I started reading Sato's book on Lévy processes. On page 6 it says that for $(X_t)_{t\ge 0}$ Lévy process
$$X_t=X_{t-},$$ for any fixed $t>0$ almost surely. It is mentioned it follows from the fact that almost sure convergence implies convergence in probability. I think it makes no sense. Can someone explain it?
On
Let $\varepsilon>0$ and denote by $\Delta X_t := X_t-X_{t-}$ the jump height at time $t$. Then
$$\begin{align} \mathbb{P}(|\Delta X_t|>\varepsilon) &= \mathbb{P} \left( \bigcup_{j \in \mathbb{N}} \bigcap_{k \geq j} |X_t-X_{t-1/k}| > \varepsilon \right) \\ &= \mathbb{P} \left( \liminf_{k \to \infty} |X_t-X_{t-1/k}|>\varepsilon \right) \\ &\leq \liminf_{k \to \infty} \mathbb{P}(|X_{1/k}|>\varepsilon) = 0 \end{align}$$
where we used Fatou's Lemma and the fact that $(X_t)_t$ has stationary increments (i.e. $X_t-X_{t-1/k} \sim X_{1/k}$) and is continuous in probability. Consequently,
$$\mathbb{P}(X_t \not= X_{t-}) = \mathbb{P}\left( \bigcup_{k \in \mathbb{N}} |\Delta X_t|> \frac{1}{k} \right) = 0$$
From the definition of a Levy process, for $s <t$, $X_t - X_s$ has the same distribution as $X_{t-s}$ and $ \lim_{u \to 0^+} X_u = 0 \ \text{ almost surely.}$ So $$ X_{t-} = \lim_{s \to t} X_s = \lim_{s \to t} \{ X_t - (X_t - X_s )\} \underset{as}{=} X_t - X_0 = X_t. $$