Nonidentifiability precludes simultaneous consistency

By [F4], nonidentifiability supplies ${\theta }_0,{\theta }_1\in \mathrm{\Theta }$ with ${\theta }_0\sim {\theta }_1$ and $\psi ({\theta }_0)\mathrm{\ne }\psi ({\theta }_1)$. Fix this pair.

Fix $n\ge 1$. By [F1], the observation laws in the model are $P_{\theta }$. Measurable rectangles in ${\mathcal{X}}^n$ form a $\pi$-system generating ${\mathcal{A}}^{\otimes n}$ by [F7]. For every rectangle $A_1\times \cdots \times A_n$, [F8] and [F2] give $P_{{\theta }_0}^{\otimes n}(A_1\times \cdots \times A_n)={\prod }_{j=1}^n{P}_{{\theta }_0}(A_j)={\prod }_{j=1}^n{P}_{{\theta }_1}(A_j)=P_{{\theta }_1}^{\otimes n}(A_1\times \cdots \times A_n)$. The full space is itself a rectangle and is an increasing exhaustion of finite mass one. Applying [F9] shows $P_{{\theta }_0}^{\otimes n}=P_{{\theta }_1}^{\otimes n}$ on all of ${\mathcal{A}}^{\otimes n}$.

Put $\mathrm{\Delta }=d(\psi ({\theta }_0),\psi ({\theta }_1))>0$ and $r=\mathrm{\Delta }\mathrm{/}3$. The two closed target balls $C_i=\{t\in \mathcal{T}:d(t,\psi ({\theta }_i))\le r\}$ are disjoint: a point in both would give $\mathrm{\Delta }\le 2r<\mathrm{\Delta }$ by [F10].

Suppose, for contradiction, that the estimator sequence is consistent at both parameters. For $i\in \{0,1\}$ let $E_{i,n}=T_n^{-1}(C_i)$. These events are measurable because each $C_i$ is closed by [F13] and Borel by [F12], while $T_n$ is measurable by [F5] and preimages of Borel sets are measurable by [F11]. By [F6], $P_{{\theta }_i}^{\otimes n}(E_{i,n})\to 1$ for each $i$.

For each $n$, step 2.1 gives one common sample law for both probabilities, and step 2.2 gives disjoint events, so their probabilities sum to at most one. Step 3.1 says both probabilities tend to one, so their sum tends to two, a contradiction. Therefore the estimator sequence cannot be consistent at both parameters.