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Let $\Bbb S^{n-1}$ denotes the unit sphere in $\Bbb R^n$ and let $O(n)$ be the group of all orthogonal linear transformations on $\Bbb R^n$. For any (outer) measure $\mu$ on $\Bbb R^n$, the pushforward of $\mu$ by $T\in O(n)$ is defined as $T_{\#}\mu(E) := \mu(T^{-1}(E))$. It is clear that the pushforward under $T\in O(n)$ maps a probability measure on $\Bbb S^{n-1}$ to another probability measure on $\Bbb S^{n-1}$.

Suppose that $f\in C(\Bbb S^{n-1})$ satisfies the condition $$ \int_{\Bbb S^{n-1}} f \,d\mu = \int_{\Bbb S^{n-1}} f \,dT_{\#}\mu \quad \text{for all } \ T\in O(n), \tag{*}\label{*} $$ is it true that $$ \int_{\Bbb S^{n-1}} f \,d\mu = \frac{1}{\mathcal H^{n-1}(\Bbb S^{n-1})} \int_{\Bbb S^{n-1}} f \,d\mathcal H^{n-1} =: \bar f, \tag{**}\label{**} $$ where $\mathcal H^{n-1}$ is the $n-1$ dimensional Hausdorff measure?

Of course, the case where $\mu$ is a Dirac mass $\delta_x$ for some $x\in\Bbb S^{n-1}$ is clear, and, in fact, we can trivially conclude that $f$ must be a constant function. One might be tempted to conjecture that the condition $\eqref{*}$ implies that $f$ must be a constant function, but that is false. We can already find a counterexample to the stronger conjecture by considering $\mu$ to be a sum of Dirac masses: $\mu = \frac1n (\delta_{e_1} + \cdots + \delta_{e_n})$ and $f(x) = x\cdot Ax$ for any $n\times n$ matrix $A$. Indeed, in this case we have $$ \int_{\Bbb S^{n-1}} f \,d\mu = \frac 1n \sum_{i=1}^n e_i\cdot Ae_i = \frac{\text{Tr}(A)}{n}, $$ and the statement that $\eqref{*}$ holds follows from the well-known fact that the trace of a matrix $A$ is invariant under a change of an orthogonal basis. Clearly, $f$ is not a constant function, but it is not hard to show that $$ \frac{\text{Tr}(A)}{n} = \frac{1}{\mathcal H^{n-1}(\Bbb S^{n-1})} \int_{\Bbb S^{n-1}} x\cdot Ax \,d\mathcal H^{n-1}(x) $$ using the divergence theorem and that fact that $\text{Tr}(A) = \text{div}\, Ax$.

This question is largely inspired by the example regarding the trace that I mentioned above. I restrict $f$ in this question to be a continuous function since I feel like there might be a beautiful proof using the duality between continuous functions and Radon measures on $\Bbb S^{n-1}$.

I am aware of the existence of the things called measures on homogeneous spaces that generalize Haar measures, but aside from that, I know next to nothing about it. I welcome a proof that uses ideas from that theory, but would like to read a proof that doesn't use it as well, even if it might be longer and less elegant.

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  • $\begingroup$ I am not sure how to use Haar measures or integrate over $O(n)$, but an idea: Since the integration of $\int_{S^{n-1}} f d \mu$ is invariant under pushforwards by $T$, we could average over $T$ and interchange the integrals by Fubini's theorem. Let $dT$ be the Haar measure on $O(n)$ and assume $\int_{O(n)} dT=1$. Then something along the lines of ... (1) $\endgroup$
    – Alex
    Dec 10, 2023 at 16:38
  • $\begingroup$ $$ \int_{S^{n-1}} f d \mu = \int_{O(n)} \int_{S^{n-1}} f(x) \, d(T_{\#} \mu)(x) \, dT = \int_{O(n)} \int_{S^{n-1}} f(T(x)) \, d \mu(x) \, dT = \int_{S^{n-1}} \int_{O(n)}f(T(x)) \, dT \, d \mu(x) = \int_{S^{n-1}} \frac{1}{\mathcal{H}^{n-1}(S^{n-1})} \int_{S^{n-1}} f(y) \, d \mathcal{H}^{n-1}(y) \, d \mu(x) = \frac{1}{\mathcal{H}^{n-1}(S^{n-1})} \int_{S^{n-1}} f(y) \, d \mathcal{H}^{n-1}(y) = \bar{f}$$ hopefully holds. The fourth equality should hold since x is fixed, so integrating over the Haar measure on $O(n)$ should average over all rotations (and reflections). (2) $\endgroup$
    – Alex
    Dec 10, 2023 at 16:39
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    $\begingroup$ @ChadK Could you elaborate on your idea a bit more? The set of admissible functions for Dirac masses are just the constant functions, but as I've mentioned in the question, the set of admissible functions for a convex combination of Dirac masses can be bigger than that, e.g. $f(x) = x\cdot Ax$. $\endgroup$
    – BigbearZzz
    Dec 10, 2023 at 17:38
  • $\begingroup$ @Alex It does feel intuitively true. What I had in mind when I made the question in the first place was similar to the step that you skipped (the 4th equality). I would like to see a rigorous proof, hence this question. PS. I'm sorry I made a typo when I defined the pushforward measure (I've just corrected it). Luckily, it doesn't matter here since any $T\in O(n)$ is an isomorphism. $\endgroup$
    – BigbearZzz
    Dec 10, 2023 at 18:26
  • $\begingroup$ Here is what I could find: $O(n)$ is the isometry group of $S^{n-1}$ (for example by [this][1] post), and so for arbitrary fixed $x \in S^{n-1}$ and evaluation map $$e_x : O(n) \to S^{n-1}, \quad T \mapsto e_x(T) := T(x),$$ we should have (for $d \nu := d T$) $$ (e_x)_{\#} \nu = \frac{\mathcal{H}^{n-1}}{\mathcal{H}^{n-1}(S^{n-1})}.$$ (For example by [this][2] post). [1]: math.stackexchange.com/questions/4147602/… [2]: mathoverflow.net/questions/345084/… $\endgroup$
    – Alex
    Dec 10, 2023 at 18:35

2 Answers 2

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Edit: Take a look at BigbearZzz' answer for a more detailed explanation!

Since the integration of $\int_{S^{n-1}} f d \mu$ is invariant under pushforwards by $T$, we can average over all $T \in O(n)$ via the (normalized) Haar measure $\nu$ on $O(n)$ and interchange the integrals by Fubini's theorem. \begin{align}\int_{S^{n-1}} f d \mu &= \int_{O(n)} \int_{S^{n-1}} f(x) \, d(T_{\#} \mu)(x) \, d \nu(T) = \int_{O(n)} \int_{S^{n-1}} f(T(x)) \, d \mu(x) \, d \nu(T) \\ &= \int_{S^{n-1}} \int_{O(n)}f(T(x)) \, d \nu(T) \, d \mu(x) = \int_{S^{n-1}} \int_{O(n)}f(e_x(T)) \, d \nu(T) \, d \mu(x) \\ &= \int_{S^{n-1}} \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{S^{n-1}} f(y) \, d \mathcal{H}^{n-1}(y) \, d \mu(x) = \avint_{S^{n-1}} f(y) \, d \mathcal{H}^{n-1}(y) \\ &= \bar{f}. \end{align} Here, the fourth equality holds since $x$ is fixed, so integrating over the Haar measure on $O(n)$ really averages over the sphere.

More precisely, $O(n)$ is the isometry group of $S^{n-1}$ (for example by this post), and so for arbitrary fixed $x \in S^{n-1}$ and evaluation map $$e_x : O(n) \to S^{n-1}, \quad T \mapsto e_x(T) := T(x),$$ we have $$ (e_x)_{\#} \nu = \frac{\mathcal{H}^{n-1}}{\mathcal{H}^{n-1}(S^{n-1})}.$$ (For example by this post).

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  • $\begingroup$ At first I thought about just adding some minor details and references to complement your answer, but I ended up expanding some parts of your answer too to make it a bit clearer to future readers as to what we have been discussing. $\endgroup$
    – BigbearZzz
    Dec 12, 2023 at 18:06
  • $\begingroup$ You did a great job! This is more than just adding minor details, it is very valuable due to the explanations on (Haar) measure theory. In my opinion, you should declare it as the accepted answer! So also visitors see it first. $\endgroup$
    – Alex
    Dec 12, 2023 at 18:14
  • $\begingroup$ Thank you for your kind words, but I think I should give credit to you for outlining the idea. Your comment at the beginning of your answer should be more than enough to lead people to my answer below if they want more details :) $\endgroup$
    – BigbearZzz
    Dec 12, 2023 at 18:36
  • $\begingroup$ Alright, cheers! :) $\endgroup$
    – Alex
    Dec 12, 2023 at 18:49
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This is a complement to Alex answer, aiming to make it a tiny bit more complete (and partly for myself if I come back to read it in a distant future). I am going to assume the existence and uniqueness of the normalized (left) Haar measure $\theta$ on $O(n)$, i.e. a probability measure such that $T_{\#}\theta(B) = \theta(B)$ for all $T\in O(n)$ and all Borel set $B$ in $O(n)$. Note that $\theta$ is actually a two-sided Haar measure, a consequence of the fact the compact groups are unimodular (see, for example, Knapp's Advance Real Analysis, chapter VI, regarding properties of Haar measures).

Now, the group $G = O(n)$ acts on the sphere $\Bbb S^{n-1}$ in an obvious way. A $G$-invariant measure $\nu$ on $\Bbb S^{n-1}$ is a measure such that $\nu(T^{-1}(E)) =: T_{\#}\nu(E) = \nu(E)$ for for all $T\in O(n)$ and all Borel set $E$ in $\Bbb S^{n-1}$. We claim that, for any fixed $x\in \Bbb S^{n-1}$, the pushforward of $\theta$ by the evaluation map $A^x\colon O(n) \to \Bbb S^{n-1}$, defined by $$ A^x(T) := Tx, $$ is the unique $G$-invariant probability measure on $\Bbb S^{n-1}$, namely the normalized surface measure $$ A^x_{\#}\theta = \frac{1}{\mathcal H^{n-1}(\Bbb S^{n-1})}\mathcal H^{n-1} \tag{i}\label{i}. $$ Note that $A^x_{\#}\theta(E) = \theta(\{ h\in G: hx \in E \})$.

Indeed, the fact that $A^x_{\#}\theta$ is a probability measure is obvious. To see that it is $G$-invariant, let $g$ be an arbitrary element of $O(n)$ and $E$ a Borel set in $\Bbb S^{n-1}$, then $$\begin{align} g_{\#}(A^x_{\#}\theta)(E) &= \theta((A^x)^{-1}g^{-1}E) = \theta(\{ h\in G: hx \in g^{-1}E \}) \\ &= \theta(g^{-1}\{ h\in G: hx \in E \}) = (g_{\#}\theta)(\{ h\in G: hx \in E \}) \\ &= \theta(\{ h\in G: hx \in E \}) = A^x_{\#}\theta(E). \end{align}$$

It is obvious (from the basic theory of Hausdorff measure) that $\mathcal H^{n-1}$ restricted to the unit sphere is $G$-invariant, hence in order to prove $\eqref{i}$, we need to show that the $G$-invariant probability measure on $\Bbb S^{n-1}$ is unique. However, this part takes more effort to show so I will refer to Knapp's book, Theorem 6.18 (taking into account the fact that $\Bbb S^{n-1}$ is homeomorphic to the quotient subgroup $G/H$ of $G$, where $H$ is a stabilizer subgroup of $G$ that fixes some element $x\in\Bbb S^{n-1}$). Alternatively, one can also derives the uniqueness from the fact that $\nu$ being $G$-invariant on $\Bbb S^{n-1}$ (and that $G$ acts transitively on $\Bbb S^{n-1}$) implies that it is uniformly distributed, and hence unique (see Mattila's Geometry of Sets and Measures in Euclidean Spaces, chapter 3). Thus $\eqref{i}$ holds.

Now, for any Borel measurable function $f:\Bbb S^{n-1} \to \Bbb R$, the function $\tilde f\colon O(n) \times \Bbb S^{n-1} \to \Bbb R$ defined by $\tilde f(T,x) = f(Tx)$ is also Borel measurable since the map $(T,x) \mapsto Tx$ is continuous. If $f$ is bounded on $\Bbb S^{n-1}$, we can apply the Fubini-Tonelli theorem to get $$\begin{align} \int_{\Bbb S^{n-1}} f(x) \,d\mu(x) &= \int_{O(n)} \int_{\Bbb S^{n-1}} f(x) \,dT_{\#}\mu(x) \,d\theta(T) \\ &= \int_{O(n)} \int_{\Bbb S^{n-1}} f(Tx) \,d\mu(x) \,d\theta(T) \\ &= \int_{\Bbb S^{n-1}} \int_{O(n)} f(Tx) \,d\theta(T) \,d\mu(x) \\ &= \int_{\Bbb S^{n-1}} \int_{O(n)} f(A^x(T)) \,d\theta(T) \,d\mu(x). \tag{ii}\label{ii} \end{align}$$ However, we knew from the above discussion that $\eqref{i}$ holds for all $x$, hence, by the change of variable formula, $$ \int_{O(n)} f(A^x(T)) \,d\theta(T) = \int_{\Bbb S^{n-1}} f(y) \,dA^x_{\#}\theta(y) = \frac{1}{\mathcal H^{n-1}(\Bbb S^{n-1})} \int_{\Bbb S^{n-1}} f(y) \,d\mathcal H^{n-1}(y). \tag{iii}\label{iii} $$ Since $\mu$ is a probability measure, we can substitute $\eqref{iii}$ into $\eqref{ii}$ and conclude that $$ \int_{\Bbb S^{n-1}} f \,d\mu = \frac{1}{\mathcal H^{n-1}(\Bbb S^{n-1})} \int_{\Bbb S^{n-1}} f \,d\mathcal H^{n-1}. $$

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