Proving that a spectral measure is $\sigma$-additive

Solution 1:

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 $S_i\times T_i$: $$ E_A\circ F_B = \sum_{j, k}E_{A_j'}\circ F_{B_k'} = \sum_i \sum_{(j, k)\in S_i\times T_i}E_{A_j'}\circ F_{B_k'} = \sum_i E_{A_i}\circ F_{B_i}.$$