ and }q\in\mathbb{N} \text{ coprime};\\
a, \text{ if }x\notin \mathbb{Q}.
\end{cases}
$$
One can show that $\hat{f}$ is discontinuous at every point. By Lebesgue's criterion for Riemann integrability, the set of discontinuous points has measure more than $0$, so $\hat{f}$ is not Riemann integrable.
However, we can still compute this:
$$
\lim_{n\rightarrow\infty}\sum_{i=1}^n\hat{f}(\frac{i}{n})\cdot\frac1n=\lim_{n\rightarrow\infty}\sum_{i=1}^nf(\frac{i}{n})\cdot\frac1n=0
$$
as $f$ and $\hat{f}$ have the same value on rational numbers.
In this example, we see that even if the Riemann integral of $\hat{f}$ on $[0,1]$ at equal-length partitions exists, this function is not Riemann integrable. I thought that this is the "problem" you were looking for.
