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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?

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  • $\begingroup$ 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 $\endgroup$
    – reuns
    Commented May 28, 2016 at 6:56
  • $\begingroup$ What's $p_\epsilon$? And why do we need that? Do you have a counter example to the statements? $\endgroup$
    – Argyll
    Commented May 28, 2016 at 6:58
  • $\begingroup$ $p_\epsilon$ 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 $\int_0^1 f(x)dx = 0$ $\endgroup$
    – reuns
    Commented May 28, 2016 at 6:59
  • $\begingroup$ same example as user342897 $\endgroup$
    – reuns
    Commented May 28, 2016 at 7:01
  • $\begingroup$ 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 $a$ $\endgroup$
    – reuns
    Commented May 28, 2016 at 7:02

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Dec 19, 2019 at 16:13
  • $\begingroup$ @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. $\endgroup$ Commented Dec 20, 2019 at 7:32
  • $\begingroup$ Thanks Tony, that's perfect. Please, do you know if it's still true for the Henstock-Kurzweil complete Riemann integral? $\endgroup$ Commented Dec 20, 2019 at 7:42
  • $\begingroup$ @FabricePautot Honestly I don't even know if there exists a Cousin's lemma for regular partitions. $\endgroup$ Commented Dec 20, 2019 at 16:00
  • $\begingroup$ Thanks Tony, it would be good to know. Any thought about my little problem please? I'm still looking for an elementary proof........................... $\endgroup$ Commented Dec 20, 2019 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.

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  • $\begingroup$ Nice. How about the first statement? $\endgroup$
    – Argyll
    Commented May 28, 2016 at 7:04
  • $\begingroup$ @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. $\endgroup$
    – user342897
    Commented May 28, 2016 at 7:29
  • $\begingroup$ Sorry, I am not following. Could you explain a little more please? $\endgroup$
    – Argyll
    Commented May 28, 2016 at 7:35
  • $\begingroup$ And I meant to work with only bounded functions. Would that change the answer for the first question? $\endgroup$
    – Argyll
    Commented May 30, 2016 at 17:12
  • $\begingroup$ @Argyll It would not. In fact, user342897 did not even mention unbounded functions at all. Compact intervals are bounded, necessarily. $\endgroup$
    – Angel
    Commented Oct 26, 2021 at 20:15

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