Joint Measurability for a collection of POVMs $\{\Omega_j\}_j$ where $j$ is the index of the POVMs with associated effects $\{\Omega^\omega_j\}_{\omega}$ is defined as $$\Omega^\omega_j = \sum_{\theta} p(\omega|j,\theta) \Theta^\theta,$$ where $\Theta^\theta$ is a POVM it self, and $p(\cdot | j, \theta)$ is a probability distribution.
The justification for this, is that in such case to measure any $\Omega_j$ one can just measure $\Theta$ and sample $\omega$ according to $p(\cdot|j,\theta)$. Yet, this justification is a statement about probability distributions of outcomes, not about the actual shape of the effects. I.e. the justification says,
$$\forall \rho, \quad \mathrm{Tr}(\Omega^\omega_j \rho)= \mathrm{Tr}(\sum_{\theta} p(\omega|j,\theta) \Theta^\theta \rho).$$
If this holds for a set of density matrices that spans linearly the set of all density matrices then this implies the previous operator equality because $\mathrm {Tr}$ is an inner product for the space of hermitian operators.
So far so good.
The question I have is, what if it is not spanning it? In other terms, does it make sense to say that two POVMs are jointly measurable wrt a set of states?
If a qubit-system is in a mixture of $|0\rangle$ and $|1\rangle$, then I take $\Theta$ to be the computational basis measurement $Z$. Now I can consider the observable $X$ on this system. I know that the probability of measuring $|+\rangle$ or $|-\rangle$ is going to be $1/2$ no matter the mixture. As a consequence, I will be able to conclude that on this set of states, $X$ and $Z$ are jointly measurable.
If you think this does not make sense, then what more are we trying to capture with the usual operator equality when we define joint-measurability?
