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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?

Michiel uit het Broek
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PhoenixPerson
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  • Riemann didn't have to define his integral that way. But isn't it interesting that we don't have to choose subintervals of equal length? –  May 20 '16 at 15:05

2 Answers2

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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 $0$.

Paramanand Singh
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  • Dear Paramanand. I just asked this question [Limit of the ratio of two non-Riemann sums] (https://math.stackexchange.com/questions/3481607/limit-of-the-ratio-of-two-non-riemann-sum) that is intimately related to your answer: the limit is trivial for regular partitions but not so trivial for irregular ones. So, please may you be kind enough to provide me with some references about "it can be proved with some effort that one of these theses choices must be arbitrary". In my case, the tag points would be arbitrary but the partitions would be regular. THANKS. – Fabrice Pautot Dec 19 '19 at 16:44
  • Of course, if you can ever answer my question, that would be great!!! – Fabrice Pautot Dec 19 '19 at 16:50
  • The first paragraph about "two levels of choice" was very insightful. Can you share some links for it's proof? That would really really really solve a lot of my problems and doubts. Thank you. – William Aug 08 '22 at 12:49
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    @William : two levels of choice is how I see the definition of Riemann sum. If we restrict both choices then the definition doesn't work as shown in the example in my answer. The proof that restricting choice of tag points (but keeping arbitrary partitions) works is difficult. Search Cauchy definition of integral where tag points are choosen as left endpoints of subintervals. – Paramanand Singh Aug 08 '22 at 15:49
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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.

GEdgar
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