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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.

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  • $\begingroup$ By Stone–Weierstrass theorem. $\{(t\mapsto e^{-\lambda t} )\mid \lambda \geq 0)\}$ is separating and thus the algebra generated by it is dense in $C_0([0,\infty),\mathbb{R})$. $\endgroup$ – Sayantan Apr 20 '18 at 5:59
  • $\begingroup$ Thanks to answer @Sayantan. I don't get your argument. I understand that such family of functions is dense because of Stone-Weierstrass, but how this ensures vague convergence? $\endgroup$ – Squird37 Apr 20 '18 at 6:07
  • $\begingroup$ By definition of vague convergence, we need to show that when $\mu_n,\mu$ are probability measures on $[0,\infty)$ and $$\int_0^{\infty} f \,d \mu_n \to \int_0^{\infty} f \,d \mu$$ for all $f \in D$, a dense subset of $C_0([0,\infty ))$, the same is true for any $g \in C_0([0,\infty))$. $\endgroup$ – Sayantan Apr 20 '18 at 6:17
  • $\begingroup$ Note that $$ |\int_0^{\infty} g \,d \mu_n - \int_0^{\infty} g\,d \mu | \leq |\int_0^{\infty} g \,d \mu_n - \int_0^{\infty} f\,d \mu_n | + |\int_0^{\infty} f \,d \mu_n - \int_0^{\infty} f\,d \mu | +|\int_0^{\infty} f \,d \mu - \int_0^{\infty} g\,d \mu |. $$ Choose $f \in D$ suitably so that $\lVert f - g\rVert _{\infty} <\epsilon/3$ and $n$ large enough so that the middle term is less than $\epsilon/3$. $\endgroup$ – Sayantan Apr 20 '18 at 6:21
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    $\begingroup$ Exactly. Any element of $C_{0}([0,\infty))$ is limit of a sequence of the dense set $D.$ My doubt is that limit is equal to $\frac{\Phi(\lambda)}{\lambda},$ which is a term that al least, at the begin, is not an integral. $\endgroup$ – Squird37 Apr 20 '18 at 6:29

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