Abel's test for improper integrals with only integrable functions I know this following formulation of Abel's test for improper integrals:

Let $f,g:[a,\infty)\to \mathbb{R}$ be continuous functions, where

*

*$\int_a^\infty f(t)dt$ converges.

*$g$ is monotone decreasing, bounded and continuously differentiable. 
Then $\int_a^\infty f(t)g(t)dt$ converges.


The proof I know uses integration by parts, so I am not sure if one can strengthen it, but I'm looking for counter-examples of stronger versions. More specifically, when $f$ is only Riemann integrable on any compact interval, but I currently can't think of any.
 A: I will assume that OP mistakingly forgot to include the boundedness of $g$. Then the proof using integration by parts carries over to a more general case with due modification:

Let $f, g : [a, \infty) \to \mathbb{R}$ be functions such that the following holds:

*

*$f$ is locally integrable and $\lim_{b\to\infty} \int_{a}^{b} f(x) \, \mathrm{d}x$ converges.


*$g$ is bounded and non-decreasing.
Then $\lim_{b\to\infty} \int_{a}^{b} f(x)g(x) \, \mathrm{d}x$ converges.

For the proof, let $F(x) = \int_{a}^{x} f(t) \, \mathrm{d}t$. Then
$$ \int_{a}^{b} f(x)g(x) \, \mathrm{d}x
= F(b)g(b) - \int_{a}^{b} F(x) \, \mathrm{d}g(x), $$
where the integral in the right-hand side is regarded as a Riemann-Stieltjes integral. Indeed, this holds when $f$ is continuous and then extends to a general case by approximating $f$ by continuous functions in $L^1$-norm on $[a, b]$. Then by the assumption, we note the following:

*

*Since $g(b)$ converges as $b\to\infty$ by the assumption, the term $F(b)g(b)$ also converges as $b\to\infty$.


*Since $F$ is bounded, we may choose a bound $M$ of $f$. Then for any $ a \leq b \leq c$,
$$ \left| \int_{a}^{c} F(x) \, \mathrm{d}g(x) - \int_{a}^{b} F(x) \, \mathrm{d}g(x) \right|
\leq M|g(c) - g(b)|. $$
This shows that the net $\{ \int_{a}^{b} F(x) \, \mathrm{d}g(x) \}_{b \in [a,\infty)}$ is Cauchy and hence converges.
This completes the proof.
A: Weak sufficient conditions are that $\int_a^\infty f(x) \, dx$ converges and $g$ is bounded and monotone (decreasing or increasing).  By the second mean value theorem for integrals, there exists $\xi \in [c_1,c_2]$ such that
$$\int_{c_1}^{c_2} f(x) g(x) \, dx = g(c_1) \int_{c_1}^\xi f(x) \, dx + g(c_2)\int_\xi^{c_2} f(x) \, dx,$$
Since $g$ is bounded with $|g(x)| \leqslant M$, we have
$$\left|\int_{c_1}^{c_2} f(x) g(x) \, dx\right|\leqslant |g(c_1)| \left|\int_{c_1}^\xi f(x) \, dx\right| + |g(c_2)|\left|\int_\xi^{c_2} f(x) \, dx\right|\\ \leqslant M \left|\int_{c_1}^\xi f(x) \, dx\right| + M\left|\int_\xi^{c_2} f(x) \, dx\right|$$
For every $\epsilon > 0$ there exists $C > a$ such that for all $c_2 \geqslant \xi \geqslant  c_1 > C$,
$$ \left|\int_{c_1}^\xi f(x) \, dx\right|, \,\,\left|\int_\xi^{c_2} f(x) \, dx\right| < \frac{\epsilon}{2M},$$
which implies
$$\left|\int_{c_1}^{c_2} f(x) g(x) \, dx\right| < \epsilon$$
Thus, $\int_a^\infty f(x) g(x) \, dx$ converges by the Cauchy criterion.

A counterexample where $\int_a^\infty f(x) \, dx$ fails to converge --  while other conditions are met -- has been given in comments.
Also, the convergence of $\int_a^\infty f(x) \, dx$ is not a necessary condition. An example is $a=1$, $f(x) = g(x) = 1/x$, where
$$\int_1^\infty f(x) \, dx = \infty, \,\,\int_1^\infty f(x)g(x) \, dx = 1$$
