This question is a related question see this post [Vague convergence VS Laplace transform convergence](https://mathoverflow.net/questions/445665/vague-convergence-vs-laplace-transform-convergence). But now we assume that
\begin{equation}
 \int\_0^\infty e^{-sx}\mu\_n(dx)\to \int\_0^\infty e^{-sx}\mu(dx),
\end{equation}


for all $s\in (0, \infty)$, where $\mu\_n$ is a finite measure. Please check this book for [vague convergence](https://www.google.com/books/edition/Foundations_of_Modern_Probability/L6fhXh13OyMC?hl=en&gbpv=0&bsq). Can we get what from the above Laplace transform convergence? From the above post, my understanding is that vague convergence is weaker than weakly convergence (please correct me, if I am wrong). Is it possible to get vague convergence and why or why not?



Which kind of convergence can we get from Laplace transform convergence?
$\newcommand{\thh}{\theta}$In general, you cannot get the vague convergence here. E.g., suppose that $\mu=0$ and $\mu\_n(dx)=x^2\,1(n<x<n+1)\,dx$. Then for each real $s>0$
\begin{equation\*}
 0\le\int\_0^\infty e^{-sx}\mu\_n(dx)=\int\_n^{n+1} e^{-sx}x^2\,dx
 \le e^{-sn}(n+1)^2\to0
\end{equation\*}
(as $n\to\infty$), so that
\begin{equation\*}
 \int\_0^\infty e^{-sx}\mu\_n(dx)\to\int\_0^\infty e^{-sx}\mu(dx). 
\end{equation\*}
However, letting $f(x):=\frac1{1+x}$ for real $x\ge0$, we have $f\in C\_0([0,\infty))$ and
\begin{equation\*}
 \int\_0^\infty f(x)\mu\_n(dx)=\int\_n^{n+1} f(x)x^2\,dx
 \ge \frac1{1+n+1}n^2\to\infty,
\end{equation\*}
so that
\begin{equation\*}
 \int\_0^\infty f(x)\mu\_n(dx)\to\infty\ne0=\int\_0^\infty f(x)\mu(dx). 
\end{equation\*}




---


However, assuming that the measure $\mu$ is finite, you can get the following:
\begin{equation\*}
 \int\_0^\infty f(x)\mu\_n(dx)\to\int\_0^\infty f(x)\mu(dx) \tag{$\*$}\label{1}
\end{equation\*}
for all $f\in C\_c([0,\infty))$, that is, for all continuous functions $f$ on $[0,\infty)$ with a compact support.


Indeed, take any such $f$, so that $f=0$ on $[a,\infty)$ for some real $a>0$. Consider first the case when $f\ge0$.


Take any real $h>0$ and let
\begin{equation\*}
 \nu\_{n,h}(dx):=e^{-hx}\mu\_n(dx),\quad \nu\_h(dx):=e^{-hx}\mu(dx). 
\end{equation\*}
Then for any real $s\ge0$
\begin{equation\*}
 \int\_0^\infty e^{-sx}\nu\_{n,h}(dx)=\int\_0^\infty e^{-(s+h)x}\mu\_n(dx)
 \to\int\_0^\infty e^{-(s+h)x}\mu(dx)=\int\_0^\infty e^{-sx}\nu\_h(dx). 
\end{equation\*}
So, by the [previous answer](https://mathoverflow.net/a/445676/36721), $\nu\_{n,h}\to\nu\_h$ weakly. So,
\begin{equation\*}
 \int\_0^\infty f(x)e^{-hx}\mu\_n(dx)=\int\_0^\infty f(x)\nu\_{n,h}(dx)
\to\int\_0^\infty f(x)\nu\_h(dx)=\int\_0^\infty f(x)e^{-hx}\mu(dx) 
\end{equation\*}
for each real $h>0$.
Next,
\begin{equation\*}
 \int\_0^\infty f(x)\mu\_n(dx)\ge\int\_0^\infty f(x)e^{-hx}\mu\_n(dx)
 =\int\_0^a f(x)e^{-hx}\mu\_n(dx) \\ 
 \ge e^{-ha}\int\_0^a f(x)\mu\_n(dx)
 =e^{-ha}\int\_0^\infty f(x)\mu\_n(dx),
\end{equation\*}
so that
\begin{equation\*}
 \int\_0^\infty f(x)\mu\_n(dx)=e^{\thh\_{n,h,f}ha}\int\_0^\infty f(x)e^{-hx}\mu\_n(dx)
\end{equation\*}
for some $\thh\_{n,h,f}\in[0,1]$ depending on $n,h,f$. Similarly,
\begin{equation\*}
 \int\_0^\infty f(x)\mu(dx)=e^{\thh\_{h,f}ha}\int\_0^\infty f(x)e^{-hx}\mu(dx)
\end{equation\*}
for some $\thh\_{h,f}\in[0,1]$ depending on $h,f$. So, for each real $h>0$,
\begin{equation\*}
\begin{aligned}
 \limsup\_n\int\_0^\infty f(x)\mu\_n(dx)&\le e^{ha}\limsup\_n\int\_0^\infty f(x)e^{-hx}\mu\_n(dx) \\ 
& =e^{ha}\int\_0^\infty f(x)e^{-hx}\mu(dx) \\ 
& \le e^{ha}\int\_0^\infty f(x)\mu(dx) \\ 
& \underset{h\downarrow0}\longrightarrow
 \int\_0^\infty f(x)\mu(dx)
\end{aligned}
\end{equation\*}
and
\begin{equation\*}
\begin{aligned}
 \liminf\_n\int\_0^\infty f(x)\mu\_n(dx)&\ge\liminf\_n\int\_0^\infty f(x)e^{-hx}\mu\_n(dx) \\ 
 &=\int\_0^\infty f(x)e^{-hx}\mu(dx) \\ 
 &\ge e^{-ha}\int\_0^\infty f(x)\mu(dx) \\ 
& \underset{h\downarrow0}\longrightarrow
 \int\_0^\infty f(x)\mu(dx). 
\end{aligned}
\end{equation\*}
So, $\limsup\_n\int\_0^\infty f(x)\mu\_n(dx)=\liminf\_n\int\_0^\infty f(x)\mu\_n(dx)=\int\_0^\infty f(x)\mu(dx)$ and hence \eqref{1} holds
for any nonnegative $f\in C\_c([0,\infty))$.


Finally, writing $f=f\_+-f\_-$, where $f\_\pm:=\max(0,\pm f)$, we get \eqref{1} for all $f\in C\_c([0,\infty))$. $\quad\Box$


