Properties of an increasing sequence of sigma-algebras. On a probability space $(\Omega,\mathcal{A},P)$ let $\mathcal{F} = (\mathcal{F}_n)$ be a filtration which converges towards $\mathcal{A}$. Is there any chance that the following statemente might be true?

For every $A$ in $\mathcal{A}$ and every $n$, there exist $B_n$ and $C_n$ in $\mathcal{F}_n$ such that
$$ B_n \subseteq A \subseteq C_n \qquad\text{and}\qquad P(C_n \setminus B_n) \to 0$$

If it is, how can I prove it?
My Attempt
I thought of considering the families
$$ \mathcal{B}_n = \{ B \in \mathcal{F}_n : B \subseteq A \}
\qquad\text{and}\quad
\mathcal{C}_n = \{ C \in \mathcal{F}_n : A \subseteq C \} $$
and to see whether I could find a maximal element in both classes, but alas I cannot even prove that these classes have a maximal element.
 A: Assuming that $\mathcal{F}$ converging towards $\mathcal{A}$ means that $\sigma(\bigcup_n \mathcal{F}_n) = \mathcal{A}$, as suggested in the comments, here's a counterexample to the statement.
Let $\Omega = [0, 1)$, let $\mathcal{A}$ be the Borel $\sigma$-algebra on $\Omega$, and let $P$ be Lebesgue measure. Also define the filtration via $\mathcal{F}_n = \sigma(\{[j/2^{n}, (j+1)/2^n) \mid 0 \leq j < 2^n \})$. In other words, $\mathcal{F}_n$ is generated by the dyadic sub-intervals of length $2^{-n}$.
One can show that this filtration does indeed converge to $\mathcal{A}$, for example by thinking about how open sets can be approximated from the inside by unions of dyadic sub-intervals.
But now consider a set $A$ that consists of one point from every dyadic sub-interval. That is, pick $p_{n,j} \in [j/2^n, (j+1)/2^n)$ for every $n$ and $j$, and let $A$ be the set of these points.
On the one hand, $A$ is countable, so $P(A) = 0$. On the other hand, the only set in $\mathcal{F}_n$ that contains $A$ is $\Omega$ itself. (Likewise, the only set in $\mathcal{F}_n$ contained in $A$ is the empty set.) So the only possibilities for the sets $B_n$ and $C_n$ are $B_n = \emptyset$, $C_n = \Omega$, and we have $P(C_n \setminus B_n) = P(\Omega) \to 1$.
A: It is possible to approximate  elements in $\sigma(\mathcal{A})$ by elements in $\mathcal{A}=\bigcup_n\mathscr{F}_n$, but not exactly the way you intend. In any case I think the arguments are enough for many applications, for instance in proving some  $0-1$ type laws in probability.
This is a sketch:
$\mathcal{A}=\bigcup_n\mathscr{F}_n$ is an algebra. Using a Carathéodory's method, you can construct an outer measure $\mathbb{P}^*$ that happens to coincide with $\mathbb{P}$ on each $\mathscr{F}$. Then yo can see that for any $A\in\sigma(\mathcal{A})$ and $\varepsilon>0$, there is $E\in\mathcal{A}$ such that $\mathbb{E}[|\mathbb{1}_A-\mathbb{1}_E|]<\varepsilon$.
I hope this works for what lies behind your need for approximation.
