If we assume that $\int\_0^\infty e^{-sx}\mu\_n(dx)\to \int\_0^\infty e^{-sx}\mu(dx), \forall s\geq0$, it is possible to show that $\mu\_n\to\mu$ vaguely. Where $\mu\_n$ is a measure. Please check here for [vague convergence](https://en.wikipedia.org/wiki/Vague_topology).
If it is true, how to prove it? Otherwise please give a counter-example? Thank in advance.



Vague convergence VS Laplace transform convergence?
$\newcommand\R{\mathbb R}$We have
\begin{equation\*}
 L\_n(s)\to L(s) \tag{1}\label{1}
\end{equation\*}
(as $n\to\infty$) for each $s\in\R\_+:=[0,\infty)$, where
\begin{equation\*}
 L\_n(s):=\int\_{\R\_+}e^{-sx}\mu\_n(dx),\quad L(s):=\int\_{\R\_+}e^{-sx}\mu(dx). 
\end{equation\*}


From the context, $\mu$ and the $\mu\_n$'s are finite measures over $\R\_+$.


If they are probability measures, then, by (say) [Theorem 5.22](https://www.google.com/books/edition/Foundations_of_Modern_Probability/L6fhXh13OyMC?hl=en&gbpv=1&bsq=Theorem%205.22), condition \eqref{1} implies the weak convergence of $\mu\_n$ to $\mu$; that is, we have
\begin{equation\*}
 \int\_{\R\_+}f(x)\mu\_n(dx)\to\int\_{\R\_+}f(x)\mu(dx) \tag{2}\label{2}
\end{equation\*}
for all bounded continuous functions $f$ on $\R\_+$ -- which of course implies the vague convergence, the latter being the convergence \eqref{2} for all continuous functions $f$ on $\R\_+$ such that $f(x)\to0$ as $x\to\infty$. (Details on the use of [Theorem 5.22](https://www.google.com/books/edition/Foundations_of_Modern_Probability/L6fhXh13OyMC?hl=en&gbpv=1&bsq=Theorem%205.22): This theorem is applicable here because (i) $L(s)\to L(0)$ as $s\downarrow0$, by the monotone convergence theorem and (ii) the measure $\mu$ is uniquely determined by its Laplace transform $L$.)




---


Suppose now that $\mu$ and the $\mu\_n$'s are any finite measures over $\R\_+$ (not necessarily probability measures) such that \eqref{1} holds for each $s\in\R\_+$. In particular, we have $\mu\_n(\R\_+)=L\_n(0)\to L(0)=\mu(\R\_+)$.


Consider first the case when $\mu(\R\_+)=0$. Take any bounded continuous function $f$ on $\R\_+$, so that $|f|\le c$ for some real $c$. Then
$$\Big|\int\_{\R\_+}f(x)\mu\_n(dx)\Big|\le c\mu\_n(\R\_+)\to c\mu(\R\_+)=0.$$
So, $\int\_{\R\_+}f(x)\mu\_n(dx)\to0=\int\_{\R\_+}f(x)\mu(dx)$. Thus, $\mu\_n$ converges to $\mu$ weakly and hence vaguely.


Finally, consider the case when $\mu(\R\_+)>0$. Then eventually (that is, for all large enough $n$) we have $\mu\_n(\R\_+)>0$. So, (for such $n$) we may introduce the probability measures
\begin{equation\*}
 \nu\_n:=\frac{\mu\_n}{\mu\_n(\R\_+)},\quad \nu:=\frac{\mu}{\mu(\R\_+)}. 
\end{equation\*}
Since $\mu\_n(\R\_+)\to\mu(\R\_+)>0$, we see that the Laplace transform of $\nu\_n$ will converge to the Laplace transforms of the $\nu$. So, by what was said in the case when $\mu$ and the $\mu\_n$'s are probability measures, $\nu\_n$ will converge to $\nu$ weakly. Again, since $\mu\_n(\R\_+)\to\mu(\R\_+)>0$, it will follow that in this case as well $\mu\_n$ converges to $\mu$ weakly and hence vaguely. $\quad\Box$


