Non-type-I groups have non-smooth irreducible disintegration

Write $A=C^{\ast }(G)$ and $q:L^1(G)\to A$. For any nondegenerate $\rho$ and its corresponding group representation $U$, [F2] gives $U(g)={\mathrm{s-lim}}_n\rho (q(L_g{u}_n))$, because $U(g)\rho (q(u_n))=\rho (q(L_g{u}_n))$ and $\rho (q(u_n))\to I$. Conversely, every $\rho (q(f))$ is the integrated operator of $U$ and hence lies in $U(G{)}^{\mathrm{\prime }\mathrm{\prime }}$: every operator commuting with every $U(g)$ commutes with the integrated operators, and density of $q(L^1(G))$ gives commutation with all $\rho (A)$. The displayed strong limits give the reverse inclusion. Thus $\rho (A{)}^{\mathrm{\prime }\mathrm{\prime }}=U(G{)}^{\mathrm{\prime }\mathrm{\prime }}$ and their commutants agree. The same limits show that an algebra intertwiner intertwines every $U(g)$; a group intertwiner intertwines the integrated operators and, by density, all $\rho (A)$. These statements apply to bounded intertwiners between different carriers as well.

If $A$ were type I, every nondegenerate factor representation of $A$ would have a type-I generated algebra. Step 1.1 would then make every separable factor representation of $G$ type I, contradicting the hypothesis and [F6]. Hence $A$ is separable and non-type-I. Equivalently its GCR condition fails, so [F6] also gives failure of countable separation of the dual. This nonsmoothness alone is not used to infer the witness.

Apply only the cited fact [F7] with $n=2$. Choose disjoint standard-Borel sets $X,Y$ of irreducible classes and measurable fields ${\rho }_x,{\tau }_y$ representing their respective classes, with nonzero spectral measures $\mu ,\nu$, so $R_X={\int }_X^{\oplus }{\rho }_{x}d\mu$ and $R_Y={\int }_Y^{\oplus }{\tau }_{y}d\nu$ are equivalent. Use the sigma-finite spectral-measure models of this separable construction, replacing a measure by an equivalent finite one if necessary: for disjoint finite-measure exhaustion sets $E_k$, the strictly positive density $h={\sum }_k{2}^{-k}(1+\mu (E_k){)}^{-1}{\mathbf{1}}_{E_k}$ has finite nonzero integral, and multiplication by $h^{-1\mathrm{/}2}$ carries $L^2(\mu )$ unitarily onto $L^2(h\mu )$ and commutes with all fibre operators. The same applies to $\nu$. Thus this harmless change preserves both the fields' disjoint class labels and the equivalent representations. Zero or exceptional fibres are removed on Borel null sets once, so the retained fields represent exactly their stated irreducible classes at every point. The only existence input in this step is [F7], not a locally asserted hyperfinite or Glimm construction.

For each retained $x$, [F1] gives a strongly continuous irreducible group representation ${\pi }_x$ corresponding to ${\rho }_x$, and for each retained $y$ it gives ${\sigma }_y$ corresponding to ${\tau }_y$. For every fixed $g$, step 1.1 yields ${\pi }_x(g)={\mathrm{s-lim}}_n{\rho }_x(q(L_g{u}_n))$. Each term has measurable fundamental coefficients: the original algebra field is measurable on a countable dense algebra, and contractivity extends this to any fixed element of $A$ by norm approximation. Coefficient limits therefore prove measurability for this fixed $g$. The identical argument applies on $Y$. Group laws and strong continuity hold for every $g$ at each point because the full correspondence was applied separately to each genuine nondegenerate fibre representation; no intersection of uncountably many group-dependent conull sets is taken.

The algebra field integrals are nondegenerate. Indeed, $\mathrm{\parallel }{\rho }_x(q(u_n))\mathrm{\parallel }\le 1$ and ${\rho }_x(q(u_n))\xi (x)\to \xi (x)$ at every retained fibre. The squared norm of their difference is bounded by $4\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2$, so [F4] gives $R_X(q(u_n))\xi \to \xi$ in the direct-integral norm; its limit lies in the closed span of $R_X(A)H_X$, proving nondegeneracy. The same holds for $R_Y$. Both carriers are separable by [F3] and nonzero: a countable fundamental family and nonzero fibres give a section nonzero on a positive-measure set; intersecting with a finite-measure exhaustion set and a bound on its norm produces a nonzero square-integrable section. For every fixed $g$, the same estimate applied to the strong limits in step 4.1 gives ${\int }_X^{\oplus }{\pi }_x(g)d\mu ={\mathrm{s-lim}}_n{R}_X(q(L_g{u}_n))$. It is therefore the group representation corresponding to $R_X$ by step 1.1, and similarly on $Y$. These direct-integral representations are strongly continuous by [F3].

Let $W:H_X\to H_Y$ be the unitary intertwining $R_X$ and $R_Y$ supplied in step 3.1. For every $g$, applying $W$ to the strong-limit formula of step 5.1 proves that it intertwines the two group direct integrals. Taking $H=H_X$ and $\pi ={\int }_X^{\oplus }{\pi }_{x}d\mu$ gives the two promised decompositions of one separable strongly continuous representation. If for any retained pair $(x,y)$ one had ${\pi }_x\cong {\sigma }_y$, the converse intertwiner assertion of step 1.1 would give ${\rho }_x\cong {\tau }_y$, contrary to their labels belonging to disjoint sets $X,Y$ of algebra irreducible classes. Thus the cross-class inequivalence holds for every pair, rather than merely almost everywhere.

Central factor decomposition and its essential uniqueness still apply to this $\pi$ by [F5]. The ambiguity just constructed concerns irreducible disintegration, so it does not contradict central uniqueness, and explicitly exhibits the existence of irreducible decompositions rather than their absence. Therefore all clauses of the Statement hold, with exactly the original witness existence cited and the correspondence, measurability, nondegeneracy, all-group equivalence and pointwise cross inequivalence proved locally.