Im reading the book "An invitation to Von Neumann algebras" from Sunder. I have two questions about two exercises there.
In the first (1.1.4) He said that for a closed densely defined operator the next conditions are equivalent (where $\eta$ means affiliated to, and $M$ is a Von Neumann Algebra)
I) $A \eta M$
II) $A^{\ast} \eta M$
III) If $A=uH$ is thepolar descomposition of $A$, then $u\in M$ and $1_{F}(H)\in M$ for every borel subset $F\subseteq[0,\infty).$
I can see that I and II are equivalent and that both implies III but only for bounded borel subsets. I'm looking for a proof for the general case.
My second question is about a part of the exercise (1.1.6), which asserts that for a densely defined closed operator $A$. If $A\eta M$ then the range projection equals $1_{(0,\infty)}(|A^{\ast}|)$, Why?.
Many thanks in advanse.
Let $F_n=F\cap[0,n]$. Then $F=\bigcup_nF_n$ and $1_{F_n}\nearrow 1_{F}$. Using monotone (or dominated, whatever you prefer) convergence, $$ 1_F(H)=\int_{\sigma(H)}1_F\,dE=\lim_n\int_{\sigma(H)}1_{F_n}\,dE\in M, $$ since the limit is in the weak operator topology (actually sot, but wot is enough).
You have $$\overline{\operatorname{ran} A}=(\ker A^*)^\perp=(\ker AA^*)^\perp=(\ker |A^*|^2)^\perp=\overline{\operatorname{ran} |A^*|}. $$ So it is enough to show that if $A$ is positive, then the range projection of $A$ is $1_{(0,\infty)}(A)$. Since $t\,1_{[0,\infty)}(t)=t$ for all $t\in[0,\infty)$, we get $A=1_{(0,\infty)}(A)\,A$. So $\operatorname{rp}(A)\leq 1_{(0,\infty)}(A)$. And from $$ 1_{(0,\infty)}(A)=\lim_n \int_{\sigma(A)} f_n\,dE $$ with $\{f_n\}$ continuous functions that converge monotonically to $1$, we get that $1_{(0,\infty)}(A)\in\overline{\operatorname{ran} A}$. Thus $1_{(0,\infty)}(A)=\operatorname{rp}(A)$.