construct a faithful normal conditional expectation on the ultraproduct of a von Neumann algebra 
The Lemma is from the paper “structure of bicentralizer algebras and inclusions of type III”. When reading the proof of the second part of the Lemma, I met with some problems.

*

*The conditional expectation can be constucted as following；
$$\forall x \in M^\omega, \quad E(x)=\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})$$.

Is it natrual to think of this construction? It is easy to check that
$$\forall y \in \theta(F)'\cap M^\omega, E(y)=y \quad and \quad E(axb)=aE(x)b \quad \forall a,b\in \theta(F)'\cap M^\omega, \forall x\in M^\omega.$$ How to check that $E$ is positive?


*According to  the above construction, we have

$(\varphi \circ E_n)^\omega((x_n)^\omega)=\lim_{n\to \omega}\varphi \circ E_n(x_n)=\lim_{n\to \omega}\varphi (\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})x_n\theta_n(e_{jk}))$
$\varphi^\omega \circ E((x_n)^\omega)=\varphi^\omega(\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})(x_n)^\omega\theta_n(e_{jk})))$.
How to verify that $\varphi^\omega(\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})(x_n)^\omega\theta_n(e_{jk})))=\lim_{n\to \omega}\varphi (\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})x_n\theta_n(e_{jk}))$.
 A: The proof seems to be using that $\theta$ is unital, and it mixes $n$ as the size of $F$ and as the index of the sequences.
With $\theta$ being unital, the matrix units $\{e_{kj}\}$ of $F$ are taken to matrix units $\{\theta(e_{kj})\}$ of $M^\omega$ with $\sum_{k=1}^n\theta(e_{kk})=1$. Then
\begin{align}
E(x)
&=\psi(1)1E(x)=\sum_i\psi(e_{ii})\sum_k\theta(e_{kk})E(x)=\sum_{i,k,j}\psi(e_{ij})\theta(e_{ki}e_{jk})E(x)\\[0.3cm]
&=\sum_{i,k,j}\psi(e_{ij})\theta(e_{ki})E(x)\theta(e_{jk})\\[0.3cm]
&=E\Big(\sum_{i,k,j}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})\Big).
\end{align}
For $e_{st}\in F$, we have
\begin{align}
\theta(e_{st})\sum_{i,k,j}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})
&=\sum_{i,k,j}\psi(e_{ij})\theta(e_{st}e_{ki})x\theta(e_{jk})\\[0.3cm]
&=\sum_{i,j}\psi(e_{ij})\theta(e_{si})x\theta(e_{jt})\\[0.3cm]
&=\sum_{i,j,k}\psi(e_{ij})\theta(e_{si})x\theta(e_{jk}e_{st})\\[0.3cm]
&=\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})\,\theta(e_{st}).\\[0.3cm]
\end{align}
Hence $\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})\in\theta(F)'\cap M^\omega$ and thus
$$
E(x)=E\Big(\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})\Big)=\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk}).
$$
Not only is $E$ positive but it is completely positive, as every conditional expectation is. But if you want to check the positivity directly here, since $\psi$ is a state on the matrices it is of the form
$$
\psi(z)=\operatorname{Tr}(Hz)=\sum_{s=1}^nw_s^*zw_s
$$
\begin{align}
\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x^*x\theta(e_{jk})
&=\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x^*x\theta(e_{jk})
\\[0.3cm]
&=\sum_{i,j,k,s} \theta(e_{ki})x^*w_s^*e_{ij}w_sx\theta(e_{jk})\\[0.3cm]
&=\sum_{i,j,k,s} \theta(e_{ki})x^*w_s^*e_{i1}e_{1j}w_sx\theta(e_{jk})\\
&=\sum_{k,s} \Big(\sum_i\theta(e_{ki})x^*w_s^*e_{1i}^*\Big)\Big(\sum_je_{1j}w_sx\theta(e_{jk})\Big)\\[0.3cm]
&=\sum_{k,s} \Big(\sum_je_{1j}w_sx\theta(e_{jk})\Big)^*\Big(\sum_je_{1j}w_sx\theta(e_{jk})\Big)\geq0.
\end{align}

How to verify that
$$\varphi^\omega(\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})(x_n)^\omega\theta_n(e_{jk})))=\lim_{n\to \omega}\varphi (\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})x_n\theta_n(e_{jk}))$$

The definition of $\varphi^\omega$ is that
$$
\varphi^\omega((x_n)^\omega)=\lim_{n\to\omega}\varphi(x_n). 
$$
