Definition $:$ Let $(Y, \mathcal A)$ be a measurable space, $\mathcal H$ be a Hilbert space and $\mathscr P = \mathscr P(\mathcal H)$ be the set of all orthogonal projections on $\mathcal H.$ Suppose $E : \mathcal A \longrightarrow \mathscr P$ be a mapping satisfying the following conditions $:$
$(1)$ Countable additivity $:$ If $\{A_n\}$ is finite or countable set of disjoint sets, $A_n \in \mathcal A$ and $A = \bigcup\limits_n A_n$ then $$E(A) = \text {s-} \sum\limits_n E(A_n)\ \ (\text {strong limit})$$
$(2)$ Completeness $:$ $E(Y) = I.$
Then $E$ is called a spectral measure on $\mathcal H$ and $(Y,\mathcal A, \mathcal H, E)$ is referred to as a spectral measure space.
Theorem $:$ Let $E$ be a spectral measure on $\mathcal H.$ Then for any $A_1,A_2 \in \mathcal A$ $$E(A_1) E(A_2) = E(A_2) E(A_1) = E(A_1 \cap A_2).$$
The proof of the above theorem goes as follows $:$
Proof $:$ We have $$A_1 \cup A_2 = (A_1 \setminus A_2) \cup (A_2 \setminus A_1) \cup (A_1 \cap A_2).$$
So by property $(1)$ we have $$E(A_1 \cup A_2) = E (A_1 \setminus A_2) + E(A_2 \setminus A_1) + E(A_1 \cap A_2).$$
Multiplying both sides of the above equation by $E(A_1)$ from the left and $E(A_2)$ from the right the result follows.
Though I can't quite follow the last sentence. If $A \subseteq B$ with $A,B \in \mathcal A$ then $B = A \cup (B \setminus A).$ Hence $E(B) = E(A) + E(B \setminus A).$ But then for all $x \in \mathcal H$ we have $$\begin{align*} \left \langle E(B) x, x \right \rangle & = \left \langle (E(A) + E(B \setminus A))x, x \right \rangle \\ & = \left \langle E(A) x, x \right \rangle + \left \langle E(B)x,x \right \rangle \\ & = \left \langle E(A)x , x \right \rangle + \left \| E(B \setminus A) x \right \|^2\ \ (\because E(B \setminus A) \in \mathscr P(\mathcal H)) \\ & \geq \left \langle (E(A)x, x \right \rangle \end{align*}$$
This shows that $E(A) \leq E(B).$ But then $\text {Range} (E(A)) \subseteq \text {Range} (E(B))$ and hence $E(A) E(B) = E(B) E(A) = E(A).$
With this what we get from the proof is $$E(A_1 \cup A_2) = E(A_1 \setminus A_2) E(A_2) + E(A_1) E(A_2 \setminus A_1) + E(A_1 \cap A_2).$$
But now I got stuck. I don't know what are $E(A_1 \setminus A_2) E(A_2)$ and $E(A_1) E(A_2 \setminus A_1).$ Could anyone give some suggestion in this regard?
Thanks in advance.
Hint $:$ Note that if $A, B \in \mathcal A$ are disjoint then $$E(A \cup B) = E(A) + E(B)\ \ \ \ (*)$$ Now multiplying both sides of $(*)$ by $E(B)$ from the right we get $$E(B) = E(A \cup B) E(B) = E(A) E(B) + E(B)^2 = E(A) E(B) + E(B)\ \ (\because E(B) \in \mathscr P(\mathcal H))$$ This shows that $E(A) E(B) = 0.$ Similarly multiplying both sides of $(*)$ by $E(B)$ from the left we get $E(B) E(A) = 0.$
Can you take it from here?