Spectral measure of a real L2 multiplication operator

[Self-adjoint multiplication and indicator PVM] Put $X_n:=\{x:\mathrm{\mid }m(x)\mathrm{\mid }\le n\}$. For $f\in H$, $f{\mathbf{1}}_{X_n}\in D(M_m)$ and $f{\mathbf{1}}_{X_n}\to f$ in $H$ by dominated convergence, so $D(M_m)$ is dense. For $f,g\in D(M_m)$, Cauchy-Schwarz makes the products integrable and real-valuedness of $m$ gives $⟨M_{m}f,g⟩=⟨f,M_{m}g⟩$, so $M_m$ is symmetric. If $h\in D(M_m^{\ast })$ and $k:=M_m^{\ast }h$, the adjoint identity holds for every $f\in D(M_m)$. For each $n$ take $f=(mh-k){\mathbf{1}}_{X_n}$; this vector lies in $D(M_m)$ because $m$ is bounded on $X_n$, and the identity gives $0=⟨M_{m}f,h⟩-⟨f,k⟩={\int }_{X_n}\mathrm{\mid }mh-k{\mathrm{\mid }}^{2}d\mu$. Thus $mh=k$ a.e. on every $X_n$, and since ${\bigcup }_n{X}_n$ has full measure, $mh=k\in H$. Therefore $h\in D(M_m)$ and $M_m^{\ast }h=M_{m}h$, so $M_m=M_m^{\ast }$; this proves (a). [F1, F7, algebra] For a Borel set $B$ the class ${\mathbf{1}}_{m^{-1}(B)}$ is real measurable and bounded, so multiplication by it is a bounded self-adjoint operator on $H$ whose square is multiplication by ${\mathbf{1}}_{m^{-1}(B)}^2={\mathbf{1}}_{m^{-1}(B)}$: hence $E(B)=M_{{\mathbf{1}}_{m^{-1}(B)}}$ is an orthogonal projection. Moreover $E(\mathrm{\varnothing })=0$ and $E(\mathbb{R})=I$ because $m$ is finite $\mu$-a.e., and $E(B)E(C)=M_{{\mathbf{1}}_{m^{-1}(B)}{\mathbf{1}}_{m^{-1}(C)}}=E(B\cap C)$ because $m^{-1}(B)\cap m^{-1}(C)=m^{-1}(B\cap C)$ up to a $\mu$-null set. For pairwise disjoint Borel $B_n$ with union $B$ the multiplicativity of classes gives $\mathrm{\parallel }E(B)h-{\sum }_{n\le N}E(B_n)h{\mathrm{\parallel }}^2=\int \mathrm{\mid }({\mathbf{1}}_B-{\sum }_{n\le N}{\mathbf{1}}_{B_n})\circ m{\mathrm{\mid }}^2\mathrm{\mid }h{\mathrm{\mid }}^{2}d\mu$ for every $h\in H$, and the integrand is bounded by $\mathrm{\mid }h{\mathrm{\mid }}^2$ and tends to $0$ pointwise, so dominated convergence [F7] makes the truncated sums converge in norm to $E(B)h$; thus $E$ is strongly countably additive and is a PVM.

[Regularity] For $x\in H$ the scalar measure of [F2] is $E_x(B)=⟨E(B)x,x⟩={\int }_{m^{-1}(B)}\mathrm{\mid }x{\mathrm{\mid }}^{2}d\mu$, a finite Borel measure on $\mathbb{R}$ of total mass $\mathrm{\parallel }x{\mathrm{\parallel }}^2$; $\mathbb{R}$ is locally compact, Hausdorff and second countable, so [F6] makes $E_x$ regular. As this holds for every $x$, $E$ is a regular PVM.

[The integral of the identity is $M_m$] Let $g$ be bounded Borel. For indicators $g={\mathbf{1}}_B$ the pairing identity $⟨{\mathrm{\Phi }}_E(g)x,y⟩=\int gd{E}_{x,y}$ and the scalar measures of [F2] give $⟨{\mathrm{\Phi }}_E({\mathbf{1}}_B)x,y⟩={\int }_{m^{-1}(B)}x\bar{y}d\mu =⟨M_{{\mathbf{1}}_B\circ m}x,y⟩$, so ${\mathrm{\Phi }}_E({\mathbf{1}}_B)=M_{{\mathbf{1}}_B\circ m}$; by linearity of ${\mathrm{\Phi }}_E$ and [F3] the identity passes to simple functions, and uniform approximation with $\mathrm{\parallel }{\mathrm{\Phi }}_E(g)\mathrm{\parallel }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ [F3] gives ${\mathrm{\Phi }}_E(g)=M_{g\circ m}$ for every bounded Borel $g$. For the unbounded identity function $f(\lambda )=\lambda$ the scalar measures give $\int {\lambda }^{2}d{E}_x=\int \mathrm{\mid }m{\mathrm{\mid }}^2\mathrm{\mid }x{\mathrm{\mid }}^{2}d\mu$, so by [F4] the domain of $f(E)$ is exactly $D(M_m)$, and for $x\in D(M_m)$ the truncations satisfy $f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le n\}}(E)x=M_{m{\mathbf{1}}_{\{\mathrm{\mid }m\mathrm{\mid }\le n\}}}x$ by the bounded case; these converge to $mx$ in $H$ by dominated convergence [F7] applied to $\mathrm{\mid }m{\mathbf{1}}_{\{\mathrm{\mid }m\mathrm{\mid }\le n\}}-m{\mathrm{\mid }}^2\mathrm{\mid }x{\mathrm{\mid }}^2\le 4\mathrm{\mid }m{\mathrm{\mid }}^2\mathrm{\mid }x{\mathrm{\mid }}^2$. Hence $f(E)x=M_{m}x$ on $D(M_m)$.

[The spectral PVM and the scalar law] By step 2.2 the regular PVM $E$ of step 1.1 satisfies $\int \lambda dE(\lambda )=M_m$ with the displayed domain, and $M_m$ is self-adjoint by [F1]; the converse uniqueness clause of [F5] therefore identifies $E$ as the spectral projection-valued measure of $M_m$. For a unit vector $\psi$ the scalar law is $\mathrm{\parallel }E(B)\psi {\mathrm{\parallel }}^2={\int }_{m^{-1}(B)}\mathrm{\mid }\psi {\mathrm{\mid }}^{2}d\mu$ by step 2.1.

[Vanishing sets and the spectrum] If $E(B)=0$ then $0=\mathrm{\parallel }E(B)h{\mathrm{\parallel }}^2={\int }_{m^{-1}(B)}\mathrm{\mid }h{\mathrm{\mid }}^{2}d\mu$ for every $h\in H$; were $\mu (m^{-1}(B))>0$, $\sigma$-finiteness would give a subset $A\subseteq m^{-1}(B)$ of positive finite measure, and $h={\mathbf{1}}_A$ would give ${\int }_{m^{-1}(B)}\mathrm{\mid }h{\mathrm{\mid }}^{2}d\mu =\mu (A)>0$, a contradiction; hence $\mu (m^{-1}(B))=0$. Conversely, if $\mu (m^{-1}(B))=0$ then the multiplier ${\mathbf{1}}_{m^{-1}(B)}$ vanishes as an $L^2$ class and $E(B)=0$. By step 3.1, $E$ is the spectral PVM of $M_m$, so [F8] identifies its support with $\sigma (M_m)$; by the vanishing criterion just proved the support is the set of $t$ with $\mu (m^{-1}(t-\epsilon ,t+\epsilon ))>0$ for every $\epsilon >0$, namely the essential range of $m$. This proves (d) and completes the argument.