Partial order - Unbounded normal operators affiliated with von Neumann algebra. Hello, I have a question which is related to a partial order in a set of self-adjoint operators.
Let $\mathcal{M}$ be a semifinite von Neumann algebra with a faithful semi-finite normal trace $\tau$. Let $T$ and $S$ be two self-adjoint operators (possibly unbounded) $\tau$-measurable (here probably the assumption that they are affiliated with $\mathcal{M}$ is enough) such that 
$0 \leq T \leq S$ i.e. $S-T$ is positive.
How to get that 
$$E_{(s, \infty)}(|T|) \preceq E_{(s, \infty)}(|S|), \ \ s \geq 0,$$
where $E_I(|T|)$ (resp. $E_I(|S|)$) stands for a spectral projection of $T$ (resp. $S$) corresponding to the interval $I$ and $\preceq$ means sub-equivalence relation in Murray-von Neumann sense.
I am looking also for some good references which describe the relation between
$U|T|$ the elements of the polar decomposition of closed densely defined (possibly unbounded) operator $T$  affiliated with some von Neumann algebra $\mathcal{M}$.
I mean that $U$ and each spectral projection of $|T|$ are in this von Neumann algebra.
Probably, I can find this in Takesaki vol 2 or vol 3. 
I will be really grateful for any help.
Thank you, VdM
 A: I assume you are following the proof in Fack-Kosaki (if you are not, we are talking here about Proposition 2.2 and 2.5 there). 
Note that there is no need for absolute value bars since both $T,S$ are positive. 
The key fact is that $E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)=0$ (to be proven afterwards). Using this, we have (using Kaplansky's formula)
\begin{align}
E_{(s,\infty)}(T)&=E_{(s,\infty)}(T)-E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)\sim
E_{(s,\infty)}(T)\vee E_{[0,s]}(S)-E_{[0,s]}(S)\\ &\leq I-E_{[0,s]}(S)=E_{(s,\infty)}(S)
\end{align}
So we only need to prove that $E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)=0$. Now, if $\xi\in
E_{(s,\infty)}(T)H \cap E_{[0,s]}(S)H$ with $\|\xi\|=1$, the following happens:
\begin{align}
\langle T\xi,\xi\rangle&=\langle TE_{(s,\infty)}(T)\xi,E_{(s,\infty)}(T)\xi\rangle
=\|T^{1/2}E_{(s,\infty)}(T)\xi\|^2>s,
\end{align}
\begin{align}
\langle T\xi,\xi\rangle&=\langle E_{[0,s]}(S)TE_{[0,s]}(S)\xi,\xi\rangle
\leq\langle E_{[0,s]}(S)SE_{[0,s]}(S)\xi,\xi\rangle=\|S^{1/2}E_{[0,s]}(S)\xi\|^2\leq s
\end{align}
The contradiction implies that $\xi$ cannot exist. 
