> [!definition] > > Let $f \in (\real^d)^{[0, \infty)}$, if > 1. $f$ is right-[[Continuity|continuous]]. > 2. For any $x \in [0, \infty)$, $f(x-) = \lim_{y \upto x}f(y)$ exists. > > then $f \in D([0, \infty), \real^d) = D([0, \infty))$ is a RCLL function. > [!theorem] > > Let $f \in D([0, \infty))$, then $f$ is bounded on bounded sets. Moreover, $\norm{f}_u$ can be captured using dyadic rational numbers as > $ > \norm{f|_{[0, t]}}_u = \lim_{n \to \infty}\max_{1 \le m \le 2^n}\norm{f(2^nmt)} > $ > > *Proof*. Let $K$ be compact. Suppose that $\norm{f|_K}_{u} = \infty$, then there exists a sequence $\seq{x_n}$ such that $f(x_n) \to \infty$ (WLOG). Let $x$ be an accumulation point of $\seq{x_n}$. If there exists a subsequence $\seq{x_{n_k}}$ such that $x_{n, k} \upto x$, then $f(x-)$ does not exist. If there exists a subsequence $\seq{x_{n_k}}$ such that $x_{n_k} \downto x$, then $f(x) = \infty$. > [!theorem] > > Let $f \in D([0, \infty))$, and $I = [0, t] \subset [0, \infty)$ be an interval, and > $ > \norm{f}_{v} = \sup\bracs{\sum_{j = 1}^p\abs{f(x_j) - f(x_{j - 1})}: 0 = x_0 \le \cdots \le x_p = t, p \in \nat} > $ > be its total variation, then > $ > \norm{f}_{v} = \sup\bracs{\sum_{j = 1}^{2^n}\abs{f(jt/2^n) - f((j - 1)t/2^n)}: n \in \nat} > $