= Solution
The <Kraus operators> of the <generalized measurement postulate> must obey
$$
\boxed{\sum_i A_i^\dagger A_i=I.}
$$
For a normalized <pure state>, the <Born rule> gives the outcome probability and the normalized conditional state:
$$
\boxed{p_i=\langle\psi|A_i^\dagger A_i|\psi\rangle=\lVert A_i|\psi\rangle\rVert^2,\qquad|\psi_i\rangle=\frac{A_i|\psi\rangle}{\sqrt{p_i}}\quad(p_i>0).}
$$
Here the dagger denotes the <Hermitian conjugate>. Completeness ensures $\sum_i p_i=1$. An outcome with $p_i=0$ never occurs for that input, so no conditional state is required there. The positive operators $E_i=A_i^\dagger A_i$ form a <POVM>, but the actual operators $A_i$ also specify the post-measurement states.
Let $\rho_{12}=|\Psi\rangle\langle\Psi|$ be the initial joint <density operator>, and $\rho_2=\operatorname{Tr}_1\rho_{12}$ its <reduced density matrix> on the accessible subsystem. If the first experimenter performs the measurement, the conditional joint state for outcome $i$ is $(A_i\otimes I)\rho_{12}(A_i^\dagger\otimes I)/p_i$. The second experimenter does not know that outcome. The relevant <nonselective quantum measurement> state is therefore the probability-weighted average
$$
\rho'_{12}=\sum_i(A_i\otimes I)\rho_{12}(A_i^\dagger\otimes I).
$$
To compare the accessible states, take any operator $B$ on the second subsystem. Cyclicity of the <trace> and the fact that operators on different tensor factors commute give
$$
\begin{aligned}
\operatorname{Tr}[(I\otimes B)\rho'_{12}]&=\sum_i\operatorname{Tr}[(A_i^\dagger A_i\otimes B)\rho_{12}]\\
&=\operatorname{Tr}[(I\otimes B)\rho_{12}].
\end{aligned}
$$
By the defining property of the <partial trace>, this says $\operatorname{Tr}(B\rho'_2)=\operatorname{Tr}(B\rho_2)$ for every $B$. Matrix units, or arbitrary Hermitian test operators, distinguish any two different <density matrices>, so
$$
\boxed{\rho'_2=\rho_2.}
$$
This proves <remote state invariance under a local trace-preserving operation> directly from completeness.
In particular, if the accessible experiment has <POVM> elements $F_k$, its probabilities are $\operatorname{Tr}(F_k\rho'_2)=\operatorname{Tr}(F_k\rho_2)$ whether the distant measurement happens or not. Attaching an independent local ancilla, evolving locally, or performing a sequence of local measurements cannot help: these operations act on the identical input <density operator> and therefore give identical outcome distributions. \b[No experiment confined to the second subsystem can distinguish whether the unannounced measurement occurred.] This is a derivation of the <no-communication theorem>, not an appeal to a signalling postulate.
The qualification is that the outcome is not communicated. A <selective quantum measurement> can give different conditional remote states, as happens when measuring one half of a <Bell state>; their probability-weighted average is still $\rho_2$. An observer supplied with the distant outcome has extra classical information and is no longer restricted to the unannounced local experiment considered here.
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