# Equivalence of three definitions of Riemann integral for improper integrals.

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).

• 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.
– RRL
Commented Aug 26, 2020 at 20:09

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.