f\chi_E \in L^\infty(\Omega,\Sigma,\mu), \
\|f\chi_E\|_{{L_\infty}(\Omega,\Sigma,\mu)}\leq\varepsilon\}.$$ Here  $\varepsilon, \ \delta $ run over all strictly positive numbers and $B\in\Sigma$,  $\mu(B)<\infty.$ It is known that $(S(M),t(M))$ is a complete topological $*$-algebra. Note that a net $\{f_\alpha\}$ converges locally in measure to $f$ (notation: $f_\alpha \stackrel{t(M)}{\longrightarrow}f$) if and only if $f_\alpha \chi_B $ converges in $\mu$-measure to $f\chi_B$ for each $B\in\Sigma$ with $\mu(B)<\infty$.

Let $M$ be any finite von Neumann algebra, $S(M)$ be the set all measurable operators affiliated to $M$. Let $Z$ be some subalgebra of the center $Z(M)$. Then one may identify $Z$ with $*$-algebra $L_\infty(\Omega,\Sigma,m)$ and do $S(Z)$ with $L_0(\Omega,\Sigma,m)$. Recall that *a center valued (i.e. $Z$-valued) trace* on the von Neumann algebra $M$ is a $Z$-linear mapping $\Phi:M\to Z$ with $\Phi(x^*x)=\Phi(xx^*)\geq 0$ for all $x\in M$. It is clear that $\Phi(M_+)\subset Z_+$. A trace $\Phi$ is said to be *faithful* if the equality $\Phi(x^*x)=0$ implies $x=0$, *normal* if $\Phi(x_{\alpha})\uparrow\Phi(x)$ for every $x_{\alpha},x\in
M_{sa}$, $x_{\alpha}\uparrow x$. Note that the existence of such kind of traces has been studied in [@CZ1].

Let now $M$ be an arbitrary finite von Neumann algebra, $\Phi$ be a center-valued trace on $M$. The locally measure topology $t(M)$ on $S(M)$ is the linear (Hausdorff) topology whose 