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If the left Riemann sum of a function over uniform partition converges, is the function integrable?

To put the question more precisely, let me borrow a few definitions first. Pardon my use of potentially non-canon definitions of convergence. Given a bounded function $f:\left[a,b\right]\to\mathbb{R}$,

  • A partition $P$ is a set $\{x_i\}_{i=0}^{n}\subset\left[a,b\right]$ satisfying $a=x_0\leq x_1\leq\cdots\leq x_n=b$.
  • The norm of a partition $\newcommand\norm[1]{\left\lVert#1\right\rVert}\norm{P}:=\max_{0\leq i\leq n}|x_i-x_{i-1}|$
  • The left Riemann sum of $f$ over partition $P$ is $l(f,P):=\sum_{i=1}^nf(x_{i-1})(x_i-x_{i-1})$
  • The left Riemann sum of $f$ is said to converge to $L$ iff $\newcommand\norm[1]{\left\lVert#1\right\rVert}\forall\epsilon>0, \exists\delta>0:\norm{P}<\delta$ implies $\left|l(f,P)-L\right|<\epsilon$
  • A uniform partition $P_n$ of $n$ divisions is defined by $x_i=a+\frac{b-a}{n}i$
  • The left Riemann sum of $f$ over uniform partitions is said to converge to $L$ iff $\forall\epsilon>0, \exists N\in\mathbb{N}: n\geq N\implies \left|l(f,P_n)-L\right|<\epsilon$

Now, is the following statement true?

If the left Riemann sum of $f$ converges to $L$, $f$ is Riemann integrable and its Riemann integral $\int_a^b f$ equals $L$.

In particular, I am curious whether the following limited case is true.

If the left Riemann sum of $f$ over uniform partitions converges to $L$, $f$ is Riemann integrable and its Riemann integral $\int_a^b f$ equals $L$.

My hunch is that the statements above are not true. But I can't come up with a counter example. Can someone give me some help here please?

Argyll
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  • you need $|l(f,p_\epsilon) - L| \to 0$ as $\epsilon \to 0$ for any partition of $[a,b]$ with $\max |x_i-x_{i-1}| < \epsilon$, and not only the rationals partitions. if $|x_i-x_{i-1}|$ is constant you get the standard Riemann integral, if it is not you get the Riemann–Stieltjes_integral – reuns May 28 '16 at 06:56
  • What's $p_\epsilon$? And why do we need that? Do you have a counter example to the statements? – Argyll May 28 '16 at 06:58
  • $p_\epsilon$ is a partition of $[a,b]$ as you wrote with $\epsilon$ the $\max_i |x_i - x_{i-1}|$. the counter-example is with $f(x) = 1$ if $x$ is rational, $0$ otherwise. if you consider only the rationals partitions you get $\int_0^1 f(x) dx = 1$, if you consider only the irrationals partitions you get $\int_0^1 f(x)dx = 0$ – reuns May 28 '16 at 06:59
  • same example as user342897 – reuns May 28 '16 at 07:01
  • this is indeed equivalent to the assertion : $g(n) \to 0$ as $n \to \infty, n \in \mathbb{N}$ doesn't mean that $g(x) \to 0$ as $x \to \infty, x \in \mathbb{R}$. you can't consider only the value of $|l(f,P_a)- L|$ at integers values of $a$ – reuns May 28 '16 at 07:02
  • So the example proves the second statement about uniform partition is not true. And nice connection. How about the first one? – Argyll May 28 '16 at 07:03
  • uniform partition is the standard Riemann integral, when you allow any partition, it is the Riemann-Steltjes integral, and again you can't look only at integers values of $a$ in $|l(f,P_a-L)|$ (except if $f$ is continuous, of course) – reuns May 28 '16 at 07:06
  • So the fact that the sample points are on the left ends of the divisions is not a problem? – Argyll May 28 '16 at 07:08
  • And would the statement be true for piecewise continuous function? Thank you for all the clarifications. – Argyll May 28 '16 at 07:08
  • what do you think ? – reuns May 28 '16 at 07:09
  • I think it would hold – Argyll May 28 '16 at 07:10
  • It is an old story: see this question. Left or right doesn't matter of course.. – Tony Piccolo May 29 '16 at 05:43
  • Why is that related? – Argyll May 29 '16 at 09:33
  • Read Gillespie's famous article. – Tony Piccolo May 29 '16 at 17:11
  • As to the the use of regular partitions, they are enough to define equivalently Riemann integral but one has problems with proving theorems as the additivity on intervals. – Tony Piccolo May 29 '16 at 17:25
  • The use of only left sums (or only right sums) with only regular partitions doesn't work. – Tony Piccolo May 29 '16 at 17:47
  • thanks for the reference. that clarifies what "Cauchy integral" means, though I find most people use that term for complex line integrals. I still dont understand your. I understand that assuming Riemann integrability, any Riemann sum over any partition will approach Riemann integral. Without integrability known first, what happens to the first statement? – Argyll May 30 '16 at 03:14
  • Be patient, next hours I'll write an answer. – Tony Piccolo May 30 '16 at 04:52
  • I will! I'll look forward to it! – Argyll May 30 '16 at 05:17
  • @reuns No, the definition of the Riemann integral does not use uniform partitions, it uses arbitrary partitions. The Riemann-Stieltjes integral is also not defined the way you say it is defined. The Riemann-Stieltjes integral is defined with respect to an integrator $g$, and $x_{i+1}-x_i$ is replaced by $g(x_{i+1})-g(x_i)$ in the definition. The partitions themselves have nothing to do with the definition. – Angel Oct 26 '21 at 20:13

2 Answers2

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In this context "Cauchy integral" has the meaning you know.

It is a fact that if a function is bounded and Cauchy integrable over $[a,b]$, then it is also Riemann integrable over that interval.

It seems that there is no elementary proof of this theorem.
The proof in Kristensen, Poulsen, Reich A characterization of Riemann-Integrability, The American Mathematical Monthly, vol.69, No.6, pp. 498-505, (theorem 1), could be considered elementary because plays only with Riemann sums but is an indigestible game.

Note that there exist unbounded functions Cauchy integrable.

Also the use of regular partitions is enough to define Riemann integral.
See Jingcheng Tong Partitions of the interval in the definition of Riemann integral, Int. Journal of Math. Educ. in Sc. and Tech. 32 (2001), 788-793 (theorem 2).

I repeat that the use of only left (or right) Riemann sums with only regular partitions doesn't work.

Tony Piccolo
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  • Dear Tony. I just asked this question: Limit of the ratio of two non-Riemann sums. Then I fell on your answer. Tong's theorem 2 might be what I'm looking for but I have no access to his paper (requested on ReseachGate). Please, may you be king enough to confirm or just to state the theorem? THANKS. – Fabrice Pautot Dec 19 '19 at 16:13
  • @FabricePautot The statement is: $f$ is Riemann integrable iff $,\lim_{n\to\infty} \frac {b-a}n \sum_{i=0}^{n-1} f(t_i),$ exists for every selection of the tags and is independent of it. – Tony Piccolo Dec 20 '19 at 07:32
  • Thanks Tony, that's perfect. Please, do you know if it's still true for the Henstock-Kurzweil complete Riemann integral? – Fabrice Pautot Dec 20 '19 at 07:42
  • @FabricePautot Honestly I don't even know if there exists a Cousin's lemma for regular partitions. – Tony Piccolo Dec 20 '19 at 16:00
  • Thanks Tony, it would be good to know. Any thought about my little problem please? I'm still looking for an elementary proof........................... – Fabrice Pautot Dec 20 '19 at 16:21
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No, one example is the function $$f(x) = \begin{cases} 1: & \; x \notin \mathbb{Q}; \\ 0: & x \in \mathbb{Q} \end{cases}$$ for which the left Riemann sum for any uniform partition of $[0,1]$ is always zero since $$f(a + (b-a)i/n) = 0$$for all integers $i$. This Riemann sum gives very little useful information about $f$ and in fact $f$ is not Riemann integrable at all.

user342897
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  • Nice. How about the first statement? – Argyll May 28 '16 at 07:04
  • @Argyll Good question. The other answer can be modified to work on a compact interval $[a,b]$ as in the question, showing that the answer to the first question is also "no". If you generalize to "improperly Riemann integrable" I'm not sure what the answer will be. – user342897 May 28 '16 at 07:29
  • Sorry, I am not following. Could you explain a little more please? – Argyll May 28 '16 at 07:35
  • And I meant to work with only bounded functions. Would that change the answer for the first question? – Argyll May 30 '16 at 17:12
  • @Argyll It would not. In fact, user342897 did not even mention unbounded functions at all. Compact intervals are bounded, necessarily. – Angel Oct 26 '21 at 20:15