I am in a situation where I was able to prove that for a stochastic process $X= \{X_t: t \in R_+\} $of relevance to me is such that for every $\delta \in (0, 1) $ I can find a number $K_\delta $ such that $$P[|X_t|> K_\delta ] < \delta \hskip 10pt \forall t \in [0, n]. $$
The process can be assumed to be a Levy process, or at least a process with independent increments and continuous in probability.
Using an argument based on downcrossings, it is possible to prove that if one considers an index set like $\{0 = r_0 < r_1 < \ldots < r_m\} \in Q\cap [0, n] $ then, $$ P[max_{1\le i \le m} |X_{r_i}| > b] < c. $$
K.Ito in his Stochastic Processes concludes that this is sufficient to conclude that
$$ P[\sup_{0 \le t \le n} |X_t| > b] < c, $$
as well.
I am convinced that the fact that $Q \cap [0, n] $ is dense in $[0, n] $ has to play a key role, but I do not seem to be able to put together a rigorous argument.
Any hint would be appreciated. Thank you.
Maurice
Suppose that the stochastic process $(X_t)_{t \geq 0}$ satisfies
$$\mathbb{P} \left( \max_{1 \leq i \leq m} |X_{t_i}| > b \right) \leq c \tag{1}$$
for any $t_i \geq 0$, $i=1,\ldots,m$ for some fixed constants $b,c>0$.
If $(t_i)_{i \in \mathbb{N}}$ is a sequence of non-negative numbers, then
$$\left\{ \max_{1 \leq i \leq m} |X_{t_i}|>b \right\} \uparrow \left\{ \sup_{i \geq 1} |X_{t_i}|>b \right\}.$$
(This follows directly from the definition of the supremum; if $\sup_{i \geq 1} |X_{t_i}(\omega)|$ is strictly larger than $b$, then we can find some index $m$ such that $\sup_{1 \leq i \leq m} |X_{t_i}(\omega)|>b$.) Using the continuity of the probability measure $\mathbb{P}$, we get
$$\mathbb{P} \left( \sup_{i \geq 1} |X_{t_i}|>b \right) = \lim_{m \to \infty} \mathbb{P} \left( \sup_{1 \leq i \leq m} |X_{t_i}|>b \right) \stackrel{(1)}{\leq} c. \tag{2}$$
This shows that $(1)$ extends to countable index sets. If we assume additionally that $(X_t)_{t \geq 0}$ has càdlàg sample paths (right-continuous with finite left-hand limits), then
$$\sup_{t \in [0,n]} |X_t| = \sup_{t \in [0,n] \cap \mathbb{Q}} |X_t|.$$
Choosing an enumeration $(t_i)_{i \in \mathbb{N}}$ of $[0,n] \cap \mathbb{Q}$, it follows from $(2)$ that
$$\mathbb{P} \left( \sup_{t \in [0,n]} |X_t|>b \right) = \mathbb{P} \left( \sup_{i \in \mathbb{N}} |X_{t_i}|>b \right) \leq c.$$