If $\int_{0}^{\infty} f(x) \, dx $ converges, will $\int_{0}^{\infty}e^{-sx} f(x) \, dx$ always converge uniformly on $[0, \infty)$? I previously asked about sufficient conditions to conclude that $$\lim_{s \to 0^{+}}\int_{0}^{\infty} e^{-sx} f(x) \, dx = \int_{0}^{\infty} f(x) \, dx$$ when $\int_{0}^{\infty} f(x) \, dx$ does not converge absolutely.
Daniel Fischer showed that a sufficient condition is if $\int_{0}^{\infty} e^{-sx} f(x) \, dx$ converges uniformly on $[0, \delta]$ for some $\delta >0$.
Recently I came across the following exercise:

Show that if $F(s) = \int_{0}^{\infty} e^{-sx} f(x) \, dx$ converges
  for $s=s_{0}$, then it converges uniformly on $[s_{0}, \infty)$.

The above excercise is exercise 27 in the first supplement to the textbook Introduction to Real Analysis by William F. Trench.
It's basically a stronger version of a theorem that states that if $f(x)$ is continuous on $[0, \infty)$ and $\int^{{\color{red}{x}}}_{0} e^{-s_{0}u} f(u) \, du$ is bounded for all $x \ge 0$, then $\int_{0}^{\infty} e^{-sx} f(x) \, dx$ will converge uniformly on $[s_{1}, \infty)$ for $s_{1} >s_{0}$. A proof of this theorem can be found on page 20 of the supplement.
But with the only condition being that $\int_{0}^{\infty} e^{-s_{0}x} f(x) \, dx$ must converge, it's hard to believe that there is not a counterexample.
Perhaps it has something to do with $e^{-sx}$ being monotonic in the parameter $s$.
 A: By replacing $f(x)$ with $f(x) e^{-s_0 x}$, we can assume that $s_0 = 0$. For $t > 0$ define
$$R(t) := \int_t^{+\infty} f(x)\,dx.$$
Since the improper Riemann integral
$$\int_0^{+\infty} f(x)\,dx$$
exists, $R \colon (0,+\infty) \to \mathbb{R}$ is a continuous function with $\lim\limits_{t\to +\infty} R(t) = 0$. Let further
$$M(u) := \sup \{ \lvert R(t)\rvert : t \geqslant u\}.$$
Since we have only minimal regularity assumptions on $f$, it's easier to justify integration by parts from the other end: For $s > 0$ and $0 < a < b < +\infty$ we have
\begin{align}
s\int_a^b R(x) e^{-sx}\,dx &= s\int_a^b R(b) e^{-sx}\,dx + s\int_a^b \bigl(R(x) - R(b)\bigr) e^{-sx}\,dx\\
&= R(b)\bigl(e^{-sa} - e^{-sb}\bigr) + s\int_a^b e^{-sx}\int_x^b f(u)\,du \,dx\\
&= R(b)\bigl(e^{-sa} - e^{-sb}\bigr) + \int_a^b f(u) s\int_a^u e^{-sx}\,dx\,du\\
&= R(b)\bigl(e^{-sa} - e^{-sb}\bigr) + \int_a^b f(u) \bigl(e^{-sa} - e^{-su}\bigr)\,du\\
&= R(b)\bigl(e^{-sa} - e^{-sb}\bigr) + e^{-sa}\bigl(R(a) - R(b)\bigr) - \int_a^b f(u) e^{-su}\,du\\
&= R(a) e^{-sa} - R(b)e^{-sb} - \int_a^b f(u)e^{-su}\,du.
\end{align}
The change of order of integration is justified since $f$ is Riemann-integrable (and in particular bounded) on $[a,b]$. Since
$$\biggl\lvert s\int_a^b R(x)e^{-sx}\,dx\biggr\rvert \leqslant s\int_a^b \lvert R(x)\rvert e^{-sx}\,dx \leqslant M(a)s\int_a^b e^{-sx}\,dx \leqslant M(a) e^{-sa},$$
rearranging yields
$$\biggl\lvert \int_a^b f(x)e^{-sx}\,dx\biggr\rvert \leqslant \lvert R(a)\rvert e^{-sa} + \lvert R(b)\rvert e^{-sb} + M(a)e^{-sa} \leqslant 3\cdot M(a),\tag{$\ast$}$$
with a bound independent of $b$ and $s > 0$. Since $(\ast)$ also holds for $s = 0$, and $M$ is monotonically decreasing (nonstrictly, in general) with $\lim\limits_{u\to \infty} M(u) = 0$, this shows the uniform convergence of the improper Riemann integrals
$$\int_c^{+\infty} f(x) e^{-sx}\,dx \tag{1}$$
for $s \in [0,+\infty)$ for any fixed $c > 0$. If $f$ is bounded on $[0,\delta]$ for some $\delta > 0$, we can even take $c = 0$ here and are done. If $f$ is unbounded at $0$, so the integral $\int_0^{+\infty} f(x)\,dx$ is improper at both ends, a small modification of the argument using
$$P(t) := \int_0^t f(x)\,dx$$
and $N(u) := \sup \{ \lvert P(t)\rvert : t \leqslant u\}$ gives the estimate
$$\biggl\lvert \int_a^b f(x) e^{-sx}\,dx\biggr\rvert \leqslant 3\cdot N(b)$$
for $0 < a < b < +\infty$, and the uniform convergence of
$$\int_0^c f(x) e^{-sx}\,dx$$
for $s\in [0,+\infty)$ given any fixed $c > 0$. Together with $(1)$, we have the uniform convergence of
$$\int_0^{+\infty} f(x) e^{-sx}\,dx$$
for $s \in [0,+\infty)$.
