= Solution
For a <continuous local martingale> $M$, its \b[<stochastic exponential>] is
$$
\boxed{Z_t=\mathcal E(M)_t=\exp\{M_t-M_0-\tfrac12[M]_t\}.}
$$
The <Itô formula> gives $dZ_t=Z_t\,dM_t$ and $Z_0=1$, so $Z$ is a strictly positive <local martingale>. The finite-horizon \b[<Girsanov theorem>] says: if $Z$ is a true <martingale> on $[0,T]$ and $d\widetilde{\mathbb P}=Z_T\,d\mathbb P$ on $\mathcal F_T$, then for every <continuous local martingale> $N$ under $\mathbb P$,
$$
\boxed{\widetilde N_t=N_t-[N,M]_t\text{ is a }\widetilde{\mathbb P}\text{-local martingale},\quad 0\leq t\leq T.}
$$
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 $\widetilde B=B-\int\theta\,ds$ is <Brownian motion> under the new <probability measure>, by the <Lévy characterization of Brownian motion>.
\b[The <quadratic variation> cannot change.] Adding the continuous <finite-variation process> $-[N,M]$ leaves $[\widetilde N]=[N]$. 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>.
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