$\mathcal{E}$ (i.e. $\mathcal{E}(x)=(bo) -
\lim\limits_{n\rightarrow\infty}\mathcal{E}(x_n)$, if $ x_n\in M$ and $\|x_n-x\|_p\stackrel{(o)}\rightarrow0.$) It is clear that $\mathcal{E}(\id)=\id$ and $\|\mathcal{E}\|=\mathbf{1}$.

Let us show $\mathcal{E}$ is positive. Let $0\leq x\in L_p(M,\Phi)$. We will chouse $0\leq x_n\in M$ such that $x_n$ is increasing and $\|x_n-x\|_p\stackrel{(o)}\rightarrow0.$ Then the sequence $\{\mathcal{E}(x_n)\}$ is increasing as well, and one has $\mathcal{E}(x)=(bo) -
\lim\limits_{n\rightarrow\infty}\mathcal{E}(x_n)$. Since $\mathcal{E}(x_n)\geq0$ and thw cone $K$ of positive elements of $L_p(M,\Phi)$ is monotone closed (see[@GaC]), we get $\mathcal{E}(x)\geq 0.$

Let $y\in L_p(N,\Phi)$ and $ y_n\in N$ and $\|y_n-y\|_p\stackrel{(o)}\rightarrow0.$ Then $\mathcal{E}(y)=(bo) -
\lim\limits_{n\rightarrow\infty}\mathcal{E}(y_n)=(bo) -
\lim\limits_{n\rightarrow\infty}y_n=y$, i.e. $\mathcal{E}$ is onto.

Now let $a,b\in N$, $x\in L_p(M,\Phi)$ and $x_n\in M$ which $\|x_n-x\|_p\stackrel{(o)}\rightarrow0.$ Then $$\begin{aligned}
\mathcal{E}(axb)&=&(bo) -
\lim\limits_{n\rightarrow\infty}\mathcal{E}(ax_nb)=(bo) -
\lim\limits_{n\rightarrow\infty}a\mathcal{E}(x_n)b\\[2mm]
&=&a((bo) -
\lim\limits_{n\rightarrow\infty}\mathcal{E}(x_n))b=a\mathcal{E}(x)b.\end{aligned}$$

The defined operator $\mathcal{E}: L_p(M,\Phi)\rightarrow
L_p(N,\Phi)$ in Theorem \[Exp2\] is called a *conditional expectation*.

\[2.3\] Let $\mathcal{E}: L_p(M,\Phi)\rightarrow L_p(N,\Phi)$ be the conditional expectation. Then there exists a measurable bundle of conditional expectations $\mathcal{E}_\omega: L_p(M(\omega),\tau_\omega)\rightarrow L_p(N(\omega),\tau_\omega)$ such that $$\mathcal{E}(x)(\omega)= \mathcal{E}_\omega(x(\omega))$$ for all $x\in L_p(M,\Phi)$ and for almost all $\omega\in\Omega$.

Let $\ell: M\rightarrow \mathcal{L}^\infty(\Omega,X)$ be the noncommutative vector-valued lifting associated with lifting $\rho$. Since $\mathcal{E}(M)\subset N$ can define a linear operator $\varphi_\omega: \{\ell(x)(\omega): x\in M\}\rightarrow
 L_p(N(\omega),\tau_\omega)$ by $$\varphi_\omega(\ell(x)(\omega))=\ell(\mathcal{E}(x))(\omega).$$ From the relations $$\begin{aligned}
\|\varphi_\omega(\ell(x)(\omega))\|_{L_p(N(\omega),\tau_\omega)}&=&
 \|\ell(\mathcal{E}(x))(\omega)\|_{L_p(N(\omega),\tau_\omega)}=\rho(\|\mathcal{E}(x)\|_{p})(\omega)\\[2mm]
 &\leq&\rho(\|x\|_p)(\omega)=\|\ell(x)(\omega)\|_{L_p(M(\omega),\tau_\omega)}
 \end{aligned}$$ follow that 