Projection valued measure: $E(A \cap B) = E(A) \cdot E(B)$ if disjoint Let $M \subseteq \mathbb{R}$. Define a projection valued measure as a function $E: \mathbb{B}(M) \rightarrow B(H)$ from the set of Borel sets of $M$ to the set of bounded linear operators on a Hilbert space $H$ s.t. for each $S \in \mathbb{B}(M)$ the operator $E(S)$ is an orthogonal projection. Furthermore for $S_1, S_2, \ldots \in \mathbb{B}(M)$ disjoint we have for each $x \in H: E(\cup_{n = 1}^{\infty} S_n) x = \sum_{n = 1}^{\infty} E(S) x$.
Now I want to show that for any disjoint sets $A, B \in \mathbb{B}(M)$ we have $E(A \cap B) = E(A) \cdot E(B)$.
I was able to show that if $A \subseteq B$ then $rg(A) \subseteq rg(B)$. Together with the fact that for orthogonal operators $S,T$ we have $rg(S) \subseteq rg(T) \Leftrightarrow ST = TS = S$ we get that
$$ E(A) \cdot E(A \cap B) = E(A \cap B), ~~~ E(B) \cdot E(A \cap B) = E(A \cap B)$$
Multiplying the two gives that $E(A) E(B) E(A \cap B) = E(A \cap B)$ which implies that $rg(E(A\cap B)) \subseteq rg(E(A) \cdot E(B))$.
If I can show the other inclusion then we are done, but I don't know how. I am almost certain that I am missing some easy argument.
 A: Lemma 1.  If $E$ and $F$ are orthogonal projections and $E+F$ is also an orthogonal projection, then $EF=0$.
Proof. For every $x$ in the range of $F$ we have
$$
  \|x\|^2 =
  \|Fx\|^2 \leq
  \|Fx\|^2 + \|Ex\|^2 =
  \langle Fx, Fx\rangle  + \langle Ex, Ex\rangle  =
  \langle (F+E)x, x\rangle  \leq  \|x\|^2,
  $$
so equality holds throughout and in particular $Ex=0$.  This says that the range of $F$ is contained in the null space
of $E$, and hence $EF=0$.
Lemma 2.  If $A$ and $B$ are disjoint then $E(A)E(B)=0$.
Proof. $E(A)+E(B)=E(A\cup B)$ so the conclusion follows from Lemma 1.
Theorem.  If $A$ and $B$ are disjoint then $E(A)E(B)=E(A\cap B)$.
Proof.  In view of Lemma 2, it suffices to prove that $E(\emptyset)=0$, but this follows immediately from
$$
  E(\emptyset) =   E(\emptyset\cup \emptyset)  =   E(\emptyset)+E(\emptyset).
  $$
Theorem.  For all Borel sets  $A$ and $B$ whatsoever one has that $E(A)E(B)=E(A\cap B)$.
Proof. Letting $A'=A\setminus B$ and $B'=B\setminus A$ we have that $A$ is the disjoint union of $A'$ and $A\cap B$,
whence
$$
  E(A)= E(A')+ E(A\cap B),
  $$
and similarly  $E(B)= E(B')+ E(A\cap B)$.   Since any two of $A'$, $B'$, and $A\cap B$ are disjoint, we have that
$$
  E(A)E(B)= (E(A')+ E(A\cap B))(E(B')+ E(A\cap B)) = E(A\cap B)^2 = E(A\cap B).
  $$
