The sum of exponential-family statistics is sufficient for an IID sample

For fixed $\eta$, each coordinate pullback of $p_{\eta }$ is measurable on the product sigma-algebra [F2], and the product density $g_{\eta }(y_{1:n})={\prod }_i{p}_{\eta }(y_i)$ is measurable by [F9]. Its integral against ${\mu }^{\otimes n}$ is one by [F2], repeated Tonelli [F3], and $\int p_{\eta }d\mu =1$. Thus $g_{\eta }d{\mu }^{\otimes n}$ is a probability measure by [F4]. On every measurable rectangle it equals ${\prod }_i{P}_{\eta }(A_i)$ by the rectangle formula and Tonelli, so the sigma-finite pi-system uniqueness theorem [F5] identifies it with the IID product law. Expanding $g_{\eta }$ gives the density displayed in the Statement.

Since $\mathrm{\Omega }$ is nonempty, choose ${\eta }_0\in \mathrm{\Omega }$. The density ratio on the common support is $\frac{g_{\eta }(y_{1:n})}{g_{{\eta }_0}(y_{1:n})}=\exp \{(\eta -{\eta }_0{)}^T{T}_n(y_{1:n})-n(A(\eta )-A({\eta }_0))\}\mathrm{.}$ If any $h(y_i)=0$, both product densities vanish, so the same positive Borel function of $T_n$ satisfies $g_{\eta }=r_{\eta }(T_n)g_{{\eta }_0}$ everywhere. Integrating this identity on each measurable set proves that it is a version of the Radon–Nikodym derivative under $P_{{\eta }_0}^{\otimes n}$.

By [F6] the finite sample space is standard Borel. If $k\ge 1$, [F6] also gives that the statistic target ${\mathbb{R}}^k$ is standard Borel. If $k=0$, ${\mathbb{R}}^0$ is the empty product, a singleton with its full two-element sigma-algebra, which is standard Borel by the empty-product case of [F6]. In this case $T_n$ is the unique constant map, the dot products in step 2.1 are empty sums equal to zero, and its likelihood-ratio expression is still a Borel function of this singleton statistic. Thus both target cases meet the standard-Borel hypothesis. The reference $P_{{\eta }_0}^{\otimes n}$ belongs to the family, and step 2.1 makes every likelihood ratio a Borel function of $T_n$. Applying [F7] gives a common conditional-law kernel, which is sufficiency by [F8]. The sole choice-axiom use is inherited from [A1] and [F7].