Two common diagonalizations differ by a bimeasurable base isomorphism and a measurable field of unitaries

For either field, if $A$ is a Borel set of positive measure, sigma-finiteness and the nonzero-fibre condition give a countable cover of $A$ by sets $E\cap \{1\mathrm{/}k\le \mathrm{\parallel }{e}_n(x)\mathrm{\parallel }\le k\}$, where $E$ has finite measure and $(e_n)$ is the fundamental family. Some such set has positive measure by [F14]. Its localized section ${\mathbf{1}}_{E\cap \{1\mathrm{/}k\le \mathrm{\parallel }{e}_n\mathrm{\parallel }\le k\}}{e}_n$ is square-integrable and nonzero by [F3], so the diagonal representation is faithful on measure classes; the same argument shows the direct integral is zero exactly when the base measure is zero. Since $W$ is unitary, either both base measures are null or both are non-null. If both are null, take $X_0=Y_0=\mathrm{\varnothing }$ and $c:\mathrm{\varnothing }\to \mathrm{\varnothing }$; then $\lambda =\nu {\mathrm{\mid }}_{Y_0}=0$, the measures are equivalent, and the unique map between the zero direct integrals is the stated $J$ (choose the empty representative of $r=1$), while the fibre formula and uniqueness are vacuous. Henceforth both bases have positive measure.

In the non-null case put $\mathrm{\Theta }(S)=WS{W}^{-1}$. Every diagonal projection is $M_f$ for a bounded Borel representative $f$ by [F5, F16]. If $M_f$ is a projection, then $M_f=M_f^{\ast }=M_f^2$; pointwise scalar multiplication and uniqueness of decomposable fields [F5] give $f=\bar{f}=f^2$ almost everywhere. Since every $H_x$ is nonzero, $f(x)\in \{0,1\}$ almost everywhere, so $M_f=M_{{\mathbf{1}}_A}$ for the Borel set $A=\{x:f(x)=1\}$. Conversely every indicator multiplier is a projection, and faithfulness from step 1.1 identifies these projections exactly modulo null sets. Thus $\mathrm{\Theta }$ induces a Boolean isomorphism $\alpha$ between the two measure algebras. It preserves countable unions: represent a union by the strong limit of the increasing finite-union projections, whose convergence is [F4], and conjugation by $W$ preserves strong limits. The inverse ${\mathrm{\Theta }}^{-1}$ gives ${\alpha }^{-1}$ with the same property.

Choose bimeasurable codes $h_X:X\to B_X\subseteq [0,1]$ and $h_Y:Y\to B_Y\subseteq [0,1]$ by [F1]. For each $q\in \mathbb{Q}$, choose a Borel representative $E_q\subseteq Y$ of $\alpha ([h_X^{-1}((-\mathrm{\infty },q))])$; this is a countable choice by [F13, F15]. The order and countable-union relations hold modulo null sets; since there are countably many such relations, remove one Borel null set $N$ so they hold pointwise on $Y\setminus N$. On $N$, reset $E_q$ to $N$ for $q>0$ and to $\mathrm{\varnothing }$ for $q\le 0$. By the density in [F13], the resulting representatives are increasing and satisfy $E_q={\bigcup }_{r<q}{E}_r$ pointwise. Define $s(y)=\min (1,\max (0,\inf \{q\in \mathbb{Q}:y\in E_q\}))$, with $\inf \mathrm{\varnothing }=+\mathrm{\infty }$. For rational $q$, $s^{-1}((-\mathrm{\infty },q))={\bigcup }_{r<q}{E}_r$; by the density in [F13], $h_X^{-1}((-\mathrm{\infty },q))={\bigcup }_{r<q}{h}_X^{-1}((-\mathrm{\infty },r))$. Since $\alpha$ preserves countable unions by step 2.1, these identities give $[s^{-1}((-\mathrm{\infty },q))]=\alpha ([h_X^{-1}((-\mathrm{\infty },q))])$. The rational left rays generate the Borel sigma-algebra of $\mathbb{R}$: for rational $q$, $(q,+\mathrm{\infty })={\bigcup }_{r\in \mathbb{Q},r>q}[r,+\mathrm{\infty })$ by density, each $[r,+\mathrm{\infty })$ is the complement of $(-\mathrm{\infty },r)$, and the union is countable; [F13] says the rational right rays generate. Hence $s$ is Borel. The class of Borel $C\subseteq [0,1]$ for which $[s^{-1}(C)]=\alpha ([h_X^{-1}(C)])$ is a sigma-algebra by step 2.1 and preimage identities; it contains the generating left rays, hence all Borel $C$. In particular $Y_1=s^{-1}(B_X)$ is conull, and $d:Y_1\to X$, $d(y)=h_X^{-1}(s(y))$, is Borel with $[d^{-1}(A)]=\alpha ([A])$ for every Borel $A\subseteq X$. Repeat with ${\alpha }^{-1}$ and $h_Y$ to obtain a conull Borel $X_1\subseteq X$ and a Borel map $c:X_1\to Y$ satisfying $[c^{-1}(B)]={\alpha }^{-1}([B])$ for every Borel $B\subseteq Y$.

Let ${\mathcal{A}}_X,{\mathcal{A}}_Y$ be countable separating Borel algebras from [F2]. For each $A\in {\mathcal{A}}_X$, the sets $\{x\in X_1:c(x)\in Y_1,x\in A\}$ and $\{x\in X_1:c(x)\in Y_1,d(c(x))\in A\}$ agree modulo a $\mu$-null set, because their classes are related by ${\alpha }^{-1}\alpha$. Countability and separation give $d(c(x))=x$ on a conull Borel subset; symmetrically $c(d(y))=y$ on a conull Borel subset of $Y_1$. Define $X_0$ and $Y_0$ by the respective countable membership equalities, together with the conditions $c(x)\in Y_1$ and $d(y)\in X_1$. They are conull Borel sets, $c:X_0\to Y_0$ and $d:Y_0\to X_0$ are inverse bimeasurable bijections, and the projection-class identities from step 3.1 remain valid after restriction.

For every Borel $B\subseteq Y_0$, the identities in step 4.1 give $\mu (c^{-1}(B))=0$ exactly when $\nu (B)=0$, so $\lambda =c_{\ast }(\mu {\mathrm{\mid }}_{X_0})$ is equivalent to $\nu {\mathrm{\mid }}_{Y_0}$. It is sigma-finite because the images under the bimeasurable bijection $c$ of a finite-measure cover of $X_0$ are Borel and have finite $\lambda$-measure. On indicators the conjugation identity is $W{M}_{{\mathbf{1}}_{c^{-1}(B)}}{W}^{-1}=M_{{\mathbf{1}}_B}$, so linearity proves it for simple functions. For any $g\in L^{\mathrm{\infty }}(Y_0,\lambda )$, [F16] gives a bounded Borel representative. Quantize its bounded real and imaginary ranges by finite grids of mesh $1\mathrm{/}n$ to obtain simple functions $g_n$ with $\mathrm{\parallel }{g}_n-g{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Pointwise multiplication gives $\mathrm{\parallel }{M}_{g_n-g}\mathrm{\parallel }\le \mathrm{\parallel }{g}_n-g{\mathrm{\parallel }}_{\mathrm{\infty }}$, and conjugation by the unitary $W$ is norm-continuous, so the identity extends to every $g\in L^{\mathrm{\infty }}(Y_0,\lambda )=L^{\mathrm{\infty }}(Y_0,\nu {\mathrm{\mid }}_{Y_0})$.

Sigma-finiteness of $\lambda$ and $\nu {\mathrm{\mid }}_{Y_0}$ and [F14] give a common increasing exhaustion $(E_m)$ by sets finite for both measures. By [F17], $r=d\lambda \mathrm{/}d(\nu {\mathrm{\mid }}_{Y_0})$ and $\tilde{r}=d(\nu {\mathrm{\mid }}_{Y_0})\mathrm{/}d\lambda$ have measurable real-valued representatives, are integrable on each $E_m$, and satisfy the measure-integral identities there; they are unique up to null sets. The reciprocal rule [F9] gives $r\tilde{r}=1$ almost everywhere. The representative $r$ is nonnegative almost everywhere: if $\{r<0\}$ had positive $\nu$-measure, then $\{r<0\}={\bigcup }_{n\ge 1}\{r\le -1\mathrm{/}n\}$ and the common exhaustion would give some $n,m$ for which $A=E_m\cap \{r\le -1\mathrm{/}n\}$ has $0<\nu (A)<\mathrm{\infty }$ and $\lambda (A)<\mathrm{\infty }$. Since $r{\mathbf{1}}_A$ is integrable by [F17], [F18, F19] give $\lambda (A)={\int }_{A}rd\nu =-{\int }_A{r}^{-}d\nu \le -\nu (A)\mathrm{/}n<0$, contradicting positivity of $\lambda$. Thus $r\ge 0$ almost everywhere; the reciprocal identity and the finite real-valued representatives make $r$ strictly positive and finite almost everywhere. Change both representatives to $1$ on a Borel null exceptional set. The identity in [F10] extends to every nonnegative measurable $g$: on each finite-$\lambda$ set $E_m$, first apply the Radon–Nikodym formula to simple functions supported in $E_m$, then use increasing simple approximation and monotone convergence to obtain ${\int }_{E_m}gd\lambda ={\int }_{E_m}grd\nu$; applying monotone convergence as $E_m\uparrow Y_0$ gives $\int gd\lambda =\int grd\nu$. By [F11], $\sqrt{r}$ exists pointwise; for every rational $q\ge 0$, $\{\sqrt{r}>q\}=\{r>q^2\}$ by [F20], and for $q<0$ the preimage is all of $Y_0$. Since rational right rays generate the Borel sets by [F13], $\sqrt{r}$ is measurable; the same argument makes $\sqrt{\tilde{r}}$ measurable. Multiplication $J\eta =\sqrt{r}\eta$ preserves measurability by [F12] and satisfies $\mathrm{\parallel }J\eta {\mathrm{\parallel }}_{L^2(\nu )}^2=\int r\mathrm{\parallel }\eta {\mathrm{\parallel }}^{2}d\nu =\mathrm{\parallel }\eta {\mathrm{\parallel }}_{L^2(\lambda )}^2$. Its inverse is multiplication by $\sqrt{\tilde{r}}=r^{-1\mathrm{/}2}$ almost everywhere; applying [F10] with the measures reversed proves it is an inverse isometry. Hence $J$ is a unitary and commutes with every diagonal multiplier.

Transport the field $(H_x)$ from $(X_0,\mu )$ to $(Y_0,\lambda )$ along $d:Y_0\to X_0$ by [F6], and write the transported field as ${\tilde{H}}_y=H_{d(y)}$. The unitary $T=J^{-1}W(d^{\ast }{)}^{-1}:{\int }_{Y_0}^{\oplus }{\tilde{H}}_{y}d\lambda (y)\to {\int }_{Y_0}^{\oplus }{K}_{y}d\lambda (y)$ intertwines all diagonal multipliers by [F6] and steps 4.1–6.1. On the direct-sum field $L_y={\tilde{H}}_y\oplus K_y$, define the block operator $\mathsf{S}(\xi ,\eta )=(T^{\ast }\eta ,T\xi )$. It is a self-adjoint unitary commuting with every scalar diagonal multiplier. The measurable frames and Borel field coordinates supplied by [F7] make $L_y$ a measurable Hilbert field and make the adjoint and product matrix coefficients of its bounded operator fields measurable (the product coefficients are limits of finite frame sums).

Apply [F5] to $\mathsf{S}$ on the field $L_y$ to obtain its unique measurable essentially bounded fibre field $({\mathsf{S}}_y)$. The adjoint and product fields are measurable by step 7.1 and [F8]; since $\mathsf{S}={\mathsf{S}}^{\ast }$ and ${\mathsf{S}}^2=I$, uniqueness in [F5] gives ${\mathsf{S}}_y={\mathsf{S}}_y^{\ast }$ and ${\mathsf{S}}_y^2=I$ almost everywhere. The two diagonal blocks of ${\mathsf{S}}_y$ vanish almost everywhere because those blocks induce the zero operators globally, again by uniqueness. By the countable Borel frame coordinates in [F7], the fibre identities and block vanishing hold on a conull Borel set $Y_2\subseteq Y_0$. Replace $Y_0$ by $Y_2$ and $X_0$ by $c^{-1}(Y_2)$, and restrict $c$; these are still conull and bimeasurably bijective, and they change $\lambda$, its Radon–Nikodym class, and the direct-integral maps only on null sets. On this restricted base, write the lower-left block as $u_y:{\tilde{H}}_y\to K_y$. Self-adjointness makes the upper-right block $u_y^{\ast }$, and the equations ${\mathsf{S}}_y^2=I$ give $u_y^{\ast }{u}_y=I_{{\tilde{H}}_y}$ and $u_y{u}_y^{\ast }=I_{K_y}$ for every $y\in Y_0$, so $u_y$ is unitary everywhere on this base. Its field is measurable by [F7, F8], and $u_x:=u_{c(x)}$ is measurable under the bimeasurable change of base. From the definition of $T$, $(J^{-1}W\xi {)}_{c(x)}=u_x{\xi }_x$ almost everywhere. Any other pair $(c^{\mathrm{\prime }},u^{\mathrm{\prime }})$ satisfying the statement's formula induces the same measure-algebra map $\alpha$: its fibre formula, the scalar-multiplier intertwining in step 5.1, and commutation of its square-root Radon–Nikodym unitary with multipliers as in step 6.1 give $W{M}_{{\mathbf{1}}_{c^{\mathrm{\prime }-1}(B)}}{W}^{-1}=M_{{\mathbf{1}}_B}$ for every Borel $B\subseteq Y$. On the countable separating algebra from [F2], the pullback sets for $c^{\mathrm{\prime }}$ and $c$ therefore agree modulo null sets; outside their countable union of symmetric differences and the two exceptional base null sets, separation gives $c^{\mathrm{\prime }}(x)=c(x)$. With $c$ fixed, $\lambda$ and the derivative class are fixed by [F17], so the square-root unitary is fixed on direct-integral classes by step 6.1; uniqueness of the decomposable fibre field in [F5] then gives $u$ almost everywhere. Finally $W{I}_{\mathcal{H}}{W}^{-1}=I_{\mathcal{K}}$ because $W$ is unitary.