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.

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.