A full exponential family is complete

Choose ${\eta }_0$ in the open subset and $ϵ>0$ so the closed cube ${\eta }_0+[-ϵ,ϵ{]}^k$ lies in $\mathrm{\Omega }$. Define the finite signed measure $\rho (A)={\int }_{A}g(T(x))P_{{\eta }_0}(dx)$ by [F4], and set $\nu (B)=\rho (T^{-1}B)$ for Borel $B\subseteq {\mathbb{R}}^k$. Since $T$ is measurable, preimages preserve disjoint countable unions, so $\nu$ is a finite signed Borel measure. The preimages of a countable partition form a countable partition, giving $\mathrm{\mid }\nu \mathrm{\mid }\le T_{\mathrm{\#}}\mathrm{\mid }\rho \mathrm{\mid }$ by [F4]. Hence $\mathrm{\mid }\nu \mathrm{\mid }({\mathbb{R}}^k)\le E_{{\eta }_0}\mathrm{\mid }g(T)\mathrm{\mid }<\mathrm{\infty }$.

Choose $0<\delta <ϵ$. Since $e^{\delta \mathrm{\parallel }s{\mathrm{\parallel }}_1}\le {\sum }_{\sigma \in \{-1,1{\}}^k}{e}^{\delta \sigma \cdot s}$ and $\mathrm{\mid }\nu \mathrm{\mid }\le T_{\mathrm{\#}}\mathrm{\mid }\rho \mathrm{\mid }$ by step 1.1, the weighted-integral bound follows first for nonnegative simple functions from that setwise domination and then for $e^{\delta \mathrm{\parallel }s{\mathrm{\parallel }}_1}$ by [F5]. For each sign vector, changing from $P_{{\eta }_0}$ to $P_{{\eta }_0+\delta \sigma }$ contributes the factor $Z({\eta }_0+\delta \sigma )\mathrm{/}Z({\eta }_0)$, so $\int e^{\delta \mathrm{\parallel }s{\mathrm{\parallel }}_1}\mathrm{\mid }\nu \mathrm{\mid }(ds)\le {\sum }_{\sigma }\frac{Z({\eta }_0+\delta \sigma )}{Z({\eta }_0)}{E}_{{\eta }_0+\delta \sigma }\mathrm{\mid }g(T)\mathrm{\mid }<\mathrm{\infty }$, the exponential moment required by [F3].

For every real $t$ with $\mathrm{\parallel }t{\mathrm{\parallel }}_{\mathrm{\infty }}<\delta$, its Laplace transform is $\int e^{t\cdot s}\nu (ds)=\frac{Z({\eta }_0+t)}{Z({\eta }_0)}{E}_{{\eta }_0+t}g(T)=0$ by the mean-zero premise. The transform-uniqueness result [F3], under axiom branch [A1], gives $\nu =0$.

For every Borel $B$, $0=\nu (B)=\rho (T^{-1}B)={\int }_{B}g(s)P_{{\eta }_0}^T(ds)$ by the definition of the pushforward measure and $\rho$. Taking $B=\{g>0\}$ and $B=\{g<0\}$ shows both positive and negative parts have zero integral, so $g(T)=0$ $P_{{\eta }_0}$-almost surely. Every family density is strictly positive on $\{h>0\}$, so all $P_{\eta }$ are equivalent to $P_{{\eta }_0}$ and the same vanishing holds for every $\eta$. By [F2], $T$ is complete.