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The title says it all.

Given two measurable functions $f:X\to \mathbb C$ and $g:Y\to \mathbb C$, where $(X,\mu)$ and $(Y,\nu)$ are two probability spaces, suppose they have the same distributions.

Is it true that there exists a measure-preserving measurable function $\phi:X\to Y$

$\nu(\phi(U)) = \mu(U)$ and $\mu(\phi^{-1}(V)) = \nu(V)$

such that $f = g \circ \phi$ almost everywhere?

Exodd
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1 Answers1

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As you have stated it, it's a tiny bit delicate. It holds if $X$ and $Y$ are standard probability spaces. If you talk about the induced measures on $\mathbb C$ given by $\mu'(A)=\mu(f^{-1}(A))$ and $\nu'=\nu(g^{-1}(A))$ for $A\subset\mathbb C,$ the answer is yes. (Because $\mathbb C$ is a standard probability space.)

kimchi lover
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  • $f,g$ have the same distribution if and only if $\mu'=\nu'$ as you've defined them. The question is exactly as I have stated it. Do you have any reference for this result? – Exodd Dec 09 '17 at 22:31
  • The wikipedia article I cited has references. I know very little about non-standard probability spaces. – kimchi lover Dec 09 '17 at 22:36
  • A bit convoluted, but I guess it goes like this: Let $F(U) := f^{-1}\circ g(U)$. By definition, it should be an homomorphism of measure algebras between $Y$ and $X$, so if they are standard probability space, there is $\varphi:X\to Y$ such that $F(U) = \varphi^{-1}(U)$ and all that's missing to prove is $f=g\circ \varphi$. Am I correct? – Exodd Dec 09 '17 at 22:49
  • Yes, I think so. – kimchi lover Dec 09 '17 at 22:54