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, $T_n<T$ on $\{T>0\}$ for all n, and $\lim_{n\rightarrow \infty}=T$ a.s..
$\hspace{2cm}$
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 $\tau_n$.
