Assume we have equivalent probability measures $\mathbb Q^{M}$ and $\mathbb Q^{N}$ and further assume that $$ \tilde{X}(T)-\tilde{X}(0) \text{ under }\mathbb Q^{M} \sim X(T)-X(0) \text{ under }\mathbb Q^{N}\; \; (*),$$
where $\tilde{X}(t):=X(t)\exp(\int_{t}^{T}\gamma(s)ds)$,
Clearly, it holds that $\tilde{X}(T) = X(T)$, and then it is stated:
$$X(T)\text{ under }\mathbb Q^{M}\sim X(T)-X(0)+\tilde{X}(0)\text{ under }\mathbb Q^ {N}$$
but unfortunately, I do not see why this is the case.
Comments/My thoughts: Using the fact that $X(T)=\tilde{X}(T)$ it is clear that $X(T)\sim\tilde{X}(T)\text{ under }\mathbb Q^{N}\text{ and }\mathbb Q^{M}$ and further from $(*)$, we obtain:
we have $X(T)-\tilde{X}(0)\text{ under }\mathbb Q^{M} \sim X(T)-X(0)\text{ under }\mathbb Q^{N}$.
If $\tilde{X}(0)$ were deterministic, i.e. has a trivial distribution under $\mathbb Q^{M}$ and $\mathbb Q^{N}$, then we can simply "add" $\tilde{X}(0)$ to both sides, but in general we cannot do this? Am I right, or what exactly am I missing here?