Probability measures on the hypersphere $\Bbb S^{n-1}$ and the $O(n)$ isomorphisms

Question

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.

Answer 1

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).

Answer 2

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 $$ \int_{\Bbb S^{n-1}} f \,d\mu = \frac{1}{\mathcal H^{n-1}(\Bbb S^{n-1})} \int_{\Bbb S^{n-1}} f \,d\mathcal H^{n-1}. $$