I would like to consider the stochastic integral
\begin{equation} \int_0^t sgn(B_s)\, dB_s \, , \end{equation} where $sng(x) = \begin{cases} 1, & x>0 \\ 0, & x= 0 \\ -1 , &x<0 \end{cases} $ .
Now I am wondering whether this integral is properly defined since $sgn$ is not left-continuous (previsible). Any thoughts on this?
Brownian motion is previsible, i.e.
$$([0,\infty)\times \Omega,\mathcal{P}) \ni (t,\omega) \mapsto f(t,\omega):=B_t(\omega) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable where $\mathcal{P}$ is the predictable $\sigma$-algebra. On the other hand, $\text{sgn}$ is Borel measurable, i.e.
$$(\mathbb{R},\mathcal{B}(\mathbb{R})) \ni x \mapsto g(x) := \text{sgn}(x) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable. As a consequence, the composition $g \circ f$ is measurable, that is,
$$([0,\infty)\times \Omega,\mathcal{P}) \ni (t,\omega) \mapsto (g \circ f)(t,\omega) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$
is measurable. By definition $(g \circ f)(t,\omega) = \text{sgn}(B_t(\omega))$, and so we have shown that the process $X_t(\omega):=\text{sgn}(B_t(\omega))$ is previsible.