Vague convergence VS Laplace transform convergence?

Question

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. If it is true, how to prove it? Otherwise please give a counter-example? Thank in advance.

Answer 1

$\newcommand\R{\mathbb R}$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, 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: 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 $L$.)

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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. $\quad\Box$