Weakly convergent finite-step output laws need not be exact

Choose $N=1$, $t_0=0$, $t_1=1$, state space $\mathbb{R}$, and observation space $(\mathsf{Y},\mathcal{Y})=(\{\ast \},2^{\{\ast \}})$ with $y=\ast$. Take the unique forward map $\mathcal{A}:\mathbb{R}\to \mathsf{Y}$, likelihood ${\mathrm{\ell }}_y\equiv 1$ relative to counting measure, terminal law ${\nu }_1={\delta }_0$ with deterministic initialization at $0$, zero score field, a singleton correction space with constant correction, the Borel constant kernel $Q_m(B\mid x,y,a)={\mu }_m(B)$, and no projection. The kernel is a probability kernel because ${\mu }_m$ has total mass one. If ${\xi }_1,\dots ,{\xi }_m$ are independent fair coin flips, the sampler that returns $1$ exactly when every ${\xi }_i=1$ and returns $0$ otherwise has law ${\mu }_m$ and uses at most $m$ flips. Thus [F2] permits the one-step instance and its returned state has law ${\mu }_m$.

For every bounded continuous $f:\mathbb{R}\to \mathbb{R}$, ${\displaystyle \int fd{\mu }_m-\int fd\pi =2^{-m}(f(1)-f(0))⟶0}$. Thus the defining test integrals converge, so ${\mu }_m\Rightarrow \pi$ by [F1].

For each finite integer $m\ge 1$, ${\mu }_m(\{1\})=2^{-m}>0=\pi (\{1\})$, hence ${\mu }_m\mathrm{\ne }\pi$. In particular this holds at the first index $m=1$ as well as every later finite index.

The output laws therefore converge weakly to the target but are not equal to it at any finite index, so this convergence statement supplies no finite-index exactness claim.