A function $f: [a,b] \to \mathbb{R}$ is called Riemann integrable, with integral $K$, if the following condition is satisfied:

For every $\epsilon > 0$, there is some $\delta > 0$, such that for all partitions $P$ with $\Vert P \Vert < \delta$ and for all tags $T$ of $P$, we have $|S(f,P,T)- K| < \epsilon$.

($S(f,P,T)$ is a Riemann sum)

I don't understand intuitively why we need to include the tag requirement in the definition, and I have two questions:

(1) Can someone explain why we intuitively need this? Can't we just capture the behaviour of the function in the partitions point or in the middle of it, by letting the partitions vary? (See question (2) for an alternative definition, which I believe is weaker)

(2) Suppose I would adapt the definition:

For every $\epsilon > 0$, there is some $\delta > 0$, such that for all partitions $P$ with $\Vert P \Vert < \delta$, we have $|S(f,P)- K| < \epsilon$

where $P = (x_0, \dots x_n); S(f,P):= \sum_{i=1}^n f((x_i+x_{i-1})/2) (x_i-x_{i-1})$

Can you give me a function that is integrable in this sense but not Riemann integrable?

  • $\begingroup$ You may find Darboux sums interesting: en.wikipedia.org/wiki/Darboux_integral#Darboux_sums $\endgroup$ – paf Aug 26 '18 at 9:59
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    $\begingroup$ Yes, I'm familiar with Darboux sums, but how it's relevant? $\endgroup$ – user370967 Aug 26 '18 at 9:59
  • $\begingroup$ Your second definition is equivalent to the first one provided $f$ is bounded, but proving this is not easy. In general you can keep one of the things (out of partition and tags) as fixed and let the other vary and you get Riemann integral. But if you fix both tag and partition then the definition breaks down. $\endgroup$ – Paramanand Singh Aug 26 '18 at 10:02
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    $\begingroup$ @Math_QED: Because it's an equivalent definition of Riemann integral where you don't have this 'tags' problem. $\endgroup$ – paf Aug 26 '18 at 10:02
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    $\begingroup$ +1 for raising a point which is not generally given much attention in textbooks. $\endgroup$ – Paramanand Singh Aug 26 '18 at 10:06

You can generalize your "weaker" definition as follows.

Let $\mathcal{I}$ denote the set of closed intervals $[a,b]$ and let $\tau : \mathcal{I} \to \mathbb{R}$ be any function such that $\tau([a,b]) \in [a,b]$.

Examples are $\tau([a,b]) = a$, $\tau([a,b]) = b$, $\tau([a,b]) = \frac{a + b}{2}$ etc.

Define $S(f,P,\tau) = \sum_{i=1}^n f(\tau([x_{i-1},x_i])) (x_i-x_{i-1})$. Let us say that $f$ is $\tau$-integrable with $\tau$-integral $K$ if the obvious condition is satisfied.

As pointed out by Paramanand Singh, the following are equivalent for a bounded function $f$:

(1) $f$ is Darboux integrable

(2) $f$ is Riemann integrable

(3) $f$ is $\tau$-integrable for all $\tau$

(4) $f$ is $\tau$-integrable for some $\tau$

$\tau$-integrability seems to be conceptually simpler than Riemann integrability because it avoids to use tags. But it should be clear that there is a substantial arbitrariness in the choice of $\tau$. It can be defined by a simple rule as in your question, but it can also be "erratic". It may even depend on $f$ if you want. For example, if $f$ is continuous, then you can take $\tau([x_{i-1},x_i]))$ to be any point of $[x_{i-1},x_i]$ at which $f \mid_{[x_{i-1},x_i]}$ attains its minimum (or maximum).

So what might be the benefit of general tags? Here are some arguments.

a) If you know that $f$ is integrable (for example if $f$ is continuous or monotonic), then you can choose suitable tags $T$ which make $S(f,P,T)$ explicitly evaluable. For example, you can prove that $\int_0^t \frac{1}{1 + x^2}dx = \arctan t$ by choosing a tag as $\xi_i \in [x_{i-1},x_i]$ such that $\frac{1}{1 + \xi_i^2} = \frac{1}{1 + x_{i-1}x_i}$. I shall not go into details and I do not claim that this is an elegant proof, but it shows that general tags can be useful.

b) If you have integrable functions $f,g$, you know that their product $fg$ is integrable. Then you may choose any two $\xi_i, \xi'_i \in [x_{i-1},x_i]$ and consider the sums

$$\Sigma_{i=1}^n f(\xi_i)g(\xi'_i)(x_i - x_{i-1}) .$$

These converge to $\int_a^b f(x)g(x)dx$ as was shown by G.A. Bliss.

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  • $\begingroup$ Thanks. I know about the equivalence between Riemann and Darboux integral. Could you provide a reference for the equivalence with $\mathcal{T}$-integrability? $\endgroup$ – user370967 Aug 26 '18 at 16:36
  • $\begingroup$ In particular, I see that $(1) \iff (2) \implies (3) \implies (4)$ so a reference to the proof of the missing implication would be useful. $\endgroup$ – user370967 Aug 26 '18 at 16:48
  • $\begingroup$ @Math_QED I know that it is true, but I a do not know where to find a good reference .In the comment by Calvin Khor you find a link to math.stackexchange.com/q/326197 which provides references to the special case $\tau([a,b]) = b$. Surely the proof generalizes to arbitrary $\tau$. Alternatively given any $\tau$, you can refine the partition $P$ to a new partition $P'$ so that $\tau$ induces a "right endpoint" $\tau'$. $\endgroup$ – Paul Frost Aug 26 '18 at 17:29
  • $\begingroup$ Thanks for your answer. It's always a pleasure reading your answers. I am relieved that my intuition seems correct: the tags in the partition definition don't matter much. $\endgroup$ – user370967 Aug 26 '18 at 19:26
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    $\begingroup$ For the last point about product of Riemann integrable functions see math.stackexchange.com/a/1944665/72031 $\endgroup$ – Paramanand Singh Aug 27 '18 at 2:53

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