I'm reading the book "Introduction to Hilbert space and the Theory of Spectral Multiplicity" by Paul R. Halmos where I have trouble to follow the proof of Theorem 36.2.
Theorem 36.2
If $E$ is a finitely additive, projection-valued set function on the class $S$ of all measurable subsets of a measurable space (in particular if $E$ is a spectral measure), then $E$ is modular and multiplicative, i.e. if $M$ and $N$ are in $S$, then
- $E(M\cup N)+E(M\cap N)=E(M)+E(N)$
- $E(M \cup N) =E(M)E(N).$
The proof uses the monotonicity of finitely additive, projection-valued set function, i.e. that for $M,N \in S$ with $M \subset N$ it holds that $E(M)\leq E(N)$. Here $\leq$ is used to indicate that $E(N)-E(M)$ is a positive (i.e. $\langle Ax,x \rangle \geq 0$ for a self-adjoint linear operator $A$ and every $x$). Furthermore $\leq$ forms a partial ordering on the set of self-adjoint operators (over the considered Hilbert space).
In the first step the modularity of $E$ is proved. Then the proof states that, due to the monotonicity, $E(M \cap N)\leq E(M)\leq E(M\cap N)$ holds for arbitrary sets $M, N \in S$. While this is clear, I do not understand the conclusion which is drawn from this without further explanation, which is that $$E(M)E(M\cap N)=E(M\cap N)$$ and $$E(M)E(M\cup N)=E(M)$$.
I'm able to conclude (using the idempotence of projections, the monotonicity of $E$ and the $transitivity$ of $\leq$) that $$E(M\cap N)=E(M\cap N)E(M\cap N)\leq E(M)E(M\cap N)$$
However I have no idea why the other direction should hold which would allow the conclusion that $E(M)E(M \cap N)=E(M \cap N)$. Any hints?