= Solution
On the finite horizon $[0,T]$, strict positivity gives equivalence of the restricted probability measures: for $A\in\mathcal F_T$, $\mathbb Q(A)=\mathbb E_{\mathbb P}(Z_T\mathbf1_A)$ vanishes if and only if $\mathbb P(A)$ vanishes. The standard change-of-measure theorem gives <semimartingale invariance under equivalent measures>: \b[a process is a semimartingale under $\mathbb P$ on this horizon if and only if it is one under $\mathbb Q$]. The sample-path <quadratic variation> is unchanged, but the local-<martingale> and finite-variation parts generally change.
More explicitly, set $N=\int Z^{-1}\,dZ$, the zero-start <stochastic logarithm> of the density. For every continuous $\mathbb P$-<local martingale> $L$, the <Girsanov theorem> states that
$$
\boxed{L_t^{\mathbb Q}=L_t-\int_0^t\frac{d[L,Z]_s}{Z_s}=L_t-[L,N]_t}
$$
is a continuous $\mathbb Q$-<local martingale>. Consequently, if the continuous <semimartingale decomposition> under $\mathbb P$ is $X=X_0+L+V$, its decomposition under $\mathbb Q$ is
$$
X=X_0+L^{\mathbb Q}+\bigl(V+[L,N]\bigr).
$$
The covariation correction is a continuous <finite-variation process>, locally well-defined because $Z$ stays strictly positive. In the Brownian case, if $dZ_t=Z_t\theta_t\,dB_t$, then
$$
B_t^{\mathbb Q}=B_t-\int_0^t\theta_s\,ds
$$
is a <Brownian motion> under $\mathbb Q$. The inverse finite-horizon density is $1/Z$, giving the reverse change of measure. These assertions apply on each horizon on which the density is strictly positive; they do not assert equivalence on a larger terminal sigma-algebra where the full density $D$ may vanish.
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