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
- 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). $${}$$