As I mentioned in my comment, there is an analogue of this for $\sigma$-fields but it is a lot more complicated. To slightly simplify the discussion, I'm going to pretend the only operations we care about are union and intersection, and ignore complements. In fact, because of the identities $(\bigcap A_i)^c=\bigcup A_i^c$ and $(\bigcup A_i)^c=\bigcap A_i^c$, you don't lose anything by ignoring complements if you just assume your generating set $\mathcal{A}$ is closed under taking complements: those identities always allow you to move complements to the "inside" of any Boolean expression.
Let's begin by sketching a solution to Billingsley's Problem 2.5(b); we'll then look at what goes wrong and makes the case of $\sigma$-fields so much harder. Intuitively, $\mathcal{F}(\mathcal{A})$ should be built from $\mathcal{A}$ by iteratively taking unions and intersections and unions and intersections and (...) any finite number of times. More precisely, let $S_0=\mathcal{A}$, let $S_1$ be the set of all finite intersections of elements of $S_0$, let $S_2$ be the set of all finite unions of elements of $S_1$, let $S_3$ be the set of all finite intersections of elements of $S_2$, and so on. It is not difficult to show that $\mathcal{F}(\mathcal{A})=\bigcup_{n\in\mathbb{N}} S_n$.
Concretely, an element of $S_1$ looks like $\bigcap_{i=1}^n A_i$, an element of $S_2$ looks like $\bigcup_i\bigcap_j A_{ij}$, an element of $S_3$ looks like $\bigcap_i\bigcup_j\bigcap_k A_{ijk}$, and so on (in each case, these unions and intersections are over a finite index set). So Problem 2.5(b) is asserting that actually, $\mathcal{F}(\mathcal{A})$ is just $S_2$ (the additional assertion that you can arrange for the sets $\bigcup_j A_{ij}$ to be disjoint is then not hard to show as well). Why is this true? It comes from the fact that intersections distribute over unions: $$\bigcap_{i=1}^n\bigcup_{j=1}^m B_{ij}=\bigcup_f\bigcap_{i=1}^n B_{if(i)},$$
where $f$ ranges over all functions $\{1,\dots,n\}\to\{1,\dots m\}$ (if you are unfamiliar with this identity, it is a nice exercise to prove it). Using this identity, whenever we have a sequence $\bigcap\bigcup$ in an expression for an element of $\mathcal{F}(\mathcal{A})$, we can replace it with $\bigcup\bigcap$ (with a change of index set on the $\bigcup$). So for instance, an element $\bigcap\bigcup\bigcap A_{ijk}$ of $S_3$ can be rewritten in the form $\bigcup\bigcap\bigcap A_{ijk}$, which can then be rewritten in the form $\bigcup\bigcap A_{ij}$ by just combining the index sets of the two intersections. Using this rule repeatedly, we can reduce any expression in any $S_n$ all the way down to $S_2$.
OK, now let's try to apply what we just did to $\sigma$-fields. We can define the sets $S_n$ just as before, except that now we want to allow countable intersections and unions, not just finite intersections and unions. We next hope to show that an element of any $S_n $ can be reduced to $S_2$ using the distributive law. Here we run into a problem: the distributive law for countably infinite intersections and unions says that
$$\bigcap_{i\in\mathbb{N}}\bigcup_{j\in\mathbb{N}} B_{ij}=\bigcup_{f:\mathbb{N}\to\mathbb{N}}\bigcap_{i\in\mathbb{N}} B_{if(i)}.$$
This is no good because there are uncountably many functions $f:\mathbb{N}\to\mathbb{N}$, so our outer union is no longer a countable union! (This wasn't an issue in the finite case because the set of all functions between two finite sets is always again a finite set.) So if we try to reduce an element of $S_3$ to an element of $S_2$ like before, we end up with an expression $\bigcup\bigcap A_{ij}$, where the union is over an uncountable index set. Such a union need not be an element of $S_2$, so this is no use.
So in the case of $\sigma$-fields, we cannot say that $\bigcup_n S_n$ is just $S_2$ and get a very simple expression for what an element of $\mathcal{F}(\mathcal{A})$ looks like. But that's not the only problem! We also face the issue that $\bigcup_n S_n$ may not even be a $\sigma$-field, so it is not all of $\mathcal{F}(\mathcal{A})$. For instance, if we take a sequence of sets $B_n$ such that $B_n\in S_n$ for each $n$, there is no reason to believe that the intersection $\bigcap_n B_n$ is in any $S_n$. (This issue didn't come up in the finite case, since if we're only taking a finite intersection, there would have to be some $S_n$ that contains all of our $B$'s, and then the intersection would have to be in $S_{n+1}$.) To get a $\sigma$-field, we have to keep going: we could define $S_\omega=\bigcup_n S_n$, and then let $S_{\omega+1}$ be the set of all countable intersections of elements of $S_{\omega}$, and then let $S_{\omega+2}$ be the set of counable unions of elements of elements of $S_{\omega+1}$, and so on. More precisely, we can use transfinite induction to define sets $S_{\alpha}$ for all ordinals as follows:
- $S_0=\mathcal{A}$.
- If $\alpha$ is a limit ordinal, $S_\alpha=\bigcup_{\beta<\alpha} S_\beta$.
- If $\alpha=\beta+1$ is an odd successor ordinal, $S_\alpha$ is the set of all countable intersections of elements of $S_\beta$.
- If $\alpha=\beta+1$ is an even successor ordinal, $S_\alpha$ is the set of all countable unions of elements of $S_\beta$.
To be sure that we actually get a $\sigma$-field (which will be the smallest $\sigma$-field containing $\mathcal{A}$), we have to iterate this all the way out to the first uncountable ordinal $\omega_1$. (We can finally stop at $\omega_1$ because any countable set of ordinals below $\omega_1$ has an upper bound below $\omega_1$.) In the end, when you look at what an arbitrary element of $\mathcal{F}(\mathcal{A})=S_{\omega_1}$ actually looks like, you find that it is some transfinitely iterated countable union of countable intersections of countable unions of (...) of elements of $\mathcal{A}$, where the iteration is indexed by some countable well-founded tree, as described in this question.
