I am working on a proof of the following:
Take $f\colon [0,\infty) \rightarrow \mathbb{R}$, Riemann integrable on every $[0,b]$, and such that there exists $M, a,$ and $T$, such that $|f(t)| \leqslant Me^{at}$ for all $t \geqslant T.$ Show that the Laplace transform of $f$ exists. That is, for every $s > a$ the following integral converges: $$F(s) := \int_0^\infty f(t)e^{-st}\mathrm{d}t.$$
I think overall I understand how to prove it: Write $\int_0^\infty f(t)e^{-st}\mathrm{d}t$ as $$\int_0^\infty f(t)e^{-st}\mathrm{d}t = \int_0^T f(t)e^{-st}\mathrm{d}t + \int_T^\infty f(t)e^{-st}\mathrm{d}t,$$ and work through some inequalities until you get $\int_0^\infty f(t)e^{-st}\mathrm{d}t$ as something less than or equal to $\int_0^T f(t)e^{-st}\mathrm{d}t$ plus some number. But what if $T$ is negative? Then $\int_0^T f(t)e^{-st}\mathrm{d}t = -\int_T^0 f(t)e^{-st}\mathrm{d}t$, and I'm not sure we know if this integral even exists given these assumptions -- $f$ is only given to be on $[0,b]$ for $b>0$.
Thank you