Write . Since and , the orthogonal reflection satisfies and . Thus, for every Borel set , invariance under the orthogonal group gives
The spherical average must use normalized surface measure without an additional area factor. Let be the normalized surface measure on a sphere, and set
The correct radial disintegration of an orthogonally invariant measure is
Indeed, take normalized Haar measure on the compact orthogonal group . Invariance and Fubini's theorem give
Right invariance of Haar measure and transitivity of the orthogonal group on each sphere show that the inner average depends only on . Averaging it once more over , using invariance of surface measure on a sphere and Fubini's theorem, identifies its value as . At the same identity reads . The definition of pushforward measure now proves the boxed formula. This argument also works for invariant probability measures with an atom of a measure at the origin.
For the printed normalization, choose a smooth cutoff function equal to one on a ball containing the support of a measure . Its left integral is one, whereas the printed right integral is . Thus the extra cannot be retained. Equivalently, with unnormalized area measure , the average is .
The reflection-averaging hint also fails in general dimension. For ,
The first equality uses , which follows from coordinate reflections, coordinate permutations, and . Taking a test function equal to the first coordinate near the unit sphere disproves the hint when . Averaging over the full orthogonal group, as above, proves the required corrected result without that hint.

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