Borel–Kolmogorov paradox

Counterexample

Let $P$ be the uniform probability measure on the unit sphere $S^2 \subset \R^3$. Describe a point by its longitude $\lambda \in (-\pi, \pi]$ and its latitude $\varphi \in [-\pi/2, \pi/2]$, as in spherical coordinates. The area element of $S^2$ is $\cos \varphi \, d\lambda \, d\varphi$, so under $P$ the pair $(\lambda, \varphi)$ has the density $\frac{1}{4\pi} \cos \varphi$. Hence $\lambda$ and $\varphi$ are independent, $\lambda$ is uniform on $(-\pi, \pi]$, and $\varphi$ has density $\frac{1}{2} \cos \varphi$.

Condition on the event $\{\varphi = 0\}$, the equator. The conditional distribution of $\lambda$ given $\varphi = 0$ is the law of $\lambda$, which is uniform. So the point is uniform on the equator.

Now let $\Theta = \lambda \bmod \pi \in [0, \pi)$ and condition on the event $\{\Theta = 0\}$, the great circle through the two poles. Since $\Theta$ is independent of $\varphi$, the conditional distribution of $\varphi$ given $\Theta = 0$ still has density $\frac{1}{2} \cos \varphi$. So the point is not uniform on this great circle: it is rarely near the poles.

A rotation of the sphere maps the equator onto the great circle through the poles and preserves $P$. Yet the two conditional laws on these two great circles are not images of each other under the rotation.