For a Levy-process $(X_t)_{t\geq 0}$ with stationary indepedent increments which is a markov process, we know that its law is defined by its one dimensional distribution. This is so because for its cumulant generating $k_1(s)=log(E[exp(sX_1)])$ we have $k_t(s)=log(E[exp(sX_t)])=t\cdot k_1(s)$, where $s\in \mathbb{R}$ so that the expression is finite. The cumulant generating function $k_t$ determines the distribution function of $X_t$. Despite that fact, we know , that $X_{t}=\sum_{i=1}^{n}X_{t_i}-X_{t_{i-1}}$ for arbitary $n$, where the terms in the sum are i.i.d.
Lets say we have a function $f((X)_{0\leq u\leq t})$. Is this $\sigma(X_t)$ measurable?