Measurable splitting of a field of type I factors into irreducible representations with multiplicity

Discard the initial Borel null exceptions and trivialize on the countably many positive-dimension strata by [F1]. Put $N_x=M_x^{\mathrm{\prime }}$ and choose WOT-dense sections $a_j(x)$ of its unit ball. In constant-space coordinates use a complete orthonormal frame $(f_k(x){)}_{k\in \mathbb{N}}$, padded with zeros on finite-dimensional fibres, and define ${\phi }_x(T)={\sum }_{k\in \mathbb{N}}{2}^{-(k+1)}⟨T{f}_k(x),f_k(x)⟩$. On positive operators this is faithful, since zero diagonal coefficients force $T^{1\mathrm{/}2}{f}_k=0$ on a basis; it is normal, since bounded increasing positive sequences have increasing coefficient sums and their limits commute with the summable series. Its value on $I$ is positive and at most one. On the unit ball it is WOT-continuous by uniform tail bounds. Operator products are jointly Borel in WOT-ball coordinates: each coefficient is the limit of finite basis-coordinate sums; adjoints are Borel.

The relation defining nonzero minimal $q\in N_x$ is Borel: impose $q=q^{\ast }=q^2$, $q\mathrm{\ne }0$, commutation with the countable generators of $M_x$, and for every $j$ impose $q{a}_j(x)q={\phi }_x(q{a}_j(x)q)q\mathrm{/}{\phi }_x(q)$. These are countably many coefficient equations using the Borel operations of step 1.1. For fixed $q$, compression is WOT-continuous and the scalar functional is WOT-continuous on bounded sets; density of the $a_j$ therefore makes these equations equivalent to $q{N}_{x}q=\mathbb{C}q$. They characterize minimality. For any Borel residual projection $r(x)\in N_x$, add $q\le r(x)$. If $r\mathrm{\ne }0$, [F3] makes its section nonempty.

Set $r_0=I$. Inductively, on $\{r_{n-1}\mathrm{\ne }0\}$ let $s_n(x)$ be the supremum of ${\phi }_x(q)$ over the minimal projections in step 2.1 below $r_{n-1}$. The functional is bounded real on projections, so [F2] gives a Borel version of $s_n$ on a conull Borel subset; there $s_n>0$ by faithfulness. Apply [F2] to the nonempty Borel relation ${\phi }_x(q)>s_n(x)\mathrm{/}2$ to select $q_n$, set $q_n=0$ on the zero-residual part, and put $r_n=r_{n-1}-q_n$. Repeat on retained bases and remove the countable union of Borel null exceptions once at the end. At each retained $x$, the $q_n$ are orthogonal. If the strong residual limit $r_{\mathrm{\infty }}$ were nonzero, [F3] would supply a minimal $q\le r_{\mathrm{\infty }}$ with $c={\phi }_x(q)>0$. Then $s_n\ge c$ at every step, hence ${\phi }_x(q_n)>c\mathrm{/}2$ for every $n$, contradicting ${\sum }_n{\phi }_x(q_n)\le {\phi }_x(I)\le 1$. Thus ${\sum }_n{q}_n=I$ strongly.

Let $K_x=q_1(x)H_x$, with fundamental sections $q_1{f}_k$; [F1] makes this a measurable nonzero subfield. Let $m(x)$ count the nonzero $q_n$. Because construction stops exactly when the residual is zero, $\{m\ge n\}=\{q_n\mathrm{\ne }0\}$ is Borel. On each such set the solutions $u_n\in N_x$ to $u_n^{\ast }{u}_n=q_n$, $u_n{u}_n^{\ast }=q_1$ form a nonempty Borel relation in the operator unit ball by [F3] and step 1.1. Use [F2] to select them conull, put $u_1=q_1$ and $u_n=0$ where $q_n=0$, and remove the countably many new null exceptions.

Define $V_x\xi =(u_n(x)\xi {)}_{n\le m(x)}$ and ${\sigma }_x(g)={\pi }_x(g)\mathrm{\mid }{K}_x$. Then ${\sum }_n\mathrm{\parallel }{u}_n\xi {\mathrm{\parallel }}^2={\sum }_n\mathrm{\parallel }{q}_n\xi {\mathrm{\parallel }}^2=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$, and $u_i{u}_j^{\ast }={\delta }_{ij}{q}_1$, so the inverse is the norm-convergent series $V_x^{-1}({\eta }_n)={\sum }_n{u}_n^{\ast }{\eta }_n$. This proves unitarity including the infinite case. Fundamental coefficients and pointwise norm limits make both fields measurable. Each $u_n$ belongs to the actual commutant ${\pi }_x(G{)}^{\mathrm{\prime }}$, so the amplification identity holds for every $g$ at each retained $x$. Restriction preserves strong continuity, and [F3] makes ${\sigma }_x$ irreducible. Its fixed-g matrix coefficients against fundamental sections are Borel, so it is a measurable representation field. Fibrewise application of the uniqueness clause in [F3] gives the final invariant pair.