Consider the abstract von Neumann algebra $M = \ell^\infty\text{-}\bigoplus_{i \in I} B(H_i)$. Moreover, we assume $\dim H_i< \infty$ for all $i \in I$. Let $x_i$ be the identity on $B(H_i)$ and let $z_F = \sum_{i\in F} x_i$ where $F$ is a finite subset of $I$. If $\omega:M \to \mathbb{C}$ is a $\sigma$-weakly functional, then is it true that $$\lim_F \sup_{\|m\| \leq 1}|\omega(m)-\omega(z_Fm)|= 0?$$
I tried to estimate $$\sup_m |\omega(m)-\omega(z_Fm)| \leq \|\omega\|\|z_F-1\|$$ so it suffices to show that $$\lim_F \|z_F-1\| = 0.$$
Let $\epsilon > 0$. I want to prove that there is a finite subset $F_0$ of $I$ such that $$F \supseteq F_0 \implies \|z_F-1\| < \epsilon$$ but I have no idea how to construct $F_0$. This approach does not use the $\sigma$-weak continuity of $\omega$ though so it might be flawed.
Notice that $1-z_F$ is a nonzero projection in $M$, so $\|z_F-1\| = 1$, and hence it is impossible to prove that $\lim_F \|z_F-1\| = 0.$ Nevertheless it is true that $$ \lim_F \sup_{\|m\| \leq 1}|\omega(m)-\omega(z_Fm)|= 0, $$ and the reason is as follows: the $\sigma $-weakly continuous linear functionals of $M$ are of the form $$ \omega (m) = \sum_{i\in I} \text{tr}(m_ih_i), \quad\forall m = (m_i)_i\in M, $$ where $$ h = (h_i)_i\in \prod_{i\in I}B(H_i) $$ is such that $$ \infty > \sum_{i\in I}\text{tr}(|h_i|) = \Vert \omega \Vert . \tag {1} $$ Given any such $\omega $, one has for every finite $F\subseteq I$, and every $m$ in $M$, that $$ |\omega(m)-\omega(z_Fm)| = \Big|\sum_{i\in I} \text{tr}(m_ih_i) - \sum_{i\in F} \text{tr}(m_ih_i)\Big| = $$$$ = \Big|\sum_{i\in I\setminus F} \text{tr}(m_ih_i) \Big| \leq \sum_{i\in I\setminus F} |\text{tr}(m_ih_i)| \leq $$$$\leq \sup_i\Vert m_i\Vert \sum_{i\in I\setminus F} \text{tr}(|h_i|) = \Vert m\Vert \sum_{i\in I\setminus F} \text{tr}(|h_i|). $$ Therefore $$ \lim_F \sup_{\|m\| \leq 1}|\omega(m)-\omega(z_Fm)|\leq \lim_F \sum_{i\in I\setminus F} \text{tr}(|h_i|) = 0, $$ by (1).