If $P$ is the set of all distributions, the only sufficient subfield is the trivial one According to an article by Bahadur, if $P=\left\\{p\right\\}$ is the set of all probability measures on the measurable space $\left(\Omega,\mathcal{A}\right)$, $\mathcal{A}$ is the only possible sufficient subfield. The claim is left unproved in the article. Any help will be appreciated. _Bahadur, R. R. Sufficiency and Statistical Decision Functions. Ann. Math. Statist. Volume 25, Number 3 (1954), 423-462. A remark following definition 5.2_

Let $\mathcal{A}'$ be a sufficient subfield of $\mathcal{A}$. We'll show that $\mathcal{A}\subseteq\mathcal{A}'$. Let $B\in\mathcal{A}$. Since $\mathcal{A}'$ is sufficient, there's some $\mathcal{A}'/\overline{\mathfrak{B}}$-measurable $\varphi_B:\Omega\rightarrow\overline{\mathbb{R}}$ such that $p(B)=\int_\Omega fdp$ for all $p\in P$.

Now let $\omega\in\Omega$ and set $p:=\delta_\omega$, where $\delta_\omega$ is the Dirac measure on $\left(\Omega,\mathcal{A}\right)$. Then $p(B)=\mathbb{1}_B\left(\omega\right)$ and $\int_\Omega fdp=f\left(\omega\right)$. Since $\omega$ was arbitrary, $f=\mathbb{1}_B$. So $B\in\mathcal{A}'$.$\square$