Can we extend a Lévy process at $t=\infty$?

Question

Let $E$ be a normed $\mathbb R$-vector space and $(X_t)_{t\ge0}$ be an $E$-valued Lévy process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$, i.e.

1. $X_0=0$;
2. $X$ is $\mathcal F$-adapted and continuous in probability;
3. $X_{s+t}-X_s$ is independent of $\sigma(X_r,r\le s)$ for all $s,t\ge0$;
4. $X_{s+t}-X_s\sim X_t$ for all $s,t\ge0$.

As usual, we can extend $(X_t)_{t\ge0}$ to a process with time scale $[0,\infty]$: Let $$\mathcal F_\infty:=\sigma(X_t,t\ge0)$$ and $X_\infty$ be an $E$-valued $\mathcal F_\infty$-measurable random variable on $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$.

Clearly, $(\mathcal F_t)_{t\in[0,\:\infty]}$ is a filtration on $(\Omega,\mathcal A)$ and $(X_t)_{t\in[0,\:\infty]}$ is $(\mathcal F_t)_{t\in[0,\:\infty]}$-adapted.

But is $(X_t)_{t\in[0,\:\infty]}$ an $(\mathcal F_t)_{t\in[0,\:\infty]}$-Lévy process?

I'm not sure if such a claim really makes sense at all, since I'm not sure what a suitable extension of property (4.) is. However, in order to extend (3.), we would need to show that $X_\infty-X_s$ is independent of $\mathcal F_s$ for all $s\in[0,\infty)$. Can we at least do this?