Class property proof 
Suppose for each $A$ in $\mathscr A$ that $A^\complement$ is a countable union of elements of $\mathscr A$. The class of intervals in $(0,1]$ has this property. Show that $\sigma(\mathscr A)$ coincides with the smallest class over $\mathscr A$ that is closed under the formation of countable unions and intersections.

How can this be proven ? should I apply good set argument, how?
 A: We need an additoinal assumption, see below.

Let $\mathscr B$ be the smallest class over $\mathscr A$ that is closed under the formation of countable unions and intersections.
Claim 1. We can construct $\mathscr B$ as follows:
Let $\mathscr B_0=\mathscr A$. Recursively, let 
$$\mathscr B_{n}=\left\{\,\bigcup_{i=1}^\infty B_i\Biggm| B_i\in\mathscr B_{k_i}, k_i<n\,\right\}\cup \left\{\,\bigcap_{i=1}^\infty B_i\Biggm| B_i\in\mathscr B_{k_i}, k_i<n\,\right\}$$
for all ordinals $n$.
Then $k<n$ implies $\mathscr B_k\subseteq \mathscr B_n$. As there are more than $|\mathcal P(\mathcal P(\Omega))|$ ordinals, there must exist ordinals $k,n$ with $k<n$ and $\mathscr B_k=\mathscr B_n$. Then  $\mathscr B=\mathscr B_{k_\min}$ where $k_\min$ is minimal among all $k$ with that property.
Proof:
Clearly, $\mathscr A=\mathscr B_0\subseteq \mathscr B$.
Also, the definition of $\mathscr B_n$ is such that if $\mathscr B_k\subseteq \mathscr B$ for all $k<n$, then by the closure property, we must also have $\mathscr B_n\subseteq \mathscr B$. It follows that also $\mathscr B_{k_\min}\subseteq \mathscr B$. Let $n>k_\min $ with $\mathscr B_n=\mathscr B_{k_\min}$. Then $B_{k_\min+1}\subseteq \mathscr B_n=\mathscr B_{k_\min}$, showing that all countable unions or intersections withnb $\mathscr B_{k_\min}$ are already in $\mathscr B_{k_\min}$. $\square$
We show by induction on $n$ the crucial
Lemma 1. If $B\in \mathscr B_n$ then $B^\complement \in\mathscr B_{n+1}$.
Proof. This is given for $n=0$. 
Suppose $n>0$ and the lemma is true for all smaller ordinals. 
Let $B\in\mathscr B_n$. Then $B$ is either the countable union or countable intersection of elements of "smaller" $\mathscr B_k$, and by induction hypothesis, their complements are in the respective $\mathscr B_{k+1}$. Hence
$$ B^\complement=\left(\bigcup_i B_i\right)^\complement=\bigcap_i B_i^\complement\qquad \text{or}\qquad B^\complement=\left(\bigcap_i B_i\right)^\complement=\bigcup_i B_i^\complement$$
is in $\mathscr B_{n+1}$. $\square$
Now we can show 
Claim 2. $\mathscr B$ is a $\sigma$-algebra.
Proof:


*

*$\Omega\in \mathscr B$: This follows under the additional assumption that $\mathscr A\ne\emptyset$: Pick $A\in\mathscr A=\mathscr B_0$. Then by lemma 1, $A^\complement\in\mathscr B_1$  and ultimately $A\cup A^\complement\in\mathscr B_1\subseteq \mathscr B$.

*If $B\in\mathscr B$ then $B^\complement\in\mathscr B$: By lemma 1, $B^\complement\in \mathscr B_{k_\min +1}=\mathscr B$.
As we already know that $\mathscr B$ is closed under countableunios, the claim follows. $\square$
This implies $\sigma(\mathscr A)\subseteq \mathscr B$.
As a $\sigma$-algebra is also closed under countable intersections, it follows that also $\mathscr B\subseteq \sigma(\mathscr A)$ and ultimately
$$\sigma(\mathscr A)= \mathscr B.$$
