Measurable fields of Hilbert spaces and their direct integrals

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 is a continuous version of a direct sum. One attaches a Hilbert space $H(\gamma)$ to each point $\gamma$ of a measure space and forms the space of square-integrable sections $\gamma\mapsto\xi(\gamma)\in H(\gamma)$. When all fibres equal one separable space $\mathcal K$, this is $L^2(\Gamma,\mu;\mathcal K)$. When the base is countable, it is a weighted direct sum. Direct integrals are the standard tool for decomposing representations and von Neumann algebras into simpler pieces. The spectral theorem, for instance, can be stated this way: a self-adjoint operator on a separable Hilbert space is unitarily equivalent to multiplication by the variable on a direct integral of Hilbert spaces over its spectrum.

The difficulty is measurability. The fibres are different spaces, their dimension may vary from point to point, and some of them may be zero. So there is no single space in which to measure a section. The remedy is to fix a space of sections that count as measurable, subject to three axioms. This lesson develops that framework over an arbitrary $\sigma$-finite measure space. Sections 2 to 5 treat fields of Hilbert spaces: the axioms, measurable orthonormal bases and the dimension function, a criterion that generates a field from a sequence of sections, fields of subspaces, and the identification of a field with a constant field on each set where the dimension is constant. Sections 6 and 7 treat measurable fields of bounded operators, conjugate fields and direct sums. Sections 8 to 10 construct the direct integral Hilbert space and study the diagonal and the decomposable operators on it. Section 11 works out four examples, among them the field of GNS spaces over the quasi-state space of a separable C*-algebra, and Section 12 has exercises with solutions.

We assume measure and integration theory and the elementary theory of Hilbert spaces. Example 11.2 uses the Hilbert tensor product from Spatial tensor products of von Neumann algebras, and Example 11.4 uses the GNS construction. The facts we use without proof are stated in full near the end, in the section Background used without proof.

Direct integrals go back to von Neumann's reduction theory (1949; see [Blackadar, Section III.1.6]), and the axioms for measurable fields used here are Dixmier's. Other basic references are [Takesaki I] and [Blackadar].

1. Conventions

Throughout, $(\Gamma,\Sigma,\mu)$ is a $\sigma$-finite measure space, and measurable means $\Sigma$-measurable. We do not assume that $\mu$ is complete or that $\Gamma$ is a standard Borel space, and we make no countability assumption on $\Sigma$, except in Theorem 9.1(3), which says so. When $\Sigma$ is complete for $\mu$, "measurable" means "$\mu$-measurable" in the usual sense.

Hilbert spaces are complex, inner products are linear in the first variable, and the zero space is allowed. The Gaussian rationals $\mathbb Q+i\mathbb Q$ form a countable dense subset of $\mathbb C$.

Let $(H(\gamma))_{\gamma\in\Gamma}$ be a family of Hilbert spaces. A section is a map $\xi$ with $\xi(\gamma)\in H(\gamma)$ for every $\gamma$. Under pointwise operations the sections form a vector space, $\prod_\gamma H(\gamma)$. For sections $\xi,\eta$ we write $\langle\xi,\eta\rangle$ for the function $\gamma\mapsto\langle\xi(\gamma),\eta(\gamma)\rangle$, and $\|\xi\|$ for the function $\gamma\mapsto\|\xi(\gamma)\|$.

2. Measurable fields of Hilbert spaces

Definition 2.1 (Measurable field of Hilbert spaces). A measurable field of Hilbert spaces over $(\Gamma,\Sigma,\mu)$ is a family $(H(\gamma))_{\gamma\in\Gamma}$ together with a linear subspace $\mathfrak M\subseteq\prod_\gamma H(\gamma)$, whose elements are called measurable sections, such that:

• (F1) for every $\xi\in\mathfrak M$, the function $\|\xi\|$ is measurable;
• (F2) (saturation) a section $\eta$ belongs to $\mathfrak M$ whenever $\langle\eta,\xi\rangle$ is measurable for every $\xi\in\mathfrak M$;
• (F3) there is a sequence $(\xi_n)_{n\geq1}$ in $\mathfrak M$ such that, for every $\gamma$, the vectors $\xi_n(\gamma)$ have dense linear span in $H(\gamma)$. Such a sequence is called fundamental.

Reference: [Takesaki I, Definition IV.8.9] works with $\mu$-measurable sections over a Borel space with a $\sigma$-finite measure, that is, with the completion of the Borel $\sigma$-algebra; Definition 2.1 allows any $\sigma$-algebra.

By (F3) every fibre $H(\gamma)$ is separable: the finite combinations of the $\xi_n(\gamma)$ with Gaussian-rational coefficients are dense. Exercise 12.1 shows that (F2) does not follow from (F1) and (F3).

Lemma 2.2 (Closure properties). Let $(H(\gamma)),\mathfrak M$ be a measurable field.

1. For $\xi,\eta\in\mathfrak M$, the function $\langle\xi,\eta\rangle$ is measurable.
2. If $f:\Gamma\to\mathbb C$ is measurable and $\xi\in\mathfrak M$, then $f\xi\in\mathfrak M$.
3. If $\xi_k\in\mathfrak M$ and $\xi_k(\gamma)\to\xi(\gamma)$ weakly in $H(\gamma)$ for every $\gamma$, then $\xi\in\mathfrak M$.
4. (Gluing) If $(\Gamma_k)$ is a countable measurable partition of $\Gamma$ and $\xi^{(k)}\in\mathfrak M$, the section equal to $\xi^{(k)}$ on $\Gamma_k$ belongs to $\mathfrak M$.

Proof. (1) Since $\mathfrak M$ is a linear subspace, $\xi+i^k\eta\in\mathfrak M$ for $k=0,1,2,3$. The polarization identity $\langle\xi,\eta\rangle=\tfrac14\sum_{k=0}^3 i^k\|\xi+i^k\eta\|^2$ expresses $\langle\xi,\eta\rangle$ through four functions that are measurable by (F1).

(2) For every $\eta\in\mathfrak M$, the function $\langle f\xi,\eta\rangle=f\langle\xi,\eta\rangle$ is measurable by (1). Saturation (F2) gives $f\xi\in\mathfrak M$.

(3) Let $\eta\in\mathfrak M$. Weak convergence gives $\langle\xi,\eta\rangle=\lim_k\langle\xi_k,\eta\rangle$ pointwise, and each $\langle\xi_k,\eta\rangle$ is measurable by (1). A pointwise limit of measurable functions is measurable, so (F2) gives $\xi\in\mathfrak M$.

(4) The partial sums $\sum_{k\leq m}1_{\Gamma_k}\xi^{(k)}$ lie in $\mathfrak M$ by (2). At each point they are eventually constant, because the point lies in exactly one $\Gamma_k$. So they converge pointwise to the glued section, and (3) applies. $\square$

3. Orthonormal fundamental sequences, dimension and generation

Theorem 3.1 (Orthonormal fundamental sequences). Let $(H(\gamma)),\mathfrak M$ be a measurable field with fundamental sequence $(\xi_j)$.

1. There is a sequence $(e_k)_{k\geq1}$ in $\mathfrak M$ with the following property. For every $\gamma$, let $n(\gamma)\in\{0,1,2,\ldots,\infty\}$ be the number of indices $k$ with $e_k(\gamma)\neq0$. Then $e_k(\gamma)\neq0$ exactly for $k\leq n(\gamma)$, and $(e_k(\gamma))_{k\leq n(\gamma)}$ is an orthonormal basis of $H(\gamma)$. In particular $n(\gamma)=\dim H(\gamma)$, and the dimension function $\gamma\mapsto\dim H(\gamma)$ is measurable.
2. Each $e_k$ has the form $e_k=\sum_j c_{kj}\xi_j$, where the $c_{kj}$ are measurable functions and, at each $\gamma$, only finitely many $c_{kj}(\gamma)$ are nonzero. The coefficients are built from the Gram functions $\langle\xi_j,\xi_i\rangle$ alone.
3. (Testing) A section $\eta$ belongs to $\mathfrak M$ if and only if $\langle\eta,\xi_j\rangle$ is measurable for every $j$.
4. (Generation) Conversely, let $(\xi_j)$ be any sequence of sections of a family $(H(\gamma))$ such that all Gram functions $\langle\xi_j,\xi_i\rangle$ are measurable and $(\xi_j(\gamma))_j$ is total in $H(\gamma)$ for every $\gamma$. Then $$\mathfrak M=\{\eta :\ \langle\eta,\xi_j\rangle\ \text{is measurable for every }j\} \tag{3.1}$$ is a measurable field with fundamental sequence $(\xi_j)$, and it is the only measurable field containing every $\xi_j$.

A sequence $(e_k)$ as in (1) is called an orthonormal fundamental sequence.

Proof. (1)–(2) The recursion. We build $e_1,e_2,\ldots$ recursively. At each stage $k$ we maintain two properties:

• at every $\gamma$, the nonzero vectors among $e_1(\gamma),\ldots,e_k(\gamma)$ are orthonormal and form an initial segment $e_1(\gamma),\ldots,e_m(\gamma)$;
• each $e_i$ with $i\leq k$ has the form stated in (2).

We start with $k=0$ and no vectors.

Suppose $e_1,\ldots,e_k$ have been built. Let $P_k(\gamma)v=\sum_{i\leq k}\langle v,e_i(\gamma)\rangle e_i(\gamma)$. By the orthonormality property, $P_k(\gamma)$ is the orthogonal projection onto the span of $e_1(\gamma),\ldots,e_k(\gamma)$. Put $r_j=\xi_j-P_k\xi_j$. These sections lie in $\mathfrak M$, because the coefficients $\langle\xi_j,e_i\rangle$ are measurable (Lemma 2.2(1)) and $\mathfrak M$ is closed under multiplication by measurable functions (Lemma 2.2(2)). By Pythagoras, $\|r_j\|^2=\|\xi_j\|^2-\sum_{i\leq k}|\langle\xi_j,e_i\rangle|^2$. Define $$E_j=\{\gamma :\ r_i(\gamma)=0\ \text{for } i<j,\ r_j(\gamma)\neq0\},$$ which are pairwise disjoint, and measurable because the norm functions $\|r_i\|$ are measurable. Set $$e_{k+1}=\sum_{j\geq1}1_{E_j}\,\|r_j\|^{-1}r_j . \tag{3.2}$$ Here the $j$-th term is read as $0$ off $E_j$. Each term lies in $\mathfrak M$ by Lemma 2.2(2), and at each point at most one term is nonzero. So the partial sums converge pointwise, and Lemma 2.2(3) gives $e_{k+1}\in\mathfrak M$.

We check the two properties at stage $k+1$.

• On $E_j$, $e_{k+1}(\gamma)$ is a unit vector orthogonal to $e_1(\gamma),\ldots,e_k(\gamma)$.
• Off $\bigcup_jE_j$, every $r_j(\gamma)$ vanishes. So every $\xi_j(\gamma)$ already lies in the span of $e_1(\gamma),\ldots,e_k(\gamma)$, and $e_{k+1}(\gamma)=0$.
• Suppose $e_k(\gamma)=0$. Then at stage $k$ the point $\gamma$ lay outside all the sets $E_j$, so every $\xi_j(\gamma)$ lay in the span of the earlier vectors $e_1(\gamma),\ldots,e_{k-1}(\gamma)$. Hence $\gamma\notin\bigcup_jE_j$ at stage $k+1$ as well, and $e_{k+1}(\gamma)=0$. So the initial-segment property persists.
• Expanding $P_k\xi_j$ by the form of $e_1,\ldots,e_k$ shows that $e_{k+1}$ again has the form (2): at each point only one $j$ contributes, and each $e_i$ has only finitely many nonzero coefficients there. The coefficients of $e_{k+1}$ involve only $1_{E_j}$, the norms $\|r_j\|$ and the functions $\langle\xi_j,e_i\rangle=\sum_l\overline{c_{il}}\langle\xi_j,\xi_l\rangle$. By the Pythagoras formula these are measurable functions of the Gram functions, and so are the sets $E_j$.

In particular, the recursion uses nothing about $\mathfrak M$ beyond the Gram functions. This is what (4) needs.

The vectors span. Fix $\gamma$. When $\gamma\in E_{j}$ at stage $k+1$, write $j_k=j$. After that stage, the vectors $\xi_1(\gamma),\ldots,\xi_{j_k}(\gamma)$ all lie in the span of $e_1(\gamma),\ldots,e_{k+1}(\gamma)$. Indeed, those with index below $j_k$ already lay in the smaller span, and $\xi_{j_k}=P_k\xi_{j_k}+r_{j_k}$, where $r_{j_k}(\gamma)$ is a multiple of $e_{k+1}(\gamma)$. Hence the indices $j_k$ strictly increase. If the recursion never produces a zero vector at $\gamma$, every $\xi_j(\gamma)$ therefore eventually lies in the span of finitely many $e_i(\gamma)$. If it produces $e_{m+1}(\gamma)=0$, every $\xi_j(\gamma)$ lies in the span of $e_1(\gamma),\ldots,e_m(\gamma)$. In both cases the closed span of the nonzero $e_i(\gamma)$ contains the total family $(\xi_j(\gamma))$, so it is $H(\gamma)$. This proves (1), apart from measurability of $n$, which follows from $\{n\geq k\}=\{\|e_k\|=1\}$.

(3) If $\eta\in\mathfrak M$, the functions $\langle\eta,\xi_j\rangle$ are measurable by Lemma 2.2(1). Conversely, suppose they are measurable. By (2), $\langle\eta,e_k\rangle=\sum_j\overline{c_{kj}}\langle\eta,\xi_j\rangle$. At each point this is a finite sum, so it is a pointwise limit of measurable functions, hence measurable. For $\zeta\in\mathfrak M$, Parseval's identity in $H(\gamma)$ gives $$\langle\eta,\zeta\rangle(\gamma)=\sum_{k\leq n(\gamma)}\langle\eta(\gamma),e_k(\gamma)\rangle\langle e_k(\gamma),\zeta(\gamma)\rangle =\sum_{k\geq1}\langle\eta,e_k\rangle(\gamma)\,\langle e_k,\zeta\rangle(\gamma),$$ since $e_k(\gamma)=0$ for $k>n(\gamma)$. The series converges absolutely at every point, so $\langle\eta,\zeta\rangle$ is measurable, and (F2) gives $\eta\in\mathfrak M$.

(4) Each $\xi_i$ belongs to $\mathfrak M$, since its inner products with the $\xi_j$ are Gram functions. Clearly $\mathfrak M$ is a linear subspace. Run the recursion of (1)–(2) on $(\xi_j)$. It uses only the Gram functions and the operations of (2), so it produces sections $e_k$ of the form (2) whose nonzero values form an orthonormal basis at every point. For $\eta\in\mathfrak M$, the functions $\langle\eta,e_k\rangle$ are measurable as in (3), hence so is $\|\eta\|^2=\sum_k|\langle\eta,e_k\rangle|^2$. This is (F1). For (F2), suppose $\langle\eta,\zeta\rangle$ is measurable for every $\zeta\in\mathfrak M$. Then it is measurable in particular for every $\zeta=\xi_j$, so $\eta\in\mathfrak M$. (F3) holds by hypothesis. Finally, let $\mathfrak M'$ be any measurable field containing all $\xi_j$. Then $\mathfrak M'\subseteq\mathfrak M$ by Lemma 2.2(1), applied in $\mathfrak M'$, and $\mathfrak M\subseteq\mathfrak M'$ by the testing criterion (3), applied to $\mathfrak M'$. $\square$

4. Fields of subspaces and measurable projections

Proposition 4.1 (Subspace fields). Let $(H(\gamma)),\mathfrak M$ be a measurable field with fundamental sequence $(\xi_n)$. Let $(\eta_j)$ be a sequence in $\mathfrak M$, and let $K(\gamma)$ be the closed linear span of $\{\eta_j(\gamma)\}$.

1. Let $P(\gamma)$ be the orthogonal projection of $H(\gamma)$ onto $K(\gamma)$. Then $P\xi\in\mathfrak M$ for every $\xi\in\mathfrak M$.
2. $(K(\gamma))$ with $\mathfrak M_K=\{\xi\in\mathfrak M:\ \xi(\gamma)\in K(\gamma)\ \text{for all }\gamma\}$ is a measurable field with fundamental sequence $(\eta_j)$. In particular $\gamma\mapsto\dim K(\gamma)$ is measurable.
3. The orthogonal complements $K(\gamma)^\perp$, with $\{\xi\in\mathfrak M:\ \xi(\gamma)\perp K(\gamma)\ \text{for all }\gamma\}$, form a measurable field with fundamental sequence $((1-P)\xi_n)$.

Proof. Run the recursion from the proof of Theorem 3.1 on the sequence $(\eta_j)$ inside $\mathfrak M$. Totality was used there only at the end, to identify the closed span of the new vectors with the whole fibre. So the recursion produces sections $f_k\in\mathfrak M$ whose nonzero values are orthonormal and have a closed span containing every $\eta_j(\gamma)$. Each $f_k(\gamma)$ lies in $K(\gamma)$, because the recursion gives $f_k$ the form of Theorem 3.1(2): at each point it is a finite combination of the $\eta_j(\gamma)$. So the nonzero values of the $f_k$ form an orthonormal basis of $K(\gamma)$ at each point. Then $P\xi=\sum_k\langle\xi,f_k\rangle f_k$ is a pointwise norm-convergent series. Its terms lie in $\mathfrak M$ by Lemma 2.2(1)–(2), so $P\xi\in\mathfrak M$ by Lemma 2.2(3). This proves (1).

For (2), $\mathfrak M_K$ is a linear subspace satisfying (F1). For (F2), let $\zeta$ be a section of $(K(\gamma))$ such that $\langle\zeta,\xi\rangle$ is measurable for every $\xi\in\mathfrak M_K$. For arbitrary $\xi\in\mathfrak M$ we have $\langle\zeta,\xi\rangle=\langle\zeta,P\xi\rangle$, because $\zeta(\gamma)\in K(\gamma)$, and $P\xi\in\mathfrak M_K$ by (1). So $\langle\zeta,\xi\rangle$ is measurable for every $\xi\in\mathfrak M$. Then (F2) for the ambient field gives $\zeta\in\mathfrak M$, hence $\zeta\in\mathfrak M_K$. The $\eta_j$ form a fundamental sequence by definition of $K(\gamma)$. The dimension function is measurable by Theorem 3.1(1), applied to this field.

Part (3) is the same argument with $1-P$ in place of $P$. The vectors $(1-P(\gamma))\xi_n(\gamma)$ are total in $K(\gamma)^\perp$, because the $\xi_n(\gamma)$ are total in $H(\gamma)$ and $1-P(\gamma)$ maps $H(\gamma)$ continuously onto $K(\gamma)^\perp$. $\square$

Remark 4.2 (Null sets). Suppose we only know that the vectors $\eta_j(\gamma)$ span a dense subspace of some intended subspace $K(\gamma)$ for $\gamma$ outside a measurable null set $N$. Apply Proposition 4.1 to the sections $1_{\Gamma\setminus N}\eta_j$. The resulting projection field is measurable, and it agrees with the projection onto $K(\gamma)$ almost everywhere. For operators on the direct integral this is enough, because measurable fields that agree almost everywhere define the same operator (Theorem 10.1(2)). This situation arises, for instance, when certain sections are known to be dense in the graphs of a field of closed operators only at almost every point.

5. Reduction to constant fields

For $d\in\{0,1,2,\ldots,\infty\}$, let $\ell^2_d$ be $\mathbb C^d$ ($\ell^2(\mathbb N)$ when $d=\infty$) with standard orthonormal basis $(\varepsilon_k)$.

Theorem 5.1 (Reduction to constant fields). Let $(H(\gamma)),\mathfrak M$ be a measurable field, let $n(\gamma)=\dim H(\gamma)$, and let $\Gamma_d=\{\gamma: n(\gamma)=d\}$. These sets form a countable measurable partition of $\Gamma$. Let $(e_k)$ be an orthonormal fundamental sequence as in Theorem 3.1(1), and define for $\gamma\in\Gamma_d$ $$U(\gamma):H(\gamma)\to\ell^2_d,\qquad U(\gamma)v=\sum_{k\leq d}\langle v,e_k(\gamma)\rangle\varepsilon_k . \tag{5.1}$$ Each $U(\gamma)$ is unitary. A section $\xi$ is measurable if and only if, for every $d$, all coordinates $\gamma\mapsto\langle U(\gamma)\xi(\gamma),\varepsilon_k\rangle$ are measurable on $\Gamma_d$. Consequently, the restriction of the field to $\Gamma_d$ is carried by the measurable unitary field $U$ onto the constant field $\ell^2_d$, whose measurable sections are the maps $\Gamma_d\to\ell^2_d$ with measurable coordinates.

Proof. The partition is measurable because the dimension function is measurable (Theorem 3.1(1)). $U(\gamma)$ maps an orthonormal basis onto an orthonormal basis, so it is unitary. On $\Gamma_d$, the coordinates of $U\xi$ are the functions $\langle\xi,e_k\rangle$, $k\leq d$. If $\xi$ is measurable, they are measurable by Lemma 2.2(1). Conversely, suppose they are measurable on every $\Gamma_d$, and fix $k$. On $\Gamma_d$ with $d\geq k$, the function $\langle\xi,e_k\rangle$ is one of these coordinates. On $\Gamma_d$ with $d<k$ it vanishes, because $e_k(\gamma)=0$ there. So $\langle\xi,e_k\rangle=\sum_d1_{\Gamma_d}\langle\xi,e_k\rangle$ is measurable on $\Gamma$. The Parseval argument in the proof of Theorem 3.1(3) then shows that $\langle\xi,\zeta\rangle$ is measurable for every $\zeta\in\mathfrak M$, and (F2) gives $\xi\in\mathfrak M$. For the constant field, the constant sections $\varepsilon_k$ form a fundamental sequence with constant Gram functions, and Theorem 3.1(4) identifies its measurable sections with the maps that have measurable coordinates. $\square$

Remark 5.2. A map into the separable space $\ell^2_d$ with measurable coordinates is weakly measurable, by Parseval's identity. By Pettis's theorem (see Background used without proof), it is then Borel measurable and a pointwise limit of measurable maps with finitely many values. So in the constant case the usual notions of measurability agree. One can also embed all fibres at once into one fixed infinite-dimensional Hilbert space by a measurable field of isometries. That form uses a Borel structure on the set of closed subspaces, and it is treated in The Effros Borel structure. The form above, on the sets of constant dimension, needs no Borel structure on spaces of subspaces, and it is the form used in later lessons.

6. Measurable fields of bounded operators

Definition 6.1 (Measurable operator field). Let $(H(\gamma)),\mathfrak M$ and $(K(\gamma)),\mathfrak N$ be measurable fields. A family $x=(x(\gamma))$ with $x(\gamma)\in B(H(\gamma),K(\gamma))$ is a measurable field of bounded operators if the section $x\xi:\gamma\mapsto x(\gamma)\xi(\gamma)$ belongs to $\mathfrak N$ for every $\xi\in\mathfrak M$. No bound on $\|x(\gamma)\|$ that is uniform in $\gamma$ is assumed.

Theorem 6.2 (Properties of measurable operator fields). Let $x$ be a family of bounded operators as in Definition 6.1.

1. (Testing) $x$ is measurable if and only if $x\xi_j\in\mathfrak N$ for every member of one fundamental sequence $(\xi_j)$ of $\mathfrak M$.
2. (Adjoints) If $x$ is measurable, so is $x^*=(x(\gamma)^*)$ from $(K(\gamma))$ to $(H(\gamma))$.
3. (Norms) If $x$ is measurable, the function $\gamma\mapsto\|x(\gamma)\|\in[0,\infty)$ is measurable.
4. (Algebra) Sums, products with measurable scalar functions, and composites of measurable operator fields are measurable. The identity field and the projection fields of Proposition 4.1 are measurable.

Proof. (1) Necessity is clear. For sufficiency, apply Theorem 3.1(2) to the fundamental sequence $(\xi_j)$. It gives $e_k=\sum_jc_{kj}\xi_j$ with measurable coefficients, only finitely many of them nonzero at each point. Hence $xe_k=\sum_jc_{kj}\,x\xi_j$ is a countable sum of sections of $\mathfrak N$, with finitely many nonzero terms at each point, and it lies in $\mathfrak N$ by Lemma 2.2(2)–(3). Now let $\xi\in\mathfrak M$. The expansion $\xi(\gamma)=\sum_k\langle\xi,e_k\rangle(\gamma)e_k(\gamma)$ converges in norm, and $x(\gamma)$ is bounded, so $$x\xi=\sum_k\langle\xi,e_k\rangle\,xe_k$$ converges pointwise in norm. Each term lies in $\mathfrak N$ by Lemma 2.2(1)–(2), so $x\xi\in\mathfrak N$ by Lemma 2.2(3).

(2) For $\eta\in\mathfrak N$ and $\xi\in\mathfrak M$, $\langle x^*\eta,\xi\rangle=\langle\eta,x\xi\rangle$, which is measurable by Lemma 2.2(1) because $x\xi\in\mathfrak N$. Saturation (F2) for $\mathfrak M$ gives $x^*\eta\in\mathfrak M$.

(3) Let $(e_k)$ be an orthonormal fundamental sequence of $\mathfrak M$ (Theorem 3.1(1)). Let $Q$ be the countable set of finitely supported sequences $q=(q_1,\ldots,q_m)$ of Gaussian rationals with $\sum|q_k|^2\leq1$, and let $v_q=\sum_kq_ke_k\in\mathfrak M$. At every $\gamma$, $\|v_q(\gamma)\|\leq1$, because the nonzero $e_k(\gamma)$ are orthonormal and the others vanish. Moreover $\{v_q(\gamma):q\in Q\}$ is dense in the closed unit ball of $H(\gamma)$. Since $x(\gamma)$ is continuous, $$\|x(\gamma)\|=\sup_{q\in Q}\|x(\gamma)v_q(\gamma)\|, \tag{6.1}$$ a countable supremum of functions that are measurable by (F1) for $\mathfrak N$. If $H(\gamma)=0$, both sides are $0$.

(4) Sums and products with measurable scalar functions are measurable because $\mathfrak N$ is a linear subspace closed under multiplication by measurable functions (Lemma 2.2(2)). Composites are measurable because $(yx)\xi=y(x\xi)$. The identity field is measurable by definition, and the projection fields are measurable by Proposition 4.1(1). $\square$

7. Conjugate fields, direct sums and block operators

Proposition 7.1 (Conjugates, direct sums and blocks). Let $(H(\gamma)),\mathfrak M$ and $(K(\gamma)),\mathfrak N$ be measurable fields with fundamental sequences $(\xi_n)$ and $(\eta_n)$.

1. (Conjugate field) Let $\overline{H(\gamma)}$ be the conjugate Hilbert space, the same set with scalar multiplication $\lambda\cdot\overline v=\overline{\bar\lambda v}$ and $\langle\overline v,\overline w\rangle=\langle w,v\rangle$. Let $\overline{\mathfrak M}=\{\overline\xi:\xi\in\mathfrak M\}$, where $\overline\xi(\gamma)=\overline{\xi(\gamma)}$. Then $(\overline{H(\gamma)}),\overline{\mathfrak M}$ is a measurable field with fundamental sequence $(\overline{\xi_n})$. The canonical antiunitaries $\kappa(\gamma):v\mapsto\overline v$ carry measurable sections exactly onto measurable sections.
2. (Direct sums) $(H(\gamma)\oplus K(\gamma))$ with $\mathfrak M\oplus\mathfrak N=\{(\xi,\eta):\xi\in\mathfrak M,\eta\in\mathfrak N\}$ is a measurable field. A fundamental sequence is obtained by interleaving $(\xi_n,0)$ and $(0,\eta_n)$.
3. (Blocks) A field of bounded operators on $(H(\gamma)\oplus K(\gamma))$, or between two such direct sums, is measurable if and only if its four matrix blocks are measurable.

Proof. (1) $\overline{\mathfrak M}$ is a complex linear subspace, since $\lambda\cdot\overline\xi=\overline{\bar\lambda\xi}$, and $\|\overline\xi\|=\|\xi\|$ gives (F1). Every section of the conjugate family has the form $\overline\zeta$ for a unique section $\zeta$ of $(H(\gamma))$. If $\langle\overline\zeta,\overline\xi\rangle=\langle\xi,\zeta\rangle$ is measurable for every $\xi\in\mathfrak M$, then so is its complex conjugate $\langle\zeta,\xi\rangle$, so $\zeta\in\mathfrak M$ and $\overline\zeta\in\overline{\mathfrak M}$. This is (F2). Totality of $(\overline{\xi_n(\gamma)})$ is the same statement as totality of $(\xi_n(\gamma))$. The last clause restates the definition of $\overline{\mathfrak M}$.

(2) (F1) holds since $\|(\xi,\eta)\|^2=\|\xi\|^2+\|\eta\|^2$. For (F2), suppose $\langle(\zeta_1,\zeta_2),(\xi,\eta)\rangle$ is measurable for all measurable pairs. Taking $\eta=0$ shows that $\langle\zeta_1,\xi\rangle$ is measurable for every $\xi\in\mathfrak M$, so $\zeta_1\in\mathfrak M$; similarly $\zeta_2\in\mathfrak N$. Totality of the interleaved sequence is clear.

(3) The inclusions $\xi\mapsto(\xi,0)$ and $\eta\mapsto(0,\eta)$ and the coordinate projections are measurable operator fields by (2). Each block is a composite of the given field with these, hence measurable by Theorem 6.2(4). Conversely, suppose the four blocks $x_{11},x_{12},x_{21},x_{22}$ are measurable. Then the field maps a measurable section $(\xi,\eta)$ to $(x_{11}\xi+x_{12}\eta,\ x_{21}\xi+x_{22}\eta)$, which is measurable by (2). $\square$

8. The direct integral Hilbert space

Definition 8.1 (Direct integral). Let $(H(\gamma)),\mathfrak M$ be a measurable field, and let $\mathcal L^2$ be the set of $\xi\in\mathfrak M$ with $\int_\Gamma\|\xi(\gamma)\|^2\,d\mu(\gamma)<\infty$. The direct integral $$\int_\Gamma^\oplus H(\gamma)\,d\mu(\gamma)$$ is $\mathcal L^2$ modulo equality $\mu$-almost everywhere, with $$\langle\xi,\eta\rangle=\int_\Gamma\langle\xi(\gamma),\eta(\gamma)\rangle\,d\mu(\gamma). \tag{8.1}$$

We use the same letter for a section in $\mathcal L^2$ and for its class, and we also write $\|\xi\|$ for the norm of $\xi$ in the direct integral. When $\|\xi\|$ appears as a function of $\gamma$, or inside a set such as $\{\|\xi\|\leq m\}$, it is the pointwise norm function of Section 1.

Theorem 8.2 (Completeness).

1. (8.1) is a well-defined inner product.
2. The direct integral is complete, hence a Hilbert space.
3. (Almost-everywhere subsequences) If $\xi_k\to\xi$ in the direct integral, some subsequence satisfies $\xi_{k_j}(\gamma)\to\xi(\gamma)$ in $H(\gamma)$ for almost every $\gamma$.

Proof. (1) The integrand is measurable by Lemma 2.2(1). It is integrable, since $|\langle\xi(\gamma),\eta(\gamma)\rangle|\leq\|\xi(\gamma)\|\|\eta(\gamma)\|$ and the right side is integrable by the Cauchy–Schwarz inequality in $L^2(\Gamma,\mu)$. $\mathcal L^2$ is a linear subspace because $\|\xi+\eta\|^2\leq2\|\xi\|^2+2\|\eta\|^2$ pointwise. Changing $\xi$ on a null set does not change (8.1), and $\langle\xi,\xi\rangle=0$ forces $\xi=0$ almost everywhere.

(2) Let $(\xi_k)$ be a Cauchy sequence, with representatives in $\mathcal L^2$. Choose $k_1<k_2<\cdots$ with $\|\xi_{k_{j+1}}-\xi_{k_j}\|\leq2^{-j}$, and put $$G(\gamma)=\sum_{j\geq1}\|\xi_{k_{j+1}}(\gamma)-\xi_{k_j}(\gamma)\|\in[0,\infty].$$ $G$ is measurable, as a countable sum of nonnegative measurable functions. The $j$-th term of $G$ has norm $\|\xi_{k_{j+1}}-\xi_{k_j}\|\leq2^{-j}$ in $L^2(\Gamma,\mu)$. By Minkowski's inequality in $L^2(\Gamma,\mu)$ and monotone convergence, $\|G\|_{L^2}\leq\sum_j2^{-j}\leq1$. So $N=\{G=\infty\}$ is a measurable null set. For $\gamma\notin N$, the sequence $(\xi_{k_j}(\gamma))_j$ is Cauchy in $H(\gamma)$, because the distances between consecutive terms have a finite sum. Let $\xi(\gamma)$ be its limit, and set $\xi(\gamma)=0$ on $N$. Then $\xi$ is the pointwise limit of the sections $1_{\Gamma\setminus N}\,\xi_{k_j}$, so $\xi\in\mathfrak M$ by Lemma 2.2(2)–(3). For each $i$, Fatou's lemma gives $$\int\|\xi-\xi_{k_i}\|^2\,d\mu\leq\liminf_{j\to\infty}\int\|\xi_{k_j}-\xi_{k_i}\|^2\,d\mu=\liminf_{j\to\infty}\|\xi_{k_j}-\xi_{k_i}\|^2 .$$ The right side is finite, and it tends to $0$ as $i\to\infty$ because the sequence is Cauchy. Hence $\xi\in\mathcal L^2$ and $\xi_{k_i}\to\xi$. A Cauchy sequence with a convergent subsequence converges, so $\xi_k\to\xi$.

(3) A convergent sequence is Cauchy. Applied to $(\xi_k)$, the proof of (2) produces a subsequence that converges at almost every point to a section representing the limit of the sequence, and that limit is $\xi$. $\square$

9. Diagonal operators, localization and separability

Theorem 9.1 (Diagonal operators, localization, separability). Let $(H(\gamma)),\mathfrak M$ be a measurable field.

1. (Diagonal operators) For $f\in L^\infty(\Gamma,\mu)$, the formula $(m_f\xi)(\gamma)=f(\gamma)\xi(\gamma)$ defines a bounded operator on the direct integral with $\|m_f\|\leq\|f\|_\infty$. The map $f\mapsto m_f$ is a unital $*$-homomorphism.
2. (Finite-measure localization) Let $(\xi_n)$ be a fundamental sequence. For every $E\in\Sigma$ with $\mu(E)<\infty$ and every $n,m$, the section $1_{E\cap\{\|\xi_n\|\leq m\}}\,\xi_n$ belongs to $\mathcal L^2$. A vector of the direct integral orthogonal to all of them is zero, so their linear span is dense.
3. (Separability) Suppose $\Sigma$ is countably generated modulo $\mu$-null sets: there is a countable algebra $\mathcal A\subseteq\Sigma$ such that every set in $\Sigma$ differs by a null set from a set in the $\sigma$-algebra generated by $\mathcal A$. This holds, for example, on a standard Borel space. Then the direct integral is separable.

Proof. (1) $f\xi\in\mathfrak M$ by Lemma 2.2(2), and $\|f\xi\|^2\leq\|f\|_\infty^2\|\xi\|^2$ pointwise almost everywhere. Changing $f$ on a null set does not change $m_f$. Pointwise, $\langle f\xi,\eta\rangle=\langle\xi,\bar f\eta\rangle$ and $(fg)\xi=f(g\xi)$, so $m_f^*=m_{\bar f}$, $m_{fg}=m_fm_g$ and $m_1=1$.

(2) The integral of $\|1_{E\cap\{\|\xi_n\|\leq m\}}\xi_n\|^2$ is at most $m^2\mu(E)$, so these sections lie in $\mathcal L^2$. Let $\eta$ be a vector of the direct integral orthogonal to all of them. Put $h_{n,m}=1_{\{\|\xi_n\|\leq m\}}\langle\eta,\xi_n\rangle$. This function is measurable, and $|h_{n,m}|\leq m\|\eta\|$ pointwise. Since the function $\|\eta\|$ lies in $L^2(\Gamma,\mu)$, $h_{n,m}$ is integrable over every set of finite measure. The orthogonality hypothesis says exactly that $\int_Eh_{n,m}\,d\mu=0$ whenever $\mu(E)<\infty$. By $\sigma$-finiteness there are sets $F_k$ of finite measure with union $\Gamma$. Apply the hypothesis to $E=F_k\cap\{\operatorname{Re}h_{n,m}>0\}$. Taking real parts, the integral of $\operatorname{Re}h_{n,m}$ over this set vanishes, while $\operatorname{Re}h_{n,m}>0$ on it, so the set is null. Hence $\{\operatorname{Re}h_{n,m}>0\}$ is null. The same argument applies to $\{\operatorname{Re}h_{n,m}<0\}$, $\{\operatorname{Im}h_{n,m}>0\}$ and $\{\operatorname{Im}h_{n,m}<0\}$, so $h_{n,m}=0$ almost everywhere. Discarding countably many null sets, we get $\langle\eta(\gamma),\xi_n(\gamma)\rangle=0$ for all $n$ at almost every $\gamma$, since every point lies in $\{\|\xi_n\|\leq m\}$ for large $m$. Totality of $(\xi_n(\gamma))$ gives $\eta(\gamma)=0$ almost everywhere.

(3) Fix an increasing sequence $(F_k)$ of sets of finite measure with union $\Gamma$. For $E$ of finite measure, $\|1_{E\setminus F_k}1_{\{\|\xi_n\|\leq m\}}\xi_n\|^2\leq m^2\mu(E\setminus F_k)\to0$ as $k\to\infty$. So it suffices to approximate the vectors of (2) with $E\subseteq F_k$ for some $k$.

Fix $k$. We show first that every measurable $E\subseteq F_k$ can be approximated by sets from $\mathcal A$: for every $\delta>0$ there is $A\in\mathcal A$ with $\mu\big((A\cap F_k)\,\triangle\,E\big)<\delta$. Call a set $B\in\Sigma$ approximable if for every $\delta>0$ there is $A\in\mathcal A$ with $\mu\big((A\,\triangle\,B)\cap F_k\big)<\delta$.

• Every set in $\mathcal A$ is approximable.
• The complement of an approximable set is approximable, because $\mathcal A$ is closed under complements and $(\Gamma\setminus A)\,\triangle\,(\Gamma\setminus B)=A\,\triangle\,B$.
• A countable union $B=\bigcup_iB_i$ of approximable sets is approximable. Indeed, since $\mu(F_k)<\infty$, there is $N$ with $\mu\big((B\setminus\bigcup_{i\leq N}B_i)\cap F_k\big)<\delta/2$. Choose $A_i\in\mathcal A$ with $\mu\big((A_i\,\triangle\,B_i)\cap F_k\big)<\delta/(2N)$ for $i\leq N$. Then $A=\bigcup_{i\leq N}A_i$ lies in $\mathcal A$, and $\mu\big((A\,\triangle\,B)\cap F_k\big)<\delta$.

So the approximable sets form a $\sigma$-algebra containing $\mathcal A$, and every set in the $\sigma$-algebra generated by $\mathcal A$ is approximable. Now let $E\subseteq F_k$ be measurable, and choose $B$ in the $\sigma$-algebra generated by $\mathcal A$ with $\mu(E\,\triangle\,B)=0$. Since $E\subseteq F_k$, we have $(A\cap F_k)\,\triangle\,E\subseteq\big((A\,\triangle\,B)\cap F_k\big)\cup(B\,\triangle\,E)$ for every $A$. So a set $A\in\mathcal A$ that approximates $B$ within $\delta$ also satisfies $\mu\big((A\cap F_k)\,\triangle\,E\big)<\delta$.

For such $E$ and $A$, $$\big\|(1_E-1_{A\cap F_k})\,1_{\{\|\xi_n\|\leq m\}}\xi_n\big\|^2\leq m^2\,\mu\big(E\,\triangle\,(A\cap F_k)\big).$$ So the countable family of vectors $q\,1_{A\cap F_k\cap\{\|\xi_n\|\leq m\}}\xi_n$, with $q$ Gaussian rational, $A\in\mathcal A$ and $k,n,m\geq1$, approximates the spanning vectors of (2). Their finite sums form a countable dense set. $\square$

10. Decomposable operators

Theorem 10.1 (Decomposable operators). Let $(H(\gamma)),\mathfrak M$ and $(K(\gamma)),\mathfrak N$ be measurable fields, and let $x$ be a measurable field of bounded operators from $(H(\gamma))$ to $(K(\gamma))$ with $\gamma\mapsto\|x(\gamma)\|$ essentially bounded. (That function is measurable by Theorem 6.2(3).) Then:

1. $(x\xi)(\gamma)=x(\gamma)\xi(\gamma)$ defines a bounded operator $$\int_\Gamma^\oplus x(\gamma)\,d\mu(\gamma):\ \int_\Gamma^\oplus H(\gamma)\,d\mu\longrightarrow\int_\Gamma^\oplus K(\gamma)\,d\mu,$$ and its norm is exactly $\operatorname*{ess\,sup}_\gamma\|x(\gamma)\|$.
2. The assignment $x\mapsto\int^\oplus x$ is linear and multiplicative on composable fields, and $\big(\int^\oplus x\big)^*=\int^\oplus x^*$. Two fields define the same operator if and only if they agree almost everywhere.
3. When $K=H$, every such operator commutes with every diagonal operator $m_f$ of Theorem 9.1(1). Operators of this form are called decomposable, and the $m_f$ are the diagonal operators.