The following was part of a problem in an old analysis qualifier.
Let $G$ be a right-continuous function on $[0,\infty)$ that has finite variation on any bounded interval in $[0,\infty)$. Suppose $$\int^\infty_0 e^{-s_0 t} G(dt)$$ converges for some $s_0\in\mathbb{C}$ with $\mathfrak{R}(s_0)=a>0$. Show that $G(t)=o(e^{at})$ as $t\rightarrow\infty$.
The integral in the problem is with respect the Lebesgue-Stieltjes Radon measure induced by $G$, which I will denote by $\mu_G$. By the Radon-Nikodym theorem $$G(t)-G(0)=\mu_G((0,t])=\int_{(0,t]}e^{s_0 x}e^{-s_0 x} \mu_G(dx)=\int_{(0,t]}e^{s_0 x} \mu_F(dx)$$ where $$ F(t)=\int_{(0,t]}e^{-s_0 x}\mu_G(dx)$$ Integrating by parts gives $$\begin{align} G(t)-G(0)&= e^{s_0t}F(t)-F(0)-s_0\int_{(0,t]}e^{s_0x}F(x-)\,dx\\ &=e^{s_0t}F(t)-s_0\int_{(0,t]}e^{s_0x}F(x)\,dx \end{align} $$ where I have used that $F(0)=0$ and that $F(x-)=F(x)$ Lebesgue almost everywhere. All this reduces to $$e^{-s_0t}(G(t)-G(0))=F(t)-s_0e^{-s_0t}\int_{(0,t]}e^{s_0x}F(x)\,dx$$
By assumption $\lim_{t\rightarrow\infty}F(t):=F(\infty)$ exists in $\mathbb{C}$.
- I am struggling to show that the integral in the RHS of the equation above converges to $F(\infty)$ ( at least I think that this will be the case). A hint will be enough, a solution (or an alternative solution) of course is also appreciated.
The idea is to split the integral in the right-hand-side of the last equation in the OP:
First, given $\varepsilon>0$, choose $T_\varepsilon>0$ such that $$|F(\infty)-F(t)|<\frac{a\varepsilon}{|s_0|}\qquad\text{if}\qquad t\geq T_\varepsilon.$$ Then \begin{align} s_0e^{-s_0t}\int^t_0 e^{s_0 x}F(x)\,dx &= s_0e^{-s_0t}\int^t_0e^{s_0x}(F(x)-F(\infty))\,dx+F(\infty)(1-e^{-s_0t})\tag{1}\label{one} \end{align} Splitting the domain of integration integral on right-hand-side of \eqref{one} as $(0,T_\varepsilon]$ and $(T_\varepsilon, t]$ yields \begin{align} \left|s_0e^{-s_0t}\int^t_0e^{s_0x}(F(x)-F(\infty))\,dx\right|&\leq|s_0| e^{-at}\left(\int^{T_\varepsilon}_0e^{ax}|F(x)-F(\infty)|\,dx +\frac{a\varepsilon}{|s_0|}\int^t_{T_\varepsilon} e^{ax}\,dx\right)\\ &\leq |s_0| e^{-at}\int^{T_\varepsilon}_0e^{ax}|F(x)-F(\infty)|\,dx\\ &\qquad\qquad \qquad +\varepsilon\big(1-e^{-a(t-T_\varepsilon)}\big)\xrightarrow{t\rightarrow\infty}\varepsilon \end{align} since $a>0$. From this, it follows that \begin{align} \lim_{t\rightarrow\infty} s_0e^{-s_0t}\int^t_0e^{s_0x}(F(x)-F(\infty))\,dx=0\tag{2}\label{two} \end{align} From \eqref{one} and the fact that $a>0$. we have that \begin{align} \lim_{t\rightarrow\infty}s_0e^{-s_0t}\int^t_0e^{s_0x}F(x)\,dx=F(\infty)\tag{3}\label{three} \end{align}
Putting things together, \begin{align} \lim_{t\rightarrow\infty} e^{-s_0t}(G(t)-G(0))=0 \end{align} whence the desired conclusion follows.