Essentially invertible measurable operators affiliated to a semifinite von Neumann algebra and commutators

Suppose that a von Neumann operator algebra $\mathcal{M}$ acts on a Hilbert space $\mathcal{H}$ and $\tau$ is a faithful normal semifinite trace on $\mathcal{M}$. If Hermitian operators $X, Y \in S(\mathcal{M}, \tau )$ are such that $-X\leq Y \leq X$ and $Y$ is $\tau$-essentially invertible then so is $X$. Let $0<p\leq 1$. If a $p$-hyponormal operator $A\in S(\mathcal{M}, \tau )$ is right $\tau$-essentially invertible then $A$ is $\tau$-essentially invertible. If a $p$-hyponormal operator $A\in \mathcal{B}(\mathcal{H})$ is right invertible then $A$ is invertible in $\mathcal{B}(\mathcal{H})$. If a hyponormal operator $A \in S( \mathcal{M}, \tau )$ has a right inverse in $S(\mathcal{M}, \tau)$ then $A$ is invertible in $S(\mathcal{M}, \tau)$. If $A, T\in \mathcal{M}$ and $\mu_t(A^n)^{\frac1n}\to 0$ as $n \to \infty$ for every $t>0$ then $AT$ ($TA$) has no right (left) $\tau$-essential inverse in $S(\mathcal{M}, \tau )$. Suppose that $\mathcal{H}$ is separable and $\dim \mathcal{H}=\infty$. A right (left) essentially invertible operator $A \in \mathcal{B}(\mathcal{H})$ is a commutator if and only if the right (left) essential inverse of $A$ is a commutator.