# Minimal sufficient statistic criterion

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?