So I'm trying to show that if the Laplace transform of a function $\mathit f$, $s \to \mathcal L(\mathit f)(s)$ exists at $s=s_{0}$, then it exists for all $s > s_{0}$. Anyway, here's what I thought.
If the Laplace of f exists at $s_{0}$, then we know that the indefinite integral
$\int_0^\infty e^{-s_{0}t}\mathit f (t) dt$ converges. Then I thought that since for all $s>s_{0}$ we have $e^{-s_{0}t}>e^{-st}$ for all $t>0$ the Laplace transform at $s$ must converge as a consequence of the integral comparison test.
However, my professor said that I needed to consider functions f that might not be absolutely convergent to show this result. So I'm looking for some clarification as to what exactly his concern is addressing and how I may go about fixing the issue.
Since $\int_0^\infty f(t) e^{-s_0 t }dt$ converges we know that $g(t) = \int_0^t f(u) e^{-s_0 u }du$ is bounded.
Then integrate by parts $$\int_0^A e^{-s t} f(t)dt = \int_0^A e^{(s_0-s) t} e^{-s_0 t} f(t)dt = e^{(s_0-s)A} g(A) - \int_0^A (s_0-s) e^{(s_0-s) t} g(t)dt$$ and it converges as $A \to \infty$ whenever $Re(s) > Re(s_0)$.
Note how it makes a huge difference with your proof assuming that $e^{-s_0t}f(t) \in L^1$, try with $s_0= 0$, $f(t) = e^{t/2} \sin(e^{t}) 1_{t > 1}$, you get $\mathcal{L}[f](0) = \int_1^\infty e^{t/2} \sin(e^{t})dt = \int_0^\infty \sqrt{u}\sin(u)\frac{du}{u}$ and it converges