For each $nin mathbb N$, let $mathcal M_n$ be a finite von Neumann algebra with a faithful trace $tau_n$. Fix a non-principal ultrafilter $omega$ on $mathbb N$. Let $mathcal M^omega$ be the tracial ultraproduct of $(mathcal M, tau_n)$ (see the definition in https://arxiv.org/abs/1212.5457).

Assume first that $A=(A_i)in mathcal M^omega$ and $A_i$ are self-adjoint (or normal). Denote by $mu_{A_i}$ and $mu_A$ the spectral measures corresponding to the operators $A_i$ and $A$ respectively. It is known that the weak ultralimit of $mu_{A_i}$ (with respect to $omega$) is $mu_A$.

If $A=(A_i)$ is not self-adjoint, one can define the Brown measure $mu_{A_i}$ and $mu_A$ (see the definition in https://arxiv.org/abs/math/0605251). We can ask whether the ultralimit of $mu_{A_i}$ is $mu_A$. The following trivial example gives a negative answer. Assume $mathcal M_i=M_i(mathbf C)$ and $A_iin mathcal M_i$ is the matrix having 1 under the diagonal and 0 in the rest of entires.

I want to modify slightly the previous question. Let $R$ be a $*$-algebra and $phi_i:Rto mathcal M_i$ a $*$-homomorphism such that for every $xin R$, $sup |phi_i(x)|<infty$. Then we can define $phi=(phi_i):Rto mathcal M^omega$. Assume, moreover that $phi$ is **injective**.

Let $xin R$. Is it true that the ultralimit of $mu_{phi_i(x)}$ is

$mu_{phi(x)}$?