Editing Borel–Kolmogorov paradox (section)

=== Measure theoretic perspective ===

To understand the problem we need to recognize that a distribution on a continuous random variable is described by a density ''f'' only with respect to some measure ''μ''. Both are important for the full description of the probability distribution. Or, equivalently, we need to fully define the space on which we want to define ''f''.

Let Φ and Λ denote two random variables taking values in Ω<sub>1</sub> = <math display="inline">\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]</math> respectively Ω<sub>2</sub> = [−{{pi}}, {{pi}}]. An event {Φ&nbsp;=&nbsp;''φ'',&nbsp;Λ&nbsp;=&nbsp;''λ''} gives a point on the sphere ''S''(''r'') with radius ''r''. We define the [[coordinate transform]]
:<math>\begin{align}
  x &= r \cos \varphi \cos \lambda \\
  y &= r \cos \varphi \sin \lambda \\
  z &= r \sin \varphi
\end{align}</math>

for which we obtain the [[volume element]]
:<math>\omega_r(\varphi,\lambda) = \left\| {\partial (x,y,z) \over \partial \varphi} \times {\partial (x,y,z) \over \partial \lambda} \right\| = r^2 \cos \varphi \ .</math>

Furthermore, if either ''φ'' or ''λ'' is fixed, we get the volume elements
:<math>\begin{align}
  \omega_r(\lambda) &= \left\| {\partial (x,y,z) \over \partial \varphi } \right\| = r \ , \quad\text{respectively} \\[3pt]
  \omega_r(\varphi) &= \left\| {\partial (x,y,z) \over \partial \lambda } \right\| = r \cos \varphi\ .
\end{align}</math>

Let
:<math>\mu_{\Phi,\Lambda}(d\varphi, d\lambda) = f_{\Phi,\Lambda}(\varphi,\lambda) \omega_r(\varphi,\lambda) \, d\varphi \, d\lambda</math>

denote the joint measure on <math>\mathcal{B}(\Omega_1 \times \Omega_2)</math>, which has a density <math>f_{\Phi,\Lambda}</math> with respect to <math>\omega_r(\varphi,\lambda) \, d\varphi \, d\lambda</math> and let
:<math>\begin{align}
          \mu_\Phi(d\varphi) &= \int_{\lambda \in \Omega_2} \mu_{\Phi,\Lambda}(d\varphi, d\lambda)\ ,\\
  \mu_\Lambda (d\lambda) &= \int_{\varphi \in \Omega_1} \mu_{\Phi,\Lambda}(d\varphi, d\lambda)\ .
\end{align}</math>

If we assume that the density <math>f_{\Phi,\Lambda}</math> is uniform, then
:<math>\begin{align}
  \mu_{\Phi \mid \Lambda}(d\varphi \mid \lambda) &= {\mu_{\Phi,\Lambda}(d\varphi, d\lambda) \over \mu_\Lambda(d\lambda)} = \frac{1}{2r} \omega_r(\varphi) \, d\varphi \ , \quad\text{and} \\[3pt]
  \mu_{\Lambda \mid \Phi}(d\lambda \mid \varphi) &= {\mu_{\Phi,\Lambda}(d\varphi, d\lambda) \over \mu_\Phi(d\varphi)} = \frac{1}{2r\pi} \omega_r(\lambda) \, d\lambda \ .
\end{align}</math>

Hence, <math>\mu_{\Phi \mid \Lambda}</math> has a uniform density with respect to <math>\omega_r(\varphi) \, d\varphi</math> but not with respect to the [[Lebesgue measure]]. On the other hand, <math>\mu_{\Lambda \mid \Phi}</math> has a uniform density with respect to <math>\omega_r(\lambda) \, d\lambda</math> and the Lebesgue measure.