Measurable and decomposable operator fields

Fix a bijection $\rho :\mathbb{Q}\to \mathbb{N}$ from [F18] and a bijection $\beta :{\mathbb{N}}^2\to \mathbb{N}$ from [F8]. Encode a term $(n,a,b)$, representing $(a+ib)e_n$, by $\kappa (n,a,b)=\beta (n,\beta (\rho (a),\rho (b)))$. Use the injective finite-sequence code [F3] on each finite list of term codes; if a natural $j$ is not in the code's range, decode it as the empty list, and otherwise use its unique decoded list. Define $q_j(x)={\sum }_r(a_r+i{b}_r)e_{n_r}(x)$ over that list, with the empty sum zero. Every rational-complex finite combination occurs, and each $q_j$ is a measurable section by [F1,F23]. These values are dense in every fibre: given $v\in H_x$ and $\epsilon >0$, choose a finite complex combination $u={\sum }_{r=1}^N{z}_r{e}_{n_r}(x)$ with $\mathrm{\parallel }v-u\mathrm{\parallel }<\epsilon \mathrm{/}2$ by [F20]. Put $C={\sum }_r\mathrm{\parallel }{e}_{n_r}(x)\mathrm{\parallel }$ and use [F19] to choose a rational $\eta$ with $0<\eta <\epsilon \mathrm{/}(4(1+C))$; approximate each real and imaginary coordinate of $z_r$ within $\eta$ by rationals $a_r,b_r$. Absolute homogeneity and the triangle inequality [F21], together with the complex modulus triangle inequality [F14], give $\mathrm{\parallel }u-{\sum }_r(a_r+i{b}_r)e_{n_r}(x)\mathrm{\parallel }\le 2\eta C<\epsilon \mathrm{/}2$. Hence $q_j(x)$ is dense. This construction fixes only individual bijections and makes no arbitrary selections, so it uses no AC.

Fix $j,m$ and write $q_j={\sum }_{r=1}^N{c}_r{e}_{n_r}$ with its finite rational-complex coefficients. First-variable linearity [F17] gives $⟨T_x{q}_j(x),e_m(x)⟩={\sum }_r{c}_r⟨T_x{e}_{n_r}(x),e_m(x)⟩$. If this combination is empty, then $T_x{q}_j(x)=0$ and is measurable. Otherwise the finitely many complex coefficient functions form a measurable map into their real-coordinate space because rational boxes give a countable basis [F15,F16]; for input tuples $z,w$, [F14] gives $\mathrm{\mid }{\sum }_r{c}_r{z}_r-{\sum }_r{c}_r{w}_r\mathrm{\mid }\le {\sum }_r\mathrm{\mid }{c}_r\mathrm{\mid }\mathrm{\mid }{z}_r-w_r\mathrm{\mid }$, so the finite-sum map is continuous in these finitely many coordinates and hence Borel [F13]; composition [F12] makes this coefficient measurable. The coefficient criterion [F2] therefore makes $x\mapsto T_x{q}_j(x)$ a measurable section.

Put $r_j(x)=1\mathrm{/}\mathrm{\parallel }{q}_j(x)\mathrm{\parallel }$ when $\mathrm{\parallel }{q}_j(x)\mathrm{\parallel }>0$ and $r_j(x)=0$ otherwise, and set $u_j(x)=r_j(x)q_j(x)$. Since $\mathrm{\parallel }{q}_j\mathrm{\parallel }$ is measurable [F23], define $g:[0,\mathrm{\infty })\to \mathbb{R}$ by $g(0)=0$ and $g(t)=1\mathrm{/}t$ for $t>0$. For every Borel $A\subseteq \mathbb{R}$, $g^{-1}(A)$ is the union of $\{0\}$ when $0\in A$ and $(0,\mathrm{\infty })\cap (t\mapsto 1\mathrm{/}t{)}^{-1}(A\cap (0,\mathrm{\infty }))$, a Borel set by continuity of reciprocal on $(0,\mathrm{\infty })$ and [F13]. Thus $g$ is Borel and $r_j=g\circ \mathrm{\parallel }{q}_j\mathrm{\parallel }$ is measurable by [F11,F12]; hence $u_j$ is a measurable section by [F23]. It has norm one when $q_j(x)\mathrm{\ne }0$ and is zero otherwise. On a nonzero fibre, the nonzero normalized $q_j(x)$ are dense in the unit sphere: for each unit vector $u$ and each $k$, take the least $j$ with $\mathrm{\parallel }{q}_j(x)-u\mathrm{\parallel }<1\mathrm{/}k$; these approximants are eventually nonzero and their normalizations converge to $u$, since $\mathrm{\parallel }{q}_j\mathrm{/}\mathrm{\parallel }{q}_j\mathrm{\parallel }-u\mathrm{\parallel }\le 2\mathrm{\parallel }{q}_j-u\mathrm{\parallel }$ when $q_j\mathrm{\ne }0$. This is a least-index construction, not a choice of arbitrary witnesses. By step 1.2, each $T_x{q}_j$ is measurable; measurable scalar closure [F23] makes $T_x{u}_j$ measurable.

Suppose first that the fundamental matrix coefficients are measurable. For any measurable section $\xi$ and $k\ge 1$, the sets $E_{j,k}=\{x:\mathrm{\parallel }\xi (x)-q_j(x)\mathrm{\parallel }<1\mathrm{/}k\}$ are measurable by [F23] and cover $X$ by step 1.1. At each $x$ take the least qualifying $j_k(x)$; the sets $E_{j,k}\setminus {\bigcup }_{i<j}{E}_{i,k}$ form a measurable partition; call its pieces $P_{j,k}$. For each $m$ and Borel $B\subseteq \mathbb{C}$, the inverse image of $B$ under the $m$th coefficient of ${\xi }_k(x)=q_{j_k(x)}(x)$ is ${\bigcup }_j(P_{j,k}\cap \{x:⟨q_j(x),e_m(x)⟩\in B\})$, which is measurable, so ${\xi }_k$ is a measurable section and $\mathrm{\parallel }{\xi }_k(x)-\xi (x)\mathrm{\parallel }<1\mathrm{/}k$. The inverse image under the $m$th coefficient of $T{\xi }_k$ is ${\bigcup }_j(P_{j,k}\cap \{x:⟨T_x{q}_j(x),e_m(x)⟩\in B\})$, which is also measurable; thus $T{\xi }_k$ is measurable. Pointwise boundedness of $T_x$ [F5] gives $\mathrm{\parallel }{T}_x{\xi }_k(x)-T_x\xi (x)\mathrm{\parallel }\le \mathrm{\parallel }{T}_x\mathrm{\parallel }\mathrm{/}k\to 0$ for every $x$, so [F23] makes $T\xi$ measurable and its pairing with every measurable $\eta$ measurable. Conversely, if all section pairings are measurable, test on $\xi =e_n$ and $\eta =e_m$, both measurable by [F1], to recover every fundamental coefficient. This proves both directions of the equivalence; no essential bound is needed because each individual $T_x$ is bounded.

For each $j,k$, the pairing $x\mapsto ⟨T_x{u}_j(x),u_k(x)⟩$ is measurable by [F23]. Modulus is continuous by the reverse triangle inequality from [F14,F15], so [F13] makes these moduli Borel and [F12] makes them measurable. For every nonzero fibre, $\mathrm{\parallel }{T}_x\mathrm{\parallel }={\sup }_{j,k\in \mathbb{N}}\mathrm{\mid }⟨T_x{u}_j(x),u_k(x)⟩\mathrm{\mid }$. The upper bound follows from [F6,F7], since $\mathrm{\parallel }{u}_j\mathrm{\parallel },\mathrm{\parallel }{u}_k\mathrm{\parallel }\le 1$. For the reverse bound, [F22] gives $\mathrm{\parallel }{T}_x\mathrm{\parallel }={\sup }_{\mathrm{\parallel }u\mathrm{\parallel }=1}\mathrm{\parallel }{T}_{x}u\mathrm{\parallel }$, and Cauchy--Schwarz with $v=T_{x}u\mathrm{/}\mathrm{\parallel }{T}_{x}u\mathrm{\parallel }$ when $T_{x}u\mathrm{\ne }0$ gives $\mathrm{\parallel }{T}_{x}u\mathrm{\parallel }={\sup }_{\mathrm{\parallel }v\mathrm{\parallel }=1}\mathrm{\mid }⟨T_{x}u,v⟩\mathrm{\mid }$. The normalized $q_j$ are dense in the unit sphere by step 2.1. First-variable linearity [F17], the modulus triangle inequality [F14], and [F6,F7] give $\mathrm{\mid }⟨T_{x}u,v⟩-⟨T_x{u}_j,u_k⟩\mathrm{\mid }\le \mathrm{\parallel }{T}_x\mathrm{\parallel }(\mathrm{\parallel }u-u_j\mathrm{\parallel }+\mathrm{\parallel }v-u_k\mathrm{\parallel })$, proving continuity in both unit vectors. On a zero fibre all $u_j$ and the operator norm are zero, so the same supremum formula holds. Use [F8] to enumerate pairs $(j,k)$ by one natural index; [F9] then makes $x\mapsto \mathrm{\parallel }{T}_x\mathrm{\parallel }$ measurable, including on zero fibres.

Since step 3.1 proves $\mathrm{\parallel }{T}_x\mathrm{\parallel }$ measurable and nonnegative, [F10] defines ${\mathrm{ess\hspace{0.17em}sup}}_x\mathrm{\parallel }{T}_x\mathrm{\parallel }$ and the field is essentially bounded exactly when that value is finite. By [F4], $S\in \mathcal{B}(\mathcal{H})$ is decomposable when there is such a field for which each pointwise action $T\xi$ belongs to the square-integrable section space and $S[\xi ]=[T\xi ]$ for every class. This states the well-definedness required by the formula; it does not presume the next theorem's existence result. If two fields agree off a measurable null set, their pointwise actions agree there, so they give the same direct-integral class. The definition and measurable-field arguments use no axiom of choice.