The trace and integrable commutators of the measurable operators affiliated to a semifinite von Neumann algebra

Assume that $\tau$ is a faithful normal semifinite trace on a von Neumann algebra ${\mathcal{M}}$, $I$ is the unit of $\mathcal{M}$, $S({\mathcal{M}},\tau )$ is the $*$-algebra of $\tau$-measurable operators, and $L_1({\mathcal{M}},\tau)$ is the Banach space of \hbox{$\tau$-integrable} operators. We present a new proof of the following generalization of Putnam's theorem (1951): No positive self-commutator $[A^*, A]$ with $A\in S({\mathcal{M}}, \tau )$ is invertible in ${\mathcal{M}}$. If $\tau$ is infinite then no positive self-commutator $[A^*, A]$ with $A\in S({\mathcal{M}}, \tau )$ can be of the form $\lambda I +K$, where $\lambda$ is a nonzero complex number and $K$ is a $\tau$-compact operator. Given $A, B \in S({\mathcal{M}}, \tau )$ with $[A, B]\in L_1({\mathcal{M}},\tau)$ we seek for the conditions that $\tau ([A, B])=0$. If $X\in S({\mathcal{M}}, \tau )$ and $Y=Y^3 \in {\mathcal{M}}$ with $[X, Y]\in L_1({\mathcal{M}},\tau)$ then $\tau ([X, Y])=0$. If $A^2=A\in S({\mathcal{M}},\tau)$ and $[A^*, A]\in L_1({\mathcal{M}},\tau)$ then $\tau ([A^*, A])=0$. If a partial isometry $U$ lies in ${\mathcal{M}}$ and $U^n=0$ for some integer $n\geq 2$ then $U^{n-1}$ is a commutator and $U^{n-1}\in L_1({\mathcal{M}},\tau)$ implies that $\tau (U^{n-1})=0$.