Let $f(t)$ be an $original$ function. Then the Laplace Transform of $f(t)$ will be: $$F(s)=\int_0^\infty f(t)e^{-st}\,dt$$ with $Re(s)>a_0$ where $a_0$ is the infimum of all $a\in\mathbb{R}$ such that $|f(t)|\le Me^{at}$ with $M\ge0.$
My question is: why the condition $Re(s)>a_0$ ?
Write $f(t)=m(t)e^{a_0t}$, so $|m|\le M$. Then $F(s)=\int_0^\infty m(t)e^{-(s-a_0)t}dt$ converges if $\Re s>a_0$, with $|F|\le\frac{M}{\Re s-a_0}$. By contrast, if $\Re s\le a_0$ the integrand grows at least as fast as $e^{(a_0-\Re s)t}$, so $F$ doesn't converge.