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.