Direct integrals of von Neumann algebras

Written by Claude Opus 5.5 (Anthropic), September 2026. Self-checked by the writing AI. Original text: CC0 1.0.

A direct integral of Hilbert spaces $\mathcal H=\int_\Gamma^\oplus H(\gamma)\,d\mu(\gamma)$ comes with two von Neumann algebras of its own. The diagonal algebra $\mathcal A$ multiplies every fibre by a scalar that varies measurably with the point. The decomposable algebra $\mathcal D$ acts on every fibre by a bounded operator. Each of the two is the commutant of the other. Many algebras lie between them. Pick, at each point $\gamma$, a von Neumann algebra $M(\gamma)$ on the fibre $H(\gamma)$, and keep the decomposable operators whose fibres lie in $M(\gamma)$ at almost every point. The resulting set is written $\int_\Gamma^\oplus M(\gamma)\,d\mu(\gamma)$.

This lesson studies that construction. When the family $\gamma\mapsto M(\gamma)$ is measurable, in the sense of Section 1, the set is a von Neumann algebra. Its commutant is obtained by integrating the commutants $M(\gamma)'$, and its centre by integrating the centres $M(\gamma)\cap M(\gamma)'$ (Theorems 3.2 and 4.3). So the centre equals the diagonal algebra exactly when almost all the $M(\gamma)$ are factors. The family can be recovered from its integral, up to a null set (Theorem 4.1). Conversely, every von Neumann algebra that contains $\mathcal A$ and is contained in $\mathcal D$ arises in this way (Theorem 5.3). Measurable families, taken up to null sets, therefore match the von Neumann algebras between $\mathcal A$ and $\mathcal D$ one to one.

These facts form the core of von Neumann's reduction theory (1949); see [Blackadar, Section III.1.6]. After a von Neumann algebra has been written as an integral over an abelian subalgebra of its centre, many questions about it become questions about the fibres. When the whole centre is used, the fibres are factors. The delicate point is measurability. That the fields of commutants and of centres are again measurable rests on the Borel space of von Neumann algebras introduced in [Effros 1965].

The base is an arbitrary $\sigma$-finite measure space throughout. We need no standard Borel structure, no completeness of the measure and no separability of $\mathcal H$. Section 6 explains where standard bases will matter later. It also gives a base on which a harmless-looking family is not measurable, and on which the commutant formula fails for it.

We assume the lessons Measurable fields of Hilbert spaces and their direct integrals, Decomposable operators and the diagonal algebra and The Effros Borel structure. Every fact we take from them is stated in full in the section Background used without proof. Example 7.2 also uses Spatial tensor products of von Neumann algebras.

Basic references are [Takesaki I, Chapter IV] and [Blackadar, Section III.1.6].

Conventions

Throughout, $(\Gamma,\Sigma,\mu)$ is a $\sigma$-finite measure space, and measurable means $\Sigma$-measurable. We assume nothing about completeness of $\mu$, standardness of $(\Gamma,\Sigma)$ or countability properties of $\Sigma$. A null set is a set $N\in\Sigma$ with $\mu(N)=0$. A statement holds almost everywhere, or for almost every $\gamma$, if it holds for all $\gamma$ outside some null set. A subset $E\subseteq\Gamma$ is measurable up to a null set if some $F\in\Sigma$ has $E\,\triangle\,F$ contained in a null set. When $\Sigma$ is complete for $\mu$, these are the familiar notions.

Hilbert spaces are complex, inner products are linear in the first variable, and the zero space is allowed. For a Hilbert space $K$, $B(K)$ is the algebra of bounded operators on $K$, and $S'$ is the commutant of a set $S\subseteq B(K)$. A von Neumann algebra on $K$ is a $*$-subalgebra $M\subseteq B(K)$ with $M''=M$. The von Neumann algebra generated by a set $S$ is $(S\cup S^*)''$. The centre of $M$ is $M\cap M'$, and $M$ is a factor if its centre is $\mathbb C1$. For von Neumann algebras $M_1,M_2$ on $K$ we put $M_1\vee M_2=(M_1\cup M_2)''$. On the zero space, $B(0)=\{0\}=\mathbb C1$ is the only von Neumann algebra, and it is a factor by this definition.

We use four elementary facts about commutants. Let $S,T\subseteq B(K)$.

• (C1) If $S\subseteq T$, then $T'\subseteq S'$. Also $S\subseteq S''$, and hence $S'''=S'$.
• (C2) $S'$ is a subalgebra of $B(K)$ containing $1$. If $S^*=S$, then $S'$ is a von Neumann algebra.
• (C3) The von Neumann algebra generated by $S$ is the smallest von Neumann algebra containing $S$, and its commutant is $(S\cup S^*)'$.
• (C4) If $M_1,M_2$ are von Neumann algebras, then $M_1\cap M_2=(M_1'\cup M_2')'$ is a von Neumann algebra. In particular the centre of a von Neumann algebra is a von Neumann algebra. Moreover $M_1\vee M_2$ is the von Neumann algebra generated by $M_1\cup M_2$, and $(M_1\vee M_2)'=M_1'\cap M_2'$.

Proof. (C1) An operator that commutes with all of $T$ commutes with all of $S$. Every element of $S$ commutes with every element of $S'$, so $S\subseteq S''$. Applying the first statement to $S\subseteq S''$ gives $S'''\subseteq S'$, and applying the second to $S'$ gives $S'\subseteq S'''$.

(C2) Sums and products of operators that commute with $S$ commute with $S$, and so does $1$. Let $S^*=S$ and $x\in S'$. For $s\in S$ we have $xs^*=s^*x$, and taking adjoints gives $sx^*=x^*s$. So $x^*\in S'$, and $S'$ is a $*$-algebra. It equals its own bicommutant by (C1).

(C3) The set $S\cup S^*$ is self-adjoint, so $(S\cup S^*)'$ is a self-adjoint set by (C2), and $(S\cup S^*)''$ is a von Neumann algebra by (C2) again. It contains $S$ by (C1). If a von Neumann algebra $P$ contains $S$, it contains $S\cup S^*$, and then $P=P''\supseteq(S\cup S^*)''$ by (C1) used twice. The commutant is $(S\cup S^*)'''=(S\cup S^*)'$ by (C1).

(C4) An operator commutes with every element of $M_1'\cup M_2'$ exactly when it lies in $M_1''\cap M_2''=M_1\cap M_2$. The set $M_1'\cup M_2'$ is self-adjoint, so $M_1\cap M_2$ is a von Neumann algebra by (C2). The set $M_1\cup M_2$ is self-adjoint too, so it generates $(M_1\cup M_2)''=M_1\vee M_2$, whose commutant is $(M_1\cup M_2)'=M_1'\cap M_2'$ by (C3). $\square$

The bicommutant theorem is not needed anywhere in this lesson.

Background used without proof

Throughout, $(H(\gamma))_{\gamma\in\Gamma}$, with its space $\mathfrak M$ of measurable sections, is a fixed measurable field of Hilbert spaces over $(\Gamma,\Sigma,\mu)$, and $\mathcal H=\int_\Gamma^\oplus H(\gamma)\,d\mu(\gamma)$ is its direct integral. The following facts are proved in the lessons named.

From Measurable fields of Hilbert spaces and their direct integrals:

• (B1) Fields. $\mathfrak M$ is a vector space of sections $\xi$, with $\xi(\gamma)\in H(\gamma)$, with three properties. The function $\gamma\mapsto\|\xi(\gamma)\|$ is measurable for $\xi\in\mathfrak M$. A section $\eta$ belongs to $\mathfrak M$ as soon as $\gamma\mapsto\langle\eta(\gamma),\xi(\gamma)\rangle$ is measurable for every $\xi\in\mathfrak M$. Some sequence in $\mathfrak M$, called fundamental, has values that span a dense subspace of every fibre; so every fibre is separable. Inner products of measurable sections are measurable functions, and $f\xi\in\mathfrak M$ for measurable $f$ and $\xi\in\mathfrak M$. Conversely, let $(\xi_j)$ be sections of a family of Hilbert spaces such that every function $\gamma\mapsto\langle\xi_j(\gamma),\xi_i(\gamma)\rangle$ is measurable and $(\xi_j(\gamma))_j$ is total in every fibre. Then exactly one measurable field contains all the $\xi_j$. Its measurable sections are the sections $\eta$ for which every $\gamma\mapsto\langle\eta(\gamma),\xi_j(\gamma)\rangle$ is measurable. (Definition 2.1, Lemma 2.2 and Theorem 3.1.)
• (B2) Dimension strata. There are sections $e_k\in\mathfrak M$, $k\geq1$, such that $e_k(\gamma)\neq0$ exactly when $k\leq n(\gamma)=\dim H(\gamma)$, and the nonzero $e_k(\gamma)$ form an orthonormal basis of $H(\gamma)$. The function $n$ is measurable, so the sets $\Gamma_d=\{n=d\}$, $d\in\{0,1,2,\ldots,\infty\}$, form a countable measurable partition of $\Gamma$. Let $\ell^2_d$ be $\mathbb C^d$, or $\ell^2(\mathbb N)$ if $d=\infty$, with standard basis $(\varepsilon_k)$. For $\gamma\in\Gamma_d$, the formula $U(\gamma)v=\sum_{k\leq d}\langle v,e_k(\gamma)\rangle\varepsilon_k$ defines a unitary $U(\gamma):H(\gamma)\to\ell^2_d$. (Theorems 3.1 and 5.1.)
• (B3) Operator fields. Let $(K(\gamma)),\mathfrak N$ be a second measurable field over the same base. A family of bounded operators $x(\gamma):H(\gamma)\to K(\gamma)$ is a measurable operator field if $\gamma\mapsto x(\gamma)\xi(\gamma)$ lies in $\mathfrak N$ for every $\xi\in\mathfrak M$. No bound on $\|x(\gamma)\|$ is assumed. It suffices that this holds for the members of one fundamental sequence of $\mathfrak M$. The norm function $\gamma\mapsto\|x(\gamma)\|$ is measurable. Sums, composites and adjoints of measurable operator fields are measurable, and so are their products with measurable scalar functions. (Definition 6.1 and Theorem 6.2.)
• (B4) Decomposable operators. For a bounded measurable function $f$, $m_f$ is the operator of multiplication by $f$ on $\mathcal H$. A measurable operator field $x$ is essentially bounded if its norm function is. Then $(\int^\oplus x\,\xi)(\gamma)=x(\gamma)\xi(\gamma)$ defines a bounded operator $\int^\oplus x=\int^\oplus_\Gamma x(\gamma)\,d\mu(\gamma)$, whose norm is $\operatorname*{ess\,sup}_\gamma\|x(\gamma)\|$. The same holds for fields between two measurable fields. The map $x\mapsto\int^\oplus x$ is linear and multiplicative on composable fields, and $(\int^\oplus x)^*=\int^\oplus x^*$. Two fields define the same operator exactly when they agree almost everywhere. Every $\int^\oplus x$ commutes with every $m_f$. If $\Sigma$ is countably generated up to null sets, for instance if it is the Borel $\sigma$-algebra of a standard Borel space, then $\mathcal H$ is separable. (Theorems 9.1 and 10.1.)
• (B5) Constant fields. Let $K_0$ be a separable Hilbert space with orthonormal basis $(\varepsilon_k)$. By (B1), the constant sections $\varepsilon_k$ generate a measurable field whose fibres all equal $K_0$: the constant field $K_0$. Its measurable sections are the maps $\xi:\Gamma\to K_0$ whose coordinates $\gamma\mapsto\langle\xi(\gamma),\varepsilon_k\rangle$ are all measurable. The map $f\otimes v\mapsto(\gamma\mapsto f(\gamma)v)$ extends to a unitary $W$ from $L^2(\Gamma,\mu)\otimes K_0$ onto its direct integral. (Example 11.2.)

From Decomposable operators and the diagonal algebra:

• (B6) The two algebras. Let $\mathcal A=\{m_f\}$ be the diagonal algebra, and let the decomposable algebra $\mathcal D$ be the set of all $\int^\oplus x$ with $x$ an essentially bounded measurable operator field on $(H(\gamma))$. Both are von Neumann algebras on $\mathcal H$, $\mathcal A'=\mathcal D$ and $\mathcal D'=\mathcal A$. More precisely, every $T\in\mathcal A'$ equals $\int^\oplus x$ for a measurable operator field $x$ with $\|x(\gamma)\|\leq\|T\|$ for every $\gamma$. We have $m_f=\int^\oplus f(\gamma)1\,d\mu(\gamma)$. If $x$ is essentially bounded and $x(\gamma)$ is a scalar for almost every $\gamma$, then $\int^\oplus x\in\mathcal A$. Finally, $m_f=0$ exactly when $f=0$ almost everywhere on $\{n\geq1\}$. (Proposition 2.2, Theorems 5.1 and 7.1.)
• (B7) Matrix units. For $k,l\geq1$, the formula $e_{kl}(\gamma)v=\langle v,e_l(\gamma)\rangle e_k(\gamma)$ defines a measurable operator field with $\|e_{kl}(\gamma)\|\leq1$ and $e_{kl}(\gamma)^*=e_{lk}(\gamma)$. (Proposition 2.2.)
• (B8) Unitary fields and matrix entries. Let $(K(\gamma)),\mathfrak N$ be a second measurable field over the same base, with direct integral $\mathcal K$. If $v$ is a measurable field of unitaries $v(\gamma):H(\gamma)\to K(\gamma)$, then $V=\int^\oplus v$ is a unitary from $\mathcal H$ onto $\mathcal K$. On a constant field $K_0$ as in (B5), a family $y$ of bounded operators is a measurable operator field exactly when every matrix entry $\gamma\mapsto\langle y(\gamma)\varepsilon_l,\varepsilon_k\rangle$ is measurable. (Proposition 8.1.)
• (B9) A finite equivalent measure. There is a measurable $w:\Gamma\to(0,1]$ with $\int_\Gamma w\,d\mu<\infty$. The finite measure $\lambda(E)=\int_Ew\,d\mu$ has the same null sets as $\mu$. (Lemma 3.1.)

From The Effros Borel structure:

• (B10) The Borel space of von Neumann algebras. Let $K$ be a separable Hilbert space, and $B(K)_*$ the Banach space of $\sigma$-weakly continuous linear functionals on $B(K)$. It is separable. The closed unit ball of $B(K)$ is compact and metrizable in the $\sigma$-weak topology, and every commutant is $\sigma$-weakly closed. For a $\sigma$-weakly closed subspace $P\subseteq B(K)$ and $\psi\in B(K)_*$, put $p_P(\psi)=\sup\{|\psi(x)|:x\in P,\ \|x\|\leq1\}$. Then $p_P$ is a seminorm with $p_P(\psi)\leq\|\psi\|$, and $P$ is determined by the values of $p_P$ on any dense subset of $B(K)_*$. The set $\mathrm{vN}(K)$ of von Neumann algebras on $K$, with the $\sigma$-algebra generated by the functions $P\mapsto p_P(\psi)$, $\psi\in B(K)_*$, is a standard Borel space. (Conventions, Definition 3.1, Proposition 3.2, Lemma 8.1 and Theorem 9.1.)

• (B11) Measurably generated families. A family $(L(\gamma))$ of von Neumann algebras on the fibres $H(\gamma)$ is measurably generated if there are measurable operator fields $x_j$, $j\geq1$, such that $L(\gamma)$ is generated by $\{x_j(\gamma):j\geq1\}$ for every $\gamma$. Let $(L(\gamma))$ and $(L_1(\gamma))$ be measurably generated. Then:

  • (a) the generating fields can be chosen with $\|x_j(\gamma)\|\leq1$, and with $\{x_j(\gamma):j\geq1\}$ $\sigma$-weakly dense in the unit ball of $L(\gamma)$, for every $\gamma$;
  • (b) the families $(L(\gamma)')$, $(L(\gamma)\cap L_1(\gamma))$, $(L(\gamma)\vee L_1(\gamma))$ and $(L(\gamma)\cap L(\gamma)')$ are measurably generated;
  • (c) the sets of those $\gamma$ for which $L(\gamma)$ is a factor, is abelian, or equals $B(H(\gamma))$ belong to $\Sigma$;
  • (d) a family $(L(\gamma))$ of von Neumann algebras on the fibres is measurably generated if and only if, for every $d$, the map $\gamma\mapsto U(\gamma)L(\gamma)U(\gamma)^*$ from $\Gamma_d$ to $\mathrm{vN}(\ell^2_d)$ is measurable, with $U(\gamma)$ as in (B2).

  None of this uses the measure. (Definition 10.1 and Theorem 10.3.)

From Spatial tensor products of von Neumann algebras, for Example 7.2 only:

• (B12) Tensor products. For von Neumann algebras $M\subseteq B(H)$ and $N\subseteq B(K)$, the spatial tensor product $M\bar\otimes N$ is the von Neumann algebra generated by the operators $x\otimes y$ with $x\in M$ and $y\in N$. If $S\subseteq B(H)$ and $T\subseteq B(K)$ are self-adjoint sets with $S''=M$ and $T''=N$, then $M\bar\otimes N=(S\otimes1\cup1\otimes T)''$. The commutation theorem says that $(M\bar\otimes N)'=M'\bar\otimes N'$. (Definition 5.1, Theorems 5.2 and 11.4.)

1. Measurable fields of von Neumann algebras

Definition 1.1. A field of von Neumann algebras on $(H(\gamma))$ is a family $M=(M(\gamma))_{\gamma\in\Gamma}$ in which each $M(\gamma)$ is a von Neumann algebra acting on $H(\gamma)$. The field is measurable if there are measurable operator fields $x_1,x_2,\ldots$ on $(H(\gamma))$ such that, for almost every $\gamma$, the algebra $M(\gamma)$ is generated by $\{x_j(\gamma):j\geq1\}$. Two fields are equivalent if they agree almost everywhere.

The generating fields need not be essentially bounded; their norms may grow without bound. Finitely many generators are allowed: repeat one of them. For a field $M$ we write $M'$ for the field $(M(\gamma)')$ of commutants and $Z_M$ for the field $(M(\gamma)\cap M(\gamma)')$ of centres. For two fields $M_1,M_2$ we write $M_1\cap M_2$ and $M_1\vee M_2$ for the fields $(M_1(\gamma)\cap M_2(\gamma))$ and $(M_1(\gamma)\vee M_2(\gamma))$. By (C2) and (C4), these are again fields of von Neumann algebras.

A measurably generated family in the sense of (B11) is a measurable field. The only difference is the null set: Proposition 1.3(1) shows that every measurable field is equivalent to a measurably generated family.

Example 1.2.

1. The field $\mathbb C1=(\mathbb C1_{H(\gamma)})$ is measurable. It is generated by the identity field, since $\{1\}''=B(H(\gamma))'=\mathbb C1$; the last equality is the computation in (2).
2. The field $(B(H(\gamma)))$ is measurable: it is generated by the matrix-unit fields $e_{kl}$ of (B7). Indeed, fix $\gamma$ with $n(\gamma)\geq1$, and let $s\in B(H(\gamma))$ commute with every $e_{kl}(\gamma)$. For $k\leq n(\gamma)$ we have $e_{k1}(\gamma)e_1(\gamma)=e_k(\gamma)$, so $$se_k(\gamma)=se_{k1}(\gamma)e_1(\gamma)=e_{k1}(\gamma)se_1(\gamma)=\langle se_1(\gamma),e_1(\gamma)\rangle\,e_k(\gamma).$$ So $s$ agrees with the scalar $c=\langle se_1(\gamma),e_1(\gamma)\rangle$ on an orthonormal basis, and $s=c1$. The set $\{e_{kl}(\gamma)\}$ is self-adjoint and its commutant is $\mathbb C1$, so by (C3) the von Neumann algebra it generates is $(\mathbb C1)'=B(H(\gamma))$. If $n(\gamma)=0$, then $B(H(\gamma))=\mathbb C1$, and there is nothing to prove.

Proposition 1.3. Let $M$, $M_1$ and $M_2$ be measurable fields of von Neumann algebras.

1. (Repair on a null set) There are a null set $N$ and a measurably generated family $\tilde M$ with $\tilde M(\gamma)=M(\gamma)$ for all $\gamma\notin N$.
2. (Generators in the unit ball) There are measurable operator fields $y_k$, $k\geq1$, with $\|y_k(\gamma)\|\leq1$ for every $\gamma$, such that for almost every $\gamma$ the operators $y_k(\gamma)$ lie in $M(\gamma)$, are $\sigma$-weakly dense in its unit ball, and generate it.
3. (Operations) The fields $M'$, $Z_M$, $M_1\cap M_2$ and $M_1\vee M_2$ are measurable.
4. (Fibrewise properties) The set of $\gamma$ for which $M(\gamma)$ is a factor is measurable up to a null set. So are the set where $M(\gamma)$ is abelian and the set where $M(\gamma)=B(H(\gamma))$.
5. (Effros form) A field $L$ of von Neumann algebras on $(H(\gamma))$ is measurable exactly when, for every $d$, some measurable map $\Phi_d:\Gamma_d\to\mathrm{vN}(\ell^2_d)$ satisfies $\Phi_d(\gamma)=U(\gamma)L(\gamma)U(\gamma)^*$ for almost every $\gamma\in\Gamma_d$.

Proof. (1) Choose generating fields $x_j$ and a null set $N$ such that $M(\gamma)$ is generated by $\{x_j(\gamma)\}$ for $\gamma\notin N$. For every $\gamma$, let $\tilde M(\gamma)$ be the von Neumann algebra generated by $\{x_j(\gamma)\}$. Then $\tilde M$ is measurably generated, and it agrees with $M$ off $N$.

(2) Apply (B11)(a) to $\tilde M$. It gives fields that, for every $\gamma$, lie in $\tilde M(\gamma)$, are dense in its unit ball and generate it. Off $N$, $\tilde M(\gamma)=M(\gamma)$.

(3) Take repairs $\tilde M,\tilde M_1,\tilde M_2$ as in (1), with null sets $N,N_1,N_2$. By (B11)(b), the families $\tilde M'$, $Z_{\tilde M}$, $\tilde M_1\cap\tilde M_2$ and $\tilde M_1\vee\tilde M_2$ are measurably generated. Off the null set $N\cup N_1\cup N_2$ they agree with $M'$, $Z_M$, $M_1\cap M_2$ and $M_1\vee M_2$. The generating fields of a family that agrees with a given field off a null set generate that field almost everywhere. So the four fields are measurable.

(4) By (B11)(c), the set $F$ of $\gamma$ for which $\tilde M(\gamma)$ is a factor lies in $\Sigma$. The set of $\gamma$ for which $M(\gamma)$ is a factor differs from $F$ only inside $N$. The same argument applies to the other two properties.

(5) Let $L$ be measurable, with a repair $\tilde L$ and a null set $N$ as in (1). By (B11)(d), the map $\Phi_d(\gamma)=U(\gamma)\tilde L(\gamma)U(\gamma)^*$ is measurable on $\Gamma_d$, and it equals $U(\gamma)L(\gamma)U(\gamma)^*$ for $\gamma\in\Gamma_d\setminus N$. Conversely, let maps $\Phi_d$ be given, and let $N_d\subseteq\Gamma_d$ be a null set off which $\Phi_d(\gamma)=U(\gamma)L(\gamma)U(\gamma)^*$. For $\gamma\in\Gamma_d$ put $\tilde L(\gamma)=U(\gamma)^*\Phi_d(\gamma)U(\gamma)$. The map $a\mapsto U(\gamma)^*aU(\gamma)$ is a $*$-isomorphism of $B(\ell^2_d)$ onto $B(H(\gamma))$ that carries commutants to commutants, so $\tilde L(\gamma)$ is a von Neumann algebra acting on $H(\gamma)$. Since $U(\gamma)\tilde L(\gamma)U(\gamma)^*=\Phi_d(\gamma)$, the family $\tilde L$ is measurably generated by (B11)(d). It agrees with $L$ off the null set $\bigcup_dN_d$, so $L$ is measurable. $\square$

Remark 1.4.

1. Every von Neumann algebra $P$ acting on a separable space $K$ is generated by a countable set. Indeed, by (B10) the unit ball of $P$ lies in a compact metrizable space, so it has a countable $\sigma$-weakly dense subset $S$. The von Neumann algebra generated by $S$ is a commutant, hence $\sigma$-weakly closed (B10). It contains $S$, hence the unit ball of $P$, hence $P$; and it lies in $P$ by (C3). So measurability is not about the number of generators at one point. It asks that the generators be chosen measurably in $\gamma$.
2. A field that is equivalent to a measurable field is measurable, since the same generating fields serve outside a null set.
3. By (B10), the maps $\Phi_d$ of Proposition 1.3(5) take values in standard Borel spaces. So a measurable field is, up to a null set, a measurable map from each dimension stratum into a standard Borel space of von Neumann algebras.

2. The direct integral of a field of von Neumann algebras

Definition 2.1. Let $M$ be a field of von Neumann algebras on $(H(\gamma))$, measurable or not. Put $$\int_\Gamma^\oplus M(\gamma)\,d\mu(\gamma)=\Big\{\int^\oplus x:\ x\ \text{an essentially bounded measurable operator field with}\ x(\gamma)\in M(\gamma)\ \text{for almost every}\ \gamma\Big\}, \tag{2.1}$$ and abbreviate it to $\int^\oplus M$. When $M$ is measurable, we call $\int^\oplus M$ the direct integral of the field $M$. By definition it is a subset of $\mathcal D$.

Proposition 2.2. Let $M$, $M_1$ and $M_2$ be fields of von Neumann algebras on $(H(\gamma))$, measurable or not.

1. (Representing fields) Let $T\in\int^\oplus M$. Every essentially bounded measurable operator field $x$ with $T=\int^\oplus x$ satisfies $x(\gamma)\in M(\gamma)$ for almost every $\gamma$. Moreover, $T=\int^\oplus x$ for a measurable operator field $x$ with $x(\gamma)\in M(\gamma)$ and $\|x(\gamma)\|\leq\|T\|$ for every $\gamma$.
2. (Algebra) $\int^\oplus M$ is a $*$-subalgebra of $\mathcal D$ that contains $\mathcal A$, and $\mathcal A\subseteq(\int^\oplus M)'$.
3. (Monotonicity) If $M_1(\gamma)\subseteq M_2(\gamma)$ for almost every $\gamma$, then $\int^\oplus M_1\subseteq\int^\oplus M_2$. In particular, equivalent fields have the same direct integral.
4. (Intersections) $\int^\oplus(M_1\cap M_2)=\int^\oplus M_1\cap\int^\oplus M_2$.
5. (The extreme fields) $\int^\oplus B(H(\gamma))\,d\mu(\gamma)=\mathcal D$ and $\int^\oplus\mathbb C1\,d\mu(\gamma)=\mathcal A$.

Proof. (1) Choose an essentially bounded $x_0$ with $T=\int^\oplus x_0$ and $x_0(\gamma)\in M(\gamma)$ off a null set $N_0$. If $T=\int^\oplus x$, then $x=x_0$ off a null set $N_1$ by (B4), so $x(\gamma)\in M(\gamma)$ off $N_0\cup N_1$. For the second claim, $T$ lies in $\mathcal D=\mathcal A'$, so (B6) gives a measurable operator field $y$ with $T=\int^\oplus y$ and $\|y(\gamma)\|\leq\|T\|$ for every $\gamma$. By the first claim, $y(\gamma)\in M(\gamma)$ off a null set $N_2$. The field $x=1_{\Gamma\setminus N_2}\,y$ is measurable by (B3), and it agrees with $y$ almost everywhere, so $\int^\oplus x=T$. At each $\gamma$, $x(\gamma)$ is either $y(\gamma)\in M(\gamma)$ or $0\in M(\gamma)$, and $\|x(\gamma)\|\leq\|T\|$.

(2) Let $S=\int^\oplus x$ and $T=\int^\oplus y$ lie in $\int^\oplus M$, and let $c\in\mathbb C$. By (B4), $S+cT=\int^\oplus(x+cy)$, $ST=\int^\oplus xy$ and $S^*=\int^\oplus x^*$. The fields $x+cy$, $xy$ and $x^*$ are measurable and essentially bounded by (B3). Their values lie in $M(\gamma)$ for almost every $\gamma$, because $M(\gamma)$ is a $*$-algebra. So $\int^\oplus M$ is a $*$-subalgebra of $\mathcal D$. For bounded measurable $f$, $m_f=\int^\oplus f(\gamma)1\,d\mu(\gamma)$ by (B6), and $f(\gamma)1\in M(\gamma)$; so $\mathcal A\subseteq\int^\oplus M$. Every element of $\int^\oplus M$ is decomposable, so it commutes with $\mathcal A$ by (B4).

(3) An essentially bounded field whose values lie in $M_1(\gamma)$ almost everywhere has its values in $M_2(\gamma)$ almost everywhere, because a union of two null sets is null.

(4) The inclusion $\subseteq$ follows from (3). Let $T\in\int^\oplus M_1\cap\int^\oplus M_2$, and write $T=\int^\oplus x$. By (1), applied to $M_1$ and to $M_2$, the operator $x(\gamma)$ lies in $M_1(\gamma)$ and in $M_2(\gamma)$ for almost every $\gamma$. So $T\in\int^\oplus(M_1\cap M_2)$.

(5) The first equality is the definition of $\mathcal D$. For the second, every $m_f=\int^\oplus f(\gamma)1\,d\mu(\gamma)$ lies in $\int^\oplus\mathbb C1$. Conversely, a decomposable operator whose field is scalar almost everywhere lies in $\mathcal A$ by (B6). $\square$

Proposition 2.3 (Unitary transfer). Let $(K(\gamma)),\mathfrak N$ be a second measurable field over $(\Gamma,\Sigma,\mu)$, with direct integral $\mathcal K$. Let $v$ be a measurable field of unitaries $v(\gamma):H(\gamma)\to K(\gamma)$, and put $V=\int^\oplus v$. For a field $M$ on $(H(\gamma))$, let $vMv^*$ be the field $(v(\gamma)M(\gamma)v(\gamma)^*)$ on $(K(\gamma))$. Then $$V\Big(\int^\oplus M\Big)V^*=\int^\oplus vMv^* .$$ If $M$ is measurable, so is $vMv^*$.

Proof. The map $a\mapsto v(\gamma)av(\gamma)^*$ is a $*$-isomorphism of $B(H(\gamma))$ onto $B(K(\gamma))$ that carries commutants to commutants. So $vMv^*$ is a field of von Neumann algebras, and the map carries the von Neumann algebra generated by a set $S$ to the one generated by $v(\gamma)Sv(\gamma)^*$. The operator $V$ is unitary by (B8). Let $T\in\int^\oplus M$, and write $T=\int^\oplus x$ with $x(\gamma)\in M(\gamma)$ for every $\gamma$ (Proposition 2.2(1)). By (B4), $VTV^*=\int^\oplus vxv^*$. The field $vxv^*$ is measurable by (B3). It has the same norm function as $x$, and its values lie in $v(\gamma)M(\gamma)v(\gamma)^*$. So $V(\int^\oplus M)V^*\subseteq\int^\oplus vMv^*$. The adjoint field $v^*$ is a measurable field of unitaries from $(K(\gamma))$ to $(H(\gamma))$, and $\int^\oplus v^*=V^*$. So the same argument gives $V^*(\int^\oplus vMv^*)V\subseteq\int^\oplus v^*vMv^*v=\int^\oplus M$. Conjugating by $V$ gives the reverse inclusion. Finally, if measurable fields $x_j$ generate $M(\gamma)$ off a null set, then the measurable fields $vx_jv^*$ generate $v(\gamma)M(\gamma)v(\gamma)^*$ off the same null set. $\square$

3. The commutant theorem

Lemma 3.1.

1. Let $x$ be an essentially bounded measurable operator field, and let $T=\int^\oplus t\in\mathcal D$. Then $T$ commutes with $\int^\oplus x$ if and only if $t(\gamma)x(\gamma)=x(\gamma)t(\gamma)$ for almost every $\gamma$.
2. (Truncation) Let $x$ be a measurable operator field and $m\geq1$. Then $x^{[m]}=1_{\{\|x\|\leq m\}}\,x$ is a measurable operator field with $\|x^{[m]}(\gamma)\|\leq m$ for every $\gamma$, and $x^{[m]}(\gamma)=x(\gamma)$ whenever $\|x(\gamma)\|\leq m$. If $M$ is a field of von Neumann algebras with $x(\gamma)\in M(\gamma)$ for almost every $\gamma$, then $\int^\oplus x^{[m]}$ and its adjoint lie in $\int^\oplus M$.

Proof. (1) By (B4), $T\int^\oplus x-(\int^\oplus x)T=\int^\oplus(tx-xt)$. The field $tx-xt$ is measurable and essentially bounded by (B3). By (B4), it defines the zero operator exactly when it vanishes almost everywhere.

(2) The norm function of $x$ is measurable (B3), so $1_{\{\|x\|\leq m\}}$ is a measurable function and $x^{[m]}$ is a measurable operator field (B3). The bound and the equality are clear. Where $x(\gamma)\in M(\gamma)$, the operator $x^{[m]}(\gamma)$, which is $x(\gamma)$ or $0$, lies in $M(\gamma)$. So $\int^\oplus x^{[m]}\in\int^\oplus M$, and its adjoint lies there too by Proposition 2.2(2). $\square$

Theorem 3.2 (Commutant theorem). Let $M$ be a measurable field on $(H(\gamma))$, as in Definition 1.1.

1. $\big(\int^\oplus M\big)'=\int^\oplus M'$.
2. $\int^\oplus M$ is a von Neumann algebra acting on $\mathcal H$, and its centre contains the diagonal algebra $\mathcal A$.
3. $\int^\oplus M'$ is also a von Neumann algebra, with commutant $\big(\int^\oplus M'\big)'=\int^\oplus M$.

Proof. (1) The inclusion $\supseteq$. Let $S\in\int^\oplus M'$ and $T\in\int^\oplus M$. Write $S=\int^\oplus s$ and $T=\int^\oplus t$, with $s(\gamma)\in M(\gamma)'$ and $t(\gamma)\in M(\gamma)$ for almost every $\gamma$. Then $s(\gamma)t(\gamma)=t(\gamma)s(\gamma)$ for almost every $\gamma$, and Lemma 3.1(1) gives $ST=TS$. This part holds for every field $M$.

The inclusion $\subseteq$. Let $T\in(\int^\oplus M)'$. Since $\mathcal A\subseteq\int^\oplus M$ (Proposition 2.2(2)), $T$ commutes with $\mathcal A$, so $T\in\mathcal D$ by (B6); write $T=\int^\oplus t$. Choose measurable operator fields $x_j$ and a null set $N_0$ such that $M(\gamma)$ is generated by $\{x_j(\gamma)\}$ for $\gamma\notin N_0$. For all $j,m\geq1$, the operators $\int^\oplus x_j^{[m]}$ and $(\int^\oplus x_j^{[m]})^*=\int^\oplus(x_j^{[m]})^*$ lie in $\int^\oplus M$ (Lemma 3.1(2)), so $T$ commutes with them. By Lemma 3.1(1), there is a null set $N_{j,m}$ outside which $t(\gamma)$ commutes with $x_j^{[m]}(\gamma)$ and with $x_j^{[m]}(\gamma)^*$. Let $N$ be the union of $N_0$ and all the $N_{j,m}$; it is a null set. Fix $\gamma\notin N$ and $j$, and take $m\geq\|x_j(\gamma)\|$. Then $x_j^{[m]}(\gamma)=x_j(\gamma)$, so $t(\gamma)$ commutes with $x_j(\gamma)$ and $x_j(\gamma)^*$. Hence $t(\gamma)$ lies in $\{x_j(\gamma),x_j(\gamma)^*:j\geq1\}'$, which is $M(\gamma)'$ by (C3). Therefore $T\in\int^\oplus M'$.