Suppose $a, b \in \mathbb{R}$ s.t. $a \leq b$ and $f: [a, b] \rightarrow \mathbb{R}$ is a bounded function.


  • $\Delta = \{x_i\}_{i=0}^n$ is a partition of $[a, b]$ if $a = x_0 < x_1 < \ldots < x_n = b$ for some $n \in \mathbb{N}$
  • $|\Delta| = \max_{i \in \{0, \ldots, n-1\}} (x_{i+1} - x_i)$
  • $\{\xi_i\}_{i=0}^{n-1}$ are representative points of $\Delta$ if $\xi_i \in [x_i, x_{i+1}]$ for all $i \in \{0, \ldots, n-1\}$

Then consider the following two definitions of Riemann Integral:

Definition 1

$f$ is Riemann integrable on $[a, b]$ if $\exists S \in \mathbb{R}, \forall \epsilon > 0, \exists \delta > 0$ s.t.

$$ \forall \text{partition: } \Delta \text{ s.t. } |\Delta| < \delta, \forall \text{representative points: }\{\xi_i\}_{i=0}^{n-1}, \left|S - \sum_{j=0}^{n-1}(f(\xi_j)(x_{j+1} - x_j))\right| < \epsilon $$

Definition 2

$f$ is Riemann integrable on $[a, b]$ if $\exists S \in \mathbb{R}, \forall \epsilon > 0, \exists \delta > 0$ s.t.

$$ \exists \text{partition: } \Delta \text{ s.t. } |\Delta| < \delta, \forall \text{representative points: }\{\xi_i\}_{i=0}^{n-1}, \left|S - \sum_{j=0}^{n-1}(f(\xi_j)(x_{j+1} - x_j))\right| < \epsilon $$

The only difference is the way to take a partition. I think the standard definition is 1, but is there an example $f$ that is Riemann Integrable by definition2 and not by definition1?


It is obvious that $$\{f \ | \ \text{integrable by definition 1}\} \subseteq \{f \ | \ \text{integrable by definition 2}\}$$ so my question is whether $$\{f \ | \ \text{integrable by definition 1}\} \overset{?}{\supseteq} \{f \ | \ \text{integrable by definition 2}\}$$ holds or not.

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    $\begingroup$ the definitions are equivalent. $\endgroup$ – Veridian Dynamics Sep 14 '17 at 16:39
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    $\begingroup$ The essetial idea is that there should some degree of freedom in choosing partition as well as the representative points. If we chose one partition with norm less than $\delta$ and all possible representative points the definition works. If we make a specific choice of representatives (say left end points of each subinterval) then we need to deal with all partitions with norm less tha $\delta $. $\endgroup$ – Paramanand Singh Sep 14 '17 at 18:55

Actually we can weaken Definition 2 and still have equivalence with Definition 1. Definition 3: $f$ is Riemann integrable on $[a,b]$ if there exists $S\in \mathbb R$ such that for all $\epsilon>0,$ there exists a partition $\Delta$ such that

$$|S - \sum_{j=0}^{n-1}f(\xi_j)(x_{j+1}-x_j)|<\epsilon$$

for all representative points $\xi_j.$ Definition 3 implies $U(\Delta,f)-L(\Delta,f)\le 2\epsilon,$ where $U(\Delta,f),L(\Delta,f)$ are the Darboux sums of $f$ relative to $\Delta.$ It is well known that the Darboux formulation and Definition 1 are equivalent. So we actually obtain Definition 1 $\implies$ Definition 2 $\implies$ Definition 3 $\implies$ Darboux $\implies $ Definition 1.

  • $\begingroup$ It's interesting... thanks! $\endgroup$ – myuuuuun Sep 14 '17 at 17:08
  • $\begingroup$ $U-L = |U-L| = |(U-S)+(S-L)| \leq |S-U| + |S-L| < 2\epsilon$ $\endgroup$ – myuuuuun Sep 14 '17 at 17:10
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    $\begingroup$ You need to be careful there. $U$ itself may not arise as a sum. However it is the limit of such sums. Hence $\le$ rather than $<.$ $\endgroup$ – zhw. Sep 14 '17 at 17:23

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