A $[0,\infty)-$random variable is said to be a $\mathcal{F}_{t}-$stopping time with if for all $t\geq 0$, we have $\{\tau\leq t\}\in\mathcal{F}_{t}$. Also, to analyze if a filtration is right continuous, we define $$\mathcal{F}_{t+}:=\bigcap_{\epsilon>0}\mathcal{F}_{t+\epsilon}=\bigcap_{s>t}\mathcal{F}_{s}.$$
Then, I am reading an online note which claims that:
A $[0,\infty)-$random variable $\tau$ is $\mathcal{F}_{t+}-$stopping time if and only if $\{\tau<t\}\in\mathcal{F}_{t}$ for each $t\geq 0$.
I have no problem showing $(\Leftarrow)$. However, I don't understand the one-line proof for $(\Rightarrow)$ given by the note:
$(\Rightarrow)$. $$\{\tau<t\}=\bigcup_{n=1}^{\infty}\{\tau\leq 1-\frac{1}{n}\}\in\mathcal{F}_{t}\subset\mathcal{F}_{t+}.$$
I have no problem with the first equality, which is a common technique we used in measure theory.
I also understand the last equality, by definition $\mathcal{F}_{t+}$ is the intersection of all the intersection of $\mathcal{F}_{s}$ for $s>t$, but $\mathcal{F}_{t}$ is a filtration, so we have $\mathcal{F}_{t}\subset\mathcal{F}_{s}$ for each $s$, and thus holds for the intersection.
However, I don't understand why we have $$\bigcup_{n=1}^{\infty}\{\tau\leq 1-\frac{1}{n}\}\in\mathcal{F}_{t}.$$ By hypothesis we only know $\{\tau\leq t\}$ is $\mathcal{F}_{t+}-$measurable, what could I do to get such a relationship?
Also, is it necessary to write $\mathcal{F}_{t}\subset\mathcal{F}_{t+}$? what does this imply?
Thank you!
You have written $\{\tau \leq 1-\frac 1 n\}$ instead of $\{\tau \leq t-\frac 1 n\}$.
For any $n \geq 1$ we have $\{\tau \leq t-\frac 1 n\} \in \mathcal F_{(t-\frac 1 n)+}$. But $\mathcal F_{u+} \leq \mathcal F_v$ for any $v >u$ by definition. Hence $\mathcal F_{(t-\frac 1 n)+} \subseteq \mathcal F_t$.