(Note: This is not about Bayesian inference, but about classical inference)

Let $\{P_\theta\}_{\theta\in \Theta}$ be a family of probability measures on $\mathbb{R}^n$ with density functions $f_\theta$.

Let $T:\mathbb{R}^n \rightarrow \mathscr{T}$ be a statistic.

Define $$D(x):=\{y\in \mathbb{R}^n: \text{ There exists a positive function } h \text{ such that } f_{\theta}(x)=f_{\theta}(y)h(x,y) \text{ for all } \theta \}$$.

(Note that $D(x)$ forms a partition of $\mathbb{R}^n$)

Assume that $T(x)=T(y)$ if and only if $x\in D(y)$.

Many textbooks and even wikipedia asserts that, with the above assumption, one can show that $T$ is minimal sufficient.

However, I do not get this Here is a proof that textbooks I have seen have:

Let $\alpha:range(T)\rightarrow \mathbb{R}^n$ be a representative function for $T$. (That is, $T(\alpha(t))=t$)

If we pick $x\in\mathbb{R}^n$, from our assumption,$f_\theta(x)=f_\theta(\alpha(T(x)))h(x,\alpha(T(x))$ holds for all $\theta$ for some positive function $h>0$.

If we take $g_\theta:=f_\theta\circ \alpha$, then by Fisher-Neymann theorem, $T$ is sufficient.

However, since $\alpha$ need not be measurable, $g_\theta$ need not be measurable. So we cannot apply Fisher-Neymann theorem. (As you can see here, $\alpha$ need mot be measurable)

Is $g_\theta$ measurable in anyways? Or is there a correct proof for this?

Thank you in advance.


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