I was reading a proof of the existence and uniqueness of Laplace exponent of subordinators:
If $\Phi$ is the Laplace exponent of a subordinator,then there exist a unique pair $(k,d)$ of nonnegative real numbers and a unique measure $\Pi$ on $(0,\infty)$ with $\int(1\land |x|)\Pi(dx)<\infty$ such that for every $\lambda\geq 0,$
$$\Phi(\lambda)=k+d\lambda+\int_{(0,\infty)}(1-e^{-\lambda x})\Pi(dx).$$
Part of the proof is the next:
Making use of the independence and homogeneity of the increments in the second equality below, we get from $$E(\exp(-\lambda\sigma_{t}))=\exp(-t\Phi(\lambda))$$ that for every $\lambda\geq 0$ \begin{eqnarray} \Phi(\lambda)&=&\displaystyle\lim_{n\rightarrow\infty}n(1-\exp\{-\Phi(\lambda)/n\})=\displaystyle\lim_{n\rightarrow\infty}nE((1-\exp\{-\lambda\sigma_{1/n}\}))\\ &=&\lambda\displaystyle\lim_{n\rightarrow\infty}\int_{0}^{\infty}e^{-\lambda x}nP(\sigma_{1/n}\geq x).\end{eqnarray}
Write $\overline{\Pi_{n}}(x)=nP(\sigma_{1/n}\geq x),$ so that $$\frac{\Phi(\lambda)}{\lambda}=\displaystyle\lim_{n\rightarrow\infty}\int_{0}^{\infty}e^{-\lambda x}\overline{\Pi_{n}}(x)dx.$$
This shows that the sequence of absolutely continuous measures $\overline{\Pi_{n}}(x)dx$ converges vaguely as $n\rightarrow\infty.$
Here comes my doubt: Why such limit ensures the vague convergence of $\overline{\Pi_{n}}(x)dx?$
I've understood every step in this part of the proof except this one.
Any kind of help is thanked in advanced.