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 measure identity $\nu {\mathrm{\mid }}_B=hd(\mu {\mathrm{\mid }}_B)$ and [A7], applied to each $\mathrm{\mid }{g}_s{\mathrm{\mid }}^2\ge 0$ and then to the increasing coordinate sums, show that $(g_s{)}_s\mapsto (g_s\sqrt{h}{)}_s$ is an isometry ${⨁}_{s\le k^{\mathrm{\prime }}}{L}^2(\nu {\mathrm{\mid }}_B)\to {⨁}_{s\le k^{\mathrm{\prime }}}{L}^2(\mu {\mathrm{\mid }}_B)$. It is onto because $h>0$ almost everywhere and its inverse is multiplication by $h^{-1\mathrm{/}2}$; the same nonnegative integral identity proves the inverse isometric. 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: write $P_r,Q_s$ for the coordinate projections and $V_{sr}:=Q_{s}V{P}_r$. Each $V_{sr}$ commutes with all multiplications, so [A4] gives $V_{sr}=M_{g_{sr}}$ for a bounded Borel representative $g_{sr}$; choose these representatives simultaneously, since the coordinate pairs form a countable set. Because $0<\mu (B)<\mathrm{\infty }$, the constant coordinate vectors $e_r$ and $e_s^{\mathrm{\prime }}$ belong to the localized Hilbert spaces. For every Borel $D\subseteq B$, the identities $V^{\ast }V=I$ and $V{V}^{\ast }=I$, localized by $M_{{\mathbf{1}}_D}$, give ${\int }_D{\sum }_s{g}_{sr}\overline{g_{st}}d\mu ={\delta }_{rt}\mu (D)$ and ${\int }_D{\sum }_r{g}_{sr}\overline{g_{ur}}d\mu ={\delta }_{su}\mu (D)$. For equal indices, the sums are nonnegative. Monotone convergence and equality for every $D$ imply ${\sum }_s\mathrm{\mid }{g}_{sr}{\mathrm{\mid }}^2=1$ and ${\sum }_r\mathrm{\mid }{g}_{sr}{\mathrm{\mid }}^2=1$ almost everywhere, for each fixed column $r$ and row $s$. Cauchy–Schwarz then makes each off-diagonal sum absolutely convergent almost everywhere and bounded by $1$; passing finite partial sums through the integrals gives the off-diagonal identities pointwise almost everywhere. Intersect the countably many resulting conull sets. At every point of that common set the columns define an isometry $\phi (z):{\mathrm{\ell }}^2(k)\to {\mathrm{\ell }}^2(k^{\mathrm{\prime }})$ first on finite-support vectors and then by completion. The row identities likewise make its adjoint an isometry, so $\phi (z)$ is onto. Thus $\phi (z{)}^{\ast }\phi (z)=I_k$ and $\phi (z)\phi (z{)}^{\ast }=I_{k^{\mathrm{\prime }}}$ there, with convergent matrix products.

A fibrewise unitary $\phi (z):{\mathrm{\ell }}^2(k)\to {\mathrm{\ell }}^2(k^{\mathrm{\prime }})$ forces $k=k^{\mathrm{\prime }}$. If both are finite, its injectivity gives $k\le k^{\mathrm{\prime }}$ and its surjectivity gives $k^{\mathrm{\prime }}\le k$. If $k<\mathrm{\infty }$ and $k^{\mathrm{\prime }}=\mathrm{\infty }$, its range has dimension at most $k$ and cannot be all of ${\mathrm{\ell }}^2$; the reverse case is excluded by injectivity. The only remaining case has $k=k^{\mathrm{\prime }}=\mathrm{\infty }$. Hence every positive-measure constant-multiplicity region has equal fibre dimensions.

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.