Show that the class of countable unions and intersections of the elements of an algebra is an algebra. I have the following problem:
Let $\mathcal{A}$ be an algebra of subsets of a set $X$. We form the class $\mathcal{B}$ of countable unions and intersections of elements of $\mathcal{A}$, that is
$\mathcal{B}=\{B\in X : B=\cup_{n\in N} B_n \;\text{with} \;B_n\; \text{in}\; \mathcal{A}, \text{or} \; B=\cap_{n\in N} C_n \;\text{with}\; C_n \text{in} \;\mathcal{A}\}$.
I would like to prove that $\mathcal{B}$ is an algebra (I saw this fact in a book).  
It is clear that $\mathcal{B}$ contains $X$ and is stable by taking complements, but I cannot prove that it is stable under finite unions or intersections. 
My problem is that if I take 2 elements $Z$ and $Y$ in $\mathcal{B}$ such that $Z=\cup_{n\in N} B_n$ and $Y=\cap_{n \in N}C_n$ then I do not know how to prove that  $Z\cup Y$ or $Z\cap Y$ is in $\mathcal{B}$.
Thank you for your help!
Alain
 A: It seems the following.
In general, $\mathcal B$ is not an algebra. As a counterexample let $\mathcal A$ be the algebra of all Jordan measurable subsets of the unit segment $[0,1]$. Let $A$ be the set  of all rational points of the segment  $[0,1/2]$ and $B$ be the set  of all irrational points of the segment  $[1/2,1]$. Then both $A$ and $B$ belongs to $\mathcal B$. Suppose that $C=A\cup B\in\mathcal B$. Then one of two cases holds. 
1.There exists a countable family $\{C_n\}\subset A$ such that $C=\bigcup C_n$. For an arbitrary index $n$ the set $C_n\cap [1/2,1]$ is a Jordan measurable subset of $B$. Then $\lambda(C_n\cap [1/2,1])=0$, where $\lambda$ is Lebesgue measure on the segment $[0,1]$. So 
$$\lambda(C\cap [1/2,1])= \lambda(\bigcup C_n\cap [1/2,1])=0\ne 1/2=\lambda(B)=\lambda(C\cap [1/2,1]),$$
a contradiction.
2.There exists a countable family $\{C_n\}\subset\mathcal A$ such that $C=\bigcap C_n$. For an arbitrary index $n$ the set $[0,1/2]\setminus C_n$ is a Jordan measurable subset of the set  of all irrational points of the segment  $[0,1/2]$. Then $\lambda([0,1/2]\setminus C_n)=0$. So 
$$\lambda([0,1/2]\setminus C)= \lambda([0,1/2]\setminus \bigcap C_n)=0\ne 1/2=\lambda([0,1/2]\setminus A)=\lambda([0,1/2]\setminus C),$$
a contradiction.
