Can it be said that $E(A_1)E(A_2) = E(A_2) E(A_1) = 0$ for $A_1 \cap A_2 = \varnothing\ $?

Question

Let $(X,\mathcal A)$ be a measurable space. Let $\mathcal H$ be a Hilbert space and $E : \mathcal A \longrightarrow \mathscr P (\mathcal H)$ be a projection valued map. For all $x \in \mathcal H$ with $\|x\| = 1$ define $\mu_x : \mathcal A \longrightarrow [0,\infty]$ by $\mu_x (A) = \left \langle x, E(A) x \right \rangle,$ $A \in \mathcal A.$ Suppose that $\mu_x$ is a probability measure for all $x \in \mathcal H$ with $\|x\| = 1.$ Then can we conclude the following $:$

If $A_1, A_2 \in \mathcal A$ with $A_1 \cap A_2 = \varnothing$ then $$E(A_1) E(A_2) = E(A_2) E(A_1) = 0.$$

Our instructor implicitly assuming this result to conclude that $E$ is countably additive but I can't figure out as to why it should be true. Do anybody have any idea about it?

Thanks for your time.

Answer 1

Note that for all $x\in H$ with norm $1$ we have $$\langle x,E(A_1\cup A_2)x\rangle=\mu_x(A_1\cup A_2)=\mu_x(A_1)+\mu_x(A_2)=\langle x,E(A_1)x\rangle+\langle x, E(A_2)x\rangle=\langle x, E(A_1)+E(A_2)x\rangle $$ so, assuming that $H$ is a complex Hilbert space, by the polarization identity, $E(A_1\cup A_2)=E(A_1)+E(A_2)$. Therefore, we have two projections that their sum is a projection, since $E(A_1\cup A_2)$ is a projection. But this yields that they have to be orthogonal.