Measurable Hilbert Fields and Direct-Integral Operators — Examples

Enumerate the fixed basis as a sequence, extending it by zero sections if $J$ is finite or empty. These constant sections $e_j(x)=b_j$ form a measurable fundamental family: their Gram coefficients are constant, and their span is dense in every fibre $K$. Thus the field's measurable sections are exactly the maps tested by these coordinates, and its almost-everywhere quotient is the stated $L^2(X,\mu ;K)$

Define $U$ by the displayed coordinate map. Each coordinate class belongs to scalar $L^2$ because Parseval and monotone convergence give ${\sum }_{j\in J}\mathrm{\parallel }⟨\xi (\cdot ),b_j⟩{\mathrm{\parallel }}_2^2={\int }_X{\sum }_{j\in J}\mathrm{\mid }⟨\xi (x),b_j⟩{\mathrm{\mid }}^{2}d\mu (x)={\int }_X\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^{2}d\mu (x)\mathrm{.}$ The identity also shows that $U$ is well-defined on almost-everywhere classes, linear, and isometric; when $J$ is finite the increasing sums stabilize.

With counting measure on a countable standard-Borel $X$, every section is measurable, the integral norm is ${\int }_X\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^{2}d\mathrm{\#}(x)={\sum }_{x\in X}\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2$, and the only null set is empty. The quotient therefore consists exactly of square-summable families $(\xi (x){)}_{x\in X}$ in $K$, with its ordinary Hilbert-sum norm.

For a finitely supported tuple $(f_j{)}_{j\in J}$, choose a Borel representative of each of its finitely many coordinates in the support; these finitely many existential instantiations require no choice axiom. The finite-coordinate section $\xi (x)={\sum }_{j\in J_0}{f}_j(x)b_j$ has measurable fundamental coefficients, and orthonormality gives $\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2={\sum }_{j\in J_0}\mathrm{\mid }{f}_j(x){\mathrm{\mid }}^2$. Additivity of the nonnegative integral makes this section square-integrable, and its coordinate image is the given tuple. Changing representatives changes the section only on a finite union of Borel null sets, so its class is independent of the representatives.

For an arbitrary square-summable tuple $(f_j{)}_{j\in J}$, its finite truncations converge in the Hilbert-sum norm. By step 2.1 each truncation has its canonical finite-coordinate section class; since $U$ is an isometry, these classes form a Cauchy sequence in the direct integral. AC supplies its completeness; the limit has coordinates $(f_j)$ because $U$ preserves distances and the truncations converge to that tuple. Hence $U$ is onto. No representatives are selected in this step: all tuples and sections are already almost-everywhere classes.

If $X=\mathrm{\varnothing }$ or $K=\{0\}$, both sides are the zero space; in the zero-fibre case the basis is empty and the coordinate sum has no terms. If $K=\mathbb{C}$ with basis $b_1=1$, $U$ is the identity on scalar $L^2$. Finite nonzero-dimensional fibres use only their finitely many coordinates, and padding the measurable fundamental sequence by zero sections changes no coordinate or norm. A null base gives only the zero almost-everywhere class. AC is used in step 3.1 for the completeness theorem and in step 1.2 through Parseval's Countable Choice hypothesis; the fixed basis and coordinate truncations require no further choice. This example has no interval endpoint or if-and-only-if assertion. [F3, F4, F7, F8, step 1.2, step 3.1, algebra] $\mathrm{\square }$

By [F1] and [F2], the base is standard Borel. By [F3] and [F4], $\lambda ([0,1])=1$, and $\lambda$ is finite and regular. By [F5], identify $\mathcal{H}$ with $L^2(\lambda )\oplus L^2(\lambda )$. This Hilbert space is nonzero and separable, since the two constant-fibre coordinates are nonzero and the direct integral theorem applies.

By [F14], every operator in ${\mathcal{D}}^{\mathrm{\prime }}$ is induced by an essentially bounded weakly measurable field $T(t)\in \mathcal{B}({\mathbb{C}}^2)$. In the fixed standard basis write $T(t)=\left(\begin{array}{cc}{a}_{11}(t)& a_{12}(t)\\ a_{21}(t)& a_{22}(t)\end{array}\right)$. The four entries are measurable exactly when the field is weakly measurable. Essential boundedness of $\mathrm{\parallel }T(t)\mathrm{\parallel }$ implies that all entries are essentially bounded; conversely, if each of the four entries is essentially bounded, then $\mathrm{\parallel }T(t)\mathrm{\parallel }\le {\sum }_{i,j=1}^2\mathrm{\mid }{a}_{ij}(t)\mathrm{\mid }$ almost everywhere, so the field is essentially bounded. The operator-field action theorem realizes every such matrix field, and the commutant theorem supplies the reverse inclusion. Therefore ${\mathcal{D}}^{\mathrm{\prime }}$ is precisely the essentially bounded measurable $2\times 2$ matrix fields modulo almost-everywhere equality.

On either scalar coordinate, $\mathrm{\parallel }{M}_{t}f{\mathrm{\parallel }}_2\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$. For $f,g\in L^2(\lambda )$, the integral pairing in [F6] gives $⟨M_{t}f,g⟩={\int }_0^{1}tf\bar{g}d\lambda ={\int }_0^{1}f\overline{tg}d\lambda =⟨f,M_{t}g⟩,$ since $t$ is real; hence $M_t$ and its direct sum $S$ are bounded self-adjoint. If $z\mathrm{\notin }[0,1]$, the bounded multiplier $(z-t{)}^{-1}$ on each coordinate is a two-sided bounded inverse of $zI-S$. If $s\in [0,1]$, let $B_n=[0,1]\cap (s-1\mathrm{/}n,s+1\mathrm{/}n)$. By [F8], $\lambda (B_n)>0$ even at the endpoints. The unit vectors $x_n=({\mathbf{1}}_{B_n}\mathrm{/}\sqrt{\lambda (B_n)},0)$ satisfy $\mathrm{\parallel }(S-sI)x_n\mathrm{\parallel }\le 1\mathrm{/}n$, so $sI-S$ cannot have a bounded inverse. Thus $\sigma (S)=[0,1]$ by [F7]. For Borel $B\subseteq [0,1]$ put $E_0(B)(f,g)=({\mathbf{1}}_{B}f,{\mathbf{1}}_{B}g)$. These are orthogonal projections with $E_0(\mathrm{\varnothing })=0$, $E_0([0,1])=I$, and $E_0(B\cap C)=E_0(B)E_0(C)$. For disjoint $B_j$ with union $B$, the squared norm of the difference between $E_0(B)(f,g)$ and the first $N$ projection terms is the integral over $B\setminus {\bigcup }_{j\le N}{B}_j$ of $\mathrm{\mid }f{\mathrm{\mid }}^2+\mathrm{\mid }g{\mathrm{\mid }}^2$, which tends to zero by [F15] and the finite $L^2$ norm. Its scalar measures $E_{0,x}(B)={\int }_B(\mathrm{\mid }f{\mathrm{\mid }}^2+\mathrm{\mid }g{\mathrm{\mid }}^2)d\lambda$ are finite regular Borel measures by [F4], so $E_0$ is a regular PVM. For $x=(f,g)$ and $y=(h,k)$, its complex scalar measure is $E_{0,x,y}(B)={\int }_B(f\bar{h}+g\bar{k})d\lambda$; the bounded PVM integral therefore gives $⟨(\int td{E}_0)x,y⟩={\int }_0^{1}t(f\bar{h}+g\bar{k})d\lambda =⟨Sx,y⟩$. The uniqueness clause in [F7] identifies $E_0$ as the spectral PVM of $S$. The same pairing calculation for bounded Borel $\psi$ gives $\psi (S)=M_{\psi }\oplus M_{\psi }$, in particular $E(B)=M_{{\mathbf{1}}_B}\oplus M_{{\mathbf{1}}_B}$.

The algebra $\mathcal{D}$ is an abelian concrete von Neumann algebra by [F10]. To prove $W^{\ast }(S)=\mathcal{D}$, first note that $S\in \mathcal{D}$ and $\mathcal{D}$ is WOT closed, giving $W^{\ast }(S)\subseteq \mathcal{D}$. For the reverse inclusion fix $f\in L^{\mathrm{\infty }}$ and put $M=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. If $M=0$, then $M_f=0\in W^{\ast }(S)$. Otherwise choose a Borel representative with $\mathrm{\mid }f\mathrm{\mid }\le M$ everywhere after changing it on a Borel null set, by [F12]. Apply [F12] separately to $u=\mathrm{Re}f$ and $v=\mathrm{Im}f$ to choose real continuous $u_n,v_n$ with $\mathrm{\parallel }{u}_n-u{\mathrm{\parallel }}_2+\mathrm{\parallel }{v}_n-v{\mathrm{\parallel }}_2<2^{-n}$. Clip each real function to $[-M,M]$; clipping is continuous, does not increase its pointwise error from $u$ or $v$, and makes ${\psi }_n=u_n+i{v}_n$ uniformly bounded by $\sqrt{2}M$. The squared approximation errors have finite sum, so [F15] gives ${\sum }_n\mathrm{\mid }{\psi }_n-f{\mathrm{\mid }}^2<\mathrm{\infty }$ almost everywhere; hence ${\psi }_n\to f$ almost everywhere. Every ${\psi }_n(S)=M_{{\psi }_n}$ lies in $C^{\ast }(I,S)\subseteq W^{\ast }(S)$ by continuous functional calculus. The PVM in step 2.1 has the same null sets as $\lambda$: if $\lambda (B)>0$, then ${\mathbf{1}}_B$ is a nonzero vector in scalar $L^2$ and $E(B)\mathrm{\ne }0$, while the converse is immediate. Thus the uniformly bounded pointwise-convergence clause of [F9] gives $M_{{\psi }_n}\to M_f$ strongly. WOT closedness now implies $M_f\in W^{\ast }(S)$. This proves $\mathcal{D}\subseteq W^{\ast }(S)$ as well, so equality holds. The Borel functional calculus identifies $S$ as the coordinate generator for $\mathcal{D}$.

Put $x_1=(\mathbf{1},0)$ and $x_2=(0,\mathbf{1})$. From the spectral projections in step 2.1, for every Borel $B\subseteq [0,1]$, ${\mu }_j(B)=⟨E(B)x_j,x_j⟩=\lambda (B)(j=1,2)\mathrm{.}$ The explicit weights $a_1=2^{-1}\mathrm{/}(1+\mathrm{\parallel }{x}_1{\mathrm{\parallel }}^2)=1\mathrm{/}4$ and $a_2=2^{-2}\mathrm{/}(1+\mathrm{\parallel }{x}_2{\mathrm{\parallel }}^2)=1\mathrm{/}8$ give the finite nonzero regular common measure $\mu =a_1{\mu }_1+a_2{\mu }_2=(3\mathrm{/}8)\lambda$ by [F4]. Its RN densities are $h_1=h_2=8\mathrm{/}3$, since ${\int }_B(8\mathrm{/}3)d\mu =\lambda (B)$. To verify the multiplicity directly, define $U:L^2(\lambda )\oplus L^2(\lambda )⟶L^2([0,1],\mu ;{\mathbb{C}}^2),U(f,g)=\sqrt{8\mathrm{/}3}(f,g)\mathrm{.}$ It is onto with inverse multiplication by $\sqrt{3\mathrm{/}8}$, and $\mathrm{\parallel }U(f,g){\mathrm{\parallel }}^2={\int }_{[0,1]}(8\mathrm{/}3)(\mathrm{\mid }f{\mathrm{\mid }}^2+\mathrm{\mid }g{\mathrm{\mid }}^2)d\mu =\mathrm{\parallel }f{\mathrm{\parallel }}_2^2+\mathrm{\parallel }g{\mathrm{\parallel }}_2^2\mathrm{.}$ It intertwines $S$ with the coordinate multiplier $M_t$. This is a spectral model over the finite nonzero measure $\mu$ with constant two-dimensional fibres, so its multiplicity is $2$; the uniqueness clause of [F13] identifies the multiplicity function as $m(t)=2$ almost everywhere. The two densities are positive everywhere, agreeing with the same active-coordinate count.

The vectors $x_1,x_2$ from step 3.2 are unit vectors by [F3]. By step 3.1, each continuous multiplier $\phi (S)=M_{\phi }$ lies in $W^{\ast }(S)=\mathcal{D}$ and sends $x_j$ to the corresponding coordinate copy of $\phi \in C([0,1];\mathbb{C})$. Thus the cyclic subspace generated by $x_j$ contains that copy. By [F12], $C([0,1];\mathbb{C})$ is dense in complex $L^2(\lambda )$, since its real and imaginary parts are separately approximated in real $L^2$. Hence each $x_j$ is cyclic, and the two orthogonal reducing summands exhaust $\mathcal{H}$.

The constant fields $E_{12}$ and $E_{21}$ belong to ${\mathcal{D}}^{\mathrm{\prime }}$ because scalar matrices commute pointwise with every $f(t)I_2$. Their products are $E_{12}{E}_{21}=E_{11}$ and $E_{21}{E}_{12}=E_{22}$; these differ, since on $x_1$ the first acts as the identity and the second acts as zero. Thus ${\mathcal{D}}^{\mathrm{\prime }}$ is nonabelian and strictly larger than the abelian algebra $\mathcal{D}$.

Steps 1.1, 2.1, and 3.1 identify the base, coordinate generator, and its diagonal von Neumann algebra; steps 3.2 and 4.1 compute the two cyclic scalar measures and multiplicity; steps 1.2 and 4.2 identify the commutant and exhibit its noncommutativity. [step 1.1, step 1.2, step 2.1, step 3.1, step 3.2, step 4.1, step 4.2] $\mathrm{\square }$

In the constant standard basis $e_1,e_2$, the four fundamental matrix coefficients of $T_t$ are the Borel functions $0,t,1-t,0$, so $T_t$ is weakly measurable.

The operator-norm and essential-supremum calculations use [F1, F4, F5, F6, F7, algebra]. For $z=(z_1,z_2)\in {\mathbb{C}}^2$, $\mathrm{\parallel }{T}_{t}z{\mathrm{\parallel }}^2=t^2\mathrm{\mid }{z}_2{\mathrm{\mid }}^2+(1-t{)}^2\mathrm{\mid }{z}_1{\mathrm{\mid }}^2\mathrm{.}$ The operator norm definition [F4], with the two standard unit vectors as witnesses for the larger coefficient, gives $\mathrm{\parallel }{T}_t\mathrm{\parallel }=\max \{t,1-t\}\mathrm{.}$ This is a Borel function bounded by $1$. For every $M\in [0,1)$, $\mathrm{\parallel }{T}_t\mathrm{\parallel }>M$ on $[0,1-M)$, whose measure is $1-M>0$ by [F6] and [F7]. So no $M<1$ is an almost-everywhere bound, while $1$ is a pointwise bound; therefore ${\mathrm{ess\hspace{0.17em}sup}}_{t\in [0,1]}\mathrm{\parallel }{T}_t\mathrm{\parallel }=1$ by [F5]. The field is essentially bounded. At $t=0$ and $t=1$ its rank is one, while for $0<t<1$ its rank is two; the norm formula holds in all cases.

The action theorem induces $T$ with norm one and identifies its adjoint and square fields. [F3, F7, F8, F9, step 1.1, step 1.2, algebra] It gives $T={\int }^{\oplus }{T}_{t}d\lambda (t)$ and $\mathrm{\parallel }T\mathrm{\parallel }=1$. Direct matrix multiplication yields $T_t^{\ast }=\left(\begin{array}{cc}0& 1-t\\ t& 0\end{array}\right),T_t^2=\left(\begin{array}{cc}t(1-t)& 0\\ 0& t(1-t)\end{array}\right)\mathrm{.}$

Pointwise commutation and the action theorem put $T$ in ${\mathcal{D}}^{\mathrm{\prime }}$. [F3, F9, F10, step 1.1, step 1.2, step 2.1, algebra] For $f\in L^{\mathrm{\infty }}([0,1],\lambda )$, the scalar field $f(t)I_2$ induces the corresponding $M_f\otimes I_2$ by [F10] and [F3]. At every $t$, $T_t(f(t)I_2)=f(t)T_t$. The product clause of [F3] therefore shows that $T(M_f\otimes I_2)=(M_f\otimes I_2)T$. Hence $T\in {\mathcal{D}}^{\mathrm{\prime }}$.

A vector witness separates $T$ from every $R\in \mathcal{D}$, proving $T\mathrm{\notin }\mathcal{D}$. [F1, step 2.1, algebra] Let $\eta =(\mathbf{1},0)\in L^2(\lambda )\oplus L^2(\lambda )$, so $\mathrm{\parallel }\eta \mathrm{\parallel }=1$ by [F1]. Then $T\eta =(0,1-t)$. For any $R=M_f\otimes I_2\in \mathcal{D}$, $R\eta =(f,0)$, and $\mathrm{\parallel }T\eta -R\eta {\mathrm{\parallel }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2+{\int }_0^1(1-t{)}^{2}d\lambda (t)\ge \frac{1}{3}>0.$ Thus $T\mathrm{\ne }R$ for every $R\in \mathcal{D}$.

Steps 1.1–1.2 prove weak measurability and the exact essential norm; step 2.1 gives the induced operator, its adjoint and square; steps 3.1 and 3.2 prove that $T\in {\mathcal{D}}^{\mathrm{\prime }}$ and $T\mathrm{\notin }\mathcal{D}$. $\mathrm{\square }$