Return to Answer

In the case of Brownian motion you can look at functions of the form $$g(\tau_n \wedge t) e^{i \sum \lambda_i(B_{t_i} - B_t)}$$ where $t_i > t$. Since the expectation is $$ \mathbb E(g(\tau_n \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ which converges to $$ \mathbb E(g(\tau \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ $g(\tau \wedge t)$ is independent of $\mathcal F (B_s - B_t), s>t$ which should make it $\mathcal F_t$ measurable. $${}$$ As pointed out below, that doesn't work, however the brownian case can be done completely differently by

1. continuous bounded functions of $B_{t_1},...,B_{t_n}$ are dense in $L^2(\mathcal F_{\infty}) $ because, e.g. approximate by functions of the form given in lemma 5.3.1 of Revuz and Yor, and once you have approximated truncate to get the bounded part. You could also use Wiener chaos.$${}$$2. Let $\epsilon > 0$. Approximate $\tau \wedge t$ to within $\frac {\epsilon} 2$ by a continuous bounded function of $B_{t_1},...,B_{t_n}$ which I'll call f. By the hypotheses $\mathbb E(f-\tau_n \wedge t)^2 \rightarrow \mathbb E(f-\tau \wedge t)^2 < \frac {\epsilon} 2$ Therefore by the triangle inequality $limsum E(t_n \wedge t - \tau \wedge t)^2 < \epsilon$. This shows that $\tau_n \wedge t \rightarrow \tau \wedge t$ in $L^2$, and therefore $\tau^t $ is a stopping time. (exercise 1.4.17 of revuz & yor). $${}$$

In the case of Brownian motion you can look at functions of the form $$g(\tau_n \wedge t) e^{i \sum \lambda_i(B_{t_i} - B_t)}$$ where $t_i > t$. Since the expectation is $$ \mathbb E(g(\tau_n \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ which converges to $$ \mathbb E(g(\tau \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ $g(\tau \wedge t)$ is independent of $\mathcal F (B_s - B_t), s>t$ which should make it $\mathcal F_t$ measurable. $${}$$ As pointed out below, that doesn't work, however the brownian case can be done completely differently by

1. continuous bounded functions of $B_{t_1},...,B_{t_n}$ are dense in $L^2(\mathcal F_{\infty}) $ because, e.g. lemma 5.3.1 of Revuz and Yor shows that linear combinations of $ sin(\sum \lambda_i B_{t_i}), cos(\sum \lambda_i B_{t_i})$ are dense.  $${}$$2. Let $\epsilon > 0$. Approximate $\tau \wedge t$ to within $\frac {\epsilon} 2$ by a continuous bounded function of $B_{t_1},...,B_{t_n}$ which I'll call f. By the hypotheses $\mathbb E(f-\tau_n \wedge t)^2 \rightarrow \mathbb E(f-\tau \wedge t)^2 < \frac {\epsilon} 2$ Therefore by the triangle inequality $limsum E(t_n \wedge t - \tau \wedge t)^2 < \epsilon$. This shows that $\tau_n \wedge t \rightarrow \tau \wedge t$ in $L^2$, and therefore $\tau^t $ is a stopping time. (exercise 1.4.17 of revuz & yor). $${}$$

In the case of Brownian motion you can look at functions of the form $$g(\tau_n \wedge t) e^{i \sum \lambda_i(B_{t_i} - B_t)}$$ where $t_i > t$. Since the expectation is $$ \mathbb E(g(\tau_n \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ which converges to $$ \mathbb E(g(\tau \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ $g(\tau \wedge t)$ is independent of $\mathcal F (B_s - B_t), s>t$ which should make it $\mathcal F_t$ measurable.

In the case of Brownian motion you can look at functions of the form $$g(\tau_n \wedge t) e^{i \sum \lambda_i(B_{t_i} - B_t)}$$ where $t_i > t$. Since the expectation is $$ \mathbb E(g(\tau_n \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ which converges to $$ \mathbb E(g(\tau \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ $g(\tau \wedge t)$ is independent of $\mathcal F (B_s - B_t), s>t$ which should make it $\mathcal F_t$ measurable. $${}$$ As pointed out below, that doesn't work, however the brownian case can be done completely differently by

1. continuous bounded functions of $B_{t_1},...,B_{t_n}$ are dense in $L^2(\mathcal F_{\infty}) $ because, e.g. approximate by functions of the form given in lemma 5.3.1 of Revuz and Yor, and once you have approximated truncate to get the bounded part. You could also use Wiener chaos.$${}$$2. Let $\epsilon > 0$. Approximate $\tau \wedge t$ to within $\frac {\epsilon} 2$ by a continuous bounded function of $B_{t_1},...,B_{t_n}$ which I'll call f. By the hypotheses $\mathbb E(f-\tau_n \wedge t)^2 \rightarrow \mathbb E(f-\tau \wedge t)^2 < \frac {\epsilon} 2$ Therefore by the triangle inequality $limsum E(t_n \wedge t - \tau \wedge t)^2 < \epsilon$. This shows that $\tau_n \wedge t \rightarrow \tau \wedge t$ in $L^2$, and therefore $\tau^t $ is a stopping time. (exercise 1.4.17 of revuz & yor). $${}$$

In the case of Brownian motion you can look at functions of the form $$g(\tau_n \wedge t) e^{i \sum \lambda_i(B_{t_i} - B_t)}$$ where $t_i > t$. Since this is 0 for all n, it is 0 for $\tau$ as well, implying that $g(\tau \wedge t)$ is independent of the future. I think this should make it $\mathcal F_t$ measurable.

In the case of Brownian motion you can look at functions of the form $$g(\tau_n \wedge t) e^{i \sum \lambda_i(B_{t_i} - B_t)}$$ where $t_i > t$. Since the expectation is $$ \mathbb E(g(\tau_n \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ which converges to $$ \mathbb E(g(\tau \wedge t)) \mathbb E(e^{i \sum \lambda_i(B_{t_i} - B_t)})$$ $g(\tau \wedge t)$ is independent of $\mathcal F (B_s - B_t), s>t$ which should make it $\mathcal F_t$ measurable.