Given the measurable space $\big(\mathbb{R},\mathcal{B(\mathbb{R})}\big)$ (where $\mathcal{B}(\mathbb{R})$ is the Borel's $\sigma$-algebra of $\mathbb{R}$), a Hilbert space $H$ and two spectral measures $E,F:\mathcal{B(\mathbb{R})}\rightarrow B(H)$ (where $B(H)$ is the space of bounded operators $T: H\rightarrow H$); Paul Halmos claims in his book: Introduction to Hilbert Space and the Theory of Spectral Multiplicity, that we can form a spectral measure $G:\mathcal{B(\mathbb{C})}\rightarrow B(H)$ by extending the following map
\begin{align}
\mathcal{R}&\longrightarrow B(H)\\
A\times B&\longmapsto E_A \circ F_B
\end{align}
to the $\sigma$-algebra $\mathcal{B(\mathbb{C}})$ ($\mathcal{R}$ is the set of products of real Borel sets).
I'm having trouble proving that this map is $\sigma$-additive on $\mathcal{R}$ in the strong operator topology on $B(H)$. That is equivalent to prove that for all $x \in H$ the function
\begin{align}
\mathcal{m}_{x}:\mathcal{R}&\longmapsto \mathbb{R}\\
A\times B&\longmapsto \langle x,E_AF_B(x)\rangle
\end{align}
is a premeasure in the sense that if $\{A_i\times B_i\}_{i \in \mathbb{N}}\subset \mathcal{R}$ is a sequence of pairwise disjoint borel rectangles and $A\times B=\bigcup_{i\in \mathbb{N}} A_i\times B_i \in \mathcal{R},$ then $$\langle x,E_AF_B(x)\rangle=\sum_{i\in \mathbb{N}} \langle x,E_{A_i}F_{B_i}(x)\rangle.$$
Could you give me an idea of how to prove this? my work done so far has only given me results that are already implied by the properties of the spectral measures $E$ and $F$.
Best Answer
You need to construct a finer decomposition of $A\times B$. Considering all pairwise intersections of $(A_i)_{i\in\mathbb{N}}$ and the intersections with their complementary, we can construct a family of pairwise disjoint borel sets $(A_j')_{j\in\mathbb{N}}$ such that $A = \bigcup_{j\in\mathbb{N}}A_j'$ and each $A_i$ is a union of some $A_j'$. Doing the same with the $(B_i)_{i\in\mathbb{N}}$, we construct the family $(B_k')_{k\in\mathbb{N}}$. Remark that $A\times B = \bigcup_{j, k}A_j'\times B_k'$ and that there exists disjoint product sets $S_i\times T_i\subset\mathbb{N}\times \mathbb{N}$ such that $A_i\times B_i = \bigcup_{(j, k)\in S_i\times T_i}A_j'\times B_k'$. Then the $\sigma$-additivity of a spectral measure gives $$ E_A = \sum_j E_{A_j'},\ \ F_B = \sum_k F_{B_k'}.$$ We then group the $j$ and $k$ in each set $S_i\times T_i$: $$ E_A\circ F_B = \sum_{j, k}E_{A_j'}\circ F_{B_k'} = \sum_i \sum_{(j, k)\in S_i\times T_i}E_{A_j'}\circ F_{B_k'} = \sum_i E_{A_i}\circ F_{B_i}.$$