Is there always a mapping between probability measures on the $n$-sphere?

Question

If $\mu,\nu\in\mathcal{P}(\mathbb{S}^{n-1})$ for $n\in\mathbb{N}$. Can we always find a measurable mapping $f:\mathbb{S}^{n-1}\to\mathbb{S}^{n-1}$ such that $\mu=f\#\nu$, or equivalently $$\int_{\mathbb{S}^{n-1}} gd\mu=\int_{\mathbb{S}^{n-1}} gd(f\#\nu)=\int_{\mathbb{S}^{n-1}} g\circ fd\nu$$ where $f\#\nu$ stands for the push-forward and $g\in\mathcal{L}^1(\mathbb{S}^{n-1})$? If not, what are the sufficient conditions for it to be true?

Answer 1

First note that the $d$-dimensional sphere $S^d$ with its standard topology is a standard measurable space. Call a Borel probability measure $\mu$ on $S^d$ continuous if $\forall x\in S^d: \mu(\{x\})=0$. Then we have the following abstract isomorphism theorem:

Theorem: Let $(X,\mathcal{B}(X))$ be a standard measurable space and for any continuous Borel probability measure $\mu$ on $(X,\mathcal{B}(X))$, there is a bimeasurable $f: (X,\mathcal{B}(X))\to ([0,1],\mathcal{B}([0,1]))$ such that $f_\ast(\mu)=\operatorname{leb}_{[0,1]}$.

(See e.g. Kechris' Classical Descriptive Set Theory, p.116, Thm.17.41 for a proof.)

(By bimeasurable I mean an isomorphism in the category of measurable spaces; so that $f:X\to Y$ is bimeasurable iff it's a measurable bijection with measurable inverse.)

Next, observe that any Borel probability measure $\mu$ on $S^d$, there are at most countably many points $x_{\mu,\bullet}:\mathbb{Z}_{\geq1}\to S^d$ such that $\mu(\{x_{\mu,n}\})=c_{\mu,n}>0$; and putting $D_\mu=\{x_{\mu,n}\,|\, n\in\mathbb{Z}_{\geq1}\}\subset S^d$ and $C_\mu=S^d\setminus D_\mu$, we have that

$$\mu=\mu(C_\mu)\mu(\bullet\,|\, C_\mu)+\mu(D_\mu)\mu(\bullet\,|\, D_\mu),$$

with $\mu(\bullet\,|\, C_\mu)$ a continuous Borel probability measure on $S^d$ and $\mu(\bullet\,|\, D_\mu)=\sum_{n\in\mathbb{Z_{\geq1}}}c_{\mu,n}\delta_{x_{\mu,n}}$ . By checking against characteristic functions of singletons, it's straightforward that if $f:S^d\to S^d$ is bimeasurable, then $f_\ast(c_{\mu,n}\delta_{x_{\mu,n}})=c_{\mu,n}\delta_{f(x_{\mu,n})}$; thus the discrete part of a measure is term-wise preserved under bimeasurable maps . Putting all this together we obtain:

Corollary: For any two Borel probability measures $\mu,\nu$ on $S^d$, there is a bimeasurable $g:S^d\to S^d$ such that $g_\ast(\mu)=\nu$ iff there is a bijection $\beta: \mathbb{Z}_{\geq1}\to\mathbb{Z}_{\geq1}$ such that $x_{\mu,n}=x_{\nu,\beta(n)}$ and $c_{\mu,n}=c_{\nu,\beta(n)}$.

Note that bimeasurability is necessary for this; e.g. a constant map pushes forward any measure to the Dirac delta at the point that is its image.