An external dominating law need not be a valid sufficiency reference

Define $K(t,A)=1_A((t,0))$ for $t\in \{0,1\}$. Each section is the Dirac set function [F5]: it vanishes on the empty set, has total mass one, and is countably additive because a disjoint family contains $(t,0)$ in at most one member. Thus each section is a probability measure [F4]. All evaluations are measurable on the finite full sigma-field, so [F6] makes $K$ a probability kernel independent of $\theta$, between the standard Borel spaces in [F2]. For every $A\subseteq \mathcal{X}$ and $C\subseteq \{0,1\}$, $P_{\theta }(A\cap T^{-1}(C))={\sum }_{t\in C}{1}_A((t,0))P_{\theta }(\{(t,0)\})={\int }_{C}K(t,A)P_{\theta }^T(dt)$. Hence $T$ is sufficient by [F1].

Every atom of $R$ has mass $1\mathrm{/}4$, so $P_{\theta }\ll R$. The density values are $d{P}_{\theta }\mathrm{/}dR(t,u)=4P_{\theta }(t,0)$ if $u=0$ and zero if $u=1$. For fixed $t$, the two points $(t,0)$ and $(t,1)$ have the same $T$ value but different derivative values. Every atom has positive $R$-mass, so no modification on an $R$-null set can change them; the derivative has no $\sigma (T)$-measurable version.

This refutes the implication with an arbitrary external reference. Taking the reference to be a member $P_{{\theta }_0}$ restores the common conditional law needed for the criterion: the conditional kernel at the reference is then the model’s shared kernel. The counterexample’s $R$ has conditional second coordinate uniform given $T$, while each model law has second coordinate zero, identifying the precise missing hypothesis.