Probability measure on the product space that is not the product measure Suppose  $(\Omega, \mathcal{A}, \mu_1)$ and $(\Omega, \mathcal{A}, \mu_2)$ are two probability spaces. What would be a good measure on $\Omega\times \Omega$, that differs from the product measure (which models iid data, see here) and models non-iid data?
In general, this probably cannot be answered; but if we assume  $\Omega$
to some $\mathbb{R}^n$ with the Lebesgue measure, are there some interesting or well-known measures to put on the product space aside from the product measure? (Or if the hypothetical interesting measure cannot be made up of the individual projections $\mu_i$, what would it be then?)
I would be interested in such measure that have a "name", i.e. not an artificial construction of a non-product measure the product space.
 A: We show that the angular Lebesgue measure on the circle is not a product measure in $\mathbb{R}^2$. More precisely, define $r : [0, 2 \pi ) \to \mathbb{R}^2$ by $r(t) = (\cos t , \sin t)$, so that the image of is the unit circle $C$ in $\mathbb{R}^2$. Define the measure $\rho$ on $E \in \mathbb{B}(\mathbb{R}^2)$ by $\rho (E) = \lambda (r^{-1} (C \cap E))$, where $\lambda$ is the Lebesgue measure in $[0,2\pi )$.
We verify that $\rho$ does not arise from a product measure. Arguing by contradiction, suppose that $\rho$ is given by the product measure $\rho = \mu\times \nu$. Let $E_1 = [-1,1]$ and $E_2 = [\sqrt{2}/2 , \sqrt{2}/2 + 2]$. Then, one readily checks that $$\mu (E_2) \nu (E_2) = \rho (E_2 \times E_2) = 0$$ so that $\mu (E_2) = 0$ or $\nu (E_2) = 0$. Now, note that $(E_1 \times E_2) \cap C$ and $(E_2 \times E_1) \cap C$ are arcs on $C$ subtended by an angle of $\frac{\pi}{2}$, so that:
$$\rho (E_1 \times E_2) = \rho(E_2 \times E_1) = \frac{\pi}{2} $$
which implies that $\mu (E_1) \nu (E_2) > 0$ and $\mu (E_2)\nu(E_1) > 0$, which contradicts that $\mu (E_2) = 0$ or $\nu (E_2) = 0$.
Thus $\rho$ does not arise from a product measure. $\blacksquare$
Notice that one can then normalise this measure via $\rho \mapsto \frac{1}{2\pi} \rho$, so that $\rho$ becomes a probability measure.
