Measurable essentially bounded operator fields act decomposably

Let $\xi$ be a measurable section. For each fundamental vector $e_m$, the function $x\mapsto ⟨T_x\xi (x),e_m(x)⟩$ is measurable by the all-sections coefficient criterion in [F5]. Thus all fundamental coefficients of $x\mapsto T_x\xi (x)$ are measurable, so this pointwise image is a measurable section.

Enumerate all finite rational-complex linear combinations of the fundamental sections as $(q_j{)}_{j\in \mathbb{N}}$. Such an enumeration exists by [F21]; each $q_j$ is a measurable section by [F6] and [F27]. The family is pointwise dense: first approximate a given vector by a finite complex combination using [F1], then approximate each of its finitely many complex coefficients by an element of $\mathbb{Q}+i\mathbb{Q}$ using [F22]. If $u={\sum }_{r=1}^N{z}_r{e}_{n_r}(x)$ is such a combination and $a_r+i{b}_r$ are the rational approximants, [F23] and [F24] bound the replacement error by ${\sum }_{r=1}^N\mathrm{\mid }{z}_r-(a_r+i{b}_r)\mathrm{\mid }\mathrm{\parallel }{e}_{n_r}(x)\mathrm{\parallel }$, which can be made arbitrarily small.

For each $x$, the fibre adjoint $T_x^{\ast }$ exists by [F16]. With inner products linear in the first variable [F25], its matrix coefficients satisfy $⟨T_x^{\ast }{e}_n(x),e_m(x)⟩=\overline{⟨T_x{e}_m(x),e_n(x)⟩}$. Conjugation is an isometry because if $z=a+bi$ and $w=c+di$ then $d_{\mathbb{C}}(\bar{z},\bar{w}{)}^2=(a-c{)}^2+(b-d{)}^2=d_{\mathbb{C}}(z,w{)}^2$. Thus it is continuous and Borel by [F17]; [F4] and [F18] then make the right side measurable. Hence $T_x^{\ast }$ is weakly measurable. Also $\mathrm{\parallel }{T}_x^{\ast }\mathrm{\parallel }=\mathrm{\parallel }{T}_x\mathrm{\parallel }$ by [F16], so it is essentially bounded with the same essential bound.

Put $s={\mathrm{ess\hspace{0.17em}sup}}_x\mathrm{\parallel }{T}_x\mathrm{\parallel }<\mathrm{\infty }$, and choose a measurable null set $N$ outside which $\mathrm{\parallel }{T}_x\mathrm{\parallel }\le s$ by [F8]. For $f(x)=\mathrm{\parallel }{T}_x\xi (x){\mathrm{\parallel }}^2$, step 1.1 and [F6] give nonnegative measurability. The pointwise operator bound in [F14] gives $f{\mathbf{1}}_{X\setminus N}\le s^2\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$. By [F9] and [F10], ${\int }_{X}fd\mu ={\int }_{X}f{\mathbf{1}}_{X\setminus N}d\mu +{\int }_{X}f{\mathbf{1}}_{N}d\mu \le s^2{\int }_X\mathrm{\parallel }\xi {\mathrm{\parallel }}^{2}d\mu <\mathrm{\infty }$, where the second integral is zero because $N$ is null. Hence the pointwise image is square-integrable. If $\xi ={\xi }^{\mathrm{\prime }}$ outside a measurable null set, then $T_x\xi (x)=T_x{\xi }^{\mathrm{\prime }}(x)$ there. If two field representatives agree outside a measurable null set, their pointwise images also agree there. Pointwise linearity and the same estimate show that $[\xi ]\mapsto [T\xi ]$ is a well-defined bounded linear operator on $\mathcal{H}$, independent of both section and field representatives, with $\mathrm{\parallel }T\mathrm{\parallel }\le s$.

Define $u_j(x)=q_j(x)\mathrm{/}\mathrm{\parallel }{q}_j(x)\mathrm{\parallel }$ when $q_j(x)\mathrm{\ne }0$ and $u_j(x)=0$ otherwise. The norm is measurable by [F6]. The function $g(0)=0$ and $g(t)=1\mathrm{/}t$ for $t>0$ is Borel, by splitting each preimage into its possible zero part and the preimage under the continuous reciprocal on $(0,\mathrm{\infty })$ using [F17]. Thus $u_j=(g\circ \mathrm{\parallel }{q}_j\mathrm{\parallel })q_j$ is measurable by [F6] and [F18]. It has norm at most one, and the nonzero $u_j(x)$ are dense in the unit sphere of each nonzero fibre: given a unit $v$ and $\epsilon >0$, choose $q_j(x)$ with $\mathrm{\parallel }{q}_j(x)-v\mathrm{\parallel }<\min (\epsilon \mathrm{/}2,1\mathrm{/}2)$ by step 1.2. Then $q_j(x)\mathrm{\ne }0$ and $\mathrm{\parallel }{q}_j(x)\mathrm{/}\mathrm{\parallel }{q}_j(x)\mathrm{\parallel }-v\mathrm{\parallel }\le 2\mathrm{\parallel }{q}_j(x)-v\mathrm{\parallel }<\epsilon$. Consequently, for every fibre, $\mathrm{\parallel }{T}_x\mathrm{\parallel }={\sup }_{j,k\in \mathbb{N}}\mathrm{\mid }⟨T_x{u}_j(x),u_k(x)⟩\mathrm{\mid }$. On a nonzero fibre, the operator norm is the supremum of $\mathrm{\parallel }{T}_{x}u\mathrm{\parallel }$ over unit $u$ by [F14], and $\mathrm{\parallel }{T}_{x}u\mathrm{\parallel }={\sup }_{\mathrm{\parallel }v\mathrm{\parallel }=1}\mathrm{\mid }⟨T_{x}u,v⟩\mathrm{\mid }$ by [F20] and the choice $v=T_{x}u\mathrm{/}\mathrm{\parallel }{T}_{x}u\mathrm{\parallel }$ when $T_{x}u\mathrm{\ne }0$. The pairing is continuous in both unit vectors: its change is at most $\mathrm{\parallel }{T}_x\mathrm{\parallel }(\mathrm{\parallel }u-u^{\mathrm{\prime }}\mathrm{\parallel }+\mathrm{\parallel }v-v^{\mathrm{\prime }}\mathrm{\parallel })$ by [F14] and [F20]. Density of both normalized families therefore gives the displayed supremum. On a zero fibre every term and $\mathrm{\parallel }{T}_x\mathrm{\parallel }$ are zero. Each displayed coefficient is measurable by [F5] and [F18], and its modulus is measurable because [F19] makes the modulus continuous and [F18] preserves measurability under composition. Thus the norm is a countable measurable supremum.

Let $S$ be another weakly measurable essentially bounded field. By step 1.1, $x\mapsto S_x{e}_n(x)$ is a measurable section. The all-sections criterion [F5], applied to $T$ and that section, shows that every fundamental coefficient $⟨T_x{S}_x{e}_n(x),e_m(x)⟩$ is measurable. Hence $(T_x{S}_x)$ is weakly measurable. The pointwise composition bound [F15] and the two essential bounds [F8] show that $\mathrm{\parallel }{T}_x{S}_x\mathrm{\parallel }$ is bounded almost everywhere by the product of the two essential suprema: remove the union of their two null exceptional sets, which is null by [F26]. Thus the product field is essentially bounded.

If $s=0$, step 2.1 gives $T=0$ and hence $\mathrm{\parallel }T\mathrm{\parallel }=s$. Suppose $s>0$, and fix $0<c<s$. The set $A_c=\{x:\mathrm{\parallel }{T}_x\mathrm{\parallel }>c\}$ has positive measure, since otherwise $c$ would be an almost-everywhere bound contradicting the leastness of $s$ in [F8]. By the countable supremum in step 2.2, $A_c={\bigcup }_{j,k\in \mathbb{N},n\in {\mathbb{N}}_{>0}}\{x:\mathrm{\mid }⟨T_x{u}_j(x),u_k(x)⟩\mathrm{\mid }>c+1\mathrm{/}n\}$. Countable subadditivity [F26] gives one such set positive measure. Intersect it with a set of a sigma-finite cover from [F12] so the resulting measurable set $E$ has finite positive measure. On $E$ both test vectors have norm one; by Cauchy--Schwarz in [F20], $\mathrm{\parallel }{T}_x{u}_j(x)\mathrm{\parallel }>c+1\mathrm{/}n$. For $\zeta ={\mathbf{1}}_E{u}_j$, the section lemma [F6] gives measurability, and it is square-integrable because $\mathrm{\parallel }\zeta (x){\mathrm{\parallel }}^2={\mathbf{1}}_E(x)$. Hence $\mathrm{\parallel }\zeta {\mathrm{\parallel }}^2=\mu (E)>0$ and $\mathrm{\parallel }T\zeta {\mathrm{\parallel }}^2={\int }_E\mathrm{\parallel }{T}_x{u}_j(x){\mathrm{\parallel }}^{2}d\mu (x)\ge (c+1\mathrm{/}n{)}^2\mu (E)$. The indicator integral is [F11], [F10] identifies it as an integral over $E$, and [F9] gives the comparison. Since $\mu (E)>0$, its defining unit-vector supremum [F14] gives $\mathrm{\parallel }T\mathrm{\parallel }\ge \mathrm{\parallel }T\zeta \mathrm{\parallel }\mathrm{/}\mathrm{\parallel }\zeta \mathrm{\parallel }\ge c+1\mathrm{/}n>c$. If $\mathrm{\parallel }T\mathrm{\parallel }<s$, rational density [F22] gives a $c$ strictly between $\mathrm{\parallel }T\mathrm{\parallel }$ and $s$, contradicting the established inequality; hence $\mathrm{\parallel }T\mathrm{\parallel }\ge s$, and step 2.1 proves equality.

Let $\tilde{T}={\int }^{\oplus }{T}_x^{\ast }d\mu$ be the induced operator from step 2.1 applied to the adjoint field. For direct-integral vectors $\xi ,\eta$, the fibre adjoint identity gives $⟨T\xi ,\eta {⟩}_{\mathcal{H}}={\int }_X⟨T_x\xi (x),\eta (x)⟩d\mu (x)={\int }_X⟨\xi (x),T_x^{\ast }\eta (x)⟩d\mu (x)=⟨\xi ,\tilde{T}\eta {⟩}_{\mathcal{H}}$. Uniqueness of Hilbert adjoints in [F16] therefore gives $\tilde{T}=T^{\ast }$. For the product field, step 2.1 and the definition of induced action yield, for every $\xi \in \mathcal{H}$, $({\int }^{\oplus }{T}_x{S}_{x}d\mu )\xi =[x\mapsto T_x(S_x\xi (x))]=T(S\xi )$, so its induced operator is $TS$. This proves adjoint and multiplication compatibility.