# For RCLL Functions > [!theorem] > > Let $f \in D([0, \infty))$ be a [[RCLL]] function, $t \ge 0$ and $r > 0$, and let $J_f(0, r) = 0$. For $t > 0$, define > $ > J_f(t, r) = \bracs{s \in [0, t]: \abs{f(s) - f(s-)} > r} > $ > to be the set of discontinuities of $f$ with jump size $r$. Then for any $r > 0$, $t > 0$, $J_f(t, r)$ is finite. Let > $ > J_f(t) = \bracs{s \in [0, t]: f(s) \ne f(s-)} > $ > then $J_f(t) = \bigcup_{n \in \nat}J_f(t, 1/n)$, and $J_f(t)$ is countable. > > *Proof*. Suppose that $J_f(t, r)$ is infinite. Let $\seq{x_n} \subset J_f(t, r)$, then there exists an accumulation point $x$. This means that $f(x-)$ or $f(x+)$ does not exist. > [!theorem] > > Let $f \in D([0, \infty))$, $t \ge 0$. If $t = 0$, define $j_f(t, dy) = \delta_0$. If $t > 0$ and $E \in \cb(\real^d)$, define > $ > j_f(t, B) = \sum_{s \in J_f(t)}\one_{B}(f(s) - f(s-)) = \sum_{s \in J_f(t)}\delta_{f(s) - f(s-)} > $ > then $j_f(t, dy)$ is a $\sigma$-finite Borel measure with $j_f(t, dy)(\bracs{0}) = 0$. For any $r > 0$, $j_f(t, \real^d \setminus B(0, r)) < \infty$, making it of class $\mathfrak{M}_\infty$. $j_f(t, dy)$ is known as the **jump measure** associated with $f$. > [!theorem] > > Let $f \in D([0, \infty))$ and $\varphi \in BC(\real^d, \complex)$ be a [[Space of Bounded Continuous Functions|bounded continuous]] function such that there exists $r > 0$ with $\varphi|_{B(0, r)} = 0$, then $\varphi \in L^1(j_f(t, dy))$ where > $ > \int \varphi j_f(t, dy) = \lim_{n \to \infty}\sum_{m = 0}^{2^n}\varphi(f(m \cdot 2^{-n}t) - f((m-1) \cdot 2^{-n}t)) > $ > and the mapping $t \mapsto \int \varphi j_f(t, dy)$ is right-continuous and piecewise constant. > > By collecting values from the dyadic filter, the mapping $f \mapsto \int \varphi j_f(t, dy)$ is measurable with respect to $\Sigma^{[0, \infty)}$. > > *Proof*. Let $t \ge 0$, $r > 0$, and $f \in D([0, \infty))$, then there exists $\delta_t > 0$ such that for all $s, s' \in [t, t + \delta_t]$, $\abs{f(s') - f(s)} \le r/2$. Therefore $J_f(t, r) = J_f(t + \delta_0, r)$. # For Measures > [!definition] > > Let $M \in \mathfrak{M}_\infty$, then there exists a mapping $\omega \mapsto j(t, dy, \omega) \in \mathfrak{M}_\infty$, known as the **jump measure driven by** $M$, such that > 1. For any Borel set $\Gamma$ with $0 \not\in \ol{\Gamma}$, $j(t, \Gamma)$ is a simple Poisson process with parameter $M(\Gamma)$. > 2. For any finite partition $\bracs{\Gamma_n}$ of $\real^d$ such that $0 \not\in \ol{\bigcup_{n}\Gamma_n}$, the family $\bracs{j(t, \Gamma_n)}$ is independent. > 3. For any $r > 0$, $M_r = M|_{B(0, r)^c} \in \mathfrak M_0$. Moreover, $\int_{\real^d \setminus B(0, r)}yj(t, dy)$ is a [[Lévy Process]] for $\pi_{0, 0, M}$. > > *Proof*. Let $A_0 = \ol{B(0, 1)}^c$ and $A_k = \ol{B(0, 2^{k + 1})} \setminus \ol{B(0, 2^k)}$. Let $M_k = M|_{A_k}$, and $X_k$ be a compound Poisson process associated with $M_k$ such that $\seq{X_k}$ is mutually independent. Let $j(t, dy) = \sum_{k \in \nat}j(t, dy, X_k)$, then $j(t, dy)$ is $\sigma$-finite and $j(t, dy)(\bracs{0}) = 0$. For any $B(0, r)$, $j(t, dy)|_{B(0, r)}$ is a finite sum after cutting off the tail. Therefore $j(t, dy)$ is of class $\mathfrak{M}_\infty$. > > Let $K \ge 0$ such that $\Gamma \subset \bigcup_{k \le K}A_k$. In which case, $j(t, \Gamma)$ comes from a compound Poisson process driven by $\sum_{k \le K}M_k$. Thus $j(t, \Gamma)$ is a simple Poisson process with parameter $M_k(\Gamma) = M(\Gamma)$.