fundamental system of neighborhoods of $0$ is given by $$V(B,\varepsilon,
\delta ) = \{x\in S(M)\colon \ \mbox{there exists } \ p\in P(M),
z\in P(Z(M))$$ $$\mbox{ such that} \ xp\in M,
\|xp\|_{M}\leq\varepsilon, \ z^\bot \in W(B,\varepsilon,\delta), \
\Phi_M(zp^\bot)\leq\varepsilon z\},$$ where $\|\cdot\|_{M}$ is the $C^*$-norm in $M.$ It is known that $(S(M),t(M))$ is a complete topological $*$-algebra [@Y].

From [@Mur_m §3.5] we have the following criterion for convergence in the topology $t(M). $

\[2.1.\] A net $\{x_\alpha\}_{\alpha\in A} \subset S(M)$ converges to zero in the topology $t(M)$ if and only if $\Phi_M(E^\bot_\lambda (|x_\alpha|) \stackrel{t(M)}{\longrightarrow} 0$ for any $\lambda>0.$

Following [@CZ] an operator $x\in S(M)$ is said to be [*$\Phi$-integrable*]{} if there exists a sequence $\{x_n\} \subset M$ such that $x_n \stackrel{t(M)}{\to} x $ and $\|x_n-x_m\|_\Phi
\stackrel{t(Z)}{\longrightarrow} 0$ as $n,m \to \infty.$

Let $x$ be a $\Phi$-integrable operator from $ S(M).$ Then there exists a $\widehat{\Phi}(x)\in S(Z)$ such that $\Phi(x_n)
\stackrel{t(Z)}{\longrightarrow}\widehat{\Phi}(x).$ In addition $\widehat{\Phi}(x)$ does not depend on the choice of a sequence $\{x_n\}\subset M,$ for which $x_n\stackrel{t(M)}{\longrightarrow}x,$ $\Phi(|x_n-x_m|)
\stackrel{t(Z)}{\longrightarrow}0$ [@CZ]. It is clear that each operator $x\in M$ is $\Phi$-integrable and $\widehat{\Phi}
(x)=\Phi(x).$

Denote by $L_1(M,\Phi)$ the set of all $\Phi$-integrable operators from $S(M).$ If $x\in S(M)$ then $x\in L_1(M,\Phi)$ iff $|x|\in
