= Solution
For the <Kraus operators> of the <quantum measurement>, the <Born rule> gives
$$
\boxed{p_i=\operatorname{Tr}(A_i\rho A_i^\dagger)
=\operatorname{Tr}(\rho A_i^\dagger A_i),\qquad
\rho_i=\frac{A_i\rho A_i^\dagger}{p_i}\quad(p_i>0).}
$$
The <density matrix> $A_i\rho A_i^\dagger$ is the unnormalized state of that outcome. It is positive, its trace is $p_i$, and completeness gives $\sum_i p_i=\operatorname{Tr}\rho=1$. There is no conditional state to assign to an event with $p_i=0$.
For a <projective measurement>, each $P_i$ is an <orthogonal projection>, satisfying $P_i=P_i^\dagger=P_i^2$. Completeness becomes $\sum_iP_i=I$. Fix $v$ in the range of $P_j$. Then
$$
\|v\|^2=\langle v,Iv\rangle
=\sum_i\langle v,P_iv\rangle
=\sum_i\|P_iv\|^2.
$$
Since $P_jv=v$, the $j$th summand already equals $\|v\|^2$. All other summands are nonnegative, so $P_iv=0$ for $i\neq j$. Consequently $P_iP_j=0$, and taking adjoints also gives $P_jP_i=0$. This proves the stronger <completeness forces orthogonality of measurement projections>:
$$
\boxed{P_iP_j=\delta_{ij}P_i,\qquad [P_i,P_j]=0.}
$$
For the <unitary dilation of a measurement instrument>, take a <quantum ancilla> with orthonormal states $|0\rangle,|1\rangle,\ldots,|n\rangle$, initially in $|0\rangle$. Define
$$
V|\phi\rangle=\sum_{i=1}^n A_i|\phi\rangle\otimes|i\rangle.
$$
The completeness equation implies
$$
\langle V\phi,V\chi\rangle
=\sum_i\langle\phi,A_i^\dagger A_i\chi\rangle
=\langle\phi,\chi\rangle,
$$
so $V$ is an <isometry>. Its action on the input subspace can be extended to a <unitary operator> $U$ on the enlarged system:
$$
U(|\phi\rangle\otimes|0\rangle)=V|\phi\rangle.
$$
Choose <orthonormal bases> in the two complementary subspaces and map one to the other. Their dimensions agree; the extra unused <quantum ancilla> level also ensures this extension is possible for an infinite-dimensional system. After the coupling, the joint <density matrix> is
$$
\Omega=U(\rho\otimes|0\rangle\langle0|)U^\dagger
=\sum_{i,j=1}^n A_i\rho A_j^\dagger\otimes|i\rangle\langle j|.
$$
Now perform a <projective measurement> on the larger system, with
$$
Q_1=I_S\otimes(|0\rangle\langle0|+|1\rangle\langle1|),
\qquad
Q_i=I_S\otimes|i\rangle\langle i|\quad(2\leq i\leq n).
$$
These are pairwise orthogonal and sum to the full identity. The unused $|0\rangle$ level has zero weight in $\Omega$, so merging it into outcome one adds no <probability>. For every $i$,
$$
Q_i\Omega Q_i=A_i\rho A_i^\dagger\otimes|i\rangle\langle i|.
$$
Its trace is $p_i$, and taking the <partial trace> over the <quantum ancilla> after normalization gives exactly $\rho_i$. Hence \b[unitary coupling to an <quantum ancilla>, a <projective measurement>, and discarding the <quantum ancilla> reproduce both the <probabilities> and the conditional states of the general measurement.] This is the instrument version of a <Stinespring dilation>, rather than a construction reproducing only the outcome <probabilities>.
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