Let $(\Omega, \mathcal{F}, \mu)$ be a finite measure space, and let $\mathcal{A}$ be an algebra generating $\mathcal{F}$.
It is well-known that $\mu$ enjoys the following approximation property: For all $\epsilon>0$ and $F\in\mathcal{F}$, there exists $A \in \mathcal{A}$ such that $\mu(F \triangle A) < \epsilon$.
My general question is
When does the following, stronger approximation property hold?
For all $\epsilon>0$ and $F \in \mathcal{F}$, there exists $A \in \mathcal{A}$, $A \subseteq F$, such that $P(F-A)< \epsilon$.
More specifically, I'm mainly interested in products of finite spaces. That is, for each $n \in \mathbb{N}$, let $A_n$ be a finite set equipped with discrete topology. Let $\Omega = \prod_n A_n$. Let $\mathcal{F}$ be the Borel $\sigma$-algebra in the product topology, and let $\mathcal{A}$ be the algebra of cylinder sets of the form $$B_1 \times B_2 \times..., \ B_n \subseteq A_n, \ \text{and} \ B_n = A_n \ \text{for all but finitely many $n$}.$$
In this case, one result that is pretty close to what I want is that $\mu$ can be approximated from within by compact sets. See this. But I don't think that compact sets are necessarily members of $\mathcal{A}$, so it doesn't quite answer my question.
What got me thinking about this was this post. In the presence of finite additivity, approximation from within by a compact class is sufficient for countable additivity. The present question, in the special case of products of finite spaces, is a sort of converse: When is countable additivity sufficient for approximation by the compact class that generates $\mathcal{F}$?
I think it is rather rare for this to be true, at least in natural examples where $\mathcal{A}$ consists of "nice sets" on which the measure is very easy to understand. In particular, it is not true in your infinite product example. Indeed, there are only countably many cylinder sets, so you can get a countable subset $Q\subseteq\Omega$ which intersects every nonempty cylinder set. If $\mu$ is the product measure for the uniform probability measure on each $A_n$ and each $A_n$ has more than one point, then $Q$ will have measure $0$, so the complement $Q^c$ has measure $1$. But $Q^c$ does not contain any nonempty cylinder sets, so its measure cannot be approximated by cylinder sets it contains.