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Does the following condition for a bounded function $f: [a, b] \to \mathbb R$ suffice that it be Riemann integrable on $[a,b]$, with Riemann integral $I$?

For every $\epsilon > 0$, there exists a positive integer $N$ such that, for every positive integer $n \geq N$: for each $i = 1, 2, \dots, n$, for each number $c_{i} \in [x_{i-1}, x_{i}]$ — the $i$th subinterval in the $n$th regular partition of $[a, b]$ whose subintervals have length $\Delta x = (b-a)/n$ — we have $\left| \sum_{i=1}^{n} f\bigl(c_{i})\,\Delta x - I \right| < \epsilon$.

The condition may be stated equivalently as: For every $\epsilon > 0$, there exists a $\delta > 0$ such that, for every positive integer $n$ with $\Delta x = (b - a)/n < \delta$: for each $i = 1, 2, \dots, n$, for each number $c_{i} \in [x_{i-1}, x_{i}]$ — the $i$th subinterval in the $n$th regular partition of $[a, b]$ — we have $\left| \sum_{i=1}^{n} f\bigl(c_{i})\,\Delta x - I \right| < \epsilon$.

In short, do regular partitions suffice for Riemann integrability?

Notes:

  1. The usual condition for Riemann integrability involves arbitrary partitions, not just those that are regular, and then of course it involves the mesh of partitions being $< \delta$. Of course for a regular partition with $n$ subintervals, that mesh is $\Delta x = (b-a)/n$.

    The only thing I intend to change in the usual condition is to restrict partitions just to those that are regular.

  2. Note that the condition as stated allows the sample points to be chosen arbitrarily in the subintervals. It's well known that it is not enough to allow just endpoints.

murray
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2 Answers2

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Regular partitions are enough !

The problem is not banal: see

  1. Charles G. Denlinger Elements of Real Analysis (2011), p. 378 ;

  2. Jingcheng Ton Partitions of the interval in the definition of Riemann's integral Journal of Math. Educ. in Sc. and Tech. 32 (2001), pp. 788-793 (theorem 3).

It is sufficient to use any fixed sequence of partitions with the norm converging to zero (the fact is not much known).

Tony Piccolo
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  • I'm not sure what the final sentence means. Does it mean the following? $f$ is integrable if: there exists some sequence of partitions $P_n$, with mesh tending to 0, for which, no matter what sample points in the subintervals of $P_n$ are used, the corresponding sequence of Riemann sums converges, and that limit is independent of the sample points chosen. (I knew the problem wasn't banal; I just didn't say what I meant the first time 'round.) – murray Jan 03 '15 at 01:53
  • I cannot access reference #2. For reference #1, I cannot access the relevant section beyond page 372. So would you mind quoting the theorem(s) to which you refer? And note that the Riemann-Darboux criterion in Theorem 2 of #2 involves controlling the difference between the Riemann sums using the maximum value and minimum value, respectively, of $f$ in each subinterval; but that's not the condition I asked about. – murray Jan 03 '15 at 02:04
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    I just obtained full-text access to reference #2, where I could read the cited Theorem 3. And yes, the condition I asked about in my preceding comment is indeed the one shown there to suffice. Thanks! – murray Jan 03 '15 at 02:12
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Take the following Dirichlet's function:

$$D(x):=\begin{cases}1\;,\;\;x\in\Bbb Q\cap [0,1]\\{}\\0\;,\;\;x\in\Bbb Q^c\cap [0,1]\end{cases}$$

Then, for any $\;\epsilon>0\;$ and any $\;\delta>0\;$ there exists an irrational number in any non-trivial subinterval of $\;[0,1]\;$ , and thus the Riemann sum you wrote is zero (and thus $\;I=0\;$ in this case), which would mean the above function is Riemann integrable in the unit interval...but it isn't, of course.

The problem here is, imo, not the regular partitions you mention but in fact that you require only "the existence" of some points $\;c_i\;$ in those regular subintervals, whereas the usual definition requires that the sum exists in the limit for any choice of these points $\;c_i\;$ in the subintervals of the partition, as you mention at the very end of your question.

Timbuc
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  • Sorry, I meant to allow arbitrary choice of sample points in the subintervals, not just existence of sample points there. I've edited to indicate that. Yes, of course the characteristic function of the rationals on the unit interval is a counterexample to what I originally wrote; in fact, as I had in mind in my 2nd note, it's the well-known counterexample to allowing only endpoints as sample points. – murray Jan 02 '15 at 16:50
  • Yeah, well: now it is a rather different question, but then I think it is easy to show the equivalence with Riemann integrability, since for any partition there exists a regular one which subpartitions it (for example, take the minimum of all the original subintervals' lengths). – Timbuc Jan 02 '15 at 18:01
  • Taking the min $m$ of the non-regular partition's lengths is what I immediately thought of before I asked the question. The trouble is that you won't necessarily get a regular partition from it: $\Delta x = (b-a)/m$ need not be an integer. So the final subinterval need not have the same length as the others. – murray Jan 02 '15 at 20:08