Show the following is a sigma algebra Let $\mathcal{A}$ be an algebra containing a pi system $\pi$, and let $\mathcal{F}=\sigma(\pi)$. Suppose 
$\mathcal{A}\subset \mathcal{F}$.

Define $\mathcal{A}'=\{A\in \mathcal{F} \mid$ there is $A_n\in\mathcal{A}$ with $\mathbb{P}[A_n\triangle A]\to 0\}$.

I am struggling to show that $\mathcal{A}'$ is a sigma algebra. Namely, the final property that it is closed under countable unions.
Any help would be appreciated.
Induction:

Let A,B $\subset$ A'. Then for any x $\in$ A or B, x $\in$ A $\bigcup$ B.  Therefore A $\bigcup$ B $\subset$ A'.

For a collection of sets $$\{A_i \}_{i \in \Lambda}$$s.t. $\Lambda$ is countable and $A_i$ $\subset$ A' $\forall$  $i\in \Lambda$ assume:
$$\bigcup_{i \in \Lambda} A_i \in A'$$.

Then for an additional set $A_j \in \Lambda$ then
$$A_j \bigcup (\bigcup_{i \in \Lambda} A_i) \rightarrow A_j \bigcup B \in A'$$
$$B = \bigcup_{i \in \Lambda} A_i$$
We know this because both A,B $\subset$ A'.
Induction:

Let A,B $\subset$ A'. Then for any x $\in$ A or B, x $\in$ A $\bigcup$ B.  Therefore A $\bigcup$ B $\subset$ A'.

For a collection of sets $$\{A_i \}_{i \in \Lambda}$$s.t. $\Lambda$ is countable and $A_i$ $\subset$ A' $\forall$  $i\in \Lambda$ assume:
$$\bigcup_{i \in \Lambda} A_i \in A'$$.

Then for an additional set $A_j \in \Lambda$ then
$$A_j \bigcup (\bigcup_{i \in \Lambda} A_i) \rightarrow A_j \bigcup B \in A'$$
$$B = \bigcup_{i \in \Lambda} A_i$$
We know this because both A,B $\subset$ A'.