Approximation property confusion I am a bit confused about this:

If $A$ is an algebra of sets then for any $B \in \sigma(A)$ there exists a sequence of $B_n \in A$ such that $P(B \Delta B_n) \to 0$ 

$A \Delta B $ is denoting the symmetric difference- $ (A \backslash B) \cup (B \backslash A) $ 
The proof begins with $\mathcal D = \{ B\in \sigma(A):\exists B_n\in A, P(B \Delta B_n)\to 0 \}$
So this is the set of all elements in the $\sigma$-algebra generated by $A$ with that property.
The proof shows that $\mathcal D$ is a $\sigma$-algebra and, since we have $A \subset \mathcal D,$ $ \sigma(A)\subset \mathcal D$, completing the proof.
Before getting to the main part of this proof, I cannot understand the joint significance of the above three things:


*

*Why the fact that $\mathcal D $ is a $\sigma$-algebra is something which is important

*How $A \subset \mathcal D$

*How (2) $\Rightarrow \sigma(A) \subset \mathcal D$ 


I get that if $\sigma(A) \subset \mathcal D$ then it proves $\forall B\in \sigma(A)$

Ad 1.: You want to prove that a property holds for every element of $\sigma(A)$, that is, for every element of the smallest $\sigma$-algebra that contains $A$. So the set of elements of $\sigma(A)$ that have that property should itself be a $\sigma$-algebra.
Ad 2.: Choose $B_n = B$ for all $n$ if $B\in A$ to see that every element of $A$ has the desired property.
Ad 3.: $\mathcal{D}$ is a $\sigma$-algebra that contains $A$ by 1. and 2. Therefore it contains the smallest $\sigma$-algebra that contains $A$.
