$VN^{\infty}$ as an example of a quantum group I'm trying to learn quantum groups and I have the following problem: Suppose that $G$ is a locally compact group and 
$$VN^{\infty}(G) = \bigg\{L_f \in \mathcal{B}(L^2(G))\ :\ f \in L^{\infty}\bigg\},$$
where $L_f:L^2(G)\rightarrow L^2(G)$ is defined by $L_f(g)(x) = f(x)g(x)$. I'd like to show that $\varphi:VN^{\infty}(G)\rightarrow [0,\infty]$ given by
$$\varphi(L_f) = \int_G\ f\ d\mu,$$
is a normal weight ($\mu$ is a left Haar measure). The first definition of normality that I read was lower semi-continuity with respect to the ultraweak topology. It wasn't very convenient for me, so I found a characterization saying that $\varphi$ is normal if for every increasing net $(a_i)_{i\in I}$ we have
$$\sup_{i\in I}\ \varphi(a_i) = \varphi\bigg(\sup_{i\in I}\ a_i\bigg).$$
That seemed a lot better, I thought 'Ok, so now I simply apply some sort of Monotone Convergence Theorem, and I'm done!' Much to my surprise, I quickly learnt (via google search) that MCT is not valid for nets, sending me back to square one. Am I missing something trivial? Could you please give me any hints on how to proceed? 
 A: In full generality (i.e. without additional topological assumptions on $G$) this is not so easy, but indeed true. As $f\mapsto L_f$ is a $\ast$-isomorphism from $L^\infty(G)$ onto $\mathrm{VN}^\infty(G)$, one can work directly on $L^\infty(G)$.
The crucial fact (see 443A here: https://www1.essex.ac.uk/maths/people/fremlin/chap44.pdf) is that a Haar measure is strictly localizable, that is, there exists a pairwise disjoint family $(A_j)_{j\in J}$ of measurable subsets of $G$ with $\mu(A_j)<\infty$ for all $j\in J$ such that $B\subset G$ is measurable if and only if $A_j\cap B$ is measurable for all $j\in J$ and in this case $\mu(B)=\sum_{j\in J}\mu(A_j\cap B)$.
For $F\subset J$ finite let
$$
\phi_F(f)=\sum_{j\in F}\int_{A_j} f\,d\mu.
$$
Since $L^\infty(G)_\ast\cong L^1(G)$ under the canonical map (for this fact we also need that $\mu$ is localizable), the functional $\phi_F$ is weak$^\ast$ continuous. Thus $\phi=\sup_{J\subset F\text{ finite}}\phi_F$ is weak$^\ast$ lower semicontinuous.
