normal trace on $M$. By $\rho$ we denote the lifting on $L^{\infty}(\Omega)$ (see [@G0]). Now define a finite trace $\varphi_\omega$ on $M$ by $\varphi_\omega(x)=\rho(\Phi_0(x))(\omega)$, where $\omega\in\Omega$. Due to [@Dik Lemma 6.4.1] the function $s_\omega(x,y)=\varphi_\omega(y^*x)$ is a bi-trace on $M$, and therefore, the equality $\langle x,y\rangle_\omega=s_\omega(x,y)$ defines a quasi-inner product on $M$.

Denote $I_\omega=\{x\in M: \ s_\omega(x,x)=0\}$. It is known that $I_\omega$ is a two-sided ideal in $M$, therefore, one considers the quotient space $\Gamma_\omega=M/I_\omega$, by $\pi_\omega:M\to\Gamma_\omega$ we denote the canonical mapping. The involution and multiplication are defined on $\G_\w$ by the usual way, i.e. $\pi_\w(x)^*=\pi_\w(x^*)$ and $\pi_\w(x)\cdot\pi_\w(y)=\pi_\w(xy)$. According to [@Dik Proposition 6.2.3] $\G_\w$ is a Hilbert algebra. By $\ch(\w)$ we denote the Hilbert space which is the completion of $\G_\w$; the inner product in $\ch(\w)$ we denote by the same symbol, i.e. $\langle\cdot,\cdot\rangle_\w$.

The mapping $\pi_\w(x)\to \pi_\w(y)\pi_\w(x)$, $x,y\in M$, can be extended by continuity to a bounded linear operator $T_\w(y)$ on $\ch(\w)$. It is known [@Dik] that $T_\w(x)$ is a representation of $M$ in $\ch(\w)$. Let $M(\w)$ be the von Neumann algebra generated by $T_\w(M)$, i.e. $M(\w)=T_\w(M)''$. By $\m_\w$ we denote the natural trace on $M(\w)$ which is defined by $\m_\w(\pi_\w(x))=\langle\pi_\w(x)\id_\w,\id_\w\rangle_\w$ for all $\pi_\w(x)\in\G_\w$. One can see that $\m_\w$ is 