If $t_n\uparrow t$, then $\bigcup_{n\in\mathbb N}\left\{\tau\le t_n\right\}=\left\{\tau\le t\right\}$

Question

Let

• $(\Omega,\mathcal A)$ be a measurable space
• $I\subseteq\mathbb R$ be bounded above and nonempty
• $\tau:\Omega\to I\cup\sup I$
• $(t_n)_{n\in\mathbb N}$ be nondecreasing with $$t_n\xrightarrow{n\to\infty}t:=\sup I\tag1$$

I want to show that $$\bigcup_{n\in\mathbb N}\left\{\tau\le t_n\right\}=\left\{\tau\le t\right\}\;.\tag2$$

Since $$\left\{\tau\le t_n\right\}\subseteq\left\{\tau\le t\right\}\;\;\;\text{for all }n\in\mathbb N\;,\tag3$$ we've clearly got "$\subseteq$". For the other inclusion, let $$\omega\in\left(\bigcup_{n\in\mathbb N}\left\{\tau\le t_n\right\}\right)^c=\bigcap_{n\in\mathbb N}\left\{\tau>t_n\right\}\;.\tag4$$ By definition, $$\tau(\omega)>t_n\;\;\;\text{for all }n\in\mathbb N\,\tag5$$ but this only yields $$\tau(\omega)\ge\sup_{n\in\mathbb N}t_n=t\;,\tag6$$ i.e. $$\omega\in\left\{\tau\ge t\right\}=\left\{\tau<t\right\}^c\;.\tag7$$

So, which argument do we need to conclude $\omega\in\left\{\tau\le t\right\}^c$?

Answer 1

Unless I'm missing something $\bigcup_n \{\tau \leq 1 - 1/n\}$ is a counter example. Note that none of the sets contain $1$ so it's union can't be $\{ \tau \leq 1\}$.