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Let $f:[a,b]\to\mathbb{R}$ be a function, and $P_n$ the equidistance partition of $[a,b]$ into $n$ subintervals of an equal length. Let $P_n^\ast$ be the set of sample points from each subinterval of $P_n$, and $S(f,P_n,P_n^\ast)$ the Riemann sum of $f$ constructed using $P_n$ and $P_n^\ast$. In Stewart's calculus book, the (Riemann) integrability of $f$ is defined as follows:

There exists a real number $L$ such that, for any $\epsilon>0$, there exists $N\in\mathbb{Z}$ such that if $n\geq N$, then $$\left|S(f,P_n,P_n^\ast)-L\right|<\epsilon$$ for any choice of $P_n^\ast$.

Note that this condition is stronger than the following condition:

There exists a real number $L$ such that the sequence $\{S(f,P_n,P_n^\ast)\}$ of Riemann sums converges to $L$ for any choice of the sequence $\{P_n^\ast\}$.

Presumably the latter condition is not sufficient to guarantee the (Riemann) integrability of $f$. Does anyone know an example of a non-integrable function satisfying the latter condition?

ashpool
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    Here it is shown that if $P_n={a+\tfrac{k(b-a)}{n}:0\leq k\leq n}$, a bounded function $f$ on $[a,b]$ is Riemann integrable iff $$\lim_{n\rightarrow\infty}\frac{b-a}{n}\sum^n_{k=1}f(t_{nk})$$ exists and is independent of the tags $\tau_n={t_{nk}:1\leq k\leq n}$ with $a+\frac{(k-1)(b-a)}{n}\leq t_{nk}\leq a+\tfrac{k(b-a)}{n}$. The common value $I=\int^b_af$. – Mittens Jun 29 '24 at 16:48
  • @Mittens I think your statement is not completely accurate: As I mentioned in my post, there are two ways to interpret the convergence of the Riemann sums "independent of the tags": 1) For any choice of the tags, the sequence of Riemann sums converges. 2) For any $\epsilon>0$, there exists an integer $N$ such that if $n\geq N$, then the Riemann sum $S(f,P_n,\tau_n)$ is within $\epsilon$ for any choice of $\tau_n$. I am aware that the Riemann integrability is equivalent to the second condition. My question is, is there a function satisfying the first condition but not the second? – ashpool Jun 30 '24 at 03:53
  • My statement is a theorem and the proof is contained in my old posting. Your statements are the same. – Mittens Jun 30 '24 at 14:16
  • @Mittens Unless I'm mistaken, you seem to use the second interpretation in your posting. By "the same", do you mean equivalent? – ashpool Jul 01 '24 at 00:16
  • @Mittens No, they are not the same; the difference between the two conditions is similar to the pointwise convergence and uniform convergence of functions. Note that "$f_n(x)$ converges to $L$ for any choice of $x$" is different from "For any $\epsilon>0$, there is $N$ such that if $n\geq N$, then $|f_n(x)-L|<\epsilon$ for all $x$. – ashpool Jul 02 '24 at 02:21
  • both imply convergence of the Riemann sum over the uniform partition. Also, by using Darboux criteria (difference of Upper and lower sums) you see that there is nothing misterious. – Mittens Jul 02 '24 at 02:46
  • @Mittens Darboux criteria is equivalent to the second version. I'm not sure if the first version implies the second, though. – ashpool Jul 03 '24 at 00:06
  • your first notion implies Darboux; in turn, Darboux implies the second notion; the second is equivalent to Riemann integrability. – Mittens Jul 03 '24 at 00:09
  • I'm awfully sorry, I just realized that the "first" and "second" notions in my original posting are switched in my second comment above... – ashpool Jul 04 '24 at 00:36

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