Finite priors approximate Gaussian-shift Bayes values

If dimension zero is admitted with the singleton empty-tuple Gaussian convention, there is only one parameter, observation and action. Every prior is the point mass at that parameter and is already finitely supported; every rule has risk $g(0)$. Taking ${\pi }_f=\pi$ gives zero discrepancy. No positive-dimension density, sphere or compactness theorem is needed. Henceforth let $d\ge 1$.

Write $A=S^{1\mathrm{/}2}$. By [F1] and [F10], $Q_h$ is the law of $h+AG$, and repeated scalar integration gives $G$ density $(2\pi {)}^{-d\mathrm{/}2}{e}^{-\mathrm{\parallel }z{\mathrm{\parallel }}_2^2\mathrm{/}2}$. Applying [F11] to $z\mapsto h+Az$ gives the density $q_h(y)=(2\pi {)}^{-d\mathrm{/}2}\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}\exp [-(y-h{)}^T{S}^{-1}(y-h)\mathrm{/}2]$. The map $h\mapsto Q_h$ is continuous in total variation: when $h_j\to h$, these densities converge pointwise. The matrix $S^{-1}$ is positive definite because $x^T{S}^{-1}x=(S^{-1}x{)}^{T}S(S^{-1}x)>0$ for $x\mathrm{\ne }0$. By [F9], $y^T{S}^{-1}y\ge c\mathrm{\parallel }y{\mathrm{\parallel }}_2^2$ for some $c>0$. On a fixed bounded set of shifts, the density formula and $\mathrm{\parallel }y-h{\mathrm{\parallel }}_2^2\ge \mathrm{\parallel }y{\mathrm{\parallel }}_2^2\mathrm{/}2-\mathrm{\parallel }h{\mathrm{\parallel }}_2^2$ bound all these densities by $C\exp (-c^{\mathrm{\prime }}\mathrm{\parallel }y{\mathrm{\parallel }}_2^2)$ for constants $C,c^{\mathrm{\prime }}>0$. This envelope is integrable by repeated scalar Gaussian integration [F4, F10, F11], so [F2] gives $L^1$ convergence; for every event, its probability difference is at most that $L^1$ distance. On a compact ball this continuity is uniform: otherwise there are $h_j,h_j^{\mathrm{\prime }}$ in the ball with $\mathrm{\parallel }{h}_j-h_j^{\mathrm{\prime }}{\mathrm{\parallel }}_2\to 0$ but TV distances bounded below. Compactness gives a subsequence with both converging to the same h; pointwise continuity and the event-supremum triangle inequality from [F4] contradict that lower bound.

For a randomized rule ${\delta }_y(da)$, write $R_{\delta }(h)=\int \int g(a-h){\delta }_y(da)Q_h(dy)$. Change the loss argument first and the sampling law second. [F4] yields $\mathrm{\mid }{R}_{\delta }(h)-R_{\delta }(h^{\mathrm{\prime }})\mathrm{\mid }\le B\mathrm{\parallel }h-h^{\mathrm{\prime }}{\mathrm{\parallel }}_2+M\mathrm{\parallel }{Q}_h-Q_{h^{\mathrm{\prime }}}{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$, independently of the rule. Each risk is between zero and $M$. The inner kernel integral is jointly measurable in $(h,y)$ by [F8]; multiplying by the jointly measurable Gaussian density and integrating in y makes $R_{\delta }(h)$ measurable.

The balls $B_r=\{h:\mathrm{\parallel }h{\mathrm{\parallel }}_2\le r\}$ increase to ${\mathbb{R}}^d$, so [F6] lets us choose a closed ball with discarded prior mass less than $\tau$. Cover this compact ball by open balls of radius $\eta \mathrm{/}2$; [F5] gives a finite subcover. Intersect the first ball with the closed ball, then successively remove the previous covering sets from each later intersection. This makes a finite Borel partition whose cells have diameter at most $\eta$. Assign one point to each nonempty cell and send points outside the closed ball to zero. The resulting finite-valued map q is Borel, and ${\pi }_f=\pi \circ q^{-1}$. On the ball, step 3.1 bounds $\mathrm{\mid }{R}_{\delta }(h)-R_{\delta }(q(h))\mathrm{\mid }$ by $B\eta +M\omega (\eta )$, where the compact-ball TV modulus satisfies $\omega (\eta )\to 0$ by step 2.1. Outside, use $M$. Thus prior risks differ uniformly over rules by at most $B\eta +M\omega (\eta )+M\tau$. Choose the ball first for small tail, then the cell diameter small on that fixed ball. This makes the bound less than $ϵ$; taking infima over the same rules preserves it.