Decomposable operators are the commutant of diagonal multiplication

Fix $f\in L^{\mathrm{\infty }}$ and take a Borel representative, available from its measurable-function quotient [F19]. The scalar field $R_x=f(x)I_{H_x}$ is weakly measurable: $f{e}_n$ is a measurable section by [F1, F7], so each matrix coefficient $⟨f(x)e_n(x),e_m(x)⟩$ is measurable. Also $\mathrm{\parallel }{R}_x\mathrm{\parallel }\le \mathrm{\mid }f(x)\mathrm{\mid }$, so [F12, F15] make it essentially bounded. The action theorem [F5] therefore defines a bounded multiplication operator $M_f$; changing the representative on a null set does not change its action on direct-integral classes by [F2, F19]. Now let $T$ be induced by any weakly measurable essentially bounded field $(T_x)$. Pointwise linearity gives $T_x(f(x)\xi (x))=f(x)T_x\xi (x)$ for every measurable section $\xi$, and the action theorem makes both operators bounded. Hence $T{M}_f=M_{f}T$ on every class, proving the decomposable inclusion.

For the converse, fix $S\in {\mathcal{D}}^{\mathrm{\prime }}$ and write $C=\mathrm{\parallel }S\mathrm{\parallel }$. Enumerate by the finite-sequence coding and the countability of $\mathbb{Q}+i\mathbb{Q}$ all finite rational-complex combinations $q_n$ of the fundamental sections $e_j$. They are measurable by [F1, F7] and pointwise dense in each fibre by [F6, F13]. Define $b(0)=0$ and $b(t)=1\mathrm{/}t$ for $t>0$. The reciprocal map on $(0,\mathrm{\infty })$ is continuous, so its Borel preimages are Borel by [F18]; separating the Borel singleton $\{0\}$ shows that $b$ is Borel. Set $u_n(x)=b(\mathrm{\parallel }{q}_n(x)\mathrm{\parallel })q_n(x)$. By [F7, F17], each $u_n$ is a measurable section; it has norm one where $q_n(x)\mathrm{\ne }0$ and is zero otherwise. Its nonzero values are dense in the unit sphere of every nonzero fibre: if $z$ is a unit vector, choose rational-complex combinations $q_n$ arbitrarily close to $z$ using [F6, F13]; eventually they are nonzero, and $\mathrm{\parallel }{q}_n\mathrm{/}\mathrm{\parallel }{q}_n\mathrm{\parallel }-z\mathrm{\parallel }\le 2\mathrm{\parallel }{q}_n-z\mathrm{\parallel }$. Thus $\mathrm{\parallel }{u}_n(x)\mathrm{\parallel }\le 1$ and their complex span is dense in every fibre. Choose an increasing Borel exhaustion $E_k\uparrow X$ with $\mu (E_k)<\mathrm{\infty }$ by [F8]. By [F9, F16, F22], ${\int }_X\mathrm{\parallel }{\mathbf{1}}_{E_k}{u}_n{\mathrm{\parallel }}^{2}d\mu \le \mu (E_k)<\mathrm{\infty }$, so $[{\mathbf{1}}_{E_k}{u}_n]\in \mathcal{H}$ by [F24]. For each pair $(k,n)$ choose a measurable square-integrable representative ${\eta }_{k,n}$ of $S[{\mathbf{1}}_{E_k}{u}_n]$ by the defining prequotient space [F24]. This is a countable choice, supplied by AC.

If $\mathrm{\ell }\ge k$, commutation with $M_{{\mathbf{1}}_{E_k}}$, which exists by step 1.1, gives $[{\mathbf{1}}_{E_k}{\eta }_{\mathrm{\ell },n}]=[{\eta }_{k,n}]$ because $S\in {\mathcal{D}}^{\mathrm{\prime }}$ by step 1.2 and [F4]. The set where these two sections differ is Borel by [F7] and null by [F2]. Take the countable union over $k,n,\mathrm{\ell }$; it is Borel null by [F11]. Off that set the representatives agree on every overlap. Define $v_n$ on the disjoint layers $E_k\setminus E_{k-1}$ by ${\eta }_{k,n}$, and set $v_n=0$ on the exceptional null set (with $E_0=\mathrm{\varnothing }$). For every fundamental index $m$, each coefficient $⟨v_n,e_m⟩$ is Borel on every layer by [F1]; the preimage of a Borel set is the countable union of its Borel preimages intersected with the layers, together with its preimage on the Borel exceptional set. Thus all fundamental coefficients of $v_n$ are Borel, so $v_n$ is a measurable section by [F1]. On every $E_k$ it represents the local action of $S$ on ${\mathbf{1}}_{E_k}{u}_n$.

Let $q={\sum }_{j=1}^m{c}_j{u}_{n_j}$, where the coefficients are in $\mathbb{Q}+i\mathbb{Q}$; these tests form a countable family by [F13]. The section $q$ is bounded by the construction in step 1.2. Its squared norm and the squared norm of $v_q$ are measurable by [F20], so the displayed integrals are defined. For every Borel $A\subseteq E_k$, step 2.1 and commutation with $M_{{\mathbf{1}}_A}$, which exists by step 1.1, show that $S[{\mathbf{1}}_{A}q]=[{\mathbf{1}}_A{v}_q]$: first apply the local representative identity to each $u_{n_j}$ on $E_k$, then use commutation and linearity. Thus the operator norm bound [F23] gives ${\int }_A\mathrm{\parallel }{v}_q(x){\mathrm{\parallel }}^{2}d\mu (x)=\mathrm{\parallel }S[{\mathbf{1}}_{A}q]{\mathrm{\parallel }}^2\le C^2\mathrm{\parallel }{\mathbf{1}}_{A}q{\mathrm{\parallel }}^2=C^2{\int }_A\mathrm{\parallel }q(x){\mathrm{\parallel }}^{2}d\mu (x)\mathrm{.}$ Both integrands are integrable on $E_k$: $q$ is bounded on a finite-measure set by [F9, F22], and $v_q$ represents the image under the bounded operator $S$ of $[{\mathbf{1}}_{E_k}q]$, so belongs to the prequotient space [F24].

Put $g=\mathrm{\parallel }{v}_q{\mathrm{\parallel }}^2$ and $h=C^2\mathrm{\parallel }q{\mathrm{\parallel }}^2$ on $E_k$. Their measurability and that of $g-h$ follow from [F20], so $A_{ϵ}=\{x\in E_k:g(x)-h(x)\ge ϵ\}$ is Borel by [F21] for each positive rational $ϵ$. The localized inequality, monotonicity, additivity, and the simple integral of an indicator give ${\int }_{A_{ϵ}}hd\mu +ϵ\mu (A_{ϵ})\le {\int }_{A_{ϵ}}gd\mu \le {\int }_{A_{ϵ}}hd\mu \mathrm{.}$ Here [F22] identifies each integral over $A_{ϵ}$ with the integral of the corresponding indicator product. Thus $\mu (A_{ϵ})=0$. Density of the rationals [F13] gives $\{g>h\}={\bigcup }_{ϵ\in \mathbb{Q},ϵ>0}{A}_{ϵ}$, so $g\le h$ almost everywhere on $E_k$. There are only countably many $q,k,ϵ$, by [F13]; one Borel null set makes all inequalities hold simultaneously. Enlarge the gluing exceptional set by this null set and replace every $v_n$ by zero on it; this preserves measurability and each local equivalence class. Consequently, outside one Borel null set $\parallel {\sum }_{j=1}^m{c}_j{v}_{n_j}(x)\parallel \le C\parallel {\sum }_{j=1}^m{c}_j{u}_{n_j}(x)\parallel$ for every rational-complex finite combination.

Off the common exceptional null set, the rational-complex span of the $u_n(x)$ is dense: their complex span is dense by step 1.2 and [F6], and rational-complex coefficients approximate every complex coefficient by [F13]. Define $T_x({\sum }_j{c}_j{u}_{n_j}(x))={\sum }_j{c}_j{v}_{n_j}(x)$ for $c_j\in \mathbb{Q}+i\mathbb{Q}$. The pointwise estimate in step 4.1 makes this assignment well-defined and bounded by $C$; it is $\mathbb{Q}+i\mathbb{Q}$-linear. Its unique continuous extension to $H_x$, which exists by fibre completeness, is complex-linear because $\mathbb{Q}+i\mathbb{Q}$ is dense in $\mathbb{C}$. Thus it is a bounded operator $T_x\in \mathcal{B}(H_x)$ with $\mathrm{\parallel }{T}_x\mathrm{\parallel }\le C$ by [F23]. Set $T_x=0$ on the exceptional set and on zero fibres. For each fundamental vector $e_n$, the fixed finite-sequence code has an index $j(n)$ for the one-term combination $e_n$; hence $T_x{e}_n(x)=\mathrm{\parallel }{e}_n(x)\mathrm{\parallel }{v}_{j(n)}(x)$, with both sides zero where $e_n(x)=0$. By [F7], these are measurable sections and their pairings with every $e_m$ are measurable. Thus $(T_x)$ is weakly measurable by [F5], and $\mathrm{\parallel }{T}_x\mathrm{\parallel }\le C$ everywhere, so it is essentially bounded by [F12].

The action theorem induces $\widehat{S}={\int }_X^{\oplus }{T}_{x}d\mu (x)$ and gives $\mathrm{\parallel }\widehat{S}\mathrm{\parallel }={\mathrm{ess\hspace{0.17em}sup}}_x\mathrm{\parallel }{T}_x\mathrm{\parallel }\le C$. For every $k,n$, the representatives agree outside a null set, so $\widehat{S}[{\mathbf{1}}_{E_k}{u}_n]=S[{\mathbf{1}}_{E_k}{u}_n]$. Both operators commute with $\mathcal{D}$ by step 1.1 and the assumption on $S$, so they agree on all bounded scalar localizations of these sections.

If two weakly measurable essentially bounded fields induce the same operator, [F5] makes their actions agree on every $[{\mathbf{1}}_{E_k}{u}_n]$. The quotient definition supplies a Borel null set for each such equality; the countable union is null by [F11]. Off it the fields agree on every $u_n(x)$, hence on their dense span by step 1.2 and [F6], and then on $H_x$ by boundedness. On zero fibres both fields are the zero operator. This proves uniqueness up to a null set.

These localizations have dense linear span in $\mathcal{H}$. Indeed, for $[\xi ]\in \mathcal{H}$ and $\epsilon >0$, [F10] lets us first restrict to some finite-measure $E_k$ and then to $F=E_k\cap \{\mathrm{\parallel }\xi \mathrm{\parallel }\le R\}$ with arbitrarily small $L^2$ error. On $F\cap \{\xi \mathrm{\ne }0\}$, the normalized section is $b(\mathrm{\parallel }\xi \mathrm{\parallel })\xi$, where the Borel reciprocal function $b$ was constructed in step 1.2; it is measurable by [F7, F17]. For any $\delta >0$, density of the $u_n(x)$ in the unit sphere, proved in step 1.2, gives a countable measurable cover by sets $\{x:\mathrm{\parallel }b(\mathrm{\parallel }\xi (x)\mathrm{\parallel })\xi (x)-u_n(x)\mathrm{\parallel }<\delta \}$. Assign each point the least qualifying $n$; this is a measurable partition. Its first $N$ pieces exhaust all but a set on which the $L^2$ norm of $\xi$ tends to zero, by [F10]. For each retained partition set $P_n$, the section ${\mathbf{1}}_{P_n}\mathrm{\parallel }\xi \mathrm{\parallel }{u}_n$ equals $M_{{\varphi }_n}[{\mathbf{1}}_{E_k}{u}_n]$ for ${\varphi }_n={\mathbf{1}}_{P_n}\mathrm{\parallel }\xi \mathrm{\parallel }$; this is a member of $L^{\mathrm{\infty }}$ by [F15, F19], since $P_n\subseteq F\subseteq E_k$ has finite measure and $\mathrm{\parallel }\xi \mathrm{\parallel }\le R$ there. Thus the finite sum over the first $N$ pieces lies in the span on which $S$ and $\widehat{S}$ agree by steps 1.1 and 6.1. On retained pieces its pointwise error is below $\delta \mathrm{\parallel }\xi \mathrm{\parallel }$; on the discarded tail its error is $\mathrm{\parallel }\xi \mathrm{\parallel }$. Taking $\delta$, the tail, and the two initial truncation errors small proves density. Since $S$ and $\widehat{S}$ are bounded by [F5] and agree on this dense span, $S=\widehat{S}$. Thus every member of ${\mathcal{D}}^{\mathrm{\prime }}$ is decomposable.

If $X=\mathrm{\varnothing }$ or every fibre is zero, then $\mathcal{H}=\{0\}$, ${\mathcal{D}}^{\mathrm{\prime }}=\{0\}$, and the zero field is the unique field. If every fibre is one-dimensional, each constructed fibre operator is scalar, while the localized density and uniqueness arguments remain valid. Finite-dimensional and varying-dimension fibres require no separate choice: the dense normalized sections and the pointwise inequality include zero and dependent fundamental vectors. Null exceptional sets are combined by a countable union, and the proof uses an arbitrary measure space rather than an interval, so there are no endpoint cases. Steps 1.1–7.1 prove both commutant inclusions, and step 6.2 proves uniqueness. AC is used for the countable representative choice in step 1.2 and to meet the commutant, Hilbert-space, and action supplier hypotheses [F3–F5]; the proof's indexing, partitions and extensions are explicit. [F3, F4, F5, F6, F11, F14, step 1.1, step 1.2, step 5.1, step 7.1, step 6.2, algebra] $\mathrm{\square }$