Mutually singular marginal distributions?

Question

Let $\nu,\theta$ be two probability measures on $\mathbb{R}^2$ that are mutually singular. Can we find a probability measure $\mathbb{P}$ on $\mathbb{R}^3$ such that $\nu$ and $\theta$ are the $(X,Y)$ and $(X,Z)$ marginals respectively?
To be more precise, we have $\mathbb{R}^2= \Omega_1 \cup \Omega_2$ where $\Omega_1\cap\Omega_2=\emptyset$ and $\nu(\Omega_1)=\theta(\Omega_2)=0$, while $\nu(\Omega_2),\, \theta(\Omega_1) >0$. Also, being marginals translates to $$\nu(A) = \mathbb{P}\left(\{ (x,y,z):\, (x,y) \in A \}\right);\qquad \theta(A)=\mathbb{P}\left(\{ (x,y,z):\, (x,z) \in A \}\right). $$
I have a feeling that disintegration theorem applied to the above will imply that such $\mathbb{P}$ can only be zero (using perhaps conditioning on the common variable), therefore there cannot be a probability measure on $\mathbb{R}^3$ with mutually singular $\mathbb{R}^2$- marginals (I actually try to impose conditions to two $\mathbb{R}^2$-probability measures such that they cannot be marginals).
Is the above reasoning reasonable or am I missing something (or perhaps mutual singularity is too much for this)?

Answer 1

Let $(X,Y,Z)$ place equal mass on the two points $(0,0,1)$ and $(1,1,0)$. Then $(X,Y)$ places equal mass on $(0,0)$ and $(1,1)$, while $(X,Z)$ places equal mass on the points $(0,1)$ and $(1,0)$, so these two distributions are mutually singular.