A question related to Girsanov’s theorem I’ve recently realised there is a subtlety in Girsanov’s theorem that I don’t really understand.
Consider a standard one dimensional Brownian motion $W$, and consider the SDE
$$dZ_t = \mu(t, Z_t) \, dt + \sigma(t, Z_t) \, dW_t \, , \,Z_0 = x_0 \, \, \, \text{(Equation 1)}$$
for some $x_0 \in \mathbb R$, where $\mu, \sigma: [0, \infty) \times  \mathbb R \to \mathbb R$ are Lipschitz continuous.
Denote by $\mathbb P$ the probability measure under which $W$ is a standard Brownian motion. Suppose we have an equivalent probability measure $\mathbb Q$ under which $W$ is no longer a standard Brownian motion, but a semimartingale.
We may still consider Equation 1 under $\mathbb Q$ as a semimartingale SDE.
Suppose $X$ solves Equation 1 under $\mathbb P$, and $Y$ solves Equation 1 under $\mathbb Q$.
Question: Is it true that we still have $X = Y$ up to indistinguishability? That is, do we have $X_t = Y_t$  for all $t \in [0, \infty)$, ($\mathbb P$, and hence $\mathbb Q$) almost surely?
It seems that this result is used implicitly in transforming SDE via Girsanov’s theorem, but it is not obvious to me at all.
 A: I managed to find an answer to this problem - indeed the answer is yes. The key ingredient is a theorem in Protter’s Stochastic Integration and Differential Equations (Chapter 2, Theorem 14) which states the following:

For every fixed process $X$, integrable with respect to a semimartingale $S$, the stochastic integrals $\int X\, dS$ under $\mathbb P$ and $\mathbb Q$ are indistinguishable as processes.

The indistinguishability as processes part is key, and allows us to conclude that solutions to SDE too are invariant, as follows:
Since $X$ solves Equation 1 under $\mathbb P$, by definition we have
$$X = X_0 + \int_0^\cdot \mu(s, X_s) \, ds + \int_0^\cdot \sigma(s, X_s) \, dW_s$$
under $\mathbb P$, almost surely.
By indistinguishability of the stochastic integral on the RHS, we have also
$$X = X_0 + \int_0^\cdot \mu(s, X_s) \, ds + \int_0^\cdot \sigma(s, X_s) \, dW_s$$
under $\mathbb Q$, almost surely.
So $X$ solves Equation 1 under $\mathbb Q$. By uniqueness to solutions of SDE, it is the solution, i.e. $X = Y$ up to indistinguishability, and we are done.
