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Why do we assume that when the Riemann integral is defined, the partition norm tends to zero? And if it did not tend to zero, but remained constant, would it happen?

Jack J.
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    In your last sentence, what exactly would happen? Your question is not very clear. – Mark Apr 16 '20 at 15:55
  • Riemann integral can be defined using two kinds of limiting processes which coincidentally turn out to be equivalent. See this answer for more details. – Paramanand Singh Apr 27 '20 at 11:41
  • When you are trying to measure the length of some object via a ruler, you can hope that a ruler with smaller units will help you measure it accurately. Similar is the case with area being defined as integral. If the partition mesh is small the quality of approximation is better. It just does not lead to a meaningful/useful idea when we have large mesh size. – Paramanand Singh Apr 27 '20 at 11:44

1 Answers1

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The definition of the Riemann integral requires the lower and upper integrals to be defined and equal. The lower integral is defined as the supremum of lower sums over all partitions, and the upper integral is defined as the infimum across all partitions of the interval.

Note that $L(f,P') \geq L(f,P)$ where $P'$ is a refinement of $P$. A similar but converse result holds for the upper Riemann sums.

Given this, the supremum across all partitions of the lower Riemann sum must have the mesh 'tend' to zero. To see why, consider that the supremum defined above is necessarily an upper bound of the lower sum of any partition. For any partition, say $P$, and any partition, say $P'$, that is both a refinement of $P$ and that has a smaller mesh than $P$, then the lower Riemann sum on $P'$ is greater than the lower Riemann sum on $P$, as above.

In defining the lower integral as the supremum of lower sums, we see then that the mesh of partitions has to 'decrease' as we 'approach' the supremum value. An analogous result holds for the upper Riemann sums.

I would say however that at least for me, I don't think of the definition I was taught as any sort of 'approach'. In my head the definition simply tells me I am taking some bound on some set (the set here being all upper/lower Riemann sums across all possible partitions), and using these bounds as a definition. It is definitely reasonable to think of the mesh shrinking, but I am not sure that it is necessary to explicitly state this.

EDIT: In response to the second part of the question, I am not sure the premise makes sense. Again, the definition of the Riemann integral uses the upper and lower integrals, which are themselves defined as supremums/infimums. Specifically, they are the supremum/infimums across the set of all partitions over an interval. As such, I do not know that it makes sense here to talk about 'tending' to zero, as it were. It is simply a set, and you could take an infinite subset of these partitions, and define a sequence of these elements where the mesh tends to zero. But you could also take an infinite subset of partitions where the mesh doesn't tend to zero. It makes no difference, because ultimately we are concerned with the supremum or infimum value across the entire set.

Because of what I explained above you can always find partitions with a smaller mesh that better approximate the supremum value. However this is not the same as the idea that we can control the size of the mesh. We simply define the set of ALL partitions over a given interval. In this way it doesn't make sense to think about what happens if the mesh does or doesn't tend to zero. Now it is different if we talk about approximations, then yes, choosing the size of the mesh will matter. But in terms of an actual integral, it doesn't make sense.

masiewpao
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