Direct Integral Decomposition and Type I Groups

Fix $g\in G$ and set $T_x={\pi }_x(g)$. Its fundamental matrix coefficients are measurable by the fixed-$g$ hypothesis, so $T$ is weakly measurable by [F1]. On a nonzero fibre $\mathrm{\parallel }{T}_x\mathrm{\parallel }=1$, and on a zero fibre $\mathrm{\parallel }{T}_x\mathrm{\parallel }=0$; hence $\mathrm{\parallel }{T}_x\mathrm{\parallel }\le 1$ for every $x$. Thus $T$ is essentially bounded, and [F2] gives a bounded decomposable operator $P_g={\int }_X^{\oplus }{\pi }_x(g)d\mu (x)$ on the Hilbert space $\mathcal{H}$ of [F3].

For $g,h\in G$, [F2] and the pointwise identities [F4] give $P_g{P}_h={\int }_X^{\oplus }{\pi }_x(g){\pi }_x(h)d\mu (x)=P_{gh}$, $P_e=I_{\mathcal{H}}$, and $P_g^{\ast }={\int }_X^{\oplus }{\pi }_x(g{)}^{\ast }d\mu (x)=P_{g^{-1}}$. Therefore $P_g{P}_g^{\ast }=P_g^{\ast }{P}_g=I_{\mathcal{H}}$, so every $P_g$ is unitary and $g\mapsto P_g$ is a group homomorphism into $U(\mathcal{H})$. These identities use the stipulated pointwise group laws for every $x$, so no group-element-dependent conull sets are intersected. Strong continuity is not established by this definition.

If the field is changed only on a measurable $\mu$-null set, then for each fixed $g$ the corresponding operator fields agree off that set by [F6]. Their pointwise actions on every measurable section therefore define the same class in the quotient [F5], so every induced operator $P_g$ is unchanged.

For a discrete group $\mathrm{\Gamma }$, let ${\mathrm{\#}}_{\mathrm{\Gamma }}$ be counting measure. Every function on $\mathrm{\Gamma }$ is measurable and a ${\mathrm{\#}}_{\mathrm{\Gamma }}$-null set is empty. By [F5], ${\int }_{\mathrm{\Gamma }}\mathrm{\mid }f{\mathrm{\mid }}^{2}d{\mathrm{\#}}_{\mathrm{\Gamma }}={\sum }_{\gamma \in \mathrm{\Gamma }}\mathrm{\mid }f(\gamma ){\mathrm{\mid }}^2$, so the identity on functions identifies ${\mathrm{\ell }}^2(\mathrm{\Gamma },\mathbb{C})$ isometrically with $L^2(\mathrm{\Gamma },{\mathrm{\#}}_{\mathrm{\Gamma }};\mathbb{C})$; the pairings agree as well, since [F4] makes $f\bar{g}$ absolutely summable for $f,g\in {\mathrm{\ell }}^2$. Counting measure is Radon on the locally compact discrete space by [F5], so [F6] makes ${\mathrm{\ell }}^2(\mathrm{\Gamma },\mathbb{C})$ complete and hence a Hilbert space. Its standard vectors ${\delta }_{\gamma }$ form an orthonormal basis: given $f\in {\mathrm{\ell }}^2$ and $\epsilon >0$, the finite-subset supremum defining its square sum gives a finite $F\subseteq \mathrm{\Gamma }$ with ${\sum }_{\gamma \mathrm{\notin }F}\mathrm{\mid }f(\gamma ){\mathrm{\mid }}^2<{\epsilon }^2$, and truncation to $F$ approximates $f$ within $\epsilon$.

Put $S=\pi (G)$. Since $\pi (g{)}^{\ast }=\pi (g^{-1})$ by [F2], $S$ is self-adjoint. By [F1], $S^{\mathrm{\prime }}$ is a weak-operator-closed unital $\ast$-subalgebra, and $M=S^{\mathrm{\prime }\mathrm{\prime }}=(S^{\mathrm{\prime }}{)}^{\mathrm{\prime }}$ is also one; hence $M$ is a concrete von Neumann algebra. The commutant identity is algebraic: $S\subseteq S^{\mathrm{\prime }\mathrm{\prime }}$, so every operator in $(S^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\prime }}$ belongs to $S^{\mathrm{\prime }}$; conversely every $T\in S^{\mathrm{\prime }}$ commutes with every element of $S^{\mathrm{\prime }\mathrm{\prime }}=(S^{\mathrm{\prime }}{)}^{\mathrm{\prime }}$, hence $T\in (S^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\prime }}$. Therefore $\pi (G{)}^{\mathrm{\prime }}=(\pi (G{)}^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\prime }}$.

If $\pi$ is irreducible, [F3] gives $\pi (G{)}^{\mathrm{\prime }}=\mathbb{C}{I}_H$. Step 1.2 identifies this with $M^{\mathrm{\prime }}$, so $Z(M)=M\cap M^{\mathrm{\prime }}=\mathbb{C}{I}_H$ because $I_H\in M$. Thus $\pi$ is factorial.

Let $\mathrm{\Gamma }$ be nontrivial and ICC, and let $\lambda (g)f(h)=f(g^{-1}h)$ and $\rho (g)f(h)=f(hg)$ on the Hilbert space ${\mathrm{\ell }}^2(\mathrm{\Gamma })$ of step 1.1. The left translations form a unitary representation, and each right translation is unitary; direct substitution shows $\lambda (a)\rho (g)=\rho (g)\lambda (a)$. Thus $\rho (g)\in \lambda (\mathrm{\Gamma }{)}^{\mathrm{\prime }}=(\lambda (\mathrm{\Gamma }{)}^{\mathrm{\prime }\mathrm{\prime }}{)}^{\mathrm{\prime }}$ by step 1.2. Put $M_{\mathrm{\Gamma }}=\lambda (\mathrm{\Gamma }{)}^{\mathrm{\prime }\mathrm{\prime }}$. For $z\in Z(M_{\mathrm{\Gamma }})$ let $\xi =z{\delta }_e$. Since $z$ commutes with both $\lambda (g)$ and $\rho (g^{-1})$, and $\rho (g^{-1}){\delta }_e={\delta }_g$, we have $\lambda (g)\xi =z{\delta }_g=z\rho (g^{-1}){\delta }_e=\rho (g^{-1})\xi$. In coordinates this is $\xi (g^{-1}h)=\xi (h{g}^{-1})$; setting $h=gk$ shows $\xi (k)=\xi (gk{g}^{-1})$. Hence $\xi$ is constant on conjugacy classes. Every nonidentity conjugacy class is infinite, so square summability forces $\xi$ to vanish off $e$: $\xi =c{\delta }_e$. For every $h\in \mathrm{\Gamma }$, $z{\delta }_h=z\lambda (h){\delta }_e=\lambda (h)z{\delta }_e=c{\delta }_h$. The standard vectors span densely by step 1.1, hence $z=cI$ and $M_{\mathrm{\Gamma }}$ is a factor.

Choose $g\mathrm{\ne }e$ in $\mathrm{\Gamma }$. The right translation $\rho (g)$ is a bounded self-intertwiner of $\lambda$ but is not scalar, since $\rho (g){\delta }_e={\delta }_{g^{-1}}\mathrm{\ne }{\delta }_e$. If $\lambda$ were irreducible, [F3] would force this self-intertwiner to be scalar. Therefore $\lambda$ is not irreducible. To see the example class is nonempty, let $\mathrm{\Gamma }$ be the finitary symmetric group on $\mathbb{N}$. For any nonidentity finite permutation $\sigma$, let $a$ be the least moved point; for each distinct $b$ outside the finite support, conjugation by the transposition $(ab)$ gives support $(\mathrm{supp}\sigma \setminus \{a\})\cup \{b\}$. These supports are distinct, so the conjugacy class is infinite. Thus this group is nontrivial ICC and supplies the promised example.

Let $\mathcal{B}$ be an at most countable basis for $G$. The relatively compact open sets form a basis by [F2], so the family of all relatively compact open sets covers $G$. By Lindelofness and AC's implication of Countable Choice in [F6], it has an at most countable subcover; enumerate that nonempty subcover as $(V_n{)}_{n\in \mathbb{N}}$, repeating terms if it is finite.

Put $K_n={\bigcup }_{j\le n}\overline{V_j}$. Each closure is compact; an ambient open cover of $K_n$ has a finite subcover on each of the finitely many closures by [F3], and their finite union covers $K_n$. Thus $K_n$ is compact, and it is closed and Borel because $G$ is Hausdorff. The sequence increases and covers $G$. If $f\in C_c(G)$, compactness of $\mathrm{supp}f$ gives a finite subcover from $(V_n)$; taking the largest index in that subcover (or $n=0$ for empty support) shows $\mathrm{supp}f\subseteq K_n$ for some $n$. Each $K_n$ has finite Haar measure by [F1].

The family $\mathcal{G}=\mathcal{B}\cup \{K_n:n\in \mathbb{N}\}$ is countable by [F4]. Set ${\mathcal{A}}_0$ specifically to the algebra of finite Boolean combinations of $\mathcal{G}$. As in the countable-algebra supplier proof in [F4], enumerate $\mathcal{G}$ and let ${\mathcal{C}}_m$ be the finite algebra generated by its first $m$ terms. Then ${\mathcal{A}}_0={\bigcup }_m{\mathcal{C}}_m$: every finite Boolean combination uses some finite prefix. These algebras increase, so their union is an algebra, and [F4] makes it countable. Its generators are Borel, and finite Boolean operations preserve Borel sets, so every member of ${\mathcal{A}}_0$ is Borel. Every open set is a union of basis members, and because $\mathcal{B}$ is countable this is a countable union; therefore $\sigma ({\mathcal{A}}_0)$ contains every open set. Conversely ${\mathcal{A}}_0$ consists of Borel sets, so minimality in [F8] gives $\sigma ({\mathcal{A}}_0)=\mathcal{B}(G)$. Each $K_n\in {\mathcal{A}}_0$ and has finite measure by step 2.1.

Fix $f\in C_c(G)$ and a target $\eta >0$, and choose $n$ with $\mathrm{supp}f\subseteq K_n$ by step 2.1. If $\mu (K_n)=0$, then $f$ is zero as an $L^1$ class because it vanishes outside the null set $K_n$; the zero function is in the required span since $\mathrm{\varnothing }\in {\mathcal{A}}_0$. Otherwise set $\delta =\eta \mathrm{/}(4\mu (K_n))$. Let ${\mathcal{U}}_{\delta }$ be the family of all $U\in \mathcal{B}$ for which there exists $x\in K_n$ such that $\mathrm{\mid }f(y)-f(x)\mathrm{\mid }<\delta$ for every $y\in U$. Continuity and the basis property show that ${\mathcal{U}}_{\delta }$ covers $K_n$; compactness gives a finite subcover $U_1,\dots ,U_m$. For each $i$, choose a witness $x_i\in K_n$ for its defining property, and set $E_1=K_n\cap U_1$ and $E_i=(K_n\cap U_i)\setminus {\bigcup }_{j<i}{U}_j$ for $i>1$. These sets partition $K_n$, belong to ${\mathcal{A}}_0$, and have finite measure by [F7]. For each nonempty $E_i$, choose $y_i\in E_i$ and $q_i\in \mathbb{Q}(i)$ with $\mathrm{\mid }{q}_i-f(y_i)\mathrm{\mid }<\delta$; these are finitely many choices, justified by [F6] and density in [F5]. Then $g={\sum }_{i:E_i\mathrm{\ne }\mathrm{\varnothing }}{q}_i{\mathbf{1}}_{E_i}$ lies in the required span and in $L^1$, since it is measurable and bounded with support in the finite-measure set $K_n$. For $y\in E_i$, $\mathrm{\mid }f(y)-f(y_i)\mathrm{\mid }<2\delta$, hence $\mathrm{\mid }f(y)-g(y)\mathrm{\mid }<3\delta$; outside $K_n$ both functions vanish. Thus $\mathrm{\mid }f-g\mathrm{\mid }\le 3\delta {\mathbf{1}}_{K_n}$ and [F7] gives $\mathrm{\parallel }f-g{\mathrm{\parallel }}_1\le 3\delta \mu (K_n)=3\eta \mathrm{/}4<\eta$.

Let $\mathcal{E}=\{A\in {\mathcal{A}}_0:\mu (A)<\mathrm{\infty }\}$ and let $D_0$ consist of all finite $\mathbb{Q}(i)$-linear combinations of ${\mathbf{1}}_A$ with $A\in \mathcal{E}$. The family $\mathcal{E}$ is countable as a subset of ${\mathcal{A}}_0$; the alphabet $C=\mathbb{Q}(i)\times \mathcal{E}$ is at most countable by [F4,F5]. Every finite power $C^m$, including the one-point $C^0$, is at most countable by [F4]. AC gives Countable Choice by [F6], so [F4] makes ${\bigcup }_{m\in \mathbb{N}}{C}^m$, the set of finite lists of coefficient/set pairs, at most countable. Its image under the finite-sum map is $D_0$, so $D_0$ is countable. Each generator indicator is integrable because its set has finite measure, and finite linear combinations remain in $L^1$. For $h\in L^1(G)$ and $ϵ>0$, choose $f\in C_c(G)$ with $\mathrm{\parallel }h-f{\mathrm{\parallel }}_1<ϵ\mathrm{/}2$ by [F1], then choose $d\in D_0$ with $\mathrm{\parallel }f-d{\mathrm{\parallel }}_1<ϵ\mathrm{/}2$ by step 4.1. The triangle inequality gives $\mathrm{\parallel }h-d{\mathrm{\parallel }}_1<ϵ$, so $D_0$ is dense in $L^1(G)$.

The set $D_0\times {\mathbb{N}}_{>0}$ is countable by [F4]. For each $(d,k)$ in it, density of $C_c(G)$ in $L^1(G)$ gives a nonempty set of $c\in C_c(G)$ with $\mathrm{\parallel }c-d{\mathrm{\parallel }}_1<1\mathrm{/}k$. AC's implication of Countable Choice [F6] selects one such $c_{d,k}$ for every pair. The resulting set of functions is countable; for any $h\in L^1(G)$ and $ϵ>0$, choose $d\in D_0$ with $\mathrm{\parallel }h-d{\mathrm{\parallel }}_1<ϵ\mathrm{/}2$ and then $k$ with $1\mathrm{/}k<ϵ\mathrm{/}2$. It follows that $\mathrm{\parallel }h-c_{d,k}{\mathrm{\parallel }}_1<ϵ$, so their image in $L^1(G)$ is a countable dense subset contained in the image of $C_c(G)$.

Step 5.1 proves separability of $L^1(G)$, and [F1] gives its completeness, so it is a separable Banach space. Steps 3.1 and 5.1 give the asserted generating Borel algebra and dense $\mathbb{Q}(i)$-linear span, while step 6.1 gives the countable dense subset from $C_c(G)$.

By [F1], $\mathcal{H}$ is a Hilbert space. For any weakly measurable essentially bounded operator field $(T_x)$ with $T_x\in M_x$ almost everywhere, [F4] supplies a bounded operator ${\int }_X^{\oplus }{T}_{x}d\mu (x)\in \mathcal{B}(\mathcal{H})$. Thus the displayed direct integral is a well-defined subset of $\mathcal{B}(\mathcal{H})$; this definition does not assert that the set is weak-operator closed.

Fix $f\in L^{\mathrm{\infty }}(X,\mu )$ and choose a measurable representative. By [F5], some finite $C$ and measurable null set $N$ satisfy $\mathrm{\mid }f(x)\mathrm{\mid }\le C$ for every $x\mathrm{\notin }N$. Define $T_x:=f(x)I_{H_x}$. Its fundamental matrix coefficients are $⟨T_x{e}_n(x),e_m(x)⟩=f(x)⟨e_n(x),e_m(x)⟩$, measurable by [F6, F7]; its operator norm is at most $\mathrm{\mid }f(x)\mathrm{\mid }$, including when $H_x=\{0\}$, so it is essentially bounded. Since every $M_x$ is unital, $f(x)I_{H_x}\in M_x$ for every $x$; at a zero fibre this is $0\in \{0\}$. Hence this field satisfies the direct-integral membership conditions.

By [F4], the field of step 1.2 induces a decomposable operator acting on square-integrable classes by $[\xi ]\mapsto [x\mapsto f(x)\xi (x)]$, which is exactly $M_f$. If $f$ is changed on a measurable null set, [F4] gives the same induced operator, so $M_f$ depends only on its $L^{\mathrm{\infty }}$ class. Pointwise action gives $M_f+M_g=M_{f+g}$, $c{M}_f=M_{cf}$, and $M_f{M}_g=M_{fg}$; the first-variable-linear integral pairing gives $⟨M_f[\xi ],[\eta ]⟩=⟨[\xi ],M_{\bar{f}}[\eta ]⟩$, so $M_f^{\ast }=M_{\bar{f}}$. Since $1\in L^{\mathrm{\infty }}$ and $M_1=I_{\mathcal{H}}$ (including $I_{\mathcal{H}}=0$ when $\mathcal{H}=\{0\}$), [F8] makes $\mathcal{D}$ a unital $\ast$-algebra. Each $M_f$ belongs to the displayed direct-integral set by step 1.2, proving $\mathcal{D}\subseteq {\int }_X^{\oplus }{M}_{x}d\mu (x)$.

For any sequence $(u_n)$ of measurable sections, set $v_n=u_n-{\sum }_{j<n}⟨u_n,f_j⟩f_j$ and $f_n=v_n\mathrm{/}\mathrm{\parallel }{v}_n\mathrm{\parallel }$ when $\mathrm{\parallel }{v}_n\mathrm{\parallel }>0$, with $f_n=0$ otherwise. Inductively, measurable-section closure [F2] makes each finite coefficient, sum, residual, and residual norm measurable; the reciprocal on $(0,\mathrm{\infty })$ extended by zero at zero is Borel, so each $f_n$ is measurable. If the earlier nonzero $f_j$ are orthonormal, then for each nonzero $f_{\mathrm{\ell }}$ with $\mathrm{\ell }<n$, $⟨v_n,f_{\mathrm{\ell }}⟩=⟨u_n,f_{\mathrm{\ell }}⟩-{\sum }_{j<n}⟨u_n,f_j⟩⟨f_j,f_{\mathrm{\ell }}⟩=0$; normalizing a nonzero residual preserves orthogonality. Also $u_n$ lies in the span of $f_0,\dots ,f_n$, and induction gives equality of the spans of the terms with indices $0,\dots ,n$ of $(u_n)$ and $(f_n)$. Thus the nonzero $f_n$ form an orthonormal family whose closed span equals that of $(u_n)$. Apply this recursion to the fundamental family $(e_n)$ to obtain a complete system in each $H_x$; [F2] also gives Borel $x\mapsto ⟨\xi (x),f_n(x)⟩$ for every measurable section $\xi$.

Let $a_n(x)={\mathbf{1}}_{\{\mathrm{\parallel }{f}_n(x)\mathrm{\parallel }=1\}}$, set $N_{-1}(x)=0$, and for $m\ge 0$ put $N_m(x)={\sum }_{0\le n\le m}{a}_n(x)$. These are Borel by [F2,F7]. For finite $p\ge 1$, $X_p=({\bigcap }_m\{N_m\le p\})\cap ({\bigcup }_m\{N_m=p\})$; also $X_{\mathrm{\infty }}={\bigcap }_{q\ge 1}{\bigcup }_m\{N_m\ge q\}$ and $X_0={\bigcap }_m\{N_m=0\}$. Sigma-algebra closure makes these sets Borel. Since each nonzero $f_n(x)$ has norm one and their family is complete, $N_m(x)$ counts the active vectors at indices $0,\dots ,m$; the formulas therefore give exactly the finite, infinite, and zero dimensions.

Apply the recursion of step 1.1 to $({\xi }_k)$, obtaining measurable $h_n$ whose nonzero values form an orthonormal basis of $S_x=\stackrel{}{\mathrm{span}}\{{\xi }_k(x):k\in \mathbb{N}\}$. Each $S_x$ is a closed Hilbert subspace of $H_x$, and the Borel Gram coefficients and dense span of $(h_n)$ make $(S_x,h_n(x))$ a measurable closed Hilbert subfield by [F1,F2]. Every $S$-measurable section $\zeta$ is $H$-measurable: its finite expansions ${\sum }_{n\le N}⟨\zeta ,h_n⟩h_n$ are measurable $H$-sections and converge pointwise in norm to $\zeta$ by [F4,F5], so [F2] applies. Conversely, every $H$-measurable section taking values in $S_x$ is $S$-measurable because its pairings with the measurable $h_n$ are Borel by [F2]. Thus inclusion induces an isometric embedding ${\mathcal{H}}_S\hookrightarrow {\mathcal{H}}_H$. The direct-integral Hilbert theorem [F3] makes its domain complete, so its image is closed.

On $X_p$, put $J_p=\{1,\dots ,p\}$ for finite $p$ and $J_{\mathrm{\infty }}={\mathbb{N}}_{>0}$. Enumerate the active frame indices increasingly: for $j\in J_p$, let ${\nu }_j(x)$ be the $j$th $n\ge 0$ with $a_n(x)=1$, and put $g_j(x)=f_{{\nu }_j(x)}(x)$. Using the convention $N_{-1}=0$, the fibers are $\{{\nu }_j=n\}=X_p\cap \{N_{n-1}=j-1\}\cap \{a_n=1\}$ for $n\ge 0$; hence each piece is Borel and the sections $g_j$ are measurable. Their values form an orthonormal basis of $H_x$. For a fixed separable $K_p$ of dimension $p$, take an at most countable dense set, enumerate it using [F5], and apply the dense-sequence Gram--Schmidt theorem to obtain an orthonormal basis $(b_j{)}_{j\in J_p}$. The map $g_j(x)\mapsto b_j$ on finite linear combinations is well-defined and isometric because both families are orthonormal. For any $v\in H_x$, choose finite combinations $v_n$ converging to $v$; their images are Cauchy, so completeness of $K_p$ defines $U_{x}v:={\lim }_n{U}_x{v}_n$, independently of the approximating sequence and preserving linearity and norm. If $U_x{v}_n\to w$ for a sequence in the range, isometry makes $(v_n)$ Cauchy; completeness of $H_x$ gives $v_n\to v$, whence $w=U_{x}v$ and the range is closed. It contains the dense span of $(b_j)$, so $U_x$ is onto. Parseval [F5] gives $⟨U_x\xi (x),b_j⟩=⟨\xi (x),g_j(x)⟩$ and convergence of the corresponding partial expansions.

Let $L$ be the closed span in ${\mathcal{H}}_H$ of all $f{\mathbf{1}}_E{\xi }_k$, where $E=D_j\cap \{x:\mathrm{\parallel }{\xi }_k(x)\mathrm{\parallel }\le r\}$, $(D_j)$ ranges over a countable finite-measure Borel cover, $r\in \mathbb{R}$ with $r\ge 1$, and $f$ ranges over bounded Borel scalar functions supported in $E$. It is enough to use integer radii: for any real $r\ge 1$, [F12] gives an integer $R\ge r$, so $E_r\subseteq E_R$; since $f$ is supported in $E_r$, $f{\mathbf{1}}_{E_r}{\xi }_k=f{\mathbf{1}}_{E_R}{\xi }_k$. Thus the real-radius and integer-radius generating families coincide. Each generator is an $S$-measurable section by step 2.2 and is square-integrable because $\mathrm{\parallel }f{\mathbf{1}}_E{\xi }_k\mathrm{\parallel }\le r\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\mathbf{1}}_{D_j}$; hence $L\subseteq {\mathcal{H}}_S$. Let $\eta \in L^{\perp }$ and fix $k,j$ and an integer $r\ge 1$. The pairing $h(x)=⟨{\xi }_k(x),\eta (x)⟩$ is Borel by [F2], and ${\mathbf{1}}_{E}h$ is integrable: ${\mathbf{1}}_E{\xi }_k$ and ${\mathbf{1}}_E\eta$ are square-integrable, so [F11] supplies the direct-integral Cauchy--Schwarz estimate. Define $f={\mathbf{1}}_E\bar{h}\mathrm{/}\mathrm{\mid }h\mathrm{\mid }$ where $h\mathrm{\ne }0$, and $f=0$ where $h=0$. This is bounded Borel and supported in $E$ by [F7]. Since the inner product is linear in its first variable, $0=⟨[f{\xi }_k],[\eta ]⟩={\int }_E\mathrm{\mid }h\mathrm{\mid }d\mu$, so [F6] gives that the Borel set $N_{k,j,r}:=E\cap \{x:h(x)\mathrm{\ne }0\}$ is null. For each $k$, the sets $E$ with $j$ varying and integer $r\ge 1$ cover $X$: the $D_j$ cover $X$, and every finite norm is bounded by some integer. There are countably many triples $(k,j,r)$ by iterating the pairing in [F9], so their null sets have a null union by [F6]. Off that union, $\eta (x)\perp {\xi }_k(x)$ for every $k$, hence $\eta (x)\perp S_x$. Thus $\eta \perp {\mathcal{H}}_S$, so $L^{\perp }\subseteq {\mathcal{H}}_S^{\perp }$. Since $L\subseteq {\mathcal{H}}_S$ and both are closed, [F10] yields ${\mathcal{H}}_S={\mathcal{H}}_S^{\perp \perp }\subseteq L^{\perp \perp }=L$. Therefore $L={\mathcal{H}}_S$.

Let $D\subseteq K_p$ be a countable dense set and enumerate it, and enumerate ${\mathbb{Q}}_{>0}$. The balls $B(d,q)$ for $d\in D$ and $q\in {\mathbb{Q}}_{>0}$ form a countable base: given $x\in B(y,\epsilon )$, put $\delta =(\epsilon -\mathrm{\parallel }x-y\mathrm{\parallel })\mathrm{/}3>0$ and choose $d\in D$ with $\mathrm{\parallel }x-d\mathrm{\parallel }<\delta$. Then $\mathrm{\parallel }d-y\mathrm{\parallel }\le \mathrm{\parallel }d-x\mathrm{\parallel }+\mathrm{\parallel }x-y\mathrm{\parallel }<\epsilon -2\delta$, so $\mathrm{\parallel }x-d\mathrm{\parallel }<\epsilon -\mathrm{\parallel }d-y\mathrm{\parallel }$. Choose rational $q$ strictly between these two bounds. It follows that $x\in B(d,q)\subseteq B(y,\epsilon )$. For the measurable section $\xi$, write $c_j(x)=⟨\xi (x),g_j(x)⟩$ for $j\in J_p$. If $\xi$ is measurable, each $c_j$ is Borel by [F2]; for every fixed $y\in K_p$, Parseval gives $\mathrm{\parallel }{U}_x\xi (x)-y{\mathrm{\parallel }}^2={\lim }_{N\to \mathrm{\infty }}{\sum }_{\begin{array}{c}j\in J_p\\ j\le N\end{array}}\mathrm{\mid }{c}_j(x)-⟨y,b_j⟩{\mathrm{\mid }}^2,$ a Borel function by [F7]. Hence inverse images of the countable basic balls are Borel, proving $x\mapsto U_x\xi (x)$ Borel. Conversely, if this map is Borel, continuity of its coordinate functionals follows from Cauchy--Schwarz [F10] and makes each $c_j$ Borel; the partial sections ${\sum }_{\begin{array}{c}j\in J_p\\ j\le N\end{array}}{c}_j{g}_j$ are measurable and converge pointwise in norm to $\xi$ by [F4,F5], so [F2] makes $\xi$ measurable. This proves both directions of the section criterion.