What is the minimum requirement on $f$ such that $\lim_{s\rightarrow 0}\int_0^{\infty}\exp(-st)f(t)dt = \int_0^{\infty}f(t)dt$? What is the minimum requirement on $f$ such that $\lim_{s\rightarrow 0}\int_0^{\infty}\exp(-st)f(t)dt = \int_0^{\infty}f(t)dt$?
I think if I take $f$ to be measurable and $f\in L^1(0, \infty)$, then by dominated convergence theorem we get that. Or is there any weaker condition? or is reasoning correct? 

If  $$F(0)=\lim_{T \to \infty} \int_0^T f(t)dt$$ converges then $$ F(s) = \lim_{T \to \infty} \int_0^T f(t)e^{-st}dt=\int_0^\infty (\int_0^T f(t)dt) s e^{-sT}dT$$ converges and is analytic for $\Re(s) > 0$ and 
$$\frac{F(s)}{s} = \int_0^\infty (F(0)+o(1)) e^{-sT}dT= \frac{F(0)}{s}+o(\frac{1}{\Re(s)})$$
Thus for $r\in [0,\pi/2)$
$$\lim_{s \to 0, arg(s) \in (-r,r)} F(s) =F(0)$$
This is the Abelian theorem for Laplace transforms.
