Let $M \in \mathfrak{M}_1(\real^d)$ be a $\sigma$-finite [[Measure Space|measure]] with $M(\bracs{0}) = 0$ and $ \int \min\bracs{\abs{x}, 1} dM(x) < \infty $ > [!theorem] > > Let $j(t, dy, \omega)$ be the [[Jump Measure|jump measure]] driven by $M$, then for any $t \ge 0$ and almost every $\omega$, > $ > V_t(\omega) = \int_{\real^d} \abs{y}j(t, dy, \omega) < \infty \quad \forall t \ge 0 > $ > Since $V_t(\omega)$ is a non-decreasing function of $t$, the conditions $\forall t \ge 0$ and $\forall \omega$ can be interchanged. > > *Proof*. Let $A_0 = \ol{B(0, 1)}^c$ and $A_k = \ol{B(0, 2^{k + 1})}$, $M_k = M|_{A_k}$, and $X^{k}$ be independent compound Poisson processes for $\pi_{0, 0, M_k}$, then > $ > V_t(\omega) = \int_{\real^d}\abs{y}j(t, dy, \omega) = \sum_{k = 0}^\infty\int \abs{y}j(t, dy, X^{(k)}(\omega)) > $ > It's sufficient to show that $\sum_{k = 1}^\infty \int \abs{y}j(t, dy, X^{(k)}(\omega))$ because the first term is finite almost surely. By the [[Monotone Convergence Theorem]], > $ > \begin{align*} > \ev\braks{V_t} &= \sum_{k = 1}^\infty \int \abs{y}j(t, dy, X^{(k)}(\omega)) \\ > &= \sum_{k = 1}^\infty \ev\braks{\norm{X^{(k)}_t}_{V}} \\ > &= \sum_{k = 1}^\infty \int_{\real^d}\abs{y}dM_k(y) = \int_{\real^d}\abs{y}dM(y) < \infty > \end{align*} > $ > [!theorem] > > Let $j(t, dy, \omega)$ be the [[Jump Measure|jump measure]] driven by $M$, then > $ > X_t(\omega) = \int_{\real^d}{y}j(t, dy, \omega) > $ > is well-defined for all $t \ge 0$. For each $K \ge 1$, let $X^{(k)}$ be independent compound Poisson processes associated with $\pi_{0, 0, M_k}$, then > $ > S_t^{(K)} = \sum_{k = 0}^K \int_{\real^d}yj(t, dy, X^{(k)}) > $ > with $S_t^{(K)} \to X_t$ by the previous theorem with $V_t(\omega) = \norm{X(\omega)|_{s \le t}}_{V}$. Furthermore, there exists a subsequence $S_t^{(K_n)}$ that converges to $X_t$, uniformly on compact sets of $t$. > > *Proof*. As $S_t^{(K_n)}$ is a [[Lévy Process]] with RCLL sample paths, $X_t$ also has RCLL sample paths. Since $M$ is in $\mathfrak{M}_1$, $\int_{B(0, 1)}\abs{y}dM(y) < \infty$. By partitioning into annuli, there exists a subsequence $\seq{k_n}$ such that > $ > \int_{\ol{B(0, 2^{-k_n})} \setminus B(0, 2^{-k_{n+1}})} \abs{y}dM(y) < 2^{-n} > $ > for each $n \in \nat$. For all $t \ge 0$ with $r \in [0, t]$, > $ > \begin{align*} > \abs{S_s^{(K_n)} - S_{s}^{(K_{n+1})}} &\le \sum_{j = k_n+1}^{k_{n+1}}\int_{\real^d}\abs{y}j(t, dy, X^{(j)}) \\ > \norm{S_s^{(K_n)} - S_{s}^{(K_{n+1})}}_{u, [0, t]}&= \sum_{j = k_n+1}^{k_{n+1}}\int_{\ol{B(0, 2^{-k_n})} \setminus \ol{B(0, 2^{-k_{n+1}})}} \abs{y}j(t, dy) > \end{align*} > $ > Now, by [[Markov's Inequality]], > $ > \begin{align*} > \bp\braks{\norm{S_s^{(K_n)} - S_{s}^{(K_{n+1})}}_{u, [0, t]} > 2^{-n/2}} &\le 2^{-n/2}\int_{\ol{B(0, 2^{-k_n})} \setminus \ol{B(0, 2^{-k_{n+1}})}}\abs{y}dM(y) \\ > &\le 2^{-n} > \end{align*} > $ > By the [[Borel-Cantelli Lemmas|first Borel-Cantelli lemma]], the uniform norms converge absolutely, so > $ > \sum_{n = 1}^\infty \norm{S_s^{(K_n)} - S_{s}^{(K_{n+1})}}_{u, [0, t]} < \infty > $ > [!theorem] > > The process > $ > X_t(\omega) = \int_{\real^d}{y}j(t, dy, \omega) > $ > is a [[Lévy Process]] associated with $\pi_{0, 0, M}^{(1)}$. > > *Proof*. Let $0 \le r \le s \le l$ and $\xi_1, \xi_2 \in \real^d$, then > $ > \begin{align*} > &\ev\braks{e^{i\angles{\xi_1, X_r}}e^{i\angles{\xi_2, X_{r - s}}}} \\ > &= \limv{k}\ev\braks{e^{i\angles{\xi_1, S_r^{(k)}}}e^{i\angles{\xi_2, S_{r - s}^{(k)}}}}\\ > &= \limv{k}\exp\braks{r\int_{\real^d}e^{i\angles{\xi_1, x}} - 1 dM_k(x)} \\ > &\times \exp\braks{(t - s)\int_{\real^d}e^{i\angles{\xi_2, x}}dM_k(x)} > \end{align*} > $