Let $A$ be a $\sigma$-weakly closed $*$-subalgebra of the $W^*$-algebra $\prod_{i\in I}^{\ell^\infty} M_{n_i}(\mathbb{C})$. I believe that we must have $A\cong \prod_{j\in J}^{\ell^\infty} M_{m_j}(\mathbb{C})$ for certain $(m_j)_{j\in J}$.
Does anyone have an idea how we can show this?
Write $M=\prod_{i\in I}^{\ell^\infty} M_{n_i}(\mathbb{C})$
Let $p\in A$ be a nonzero projection. Let $A_0\subset pAp$ be a masa. If $A_0$ is not diffuse, then it has a minimal projection $p_0\leq p$. Otherwise, $A_0$ is diffuse, which implies that it has a separable diffuse subalgebra $A_1$; hence $A_1\simeq L^\infty[0,1]$ [Theorem III.1.22, Takesaki I]. The identity function in $L^\infty[0,1]$ gives us a selfadjoint element $x\in A_1$ with empty point-spectrum. But $x\in M$, so $x=\prod_i x_i$, with each $x_i$ a selfadjoint matrix. We can write $x_i=\sum_{k=1}^{n_i} \lambda_{i,k}p_{i,k}$ where each $p_{i,k}$ is a minimal projection. Since $x\ne0$, there exist $i,k$ such that $\lambda=\lambda_{i,k}\ne0$. Then $1_{\{\lambda\}}(x)\geq p_{i,k}$. As Borel functional calculus stays within a von Neumann algebra, this shows that $1_{\{\lambda\}}(x)\in A_1$, a contradiction since $1_{\{\lambda\}}(x)=0$ in $L^\infty[0,1]$. So $p$ majorizes a nonzero minimal projection.
The previous paragraph shows that $A$ cannot have type II nor type III parts. Hence $A$ is type I. The centre $Z(A)$ of $A$ is an abelian von Neumann algebra with no diffuse part; by constructing a maximal family of pairwise orthogonal minimal projections in $Z(A)$ we deduce that $Z(A)=\ell^\infty(J)$ for some set $J$. Given one of those minimal projections $q\in Z(A)$, the minimality gives us that $qA$ has to be a factor. As $A$ is type I, so is $qA$; and not having diffuse projections, $qA\simeq M_{n(q)}(\mathbb C)$ for some $n(q)\in\mathbb N$ (we can disregard the case $qA\simeq B(\ell^2(\mathbb N))$ because in that case $qA$ would contain a diffuse subalgebra). It follows that $$ A\simeq\prod_{j\in J}M_{n(j)}(\mathbb C). $$