Given that a a Levy process $L=L(t)_{0 \leq t \leq T}$ is stochastic continuous i.e $\forall t $ and $\forall \epsilon>0$ we have that $$\lim_{s \to t} P(|L(t)-L(s)|>\epsilon) =0$$
then my classnotes on computational finance claim that it is clear that $\Delta L(t)=0$ P-a.s where $\Delta L(t)=L(t)-L(t_{-}$ where $L(t_{-})=\lim_{s\uparrow t} L(s)$ from the left(Please note that L(t) is a levy process and therefore cadlag). It says this is obviously true but it is far from obvious. Any hints on how could i show this? My attempt:
Could I just say that \begin{align} &P(|\Delta L(t)|\leq\epsilon )=P(|L(t)-L(t_{-})|\leq\epsilon)=P(|L(t)-\lim_{s \uparrow t}L(s)|\leq\epsilon)=\\ &\qquad=\lim_{s\uparrow t}P(|L(t)-\lim_{s \uparrow t}L(s)|\leq\epsilon)=1-\lim_{s\uparrow t}P(|L(t)-\lim_{s \uparrow t}L(s)|>\epsilon)=\\ &\qquad=1-\lim_{s\to t}P(|L(t)-\lim_{s \uparrow t}L(s)|>\epsilon)=1-0=1 \end{align}
The last equality follows from the definition of stochastic continuity and it implies that $P(\Delta L(t))=0$ P.as as $\epsilon$ can be as small as we want.
This to me looks like a fake forced proof . Could someone show me how to prove this? Thanks
How do you conclude that
$$\lim_{s \uparrow t} \mathbb{P} \left( \left|L_t- \lim_{s \uparrow t} L_s \right|>\epsilon \right)=0$$
...? What we do know is that
$$\lim_{s \uparrow t} \mathbb{P}(|L_t-L_s|\epsilon) = 0,$$
but that's something totally different.
Since $(L_t)_{t \geq 0}$ is right-continuous with left lomits, we have $L_{t-} = \lim_{s \uparrow t} L_s$ almost surely; hence, in particular $L_{t-} = \lim_{s \uparrow t} L_s$ in probability. On the other hand, the identity
$$\lim_{s \uparrow t} \mathbb{P}(|L_t-L_s|>\epsilon)=0 \qquad \text{for all $\epsilon>0$}$$
shows that $\lim_{s \uparrow t} L_s = L_t$ in probability. Now it follows from the uniqueness of the limit that $L_t = L_{t-}$ almost surely. This implies $\Delta L_t=0$ almost surely.