I have a serious concern regarding the proof of the Cauchy criterion implying the Riemann integrability
$\text{Cauchy Criterion}:$ $\space \space$ A function: $[a,b] \to \mathbb{R}$ belongs to $\mathcal{R}[a,b]$ if and only if for every $\epsilon>0$ there exists $n_\epsilon>0$ such that if $\dot P$ and $\dot Q$ are any tagged partitions of $[a,b]$ with $||\dot P||<n_\epsilon$ and $||\dot Q||<n_\epsilon,$ then $$|S(f;\dot P)-S(f;\dot Q)|<\epsilon$$ The forward implication is clear.
$(\Leftarrow)$ For each $n\in \mathbb{N},$ let $\delta_n>0$ be such that if $\dot P$ and $\dot Q$ are tagged partitions with norms $<\delta_n,$ then $$|S(f;\dot P)-S(f;\dot Q)|<\frac{1}{n}.$$ Evidently we may assume that $\delta_n\ge \delta_{n+1}$ for $n\in \mathbb{N};$ otherwise, we replace $\delta_n$ by $\delta'_n:=\min\{\delta_1,..,\delta_n\}.$
$\space$ For each $n\in \mathbb{N},$ let $\dot P_n$ be a tagged partition with $||\dot P_n||<\delta_n$. Clearly, if $m>n$ then both $\dot P_m$ and $\dot P_n$ have norms $<\delta_n$, so that $$|S(f;\dot P_n)-S(f;\dot P_m)|<\frac{1}{n} \space \space \space ,m>n \tag1$$ And the proof goes ..

How can we assume that $\delta_n\ge \delta_{n+1}?$ I get the whole point of assuming it because only then can we get $(1)$ as a Cauchy sequence. But how can we assume it?.. What would go wrong if $\delta_n<\delta_{n+1}?$ Also this whole proof is implying that $\epsilon \to 0$ implies $\delta \to 0$. Thinking about this graphically makes sense, but nowhere in the definition is this obvious to me.

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    $\begingroup$ Very good question. This criterion of integrability is not very popular and I am happy that someone cares about it. +1 $\endgroup$
    – Paramanand Singh
    Jul 24, 2016 at 12:55
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    $\begingroup$ if $\delta_{n} < \delta_{n + 1}$ then it would be somewhat difficult to show that $S(f; P_{n})$ forms a Cauchy sequence. And it is not necessary that $\delta_{n} \to 0$ as $n \to \infty$. Only $\delta_{n}$ should decrease. For a constant function $f$ any Riemann sum has the same value and the norm of partition need not tend to $0$. $\endgroup$
    – Paramanand Singh
    Jul 24, 2016 at 13:04
  • $\begingroup$ thank you sir, noted your point @ParamanandSingh $\endgroup$
    – Bijesh K.S
    Jul 26, 2016 at 15:22

1 Answer 1


I'll answer your question about how we can assume $\delta_{n} \geq \delta_{n+1}$. (You say that you understand why we want it to be true.)

Let's fix $n$ for the moment. Let's say we already found $\delta_1,\ldots, \delta_n$. According to the hypothesis, we can find $\delta_{n+1}^{\ast} > 0$ such that $$ |S(f,P) - S(f,Q)| < \frac{1}{n+1} $$ for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}^{\ast}$ and $\|Q\| < \delta_{n+1}^{\ast}$.

I used the notation $\delta_{n+1}^{\ast}$ because it is our first try for $\delta_{n+1}$. It could be that $\delta_{n+1}^{\ast}$ is larger than $\delta_{n}$.

To fix that, for our second attempt we define $$ \delta_{n+1} = \min(\delta_n,\delta_{n+1}^{\ast}). $$ Let's check that $\delta_{n+1}$ does what we want.

Clearly $\delta_{n} \geq \delta_{n+1}$. So that's taken care of.

We also want $\delta_{n+1}$ to have the property that $$ |S(f,P) - S(f,Q)| < \frac{1}{n+1} $$ for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}$ and $\|Q\| < \delta_{n+1}$. Let's check it.

From the definition of $\delta_{n+1}$, it's clear that $\delta_{n+1}^{\ast} \geq \delta_{n+1}$. So for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}$ and $\|Q\| < \delta_{n+1}$, we have $\|P\| < \delta_{n+1}^{\ast}$ and $\|Q\| < \delta_{n+1}^{\ast}$, and therefore (by the original property of $\delta_{n+1}^{\ast}$) we have $$ |S(f,P) - S(f,Q)| < \frac{1}{n+1}. $$

  • $\begingroup$ You absolutely convinced me on the first part, great job! If you could answer the question completely this could be an accepted answer, thanks @RitterSport $\endgroup$
    – Bijesh K.S
    Jul 24, 2016 at 7:01
  • $\begingroup$ Just to clarify, your second question is: What would go wrong if $\delta_n < \delta_{n+1}?$ @BijeshK.S $\endgroup$ Jul 24, 2016 at 22:18
  • $\begingroup$ never mind, one of the comments made this clear. Thank you $\endgroup$
    – Bijesh K.S
    Jul 26, 2016 at 15:23
  • $\begingroup$ You're welcome. :) $\endgroup$ Jul 26, 2016 at 22:40

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