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Let $X, Y, Z$ be measurable spaces with measures $\mu_X, \mu_Y, \mu_Z$ respectively. Let $\pi_Y : Y \times Z \rightarrow Y$ be the projection on $Y$ and $\pi_Z : Y \times Z \rightarrow Z$ the projection on $Z$. Let $\psi : X \rightarrow Y \times Z$ be a surjective map such that $(\pi_Y \circ \psi)_{*} \mu_X = \mu_Y$ and $(\pi_Z \circ \psi)_{*} \mu_X = \mu_Z$. Can we deduce that $\psi_{*}\mu_X = \mu_Y \otimes \mu_Z$ (this is a notation for the product measure of $\mu_Y$ and $\mu_Z$)? If yes how? If not, what other conditions are required?

In case the answer to the above question is no, the following is a more precise account of my problem: Let $G$ be a Lie group and $H_i, K_i \leq G$ Lie subgroups for $i = 1,2,3$. Let $X_i = K_i \setminus G /H_i$ be the double coset space given by quotienting from the right by $H_i$ and from the left by $K_i$,i.e. an element of $X_i$ is a double coset of the form $K_igH_i$. Endow $X_i$ with the unique $G$ invariant measure and denote this measure by $\mu_{X_i}$. I have a map $\psi: X_1 \rightarrow X_2 \times X_3$ which satisfies the properties above. Namely, $\psi$ is surjective and $(\pi_{X_i} \circ \psi)_{*}\mu_{X_1} = \mu_{X_i}$ for $i =2,3$. I would like to show that $\psi_{*}\mu_{X_1} = \mu_{X_2} \otimes \mu_{X_3}$. Thank you in advance for your answer.

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  • $\begingroup$ For the identity $\psi$ on $Y\times Z$, you ask whether a measure $\mu$ on the product with given marginals $\mu_Y=(\pi_Y)_*\mu$ and $\mu_Z=(\pi_Z)_*\mu$ is necessarily the product measure which, of course, is wrong. $\endgroup$ Commented Jan 17 at 9:36

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The answer to the first part is no. The point is that surjectivity is not as strong a condition in this context as one might wish for.

Let $X=Y=Z=[0,1]$ with the Borel $\sigma$-algebra and $\mu_X=\mu_Y=\mu_Z$ be the uniform distribution. Consider the non-surjective function $x\mapsto (x,x)$. Its push-forward is clearly not $\mu_Y\otimes\mu_Z$, it is the uniform distribution on the diagonal. But you can take an uncountable subset $N$ of $X=[0,1]$ of measure zero (the Cantor ternary set will do). Then $N$ will actually be measurably isomorphic to the set $\{(x,x)\mid x\in N\}\cup\{(x,y)\mid x,y\in [0,1], x\neq y\}$ by Kuratowski's Borel isomorphism and the fact that all uncountable Borel sets in Polish spaces have the cardinality of the continuum. Let $\phi$ be such a measurable isomorphism. Now define $\psi:X\to Y\times Z$ by $\psi(x)=(x,x)$ for $x\notin N$ and $\psi(x)=\phi(x)$ for $x\in N$. The pushforward of $\mu_X$ under the surjection $\phi$ is the uniform distribution on the diagonal.

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