Suppose System A has a dynamic, continuous variable $O_A(t)$, which we wish to measure via a quantum probe. Assume system A has a self-Hmiltonian, $H_{SA}$ so that the evolving wave function of system A, in the O basis, is $|\Psi_{O_A} (t)\rangle = exp (-i\int_0^t H_{SA}dt )|\Psi_{O_Ai}\rangle$, where $|\Psi_{O_Ai} \rangle$ is the initial wave function for System A, at $t=0$.
Suppose System B is a quantum probe with initial wave function $|\Psi_{P_Bi}\rangle$ in the momentum basis, and has no self-Hamiltonian. Also, suppose its momentum becomes entangled with $O_A$ at $t_1 > 0$, via impact interaction-Hamiltonian $H_{int}$. This creates a composite wave function of both systems for which not all states of system B's momentum are possible for all states of $O_A$ (due to the entanglement). Further, suppose $Y_B$ is the conjugate variable to $P_B$.
The composite wave function (for $t > t_1 + \tau$) then
\begin{align}|\Psi_C(Y_B, O_A, t)\rangle = exp(-i\int_{t_1}^{t_1 +\tau} H_{int} dt)|\Psi_{Y_Bi}\rangle exp(-i\int_0^t H_{SA}dt)|\Psi_{0_Ai}\rangle \\= exp(-i\int_{0}^{t} (H_{int} + H_{SA}) dt)|\Psi_{Y_Bi}\rangle |\Psi_{0_Ai}\rangle \end{align} where $H_{int} = 0$ for $t_1 > t > t + \tau$, and $\tau$ is the short duration of impact interaction between A and B, and Y is the conjugate variable to P.
\begin{equation}|\Psi_C(P_B, O_A, t)\rangle = F.T. (exp(-i\int_{0}^{t} (H_{int} + H_{SA}) dt)|\Psi_{Y_Bi}\rangle |\Psi_{0_Ai}\rangle) \end{equation}
where F.T. is the Fourier Transform, over $Y_B$.
Does the above sound correct, especially the inclusion of $H_{SA}$, so that $|\Psi_C\rangle$ will evolve properly over time?
The probability that a measurement of the probe's momentum, at $t_2 > t_1 + \tau$, will yield momentum eigenstate $|p_j\rangle$, if System A is in eigenstate $|o_k\rangle$, is \begin{equation}Pr(|p_j o_k\rangle) = \langle p_j o_k|\Psi_C(P_B, O_A, t_2)\rangle\end{equation}
Therefore, the total probability of the probe's momentum measurement yielding $|p_j\rangle$ is \begin{equation}Pr(|p_j\rangle) = \int_{-\infty} ^\infty \langle p_j o|\Psi_C(P_B, O_A,t_2)\rangle do\end{equation} We need the integral because we cannot generally assume that each $|p\rangle$ state will only have a single $|o\rangle$ state entangled with it. In fact, we may have a continuous range of $|o\rangle$ states that are entangled with $|p_j\rangle$
Of course, a measurement result of $|p_j\rangle$ would also pick out the corresponding set of $|o\rangle$ states that are entangled with it. Hence, a measurement of the probe's momentum yielding $|p_j\rangle$ results in an (indirect) measuremeant of $O_A$.
I find that interesting because then, at $t_2$, we are measuring $O_A$ based on how it evolved up to $t_2$, even though the probe, by which we are measuring it, had no interaction with system A after $t_1 + \tau$, when the entanglement completed. Of course, the reason is that we included $exp(-i\int_0^t H_{SA}dt)$ in the formula for $|\Psi_C(P_B, O_A,t)\rangle$, above.
Does that sound right?
In fact, if the entanglement at $t_1 + \tau$ also caused a back action to $O_A$'s conjugate variable and that, in turn, caused a back action to $O_A$ (as would be the case if $O_A$ were, for instance, position and it's conjugate were momentum), then that back-action would also be modeled in $|\Psi_C(P_B, O_A,t)\rangle$. Therefore, the measurement of the probe at $t_2$ would be a measurement of $O_A$'s complete evolution, up to $t_2$, including evolution caused by its self-hamiltonian and any back action upon it, up until $t_2$.