In these notes, Pg. 23, they show: $$\text{avg}_U \left(\int_{S^{n-1}} \frac{1}{\|U\theta\|^n_K\|\theta\|^n_K} d\sigma(\theta) \right) = R^{2n}$$
Here's the complete argument, and the line marked $(\star\star\star)$ is the one that I'm talking about:
The group of orthogonal transformations carries an invariant probability measure. This means that we can average a function over the group in a natural way. In particular, if $f$ is a function on the sphere and $θ$ is some point on the sphere, the average over orthogonal $U$ of the value $f(Uθ)$ is just the average of $f$ on the sphere: averaging over $U$ mimics averaging over the sphere: $$\text{avg}_U f(Uθ) = \int_{S^{n-1}}f (\phi) dσ(\phi)$$ Hence, \begin{align*} \text{avg}_U \left(\int_{S^{n-1}} \frac{1}{\|{U\theta}\|^n_K\|{\theta\|}^n_K} d\sigma(\theta) \right) &= \int_{S^{n-1}} \left( \text{avg}_U\frac{1}{\|{U\theta}\|^n_K} \right)\frac{1}{\|{\theta}\|^n_K}d\sigma(\theta)\ (\star\star\star)\\ &= \int_{S^{n-1}} \left( \int_{S^{n-1}} \frac{1}{\|{\phi}\|^n_K} d\sigma(\phi) \right)\frac{1}{\|{\theta}\|^n_K}d\sigma(\theta) \\ &= \left(\int_{S^{n-1}}\frac{1}{\|{\theta}\|^n_K}d\sigma(\theta)\right)^2 = R^{2n} \end{align*}
How did they take the average inside the integral? Is that a mistake? If not, why? Can we always swap averages and integrals like this?
I'd appreciate any help with understanding the above work!
An average is an integral, so Fubini-Tonelli would be the most applicable theorem. In this case, the integrand is non-negative, so we can interchange the order of integration.