Must states on local algebras of QFT be “backwards-compatible”?

[I've rephrased this question in terms of normal states to make it more succinct.]

There is a natural sense in which a state $\omega$, on a local QFT algebra of some spacetime region $O_1$, can be "backwards-compatible" with $O_1$'s sub-regions. I will define it below ... has this class been identified and studied? I can't find any references to it; any pointers would be hugely helpful in my research, and greatly appreciated!

Intuitively, the local algebra of a spacetime region $O_1$ is in a backwards-compatible state if any measurement, designed in any sub-region $O_2 \subseteq O_1$ to get strictly more information about $O_2$, would have put $O_2$ in a state compatible with what is known now about the larger region $O_1$. Perhaps this is a reasonable property to require of physically meaningful states. In any case we can formalize it (for normal states) as follows.

Order the normal states of local von Neumann algebras by how "informative" they are, in terms of their support projections:

$\gamma \leq \omega \iff$ range$(s_\gamma) \subseteq$ range$(s_\omega)$.

Note 1: $s_\gamma, s_\omega$ are computed in the respective algebras on which $\gamma, \omega$ are states; the algebras may be distinct, but so long as they act on the same Hilbert space, the ranges can be meaningfully compared.

Note 2: if $\omega$ is a normal state on $\mathcal{S}$ and $\mathcal{R} \subseteq \mathcal{S}$, then $\omega$ is $\leq$ ("at least as informative as") its restriction $\omega \upharpoonright \mathcal{R}$; the inequality may or may not be strict.

Note 3: we assume each local algebra $\mathcal{R}$ is type III and so has no pure normal states; thus for each normal state $\omega$ on $\mathcal{R}$, there exist others that are strictly $< \omega$.

Call normal states $\omega, \omega'$ compatible in $\mathcal{R}$ if there exists a normal state $\tau$ on $\mathcal{R}$ such that $\tau \leq \omega$ and $\tau \leq \omega'$. (By Note 1, $\omega, \omega'$ need not be states on $\mathcal{R}$.)

When $\mathcal{R} \subseteq \mathcal{S}$ and $\omega$ is a normal state on $\mathcal{S}$, call $\omega$ backwards-compatible to $\mathcal{R}$ if every normal state $\gamma$ on $\mathcal{R}$ satisfying $\gamma \leq (\omega \upharpoonright \mathcal{R})$ is compatible with $\omega$ in $\mathcal{S}$.

Now let $\mathcal{O}$ be the set of open regions of some spacetime, and let $\{ \mathcal{R}(O) : O \in \mathcal{O} \}$ be our "typical net of local algebras."

Call a normal state $\omega$ on $\mathcal{R}(O_1)$ backwards-compatible for our net if it is backwards-compatible to $\mathcal{R}(O_2)$ for every $O_2 \subseteq O_1$ in $\mathcal{O}$.

Has anyone identified or studied backwards-compatible states (presumably using a different term)?

Has anyone proved that nontrivial examples exist in typical AQFT models? Nontrivial meaning a state on a region, that is strictly more informative than some (preferably all) of its restrictions to proper subregions. The vacuum state is trivially backwards-compatible but all its restrictions are equally informative (having the same support projection, namely the identity). I can prove, using the "split inclusion" property of AQFT models, that for distinct $O_1 \subseteq O_2$, there exists a normal state $\omega$ on $\mathcal{R}(O_2)$, backwards-compatible to $\mathcal{R}(O_1)$, such that $\omega < (\omega \upharpoonright \mathcal{R}(O_1))$. But it seems harder to find $\omega$ having these properties towards all proper sub-regions of $\mathcal{R}(O_2)$ simultaneously.