$B \subset \mathcal{U}$ is called *bounded*, if set $\{\|x\|: x \in B\}$ is order bounded in $ L_{0}$. An operator $T: \mathcal{U} \rightarrow \mathcal{U}$ is called $L_{0}$-linear, if $T(\alpha x + \beta y)= \alpha T(x) + \beta T(y)$ for all $\alpha,\beta \in L_{0}$ and $x,y \in \mathcal{U}$. An $L_{0}$-linear operator $T$ is called $ L_{0}$–bounded, if for any bounded set $B$ in $\mathcal{U}$, the set $T (B)$ bounded in $\mathcal{U}$. For an $L_{0}$-bounded operator $T$ we put $$\|T\|=\sup\{\|T(x)\|: \|x\|\leq \bf{1}\},$$ where $\bf{1}$ - identity element in $L_{0}$

Let $H$ be a Hilbert space, let $B(H)$ be the $*$-algebra of all bounded linear operators on $H,$ and let $I$ be the identity operator on $H.$ Given a von Neumann algebra $M$ acting on $H,$ denote by $Z(M)$ the center of $M$ and by $P(M)$ the lattice of all projections in $M$. Let $P_{fin}(M)$ be the set of all finite projections in $M.$ A densely-defined closed linear operator $x$ (possibly unbounded) affiliated with $M$ is said to be *measurable* if there exists a sequence $\{p_n\}_{n=1}^{\infty}\subset P(M)$ such that $p_n\uparrow
\mathbf{1}$,  $p_n(H)\subset \mathfrak{D}(x)$ and $p_n^\bot=\mathbf{1}-p_n \in P_{fin}(M) $ for every $n=1,2,\ldots$ (here $\mathfrak{D}(x)$ is the domain of $x$). Let us denote by $S(M)$ 