I am confused about the difference between $Q^{s,x}$ and $P$ for the following ito diffusion: $$dX_t=b(X_t)dt+\sigma(X_t)dB_t,\quad t\ge s;\quad X_s=x.$$
Followings are from most books: given the unique time homogenous weak solution to the above problem. Denote $\mathcal{M}_{s,x}$ be a $\sigma$-algebra on $\Omega$ generated by random variables $X_t:\omega\in \Omega \to X_t^{s,x}(\omega)\in \mathbb{R}^n$ with $t\ge s$. Define a transition probability measure $Q^{s,x}$ on $\mathcal{M}_{s,x}$ such that $$Q^{s,x}(\{X_{t_1}\in A_1,\cdots,X_{t_k}\in A_k\})= P(\{X_{t_1}^{s,x}\in A_1,\cdots,X_{t_k}^{s,x}\in A_k\}),$$ for $A_i\in \mathcal{B}(\mathbb{R}^n)$ with $i=1,2,\cdots,k$ and $k\ge 1$. And let $E^{s,x}$ be the expectation operator with respect to $Q^{s,x}$.
Then $E^{s,x}[f(X_t)]=E[f(X_t^{s,x})]$.
My question is, is $X_{t_1}(\omega)=X_{t_1}^{s,x}(\omega)$ in the above paragraph? Then if it's true, isn't it $Q^{s,x}$ just a restriction of $P$ on $\mathcal{M}_{s,x}$? Then why do we need another notation for it instead of just using $P$?