This is a reframing of my previous question from a Banach lattice perspective: Does weak $L^2$ approximation implies $L^2$ approximation under a condition similar to convexity? The previous question has a negative answer in general, but the case for real-valued functions seems still open. Given this distinction between real- and complex-valued cases, I'm wondering if there is a Banach lattice perspective on this that I previously ignored. The following is the reframed question:
Let $(\Omega, \mu)$ be a probability space. Then $L^\infty((\Omega, \mu), \mathbb{R})$ is a Banach lattice and its unit ball is a complete lattice. Let $S$ be a subset of the unit ball s.t. for any countably many elements of $S$, their join and meet are in $S$. Furthermore, assume that arbitrary joins and meets can be approximated arbitrarily well, i.e., for any nonempty $S_0 \subseteq S$ and $\epsilon > 0$, there exists $s \in S$ s.t. $\wedge S_0 \leq s \leq \wedge S_0 + \epsilon$ and similarly for joins. Does this imply that the weak$^\ast$ closure of $S$ and its norm closure coincide? If not, does the weak$^\ast$ closure of $S$ coincide with the $p$-norm closure of $S$ for some (and therefore for all) $1 \leq p < \infty$? If the first question has a positive answer, does something similar holds in other Banach lattices (or perhaps Banach lattices whose unit balls are complete lattices) as well?
As pointed out by @erz's comment, for a subset of $L^\infty((\Omega, \mu), \mathbb{R})$, $S$ is countably complete as a lattice already implies it is complete, so the condition I previously wrote down, that arbitrary joins and meets can be approximated, is redundant.
