Unitary intertwiners preserve direct-integral fiber dimension

$U$ preserves all multiplications: $U{M}_z=M_{z}U$ gives $U{M}_z^{\ast }=M_z^{\ast }U$ by taking adjoints, so $U$ commutes with every $\ast$-polynomial in $z,\bar{z}$; $\ast$-polynomials are uniformly dense in the continuous functions and both sides are bounded, so $U{M}_h=M_{h}U$ for continuous $h$; for fixed vectors $p,q\in X$ the measure ${\rho }_{p,q}(E):=⟨M_{{\mathbf{1}}_E}p,q⟩={\int }_E{\sigma }_{p,q}d\mu$ with ${\sigma }_{p,q}={\sum }_r{p}_r\overline{q_r}\in L^1(\mu )$ is a finite regular complex measure, and for continuous $h$ one has $\int hd{\rho }_{p,q}=⟨M_{h}p,q⟩=⟨U{M}_{h}p,Uq⟩=⟨M_{h}Up,Uq⟩=\int hd{\rho }_{Up,Uq}$, so the two regular measures coincide and hence $⟨M_{h}p,q⟩=⟨M_{h}Up,Uq⟩$ for every bounded Borel $h$ and all $p,q$; therefore $U{M}_h=M_{h}U$ for every bounded Borel $h$.

Absolute continuity: the field $e:=({\mathbf{1}}_{A_1},0,0,\dots )\in X$ has scalar measure $⟨M_{h}e,e⟩=\int h{\mathbf{1}}_{A_1}d\mu =\int hd\mu$ because $A_1$ is $\mu$-conull, while the scalar measure of $Ue$ in $Y$ is $\int hwd\nu$ with $w:={\sum }_s\mathrm{\mid }(Ue{)}_s{\mathrm{\mid }}^2\in L^1(\nu )$; the identity $⟨M_{h}Ue,Ue⟩=⟨U{M}_{h}e,Ue⟩=⟨M_{h}e,e⟩$ for all bounded Borel $h$ gives ${\int }_{E}wd\nu =\mu (E)$ for every Borel $E$, hence $\mu \ll \nu$.

Localisation to constant multiplicity: let $B\subseteq \mathrm{\Lambda }$ be Borel with $m\equiv k$ and $m^{\mathrm{\prime }}\equiv k^{\mathrm{\prime }}$ on $B$ for constants $k,k^{\mathrm{\prime }}\in \{1,2,\dots \}\cup \{\mathrm{\infty }\}$ and $\mu (B)>0$; since $U{M}_{{\mathbf{1}}_B}=M_{{\mathbf{1}}_B}U$, the unitary restricts to a unitary $U_B$ between the localised spaces $X_B:=\{p\in X:p_r=0\mu \text{-a.e. off }B\}={⨁}_{r\le k}{L}^2(\mu {\mathrm{\mid }}_B)$ and $Y_B={⨁}_{s\le k^{\mathrm{\prime }}}{L}^2(\nu {\mathrm{\mid }}_B)$.

The same argument applied to the unitary $U^{\ast }$, which also intertwines the multiplications, shows $\nu \ll \mu$; hence $\mu$ and $\nu$ are mutually absolutely continuous.

Transferring $Y_B$ to the measure $\mu {\mathrm{\mid }}_B$ by the Radon–Nikodym factor: because $\nu {\mathrm{\mid }}_B\sim \mu {\mathrm{\mid }}_B$, A2 supplies a positive almost-everywhere density $h=d\nu \mathrm{/}d\mu$. The map $(g_s{)}_s\mapsto (g_s\sqrt{h}{)}_s$ is an isometry ${⨁}_{s\le k^{\mathrm{\prime }}}{L}^2(\nu {\mathrm{\mid }}_B)\to {⨁}_{s\le k^{\mathrm{\prime }}}{L}^2(\mu {\mathrm{\mid }}_B)$ by the defining integral identity, and it is onto because $h>0$ almost everywhere and its inverse is multiplication by $h^{-1\mathrm{/}2}$. It commutes with all multiplications, so composing it with $U_B$ gives a unitary $V:{⨁}_{r\le k}{L}^2(\mu {\mathrm{\mid }}_B)\to {⨁}_{s\le k^{\mathrm{\prime }}}{L}^2(\mu {\mathrm{\mid }}_B)$ commuting with all multiplications.

Constant-fibre rigidity: writing $P_r,Q_s$ for the coordinate projections and $V_{sr}:=Q_{s}V{P}_r$, each $V_{sr}$ commutes with all multiplications, so $V_{sr}=M_{g_{sr}}$ for a bounded Borel function $g_{sr}$; hence $V$ is given fibrewise by the measurable matrix field $\phi (z):=(g_{sr}(z))$, and $V^{\ast }V=I$, $V{V}^{\ast }=I$ force $\phi (z{)}^{\ast }\phi (z)=I_k$ and $\phi (z)\phi (z{)}^{\ast }=I_{k^{\mathrm{\prime }}}$ for $\mu$-almost every $z$.

A measurable field of linear maps satisfying both identities exists only if $k=k^{\mathrm{\prime }}$: if $k<\mathrm{\infty }$ then $\phi (z{)}^{\ast }\phi (z)=I_k$ shows the range of $\phi (z{)}^{\ast }$ spans a $k$-dimensional space, so $\phi (z)\phi (z{)}^{\ast }$, whose rank is at most $k$, cannot equal $I_{k^{\mathrm{\prime }}}$ when $k^{\mathrm{\prime }}=\mathrm{\infty }$; symmetrically $k^{\mathrm{\prime }}<\mathrm{\infty }<k$ is impossible; and $k=k^{\mathrm{\prime }}\in \{1,2,\dots \}\cup \{\mathrm{\infty }\}$ is consistent. Hence $k=k^{\mathrm{\prime }}$.

The Borel sets $\{m=k\}\cap \{m^{\mathrm{\prime }}=k^{\mathrm{\prime }}\}$ over $k,k^{\mathrm{\prime }}$ cover a conull set, and by the rigidity steps above every one of them with positive measure satisfies $k=k^{\mathrm{\prime }}$; therefore $m=m^{\mathrm{\prime }}$ $\mu$-almost everywhere, and by the class equality also $\nu$-almost everywhere.