Let $f(x) > 0$ on $[a,\infty)$, are integrable on $[a,b]$ whenever $0<a<b<\infty$. Suppose there is a constant $M$ s.t. $\forall t > 0$ $$ \int_{a}^{\infty} f(x)e^{-tx}dx \le M \quad $$
Then how to prove that $\int_{a}^{\infty} f(x)dx$ converges as well?
The uniform upper bound $M$ must be the key point, but I don't know how to handle it. Any help will be appreciated.
So this is true by Fatou's Lemma for instance (or the monotone convergence theorem). Indeed the sequence of functions $f_t(x) = f(x)\,e^{-tx}$ is a sequence of positive functions that converges pointwise to $f(x)$ when $t\to 0$ (you can take $t=1/n$ if you want a true sequence). Therefore, $f(x)$ is measurable and $$ 0\leq \int f(x) \,\mathrm d x \leq \liminf_t \int f_t(x) \,\mathrm d x \leq M $$ so $f = |f|$ is integrable.
