Projection valued measure: $E(A \cap B) = E(A) \cdot E(B)$ if disjoint

Question

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.

Answer 1

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 $$ E(A)E(B)= (E(A')+ E(A\cap B))(E(B')+ E(A\cap B)) = E(A\cap B)^2 = E(A\cap B). $$