3

I am trying to prove that the Lebesgue measure on $S^n$ is the unique countably additive, rotation-invariant measure of total measure 1 defined on Lebesgue-measurable sets.

I know the proof of the analogous statement for the Lebesgue measure with $\mathbb{R}^n$. However, in that case we start from rectangles, which have the nice property that a disjoint union of countably many rectangles cover the whole space and that the intersection of two rectangles is again a rectangle. I don't know how to define the analogue of a rectangle on the sphere, even if in $S^1$ and $S^2$ I can sort of picture what they should look like...

Also I have found no reference apart from one in Russian so if someone knows of something in English French Italian or German please answer with a link!

frafour
  • 3,015
  • 10
  • 22
  • Possible duplicate of https://math.stackexchange.com/questions/1183734/haars-theorem-for-the-rotation-invariant-distribution-on-the-sphere – user10354138 Oct 05 '18 at 08:06
  • 1
    You can deduce this from the uniqueness of the Haar measure on $SO(n)$ and basic theory of $G$-nvariant measures on homogeneous $G$-spaces as described in the second chapter of Folland's "A Course in Abstract Harmonic Analysis" – Adrián González Pérez Oct 12 '18 at 13:23

1 Answers1

1

For $S^1$ you can copy the proof for $\mathbb{R}$: for every $n\ge1$ segments of length $2\pi/n$ will have measure $1/n$th of the full measure of the circle. Once you have that you can use approximations to show that a segment of length $a$ should have measure $a/(2\pi)$ of the full measure. Then you are done by uniqueness of the extension.

For $S^2$ you can do similar things: hemispheres have measure $1/2$ of the full measure. The `strips' between meridians with $a\le \text{longitude}<b$ will have measure $|b-a|/(2\pi)$ of the full measure. Using rotations this holds in all directions. Then you can deal with bands between latitudes and show that the measure of a rectangle between longitudes $a$ and $b$ will have measure $|b-a|/(2\pi)$ of the measure of the band. In this way you can build up a ring on which the measure behaves like Lebesgue measure and which is large enough to apply the uniqueness-of-extension to.

hartkp
  • 2,382
  • Thank you for your answer! That's a very nice argument. Do you know if it might be generalizable to all dimensions? – frafour Oct 05 '18 at 08:24
  • I did not go through all dimensions but this rotation argument can be used to show that many block-like sets have the correct relative measure, usually that is enough. – hartkp Oct 05 '18 at 08:31
  • The idea is nice but I am a little bit unconvinced. In $S^1$ the group of rotations admits arbitrarily large finite subgroups and so you can partition $S^1$ in $n$ sets of measure $1/n$. The Jordan theorem prevents you from doing this in $SO(3)$ or any other simple compact group. Basically, you can only tile $S^2$ as given by the Platonic solids or as $2n$-wedges between meridians and the equator. I do not see how your argument deals with the measure as the latitude varies. – Adrián González Pérez Oct 12 '18 at 13:38
  • I do not use any tilings. I did not flesh out all possible details but using the rotations around the axis through the poles you can show what I mentioned in my answer: all strips between meridians have the measure you'd expect from Lebesgue measure. The same argument works for every ring parallel to the equator: fix two latitudes $a$ and $b$ say then any 'rectangle' determined by $a$ and $b$ and two longitudes $c$ and $d$ has measure $|d-c|/(2\pi)$ times the measure of the ring. This holds for every ring and implies that a density function is constant on all parallels. – hartkp Oct 12 '18 at 17:32
  • (Continued): this holds in all possible directions. If the measure is not uniform then there will be two open sets $U$ and $V$ that will have measure larger and smaller than their relative Lebesgue measure respectively. These can have sufficiently small diameter and, upon a rotation can be situated inside a band around the equator. Then the uniformity result for rings will lead to a contradiction. – hartkp Oct 12 '18 at 17:40