# Prove the following stopping time Cannot be a predictable stopping time.

Fix a filtered probability space $$(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq 0},\mathbb{P})$$ and let $$b$$ be a measurable Bernoulli random varaible independent from the filtration $$(\mathcal{F}_t)_{t\geq 0}$$. Define random time $$\tau:= \infty I_{\{b=0\}}$$ and the enlarged filtration $$\mathcal{G}_t:=\mathcal{H}_{t+}$$ where $$\mathcal{H}_t:=\mathcal{F}_t\vee \sigma(\tau\wedge t)$$ for all $$t\geq 0$$. (I edited filtration $$\mathcal{G}$$ to be right-continuous)

Question: Show that random time $$\tau$$ $$\textbf{cannot}$$ be a $$(\mathcal{G}_t)_{t\geq 0}$$-predictable stopping time.

Recall the definition of predictable stopping time: A stopping time $$T$$ is said to be predictable if there exists a sequence of stopping times $$T_n$$ such that $$T_n$$ is increasing, $$T_n on $$\{T>0\}$$ for all n, and $$\lim_{n\rightarrow \infty}=T$$ a.s..

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However, my following example shows that $$\tau$$ is a $$(\mathcal{G}_t)_{t\geq 0}$$-predictable stopping time and I don't know where I did wrong.

First, it is clear that $$\tau$$ is a $$(\mathcal{G}_t)_{t\geq 0}$$-stopping time because $$\{\tau\leq t\}\in \sigma(\tau\wedge t)\subset\mathcal{G}_t$$ for all $$t> 0$$. At $$t=0$$, $$\{\tau=0\}=\{b=1\}\in \mathcal{H}_s$$ for any $$s>0$$. Since $$\mathcal{G}_0=\mathcal{H}_{0+}$$, hence, $$\{\tau=0\}\in \mathcal{G}_0$$.

Then, consider the random times $$\tau_n:=n I_{\{b=0\}}$$ for $$n\geq 1$$. They are also $$(\mathcal{G}_t)_{t\geq 0}$$-stopping times. Moreover, $$\lim_{n\rightarrow\infty}\tau_n=\tau$$ and $$\tau_n=n<\tau=\infty$$ when $$\tau>0$$. Hence, $$\tau$$ is announced by $$\tau_n$$.