$W(t)$ is Brownian motion. Then $E\{\exp[\int_0^t u\Delta(s)dW(s)-1/2\int_0^t(u\Delta(s))^2ds]\}=1$ where $\Delta(s)$ is some appropriate function to make sense above expectation value for some $u$ over small interval.
$\textbf{Q:}$ Why $E\{\exp[\int_0^t u\Delta(s)dW(s)-1/2\int_0^t(u\Delta(s))^2ds]\}=1$ holds even for $\Delta(s)$ non-deterministic? (If $\Delta(s)$ is deterministic, then it is easy to see this holds by martingale property of generalized geometric Brownian motion.)
Ref. S. Shreve Stochastic Calculus for Finance Vol. II chapter 4, equation 4.4.31 and comment right after the proof of Theorem 4.4.9