In proposition 13.1 of the book "stochastic calculus and financial applications" by Steele,
It says that
for all $\lambda \leqslant 0$, we have the identity $E[M_{\tau_a}(\lambda\mu) = 1$
Where $$M_t(\mu) = exp(\int^t_0\mu(\omega,s)dB_s-\frac{1}{2}\int^t_0\mu^2(\omega,s)ds)\text{ for }\mu\in L^2_{Loc}$$
$$\tau_a = inf\{t\ \vert\ Y_t = -a \text{ or } t>T\}$$ is a stopping time for the following $$Y_t = \int^t_0\mu(\omega,s)dB_s -\int_0^t\mu^2(\omega,s)ds$$
and $\lambda$ smaller than 1
The proof says that $$M_{\tau_a}(\lambda\mu) = 1+\int^{\tau_a}_0\lambda\mu(\omega,s)M_s(\lambda\mu)dB_S\ \ \ \ \ \ \ (1)$$
and then it says that to prove the result, it suffices to show that the integrand is in $H^2$ i.e. $$E\left[\int^{\tau_a}_0\mu^2(\omega,s)M_s^2(\lambda\mu)ds\right] < \infty\ \ \ \ \ \ \ (2)$$
Why is this a sufficient condition?
All it says is that the integral in the right side of (1) is finite but we want it to be 0 whe, we take the mean.