$t \wedge \tau \uparrow \tau$ also if $\tau$ is not a.s. finite?

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I need a check on the following reasoning, that happens everytime I do an exercise about the stopping theorem.


Let $\tau$ be a stopping time w.r.t some filtration

Usual in the exercises about the optional stopping theorem I read:

$$ t \wedge \tau \uparrow \tau$$ And that's true if $\tau$ is a.s. finite.


BUT if $\tau$ not a.s. finite, can I still say $t \wedge \tau \uparrow \tau$?

I can note that if $\tau = \infty$ a.s., then for every $ t \in \mathbb{R}$ I have $t \wedge \tau =t$ and hence taking the limit I just can say that $$\lim_t t \wedge \tau = \infty$$ (which is a.s. the value of $\tau$)

All in all, I would say that it's true that $t \wedge \tau \uparrow \tau$ also if $\tau$ is NOT a.s. finite

Is it correct or am I missing something?