If $\nu$ is a finite measure on $(\mathbb R,\mathcal B(\mathbb R))$, let $\nu^{\ast k}$ denote the $k$-fold convolution¹ of $\nu$ with itself for $k\in\mathbb N\_0$, $$\exp(\nu)\mathrel{:=}\sum\_{k=0}^\infty\frac{\nu^{\ast k}}{k!}$$ and $$\operatorname{CPoi}\_\nu\mathrel{:=}\frac{\exp(\nu)}{\exp(\nu)(\mathbb R)}.$$ Moreover, let $$\mathcal L\_\nu(t)\mathrel{:=}\int\nu({\rm d}x)e^{-tx}\;\;\;\text{for $t\in\mathbb R$}$$ denote the Laplace transform of $\nu$.



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> Let $\mu$ be a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$. Remember that $\mu$ is called **infinitely divisible** if for all $k\in\mathbb N$, there is a probability measure $\nu$ on $(\mathbb R,\mathcal B(\mathbb R))$ with $\mu=\nu^{\ast k}$.
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> We can show that
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> 1. $\mu$ is infinitely divisible;
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> and
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> 2. There is a sequence $(\nu\_n)\_{n\in\mathbb N}$ of finite measures on $(\mathbb R,\mathcal B(\mathbb R))$ with $\operatorname{CPoi}\_{\nu\_n}\xrightarrow{n\to\infty}\mu$ weakly
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> are equivalent.
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> 



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> Now assume $\mu$ is a probability measure on $([0,\infty),\mathcal B([0,\infty)))$. By the Lévy-Khinchin formula,
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> 3. $\mu$ is infinitely divisible;
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> and
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> 4. There is a $\alpha\ge0$ and a $\sigma$-finite measure $\nu$ on $((0,\infty),\mathcal B((0,\infty)))$ with $$-\ln\mathcal L\_\mu(t)=\alpha t+\int1-e^{-tx}\:\nu({\rm d}x)\;\;\;\text{for all $t\ge0$}\tag4$$
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> are equivalent.
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> **Question**: How can we show that if (3.) holds, then the $\nu$ from (4.) is equal to the vague limit² of $\left.n\mu^{\ast1/n}\right|\_{(0,\:\infty)}$?
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¹ I.e. $\nu^{\ast 0}:=\delta\_0$ is the Dirac measure on $(\mathbb R,\mathcal B(\mathbb R))$ at $0$ and if $k\in\mathbb N$, then $\nu^{\ast k}$ is the pushforward $\tau\_k\left(\nu^{\otimes k}\right)$ of the $k$-fold product measure $\nu^{\otimes k}$ of $\nu$ with itself under the map $$\tau\_k:\mathbb R^k\to\mathbb R\;,\;\;\;x\mapsto x\_1+\dotsb+x\_k.$$


² I.e. $$n\int\_{(0,\:\infty)}f\:{\rm d}\mu^{\ast1/n}\xrightarrow{n\to\infty}\int f\:{\rm d}\nu\;\;\;\text{for all }f\in C\_0((0,\infty))\tag5.$$



If $\mu$ is an infinitely divisible probability measure on $[0,\infty)$, then the Lévy measure of $\mu$ is the vague limit of $n\mu^{*1/n}$
There are at least three ways to show that $n \mu^{\*1/n}$ converges to $\nu$ vaguely in $\mathbb{R} \setminus \{0\}$. Let $X\_t$ be the Lévy process such that $X\_1$ has distribution $\mu$, and let $f$ be a smooth, compactly supported $f$ on $\mathbb{R} \setminus \{0\}$. It is sufficient to show that $t^{-1} \mathbb{E} f(X\_t) \to \int f(x) \nu(dx)$ as $t \to 0^+$. Here are brief descriptions of the three methods.


1. Write $X\_t = Y\_t + Z\_t$, where $Y\_t$ is a compound Poisson process with Lévy measure $\nu$ restricted to $\mathbb{R} \setminus (-\varepsilon, \varepsilon)$ for a sufficiently small $\varepsilon > 0$ (such that $f = 0$ on $[-2\varepsilon, 2\varepsilon]$), and $Z\_t$ is the "remaining part" of $X\_t$ (the existence of such decomposition follows from the Lévy–Itô theorem — see Theorem 19.2 in Sato's book on Lévy processes). Then $$t^{-1} \mathbb{E} |f(Y\_t)| \to \int f(x) \nu(dx)$$ as $t \to 0^+$, as can be easily proved by conditioning on the number of jumps (only the term with a single jump has positive contribution). Furthermore, $$t^{-1} \mathbb{E} |f(X\_t) - f(Y\_t)| \leqslant t^{-1} \mathbb{P}[|Y\_t| < \varepsilon] \mathbb{P}[|Z\_t| > \varepsilon] + t^{-1} \mathbb{P}[|Y\_t| \geqslant \varepsilon] \|f'\|\_\infty \mathbb{E} |Z\_t|$$ and both terms can be verified to converge to zero as $t \to 0^+$.
2. Use an appropriate variant of Plancherel's theorem: if $\psi = -\ln \varphi\_\mu$ is the characteristic (Lévy—Khintchine) exponent, then $$t^{-1} \mathbb{E} f(X\_t) = t^{-1} \mathbb{E} (f(X\_t) - f(0)) = \int \hat{f}(z) t^{-1} (e^{-t \psi(z)} - 1) dz \to -\int \hat{f}(z) \psi(z) dz $$ by the dominated convergence theorem. Using the explicit form of $\psi$, the facts that $f'(0) = f''(0) = 0$, and again an appropriate variant of Plancherel's theorem, we can show that $$-\int \hat{f}(z) \psi(z) dz = \int f(x) \nu(dx),$$ as desired.
3. My favourite one, using semigroup theory. Let $L$ be the generator of the transition semigroup $P\_t$ of $X\_t$. We have $$t^{-1} \mathbb E f(X\_t) = t^{-1} (\mathbb E f(X\_t) - f(0)) = t^{-1} (P\_t f(0) - f(0)) \to L f(0) $$ as $t \to 0^+$. By the expression for $L$ and the fact that $f(0) = f'(0) = f''(0) = 0$, we have $$L f(0) = \int f(x) \nu(dx).$$


I am pretty much sure one can find the above arguments in the literature, but I do not have a reference off the top of my head.


