Let $(X_t)_t$ be a standard Levy process. We assume $t\mapsto X_t$ is continuous in probability or a.s. cadlag.
In this case, why does the continuity of $x\mapsto e^{i\xi \cdot x}$ implies that there is for every $\epsilon >0$ some $\delta>0 $ such that $$E|e^{i\xi\cdot (X_t - X_s)}-1|\le \epsilon + 2P(|X_t - X_s|\ge \delta)?$$
I think I have to use that $e^{i\xi\cdot x}$ is $1$ if $x $ is near zero but I've been thinking about a way to understand this but could not come up with a clear solution. I would greatly appreciate some help.
Given $\epsilon >0$ there exists $\delta >0$ such that $$E|e^{i\xi\cdot (X_t - X_s)}-1|\le \epsilon + 2P(|X_t - X_s|\ge \delta.$$ (for fixed $\xi$. In fact we can make this uniform for $\xi$ bounded).
To see that write $$E|e^{i\xi\cdot (X_t - X_s)}-1|=E|e^{i\xi\cdot (X_t - X_s)}-1|1_{|X_t-X_s| \leq \delta}+E|e^{i\xi\cdot (X_t - X_s)}-1|1_{|X_t-X_s|> \delta}.$$ In the first term use continuity of the exponential function at $0$ and in the second term use the fact that $|e^{i\xi\cdot (X_t - X_s)}-1|\leq 2$