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 $$\mathbb E f(X_t) = \mathbb E f(X_t) - f(0) = P_t f(0) - f(0) = \int_0^t P_s L f(0) ds, $$ and by continuity of the semigroup and the dominated convergence theorem we have $$t^{-1} \mathbb E f(X_t) = t^{-1} \int_0^t P_s L f(0) ds = \int_0^1 P_{tu} L f(0) du \to L f(0) $$ as $t \to 0^+$. Finally, 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.

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.

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 $\varepsilon > 0$, 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 $$t^{-1} \mathbb{E} |f(X_t) - f(Y_t)| \le t^{-1} \|f''\|_\infty \operatorname{Var} Z_t = t \|f''\|_\infty \operatorname{Var} Z_1 \to 0$$ as $t \to 0^+$. Furthermore, $$t^{-1} \mathbb{E} |f(Y_t)| \to \int f(x) \nu(dx)$$ 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).

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 $$\mathbb E f(X_t) = \mathbb E f(X_t) - f(0) = P_t f(0) - f(0) = \int_0^t P_s L f(0) ds, $$ and by continuity of the semigroup and the dominated convergence theorem we have $$t^{-1} \mathbb E f(X_t) = t^{-1} \int_0^t P_s L f(0) ds = \int_0^1 P_{tu} L f(0) du \to L f(0) $$ as $t \to 0^+$. Finally, 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.

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 $$\mathbb E f(X_t) = \mathbb E f(X_t) - f(0) = P_t f(0) - f(0) = \int_0^t P_s L f(0) ds, $$ and by continuity of the semigroup and the dominated convergence theorem we have $$t^{-1} \mathbb E f(X_t) = t^{-1} \int_0^t P_s L f(0) ds = \int_0^1 P_{tu} L f(0) du \to L f(0) $$ as $t \to 0^+$. Finally, 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.