Unitary equivalence classified by measure class and multiplicity

Unitary invariance of the spectrum: if $V:H\to H^{\mathrm{\prime }}$ is unitary with $VT{V}^{\ast }=T^{\mathrm{\prime }}$, then for every $\lambda$ the operator $\lambda I-T$ is bijective with bounded inverse exactly when $\lambda I-T^{\mathrm{\prime }}=V(\lambda I-T)V^{\ast }$ is, so $\rho (T)=\rho (T^{\mathrm{\prime }})$ and $\sigma (T)=\sigma (T^{\mathrm{\prime }})$; in particular the two spectra are compared on a common compact subset of $\mathbb{C}$.

Converse: suppose $\sigma (T)=\sigma (T^{\mathrm{\prime }})=:\mathrm{\Lambda }$, $[\mu ]=[\nu ]$ and $m=m^{\mathrm{\prime }}$ $\mu$-almost everywhere, and put $\lambda :=\mu$. Then $\nu \sim \lambda$ and, by clause (2), $m=m^{\mathrm{\prime }}$ $\lambda$-almost everywhere, so the standard models $L^2(\lambda ,m)$ and $L^2(\lambda ,m^{\mathrm{\prime }})$ are the same space with the same multiplication operators: the level sets $A_r=\{m\ge r\}$ and $A_r^{\mathrm{\prime }}=\{m^{\mathrm{\prime }}\ge r\}$ differ by $\lambda$-null sets, so each $L^2(\lambda {\mathrm{\mid }}_{A_r})$ equals $L^2(\lambda {\mathrm{\mid }}_{A_r^{\mathrm{\prime }}})$ as a subspace of $L^2(\lambda )$.

Direct implication: suppose $VT{V}^{\ast }=T^{\mathrm{\prime }}$. Since $m\ge 1$ $\mu$-almost everywhere and $m^{\mathrm{\prime }}\ge 1$ $\nu$-almost everywhere, replace their values by $1$ on the respective null sets where they are $0$, obtaining Borel representatives $\tilde{m},{\tilde{m}}^{\mathrm{\prime }}:\mathrm{\Lambda }\to \{1,2,\dots \}\cup \{\mathrm{\infty }\}$. Their level sets differ from those of $m,m^{\mathrm{\prime }}$ only by null sets, so the resulting $L^2$ models and coordinate multiplications are unchanged. By the definitional identification, $T$ is unitarily equivalent to $M_z$ on $L^2(\mu ,\tilde{m})$ and $T^{\mathrm{\prime }}$ to $M_z$ on $L^2(\nu ,{\tilde{m}}^{\mathrm{\prime }})$; composing these equivalences with $V$ gives a unitary $U:L^2(\mu ,\tilde{m})\to L^2(\nu ,{\tilde{m}}^{\mathrm{\prime }})$ with $U{M}_z=M_{z}U$.

The model of $T^{\mathrm{\prime }}$ with dominating measure $\nu$ is canonically unitarily equivalent to the model with dominating measure $\lambda$; hence $T$ and $T^{\mathrm{\prime }}$ are both unitarily equivalent to $M_z$ on $L^2(\lambda ,m)$, and composing one equivalence with the inverse of the other gives a unitary $H\to H^{\mathrm{\prime }}$ conjugating $T$ to $T^{\mathrm{\prime }}$.

Applying the intertwiner lemma to the everywhere-positive representatives in step 2.1 gives $[\mu ]=[\nu ]$ and $\tilde{m}={\tilde{m}}^{\mathrm{\prime }}$ almost everywhere for that class. Since each representative differs from the original multiplicity only on a null set, $m=m^{\mathrm{\prime }}$ almost everywhere as well; together with step 1.1 this proves the "only if" implication.

In particular, take $T^{\mathrm{\prime }}=T$ and let the two countable cyclic decompositions in the Statement be arbitrary choices for this same operator. The identity unitary gives the direct implication just proved, so their scalar measures have the same class and their multiplicity functions agree almost everywhere. This establishes the decomposition independence deferred by the definition.

Therefore $T$ and $T^{\mathrm{\prime }}$ are unitarily equivalent exactly when the spectra agree and, on the common spectrum, the scalar measure classes agree and the multiplicity functions agree almost everywhere; the zero space is the excluded trivial case carrying the zero operator and no PVM.