Direct integrals of measurable Hilbert fields are Hilbert spaces

Given a Cauchy sequence $(z_n)$ in $\mathcal{H}$, for each $k\ge 1$ let $N_k$ be the least index after which every pair of terms is less than $2^{-k}$ apart. [given, F1, F3, F16, construct] Set $n_1=N_1$ and $n_{k+1}=\max (N_{k+1},n_k+1)$. Then $(n_k)$ is strictly increasing and $\mathrm{\parallel }{z}_{n_{k+1}}-z_{n_k}{\mathrm{\parallel }}_{\mathcal{H}}<2^{-k}$. Each class $z_{n_k}$ has a nonempty set of square-integrable measurable representatives; AC [F16] selects one representative ${\xi }_k$ for each $k$. Put $d_k={\xi }_{k+1}-{\xi }_k$ and $h_k(x)=\mathrm{\parallel }{d}_k(x)\mathrm{\parallel }$. By [F1, F3, F16, construct], $d_k$ is a square-integrable measurable section, $h_k$ is measurable, and $\mathrm{\parallel }{h}_k{\mathrm{\parallel }}_2=\mathrm{\parallel }[d_k]{\mathrm{\parallel }}_{\mathcal{H}}<2^{-k}\mathrm{.}$

By [F12, F16], fix a countable algebra $\mathcal{A}$ generating $\mathcal{B}$. [given, F9, F10, F12, F16] The cited real coding identifies the standard-Borel structure with a Borel subset of $[0,1]$, and the corollary takes the pullback algebra generated by rational cuts. By sigma-finiteness [F9], choose a sequence $(E_j)$ of finite-measure measurable sets covering $X$ and set $X_m={\bigcup }_{j\le m}{E}_j$. Subadditivity [F10] makes each $X_m$ finite measure, and the sequence increases to $X$.

Recursively construct measurable fibrewise Gram--Schmidt sections. [given, F2, F3, F4, F20, construct] Put $v_0=e_0$ and, for $n>0$, $v_n(x)=e_n(x)-{\sum }_{j<n}⟨e_n(x),u_j(x)⟩u_j(x)\mathrm{.}$ Set $u_n(x)=r(\mathrm{\parallel }{v}_n(x)\mathrm{\parallel })v_n(x)$, where $r(0)=0$ and $r(t)=1\mathrm{/}t$ for $t>0$. By [F3] each $v_n$ is measurable and its norm is measurable. The scalar map $r$ is Borel, being continuous on $(0,\mathrm{\infty })$ and defined separately on the Borel singleton $\{0\}$; [F20] makes $u_n$ measurable. At each fibre the nonzero $u_n(x)$ are orthonormal: for $\mathrm{\ell }<n$ with $u_{\mathrm{\ell }}(x)\mathrm{\ne }0$, the induction hypothesis gives $⟨v_n(x),u_{\mathrm{\ell }}(x)⟩=⟨e_n(x),u_{\mathrm{\ell }}(x)⟩-{\sum }_{j<n}⟨e_n(x),u_j(x)⟩⟨u_j(x),u_{\mathrm{\ell }}(x)⟩=0;$ when $u_{\mathrm{\ell }}(x)=0$ the pairing vanishes directly. Normalizing a nonzero $v_n$ preserves these orthogonality relations. Inductively, each $e_n(x)$ lies in the span of $u_0(x),\dots ,u_n(x)$: when $v_n(x)=0$ it lies in the previous span, and otherwise $v_n(x)=\mathrm{\parallel }{v}_n(x)\mathrm{\parallel }{u}_n(x)$. Their nonzero subfamily therefore has dense span in $H_x$ by [F2]. This handles dependent vectors without ever dividing by zero.

For $N\ge 1$ put $g_N={\sum }_{k=1}^N{h}_k$. [step 1.1, F5] These are nonnegative measurable functions increasing in $N$. Minkowski gives $\mathrm{\parallel }{g}_N{\mathrm{\parallel }}_2\le {\sum }_{k=1}^N\mathrm{\parallel }{h}_k{\mathrm{\parallel }}_2<{\sum }_{k=1}^N{2}^{-k}<1.$ Let $g={\lim }_N{g}_N={\sup }_N{g}_N$. Scalar pointwise-limit measurability [F19] makes $g$ measurable. Monotone convergence [F6] applied to $g_N^2$ gives ${\int }_X{g}^{2}d\mu ={\lim }_N{\int }_X{g}_N^{2}d\mu \le 1.$ Thus $g$ is finite outside the measurable null set $E=\{x:g(x)=+\mathrm{\infty }\}$ by [F8].

Fix $m$. [step 1.2, F13] ${\mathcal{A}}_m=\{A\cap X_m:A\in \mathcal{A}\}$ is a pi-system generating the trace sigma-algebra on $X_m$. Let ${\mathcal{D}}_m$ consist of measurable $B\subseteq X_m$ such that for every $\epsilon >0$ some $A_m\in {\mathcal{A}}_m$ satisfies $\mu (B\mathrm{△}{A}_m)<\epsilon$. It contains ${\mathcal{A}}_m$. It is a lambda-system: relative complements preserve symmetric-difference measure; for pairwise disjoint $B_j\in {\mathcal{D}}_m$, continuity from below and finite measure of $X_m$ give $\mu (({\bigcup }_j{B}_j)\setminus ({\bigcup }_{j\le J}{B}_j))<\epsilon \mathrm{/}2$ for some $J\ge 1$. Approximate each of the first $J$ sets within $\epsilon \mathrm{/}(2J)$ and take their finite union in ${\mathcal{A}}_m$; finite subadditivity [F10] bounds the resulting symmetric difference by $\epsilon$. Dynkin's theorem [F13] now gives that every measurable subset of $X_m$ belongs to ${\mathcal{D}}_m$. If $B$ has finite measure in $X$, then $B\cap X_m\uparrow B$, so [F11] gives $\mu (B\setminus X_m)\to 0$. It follows that every finite-measure measurable $B$ is approximable in measure by some $A\cap X_m$ with $A\in \mathcal{A}$: given $\epsilon >0$, choose $m$ so that $\mu (B\setminus X_m)<\epsilon \mathrm{/}2$, then apply $B\cap X_m\in {\mathcal{D}}_m$ to choose $A_m=A\cap X_m$ with $\mu ((B\cap X_m)\mathrm{△}{A}_m)<\epsilon \mathrm{/}2$. The inequality $\mu (B\mathrm{△}{A}_m)\le \mu (B\setminus X_m)+\mu ((B\cap X_m)\mathrm{△}{A}_m)$ proves the required approximation.

At each $x$, let $I_x=\{n:u_n(x)\mathrm{\ne }0\}$. [step 1.3, F15, F16] It is an orthonormal family with dense span in $H_x$ by step 1.3, so Parseval gives, for each $w\in H_x$, $\mathrm{\parallel }w{\mathrm{\parallel }}^2={\sum }_{n\in I_x}\mathrm{\mid }⟨w,u_n(x)⟩{\mathrm{\mid }}^2,$ where the sum is the increasing limit of finite partial sums. In particular, for a measurable square-integrable section $\xi$, the functions $a_n(x)=⟨\xi (x),u_n(x)⟩$ are measurable by [F3] and lie in complex scalar $L^2$ by Cauchy--Schwarz [F21] and monotonicity [F22]. The finite coordinate sections ${\xi }_N={\sum }_{n<N}{a}_n{u}_n$ are measurable. Finite orthogonality gives $0\le \mathrm{\parallel }\xi (x)-{\xi }_N(x){\mathrm{\parallel }}^2=\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2-{\sum }_{n<N}\mathrm{\mid }{a}_n(x){\mathrm{\mid }}^2\le \mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2\mathrm{.}$ The initial finite subsets $I_x\cap \{0,\dots ,N-1\}$ exhaust the finite subsets of $I_x$, so Parseval makes this residual tend pointwise to zero. Also $\mathrm{\parallel }{\xi }_N(x){\mathrm{\parallel }}^2={\sum }_{n<N}\mathrm{\mid }{a}_n(x){\mathrm{\mid }}^2\le \mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2$, so each ${\xi }_N$ is square-integrable. Dominated convergence [F7] proves $\mathrm{\parallel }[\xi ]-[{\xi }_N]{\mathrm{\parallel }}_{\mathcal{H}}\to 0$.

For $x\mathrm{\notin }E$, ${\sum }_k\mathrm{\parallel }{d}_k(x)\mathrm{\parallel }=g(x)<\mathrm{\infty }$, so completeness of $H_x$ gives a limit of ${\xi }_1(x)+{\sum }_{k\ge 1}{d}_k(x)$. [step 2.1, F2] Define $\xi (x)$ to be that limit off $E$ and $0$ on $E$. For each $N$ the section ${\zeta }_N={\mathbf{1}}_{X\setminus E}({\xi }_1+{\sum }_{k=1}^N{d}_k)$ is measurable by [F3, F20]. The sequence ${\zeta }_N(x)$ converges in norm to $\xi (x)$ for every $x$, so $\xi$ is measurable by [F3]. Off $E$, $\mathrm{\parallel }\xi (x)\mathrm{\parallel }\le \mathrm{\parallel }{\xi }_1(x)\mathrm{\parallel }+g(x)$, while on $E$ it is zero. Therefore $\mathrm{\parallel }\xi (x){\mathrm{\parallel }}^2\le 2\mathrm{\parallel }{\xi }_1(x){\mathrm{\parallel }}^2+2g(x{)}^2$ everywhere, and the right side has finite integral. Thus $\xi$ is square integrable and $[\xi ]\in \mathcal{H}$.

The set $\mathcal{T}=\{A\cap X_m:A\in \mathcal{A},m\in \mathbb{N}\}$ is countable and consists of finite-measure sets. [step 2.2, F17] To see countability, enumerate the nonempty countable algebra and pair its indices with $m$ using [F17]. Every $T\in \mathcal{T}$ has finite measure. Let $\mathcal{S}$ be the scalar functions that are finite sums ${\sum }_{r<N}{q}_r{\mathbf{1}}_{T_r}$ with $q_r\in \mathbb{Q}+i\mathbb{Q}$ and $T_r\in \mathcal{T}$, including the empty sum. This family is countable by pairing the natural indices for $T_r$ and the rational real and imaginary parts, then using the finite-sequence code [F17]. Each member is measurable and lies in $L^2$, since its support is a finite union of finite-measure sets. To prove density, fix $f\in L^2(\mu ;\mathbb{C})$ and $\epsilon >0$. By [F14], choose a finite-measure-support simple function $s={\sum }_{r<N}{c}_r{\mathbf{1}}_{B_r}$ with $\mu (B_r)<\mathrm{\infty }$ and $\mathrm{\parallel }f-s{\mathrm{\parallel }}_2<\epsilon \mathrm{/}3$. If $N=0$, the empty sum belongs to $\mathcal{S}$ and already approximates $f$ within $\epsilon$. Suppose $N\ge 1$ and put $\delta =\epsilon \mathrm{/}(3N)$. For each $r$, write $c_r=a_r+i{b}_r$. If $\mu (B_r)=0$, take $q_r^{\mathrm{\prime }}=0$. Otherwise rational density [F18] gives $p_r,q_r\in \mathbb{Q}$ such that, with $q_r^{\mathrm{\prime }}=p_r+i{q}_r$, $\mathrm{\mid }{c}_r-q_r^{\mathrm{\prime }}\mathrm{\mid }\sqrt{\mu (B_r)}<\delta \mathrm{/}2$; here $\mathrm{\mid }{c}_r-q_r^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }{a}_r-p_r\mathrm{\mid }+\mathrm{\mid }{b}_r-q_r\mathrm{\mid }$ by [F18]. For these fixed $q_r^{\mathrm{\prime }}$, approximate $B_r$ in measure by $T_r\in \mathcal{T}$ using step 2.2, choosing $\mu (B_r\mathrm{△}{T}_r)<(\delta \mathrm{/}(2(1+\mathrm{\mid }{q}_r^{\mathrm{\prime }}\mathrm{\mid })){)}^2$. Minkowski [F5] gives $\mathrm{\parallel }{c}_r{\mathbf{1}}_{B_r}-q_r^{\mathrm{\prime }}{\mathbf{1}}_{T_r}{\mathrm{\parallel }}_2\le \mathrm{\mid }{c}_r-q_r^{\mathrm{\prime }}\mathrm{\mid }\sqrt{\mu (B_r)}+\mathrm{\mid }{q}_r^{\mathrm{\prime }}\mathrm{\mid }\sqrt{\mu (B_r\mathrm{△}{T}_r)}<\delta \mathrm{.}$ Thus $t={\sum }_{r<N}{q}_r^{\mathrm{\prime }}{\mathbf{1}}_{T_r}\in \mathcal{S}$, and Minkowski gives $\mathrm{\parallel }s-t{\mathrm{\parallel }}_2\le {\sum }_{r<N}\mathrm{\parallel }{c}_r{\mathbf{1}}_{B_r}-q_r^{\mathrm{\prime }}{\mathbf{1}}_{T_r}{\mathrm{\parallel }}_2<N\delta =\epsilon \mathrm{/}3$. Hence $\mathrm{\parallel }f-t{\mathrm{\parallel }}_2<2\epsilon \mathrm{/}3<\epsilon$, so $\mathcal{S}$ is dense in scalar $L^2(\mu ;\mathbb{C})$.

For $x\mathrm{\notin }E$, [step 1.1, step 3.1, F1, F4, F7] $\mathrm{\parallel }\xi (x)-{\xi }_k(x)\mathrm{\parallel }\le {\sum }_{j\ge k}{h}_j(x)\le g(x),$ and the left side tends to zero as $k\to \mathrm{\infty }$. Its square is measurable, converges to zero almost everywhere, and is dominated by $g^2\in L^1(\mu )$. Dominated convergence [F7] yields $\mathrm{\parallel }[\xi ]-z_{n_k}{\mathrm{\parallel }}_{\mathcal{H}}^2={\int }_X\mathrm{\parallel }\xi (x)-{\xi }_k(x){\mathrm{\parallel }}^{2}d\mu (x)⟶0.$ Since $(z_n)$ is Cauchy, for $\epsilon >0$ choose $K$ so that $\mathrm{\parallel }{z}_n-z_m\mathrm{\parallel }<\epsilon \mathrm{/}2$ for $n,m\ge K$, then choose $k$ with $n_k\ge K$ and $\mathrm{\parallel }{z}_{n_k}-[\xi ]\mathrm{\parallel }<\epsilon \mathrm{/}2$. The triangle inequality [F4] gives $\mathrm{\parallel }{z}_n-[\xi ]\mathrm{\parallel }<\epsilon$ for every $n\ge K$. Thus the entire sequence converges, proving completeness.

Define the countable candidate family $\mathcal{Q}=\{{\sum }_{n<N}{s}_n{u}_n:N\in \mathbb{N},s_n\in \mathcal{S}\}$. [step 2.3, step 3.2, F17] Its elements are coded by finite sequences from the countable set of pairs $(n,s)$, using [F17]. Every member is a square-integrable measurable section: pointwise triangle inequality and complex $L^2$ Minkowski give $\mathrm{\parallel }{\sum }_{n<N}{s}_n{u}_n{\mathrm{\parallel }}_{\mathcal{H}}\le {\sum }_{n<N}\mathrm{\parallel }{s}_n{\mathrm{\parallel }}_2<\mathrm{\infty }$. Given $[\xi ]\in \mathcal{H}$ and $\epsilon >0$, choose $N\ge 1$ so that $\mathrm{\parallel }[\xi ]-[{\xi }_N]\mathrm{\parallel }<\epsilon \mathrm{/}2$ by step 2.3. For each $n<N$, density of $\mathcal{S}$ [step 3.2] supplies $s_n$ with $\mathrm{\parallel }[a_n]-[s_n]{\mathrm{\parallel }}_{L^2(\mu ;\mathbb{C})}<\epsilon \mathrm{/}(2\sqrt{N})$. Put $Y_n=\{x:u_n(x)\mathrm{\ne }0\}$; it is measurable because $\mathrm{\parallel }{u}_n(\cdot )\mathrm{\parallel }$ is measurable. Since $a_n=0$ off $Y_n$ and $u_n$ has norm one on $Y_n$, pointwise orthogonality gives ${\parallel [{\xi }_N]-\left[{\sum }_{n<N}{s}_n{u}_n\right]\parallel }_{\mathcal{H}}^2={\sum }_{n<N}{\int }_{Y_n}\mathrm{\mid }{a}_n-s_n{\mathrm{\mid }}^{2}d\mu \le {\sum }_{n<N}\mathrm{\parallel }[a_n]-[s_n]{\mathrm{\parallel }}_{L^2}^2<{\epsilon }^2\mathrm{/}4.$ The triangle inequality proves that $\mathcal{Q}$ is dense in $\mathcal{H}$. The only selections here are finitely many scalar approximants for a fixed $\xi ,N$; no choice principle is used in constructing the countable set $\mathcal{Q}$.

Step 4.1 proves completeness, and step 4.2 provides a countable dense family. Together with the inner-product structure of [F1], this proves that $\mathcal{H}$ is a separable Hilbert space.