Assume that $ \tau $ is a faithful normal semifinite traceon 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 $ \tau $ -integrable operators.We present a new proof of the following generalizationof 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 infinitethen 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 commutatorand $ U^{n-1}\in L_{1}({\mathcal{M}},\tau) $ implies that $ \tau(U^{n-1})=0 $ .
