Disintegration of a separable group representation over a commuting diagonal algebra

Replacing $\pi$ by the unitarily equivalent representation $U\pi U^{-1}$ and $A$ by $\mathcal{D}$ does not change any assertion, so assume $H=\mathcal{H}$ and $A=\mathcal{D}$; then $\pi (G)\subseteq {\mathcal{D}}^{\mathrm{\prime }}$, hence $\pi (a)\in {\mathcal{D}}^{\mathrm{\prime }}$ for every $a$ in the unitisation of $C^{\ast }(G)$ because ${\mathcal{D}}^{\mathrm{\prime }}$ is a weak-operator-closed algebra, and the extension of $\pi$ to a nondegenerate star-representation of $C^{\ast }(G)$ is given by [F3].

Choose a countable norm-dense $\mathbb{Q}(i)$-star-subalgebra $(a_k)$ of $C^{\ast }(G)$ by [F4]. For each $k$ the operator $\pi (a_k)\in {\mathcal{D}}^{\mathrm{\prime }}$ is decomposable, so by [F2] there are weakly measurable essentially bounded fields $x\mapsto T_k(x)$ with $\pi (a_k)={\int }_X^{\oplus }{T}_k(x)d\mu (x)$; replacing $T_k(x)$ by $0$ on the null set where $\mathrm{\parallel }{T}_k(x)\mathrm{\parallel }>\mathrm{\parallel }{a}_k\mathrm{\parallel }$ keeps the field measurable without changing the operator and gives $\mathrm{\parallel }{T}_k(x)\mathrm{\parallel }\le \mathrm{\parallel }{a}_k\mathrm{\parallel }$ on a conull set.

Let $N_0$ be a conull Borel set meeting the countably many conull sets on which the relations $T_{k+l}(x)=T_k(x)+T_l(x)$ (for rational scalars), $T_{kl}(x)=T_k(x)T_l(x)$ and $T_{k^{\ast }}(x)=T_k(x{)}^{\ast }$ hold; such a set exists because each relation holds almost everywhere by the uniqueness in [F2]. For $x\in N_0$ the assignment $a_k\mapsto T_k(x)$ is a contractive star-homomorphism of the dense subalgebra $(a_k)$ into $\mathcal{B}(H_x)$, so it extends uniquely to a contractive star-homomorphism ${\rho }_x:C^{\ast }(G)\to \mathcal{B}(H_x)$; for each fixed $a\in C^{\ast }(G)$ the function $x\mapsto {\rho }_x(a)$ is a weak-operator limit of the measurable fields $T_k$ along a sequence $a_k\to a$, hence is a weakly measurable field.

Use measurable Gram–Schmidt [F7] to obtain a countable orthonormal frame $f_j$, allowing zero vectors on finite-dimensional fibres. Let $(E_l)$ be a countable finite-measure Borel cover of $X$. The localized vectors ${\xi }_{lj}={\mathbf{1}}_{E_l}{f}_j$ are square-integrable, since $\mathrm{\parallel }{f}_j\mathrm{\parallel }\le 1$. Enumerate them as $({\xi }_m)$; their values span every fibre. For each $m$, global strong convergence $\pi (u_n)\to I$ gives $\int \mathrm{\parallel }({\rho }_x(u_n)-I){\xi }_m(x){\mathrm{\parallel }}^{2}d\mu \to 0$. Inductively choose increasing indices $n_k$ such that the sum of these integrals for $m\le k$ is less than $2^{-k}$. For fixed $m$ the sum over $k\ge m$ of the nonnegative error integrals is finite. By monotone convergence [F6], the pointwise sum of squared errors is finite almost everywhere, so the errors tend to zero there. Remove the countable union of null exceptions for all $m$. At every remaining $x$, ${\rho }_x(u_{n_k}){\xi }_m(x)\to {\xi }_m(x)$ for all $m$. These vectors span a dense fibre subspace, proving $\overline{{\rho }_x(C^{\ast }(G))H_x}=H_x$. Thus ${\rho }_x$ is nondegenerate almost everywhere.

On the conull set of step 4.1, apply the fibrewise correspondence [F3] to obtain strongly continuous unitary representations ${\pi }_x$ of $G$ on $H_x$ with ${\rho }_x(a)={\pi }_x(a)$ for every $a\in C^{\ast }(G)$; on the null complement define ${\pi }_x$ to be the trivial representation on $H_x$, making the field defined for every $x$ by an arbitrary choice on a null set.

Fix $g\in G$. The integrated operators ${\rho }_x(g{u}_n)={\pi }_x(g{u}_n)$ are measurable by step 3.1 and uniformly contractive. On every nondegenerate fibre, the support condition of [F5] and strong continuity of ${\pi }_x$ give ${\pi }_x(g{u}_n)\eta \to {\pi }_x(g)\eta$ for every $\eta \in H_x$: the norm of the difference is bounded by the supremum of $\mathrm{\parallel }{\pi }_x(gh)\eta -{\pi }_x(g)\eta \mathrm{\parallel }$ over $h$ in the shrinking support of $u_n$. Therefore fixed-g matrix coefficients of ${\pi }_x(g)$ are measurable, and its norm is at most one. For every square-integrable section $\xi$, the pointwise difference norm between ${\rho }_x(g{u}_n)\xi (x)$ and ${\pi }_x(g)\xi (x)$ is bounded by $2\mathrm{\parallel }\xi (x)\mathrm{\parallel }$ and tends to zero almost everywhere. Dominated convergence [F6] makes the induced operators converge strongly to ${\int }_X^{\oplus }{\pi }_x(g)d\mu$. On the other hand, those operators are $\pi (g{u}_n)=\pi (g)\pi (u_n)$ by step 3.1, and converge strongly to $\pi (g)$ by [F5]. Uniqueness of strong limits proves the required equality. No simultaneous exceptional set indexed by $G$ is needed: the representations themselves were constructed on one conull set in step 5.1, and each global fixed-g operator identity follows from this limit argument.

The field of generated von Neumann algebras is measurable: for almost every $x$, ${\pi }_x(G{)}^{\mathrm{\prime }\mathrm{\prime }}$ is generated by the operators ${\pi }_x(a_k)={\rho }_x(a_k)=T_k(x)$, which are weakly measurable fields bounded by $\mathrm{\parallel }{a}_k\mathrm{\parallel }$, so the sequence $(T_k)$ satisfies the defining condition of a measurable field of von Neumann algebras in [F1]. Combining this with steps 5.1 and 6.1 proves all the assertions, and in particular ${\pi }_x$ is nondegenerate for almost every $x$.