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In the definition of Riemann integral or Darboux-integral we first study partitions (or tagged partition) of the given interval determined by finitely many points. To each partition and a function $f$ we assign a number, which serves as an "approximation" of the integral. And the integral is then defined as some kind of limit of these approximations.

Would it be possible to define something similar with partitions consisting of infinitely many points instead of just finitely many? If there is some reasonable definition, would it bring something new, or do we just get another - and more complicated - definition of the same thing? Was something along these lines studied?


To make a bit clearer what I have in mind, let me add a few comments and some examples.

We want to have partition with infinitely many points. One thing we need is to know how to sum infinitely many values induced by the partition. But another thing is that the order of the points in the partition seems also to be important. (In the usual definition, we take finitely many points in the interval, but their ordering is also important. The Darboux sums and Riemann sums are defined by adding terms of the form $f(x)(a_{i+1}-a_i)$, so it is important to know which point is an immediate successor of $a_i$. This is clear for finitely many points, but less clear if we have an infinite set.)

  • Perhaps the simplest generalization would be to take a partition of the interval $[x,y]$ determined by an increasing sequence $(a_n)$ such that $a_0=x$ and $\lim\limits_{n\to\infty} a_n=y$. Then we could use infinite sums of the form $$\sum_{i=1}^\infty f(x_i)(a_{i+1}-a_i),$$ where $x_i$ belongs to the interval $[a_i,a_{i+1}]$. (My guess would be that this should give the usual Riemann/Darboux integral, at least for bounded functions. However, this definition is "not nice", since it treats the right endpoint of the interval differently from the left one. From a reasonable definition we would expect to be symmetric.)
  • We could do something like that for any ordinal $\alpha$. So the partition would be an increasing transfinite sequence $(a_\gamma)_{\gamma<\alpha}$. We probably also want $\lim\limits_{\beta<\gamma} a_\beta=a_\gamma$ for any limit ordinal $\gamma$. Sums of sequences defined by ordinals can be reasonably defined, see Wikipedia. So we could use something like $$\sum_{\gamma<\alpha} f(x_\gamma)(a_{\gamma+1}-a_\gamma).$$ The preceding example would correspond to $\alpha=\omega$.
  • There are also sums over arbitrary index sets. Such sums were discussed here or in some other posts linked there. But since here we simply take limit of sums of finite subsets, I think that this yields the same notion of integral, only described in a more convoluted way.
  • Of course, there may be many other approaches I did not think of.

I have tried to search whether similar question was asked in the past. The closest I was able to find was this: Lebesgue integration and countable partitions. This question mentions an approach to Lebesgue integral using countable partitions. Lebesgue integral is defined in this way in the book A Primer of Lebesgue Integration by Herbert Stanley Bear, see page 62. However, it is slightly different from what I described above. They use partition into countably many measurable sets, not countably many intervals. (But still, this relates to my questions, since it shows that somewhat similar approach can lead to Lebesgue integral.)

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    $\begingroup$ In the case of a countable partition, I suggest requiring that the sums be absolutely convergent. Any rearrangement of an absolutely convergent sum converges to the same value, so this allows you to remove the condition that the partition points accumulate at an endpoint. For bounded functions, I think you will get an integral equivalent to the Riemann integral. This is because I think (though I didn't check carefully) that any absolutely convergent sum corresponding to a countable partition can be approximated arbitrarily well by a sum corresponding to some finite partition. $\endgroup$ – LucasSilva Feb 3 '18 at 19:17

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