Suppose $X$ is a cadlag process adapted to $\{\mathcal F_t\}$ and $H$ is a closed set.Verify $\sigma_H\triangleq\inf\{t\ge0:X_t(\omega)\in H\}$ is a stopping time .
The first step is:
$$\{\sigma_H\le t\}=\{X_0\in H\}\cup\{X_s\in H\text{ or }X_{s-}\in H \text{ for some }s\in\color{red}{(}0,t]\}$$
I wonder if I can omit the "$\text{ or }X_{s-}\in H$"?i.e. $$\{\sigma_H\le t\}=\{X_0\in H\}\cup\{X_s\in H\text{ for some }s\in\color{red}{(}0,t]\}$$
The equality
$$\{\sigma_H \leq t\} = \{X_0 \in H\} \cup \{X_s \in H \, \text{or} \, X_{s-} \in H \, \text{for some} \, s \in (0,t]\} \tag{1}$$
does not hold. Consider the following counter example:
Consider the (deterministic) cádlág process $$X_t := \begin{cases} t & t<1 \\ 0 & t \geq 1. \end{cases}$$ and set $H:=\{1\}$. Then it follows easily from the definition of $\sigma_H$ that $\sigma_H = \infty$ and therefore
$$\{\sigma_H \leq t\} = \emptyset$$
for all $t \geq 0$. On the other hand, the right-hand side of $(1)$ does not equal $\emptyset$ for $t \geq 1$; this follows from the fact that $X_{1-} = 1 \in H$.