Suppose we define a stochastic process $X_t$ and a sequence $X^n_t$ as follows.
\begin{align*} X_t =x_0 + \int_0^t \sigma_s dB_s \\ X^n_t = x_0 + \int_0^t (\sigma_s \wedge n) dB_s \end{align*} where $B_t$ is a standard Brownian motion, $x_0 \in (-1,1)$ and $n \in \mathbb{N}$. Also, $\sigma_s$ is a progressively measurable non-negative stochastic process.
From what I understand, the above implies that $X^n_t \xrightarrow{\text{ ucp}} X_t$.
Consider $\tau:= \inf\{t\ge 0: X_t \in \{-1,1\}\}$ and $\tau^n := \inf\{t \ge 0 : X^n_t \in \{-1,1\}\}$.
I conjecture that $\tau^n \to \tau$ and also, $\mathbb{P}(\tau^n \le \tau) \to 0$ as $n \to \infty$. Are both true? And if yes, how can I approach a proof?