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Given two positive $\sigma$-finite measures $\mu_{1/2}$ on the spaces $X_{1/2}$ one can define the product measure $\mu_1\otimes\mu_2$ on the product space $X_1\times X_2$. It can be proved that the following representation holds $$ \mu_1\otimes\mu_2(A)=\int_{X_2}\mu_1(A^{x_2})d\mu_2(x_2)=\int_{X_1}\mu_2(A_{x_1})d\mu_1(x_1) $$ where $A_{x_1}$ and $A^{x_2}$ are the usual slices of $A$ in the two "directions" for fixed points $x_1,x_2$.

Now suppose to have two commuting projection valued measures $E_{1/2}:\mathcal{B}(X_{1/2})\rightarrow P(H)$ where $H$ is a complex separable Hilbert space, then there exists a unique joint projection valued measure $E:\mathcal{B}(X_1\times X_2)\rightarrow P(H)$ such that $$ E(A_1\times A_2)=E_1(A_1)E_2(A_2). $$

My question is: does there exist some "integral representation" of $E$ in terms of $E_1$ and $E_2$ similarly to the $\mu_1\otimes\mu_2$ case?

I would like to have some relation like $$ E(A)=\int_{X_2} E_1(A^{x_2})d E_2(x_2)=\int_{X_1} E_2(A_{x_1})d E_1(x_1) $$ although I don't know how to define this rigorously.

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A standard way to prove the first result is to consider the family of subsets of the product for which the result is valid. This contains the rectangles, i.e., products of measurable subsets (by the definition), is closed under the formation of disjoint finite unions (trivial) and formations of unions of increasing sequences (Beppo-Levi). It follows from various descriptions of the generated $\sigma$-algebra that it is valid as stated. This proof can, with a bit of work, be extended to your case.

Two remarks.

  1. One uses the ultraweak topology on $L(H);

  2. Since everything in sight commutes, one can work with an $L^\infty$-space as range.

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