I know three equivalent definitions for the Riemann integral. Let $f:[a,b] \to \mathbb R$ a bounded function. We say $f$ is Riemann integrable with integral $I$ if either of the three following definition is satisfied:

  1. For all $\epsilon>0$ there exists $\delta >0$ such that if $\mathcal{P}$ is a partition of $[a,b]$ with $\mathrm{norm}(\mathcal{P}) < \delta$, then $|S(f,\mathcal{P}) - I| < \epsilon$.

  2. For all $\epsilon >0$ there exists a partition $\mathcal{P}$ of $[a,b]$ such that for any refinement $\mathcal{Q}$ of $\mathcal{P}$ we have $|S(f,\mathcal{Q})-I| < \epsilon$.

  3. We have that $\inf\{\overline{S}(f,\mathcal{P}) : \mathcal{P} \text{ partition of }[a,b] \} = \sup\{\underline{S}(f,\mathcal{P}) : \mathcal{P} \text{ partition of }[a,b] \}$ (the upper and lower Darboux integrals are equal).

What happens for improper integrals?

I'm thinking the ones where the interval is still bounded ($f:[a,b] \to \mathbb{R})$ but $f$ is not bounded.

For example, in that case the definition 3 doesn't work because the upper or lower Darboux integral may not be finite (and in this case i don't know if there's a work around to define it).

  • $\begingroup$ Your argument with definition 3 is correct. There is no workaround. Without explicity considering the supremum of the function on a subinterval as in definition (1) or (2) you can argue as below. $\endgroup$
    – RRL
    Commented Aug 26, 2020 at 20:09

1 Answer 1


Under definition (1) or (2) we can show that a function $f$ cannot be both unbounded and Riemann integrable.

This can be shown by producing an $\epsilon > 0$ such that for any real number $A$, no matter how fine the partition, there is a Riemann sum with

$$|S(f,P) - A| > \epsilon$$

Given any partition $P$, since $f$ is unbounded, it must be unbounded on at least one subinterval $[x_{j-1},x_j]$ of P. Using the reverse triangle inequality we have

$$|S(f,P) - A| = \left|f(t_j)(x_j - x_{j-1}) + \sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - A \right| \\ \geqslant |f(t_j)|(x_j - x_{j-1}) - \left|\sum_{k \neq j}f(t_k)(x_k - x_{k-1} - A \right|$$

Since $f$ is unbounded on $[x_{j-1},x_j]$, choose a partition tag $t_j$ such that

$$|f(t_j)| > \frac{\epsilon + \left|\sum_{k \neq j}f(t_k)(x_k - x_{k-1}) - A \right|}{x_j - x_{j-1}},$$

and it follows that no matter how fine the partition $P$ we have

$$|S(f,P) - A| > \epsilon.$$

Thus, when $f$ is unbounded, it is impossible to find $A$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(f,P) - A| < \epsilon$ holds. We can always select the tags so that the inequality is violated.


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