A Levy process is a.s. continuous I have to proof this:
If  is a Levy process then for each  the sample path  is, with probability 1, continuous as s=t.
This is the proof:

I don't understand the conclusion. Can someone explain to me how the proof ends?
Is there an alternative way to do this?
Thanks
 A: $\newcommand{\R}{\mathbb R}\newcommand{\Q}{\mathbb Q}$This proof is insufficient (where did you take it from?).
Indeed, the logic at the end of the proof boils to down to

Proposition 1: Suppose that for some real $t>0$, some function $[0,\infty)\ni u\mapsto x_u\in\R$, and all $q\in\Q$ we have
\begin{equation*}
    e^{iqx_s}\to e^{iqx_t} \tag{1}\label{1}
\end{equation*}
(as $s\uparrow t$). Then $x_s\to x_t$.

However, this proposition is false in general. E.g., suppose that
\begin{equation*}
    x_s=x_t+2\pi\Big(\Big\lceil\frac1{t-s}\Big\rceil\Big)!
\end{equation*}
for some real $t>0$ and all $s\in[0,t)$. Then \eqref{1} holds for each $q\in\Q$, whereas $x_s\not\to x_t$ (as $s\uparrow t$).
(The rate of growth of $x_s$ to $\infty$ can be made arbitrarily slow by replacing $\frac1{t-s}$ by $g(t-s)$, where $g(u)$ grows to $\infty$ arbitrarily slowly as $u\downarrow0$.)

However, with the additional assumption that the function $(x_u)$ be locally bounded, Proposition 1 would be true. (See also the remark at the end of this answer.) Indeed, let
\begin{equation*}
    y_s:=\frac{x_s-x_t}{2\pi}. 
\end{equation*}
Then \eqref{1} can be rewritten as
\begin{equation*}
    qy_s-k_{q,s}\to0 \tag{1a}\label{1a}
\end{equation*}
for each $q\in\Q$ (as $s\uparrow t$), where $k_{q,s}$ is an integer depending only on $q$ and $s$. Since the function $(x_u)$ is locally bounded, \eqref{1a} implies that without loss of generality (wlog) $2|k_{q,s}|<k_q$ for some integer $k_q$ depending only on $q$.
We want to show that $y_s\to0$. To obtain a contradiction, suppose the contrary. Then wlog $y_s\to c$ for some real $c\ne0$ (because the function $(x_u)$ be locally bounded). So, for any nonzero $q\in\Q$ wlog $k_{q,s}\ne0$ and hence
\begin{equation*}
    k_{q,s}\sim qy_s
\end{equation*}
and, similarly and therefore,
\begin{equation*}
    k_{q/k_q,s}\sim (q/k_q)y_s\sim k_{q,s}/k_q\in[-1/2,1/2],
\end{equation*}
which contradicts the condition that $k_{q/k_q,s}$ is a nonzero integer. $\quad\Box$
Remark: For $(x_u)$ to be locally bounded, it is enough that $(x_u)$ be right-continuous in $u\ge0$ and have finite left limits in $u>0$. Indeed, then for each real $t\ge0$ there is a neighborhood $U_t$ of $t$ in $[0,\infty)$ such that $(x_u)$ is bounded in $U_t$. Now the local boundedness of $(x_u)$ follows by the Heine--Borel lemma.
