The unilateral Laplace transform of an $f:[0,\infty]\rightarrow \mathbb{C}$ is defined as
$$F(s)=\int_{0}^{\infty}e^{-st}f(t)dt$$
My lecturer didn't go into detail on the domain of the transform, but often it is said that '$\Re (s) >0$', for instance with the transform of $\sin t$. But what's the maximal domain of the transform? I figured that it would be the $\mathbb{C} \setminus A$, where $$A = \{ z \in \mathbb{C} : \Re (z) > r \}$$
$$r=\inf_{\rho \ge 0} (\rho:\exists M \ge 0 : |f(t)|\le Me^{\rho t})$$
But this is purely my speculation - is any of this correct?
By domain, it depends for which space of functions you want it to work. The idea, as you understood, is to ensure the integral to be absolutely convergent.
What you teacher wrote is: if we take a bounded measurable function, then we are sure the integral defining the Laplace transform makes sense.
Of course, in some particular cases, there are other $s$ for which the definition make sense, e.g. $f(t)=e^{-t^2/2}$.
You definition fits with this (up to a minus sign, I think).