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Given a statistical model $\{\mathbb{P}_\theta\,|\,\theta\in\Theta\}$ on $(\Omega,\mathscr{F})$, and given a real-valued random variable $X$, we say a real-valued random variable $T=T(X)$ is a sufficient statistic if $X_*\mathbb{P}_\theta(dx|T)$ (the conditional distribution of $X$ given $T$) is independent of $\theta$. We also say a sufficient statistic $S$ is minimal if for all other sufficient statistics $T$ there is a measurable $\phi:(\mathbb{R},Bor(\mathbb{R}))\to(\mathbb{R},Bor(\mathbb{R}))$ such that $S=\phi(T)$ $\mathbb{P}_\theta$-a.e. for any $\theta$.

Suppose that all $X_*\mathbb{P}_\theta(dx)$ are absolutely continuous with respect to the Lebsegue measure with density $f(x|\theta)$, and that $X,S$ are real-valued random variables. A well-known theorem says that if "for all $x,y\in \mathbb{R}$ $\frac{f(x|\theta)}{f(y|\theta)}$ is independent of $\theta$ if and only if $S(x)=S(y)$" then $S$ is a minimal sufficient statistic.

A proof of this theorem can be found here (Thm 3.4). This proof (which is the same as all other proofs that I saw) however only shows that, given another sufficient statistic $T$, there exists a function $\phi:\mathbb{R}\to\mathbb{R}$ such that $S=\phi(T)$ $\mathbb{P}_\theta$-a.e. for any $\theta$, but doesn't show it is measurable.

My question is simple: how does one show $\phi$ is measurable?

UPDATE : The answer to my question seems to be that $\phi$ need not be measurable in the general case and that the theorem as I stated it is wrong. See here for more details.

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    $\begingroup$ The theorem based on the ratio of densities implies that the property is true almost everywhere since these densities are uniquely defined a.e.. This means that $S$ is also defined a.e. $\endgroup$
    – Xi'an
    Jun 8, 2023 at 11:02
  • $\begingroup$ Thank you, I have now changed the equalities to a.e. equalities. $\endgroup$
    – No-one
    Jun 8, 2023 at 12:06
  • $\begingroup$ My point was that the a.e. definition could prove enough to bypass measurability issues. $\endgroup$
    – Xi'an
    Jun 8, 2023 at 12:17
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    $\begingroup$ @Titti, check Shao where he used the existence of minimal sufficient statistic when the family is dominated by a sigma-finite measure and the usual argument to show that the function is one-one. He then resorts to a result in Parthasarathy which basically says the inverse function is measurable. $\endgroup$ Jun 8, 2023 at 19:51
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    $\begingroup$ @User1865345-solidarityMods thank you! The way Shao does it is very interesting and works perfectly fine. However, the fact that he is able to prove the result only for $\mathbb{R}^n$-valued $X$ and has to appeal to that deep result in Parthasarathy (whose prove requires descriptive set theory) suggests that this is in fact the best that can be done and that for a general $X$ the function $\phi$ need not be measurable. Do you know if this is indeed the case? $\endgroup$
    – No-one
    Jun 8, 2023 at 21:07

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