Is the Riemann integral defined with partitions with subintervals of the same length different from the general case?

Question

When considering Riemann sums we partition the closed interval $[a,b]$ in subintervals that don't necessarily have to have the same length. Then the Riemann integral is defined by taking the supremum of the lengths of the intervals go to zero. My question is, what is gained by not having the intervals in the partition not have the same lenght? Is there is some function that is Riemann integrable in the if the subintervals are of the same lenght, but not otherwise?

Answer 1

You are doing this by upper and lower sums? Then no, there is no difference.

But there are cases where the computation of the integral is easier if you use a partition with geometric division, rather than arithmetic division.

Also varying size partitions are good for many proofs, for example $\int_a^b = \int_a^c + \int_c^b$. So even if your original definition is with equal-size partitions, you will soon have to show it is equivalent to the varying-size partition definition.

Answer 2

There are two levels of choice in defining Riemann integral. The first choice is about partitioning : whether to choose the subintervals of equal lengths or follow some pattern like points of partition in geometric progression or to keep them totally arbitrary. The second choice is the tag points which is done after the choice of partition. The tag points may be left (or right) end points of the subintervals or may follow some other pattern or be totally arbitrary.

It can be proved with some effort that one of these choices must be arbitrary. Restricting both the choices to a specific pattern does not work. Thus if you wish to use subintervals of equal length then your definition must allow for arbitrary choice of tags. As as example consider the function $f$ which takes value $0$ at rational points and $1$ at irrational points and consider its Riemann integral on $[0,1]$. It is easy to prove that the function is not Riemann integrable. But if we use partitions with subintervals of equal length and restrict the tags to left end points of subintervals then each Riemann sum is $0$ and thus they trivially converge to $0$.