Joint PVM construction for strongly commuting regular projection-valued measures

[The finitely additive projection on the rectangle algebra] For a finite disjoint union $A={⨆}_{i=1}^n(B_i\times C_i)$ of measurable rectangles put $P(A):={\sum }_{i}E(B_i)F(C_i)$. This is well defined: two finite rectangle decompositions have a common refinement into disjoint rectangles (intersections of the pieces), and on a common refinement both sums agree by additivity of $E$ and $F$; in particular $P(\mathrm{\varnothing })=0$ and $P({\mathbb{R}}^2)=I$. Each $E(B_i)F(C_i)$ is an orthogonal projection because $(E(B_i)F(C_i){)}^2=E(B_i{)}^{2}F(C_i{)}^2=E(B_i)F(C_i)$ and $(E(B_i)F(C_i){)}^{\ast }=F(C_i)E(B_i)=E(B_i)F(C_i)$ by [F1] and strong commutation; summands belonging to disjoint rectangles are orthogonal, since $E(B_i)F(C_i)E(B_j)F(C_j)=E(B_i\cap B_j)F(C_i\cap C_j)=0$. Hence $P(A)$ is a sum of pairwise orthogonal projections, so $P(A{)}^2=P(A)=P(A{)}^{\ast }$; and from the refinement description $P$ is finitely additive on disjoint unions, $P(A\cap A^{\mathrm{\prime }})=P(A)P(A^{\mathrm{\prime }})$ and $P(A\cup A^{\mathrm{\prime }})=P(A)+P(A^{\mathrm{\prime }})$ for disjoint $A,A^{\mathrm{\prime }}\in {\mathcal{A}}_0$.

[Uniqueness for finite complex measures] Let $\mu ,\nu$ be finite complex measures on $\mathcal{B}({\mathbb{R}}^2)$ that agree on ${\mathcal{A}}_0$, which contains ${\mathbb{R}}^2$. Then $\mathcal{D}:=\{S\in \mathcal{B}({\mathbb{R}}^2):\mu (S)=\nu (S)\}$ contains ${\mathbb{R}}^2$ and the $\pi$-system ${\mathcal{A}}_0$; it is a $\lambda$-system because equality on the whole space gives closure under complements and countable additivity gives closure under disjoint countable unions. By Dynkin's $\pi$-$\lambda$ theorem [F8], $\mathcal{D}$ contains $\sigma ({\mathcal{A}}_0)=\mathcal{B}({\mathbb{R}}^2)$ by [F5], so $\mu =\nu$ on every Borel set.

[The scalar set functions are premeasures] For $\psi \in H$ define ${\mu }_{\psi }(A):=⟨P(A)\psi ,\psi ⟩=\mathrm{\parallel }P(A)\psi {\mathrm{\parallel }}^2$ for $A\in {\mathcal{A}}_0$; by step 1.1, ${\mu }_{\psi }(\mathrm{\varnothing })=0$, ${\mu }_{\psi }$ is finitely additive, and ${\mu }_{\psi }\le \mathrm{\parallel }\psi {\mathrm{\parallel }}^2$. To verify countable additivity, fix pairwise disjoint $A_n\in {\mathcal{A}}_0$ with union $A\in {\mathcal{A}}_0$ and put $R_N:=A\setminus {\bigcup }_{n\le N}{A}_n\in {\mathcal{A}}_0$. Finite additivity gives ${\mu }_{\psi }(A)={\sum }_{n\le N}{\mu }_{\psi }(A_n)+{\mu }_{\psi }(R_N)$, and $R_N\downarrow \mathrm{\varnothing }$, so continuity from above at $\mathrm{\varnothing }$ implies ${\mu }_{\psi }(A)={\sum }_n{\mu }_{\psi }(A_n)$. It remains to prove that continuity: for $A_n\downarrow \mathrm{\varnothing }$ in ${\mathcal{A}}_0$, show ${\mu }_{\psi }(A_n)\to 0$. Given $\epsilon >0$, approximate each $A_n$ from inside by a compact union of rectangles: for a rectangle $B\times C$ one has $P(B\times C)-P(K_B\times K_C)=E(B\setminus K_B)F(C)+E(K_B)F(C\setminus K_C)$ with the two summands orthogonal, so ${\mu }_{\psi }(B\times C)-{\mu }_{\psi }(K_B\times K_C)\le E_{\psi }(B\setminus K_B)+F_{\psi }(C\setminus K_C)$ for $E_{\psi }(B)=⟨E(B)\psi ,\psi ⟩$, $F_{\psi }(C)=⟨F(C)\psi ,\psi ⟩$; by regularity [F1] compact $K_B\subseteq B$, $K_C\subseteq C$ make this arbitrarily small, and finite unions are handled by summing. Choose compact $L_n\subseteq A_n$ with ${\mu }_{\psi }(A_n\setminus L_n)<\epsilon 2^{-n}$. If ${\mu }_{\psi }(A_n)\ge \epsilon$ for all $n$, then for every $N$ additivity and $L_1\cap \cdots \cap L_N\supseteq A_N\setminus {\bigcup }_{n\le N}(A_n\setminus L_n)$ give ${\mu }_{\psi }(L_1\cap \cdots \cap L_N)\ge {\mu }_{\psi }(A_N)-{\sum }_{n\le N}{\mu }_{\psi }(A_n\setminus L_n)\ge \epsilon -{\sum }_{n\le N}\epsilon 2^{-n}>0$; hence every finite intersection $L_1\cap \cdots \cap L_N$ is nonempty, and compactness gives a point of ${\bigcap }_n{L}_n\subseteq {\bigcap }_n{A}_n=\mathrm{\varnothing }$, a contradiction. Thus ${\mu }_{\psi }(A_n)\to 0$, and ${\mu }_{\psi }$ is a finite premeasure on ${\mathcal{A}}_0$.

[Carathéodory extension and polarization] By steps 1.1 and 2.1 the algebra ${\mathcal{A}}_0$ of finite unions of measurable rectangles is a $\pi$-system containing ${\mathbb{R}}^2$ and generates $\mathcal{B}({\mathbb{R}}^2)$ by [F5]; hence ${\mu }_{\psi }$ extends to a finite measure ${\nu }_{\psi }$ on $\mathcal{B}({\mathbb{R}}^2)$, unique by [F3], and ${\nu }_{\psi }(S)={\mu }_{\psi }(S)$ for $S\in {\mathcal{A}}_0$. Uniqueness also extends the scalar homogeneity ${\nu }_{\alpha x}=\mathrm{\mid }\alpha {\mathrm{\mid }}^2{\nu }_x$ and the parallelogram identity from ${\mathcal{A}}_0$ to every Borel set: the corresponding finite positive measures agree on ${\mathcal{A}}_0$ and have the same total mass. For $x,y\in H$ define ${\nu }_{x,y}(S):=\frac{1}{4}{\sum }_{k=0}^3{i}^k{\nu }_{x+i^{k}y}(S)$; this is a finite complex measure and for $S\in {\mathcal{A}}_0$ it coincides with $⟨P(S)x,y⟩$ by the polarization identity [F2]. The scalar homogeneity and parallelogram identity imply, by expansion of the four polarization terms, that $(x,y)\mapsto {\nu }_{x,y}(S)$ is sesquilinear and Hermitian for every Borel $S$, with ${\nu }_{x,x}(S)={\nu }_x(S)\ge 0$. Positivity gives Cauchy–Schwarz for this form (apply $0\le {\nu }_{x+\lambda y}(S)$ for all $\lambda \in \mathbb{C}$), so $\mathrm{\mid }{\nu }_{x,y}(S)\mathrm{\mid }\le {\nu }_x(S{)}^{1\mathrm{/}2}{\nu }_y(S{)}^{1\mathrm{/}2}\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$.

[The operators $G(S)$] Fix a Borel $S\subseteq {\mathbb{R}}^2$. For fixed $x$, the map $y\mapsto \overline{{\nu }_{x,y}(S)}$ is a bounded linear functional of norm at most $\mathrm{\parallel }x\mathrm{\parallel }$ by step 3.1, so by [F7] there is a unique $G(S)x\in H$ with $\overline{{\nu }_{x,y}(S)}=⟨y,G(S)x⟩$, that is ${\nu }_{x,y}(S)=⟨G(S)x,y⟩$, and $\mathrm{\parallel }G(S)x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }$. The assignment $x\mapsto G(S)x$ is linear because the form is linear in its first argument and the representative is unique, so $G(S)\in \mathcal{B}(H)$ with $\mathrm{\parallel }G(S)\mathrm{\parallel }\le 1$; moreover $⟨G(S)x,y⟩={\nu }_{x,y}(S)=\overline{{\nu }_{y,x}(S)}=\overline{⟨G(S)y,x⟩}=⟨x,G(S)y⟩$, so $G(S)$ is self-adjoint, and $⟨G(S)x,x⟩={\nu }_x(S)\in [0,\mathrm{\parallel }x{\mathrm{\parallel }}^2]$ gives $0\le G(S)\le I$. For every $A\in {\mathcal{A}}_0$, step 3.1 gives ${\nu }_{x,y}(A)=⟨P(A)x,y⟩$, so the defining pairings imply $G(A)=P(A)$; in particular $G(B\times C)=E(B)F(C)$.

[Multiplicativity and the PVM axioms] First observe that for arbitrary Borel $S$ and rectangle $R=B^{\mathrm{\prime }}\times C^{\mathrm{\prime }}$, for every $x,y$ the scalar set functions $S^{\mathrm{\prime }}\mapsto ⟨G(S^{\mathrm{\prime }}\cap R)x,y⟩={\nu }_{x,y}(S^{\mathrm{\prime }}\cap R)$ and $S^{\mathrm{\prime }}\mapsto ⟨G(S^{\mathrm{\prime }})G(R)x,y⟩={\nu }_{G(R)x,y}(S^{\mathrm{\prime }})$ are finite complex measures; they agree on rectangles $S^{\mathrm{\prime }}=B\times C$ because $G(S^{\mathrm{\prime }}\cap R)=E(B\cap B^{\mathrm{\prime }})F(C\cap C^{\mathrm{\prime }})=E(B)E(B^{\mathrm{\prime }})F(C)F(C^{\mathrm{\prime }})=E(B)F(C)E(B^{\mathrm{\prime }})F(C^{\mathrm{\prime }})=G(S^{\mathrm{\prime }})G(R)$ by [F1] and the commutation hypothesis, and they agree on ${\mathbb{R}}^2$ since each value is $G(R)$. By step 1.2 [F8], these scalar measures agree on all Borel $S^{\mathrm{\prime }}$, so $G(S^{\mathrm{\prime }}\cap R)=G(S^{\mathrm{\prime }})G(R)$ for all Borel $S^{\mathrm{\prime }}$. Now fix arbitrary Borel $S$ and compare, for every $x,y$, the finite complex measures $T\mapsto ⟨G(S\cap T)x,y⟩={\nu }_{x,y}(S\cap T)$ and $T\mapsto ⟨G(S)G(T)x,y⟩={\nu }_{x,G(S)y}(T)$; by the previous result they agree on every rectangle $T$, and by step 1.2 [F8] they agree for all Borel $T$. Taking $S=T$ yields $G(S{)}^2=G(S)$, so every $G(S)$ is an orthogonal projection; $G(\mathrm{\varnothing })=0$ and $G({\mathbb{R}}^2)=I$ follow from the rectangle values. Strong countable additivity holds because for pairwise disjoint $S_n$ with union $S$ the scalar measures satisfy ${\nu }_x(S)={\sum }_n{\nu }_x(S_n)$, so ${\sum }_n\mathrm{\parallel }G(S_n)x{\mathrm{\parallel }}^2={\sum }_n{\nu }_x(S_n)={\nu }_x(S)\le \mathrm{\parallel }x{\mathrm{\parallel }}^2<\mathrm{\infty }$ and the orthogonal series ${\sum }_{n}G(S_n)x$ converges in norm to a vector whose pairings with every $y$ equal ${\sum }_n{\nu }_{x,y}(S_n)={\nu }_{x,y}(S)=⟨G(S)x,y⟩$, hence to $G(S)x$. Finally the scalar measures of $G$ are the ${\nu }_{\psi }$, finite Borel measures on ${\mathbb{R}}^2$ and hence regular by [F6]; so $G$ is a regular PVM with $G(B\times C)=E(B)F(C)$.

[Marginals and uniqueness] For the marginal, $G(B\times \mathbb{R})=E(B)F(\mathbb{R})=E(B)$ and $G(\mathbb{R}\times C)=F(C)$ by [F1]. If $G^{\mathrm{\prime }}$ is another regular PVM with $G^{\mathrm{\prime }}(B\times C)=E(B)F(C)$, then for every $x,y$ the finite complex measures $S\mapsto ⟨G^{\mathrm{\prime }}(S)x,y⟩$ and ${\nu }_{x,y}$ agree on rectangles, including ${\mathbb{R}}^2$, so step 1.2 [F8] gives their equality on $\mathcal{B}({\mathbb{R}}^2)$. Taking $y=G^{\mathrm{\prime }}(S)x-G(S)x$ gives $G^{\mathrm{\prime }}(S)=G(S)$ for every Borel $S$. Hence $G$ is unique, completing the proof.