I complete Tim Goodman's answer in answer to get something more systematic. A local measurement has to be written as a tensorial product of two observables A⊗B. And it can only distinguish (with probability 1) its eigenstates. The states |ϕ⟩ and |ψ⟩ of Tim's example cannot be written as eigenstates of a tensorial product.

Note that this does not correspond exactly to states which cannot be written as tensorial product of states. For example, if let's use the following 4 states : $|\psi_0\rangle=|00\rangle$ , |\psi_1\rangle=|1+\rangle$, $|\phi_0\rangle=|01\rangle$ , |\phi_1\rangle=|1-\rangle$ and let's try to distinguish the ψs from the ϕs.Tim's arguments are still valid, even if each the 4 states is a product state and is orthogonal to all the 3 others. Furthermore, each pair of state is locally distinguishable.

I think locally distinguishable subspaces has something to do with to the direct sum of locally orthogonal subspaces, but I don't exactly know how to write it.

(Edited to correct TeX typo)

I complete Tim Goodman's answer in answer to get something more systematic. A local measurement has to be written as a tensorial product of two observables A⊗B. And it can only distinguish (with probability 1) its eigenstates. The states |ϕ⟩ and |ψ⟩ of Tim's example cannot be written as eigenstates of a tensorial product.

Note that this does not correspond exactly to states which cannot be written as tensorial product of states. For example, if let's use the following 4 states : $|\psi_0\rangle=|00\rangle$ , $|\psi_1\rangle=|1+\rangle$,
$|\phi_0\rangle=|01\rangle$ , $|\phi_1\rangle=|1-\rangle$ and let's try to distinguish the ψs from the ϕs.Tim's arguments are still valid, even if each the 4 states is a product state and is orthogonal to all the 3 others. Furthermore, each pair of state is locally distinguishable.

I think locally distinguishable subspaces has something to do with to the direct sum of locally orthogonal subspaces, but I don't exactly know how to write it.