Let $(\mu_n)_{n\in\mathbb{N}}$ be a sequence of probability measures on $[0,\infty)$. Assume we know that
$\lim_{n\to\infty}\int e^{-tx}\mu_n(dx) = \psi(t)$ for $t\geq0$, $n\in\mathbb{N}$, and $\psi:[0,\infty)\to[0,1]$ is a map continuos in $0$.
How can I show that $\psi$ is the Laplance transform of a probability measure $\mu$ on $[0,\infty)$? (This is Ex 15.3.4 of Probability Theory by A. Klenke, 3rd version).
My main doubt is that, if I am able to show that the sequence of $\mu_n$ is tight, how can I check that the limit function $\psi$ is a Laplace transform and not something else? (Basically I am trying to apply Theorem 13.34 of the same book for the interested ones).
Edit
How would you check the tightness of the sequence?
Thank you.
By tightness there is a subsequence $(\mu_{n_k})$ converging weakly to some probability measure $\mu$. Since $e^{-tx}$ is a bounded continuous function we get $\psi (t)=\int e^{-tx}d\mu(x)$ for all $t \geq 0$.
Let $\epsilon >0$. Since $\psi (0)=1$ and $\psi$ is continuous at $0$ we can find $r$ such that $\int e^{-rx}d\mu_n(x) >1-\epsilon$ for all $n$. Now, $$1-\epsilon \leq \mu_n ([0,M])+\int_M^{\infty} e^{-rx}d\mu_n(x) \leq \mu_n ([0,M])+ e^{-rM}\mu_n(M,\infty).$$ This can be written as $\mu_n(M,\infty) \leq \epsilon +e^{-rM}\mu_n(x)(M,\infty)$. I will let you finish the proof from here.