Show the existence of a right-continuous modification Suppose ($X_t$)$_{t \geq 0}$ is a stochastic process with independent increments and the function $t \rightarrow \mathbb{E}X_t$ is continuous. Prove that $(X_t)$ has a right-continuous modification.
I have no idea how to tackle this problem, I know about the Kolmogorov's continuity criterion but it only works to show the existence of continuous modifications and here we can give the Poisson process as a counterexample - it has independent increments and $\mathbb{E}X_t \equiv t$ is continuous, but it doesn't have a continuous modification.
 A: Because of the independent increments of $(X_t)$, the process $M_t:=X_t-\Bbb E[X_t]$ is a martingale and $t\mapsto \Bbb E[X_t]$ is continuous (even constant). Consequently, $(M_t)$ admits a right-continuous modification, hence so does $(X_t)$. (For this argument it suffices that $t\mapsto \Bbb E[X_t]$ be right continuous.)
(More detail added as follow up to the  comment by Byron Schmuland.) Let me assume that $X_0=0$. Then $(M_t)$ defined above is a martingale with respect to the filtration $\mathcal F_t:=\sigma(X_v-X_u: 0\le u\le v\le t)$. By a cousin of Kolmogorov's zero-one law, the $\sigma$-algebra $\mathcal F_{t+}:=\cap_{s>t}\mathcal F_s$ differs from $\mathcal F_t$ only by events of probability $0$, for each $t\ge 0$. (See p. 363 in Doob's Stochastic Processes, and reverse time.) The process $M^+_t:=\limsup_{s>t,s\in\Bbb Q}M_s$ is then almost surely right continuous, is an $(\mathcal F_{t+})$-martingale, and (by the zero-one law just noted) is a modification of $(M_t)$, by the standard regularization theory for martingales (for example, by Theorem 7.27 in Kallenberg's Foundations of Modern Probability).
