 Solution

Write the integral with an explicit dummy time:On every compact time interval, strict positivity and continuity of $Z$ make its reciprocal locally bounded; the covariation has locally finite variation. Thus $X$ is a well-defined continuous semimartingale.

Apply the Itô product rule. Subtracting a finite-variation process does not change covariation, so $[Z,X]=[Z,M]$. The finite-variation terms cancel:Thus $ZX$ is a local martingale under $\mathbb{P}$.

For the transfer step, choose stopping times increasing to the horizon that bound $X$, $M$, reciprocal $Z$, and the brackets needed for the stochastic integrals. For each such time $\sigma$, the same product calculation with $X^{\sigma }$ shows that $Z_t{X}_{t\wedge \sigma }$ is a true $\mathbb{P}$ martingale. The part after $\sigma$ is simply the bounded held value $X_{\sigma }$ times the density martingale. By the Bayes formula for conditional expectation, for $s\le t$,The stopping times exhaust the horizon also under $\mathbb{Q}\ll \mathbb{P}$. Hence $X$ is a continuous $\mathbb{Q}$ local martingale, proving this form of the Girsanov theorem without assuming a positive terminal density everywhere.