If $P$ is the set of all distributions, the only sufficient subfield is the trivial one

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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

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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 $B\in\mathcal{A}'$.$\square$