Consider random variables $Z_t>0 \ \forall t \in [0,T] $, where $E[Z_T]=1$ and $Z_t=E[Z_T|F_t]$, where $F_t$ is the Brownian filtration and $H$, which is integrable with respect to Brownian motion $W$, i.e. $\int_0^T |H(s)|^2 dt< \infty$ and it holds $Z_t = 1 + \int_0^t H(s) dW(s)$
Is there a way to show that $\frac{H}{Z}$ is $W-$integrable, i.e. $$\int_0^T \frac{|H(s)|^2}{|Z(s)|^2} ds < \infty$$
Since $Z$ is continuous, the condition $Z_t > 0$ for all $t \in [0,T]$ implies $\inf_{t \in [0,T]} Z_t = z > 0$. Hence $$\int_0^T \frac{|H(s)|^2}{|Z(s)|^2} ds \le \frac{1}{z^2} \int_0^T |H(s)|^2 < \infty.$$