Spectral Measures: Property

Question

Given a Hilbert space $\mathcal{H}$.

Consider a spectral measure: $$E:\Sigma(\Omega)\to\mathcal{B}(\mathcal{H})$$

Can you give me a hint for: $$E(A)E(B)=E(A\cap B)$$

So far for disjoints I checked: $$A\cap B=\varnothing:\quad E(A)E(B)=\mathbb{0}$$

Answer 1

Consider the partitions $$ A=(A\setminus B)\cup (A\cap B), $$ $$ B=(B\setminus A)\cup (A\cap B). $$ Also, check out the excellent book by Konrad Schmüdgen for these types of elementary things.

Answer 2

Just an elaboration of user161825's answer.

The spectral measure maps disjoint Borel subsets to projections with orthogonal images.

Therefore the projectors $E(A\backslash B)$, $E(A \cap B)$ and $E(B\backslash A)$ have zero pairwise products.

We have $E(A) = E(A\backslash B) + E(A \cap B)$ and $E(B) = E(B\backslash A) + E(A\cap B)$. Now multiply.