Let $W$ be a Brownian motion, then is the following a stopping time with respect to its filtration: $$\tau=\sup [t ∈ [0, 1] : W(t) \ge b]$$ for $b \in \mathbb{R}$.
2026-04-02 16:00:16.1775145616
Supremum of stopping times
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Think of it this way. The random time $\tau$ is a stopping time if at time $t$, you know whether $\tau$ has happened or not. What is $\tau$ intuitively? It is the latest time for which $W(t) \geq b$. How can you know whether $\tau$ has happened (i.e. $W(t) < b$ for $t > \frac{1}{4}$) at $t=\frac{1}{4}$?. You can't, so intuitively you know that $\tau$ should not be a stopping time.
More rigorously, you need to show that $\left\{\tau \leq t\right\} \not \in \mathcal{F}_t$ for some $t$. Think about the case when $t = 0$ and see if there is an obvious way to show this.