Question on whether or not a given collection is a $\sigma$-algebra. 
Let $\mathscr F$ be a collection of subsets of $X.$ Let us define $\mathscr F_1,\mathscr F_2, \mathscr F_3$ and $\mathscr F_4$  in the following way.
$\mathscr F_1 : = \mathscr F \cup \{\varnothing\}.$
$\mathscr F_2 : = \mathscr F_1 \cup \left \{A\ |\ A^c \in \mathscr F_1 \right \}.$
$\mathscr F_3 : = \left \{\bigcap\limits_{i=1}^{\infty} A_i\ \bigg |\ A_i \in \mathscr F_2 \right \}.$
$\mathscr F_4 : = \left \{\bigcup\limits_{i=1}^{\infty} A_i\ \bigg |\ A_i \in \mathscr F_3 \right \}.$
Is $\mathscr F_4$ a $\sigma$-algebra of subsets of $X\ $?

First I observe that $$\mathscr F \subseteq \mathscr F_1 \subseteq \mathscr F_2 \subseteq \mathscr F_3 \subseteq \mathscr F_4.$$ From here I have managed to prove that $\mathscr F_4$ contains $\varnothing$ and $X$ and it is closed under countable unions. But I am unable to show that $\mathscr F_4$ is closed under complements. What I did myself is as follows $:$
Let $A_i \in \mathscr F_3,$ for $i = 1,2, \cdots.$ Then we need to show that $\left (\bigcup\limits_{i=1}^{\infty} A_i \right )^c \in \mathscr F_4.$ Since $A_i \in \mathscr F_3,$ for $i = 1,2, \cdots$ there exists $B_{ij} \in \mathscr F_2,$ for $j = 1,2, \cdots$ such that $A_i = \bigcap\limits_{j=1}^{\infty} B_{ij}.$ Then $$\left (\bigcup\limits_{i = 1}^{\infty} A_i \right )^c = \left  (\bigcup\limits_{i = 1}^{\infty} \left ( \bigcap\limits_{j=1}^{\infty} B_{ij} \right ) \right )^c.$$ Now since $B_{ij} \in \mathscr F_2$ there exists $C_{ij} \in \mathscr F_1$ such that $B_{ij} = C_{ij}^c.$ So we have $$\begin{align*} \left (\bigcup\limits_{i = 1}^{\infty} A_i \right )^c & = \left  (\bigcup\limits_{i = 1}^{\infty} \left ( \bigcap\limits_{j=1}^{\infty} C_{ij}^c \right ) \right )^c \\ & = \left  (\bigcup\limits_{i = 1}^{\infty} \left ( \bigcup\limits_{j=1}^{\infty} C_{ij} \right )^c \right )^c \\  & =  \bigcap\limits_{i = 1}^{\infty} \left ( \bigcup\limits_{j=1}^{\infty} C_{ij} \right ) \end{align*}$$
Now I got stuck at this stage and couldn't able to proceed further. Would anybody please help me in this regard?
Thanks in advance.
 A: In general, $\mathscr F_4$ will not be a $\sigma$-algebra of subsets of $X$.
Here is a counter-example.
Let $X= \Bbb R$ and $\mathscr F$ be the set of semi-finite intervals in $\Bbb R$. That is, an element of $\mathscr F$ is of the form $( -\infty, a)$,  $(- \infty, a]$, $(a, \infty)$ or $[a, \infty)$.
Then $\mathscr F_1 : = \mathscr F \cup \{\emptyset\}$  and then
$$\mathscr F_2 : = \mathscr F_1 \cup \left \{A\ |\ A^c \in \mathscr F_1 \right \}= \mathscr F \cup \{\emptyset, \Bbb R\} $$
Claim: $\mathscr F_3 : = \left \{\bigcap\limits_{i=1}^{\infty} A_i\ \bigg |\ A_i \in \mathscr F_2 \right \}$ is the set of all intervals in $\Bbb R$ (including the single-element sets as intervals).
Proof: Note that all elements of $\mathscr F_2 $ are connected sets in $\Bbb R$ (intervals), so all the elements of $\mathscr F_3$ must be connected sets in $\Bbb R$, that means, intervals. It also easy to see that any interval is the intersection of two semi-finite intervals. $\square$
Now $\mathscr F_4 : = \left \{\bigcup\limits_{i=1}^{\infty} A_i\ \bigg |\ A_i \in \mathscr F_3 \right \}$ is just the set of countable unions of intervals in $\Bbb R$, which clearly is not a $\sigma$-algebra.
One easy way to see that  $\mathscr F_4 $ is not a $\sigma$-algebra is to note that $\Bbb Q \in \mathscr F_4 $ ($\Bbb Q $ is a countable union of single-point intervals), but $\Bbb R \setminus \Bbb Q \notin \mathscr F_4 $( the set of  irrational numbers can not be written as a countable union of intervals).
