Fix $T>0.$ Suppose $h_t$ is an adapted process such that $\int_0^th_s^2ds<\infty$ a.e for all $t\leq T.$ Define $$Z_t=\exp \left(\int_0^th_sdB_s-\frac{1}{2}\int_0^th_s^2ds\right)$$
From Ito's formula, $$dZ_t=Z_th_tdB_t.$$ I want to show that $Z_t$ is a local martingale. Since the ito integral is continuous a.e. $Z_t$ is continuous a.e and therefore $Z_t^2$ is continuous a.e. The goal is to show that $\int_0^t Z_s^2h_s^2ds<\infty$ a.e. Since $Z_t^2$ is continous, it is bounded on $[0,T].$ Therefore, $\int_0^tZ_s^2ds\leq\sup_{t\in [0,T]}Z_t^2=M<\infty$. Therefore, $$\int_0^tZ_s^2h_s^2 ds\leq M\int_0^t h_s^2<\infty \text{ a.e.} $$ Is this proof fine?I think I should take the $\sup$ by removing a measure 0 set from $[0,T]$.