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Write a positive continuous local martingale as $M=\exp (X-⟨X⟩/2)$ using its stochastic logarithm. If $M_t\to 0$, its logarithmic bracket must diverge. Otherwise the finite-bracket convergence lemma makes $X$ converge finitely and the exponential has a positive limit.

Since the density martingale is uniformly integrable, $Z_t={\mathbb{E}}_{\mathbb{P}}[Z_{\infty }\mid {\mathcal{F}}_t]$. Equivalence of measures, as stipulated, means $Z_{\infty }>0$ almost surely. Write $L=\int Z^{-1}dZ$ for the stochastic logarithm. The Itô formula gives $\log Z=L-⟨L⟩/2$, and hence the quadratic covariation in the drift correction isIt involves the local-martingale part of $\log Z$; its continuous finite-variation part has zero covariation.

With $Y=X-C$, the Itô product rule now gives an exact cancellation:Thus $ZY$ is a P-local martingale. To transfer this conclusion rigorously, set ${\tau }_n=\inf \{t:|Y_t|\ge n\}$. The same computation for the stopped $Y$ givesThis is again a P-local martingale. Moreover its absolute value is at most $n{Z}_t$. Uniform integrability of $Z$ makes the family over bounded stopping times uniformly integrable, so the product is a true martingale. This is the bounded-process density-product criterion.

The Bayes formula for conditional expectation therefore gives, for $s\le t$,Continuity gives ${\tau }_n\uparrow \infty$, proving $X-⟨\log Z,X⟩$ is a Q-local martingale. This proves the needed Girsanov theorem rather than invoking it.

For the Brownian-filtration conclusion, use this precise Brownian martingale representation theorem: every continuous local martingale in the usual augmentation of the natural Brownian filtration is an Itô integral with a predictable integrand locally square integrable in time. In particularDefineOn each finite horizon a strictly positive continuous $Z$ has a positive pathwise minimum. Thus ${\int }_0^t{\alpha }_s^{2}ds<\infty$ almost surely, andThe already proved measure-change result makes $\hat{W}=W-\int \alpha ds$ a continuous Q-local martingale. Its quadratic variation is $t$, unchanged by its finite-variation correction. The Lévy characterization of Brownian motion provesNo additional exponential-integrability condition is needed, since the equivalent uniformly integrable density process is already given.