I suspect this is the obvious result of something in operator algebras, but that's far outside my field.

Recall that a projection-valued measure is a map $E$ from a sigma-algebra $\mathcal{F}$ on measure space $\Omega$ to the set of orthogonal projections on a Hilbert space $H$ that is countably additive and had $E(\Omega)=Id$. It is a special case of a positive operator-valued measure (POVM), in which the operators $E(F)$ need only be positive-semidefinite. (In quantum mechanics, the intuition is that a state is given by a trace-class operator $\rho$ on $H$, and the physical "probability of measuring $F$" is defined to be $tr(E(F)\rho)$.)

More generally, to describe what happens to a state after it is measured, the notion of a quantum instrument is used. It is defined as a map $\mathcal{I}$ from $\mathcal{F}$ to the set of completely positive operators on $\mathcal{L}_1(H)$ (the trace-class operators on $H$ which intuitively represent states), such that

1. $\mathcal{I}$ is countably additive
2. $tr(\mathcal{I}(F)(\rho))\le tr(\rho)$ for $\rho$ positive-semidefinite, $F\in\mathcal{F}$
3. $tr(\mathcal{I}(\Omega)(\rho))=tr(\rho)$.

It is easy to see that any instrument induces a POVM (not necessarily projection-valued), given by $E_{\mathcal{I}}(F)=\mathcal{I}(F)^*(Id)$. However, the map from instruments to positive operator-valued measures is not injective, as different instruments can induce the same POVM.

My question is: Is that map surjective? Namely, given a POVM $E$ can we find an instrument $\mathcal{I}$ such that $E_{\mathcal{I}}=E$? I would even be interested in just the case when $E$ is a projection-valued measure.

Notes:

1. When $H$ is finite-dimensional, any projection-valued measure is discrete, and given (wlog) by $E(\{x_k\})|\psi\rangle=|k\rangle\langle k|\psi\rangle$. Then a valid instrument is generated by $\mathcal{I}(\{x_k\})\rho=|k\rangle\langle k|\rho|k\rangle\langle k|.$ (Here $|k\rangle$ are the eigenvectors.)
2. The above construction does not generalize directly to the infinite-dimensional case. Indeed there is a theorem (Theorem 4.1.2 in the book by Halmos), that says an infinite-dimensional instrument can never be repeatable in the sense that $\mathcal{I}(F_1)\mathcal{I}(F_2)=\mathcal{I}(F_1\cap F_2)$ unless the measure is discrete.

But my question doesn't need it to be repeatable (i.e. we don't care what happens to the state afterwards), we just want $\mathcal{I}(F)^*(Id)$ to give $E_{\mathcal{I}}(F)$. Can we "extend" the measure from its value on the identity?

I suspect this is the obvious result of something in operator algebras, but that's far outside my field.

Recall that a projection-valued measure is a map $E$ from a sigma-algebra $\mathcal{F}$ on measure space $\Omega$ to the set of orthogonal projections on a Hilbert space $H$ that is countably additive and had $E(\Omega)=Id$. It is a special case of a positive operator-valued measure (POVM), in which the operators $E(F)$ need only be positive-semidefinite. (In quantum mechanics, the intuition is that a state is given by a trace-class operator $\rho$ on $H$, and the physical "probability of measuring $F$" is defined to be $tr(E(F)\rho)$.)

More generally, to describe what happens to a state after it is measured, the notion of a quantum instrument is used. It is defined as a map $\mathcal{I}$ from $\mathcal{F}$ to the set of completely positive operators on $\mathcal{L}_1(H)$ (the trace-class operators on $H$ which intuitively represent states), such that

1. $\mathcal{I}$ is countably additive
2. $tr(\mathcal{I}(F)(\rho))\le tr(\rho)$ for $\rho$ positive-semidefinite, $F\in\mathcal{F}$
3. $tr(\mathcal{I}(\Omega)(\rho))=tr(\rho)$.

It is easy to see that any instrument induces a POVM (not necessarily projection-valued), given by $E_{\mathcal{I}}(F)=\mathcal{I}(F)^*(Id)$. However, the map from instruments to positive operator-valued measures is not injective, as different instruments can induce the same POVM.

My question is: Is that map surjective? Namely, given a POVM $E$ can we find an instrument $\mathcal{I}$ such that $E_{\mathcal{I}}=E$? I would even be interested in just the case when $E$ is a projection-valued measure.

Notes:

1. When $H$ is finite-dimensional, any projection-valued measure is discrete, and given (wlog) by $E(\{x_k\})|\psi\rangle=|k\rangle\langle k|\psi\rangle$. Then a valid instrument is generated by $\mathcal{I}(\{x_k\})\rho=|k\rangle\langle k|\rho|k\rangle\langle k|.$ (Here $|k\rangle$ are the eigenvectors.)
2. The above construction does not generalize directly to the infinite-dimensional case. Indeed there is a theorem (Theorem 4.1.2 in the book by Holevo), that says an infinite-dimensional instrument can never be repeatable in the sense that $\mathcal{I}(F_1)\mathcal{I}(F_2)=\mathcal{I}(F_1\cap F_2)$ unless the measure is discrete.

But my question doesn't need it to be repeatable (i.e. we don't care what happens to the state afterwards), we just want $\mathcal{I}(F)^*(Id)$ to give $E_{\mathcal{I}}(F)$. Can we "extend" the measure from its value on the identity?