Prove that $\inf\left\{t>\sigma:X_t=\varepsilon\right\}$ is a stopping time

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$
• $X$ be a real-valued continuous $\mathcal F$-adapted stochastic process on $(\Omega,\mathcal A,\operatorname P)$
• $\sigma:\Omega\to[0,\infty)$ with $X_{\sigma}=0$
• $\varepsilon>0$

I want to show that $$\tau:=\inf\left\{t>\sigma:X_t=\varepsilon\right\}$$ is an $\mathcal F$-stopping time.

I know how I would prove this, but there are two things that I'm unsure of. Please take a look at the proof:

• $X$ is continuous $\Rightarrow$ $$Y_t:=\sup_{s\in[0,\:t]}1_{\left\{\:\sigma\:<\:s\:\right\}}X_s\stackrel{\color{red}{(1)}}=\sup_{s\in[0,\:t]\:\cap\:\mathbb Q}1_{\left\{\:\sigma\:<\:s\:\right\}}X_s\;\;\;\text{for }t\ge 0$$ is $\mathcal F$-adapted
• Claim 1: Let $t\ge0$ $\Rightarrow$ $$\left\{Y_t\ge\varepsilon\right\}\subseteq\left\{\tau\le t\right\}$$ Proof:
  • Let $\omega\in\left\{Y_t\ge\varepsilon\right\}$ $\Rightarrow$ $$\sigma(\omega)<t\tag2$$ (otherwise $Y_t=0<\varepsilon$)
  • $X(\omega)$ is continuous and $X_{\sigma(\omega)}(\omega)=0<\varepsilon\le Y_t(\omega)$ $\Rightarrow$ $\exists t_{\text{max}}\in\sigma(\omega),t]\ne\emptyset$ with $$X_{t_{\text{max}}}(\omega)=\sup_{s\in(\sigma(\omega),\:t]}X_s(\omega)=Y_t(\omega)\tag3$$
  • Intermediate value theorem $\Rightarrow$ $\exists s\in(\sigma(\omega),t_{\text{max}}]$ with $$X_s(\omega)=\varepsilon\tag4$$
• Claim 2: Let $t\ge0$ $\Rightarrow$ $$\left\{\tau\le t\right\}\subseteq\left\{Y_t\ge\varepsilon\right\}$$ Proof:
  • Let $$\omega\in\left\{\tau\le t\right\}=\left\{\exists s\in(\sigma,t]:X_s=\varepsilon\right\}\tag5$$
  • Then, $$\sigma(\omega)<t\tag6$$ (otherwise $(\sigma(\omega),t=\emptyset$) and hence $$Y_t(\omega)=\sup_{s\in(s\in\sigma(\omega),\:t]}X_s(\omega)\ge\varepsilon\tag 7$$
• Claim 3: Let $t\ge0$ $\Rightarrow$ $$\left\{\tau\le t\right\}\in\mathcal F_t$$ Proof:
  • $Y$ is $\mathcal F$-adapted $\Rightarrow$ $$\left\{\tau\le t\right\}=\left\{Y_t\ge\varepsilon\right\}\in\mathcal F_t\tag8$$ by Claim 1 and Claim 2

Question: (1.) I'm unsure if $\color{red}{(1)}$ is really correct. I know that it is correct, if the supremum over $[0,t]\cap\mathbb Q$ exists, but I don't think that this is guaranteed. For example, $$\inf\left([\sqrt 2,\:2]\cap\mathbb Q\right)$$ doesn't exist. Am I correct? If so, how can we fix that?

(2.) I didn't said that $\sigma$ should be an $\mathcal F$-stopping time. In fact, I don't think that we need this or any other measurability assumption on $\sigma$. Am I missing something?