= Solution
Let $N(ds,dy)$ be a Poisson random measure on $(0,\infty)^2$ with intensity $ds\,K(dy)$ and define
$$
X_t=\int_{(0,t]\times(0,\infty)}y\,N(ds,dy).
$$
The assumption $\int yK(dy)<\infty$ makes this integral finite on compact time intervals. The exponential formula for a Poisson random measure gives
$$
\mathbb Ee^{iuX_t}
=\exp\left\{t\int_{(0,\infty)}(e^{iuy}-1)K(dy)\right\},
$$
so $X$ is a Lévy process with exponent $\psi$.
Choose finite-valued measurable functions $q_n\geq0$ which vanish off $[1/n,n]$ and satisfy
$$
\int|q_n(y)-y|K(dy)\longrightarrow0.
$$
This is possible by truncation followed by approximation by simple functions. Put
$$
X_t^n=\int_{(0,t]\times(0,\infty)}q_n(y)\,N(ds,dy).
$$
The measure of the support of $q_n$ is finite, and $q_n$ takes finitely many values, so $X^n$ is a simple pure-jump Lévy process. Under this common coupling,
$$
\mathbb E\sup_{s\leq t}|X_s^n-X_s|
\leq\mathbb E\int_{(0,t]\times(0,\infty)}
|q_n(y)-y|\,N(ds,dy)
=t\int|q_n-y|\,dK\longrightarrow0.
$$
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