I came accross the following problem in the real analysis book I am following: show that if $f: (0, +\infty) \rightarrow \mathbb{R}$ is integrable then $g: (0,+\infty) \rightarrow \mathbb{R}$ given by $g(x) = \int_{(0,+\infty)} f(u)e^{-xu} du$ is continuous.
I think the idea is to use either the dominated convergence theorem, since it is one of the main results the book presents in this section. However I am not sure, since I don't see what would be a sequence meeting any of the theorem's conditions.
Let $x_0 \in (0, +\infty)$, given $\epsilon > 0$ we need to show that $\exists \delta > 0: |x - x_0| < \delta \Rightarrow |g(x) - g(x_0)| < \epsilon$. Now if I find some convergent sequence of functions $g_n \rightarrow g$ then $|g(x) - g(x_0)| \leq |g(x) - g_n(x)| + |g_n(x) - g_n(x_0)| + |g_n(x_0) - g(x_0)| < \epsilon $. I would like to apply the dominated convergence theorem to show that $g_n \rightarrow g$.
Have you tried pushing the triangle inequality as far as it goes? $$ |g(b)-g(a)|=\left|\int f(u)(e^{-bu}-e^{-au})du\right|\leq\sup_u\{|e^{-bu}-e^{-au}|\}\int|f(u)|du\leq\ldots $$ I think optimizing the term out front should work, but I'll leave it to you.