When reading the proof of the Proposition2.6.1, I met with some problems.
The first one is that how to prove that $\tau(T)=\infty$?
My thought: we can take $R=\chi_{(t,\|B\|]}(B)$. In a semi-finite factor, a projection $T$ is infinite iff $\tau(T)=\infty$. We need to check that $T$ is infinite.
The second one is that how to get the decomposition of $T$.
My thought: For any $z\in \sigma(T)$, we have $z=(z-1)\chi_{[0,\infty)}(z-1)-(\chi_{(0,\|T\|]}(z)-z)\chi_{[0,\infty)}(\chi_{(0,\|T\|]}(z)-z)+\chi_{(0,\|T\|]}(z)$, then we use the functional calculus.
The operator $T$ is an infinite sum of equivalent projections. You have $$ T\geq\sum_{k=1}^nR_k,\qquad\qquad n\in\mathbb N, $$ so $$ \tau(T)\geq\tau\big(\sum_{k=1}^nR_k\big)=\sum_{k=1}^n\tau(R_k)=n\,\tau(R),\qquad\qquad n\in\mathbb N, $$ so $\tau(T)=\infty$.