Statistical inverse experiments and white noise

For $N$ let $V_N=\mathrm{span}\{u_j:j\in J,j\le N\}$ and let $P_{N}g={\sum }_{j\in J,j\le N}⟨g,u_j{⟩}_H{u}_j$. Orthonormality gives $g-P_{N}g\perp V_N$. For any $v\in H$, expanding the inner product of $v=P_{N}v+(v-P_{N}v)$ gives $\mathrm{\parallel }v{\mathrm{\parallel }}^2=\mathrm{\parallel }{P}_{N}v{\mathrm{\parallel }}^2+\mathrm{\parallel }v-P_{N}v{\mathrm{\parallel }}^2$, so $\mathrm{\parallel }{P}_{N}v\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }$. Given $\eta >0$, density of the finite span in [F2] gives $w\in V_M$ with $\mathrm{\parallel }g-w\mathrm{\parallel }<\eta$; for $N\ge M$, $P_{N}w=w$, so $\mathrm{\parallel }g-P_{N}g\mathrm{\parallel }\le \mathrm{\parallel }g-w\mathrm{\parallel }+\mathrm{\parallel }{P}_N(w-g)\mathrm{\parallel }\le 2\mathrm{\parallel }g-w\mathrm{\parallel }<2\eta$. Thus $P_{N}g\to g$ in norm, with the finite-basis case eventually equal to $g$.

For distinct indices $j_1,\dots ,j_m$, [F1] gives independent standard-normal variables $Z(u_{j_1}),\dots ,Z(u_{j_m})$. Since $ϵ>0$, the affine map $z\mapsto {\mu }_{j_r}+ϵz$ and its inverse are continuous, so $Y_x(u_{j_r})$ and its noise coordinate generate the same sigma-algebra; independence therefore transfers. The affine-image definition of $N$ gives each law $N({\mu }_{j_r},{ϵ}^2)$.

Put $c_j=⟨g,u_j{⟩}_H$. The linearity clause in [F1] applied to the finite sum defining $P_{N}g$ gives $Z(P_{N}g)={\sum }_{j\in J,j\le N}{c}_{j}Z(u_j)$; linearity of the inner product gives $⟨Kx,P_{N}g{⟩}_H={\sum }_{j\in J,j\le N}{c}_j{\mu }_j$. Combining these identities yields $Y_x(P_{N}g)={\sum }_{j\in J,j\le N}{c}_j{Y}_x(u_j)$.

In the carrier case covered by [F4]–[F5], [F1] supplies independent coordinates with the common standard-normal density and [A1] supplies the countable-choice hypothesis of [F5]. Hence [F5] makes the noise coefficient sequence a Borel Gaussian random element of $H(\delta )$; [F4] puts each deterministic signal $Kx$ in the same space. Their sum $Kx+ϵW$ is therefore an $H(\delta )$-valued Borel random element, and every continuous linear functional of it is a translate and positive rescaling of the corresponding centered Gaussian functional.

In the infinite-dimensional function-space case, if $Kx+ϵW$ were an $H$-valued sample on an event of probability one, then $W=((Kx+ϵW)-Kx)\mathrm{/}ϵ$ would also be $H$-valued there, since $Kx\in H$ and $ϵ>0$. This contradicts [F5], so the sample is not $H$-valued.

With $r_N=g-P_{N}g$, linearity gives $Y_x(g)-Y_x(P_{N}g)=⟨Kx,r_N{⟩}_H+ϵZ(r_N)$. Centering and the covariance identity in [F1] give $\mathbb{E}Z(r_N)=0$ and $\mathbb{E}[Z(r_N{)}^2]=\mathrm{\parallel }{r}_N{\mathrm{\parallel }}_H^2$; expanding the square therefore shows that the squared mean-square error is $\mathrm{\mid }⟨Kx,r_N{⟩}_H{\mathrm{\mid }}^2+{ϵ}^2\mathrm{\parallel }{r}_N{\mathrm{\parallel }}_H^2$. By [F3] and step 1.1, both terms tend to zero, proving the asserted mean-square reconstruction. For any finite list of test vectors, the sum of their finitely many squared errors also tends to zero, so the coordinate reconstructions converge jointly in product mean square.

By step 1.1, for each $g$ and $k\ge 1$ the set of integers $N\ge k$ with $\mathrm{\parallel }g-P_{N}g{\mathrm{\parallel }}_H\le 2^{-k}$ is nonempty. Its least element $N_k(g)$ is uniquely defined, with no choice and no dependence on $x$, $Kx$ or the observations. For fixed $g,k$, $a_{g,k}$ is a finite linear combination of coordinate evaluations on ${\mathbb{R}}^J$, hence is measurable. By [F8], its limsup and liminf are measurable; their equality set, intersected with the set where their common value is finite, is the measurable real-convergence set $A_g$. Defining $T(z)(g)$ to equal that limit on $A_g$ and zero on its complement is measurable: for a Borel set $B\subseteq \mathbb{R}$ its preimage is $A_g\cap \{{\mathrm{lim\hspace{0.17em}sup}}_k{a}_{g,k}\in B\}$, together with $A_g^c$ if $0\in B$. Thus every evaluation of $T$ is measurable, which makes $T$ measurable into ${\mathcal{C}}_H$. Each coordinate of $S$ is a field evaluation, so $S$ is measurable into ${\mathcal{C}}_J$ as well.

Fix $x$ and $g$. By steps 1.3 and 2.1, the real random variable $D_k=a_{g,k}(S(Y_x))-Y_x(g)$ satisfies $\mathbb{E}\mathrm{\mid }{D}_k{\mathrm{\mid }}^2\le (\mathrm{\parallel }Kx{\mathrm{\parallel }}_H^2+{ϵ}^2)4^{-k}$. For every positive integer $r$, [F6] applied to $\mathrm{\mid }{D}_k{\mathrm{\mid }}^2$ at threshold $r^{-2}$ gives $\mathbb{P}\{\mathrm{\mid }{D}_k\mathrm{\mid }\ge 1\mathrm{/}r\}\le r^2(\mathrm{\parallel }Kx{\mathrm{\parallel }}_H^2+{ϵ}^2)4^{-k}$. The right-hand side is summable in $k$, so [F7] makes $\mathrm{\mid }{D}_k\mathrm{\mid }<1\mathrm{/}r$ eventually almost surely for each $r$. The union over $r\ge 1$ of these null exceptional events is null: disjointifying the union and using countable additivity bounds its probability by the sum of zeros. Off that union $D_k\to 0$, proving $T(S(Y_x))(g)=Y_x(g)$ almost surely. All limits here concern the explicitly specified representatives at a fixed $x,g$; the $L^2$ equalities used to compute the error hold almost surely and therefore give exactly the same second moments.

The coordinate law $Q_x=S_{\ast }{P}_x$ and the reconstructed law $T_{\ast }{Q}_x$ are probability measures by [F10]; they use only pushforwards of the given process, with no new product-existence assumption. For any finite list $g_1,\dots ,g_n$, intersect the finitely many probability-one events in step 3.1. On that intersection the reconstructed and original evaluation vectors agree, so $T_{\ast }{Q}_x$ and $P_x$ agree on every finite cylinder $\{y:(y(g_1),\dots ,y(g_n))\in B\}$ for Borel $B\subseteq {\mathbb{R}}^n$. These cylinders form a generating pi-system for ${\mathcal{C}}_H$, and both measures have total mass one, so [F9] gives $T_{\ast }{Q}_x=P_x$ on ${\mathcal{C}}_H$. This proves equality of laws without any uncountable intersection of probability-one events.

For every basis index $j$, $N_k(u_j)\ge k\ge j$ once $k\ge j$, so orthonormality gives $a_{u_j,k}(z)=z_j$ then, for every $z\in {\mathbb{R}}^J$. Consequently $T(z)(u_j)=z_j$ and $S\circ T={\mathrm{id}}_{{\mathbb{R}}^J}$ everywhere. If $J=\mathrm{\varnothing }$, completeness forces $H=\{0\}$, every sum defining $a_{0,k}$ is zero, and $T$ returns the zero field; linearity gives $Y_x(0)=0$ almost surely. If $J$ is nonempty and finite, $P_N=I$ after its last index, so the same construction reduces to the finite linear reconstruction and all preceding arguments apply. Thus the maps in the Statement establish both sampling-law identities in every allowed dimension. The least-index construction spends no choice principle; the only countable-choice use remains step 1.4.