Diagonal multipliers form a von Neumann algebra

For each $n$, define $u_n(x)=\{\begin{array}{ll}{e}_n(x)\mathrm{/}\mathrm{\parallel }{e}_n(x)\mathrm{\parallel },& \mathrm{\parallel }{e}_n(x)\mathrm{\parallel }>0,\\ 0,& \mathrm{\parallel }{e}_n(x)\mathrm{\parallel }=0.\end{array}$ The norm is measurable by [F2]. The scalar function $r(0)=0$ and $r(t)=1\mathrm{/}t$ for $t>0$ is Borel: its inverse image of any Borel set is the possible singleton contribution at $0$ together with the inverse image under the continuous reciprocal map on $(0,\mathrm{\infty })$, using [F21]. Thus $u_n=(r\circ \mathrm{\parallel }{e}_n\mathrm{\parallel })e_n$ is a measurable section by [F1, F2, F3, F20]. Each $u_n$ is either zero or a unit vector. Its pointwise linear span is dense in $H_x$, since each $e_n(x)$ is a scalar multiple of $u_n(x)$ and [F1] gives dense span. In particular, if every $u_n(x)$ is zero then $H_x=\{0\}$.

For $n,m\in \mathbb{N}$, define the rank-one field $R_{n,m}(x)\xi =⟨\xi ,u_n(x)⟩u_m(x),\xi \in H_x\mathrm{.}$ By the first-variable linearity in [F11], this is a bounded linear operator of norm at most one, including on zero fibres, by [F9, F18]. For each pair of fundamental sections $e_i,e_j$, the section $R_{n,m}{e}_i=⟨e_i,u_n⟩u_m$ is measurable by [F2]; pairing it with $e_j$ is measurable by [F2] again. Hence every fundamental matrix coefficient of $R_{n,m}$ is measurable, so [F4] makes it a weakly measurable, essentially bounded field. By [F5] it induces a decomposable operator $Q_{n,m}={\int }_X^{\oplus }{R}_{n,m}(x)d\mu (x)$, which belongs to ${\mathcal{D}}^{\mathrm{\prime }}$ by [F6].

Let $f,g\in L^{\mathrm{\infty }}(X,\mu ;\mathbb{C})$ and choose measurable representatives. Their moduli have finite essential suprema by [F14], so [F17] gives finite bounds outside null sets; [F8] combines the two exceptional sets. The inequalities in [F16] show that $f+g$, $fg$, $af$, and $\bar{f}$ are essentially bounded for every $a\in \mathbb{C}$. Their measurability follows from [F14,F15], so all belong to $L^{\mathrm{\infty }}(X,\mu ;\mathbb{C})$. Pointwise action on direct-integral classes [F13] and product/adjoint compatibility [F5,F11] give $M_f{M}_g=M_{fg}$, $M_f+M_g=M_{f+g}$, $a{M}_f=M_{af}$, and $M_f^{\ast }=M_{\bar{f}}$. Also $M_1=I_{\mathcal{H}}$, including when $\mathcal{H}=\{0\}$. Thus $\mathcal{D}$ is a unital abelian $\ast$-subalgebra. Consequently $\mathcal{D}\subseteq {\mathcal{D}}^{\mathrm{\prime }}$ and $\mathcal{D}\subseteq {\mathcal{D}}^{\mathrm{\prime }\mathrm{\prime }}$ by [F7].

Let $T\in {\mathcal{D}}^{\mathrm{\prime }\mathrm{\prime }}$. Since $\mathcal{D}\subseteq {\mathcal{D}}^{\mathrm{\prime }}$, $T$ commutes with every element of $\mathcal{D}$, so $T\in {\mathcal{D}}^{\mathrm{\prime }}$. By [F6], there is a weakly measurable essentially bounded field $(T_x)$ inducing $T$. For each $n,m$, $T$ commutes with $Q_{n,m}$ because $Q_{n,m}\in {\mathcal{D}}^{\mathrm{\prime }}$. By the product clause in [F5], the fields $(T_x{R}_{n,m}(x))$ and $(R_{n,m}(x)T_x)$ are weakly measurable and essentially bounded and induce $T{Q}_{n,m}$ and $Q_{n,m}T$, respectively. These induced operators are equal; the uniqueness clause in [F6] therefore makes the two fields equal outside a Borel null set. There are only countably many pairs $(n,m)$, so [F8] gives one Borel null set $N$ outside which all fibrewise commutation identities hold simultaneously.

Fix $x\mathrm{\notin }N$. If $H_x=\{0\}$, then $T_x=0$. Otherwise some $u_n(x)$ is nonzero by step 1.1. For every nonzero $u_n(x)$, $R_{n,n}(x)$ is the orthogonal projection onto its one-dimensional span. Commutation with this projection shows that $T_x{u}_n(x)={\lambda }_n{u}_n(x)$, where ${\lambda }_n=⟨T_x{u}_n(x),u_n(x)⟩$. If $u_n(x)$ and $u_m(x)$ are both nonzero, apply $T_x{R}_{n,m}(x)=R_{n,m}(x)T_x$ to $u_n(x)$: the left side is ${\lambda }_m{u}_m(x)$ and the right side is ${\lambda }_n{u}_m(x)$, so ${\lambda }_m={\lambda }_n$. Thus a single scalar ${\lambda }_x$ satisfies $T_x{u}_n(x)={\lambda }_x{u}_n(x)$ for every $n$ (also when $u_n(x)=0$). Their span is dense by step 1.1, and boundedness of $T_x$ extends this identity to all of $H_x$. Hence $T_x={\lambda }_x{I}_{H_x}$.

Let $Z=\{x:H_x=\{0\}\}$, which is Borel because $Z={\bigcap }_n\{x:\mathrm{\parallel }{e}_n(x)\mathrm{\parallel }=0\}$ by [F1,F2,F3]. On $X\setminus (N\cup Z)$, partition into the Borel sets $B_n$ on which $n$ is the least index with $u_n(x)\mathrm{\ne }0$. Define $\lambda (x)=\{\begin{array}{ll}⟨T_x{u}_n(x),u_n(x)⟩,& x\in B_n,\\ 0,& x\in N\cup Z\mathrm{.}\end{array}$ For each $n$, the displayed pairing is measurable by the all-section coefficient criterion in [F4] and the measurable-section pairing result [F2]. The countable Borel partition therefore makes $\lambda$ Borel. Cauchy–Schwarz and the operator-norm bound [F9, F18] give $\mathrm{\mid }\lambda (x)\mathrm{\mid }\le \mathrm{\parallel }{T}_x\mathrm{\parallel }$ on the $B_n$; since the field $(T_x)$ is essentially bounded by [F4,F10], [F17] gives a finite a.e. bound for its norm, so $\lambda$ represents an element of $L^{\mathrm{\infty }}(X,\mu ;\mathbb{C})$ by [F14]. Step 3.1 gives $T_x=\lambda (x)I_{H_x}$ outside $N$, including zero fibres. The action definition then implies $T=M_{\lambda }$ on direct-integral classes by [F5,F13]. This proves ${\mathcal{D}}^{\mathrm{\prime }\mathrm{\prime }}\subseteq \mathcal{D}$, and step 1.3 gives the reverse inclusion. Finally, [F19] says ${\mathcal{D}}^{\mathrm{\prime }\mathrm{\prime }}$, being a commutant, is WOT closed. Thus $\mathcal{D}={\mathcal{D}}^{\mathrm{\prime }\mathrm{\prime }}$ is a concrete von Neumann algebra.