The improper Beta(0,0) prior has a divergent binomial endpoint posterior

By [A1] and [F7], countable choice is available for [F3], [F4], [F10], and [F11]. Write $f(p)=1\mathrm{/}[p(1-p)]$ on $(0,1)$. For any continuous $g:(0,1)\to \mathbb{R}$ and open $U\subseteq \mathbb{R}$, continuity makes $g^{-1}(U)$ open in the subspace $(0,1)$, so $g^{-1}(U)=O\cap (0,1)$ for some open $O\subseteq \mathbb{R}$. This intersection is open in $\mathbb{R}$, hence Borel by [F9] and Lebesgue measurable by [F10]. Thus $g$ is measurable for the restricted Lebesgue sigma-algebra. This applies to $f$ and to every binomial likelihood section, which is a polynomial. By [F11], Lebesgue measure on $\mathbb{R}$ is a measure; its restriction to the measurable interval $(0,1)$ is a measure as well. Thus [F5] defines the prior measure $\mathrm{\Pi }(A)={\int }_{A}fd\lambda$. For each $p\in (0,1)$, let $\epsilon =\min \{p,1-p,1\mathrm{/}3\}>0$; by [F8] choose $k\ge 1$ with $1\mathrm{/}k<\epsilon$. Then $k\ge 3$ and $p\in E_k=[1\mathrm{/}k,1-1\mathrm{/}k]$, so these sets cover $(0,1)$. On $E_k$, $f(p)\le k^2$. Their Lebesgue lengths are at most one by [F4], so homogeneity and monotonicity [F6] give $\mathrm{\Pi }(E_k)\le k^2<\mathrm{\infty }$; [F3] makes this a sigma-finite prior. It is not finite: on each disjoint interval $I_j=(2^{-j-1},2^{-j}]$, $j\ge 1$, one has $f(p)\ge 1\mathrm{/}p\ge 2^j$ and $\lambda (I_j)=2^{-j-1}$ by [F4]. Homogeneity and monotonicity [F6] give $\mathrm{\Pi }(I_j)={\int }_{I_j}fd\lambda \ge 1\mathrm{/}2$. By [F5], $\mathrm{\Pi }$ is a measure, so the prior mass of the union of the first $m$ such intervals is at least $m\mathrm{/}2$ for every $m$. Hence $\mathrm{\Pi }((0,1))=+\mathrm{\infty }$ and the prior is improper.

For $y=0$, put ${\nu }_0(A):={\int }_A{L}_{0}d\mathrm{\Pi }$; it is a measure by [F5], and $Z(0)={\nu }_0((0,1))$ by [F1]. The likelihood is $L_0(p)=\left(\genfrac{}{}{0px}{}{n}{0}\right)(1-p{)}^n=(1-p{)}^n$, so on each $I_j$ from step 1.1, $p\le 1\mathrm{/}2$ gives $L_0(p)\ge 2^{-n}$. Also $\mathrm{\Pi }(I_j)\ge 1\mathrm{/}2$ by step 1.1. Monotonicity and homogeneity [F6] therefore give ${\nu }_0(I_j)\ge 2^{-n-1}$. The intervals are pairwise disjoint, so measure additivity gives ${\nu }_0({\bigcup }_{j=1}^m{I}_j)\ge m{2}^{-n-1}$; since this union is contained in $(0,1)$, countable additivity gives ${\nu }_0((0,1))\ge m{2}^{-n-1}$ for every $m$. Thus $Z(0)=+\mathrm{\infty }$ and [F1] leaves the quotient undefined.

For $y=n$, put ${\nu }_n(A):={\int }_A{L}_{n}d\mathrm{\Pi }$; it is a measure by [F5], and $Z(n)={\nu }_n((0,1))$ by [F1]. Here $L_n(p)=\left(\genfrac{}{}{0px}{}{n}{n}\right)p^n=p^n$. On each reflected interval $J_j=[1-2^{-j},1-2^{-j-1})$, $j\ge 1$, one has $p\ge 1\mathrm{/}2$, so $L_n(p)\ge 2^{-n}$. Also $f(p)\ge 1\mathrm{/}(1-p)\ge 2^j$ there and $\lambda (J_j)=2^{-j-1}$ by [F4], so $\mathrm{\Pi }(J_j)\ge 1\mathrm{/}2$ by [F6]. Monotonicity and homogeneity [F6] give ${\nu }_n(J_j)\ge 2^{-n-1}$. The intervals are pairwise disjoint, and additivity followed by nonnegativity on their complement gives $Z(n)={\nu }_n((0,1))\ge m{2}^{-n-1}$ for every $m$. Thus $Z(n)=+\mathrm{\infty }$ and [F1] leaves this quotient undefined. By [F2], neither endpoint fiber admits a normalized posterior probability.