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If we replace ( the standard definition of the Riemann integral)

Definition 1. $$\int_a^b f(x)\,dx:=\lim_{\lambda \to 0} \sum_{j=1}^{j=n}f(\xi_j)\Delta x_j $$ with $a=x_0<x_1<x_2<\dots<x_{n-1}<x_n=b, \Delta x_j:=x_{j}-x_{j-1},j=1,\dots,n$,$\lambda:=\max_{1\le j\le n} \Delta x_j,$ and $\xi_j$ being an arbitrary point of the closed interval $[x_{j-1},x_j]$ by

Definition 2. $$\int_a^b f(x)\,dx:=\lim_{ \lambda \to 0} \sum_{j=1}^{j=n}f(x_j)\Delta x_j $$ (pay your attention to $f(x_j)$ instead of $f(\xi_j)$),is the same set of integrable functions obtained? It's clear that definition 1 implies definition 2. It is not clear whether definition 2 implies standard definition 1.

A proof, a counterexample, and a reference are welcome.

Many thanks from me to @Matematleta for the references to the answers. For completeness, I know how to prove the equivalence of definition 1 and

Definition 3. $$\int_a^b f(x)\,dx:=\lim_{n \to \infty} \sum_{j=1}^{j=n}f(\xi_j)\frac {b-a} n $$ with $\xi_j$ being an arbitrary point of the closed interval $[a+(j-1)(b-a)/n,a+j(b-a)/n]$ and it is not difficult.

I don't think that is new. A reference is welcome.

user64494
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  • @Mathematleta: The definitions in the linked question differ from definition 2. Thank you anyway. – user64494 Nov 24 '19 at 16:42
  • @Mathematleta: Can you kindly elaborate you comment as an answer, grounding your statement? TIA. – user64494 Nov 24 '19 at 16:51
  • If the function is integrable, then the sum of infimum and superior sums is the same, and the $\xi$ are by definition between them. – Matthias Nov 24 '19 at 17:35
  • @Mathias: All we know is the existence of the limit $$\lim_{\lambda \to 0} \sum_{j=1}^{j=n}f(x_j)\Delta x_j .$$ Does this imply the Riemann integrability of the function $f(x)$ on $[a,b]$? – user64494 Nov 24 '19 at 17:42
  • I misread your question sorry! . The answer is yes, the definitions are equivalent, but the proof is non-trivial. – Matematleta Nov 24 '19 at 22:39

1 Answers1

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Well, if we define $f(x) = \begin{cases} 1 & \; x \notin \mathbb{Q}\cap [0,1] \\ 0 & x \in \mathbb{Q}\cap[0,1]\end{cases}$

then, $f$ is not Riemann integrable, because its set of discontinuities has measure $1>0$.

And yet, the points in $\textit{any}$ uniform partition $P_n=\{x_k\}^n_{k=1}$ of $[0,1]$ are rational so $f(x_k)=0$ and so the right-endpoint Riemann sum is $0$ for all such partitions, and clearly $|P_n|\to 0$ as $n\to \infty.$

Matematleta
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  • But then you have tested only rational points partition, but OP seemed to assume that every partition that goes to zero. – user284331 Nov 24 '19 at 20:38
  • Yes, I misread the OPs question. The answer is yes, the definitions are equivalent, but the proof is non-trivial – Matematleta Nov 24 '19 at 22:38
  • Anyway, the proof that you provided includes a counter-example, and it is exactly my concern about the boundedness around the initial point. – user284331 Nov 24 '19 at 22:47
  • Yes, boundeness is definitely necessary. – Matematleta Nov 24 '19 at 22:48
  • Okay now I see what's wrong with my proof: The so called left endpoint version on $[a,b]$ is actually not equal to the right endpoint version on $[a-1,b]$. The former requires grid must include the point $a$ but the later $a$ is not an endpoint of $[a-1,b]$ and hence is not required to be in any partition. – user284331 Nov 24 '19 at 23:12
  • If you know that the set of discontinuities of a non-Riemann integrable function has measure >0, the proof is easier.See, for example ([G] D. C. Gillespie, The Cauchy definition of a definite integral, Annals of Mathematics (2) 17 (1915), 61–63.) – Matematleta Nov 25 '19 at 01:32
  • @Matematleta: Many thanks from me to you for your references. Can you make those in your answer, (not comment of you) to close the question? TIA. – user64494 Nov 26 '19 at 06:39