An improper variance hyperprior can give an improper posterior

The prior is sigma-finite: the sets $E_n=[1\mathrm{/}n,n]$ increase to $(0,\mathrm{\infty })$ and $\mathrm{\Pi }(E_n)={\int }_{1\mathrm{/}n}^{n}dt\mathrm{/}t=2\log n<\mathrm{\infty }$. Integrating out $\theta$ gives the evidence at fixed $t$, $m(0\mid t)={\int }_{\mathbb{R}}\frac{e^{-{\theta }^2\mathrm{/}(2t)}}{\sqrt{2\pi t}}\frac{e^{-{\theta }^2\mathrm{/}2}}{\sqrt{2\pi }}d\theta =\frac{1}{\sqrt{2\pi (1+t)}}$. Indeed, the exponent is $-(1+t^{-1}){\theta }^2\mathrm{/}2$; the substitution $z=\sqrt{1+t^{-1}}\theta$ reduces the integral to the unit normal integral [F2]. It is positive and finite for every $t>0$.

The evidence-weighted density of $t$ is $q(t)=1\mathrm{/}(t\sqrt{2\pi (1+t)})$ by [A2] and step 1.1. For each $k\ge 0$, let $I_k=(2^{-(k+1)},2^{-k})$. On $I_k\subset (0,1)$, $q(t)\ge 1\mathrm{/}(t\sqrt{4\pi })$ and $1\mathrm{/}t\ge 2^k$, while the interval has length $2^{-(k+1)}$. Hence ${\int }_{I_k}q(t)dt\ge 1\mathrm{/}(2\sqrt{4\pi })$, uniformly in $k$. The intervals are disjoint, so summing their lower bounds shows ${\int }_0^{1}q(t)dt=\mathrm{\infty }$. By [F3], no finite positive scaling makes this likelihood-weighted measure a posterior probability.