The posterior mean is the unique squared-loss Bayes action in a Hilbert target

The function $(y,h)\mapsto \mathrm{\parallel }h{\mathrm{\parallel }}_H^2$ is product-measurable because the norm is continuous, so kernel integration [F8] makes $s_2(y)={\int }_H\mathrm{\parallel }h{\mathrm{\parallel }}_H^2{\mathrm{\Pi }}_y(dh)$ measurable and $D_2=\{s_2<\mathrm{\infty }\}$ measurable. For each $j$, $X_j(h)=⟨h,e_j{⟩}_H$ is continuous by [F15] and $\mathrm{\parallel }{e}_j{\mathrm{\parallel }}_H=1$. The decomposition $h=X_j(h)e_j+(h-X_j(h)e_j)$ is orthogonal, so $X_j(h{)}^2\le \mathrm{\parallel }h{\mathrm{\parallel }}_H^2$; hence $2\mathrm{\mid }{X}_j(h)\mathrm{\mid }\le 1+\mathrm{\parallel }h{\mathrm{\parallel }}_H^2$. Thus each signed coordinate integral defining $m_j(y)$ is finite on $D_2$, and the signed-kernel clause of [F8] makes $m_j$ measurable there.

Fix $y\in D_2$. Under the probability space $(H,\mathcal{B}(H),{\mathrm{\Pi }}_y)$, each $X_j$ is square-integrable. The constant $m_j(y)=\int X_{j}d{\mathrm{\Pi }}_y$ satisfies the conditional-expectation identities for the trivial sigma-algebra by [F6], so [F7] makes it the projection of $X_j$ onto constants and gives ${\int }_H(X_j-m_j(y))d{\mathrm{\Pi }}_y=0$. For every real $b$, expand $X_j-b=(X_j-m_j(y))+(m_j(y)-b)$; the cross term integrates to zero, yielding ${\int }_H(X_j-b{)}^{2}d{\mathrm{\Pi }}_y={\int }_H(X_j-m_j(y){)}^{2}d{\mathrm{\Pi }}_y+(b-m_j(y){)}^2$. Taking $b=0$ gives $m_j(y{)}^2\le {\int }_H{X}_j^{2}d{\mathrm{\Pi }}_y$. Let $J_N$ be the first $N$ basis indices, taking all indices once $N$ reaches the finite endpoint when $J$ is finite. Summing over $J_N$ and applying Parseval [F5] pointwise, followed by monotone convergence [F9], yields ${\sum }_{j\in J}{m}_j(y{)}^2\le {\int }_H{\sum }_{j\in J}{X}_j(h{)}^2{\mathrm{\Pi }}_y(dh)={\int }_H\mathrm{\parallel }h{\mathrm{\parallel }}_H^2{\mathrm{\Pi }}_y(dh)<\mathrm{\infty }$. Orthonormality gives $\mathrm{\parallel }{\sum }_{j\in J_N}{m}_j(y)e_j{\mathrm{\parallel }}_H^2={\sum }_{j\in J_N}{m}_j(y{)}^2$; for $M>N$, the squared norm of the difference of the two partial sums is ${\sum }_{j\in J_M\setminus J_N}{m}_j(y{)}^2$, which tends to zero as $M,N\to \mathrm{\infty }$ because the series converges. Thus the partial sums are Cauchy, and completeness [F4] gives their norm limit $\mu (y)$. Since the sum in the statement is over those same initial segments, it is this norm limit. By continuity of the inner product [F15], $⟨\mu (y),e_j{⟩}_H=m_j(y)$ for every $j$.

Let $C$ be a countable dense subset of $H$ from [F3]. The balls with centers in $C$ and radii $1\mathrm{/}n$ form a countable base by [F14]: for any point in an open set, choose $n$ with $3\mathrm{/}(2n)$ smaller than the radius of a ball about that point contained in the open set, then choose a center in $C$ within $1\mathrm{/}(2n)$ of the point. The resulting radius-$1\mathrm{/}n$ ball contains the point and lies in the open set. For $c\in C$ and a basis ball radius $r=1\mathrm{/}n$, the finite sums $s_N(y)={\sum }_{j\in J_N}(m_j(y)-⟨c,e_j{⟩}_H{)}^2$ are measurable by step 1.1. By Parseval [F5] and step 2.1, $\mathrm{\parallel }\mu (y)-c{\mathrm{\parallel }}_H^2={\sup }_N{s}_N(y)$, where finite bases use the full sum once all coordinates are included. The preimage of the open ball is the countable union ${\bigcup }_{k\ge 2}{\bigcap }_N\{s_N<r^2(1-1\mathrm{/}k)\}$; this equals $\{{\sup }_N{s}_N<r^2\}$. Hence every basis-ball preimage is measurable, and the countable base makes $\mu$ measurable on $D_2$.

Fix $y\in D_2$ and $a\in H$. Write $a_j=⟨a,e_j{⟩}_H$. Applying the scalar identity established in step 2.1 with $b=a_j$ gives ${\int }_H(X_j-a_j{)}^{2}d{\mathrm{\Pi }}_y={\int }_H(X_j-m_j(y){)}^{2}d{\mathrm{\Pi }}_y+(a_j-m_j(y){)}^2$. Sum over the initial segments $J_N$ from step 2.1 and pass to their increasing limits with [F9]; Parseval [F5] identifies the three sums as ${\int }_H\mathrm{\parallel }h-a{\mathrm{\parallel }}_H^{2}d{\mathrm{\Pi }}_y$, ${\int }_H\mathrm{\parallel }h-\mu (y){\mathrm{\parallel }}_H^{2}d{\mathrm{\Pi }}_y$, and $\mathrm{\parallel }a-\mu (y){\mathrm{\parallel }}_H^2$, respectively. The first integral is finite because $\mathrm{\parallel }h-a{\mathrm{\parallel }}_H^2\le 2\mathrm{\parallel }h{\mathrm{\parallel }}_H^2+2\mathrm{\parallel }a{\mathrm{\parallel }}_H^2$; the same bound with $\mu (y)$ gives finite risk at the mean. The resulting identity proves that only $a=\mu (y)$ attains the minimum. By [F10] and step 3.1, $\mu$ is a measurable Bayes action rule on $D_2$.

Now suppose $\mathrm{\Pi }$ is a supplied RCD of the measurable random element $T:\mathrm{\Omega }\to H$ given $Y:\mathrm{\Omega }\to \mathsf{Y}$, with $E\mathrm{\parallel }T{\mathrm{\parallel }}_H^2<\mathrm{\infty }$. Applying [F12] to $h\mapsto \mathrm{\parallel }h{\mathrm{\parallel }}_H^2$ shows that $s_2(Y)$ has finite expectation $E\mathrm{\parallel }T{\mathrm{\parallel }}_H^2$; a nonnegative random variable with finite expectation is finite almost surely, so $P_Y(D_2)=1$. By [F15], $\mathrm{\mid }⟨T,e_j{⟩}_H\mathrm{\mid }\le \mathrm{\parallel }T{\mathrm{\parallel }}_H$, and $2\mathrm{\parallel }T{\mathrm{\parallel }}_H\le 1+\mathrm{\parallel }T{\mathrm{\parallel }}_H^2$, so every coordinate is integrable. For every $j$, [F12] now identifies $m_j(Y)$ as a version of $E[⟨T,e_j{⟩}_H\mid \sigma (Y)]$. Extend $\mu$ by zero outside $D_2$; because $D_2$ is measurable by step 1.1 and $\mu$ is measurable on $D_2$ by step 3.1, this piecewise extension is measurable, and step 4.1 at $a=0$ gives $\mathrm{\parallel }\mu (y){\mathrm{\parallel }}_H^2\le s_2(y)$ on $D_2$, so the extension evaluated at $Y$ is square-integrable. If another square-integrable $H$-valued random variable has each basis coordinate as a version of the corresponding scalar conditional expectation, [F13] makes each coordinate equal to $m_j(Y)$ outside a null set. There are at most countably many coordinates, so the union of these exceptional null sets is null; completeness of the basis [F4] then makes the two vectors equal there. The RCD's values on $P_Y$-null data sets remain version choices by [F11].