In the book Kadison-Ringrose II in proposition 8.5.2 (image attached below) it seems that the authors used the fact:
Given a non-zero positive operator $T$ in a von Neumann algebra there exists a suitable spectral projection $G$ of $T$, and some positive real number $c$ such that $cG\le T$.
I tried to verify this statement using Spectral Theorem (KR I, Theorem 5.2.2). By part (iv) of the spectral theorem we can get $c>0$ such that $c(I-E)\le T(I-E)$ for some spectral projection of $T$. But this doesn't imply the above fact.
Question 1. How does the above fact follow from Theorem 5.2.2?
Question 2. Is the converse of Proposition 8.5.2(i) true? What I mean is suppose $\mathcal{R}$ is a vNa and $\rho$ is a tracial weight on it such that for every non-zero $A\in \mathcal{R}^+$, there is a non-zero projection $G$ in $F_\rho$, and a positive number $a$ s.t. $A\ge aG$. Then can we say that $\rho$ is semi-finite (see definition 7.5.1)?
Thank you.
Theorem 5.2.2 in Kadison-Ringrose is not my favourite formulation of the Spectral Theorem, partly because it obscures the fact that $(-\infty,\lambda)\longmapsto E_\lambda$ induces a projection-valued measure on the Borel sets of $\mathbb R$.
Since $T$ is positive and nonzero, there exists $c>0$ such that $$[c,\|T\|]\cap\sigma(T)\ne\varnothing.$$ Let $G=E([c,\|T\|])$, that is $$ G=\int_{\sigma(T)}1_{[c,\|T\|]}(\lambda)\,dE_\lambda. $$ The fact that $G$ is a projection follows from typical computations in functional calculus, and $$ cG\leq\int_{\sigma(T)\cap[c,\|T\|]}\lambda\,dE_\lambda\leq\int_{\sigma(T)}\lambda\,dE_\lambda=T. $$
The above said, the theorem does give you the inequality directly. As you say, $$ c(I-E_c)\leq T(I-E_c). $$ Another fact that is obscured by the statement of the theorem (though it follows straighforwardly from it) is that $TE_\lambda=E_\lambda T$ for all $\lambda $ (in fact, $E_\lambda$ belongs to the von Neumann algebra generated by $T$). Then $$ T(I-E_c)=T^{1/2}(I-E_c)T^{1/2}\leq T^{1/2}IT^{1/2}=T. $$
As for 8.5.2.(i), the converse is true, see this question and answer. Note that this only makes sense if the weight is faithful. If $\rho(A)=0$, no nonzero $G\in F_\rho$ as in the statement can exist.