 Solution

For a continuous local martingale $M$, its stochastic exponential isThe Itô formula gives $d{Z}_t=Z_{t}d{M}_t$ and $Z_0=1$, so $Z$ is a strictly positive local martingale. The finite-horizon Girsanov theorem says: if $Z$ is a true martingale on $[0,T]$ and $d\tilde{\mathbb{P}}=Z_{T}d\mathbb{P}$ on ${\mathcal{F}}_T$, then for every continuous local martingale $N$ under $\mathbb{P}$,For an infinite-horizon change of measure, assume $Z$ is a uniformly integrable martingale and use its terminal density $Z_{\infty }$. In particular, if $M=\int \theta dB$, the process $\tilde{B}=B-\int \theta ds$ is Brownian motion under the new probability measure, by the Lévy characterization of Brownian motion.

The quadratic variation cannot change. Adding the continuous finite-variation process $-[N,M]$ leaves $[\tilde{N}]=[N]$. Also the same pathwise partition sums defining $[N]$ have the same limit under an absolutely continuous change of measure; convergence in probability transfers by integrability of the density. Thus the bracket of the unchanged process $N$, now a semimartingale, is also unchanged. The local martingale property itself need not survive. This is quadratic covariation under an absolutely continuous measure change.