Apologies if this question is too basic for MO.

I think it should be the case that

for any decreasing $f \colon [A,\infty) \to [0,\infty)$ and $k \geq 0$, if $\int_A^\infty f(x) e^{kx} \, dx < \infty$ then $f(x)e^{kx} \to 0$ as $x \to \infty$.

**Proof** [I think]**.** Write $g(x)=f(x)e^{kx}$. If $f$ is differentiable, argue by contrapositive that if $g$ is integrable but $\{x:g(x) \geq \varepsilon\}$ is unbounded for some $\varepsilon>0$, then $f$ is not decreasing: Take sequences $x_n,y_n \to \infty\,$ with $\,x_n - y_n \downarrow 0$ such that $g(x_n) = \varepsilon$, $g \leq \varepsilon$ on $[y_n,x_n]$ and $g(y_n) \leq \frac{\varepsilon}{2}$; then we can find $z_n \in (y_n,x_n)$ such that $g'(z_n) \geq \frac{\varepsilon}{2(x_n-y_n)}$ and hence $f'(z_n)=e^{-kz_n}(g'(z_n)-kg(z_n)) \geq \varepsilon e^{-kz_n}\!\left(\frac{1}{2(x_n-y_n)}-k\right)$, which is clearly positive for sufficiently large $n$, and so $f$ is not decreasing.

If $f$ is non-differentiable, let $\varphi$ be a nonnegative unit-integral $C^1$ bump function supported on an interval $[0,\delta]$, and define $\tilde{f} \colon [A+\delta,\infty) \to [0,\infty)$ by $\tilde{f}= f \ast \varphi$; then writing $\tilde{g}(x)=\tilde{f}(x)e^{kx}$, we have $$ g(x) \leq \tilde{g}(x) \leq e^{k\delta}g(x-\delta)\text{,} $$ and so the differentiable case applied to $\tilde{f}$ gives the result. $\ \square$

If the result is correct, it seems like it should be a standard result: either a result with a name, or an immediate special case of a result with a name. Is it so?